Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 8th Edition [8 ed.] 0132885514, 9780132885515

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SOLUTIONS MANUAL

Basic Principles and Calculations in Chemical Engineering Eighth Edition

David M. Himmelblau James B. Riggs

Upper Saddle River, NJ • Boston • Indianapolis • San Francisco New York • Toronto • Montreal • London • Munich • Paris • Madrid Capetown • Sydney • Tokyo • Singapore • Mexico City

The authors and publisher have taken care in the preparation of this book, but make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of the use of the information or programs contained herein. Visit us on the Web: InformIT.com/ph Copyright © 2012 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. ISBN-10: 0-13-288551-4 ISBN-13: 978-0-13-288551-5

ACKNOWLEDGMENTS We want to thank Christine Bailor for preparing this Solutions Manual, and for the many students and graders who have contributed to the solutions it contains.

David M. Himmelblau James B. Riggs

iii

TABLE OF CONTENTS Page 1. 2. 3. 4. 5. 6. 7. 8. 9.

To the Instructor ............................................................................................ Example Course Syllabus ............................................................................. Course Objectives ......................................................................................... Exam and Recitation Section Schedules ....................................................... Suggestions for Taking Exams ..................................................................... What You Should Know About This Course ............................................... Standards for Chemical Engineering Homework ......................................... Typical Assignments for One Semester ........................................................ Typical Examinations for a One Semester Course ........................................

iv

v vi viii ix x xi xii xiv xvii

To the Instructor This Solutions Manual accompanies the book Basic Principles and Calculations in Chemical Engineering, Eighth Edition, published by Prentice Hall. In addition to the detailed, worked-out solutions for all the problems that follow each chapter in the textbook and answers to the thought problems, you will find in what follows a number of useful components of a syllabus for students, information that usually are handed out during the first day of class: 1. 2. 3. 4. 5.

Class grading policies, homework and reading assignments, and examination information. Class objectives. Schedule of topics covered. Suggestions for taking examinations. Format standards for submitting homework.

Suggested Content for the Introductory Course in Chemical Engineering The introductory course in chemical engineering is usually taught over an interval of one or two semesters, or one or three quarters. The textbook contains more material than can be successfully presented in one quarter and probably in one semester (depending on the background and previous coursework of students). Although an instructor would like to assume that a student has learned all of the material covered in earlier courses in chemistry and physics, it takes just one time in teaching the introductory course to abandon that expectation. The textbook is organized into four parts comprised of 11 chapters plus 6 additional chapters on the accompanying CD that treat material usually not included in a one semester course. The following list suggests the chapters to include in courses of various duration: One quarter One semester Two quarters Two semesters

1–6, 8, 9–10 1–11 1–7 followed by 8 and 11 1–11 followed by 12–17

v

Example Course Syllabus Information for ChE 317 Introduction to Chemical Engineering Instructor: D.M. Himmelblau Office hours: M-F 10-11 a.m. 1.

Office: CPE 5.410

GENERAL a. The prerequisites for ChE 317 are Ch 302 and Math 808. If you have not completed these two courses, you will have to drop ChE 317 and should do so at once. b. Class conduct is informal. Feel free to raise your hand at any time to ask a question or for an explanation.

2.

EXAMINATIONS a. Five two-hour examinations plus a final exam will be held at specified announced dates as shown on the assignment sheets. The last examination will be scheduled during the final exam period (refer to the course schedule for details). The lowest exam of the first 5 (excluding the final exam) will be omitted in calculating your final grade. You must take the final. If you will miss an exam, notify me prior to the exam, not afterwards, to arrange for a makeup exam.

3.

GRADING a. The grading is based on scores on the examinations, each of which is weighted equally (90%), plus class discussion and homework (10%). The grades are assigned on an absolute basis, not a curve: A B C D F

> 82 71-82 61-70 51-60 < 51

hence there is no penalty for working together and helping each other. b. You will have a grader assigned to this course whose name is ___________, office number is Room ______, and office hours are __________. c. The recitation session assistant is ___________________, office number is Room _____, and office hours are __________. d. If you disagree with the grader’s method of grading and with the total points he has given you on a particular problem, discuss it with the grader first, but if you cannot reach a decision, I will be the referee. Bring exam grade questions to me. e. Prepare a grade sheet on which you can keep account of your homework and exam grades so that you will be able to compute your status at any time you wish.

vi

f. A grade of at least a C is required in this course for subsequent courses in chemical engineering. 4.

HOMEWORK PROBLEMS a. CHEMICAL ENGINEERING STANDARDS WILL BE REQUIRED AND ENFORCED. (Capital letters intended.) b. Problems are due at the beginning of each class according to the assignment. No late problems can be accepted. c. Turn in a much of a problem as you can get. It is better to get a low grade than a “miss.” d. Working together is an important part of professional practice. After the second week of class, students will be assigned to work on homework (not exams!) in pairs. During the first two weeks of class look for a possible compatible partner. You will receive a list of all of the class members with their phone numbers to help in the selection. Exceptions can be made for individuals who insist on working alone. e. After each scheduled homework assignment has been turned in, the solution(s) will be placed in a file located in the ChE stockroom that may be checked out for 2 hours at a time.

5.

If you have difficulty in the early part of the course, confer with me before you get into trouble.

vii

Course Objectives The objectives for Chemical Engineering 317 are as follows: 1.

To introduce you to the principles and calculation techniques used in the field of chemical engineering.

2.

To acquaint you with the fundamentals of material and energy balances as applied to chemical engineering.

3.

To acquaint you with efficient methods of problem solving so that you can effectively solve problems you will encounter after leaving school.

4.

To offer practice in defining problems, collecting data, analyzing the data, and breaking it down into basic patterns, and selection of pertinent information for application.

5.

To review certain principles of applied physical chemistry.

6.

To help you decide you have chosen the right field. Contributions to Program Outcomes

By graduation a chemical engineering student should have achieved certain knowledge, skills, and abilities known as Program Outcomes. ChE 317 contributes to five significant outcomes, namely an ability to: 1.

Apply knowledge of mathematics, chemistry, physics computing, safety, ethical practice, and technology to solve engineering problems.

2.

Apply and integrate elements of chemical engineering to solve problems in design, operation, and control of processes.

3.

Participate in team activity effectively and demonstrate leadership.

4.

Communicate effectively via oral, written, and graphic means.

5.

Appreciate the societal and economic impact of engineering decisions locally and globally.

viii

Fall Semester Exam Schedule The first 5 exams are evening exams on Thursday, open book, of 2 hours duration, specific times to be arranged (such as 5-7, 6-8, 7-9, etc.): Exam 1 Exam 2 Exam 3 Exam 4 Exam 5

September 14 October 1 October 16 November 1 November 15

The final exam is listed in the University final exam schedule (that will appear about December 1). Fall Semester Schedule for the Recitation Section The recitation section will meet on Thursday, 2–3:30 p.m. in CPE 2.220. The objective of the recitation section is to let you ask questions and provide assistance in problem solving for old or new ChE 317 homework and exam problems on a one-to-one basis. Attendance is not required, but you will miss the unique opportunity to get personal attention if you want it. You will also miss questions asked by other students that you have not considered. You will also have the chance to meet other students in the class, and discuss anything (!) with them. No assignments are made, no grades given, no lectures presented, and no formal structure exists for the recitation section. It’s up to you to make use of it.

ix

Suggestions for Taking Exams 1.

Bring what you want to the exams—they are open book. Be sure to have adequate pencils, batteries, etc.

2.

Read the entire examination through quickly before starting to work any one problem. Then work first on those problems which seem the simplest or about which you are most confident in solving.

3.

Be sure to allot your working times to the questions roughly according to the grade value of each. If a problem is not completed in the time allotted, it is usually better to discontinue work on it and spend time on the other problems. Be sure to spend at least some time on each problem. Partial solutions to all problems usually result in a higher overall grade than complete solutions to only a small portion of the problems (provided you do enough work on a problem to indicate that the correct method of attack is being used).

4.

When starting work on a problem read it through carefully and be certain you understand it. Spend a short time thinking about the method of solution instead of writing down what first comes to mind.

5.

When writing down the solution, organize your work in a neat and logical manner in spite of the time constraints. This step not only impresses the grader but also permits him or her to follow the work closely enough so that if a mistake is made he or she can still evaluate the succeeding work. Neatness and organization also permit you to check your work more easily and to find quickly information needed later in the problem.

6.

In answering a question write enough so that the grader does not have to guess what you had in mind. For example, when using equations, write down the equation first and then substitute numbers. A group of numbers alone may confer little information to the grader, especially if they are the wrong numbers. When using data obtained from tables or charts, state the source—and in some cases the method of using the source. Draw pictures, and separate subproblems from each other.

7.

If it is obvious that you are not going to finish a problem, carefully outline the remainder of the solution by numbered steps, and include sufficient details, such as pertinent equations and methods of solving them, sources for remaining necessary data, etc.

8.

If you start to get rattled, slow down a bit— perhaps even think of something besides the examination for a minute or two. Remember that this one examination is not going to make or break you whatever success you have on it. View the problem bothering you as you would a bridge hand, crossword puzzle, or other game that involves solving a problem based on a given set of facts with available information.

9.

Sample old exams are located in the ChE Stockroom, and can be taken out and copied. Practice solving old exams two or three days in advance of each exam to isolate your weaknesses in subject material and exam taking skills.

x

What you should know about this course at the beginning that will be clear by final exam time 1.

You no longer are a freshman so that the material covered proceeds at a rapid pace.

2.

Your notions of teaching and learning will require substantial adjustment. Our goal is not for you to reproduce what was told to you in the classroom or you read in the text. Your study habits probably must change.

3.

Lecture time is at a premium and must be used efficiently. Listening is not learning any more than lecturing is teaching. You are responsible for learning the material, a phase that will occur primarily outside the classroom. The instructor cannot “teach” all the skills you need in the short time of a class. It will take you two or three hours on the average per hour of class time to become proficient.

4.

The instructor’s job is to provide a framework of the topic along with demonstrations to guide you in your learning of concepts, methods, and efficient problem solving skills. It is not to imprint you with isolated facts and problem types.

5.

If you read the material in the assigned section for the next period before coming to class, the lecture will make more sense, and you can ask questions to clarify any uncertain issues.

xi

Standards for Chemical Engineering Homework Assignments 1.

Engineering paper must be used (paper ruled on the back with a grid).

2.

Use the unruled side of the sheet for the calculations, and the back side for drawings (rarely required).

3.

Use the sheet with the holes to the left.

4.

Turn in your work with the paper folded vertically.

5.

Write neatly. Make the text in your calculations in letters 0.20 inches high—these match the horizontal grid spacing on the back of the paper. Leave 0.20 inches (one grid interval) between lines. The idea is to be professional in presenting your work.

6.

Use engineering/scientific notation for numbers such as 0.341 and 1.453 x 105. Use judgment as to how many zeros you put after the decimal point or before the first significant figure. Note: always put a zero before the decimal point for a number less than unity.

7.

Indicate multiplication and division using units as seen below. (Note: Use vertical and horizontal rules as necessary.)

3.45 lb NaCl 4 gal soln 1 ft 3 = 7.48 gal 1 ft 3 soln As you gain experience, you can suppress the units for simple problems and show multiplication and division thus (use parenthesis rather than centered dots as the dots get confused with periods, dust specks, etc.) 1 $ (3.45)( 4) !#" 7.48 &% =

8.

Use a solid line across the page between the vertical rules on the engineering paper to demark the end of a part of a problem with multiple parts. Denote the end of the entire problem by a line across the page from the left hand rule to the far right edge of the paper.

9.

Always show the units of your answer, underline the numbers and units, and draw an arrow from the right edge of the paper to the answer so that the answer is easy to pick out on the page thus

8.5 lbH 2 1 lb mol H 2 1 lb mol Zn 4.21 lb mol Zn = lb F lb F 2.02 lb H 1 lb mol H 2

!" "

part (a) of the problem

10.

Indicate a new problem by placing the problem number in the left hand margin.

11.

Always show the basis of your calculations thus Basis: 100 lb feed xii

12.

On each submission place your class number, the date, the assignment number, your name, and the page numbering at the top of each and every page, even if you staple the pages together, thus 317

↑ Class

Sept. 10, 2003 ↑ Date

Assignment No. 5

Jones, Robert

2/3

↑ Assignment Identification

↑ Your name

↑ Page 2 of 3 pages submitted

xiii

TYPICAL ASSIGNMENTS FOR ONE SEMESTER Topic and Problem Assignments Due

All assignments are in the 8th edition. Study:

1.

First Class meeting. No assignments due

2.

UNITS, DIMENSIONS, UNIT CONVERSION 2.1.1, 2.2.2, 2.2.6

Chapter 2

3.

DIMENSIONAL CONSISTENCY, SIGNIFICANT FIGURES, VALIDATION, MOLES 2.3.1, 2.3.3, 2.3.8

Chapter 2

4.

METHODS OF ANALYSIS AND MEASUREMENT 2.6.1a, 2.6.4a, 2.9.1

Chapter 2

5.

BASIS, TEMPERATURE, PRESSURE 2.7.1a,b,c; 2.11.2, 2.11.5

Chapter 2

6.

PRESSURE MEASUREMENT 2.11.9

Chapter 2

7.

INTRODUCTION TO MATERIAL BALANCES 3.1.11, 3.1.19, 3.1.16, 3.1.7, 3.1.8

Chapter 3

8.

STRATEGY FOR SOLVING MATERIAL BALANCES 3.2.2, 3.2.4, 3.2.5, 3.2.9, 3.2.14

Chapter 3

9.

No class meeting. Exam No. 1 in the evening.

10.

MATERIAL BALANCES WITHOUT REACTION— SINGLE UNITS 4.1.7, 4.1.8, 4.1.10, 4.1.12

Chapter 4

11.

MATERIAL BALANCES (CONTINUED) 4.1.18, 4.1.20, 4.1.23, 4.1.25

Chapter 4

12.

STOICHIOMETRY 5.1.2a,e; 5.1.5; 5.2.14; 5.2.15

Chapter 5

13.

MATERIAL BALANCES WITH REACTION— SINGLE UNITS 5.3.1, 5.3.2, 5.3.6, 5.3.7

Chapter 5

14.

MATERIAL BALANCES WITH REACTION— SINGLE UNITS (CONTINUED) 5.5.5, 5.5.7, 5.5.10, 5.5.13

Chapter 5

xiv

Topic and Problem Assignments Due

All assignments are in the 8th edition. Study:

15.

Review for Exam No. 2.

Chapters 4–5

16.

No class meeting. Exam No. 2 in the evening.

17.

MATERIAL BALANCE PROBLEMS WITH MULTIPLE UNITS 6.1.2, 6.1.5, 6.2.1, 6.2.7

Chapter 6

18.

MATERIAL BALANCE PROBLEMS WITH RECYCLE (NO REACTION) 6.3.1b,d; 6.3.2; 6.3.16, 6.3.17

Chapter 6

19.

MATERIAL BALANCE PROBLEMS WITH RECYCLE (WITH REACTION) 6.3.8, 6.3.13, 6.3.21

Chapter 6

20.

IDEAL GAS AND PARTIAL PRESSURE 7.1.1, 7.1.5, 7.1.12c, 7.1.22, 7.1.31

Chapter 7

21.

MATERIAL BALANCES WITH IDEAL GASES 7.1.52, 7.1.44, 7.1.55

Chapter 7

22.

REAL GASES—COMPRESSIBILITY 7.3.1, 7.3.3, 7.3.13

Chapter 7

23.

REAL GASES—EQUATIONS OF STATE 7.2.8, 7.2.5, 7.2.17, 7.2.13

Chapter 7

24.

Review for Exam No. 3.

Chapters 6–7

25.

No class meeting. Exam No. 3 in the evening.

26.

SINGLE COMPONENT-TWO PHASE SYSTEMS (VAPOR PRESSURE) 8.2.1, 8.2.9, 8.3.2 a to e, 8.3.4, 8.3.5b, 8.3.19, 8.3.20

Chapter 8

27.

TWO PHASE GAS-LIQUID SYSTEMS 8.3.17, 8.3.22a, 8.4.2, 8.4.4, 8.4.7

Chapter 8

28.

TWO PHASE GAS-LIQUID SYSTEMS (CONTINUED) 8.4.14, 8.4.16, 8.4.18, 8.4.27

Chapter 8

xv

Topic and Problem Assignments Due

All assignments are in the 8th edition. Study:

29.

VAPOR-LIQUID EQUILIBRIA AND THE PHASE RULE 8.2.4, 8.2.5, 8.5.3, 8.5.6, 8.5.14, 8.5.18

Chapter 8

30.

Review for Exam No. 4

Chapter 8

31.

No class meeting. Exam No. 3 in the evening.

32.

ENERGY: TERMINOLOGY, CONCEPTS, AND UNITS 9.1.1, 9.1.5, 9.1.7, 9.1.11, 9.2.4, 9.2.21, 9.2.22

Chapter 9

33.

ENERGY BALANCES—CLOSED SYSTEMS 9.3.31a,b; 9.3.9; 9.3.25; 9.3.45

Chapter 9

34.

ENERGY BALANCES—OPEN SYSTEM 9.3.31, 9.3.32, 9.3.17a, 9.3.33

Chapter 9

35.

CALCULATING ENTHALPY CHANGES 9.2.26, 9.2.37, 9.2.46

Chapter 9

36.

APPLICATIONS OF ENERGY BALANCES WITHOUT REACTION-CLOSED SYSTEMS 9.3.19, 9.3.39, 9.3.40

Chapter 9

37.

APPLICATIONS OF ENERGY BALANCES WITHOUT REACTION-OPEN SYSTEMS 9.3.21, 9.3.41

Chapter 9

38.

Review for Exam No. 5

Chapter 9

39.

No class meeting. Exam No. 5 in the evening.

40.

ENERGY BALANCES WITH REACTION 10.1.6; 10.1.8a,b; 10.2.1a; 10.2.3; 10.2.4

Chapter 10

41.

CONTINUED 10.2.17, 10.2.6, 10.2.10, 10.4.6

Chapter 10

42.

PARTIAL SATURATION AND HUMIDITY 11.1.1

Chapter 11

43.

HUMIDITY CHARTS 11.2.1, 11.3.1, 11.3.7

Chapter 11

44.

Exam No. 6 is the final exam held on the scheduled final exam period (3 hours).

xvi

TYPICAL EXAMS FOR A ONE SEMESTER COURSE (scheduled in the evening to avoid time constraints on students Exam No. 1 (Open Book, 1 1/2 hours) PROBLEM 1 (5%) Hydrogen can be separated from natural gas by diffusion through a round tube. The rate of separation is given by N = 2!D"R where N = rate of transport of H2 from the tube, g moles/(sec)(cm of length of tube) D = diffusion coefficient ! = molar density of H2, g moles/cm3 R

=

!r $ log mean radius of tube, r2 –r1 / l n # 2 & , with r in cm. " r1 %

What are the units of D? PROBLEM 2 (10%) A pallet of boxes weighing 10 tons is dropped from a lift truck from a height of 10 feet. The maximum velocity the pallet attains before hitting the ground is 6 ft/sec. How much kinetic energy does the pallet have in (ft)(lbf) at this velocity? PROBLEM 3 (5%) The specific gravity of a fuel oil is 0.82. What is the density of the oil in lb/ft3? Show all units. PROBLEM 4 (10%) Sulfur trioxide (SO3) can be absorbed in sulfuric acid solution to form more concentrated sulfuric acid. If the gas to be absorbed contains 55% SO3, 41% N2, 3% SO2, and 1% O2, how many parts per million of O2 are there in the gas? (b) What is the composition of the gas on a N2 free basis? PROBLEM 5 (15%) You have 100 kilograms of gas of the following composition: CH4 H2 N2

30% 10% 60%

What is the average molecular weight of this gas?

xvii

PROBLEM 6 (15%) If the heat capacity of a substance is 5.32 J/(g)(°C) and its molecular weight is 37.4, what is its heat capacity in (a) (b) (c)

J/(g)(°F) J/(lb)(°R) J/(gmol)(K)

PROBLEM 7 (20%) A rock containing 100% BaSO4 is burned with coke (94%C, 6% ash) and the composition of the product is BaSO4 (11.1%) BaS (72.9%), C (13.9%, ash (2.2%). The reaction is BaSO 4 + 4C ! BaS + 4CO Calculate the percent excess reactant, and the degree of completion of the reaction. PROBLEM 8 (20%) A gas cylinder to which is attached a Bourden gauge appears to be at a pressure of 27.38 in. Hg at 70°F. the barometer needs 101.8 kPa. A student claims that the pressure in the tank is 1.3 psia, but another student points out that this is impossible—the pressure is really 28.2 psia. Can 1.3 psia be correct? Explain and show calculations to back up your explanation.

xviii

EXAM NO. 2 (Open Book, 2 hours) PROBLEM 1 (25%) A chemist attempts to prepare some very pure crystals of Na2SO4‑10H2O by dissolving 200 g of Na2SO4 (MolWt=142.05) in 400 g of boiling water. He then carefully cools the solution slowly until some Na2S04·10H20 crystallizes out. Calculate the g of Na2SO4·10H2O recovered in the crystals per 100 g of initial solution, if the residual solution after the crystals are removed contains 28% Na2SO4. Right answer but:

–10 if answer is in g of Na2SO4 and not g of Na2SO4·10H2O

–10 if answer not per 100 g of initial soln

PROBLEM 2 (25%) Water pollution in the Hudson River has claimed considerable recent attention, especially pollution from sewage outlets and industrial wastes. To determine accurately how much effluent enters the river is quite difficult because to catch and weigh the material is impossible, weirs are hard to construct, etc. One suggestion which had been offered is to add a trace Br– ion to a given sewage stream, let it mix well, and sample the sewage stream after it mixes well. On one test of the proposal you add ten pounds of NaBr per hour for 24 hours to a sewage stream with essentially no Br– in it. Somewhat downstream of the introduction point a sampling of the sewage stream shows 0.012% NaBr. The sewage density is 60.3 lb/ft3 and river water density is 62.4 lb/ft3. What is the flow rate of the sewage in lb/min? –10 if answer based on 0.012 fractin and not 0.00012.

–15 if 24 hr basis was used and then not converted back to per hour basis.

PROBLEM 3 (25%) In preparing 5.00 moles of a mixture of three gases (SO2, H2S, and CS2), gases from three tanks are combined into a fourth tank. The tanks have the following compositions (mole fractions): Gas SO2 H2 S CS2

Tank 1 0.10 0.40 0.50

Tank2 0.20 0.20 0.60

Tank3 0.25 0.25 0.50

Tank 4 0.20 0.26 0.54

How much of Tanks 1, 2, and 3 must be mixed to give a product with composition of Tank4? –10 for correct answer but wrong basis

–15 A lot of people said no soln. They used wrong basis, etc. No soln but correct mat’l balance xix

PROBLEM 4 (25%) 10%

a)

For the given distillation process, calculate the composition of the bottoms stream.

15%

b)

If steam leaked into the column at 1000 mole/sec and all else was constant, what would the new bottoms composition be? –5 (should be gmole), if assumed to k-mole and not stated.

xx

Exam No. 3 (Open Book Exam, 2 hours) PROBLEM 1 (35%) A company burns an intermediate product gas having the composition 4.3% CO2, 27% CO, 10% H2, 1.0% CH4, and the residual N2 together with a waste oil having the composition 87% C, 13% H2. Analysis of the stack gas gives an Orsat analysis of 14.6% CO2, 0.76% CO, and 7.65 O2 and the rest N2. Calculate the fraction of the total carbon burned that comes from the product gas. PROBLEM 2 (35%) Benzene, toluene and other aromatic compounds can be recovered by solvent extraction with sulfur dioxide. As an example, a catalytic reformate stream containing 70% by weight benzene and 30% non-benzene material is passed through the counter-current extractive recovery scheme shown in the diagram below. One thousand pounds of the reformate stream and 3000 pounds of sulfur dioxide are fed to the system per hour. The benzene product stream (the extract) contains 0.15 pound of sulfur dioxide per pound of benzene. The raffinate stream contains all the initially charged non-benzene material as well as 0.25 pound of benzene per pound of the non-benzene material. The remaining component in the raffinate stream is the sulfur dioxide. (a)

How many pounds of benzene are extracted per hour, i.e. are in the extract?

(b) If 800 pounds of benzene containing in addition 0.25 pound of the non-benzene material per pound of benzene are flowing per hour at point A and 700 pounds of benzene containing 0.07 pound of the non-benzene material per pound of benzene are flowing at point B, how many pounds (exclusive of the sulfur dioxide) are flowing at points C and D?

xxi

PROBLEM 3 (30%)

Reactant A is polymerized as shown in the figure. It is mixed with fresh catalyst and recycled catalyst. Conversion of A is 40% on one pass through the reactor. Fresh catalyst (G) enters at the rate of 0.40 lb G per lb of A in stream H. The separator removes 90% of the catalyst and recycles it as well as recycling unreacted A. Nevertheless, the product stream P constraints 15% of the unreacted A and 10% of the catalyst exiting in stream K as well as the polymer product. Determine the ratio of stream R to G. Note: catalyst does not react in the process!

xxii

Exam No. 4 (Open Book, 2 hours) PROBLEM 1 (25%) In the vapor-recompression evaporator (not insulated) shown in the figure below, the vapors produced on evaporation are compressed to a higher pressure and passed through the heating coil to provide the energy for evaporation. The steam entering the compressor is 98% quality at 10 psia, the steam leaving the compressor is at 50 psia and 400°F, and 6 Btu of heat are lost from the compressor per pound of steam throughput. The condensate leaving the heating coil is at 50 psia, 200°F. The replacement liquid is at the temperature of the liquid inside the evaporator. Computer: (a)

the work in Btu needed for compression per pound of H2O going through the compressor.

(b)

the Btu of heat transferred from the heating coil to the liquid in the evaporator per pound of H2O through the coil.

(c)

Bonus of 5 points for correct answer to the question: What is the total heat gained or lost by the entire system?

PROBLEM 2 (25%) An insulated, sealed tank that is 2 ft3 in volume holds 8 lb of water at 100°F. A 1/4 hp stirrer mixes the water for 1 hour. What is the fraction vapor at the end of the hour? Assume all the energy from the motor enters the tank. For this problem you do not have to get a numerical solution. Instead list the following in this order: 1. 2. 3. 4. 5. 6.

State what the system you select is. Specify open or closed. Draw a picture. Put all the known or calculated data on the picture in the proper place. Write down the energy balance (use the symbols in the text) and simplify it as much as possible. List each assumption in so doing. Calculate W.

xxiii

7. 8.

Lists the equations with data introduced that you would use to solve the problem. Explain step by step how to solve the problem (but do not do so).

PROBLEM 3 (15%) What is the enthalpy change in Btu when 1 pound mole of air is cooled from 600°F to 100°F at atmospheric pressure. Compute your answer by two ways: 1) Use the tables of the combustion gases. 2) Use the heat capacity equation for air. PROBLEM 4 (10%) Answer the following questions by placing T fro true and F for false on your answer page. Grading: +2 if correct, 0 if blank, -1 if wrong. (a) (b)

Heat and thermal energy are synonymous terms used to express one type of energy. You can find the enthalpy change at constant pressure of a Substance such as CO2 from the solid to the gaseous state by integrating T2

" C dT !

from T1 (solid temperature) to T2 (gas temperature) for a constant pressure path.

T1

(c) (d (e)

The enthalpy change of a substance can never be negative. Heat and work are the only methods of energy transfer in a non-flow process. Both Q and ∆H can be classed as state functions (variables)

PROBLEM 5 (25%) Hot reaction products (assume they have the same properties as air) at 1000°F leave a reactor. In order to prevent further reaction, the process is designed to reduce the temperature of the products to 400°F by immediately spraying liquid water into the gas stream. 400°F?

How many lb of water at 70°F are required per 100 lb of products leaving at

xxiv

For this problem you do not have to get a numerical solution. Instead list the following in this order. 1. 2. 3. 4. 5. 6.

State what the system you select is. Specify open or closed. Draw a picture. Put all the known or calculated data on the picture in the proper places. Write down the material and energy balances (use the symbols in the text) and simplify them as much as possible, list each assumption in so doing. Insert the known data into the simplified equation(s) you would use to solve the problem.

xxv

Exam No. 5 (Open Book, 2 hours) PROBLEM 1 (10%) Answer the following questions briefly (no more than 3 sentences); a.

Does the addition of an inert dilutent to the reactants entering an exothermic process increase, decrease, or make no change in the heat transfer to or from the process?

b.

If the reaction in a process is incomplete, what is the effect on the value of the standard heat of reaction? Does it go up, down, or remain the same?

c.

How many properties are needed to fix the state of a gas so that all of the other properties can be determined? 1 Consider the reaction H 2 ( g ) + O 2 ( g ) ! H 2O ( g ) . 2 Is the heat of reaction with the reactants entering and the products leaving at 500K higher, lower, or the same as the standard heat of reaction?

d.

PROBLEM 2 (10%) Explain how you would calculate the adiabatic reaction temperature for Problem 5 below if the outlet temperature is not specified. List each step. (You can cite some of the steps listed in Problem 5 if you list them by number in Problem 5.) PROBLEM 3 (20%) A flue gas at 750°F and 1 atm of composition 14.0% CO2, 1.0% CO, 6.4% O2, and the balance N2 is the product of combustion with excess air at 750°F and 1 atm that is used to burn coke (C). a.

What is the volume in ft3 of the flue gas leaving the furnace per pound of carbon burned?

b.

What is the volume of air in ft3 entering the furnace per pound of C burned?

PROBLEM 4 (20%) Seven pounds of N2 are stored in a cylinder 0.75 ft3 volume at 120°F. Calculate the pressure in the cylinder in atmosphere: a.

Assuming N2 to be an ideal gas.

b.

Assuming N2 is a real gas and using compressibility factors.

PROBLEM 5 (40%—one half each for the material and energy balances) Pyrites (FeS2) is converted to sulfur dioxide (SO2) gas according to the reaction

4FeS2 ( s ) + 11 O 2 ( g ) ! 2 Fe 2O3 ( s ) + 8 SO 2 ( g )

xxvi

The air, which is 35% excess (based on the above reaction) for combustion, enters at 27°C, the ore at 18°C, and the products leave at 900K. Because of equipment degradation, unburned FeS2 exits from the process. In one hour 8000 kg of pyrites are fed to the process, and 2000 kg of Fe2O3 are produced. What is the heat added or removed from the process? Data: For FeS2, C p = 44.77 + 5.590 ! 10 –2 T where T is in Kelvin and Cp is in J/(g mol)(K). For F2O3, C p = 103.4 + 6.71! 10"2 T with the same units.

xxvii

Exam No. 6 (Open Book, 2 hours) PROBLEM 1 (20%) A high pressure line carries natural gas (all methane) at 10,000 kPa and 40°C. How would you calculate the volume of the gas under these conditions that is equivalent to 0.03 m3 of CH4 at standard conditions using an equation of state? Select one equation other than van der Waal's equation, and list it on your solution page. Give a list of steps to complete the calculations. Include all the proper equations, and include a list of data involved, but you do not have to obtain a solution for the volume. PROBLEM 2 (20%) From the following data estimate the vapor pressure of sulfur dioxide at 100°C. Temperature (°C) Vapor pressure (atm)

–10 1

6.3 2

32.1 5

55.5 10

PROBLEM 3 (20%) Dry atmospheric air at the ambient conditions of 90°F and 29.42 in. Hg absolute passes through a small blower and is bubbled up through water so that the air leaving the water is saturated. The temperature of the water is constant at 80°F, and because of the back pressure in the system, the pressure in the vapor space in the top of the bottle is 2.7 in.

H2O greater than atmospheric pressure. The bottle is weighted after the air is blown for 2 hours, 13 minutes, 47 seconds, and the decrease in weight was found to be 8.73 lb. What was the hourly rate of flow of air at ambient conditions in ft3? PROBLEM 4 (20%) A vessel with a volume of 2.83 m3 contains a mixture of nitrogen and acetone at 44.0°C and 100.0 kPa. The dew point of the mixture is 20.0°C and the relative saturation of the acetone in the mixture is 58.39%. The vapor pressure of acetone at 44.0°C is 65.35 kPa and it is 24.62 kPa at 20.0°C. a.

What is the partial pressure of acetone vapor in the original mixture, in kPa?

b.

How many kg moles of acetone does the original mixture contain?

c.

If the nitrogen-acetone mixture is cooled with the volume remaining at 2.83 m3 constant so that 27.0 percent of the acetone condenses, what is the final temperature of the mixture in °C?

xxviii

PROBLEM 5 (20%) A wet sludge contains 50 percent water by weight. This sludge is first centrifuged, and 0.1 kilograms of water are removed per kilogram of wet sludge feed. The sludge is dried further using air so that the final product contains 10 percent by weight water. The air for drying is heated, passed into an oven drier, and vented back into the atmosphere. On a day when atmospheric pressure is 760mm Hg, the temperature is 70°F and the relative humidity is 50%, calculate the cubic meters of wet air required to dry one kilogram of sludge fed to the process. The air vented from the oven drier is at 100°F and 780mm Hg. It has a dew point of 94°F.

xxix

Solutions Chapter 2

2.1.1

2.2.1

(a) (b) (c) (d) (e) (f) (g) (h)

a.

N/mm or nm (nanometer) °C/M/s 100 kPa 273.15 K 1.50m, 45 kg 250°C J/s 250 N Basis: 1 mi3

1mi3 ⎛ 5280 ft ⎞ 3 ⎛ 12 in ⎞ 3 ⎛ 2. 54 cm ⎞ 3 ⎛ 1 m ⎞ 3 ⎝ 1 mi ⎠ ⎝ 1 ft ⎠ ⎝ 1 in ⎠ ⎝ 100 cm ⎠ 9 3 = 4.17 × 10 m b.

Basis: 1 ft3/s

a.

0.04 g 60 min (12 in )3 1 lb m lb m = 9.14 3 3 (min )( m ) 1 hr 1 ft 454g (hr ) (ft 3 )

b.

2L 3600 s 24 hr 1 ft ft 3 = 6.1× 10 3 s 1 hr 1 day 28.32L day

2.2.2

1 ft 3 60 s 7.48 gal = 449 gal / min 1 s 1 min 1 ft 3

3

c.

1 ft 2 (1 in)2 1 yr 1 day 1 hr 2.2 lbm 6 (in.)(cm 2 ) 2 2 2 (yr)(s)(lbm )(ft ) (12 in) (2.54 cm) 365 days 24 hr 3600s 1 kg

m 1 ft 0.3048 m –11 = 1.14 × 10 12 in 1 ft ( kg) (s2 )

2–1

Solutions Chapter 2

2.2.3

a. Basis: 60.0 mile/hr

60.0 mile 5280 ft 1 hr ft = 88 hr 1 mile 3600 sec sec b. Basis: 50.0 lbm/(in)2

50.0/lbm 454 g 1 kg 1(in)2 (100 cm)2 kg = 3.52 × 104 2 2 2 2 (in) 1 lb 1000 g (2.54 cm) (1 m) m c. Basis: 6.20 cm/(hr)2

6.20 cm 1 m 109 nm 1(hr) 2 nm = 4.79 2 2 2 (hr) 100 cm 1 m (3600 sec) sec

2.2.4

20 hp 0.7457 kW = 14.91 kW 1 hp

No , not enough power even at 100% efficiency; 68 kW = 91.2 hp.

2.2.5

1 hr 2200 gal 1000 mile = 4190.5 gal 525 mile 1 hr

1 hr 2000 gal 1000 mile 4210 gal = 475 mile 1 hr (20 gal) None: 20 gal more are needed.

2.2.6

Let tA be the time for A to paint one house; tB for B A does a house in 5 hours, or 1 house/5 hr. B does one house in 3 hours, or 1 house/3 hr.

1 house tA hr 1 house tB hr + = 1 house 5 hr 3 hr Also tA = tB so that

tA =

3 5 8 tA + tA = 1 or tA =1 15 15 15

15 hr = tB = 1.875 hr or 112.5 min 8

2–2

Solutions Chapter 2

2.2.7

2.2.8

(a) mass, because masses are balanced (b) weight, because the force exerted on the mass pushes a spring

2 1(hr)2 20.0g 1 lbm 0.3048m 3600s 1 (lbf )(s ) 1 hr 32.174(lbm )(ft) (3600)2s 2 (m)(s) 453.6 g 1 ft

= 1.16 × 10−7

2.2.9

(lbf )(hr) ft 2

1.0 Btu 24 hrs 1 ft 2 1 in 2 (100 cm)2 1.8o F 2.54 cm ⎛ o F ⎞ 1 day (12 in)2 (2.54 cm 2 ) 1 m 2 in 1oC (hr)(ft 2 ) ⎜ ⎟ ⎝ ft ⎠ kJ 252 cal 12 in 4.184 J 1 kJ = 1.49 × 104 2 1 Btu 1 ft 1 cal 1000 J (day)(m )(°C / cm)

2.2.10

Basis: 1 lb H2O 3

a.

1 2 1 1 lb m ⎛ 3 ft ⎞ (s 2 )(lb f ) KE= mv = = 0.14(ft)(lbf ) ⎜ ⎟ 2 2 ⎝ s ⎠ 32.174(ft)(lb m )

b.

Let A = area of the pipe and v = water velocity. The flow rate is

⎛ πD 2 ⎞ (v) q = Av = ⎜ ⎝ 4 ⎟⎠ π (2 in)2 (1 ft)2 3 ft 60 s 7.48 gal = 29.37 gal/min 4 (12 in)2 S 1 min 1 ft 3

2.2.11

PE =

75 gal 8.33 lbm 32.2 ft 100 60 s s 2 -lbf Btu = 4818 Btu/hr 2 min ft hr 32.2 lbm -ft 778 ft-lbf gal sec

2 hp 2545 Btu = 5090 Btu/hr hp-hr Rate of energy input for heating = PW - PE =5090 - 4818 = 272 Btu/hr Pump Work =

2–3

Solutions Chapter 2

2.2.12

The object has a mass of 21.3 kg (within a precision of ± .1 kg). The weight is the force used to support the mass.

2.2.13

In American Engineering System Power =

=

FV

800 lbf 300 ft 1 min

=

2.4 ×105

(lbf )(ft) or 7.27 hp min

In SI Power =

4000 N 1.5 m 1 (watt)(s) 1 s 1(N)(m)

=

6000 watts

=

1 m v2 2

2.2.14 KE

=

1 2300 kg 1 lb m ⎛ 10.0 ft ⎞ 2 1 1 Btu 2 0. 454 kg ⎝ s ⎠ 32. 2 lb m ft 778.2(ft )(lb f ) 2 lb f sec = 10.11 Btu

2.2.15

Basis: 10 tons at 6 ft/s 2

KE =

2.2.16

1 1 20,000 lbm ⎛ 6 ft ⎞ 1(s 2 )(lbf ) 2 mv = = 11,200(ft)(lbf ) ⎜ ⎟ 2 2 ⎝ s ⎠ 32.2(ft)(lbm )

Basis: 1 mRNA 1 nRNA 3x ribonucleotides 1 active ribosome nRNA

1200 amino acids

protein

264 ribonucleotides min-active ribosome x amino acids

13.6 protein molecules formed per min per nRNA

2–4

=

Solutions Chapter 2

2.3.1

2.3.2

A is in g/cm3 B is in g/(cm3)(°C) Since the exponent of e must be dimensionless C is in atm-1

a 1. a 2. a 3.

b1

A=

b2

B=

b3

C=

lb 1.096 g (30.48)3cm3 1 lb m = 68.4 m3 cm3 ft 3 454g ft lb 0.00086 g (30.48)3cm3 1 lbm 1oC = 0.0298 3 mo 3 o 3 o (cm )( C) ft 454 g 1.8 R (ft )( R) 0.000953 1 atm 1 = 0.0000648 2 atm 14.7 lbf /in lb f /in 2

Introduce the units. The net units are the same on both sides of the equation.

⎛ ft ⎞ (ft)(ft) ⎜ 2 ⎟ ⎝s ⎠ 1.5

1/2

=

ft 3 s 1/ 2

2.3.3 No.

⎡ ⎤ ⎢ ⎥ -3 3 1 ⎥ 2 ⎢ (2) 9.8m 10 m 50×10 (kg)(m) q = 0.6(2m ) ⎢ 2 ⎥ s2 kg (s 2 )(m 2 ) ⎛ 2 ⎞ ⎢ 1- ⎜ ⎟ ⎥ ⎢ ⎝ 5 ⎠ ⎥⎦ ⎣

The net units on the right hand side of the equation are 1/2

⎡ m3 ⎤ m ⎢ 4 ⎥ ⎣ s ⎦ 2

≠

m3 s

Consequently, the formula will not yield 80.8 m3/s, presumably in the formula the g should be gc for use in the AE system.

2–5

Solutions Chapter 2 1.19 2.3.4

Q = 0.61S

(2Δp)/ρ

assume hole is open to atmosphere

⎡ lbf ⎤ 14.7 ⎢ ⎥ 73 in gas. 0.703 H 2O 1 ft 144 in ⎢ 23 lbf in 2 ⎥ + Δp = 1 gas 12 in 33.91 ft H 2O ⎥ 1 ft 2 ⎢ in 2 ⎢ ⎥ ⎣ ⎦ 2 = 3,579 lbf /ft 2

62.4

lb

ρ=

0.703ft3 3

1 lb/ft H 2O

lbH 2O ft 3H 2O

=43.87

lbm ft 3

2

⎛ 1 ⎞ S = π⎜ = 3.41× 10-4 ft 2 ⎟ ⎝ (4)12 ⎠ Q = (3600)(0.61)(3.41× 10−4 ) (2)(3579)g c / 43.87 = 54

2–6

ft 3 hr

Solutions Chapter 2

2.3.5

a.

b.

Z

= 1 + ρB + ρ2C + ρ3D

B C D

Units cm3 / g mol (cm3/ g mol)2 (cm3/ g mol)3

Z

= 1 + ρ* B* +(ρ* )2C* +(ρ* )3D*

B* C* D*

Units ft3 / lbm (ft3/ lbm)2 (ft3/ lbm)3

If B is the original coefficient, B* is obtained by multiplying B by conversion factors. Let MW is the molecular weight of the compound.

ft 3 B cm3 B = lb m g mol *

2

3

0.016 ⎛ 1 ft ⎞ 1 g mol 454 g = B ⎜ ⎟ MW ⎝ 30.48 cm ⎠ MW g 1 lb m 2

6

2

2

⎛ ft 3 ⎞ ⎛ cm3 ⎞ ⎛ 1 ft ⎞ ⎛ 1 g mol ⎞ ⎛ 454g ⎞ 2.57 × 10-4 C ⎜ ⎟ = C ⎜ ⎟ = C ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ MW 2 ⎝ g mol ⎠ ⎝ 30.48 cm ⎠ ⎝ MWg ⎠ ⎝ 1 lbm ⎠ ⎝ lb m ⎠ *

3

⎛ ft 3 ⎞ 4.096×10-6 D ⎜ ⎟ = D MW 3 ⎝ lb m ⎠ *

2–7

Solutions Chapter 2

2.3.6

⎡m ⎤ u ⎢ ⎥ = k[1] ⎣s⎦

⎡⎛ τ ⎞ 1/2 ⎛ N ⎞ 1/2 ⎛ m 3 ⎞ 1/2 ⎤ ⎢⎜ ⎟ ⎜ 2 ⎟ ⎜ ⎥ ⎢⎝ ρ ⎠ ⎝ m ⎠ ⎝ kg ⎟⎠ ⎥ ⎣ ⎦

To get u in ft/s, substitute for τ and for ρ , and multiply both sides of the equation by 3.281 ft/1 m (k is dimensionless). 2

τ

τ′lbf ⎛ 3.281 ft ⎞ N 1N = 2 2 ⎜ ⎟ m ft ⎝ 1 m ⎠ 0.2248 lbf 3

ρ′ lbm ⎛ 3.281 ft ⎞ kg 1 kg ρ 3 = m ft 3 ⎜⎝ 1 m ⎟⎠ 0.454 lbm ⎛ τ′ ⎞ u ′ = 2.57 k ⎜ ⎟ ⎝ ρ′ ⎠

2.3.7

Place units for the symbols in the given equation, and equate the units on the left and right hand sides of the equation by assigning appropriate units to the coefficient 0.943. LHS

RHS

3 ⎡⎛ ⎤ ⎞⎟ ⎛ lb m ⎞ 2 ft Btu (hr)(ft ) Btu Btu ⎜ ? (hr ) ft 2 (Δ ° F) ⎢⎣⎝ (hr )(ft )(Δ ° F)⎠ ⎝ ft 3 ⎠ (hr)2 lb m ft lb m Δ ° F ⎥⎦

1

4

( )

The units are the same on the right and left so that 0.943 has no units associated with it.

2.3.8

η

=

numerator denominator

Numerator

=

Yxc/ s γ b ΔHbc/ e− − ⎛ mol cell C ⎞ ⎛ e equiv. ⎞⎛ energy ⎞ ⎟⎜ ⎟ ⎟ ⎜ − ⎝ mole substrate C ⎠ ⎝ mol cell C ⎠ ⎝ mol e equiv. ⎠

= ⎜

=

energy mol substrate C

(1)

2–8

Solutions Chapter 2

Denominator

η

= ΔH ccat =

energy mol substrate C

=

(1) energy = (2) energy

(2)

There is no missing conversion factor What the author claimed about the units is correct. For dimensional consistency: B – absolute temperature in either ºR or K C – absolute temperature in either ºR or K A – dimensionless

2.3.9

In most cases the argument of a logarithm function should be dimensionless, but in this case it will not be. Therefore, the numerical values of A, B, and C will depend upon the units used for temperature and the units used for p*. 2.3.10

The equation is

Δp = 4fρ ⎡⎣(v 2 / 2g) (L/D) ⎤⎦ The units on the right hand side (with f dimensionless) in SI are

ρ

⎞ kg ⎛ m 2 ⎜ m ⎟⎟ 1 m 3 ⎜ s2 → 2 ⎜ (kg)(m) m ⎟ m ⎟ ⎜ s2 ⎠ ⎝

hence the equation is not dimensionally consistent because Δp has the units of N/m2. If g is replaced with g, the units would be correct.

2.4.1

Two because any numbers added to the right hand side of the decimal point in 10 are irrelevant.

2.4.2

The sum is 1287.1430. Because 1234 has only 4 significant figures to the left of an implied decimal point, the answer should be 1287 (no decimal point).

2–9

Solutions Chapter 2

2.4.3

The number of significant figures to the right of the decimal point is 1 (from 210.0m), hence the sum of 215.110 m should be rounded to 215.1 m.

2.4.4

A calculator gives 569.8269 000, but you should truncate to 4 significant figures, or 569.8 cm2.

2.4.5

Two significant figures (based on 6.3). Use 4.8 × 103.

2.4.6

Step 1: The product 1.3824 is rounded off to 1.4 Step 2: Calculate errors. For absolute error, the product 1.4 means 1.4 + 0.05 Thus

0.05 × 100% = 3.6% error 1.4

Similarly 3.84 has

0.005 × 100% = 0.13% error 3.84

and 0.36 has

0.005 × 100% = 1.4% error 0.36 Total 2.7% error

2.6.1

a)

4 g mol mg Cl 2 (95.23)g MgCl 2 = 380.9 g gmol MgCl 2

b)

2 lb mol C3H8 (44.09 )lb C 3H8 454g C3 H8 4 = 4 × 10 g C3 H8 lb mol C3 H8 1 lb C3 H8

c)

(d)

16 g N2

gmol N 2 1 lb mol N 2 = 1.26 × 10−3 lb mol N 2 (28.02)g N2 454 gmol N 2

3 lb C2 H6O 1 lb mol C2 H6O 454 g mol = 29.56 g mol C2 H6O (46.07) lb C2 H6O 1 lb mol

2–10

Solutions Chapter 2

2.6.2

2.6.3

(a)

16.1 lb mol HCl 36.5 lb HCl = 588 lb HCl 1 lb mol HCl

(b)

19.4 lb mol KCl 74.55 lb KCl = 1466 lb KCl 1 lb mol KCl

(c)

11.9 g mol NaNO3

(d)

164 g mol SiO2 60.1 g SiO2 2.20 ×10-3lb = 21.7 lb SiO2 1 g mol SiO2 1g

85 g NaNO3 2.20 ×10-3lb = 2.23 lb NaNO3 1 g mol NaNO3 1g

Basis: 100 g of the compound comp. C O H

g 42.11 51.46 6.43

MW 12 16 1.008

g mol 3.51 3.22 6.38 13.11

Ratio of Atoms 1.09 1 2

Multiply by 11 to convert the ratios into integers The formula becomes C12O11H22 Checking MW: 12(12) + 11(16) + 22(1.008) = 342 (close enough)

2–11

Solutions Chapter 2

2.6.4

Vitamin A, C20O H30, Mol Wt.: 286

Vitamin C, C6 H8O6, mol. wt: 176

Vitamin A: a.

Vitamin A =

2.00 g mol 286 g 1 lb = 1.26 lb 1 g mol 454 g

16 g 1 lb = 0.0352 lb 454 g Vitamin C =

2.00 g mol 176 g 1 lb = 0.775 lb g mol 454 g

16 g 1 lb = 0.0352 lb 454 g b.

Vitamin A =

1.00 lb mol 286 lb 454 g = 130,000 g 1 lb mol 1 lb

Vitamin C =

1.00 lb mol 176 lb 454 g = 79,900 g 1 lb mol 1 lb

For both

12 lb 454 g = 5450 g 1 lb

2.6.5

1 kg 1160 m 3 = 223.1 kg/kg mol 5.2 m 3 kg mol

2.6.6

Mass fraction to mole fraction:

ω1 MW1 x1 = ω1 (1− ω1 ) + MW2 MW1 Mole fraction to mass fraction

ω1 =

x1MW1 (x1 )(MW1 ) + (1− x1 )(MW2 )

2–12

Solutions Chapter 2

2.6.7

Basis: 100 kg mol gas Comp. CH4 H2 N2 Total

Mol % = mol 30 10 60 100

MW 16 2 28

kg 480 20 1680 2180

21.8 kg kg mol

Basis: 100 g mol gas

2.6.8

Comp. CO2 N2 O2 H 2O

mol = % 19.3 72.1 6.5 2.1 100.0

Avg. mol. wt =

MW 44 28 32 18

g 849.2 2018.8 208.0 37.8 3113.8

3113.8 = 31.138 100.0 Basis: 100 lb mol

2.6.9

Comp. CO2 CO N2

% = mol 16 10 30

MW 44 28 28

lb 2640 280 840 3760

Avg. MW = 37.6 lb/lb mol

2.7.1

(a) A gas requires a convenient basis of 1 or 100 g moles or kg moles (if use SI units). (b) A gas requires a convenient basis of 1 or 100 lb moles (if use AE units). (c) Use 1 or 100 kg of coal, or 1 or 100 lb of coal because the coal is a solid and mass is a convenient basis. (d) Use 1 or 100 moles (SI or AE) as a convenient basis as you have a gas. (e) Same answer as (e).

2–13

Solutions Chapter 2

2.7.2

Since the mixture is a gas, use 1 or 100 moles (SI or AE) as the basis.

2.7.3

Pick one day as a basis that is equivalent to what is given—two numbers: (a) 134.2 lb C1

2.8.1

(b) 10.7 × 106 gal water.

Basis: 1000 lb oil

lb oil lb H2 O 62.4 3 ft ft 3 = 57.78 lb oil lb H2 O ft 3 oil 1.00 ft 3 0.926

1000 lb oil

2.8.2

1 ft 3 oil 7.48 gal 57.78 lb oil 1 ft 3 = 129.5 gal

Basis: 10,010 lb

10,010 lb 1 gal 0.134 ft3 = 3 8.80 lb gal 152 ft Basis: 1 g mol each compound

2.8.3

g mol

mw

g

V (cm3)

Pb

1

207.21

207.21

11.33

18.3

Zn

1

65.38

65.38

7.14

9.16

C

1

12.01

12.01

2.26

5.31

ˆ = V

2.9.1

ρ (g/cm3)

Component Na C1 O

g mol 1 1 3 5

mass (g) density (g/cm 3 )

mol fraction 0.20 0.20 0.60

2–14

Set 2 is the correct one

MW 23 35.45 16

g 23 35.45 48 106.45

mass fraction 0.22 0.33 0.45 1.00

Solutions Chapter 2

2.9.2

Basis: 1 gal of solution. Mass of solution:

1.0824 lb soln lb H O 62.4 3 2 3 ft H 2O ft H2 O 1 ft 3 1 gal = 9.03 lb soln . lb H2 O 7. 481 gal 1.00 3 ft H 2O Mass fraction KOH =

0.813 lb 9.03 lb

Mass fraction H2O = 1 – 0.09

0.09

0.91 1.00 Total

Basis: 30 lb gas

2.9.3

Comp. CO2 N2

2.9.4

=

=

lb 20 10 30

a)

1000

b)

No

MW 44 28

lb mol 0.455 0.357 0.812

mol fr. 0.56 0.44 1.00

c) Yes, because for solids and liquids the ratio in ppb is mass, whereas for gases the ratio is in moles.

2–15

Solutions Chapter 2

2.9.5

On a paper free basis the total ppm are: Brand A: 6060 ppm

Brand B: 405 ppm

The respective mass fractions are: The other entries are similar

Brand A:

Fe

Cu

1310 = 0.216 6060

2000 = 0.330 6060

350 = 0.864 405

50 = 0.123 405

2750 = 0.454 6060 Brand B:

5 = 0.0123 405

Vs =

5 m 30.2 m 200 mm

1m 2 sides 3 = 60.4 m 1000 mm

5 m 27.2 m 200 mm

1m 2 ends = 54.4 m3 1000 mm

Ends

Ve =

2–16

Pb

Solutions Chapter 2

Floor

Vf =

27.4 m 30.4 m 200 mm

1m 1000 mm

= 166.592 m

Total volume = 281.392 m3 Mass of concrete =

281.392 m3 2080 kg = 585,295 m3

Volume of displaced water required to float:

586,543.36 kg 1 m 3 H2 O = 586.543 m 3 1000 kg H2 O V = LWh → h = h=

2.9.6

V LW

586.54 m 3 27.4m 30.4 m

Basis: 190,000 ppm

190,000 g PCB × 100 = 19% 106

2–17

= 0.703 m

3

Solutions Chapter 2

Basis: 100 g of the sample

2.9.7

The biomass sample is g = % dry weight of cells C O N H P

50.2 20.1 14.0 8.2 3.0 95.5 4.5 100.0

other Total

10.5 g cells 50.2 g C 1 g mol C = 0.439 g mol C/g mol ATP g mol ATP 100 g cells 12 g C 2.9.8

MMM(s) → NN(s) + 3CO2 (g) a.

b.

2.9.9

2×107 disintegrations 1 min min

60

1 curie

106µ curie

3×1010 disintegrations/s

1 curie

= 11µ curie

2 ×107 disin tegrations 0.80 = 1.6 ×107 cpm min

The relation to use is t1/ 2 = l n(2) /(k)(OH − ) with (OH − ) = 1.5 ×106 k

t1/ 2 (sec onds)

Methanol Ethanol MTBE

30.8 ×105 4.6 ×105 7.7 ×105

0.15 ×10−12 1×10−12 0.60 ×10−12

The order is in increasing persistence ethanol, MTBE, and methanol.

2–18

Solutions Chapter 2

2.9.10

Yes.

Bases are first entries.

4800 g mol CC14 103 g mol air kg mol air 154 g CC14 103 mg CC14 = 109 g mol air kg mol air 22.8 m3 g mol CC14 g CC14

32.4 mg CC14 /m3 which exceeds the NIOSH standards 2.9.11

a.

25,600 ton P 2.6 yr 1 1 gal 2000 lb 454 g 106 µ g yr 1.2 × 1014 gal 3.785 L 1 ton 1 lb 1 g

= 133.1µ g / L b.

19,090 lb(P )municipal = 63. 4% 30,100 lb(P)total

c.

19,090 lb(P )municipal 0.70 lb(P)det. 100 lb(P )tota l = 44.4% 30,100 lb(P)total 1 lb(P)municipal 30,100 lb(P)tota l

For d. and e. assume that (P)outflow remains unchanged. d.

P retained/yr = 2,240 + 6,740 + 0.7

× 19,090 – 4,500 = 17,843 ton/yr.

P conc. in ppb =

17,843 ton 2.6 yr 1 2000 lb 1 g / cm 3 109 g = 92.6 ppb yr 1.2 × 1014 gal ton 8.345 lb / gal 1billion g This is greater than 10 ppb. Eutrophication will not be reduced. e.

P retained/yr = 2,240 + 6,740 + 0.3

P conc. in ppb =

× 19,090 – 4,500 = 10,207

ton yr

10,207 ton 2.6 yr 2000 lb 1 g / cc 109 g = 53.2 ppb 1.2 × 1014 gal ton 8.345 lb / gal 1 billion g yr

This will not help.

2–19

Solutions Chapter 2

Basis: 106 g mol gas

2.9.12

350 g mol H 2S 1 g mol gas 34g H 2S 1 g mol CO2 106g mol gas 1 g mol CO2 1 g mol H 2O 44 g CO2 =

270 g H 2S 270 g H 2S = 6 6 10 g CO2 10 g total liquid

mass fraction H2S = 2.70 ×10−4

2.9.13

(a)

1 mol O2 104 mol O2 ⇒ or 104 ppm 6 100 mol gas 10 mol gas

(b)

Basis: 100 mol gas

Comp. SO3 SO2 O2 Total

2.9.14

MW

answer

% = mol 55 3 1 59 CaCO3: Ca Mg C O

mol fr. 0.932 0.051 0.017 1.000

or mol % 93.2 5.1 1.7 100.0

100.06 40.05 24.3 12.01 16.00

100.06 g CaCO3 1 g mol CaCO3 1 g mol Ca g CaCO3 = 2.50 g mol Ca CO3 1 g mol Ca 40.05 g Ca g Ca Similarly

g CaCO3 g CaCO3 = 4.118 g Mg g Mg

Total alkalinity = 2.50 (56.4) + 4.118 (8.8) = 177

2–20

mg CaCO3 L

Solutions Chapter 2

1 molecule in 1023 or more is not 13-20 ppb

2.9.15

No.

2.9.16

On a mol basis, the carbon dioxide concentration in air is about 350 parts per million (ppm), while that of oxygen is about 209,500 ppm. If the atmospheric concentration of carbon dioxide is increasing at about 1% per year (i.e., from 350 ppm this year to 353.5 ppm next year), and not to 1%, the 3.5-ppm change in dioxide concentration causes the oxygen concentration to fall from 209,500 to about 209,497 ppm, which is less than a 0.002% decrease. So, there is no need to worry about an oxygen deficit at present.

2.10.1

TK = –10 + 273 = 263K T°F = –10 (1.8) + 32 = 14°F T°R = 14 + 460 = 474°R

2.10.2

Yes, if the temperature scale is a linear relative one (°C, °F), or a logarithmic scale (ln 1° is zero). No, if the scale is absolute, but read J. Wisniak, J. Chem. Educ., 77, 518-522 (2000) for a different conclusion.

2.10.3

C p = 8.41 +

2.4346 × 10-5TK 

⎛ ⎞ J ⎜ 2.4346×10-5 ⎟ (T ) ⎜⎝ (gmol)(K)2 ⎟⎠ K

Substitute

Cp

= =

1.8 TK = T°R

8.41 + 2.4346 × 10œ5

J ⎛ T°R ⎞ 2 ⎝ (gmol)(K) 1.8 ⎠

8.41 + 1.353 ×10−5 To R

2–21

Solutions Chapter 2

a)

10° C 1.8° F 1.0° C + 32 = 50°F

b)

10°C 1.8°F 1.0°C + 32 + 460°R = 510°R

2.10.4

c)

d)

−25o F – 32o F 1.0oC + 273K = 241.3K 1.8 1.8o F

150K 1.8° R 1.0K = 270°R

2.10.5 First multiply the RHS of the equation so that

Btu (lbmol)(°F) = 4.182

1054.8J 1 Btu

1 lb mol 454 gmol

1.8°F 1.0°C

1°C 1K

J (gmol)(°K)

and substitute T°F = 1.8T°C + 32

C p = [8.488 + 0.5757 × 10−2 (1.8T°C + 32) − 0.2159 × 10−5 (1.8T°C + 32)2 + J 0.3059 × 10−9 (1.8T°C + 32)3 ]4.182 (gmol)(°K) Simplifying, Cp = 36.05 + 0.0447T – 0.2874

2.10.6

× 10–4 T2 + 0.7424 × 10–8 T3

The instrument does not contain mercury, but has to contain a fluid that responds at –76°C and can be calibrated to measure temperature.

2–22

Solutions Chapter 2

Basis: 15cm3 water

2.11.1

1000 kg 10 m 10 m 0.15 m = 15,000 kg m3

a.

m = ρV=

b.

F m g 15,000 kg 9.80 m 1 N 1 Pa = = = 1470 Pa 2 A A s 10 m 10 m 1 (kg)(m) N s2 m2

= 1.47 kPa 1.470 kPa

or

2.11.2

1 atm 14.7 psi = 0.21 psi 101.3 kPa atm

15 cm H2O

1 in. 1 ft 14.696 psi = 0.21 2.54 cm 12 in. 33.91 ft H2 O

ρconcrete = 2080 kg/m 3 ρwater = 1000 kg/m 3 Volume of concrete: Sides

Vs =

5 m 30.2 m 200 mm

1m 2 sides 3 = 60.4 m 1000 mm

5 m 27.2 m 200 mm

1m 2 ends = 54.4 m3 1000 mm

Ends

Ve = Floor

Vf =

27.4 m 30.4 m 200 mm

1m 1000 mm

2–23

= 166.592 m

3

Solutions Chapter 2 Total volume = 281.392 m3 Mass of concrete =

281.392 m3 2080 kg = 585,295 m3

Volume of displaced water required to float:

586,543.36 kg 1 m 3 H2 O = 586.543 m 3 1000 kg H2 O V = LWh → h = h= 2.11.3

V LW

586.54 m 3 27.4m 30.4 m

= 0.703 m

The pressure is a gauge pressure.

60°F

Basis: 50.0 psig a.

50.0 psig 33.91 ft H2 O = 115 ft 14.7 psia (difference)

b.

No. Insufficient height.

Alternate solutions can be applying ∆p = ρΔhg 2.11.4

lbf in.2 = lbf → lbm in the AE system in.2 so the procedure is ok, although the unit conversion is ignored.

2–24

148 ft

Solutions Chapter 2

2.11.5

Basis: Dept = 1000 m

p = p o + ρgh p = 1 atm +

1000 m 1.024 g 9.8 m 1 kg ⎛ 100 cm ⎞ cm 3 s2 103 g ⎝ 1 m ⎠

=

3

1 N 1 kPa 1 atm (kg)(m ) 1N 1.013 × 10 5 N / m 2 (kg)(m ) s2 2 s

100.1 atm 101.3 kPa 4 = 1.014 × 10 kPa 1 atm

Alternative solution: 101.3 kPa +

1000 m sea H 2 O 1.024 g H 2 O 3.28 ft 1.00 g sea H 2 O 1 m

101.3 kPa 33.91 ft H 2 O

= 1.003 × 104 + 0.01× 104 = 1.013 × 104

2.11.6

(a) p ABS = pGauge + patm ⎛ 22.4 lbf 144 in.2 ⎞ ⎛ 28.6 in. Hg 14.7 psia 144 in.2 ⎞ 2 ⎟ = 5250 lbf /ft ⎜ ⎟ + ⎜ 2 2 2 1 ft ⎠ ⎝ 1 29.92 in. Hg 1 ft ⎠ ⎝ 1 in. ⎛ 22.4 psig 29.92 in. Hg ⎞ + 28.6 in. Hg = 74.2 in. Hg 14.7 psia ⎟⎠

(b) ⎜ ⎝

⎛ 22.4 psig 1.013 × 105 N/m 2 ⎞ ⎛ 28.6 in. Hg 1.013 × 105 N/m 2 ⎞ +⎜ = 2.51× 105 N/m 2 ⎟ ⎟ 14.7 psia 27.92 in. Hg 1 1 ⎝ ⎠ ⎝ ⎠

(c) ⎜

⎛ 22.4 psig 33.91 ft H2 O ⎞ ⎛ 28.6 in. Hg 33.91 ft H 2O ⎞ ⎟ + ⎜ ⎟ = 84.1 ft H 2 O 1 14.7 psia ⎠ ⎝ 1 29.92 in. Hg ⎠ ⎝

(d) ⎜

2.11.7

Neither John is necessarily right. The pressure at the top of Pikes Peak is continually changing.

2–25

Solutions Chapter 2

2.11.8

Δp = ρgh a.

=

1 g 9.8 m 13.1 m 1 kg (100 cm )3 Pa 1 kPa 1 kg 1000 Pa cm 3 s2 1000 g m3 m s2

= 128. 4 kPa b.

mg ρVg ρ(S.A.)(t )g F = = = A A Bottom A Bottom ABottom

A Bottom

2

π 2 π (30.5 m ) 2 = D = = 730.62 m 4 4 2

ATop = 730.62 m

Aside = πDh = π(30.5 m)(13.1m) = 1255.2 m

2

S.A.= ATop + A Bottom + ASide 2

= (730.62 + 730.62 +1255.22 ) m = 2716.46 m

2

ρ(S.A.)(t )g = A

(

)

3 7.86 g 100 cm 2,716.46 m 2 9.35 × 10 œ3 m 9.8 m 3 3 2 cm s2 m 730.6 m

1 kg 1 Pa 1 kPa = 2 1000 g 1 kg / ms 1000 Pa

2–26

2.68 kPa

Solutions Chapter 2

2.11.9

p A + ρ1gh1 + ρ 2gΔh = PD p B + ρ1gh1 + ρ 2gΔh = PD

pA -pB = ρ1gh1 +ρ2gΔh − ρ 1gh1ρ− 1gΔ h = ρ2gΔh1 − ρ1gΔh = gΔh(ρ2 − ρ1 ) =

9.8 m 0.78 in 1 m 13.546 œ0.91g 1 kg s2 3.28 ft cm 2 1000 g

1 N 1 ft ⎛ 100 cm ⎞ 3 m 2 1 Pa 760 mm H2 12 in ⎝ 1 m ⎠ 1 kg 1 N 101.3 × 10 3 Pa (m )(s)2 m 2

=18.4 mm Hg

The pressure at A is higher than the pressure at B. Alternate solution: Δp =

(0.78 in.) (13.546 − 0.91) 760 mm Hg = 18.5 mm Hg 13.546 29.92 in. Hg

2–27

Solutions Chapter 3

3.1.1

Boundary:

Around both pumps and include the soil at the end of the pipes. It’s an open system. It’s at steady-state except at startup (note system boundary limits fluid), or if fluid enters, unsteady state.

3.1.2

Either open (flow) or closed (batch) is acceptable if explanation is given (1)

flow – material comes in and out continuously over a suitably long period of time

(2)

batch – material is injected into the system, and then in a short period of time, a reaction occurs with the system valves closed.

(a)

open if you have to replace water, and water evaporates; otherwise closed

(b)

open

3.1.3

3.1.4

If the overall flows in and out over a time period for several batches are considered, and the local batches ignore, the process can be treated as continuous.

3.1.5

a) closed b)open c)open d)closed

3–1

Solutions Chapter 3

3.1.6

The system is the radiator. Open System

Closed System

Steady-State

Unsteady State

a)

X

X

b)

X

X

c)

X (Before filling)

d)

X (After filling)

X (After filling)

X

X

X (Before Filling)

3.1.7

system contains ice + H2O

Ice

boundary

boundary (c)

Unsteady state for any assumptions

b ⎫ ⎧open ⎬ assume water (melted ice) leaves the system ⎨ a ⎭ ⎩flow b ⎫ ⎧closed ⎬ assume water stays in sytem because the ice and water ⎨ a ⎭ ⎩batch remain on melting

3.1.8

P

F1

F 2

F

unsteady state open

P unsteady state open

steady state open

(a)

(b)

(c)

3–2

Solutions Chapter 3

3.1.9

(a) (b) (c) (d) (e) (f) (g) 1. x x 2. x x ____________________________________________________ 3. x x 4. Depends on the time period considered ____________________________________________________ 5. x x 6. x x x ____________________________________________________ 7. x x

3.1.10

If the overall flows in and out over a time period for several batches are considered, and the local batches ignore, the process can be treated as continuous.

3.1.11

Basis: 1 minute Accumulation (kg)

mfinal − 0

In (kg) = 300 + 100

Out (kg) −

380

n final = 20 kg 20 kg 60 min = 1200 kg min 1 hr

3.1.12

Basis: 600 kg solution Accumulation (kg)

n final − 0

In (kg) = 100 + 500

Out (kg) −

300

n final = 300 kg if no water evaporates in which case less than 300 kg would remain

3–3

Solutions Chapter 3

3.1.13

(c)

27.0 g

3.1.14

Based on the process measurements, there are 5,000 lb/h more flow for the process leaving the heat exchanger than the feed rate to the heat exchanger; therefore, the material balance for the process fluid does not close. The reason for this discrepancy could be faulty flow sensor readings or possibly a leak of the condensate or steam into the process stream.

3.1.15

Basis: Data shown on flowsheet units are MTA In 2.5 × 106

Out 404 ×103 228 ×103 152 ×103 101 ×103 67 ×103 36 ×103 100 ×103 1190 ×103 2,278 × 103

mass in does not equal to mass out Reasons:

(1) Some of the material was burned as fuel (2) Some of the material formed gases that were exhausted to atmosphere (such as H2O, CO2). (3) errors in measurement. (4) Some process streams are not shown.

3–4

Solutions Chapter 3

3.1.16

In = 250,000 ton/yr. In (ton/yr) 250,000

Out = 244,500 ton/yr. Out (ton/yr) Combustibles Combustibles Polyethylene Polystyrene PVC Acrylonitorile DB Phenol Acetone Rubber LPG Aromatics Total

3,800 39,500 30,000 5,000 40,000 20,000 8,000 10,000 5,750 10,000 24,000 48,000 244,050

Balance is not exact but very good for an operating plant.

3.1.17

Basis: 1 week in (tons) = 920 + 0.6 = 920.6 out (tons) = 3.8 + 620 + 0.01 + 0.01 + 0.08 + 1.1 + 275 + 20 = 920 Yes

3–5

Solutions Chapter 3

3.1.18

The density of the crystalline silicon in the cylinder is 2.4 g/cm3. Basis: 62 kg silicon The system is the melt, and there is no generation or consumption. Let Δmt be the accumulation. Accumulation Δmt

=

Input 0

-

Output 0.5(62 kg)

Δm t = − 31 kg

Let t be the time in minutes to remove one-half of the silicon 2.4 g π(17.5cm)2 0.3 cm t min 1 = (62,000 g) cm3 4 min 2

t = 179min

3.1.19

Basis: 1 hr Overall material balance (kg): 13,500 + 26,300 ? 39,800 Yes. The balance is satisfied NaC1 balance: 0.25 (13,500) + 0.05 (26,300) ? 0.118 (39,800) 4690 ≅ 4696 The balance is closely satisfied but not exactly. The closure is good for industrial practice.

3–6

Solutions Chapter 3

3.1.20

Basis: 1 hour The overall material balance is In (lb) 106,000

?

Out (lb) 74,000 + 34,000 = 108,000

The error in the overall material balance is 2000 lb/h or 1.9%; therefore, the overall material balance is within the expected error for industrial flow sensors. Propylene balance:

0.7 × 106,000 ? (0.997) (74,000) + (0.1) (34,000) 74,200 ? 73,778 + 3,400 = 77,178 (3.8% error)

Propane balance:

0.4 × 106,000 ? (0.003) (74,000) + (0.9) (34,000) 31,800 ? 222 + 30,600 = 30,622 (3.8% error)

Note that the relative error for the component balances are twice as large as the relative error for the overall material balance, indicating that there is additional error in the composition measurements used for the component balances.

3.1.21

Basis: 100 kg wet sludge The system is the thickener (an open system). No accumulation, generation, or consumption occur. The total mass balance is In Out = 100 kg 70 kg + kg of water

Consequently, the water amounts to 30 kg.

3.2.1

(a) (b) (c) (d)

water and air. Insulation, air and what is in the atmosphere. Yes (cold water in, hot water out) Yes

3–7

Solutions Chapter 3

3.2.2

Four balances are possible, 3 components plus 1 total.

0.10F1 + 0.50F 2 + 0.20F 3 = 0.35P 0.20F1 + 0F2

+ 0.30F 3 = 0.10P

0.70F1 + 0.50F 2 + 0.50F 3 = 0.55P Total balance F1 + F2 +F3 = P Only 3 of the equations are independent.

3.2.3

If you specify F, P, W, you can calculate all of the stream variables. a)

Unknown: three stream values F, P, W (plus two compositions if you take into account all of the variables).

b)

The two known compositions are not given but may be calculated from Σxi = 1,

c)

Two components exist; hence two independent material balances can be written. The problem cannot be solved unless one stream value is specified.

3.2.4

(a)

No. The equations have no solution – they are parallel lines. The rank of the coefficient matrix is only 1 because the ⎡1 2⎤ det ⎢ =0 ⎣1 2⎥⎦ The rank of the augmented matrix is 2

⎡1 2 1⎤ Largest non zero det. of ⎢ is of order 2 ⎣1 2 3⎥⎦ Thus, although the 2 equations are independent and the number of variables is 2 (the necessary conditions), the sufficient conditions are not met.

3–8

Solutions Chapter 3 x2 x +2x =1 1 2

x + 2x = 3 1 2

x1

(b)

x2 •

(x1 − 1)2 + (x2 − 1)2 = 0 No; 2 solutions • The equations are independent x1 x1 + x 2 = 1

Two solutions exist as can be seen from the plot hence no unique solution exists.

3.2.5

The number of independent equations is just 3. The number of unknown quantities is 3, hence a unique solution is possible.

3.2.6

a)

No for both ⎡1 1 1 ⎤ ⎢1 2 3 ⎥ rank of coefficient matrix is2. ⎢⎣3 5 7 ⎥⎦ rank of augmented matrix is2. r = 2, n = 3

b)

multiple solutions exist

⎡1 0 1⎤ ⎢5 4 9⎥ 3d column issum of1st two columns,so rank is2. ⎢⎣2 4 6⎥⎦ rank of augmented matrix is2, also. r = 2, n = 3

multiple solutions exist

3–9

Solutions Chapter 3

3.2.7

(a) F; (b) F; (c) F (the maximum can be more than the number of independent equations).

3.2.8 P = 16 kg

F = 10 kg ωF1= 0.10 ωF2= ? ωF3= 0 (assumed)

ωP1= 0.175 ωP2= ? ωP3= ?

A = 6 kg ωA1= 0.30 ωA2= ? ωA3= 0.20

Unknowns (4): ω2F ,ω2A ,ω2P ,ω3P Equations: Mass balances: (1): F(0.10) + A(0.30) = P(0.175) or 10(0.10) + 6(0.30) = 16(0.175) (2): F (ω 2 F ) + A (ω 2 A ) = P (ω 2 P ) or

10 (ω 2 F ) +6 (ω 2 A ) = 16 (ω 2 P )

(3):

F(0)

+A(0.20) = P (ω 2 P )

or

10(0)

+ 6(0.20) = 16 (ω 2 P )

redundant

ω 2 F ,ω 2 A ,ω 2 P ,ω3P 10

6 -16 0

0

0 -0 16

coeff. matrix

Two independent material balances exist (the rank of the coefficient matrix is 2). In addition three sum of mole fraction equations exist. The total number of equations is 5, hence the degrees of freedom = -1. The problem is overspecified, and will require at least squares solution.

3–10

Solutions Chapter 3

3.2.9

Examine the row of C3H8. None of the concentrations are greater than the desired 50% so 50% is not achievable by any combination of A, B, or C. Or look at the CH4 row. 3.2.10

A x1

Basis: D =100 lb

B x2

C x3

D

0 M 1.4 ⎤ ⎡ 5.0 0 ⎢90.0 10.0 0 M 31.2 ⎥ ⎢ ⎥ Coefficient matrix is ⎢ 5.0 85.0 8.0 M 53.4 ⎥ ⎢ ⎥ ⎢ 0 5.0 80.0 M 12.6 ⎥ ⎢⎣ 0 0 12.0 M 14 ⎥⎦

No unique solution exists with 5 equations and 4 variables. A least squares solution could be determined, or the 4 equations with the most accurate data solved. The rank of the coefficient matrix is 3 The rank of the augmented matrix is 4 Hence no unique solution exists.

3.2.11

You can see by inspection that no combination of tanks 1,2 and 3 will give a mole fraction of 0.52 for the mixture. Basis: 2.50 mol of tank 4 Let xi = be the total moles of tank i The balances are

0.23x1 + 0.20x2 + 0.54x3 = 0.25(2.5) = 0.625 0.36x1 + 0.33x2 + 0.27x3 = 0.23(2.5) = 0.575 0.41x1 + 0.47x2 + 0.19x3 = 0.52(2.5) = 1.300

The coefficient matrix has a rank of 3 as does the augmented matrix so the set of equations has a solution. However, the solution is

x1 = −4.00 x 2 = 6.36 x 3 = − 0.327 The values of x1and x3 are not physically realizable! 3–11

Solutions Chapter 3

3.2.12

Number of unknowns: F, W, P and 9 compositions: Equations W Specifications: x CF =0 Sum of mole fractions = 1 Material balances (3 species)

12

1 3 3

Degrees of freedom =

7 5

You can make any set of measurements that results in independent equations (assuming equal accuracy). For example, you cannot use all the flows, or all the compositions in one stream, as the resulting set of material balances will not consist of 5 independent balances, but a lesser number.

3.2.13

The inerts are treated as a compound. Unknowns: 5 compounds × 5 streams plus 5 streams =

30

Equations: Specifications: Concentrations not shown in diagram assumed to be zero 8 Concentrations with % given 7 E = 11 kg 1 Material balances: 5 5 Implicit equations (∑ ωi = 1) 5 26 a.

Degrees of freedom (additional specifications) =

4

You must make 4 measurements that result in independent equations b.

No.

3–12

Solutions Chapter 3

3.2.14a

a.

b.

The remaining gas is 100% minus the N2, and was put on the figure as R.

c.

Basis: 100 mol A

d.

Unknowns: A, B, C Equations: N2 and R material balances Basis = A = 100 mol Note: You could treat the values of R as unknowns in each stream, and then there would be 3 more unknowns and 3 more independent equations ( ∑ x i = 1 or ∑ mi = total mass flow i )

100(0.90) + B(0.30) = C(0.65) e. & f. N2: R: 100(0.10) + B(0.70) = C(0.35) Total: 100 + B =C Two of the above equations are independent g.

Solution:

B = 71.4 mol

C = 171.4 mol

A 100 = = 1.40 B 71.4

3–13

Solutions Chapter 3

3.2.14b

a.

DT = dry timber b.

The DT is the balance of each stream.

c.

Basis: F = 100 kg

d.

Unknowns: F, W, P

OR

Equations: Basis: F = 100 kg Material Balances: Water, DT (total); 2 independent

Degrees of freedom = 0

F, W, P, DTF, DTW, DTP, H2OF, H2 OP , H2 OW Equations: F = 100 kg Material Balances: Water, DT, (total); 2 independent Specifications: 3 for water Implicit equations: 3 of ∑ ω i = 1 Degrees of freedom = 0

e.&f. Introducing the specifications and basis into the material balances: Water: (0.201)100 = (1)W + (0.086)P DT: (0.80) 100 = 0 + (0.914)P a tie element Total: 100 = W + P g.

W = 12.5 kg

P = 87.5 kg

W 12.5 kg = = 0.125kg/kg F 100 kg

3–14

Solutions Chapter 3

3.2.14c

a. & b. A N2 1.00

mol. fr. N2 0.70 CH4 0.30 1.00

P

F

d.

CH4 n CH 4 C 2H 6 n C 2 H 6 N2 n N2

B

c.

or

N2 C 2H 6

0.90 0.10 1.00

∑=P ∑

x CH 4 x C2 H 6 x N2 =1.00

Basis: B = 100 mol

Unknowns: Assume all of the variables except those that are specified as zero are included in the analysis. F A P B

stream variable 1 1 1 1 4 +

component variables 2 1 3 2 8 = 12

Equations: Basis: Material balances: N2, CH4, C2H6 Specified ratio: n CH4 /n C2H6 = 1.3 Specified values of n: 2 + 1 + 0 + 2 = Implicit equations (∑ xi = 1 in P) (Note: the implicit equations of streams F, A, and B are redundant with the specifications) Total Degrees of freedom =

1 3 1 5 1

11 1

More information is needed to solve the problem uniquely.

3–15

Solutions Chapter 3

3.2.15

Here are some possibilities. Consult the tables at the end of the Chapter for more suggestions. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Rephrase the problem to make sure you understand it? Draw a simple diagram of what was happening? Think about what was going into the tank and what was coming out? Imagine yourself inside the tank, and ask what was going on around you? Ask whether there were any physical laws to consider (such as conservation of matter or energy)? Try to imagine the answer as a number, graph, table, or whatever? Try to identify essential variables? Choose a notation? Look for a ready-made formula for the answer? Look for simplifying assumptions? Try to find an easier version of the problem? Look for bounds (simple models that would definitely underestimate or overestimate the answer)?

3–16

Solutions Chapter 4

4.1.1

Basis: 100 kg cucumbers P − 100 = evaporated Cucumber

Initial 1 kg

Final (P) 2%

Water

99 kg

98%

100%

100 kg

100%

100%

Cucumber balance:

Evaporated −

0.02P − 0.01(100) = 0

P = 50 kg 4.1.2

Answer is: Yes

Initial

Final

plant 5 ppm soil 6 ppm

Plant 755 ppm Soil 5 ppm

Final − Initial

Final − Initial

Arsenic balance: S (5) − S(6) = 750P = S or

P(755) − P(5)

S = 750 P

4.1.3

Step 5: Basis is 1 ton (2000 lb) sludge Steps 1, 2, 3, 4:

Sludge 0.70 H 2 O 0.30 Solids 1.00

F

Drier W H2 O 1.00 4–1

P

Dried Sludge 0.25 H 2O 0.75 Solids 1.00

Solutions Chapter 4

Steps 6 and 7:

Basis: 2000 lb F

Balances: H2O Solids

Unknowns: P, W

Steps 8 and 9: Total: 2000 = P + W Solids: 2000 (0.30) = P (0.75) P = 800 lb

W = 2000 − 800 = 1200 lb

4.1.4 Step 5:

Basis: 1 min

Steps 2, 3, 4: D 220 mL

V 215 mL

B

Urea 2.30 mg/mL H2O 220 mL

P

1.70 mg/mL H2O 215 mL

Steps 6 and 7: Unknowns: mP urea and mP H2O with two equations: urea and H2O. Degrees of freedom = 0. Steps 8 and 9: (a)

H2O balance: 220 mL − 215 mL = 5 mL

Urea balance:

(b)

2.30 mg 220 mL 1.70 mg 215 mL − = 141 mg mL mL

Dialisate: 1500 + 5 = 1505 mL Urea: 141 mg Concentration: 141/1505 = 0.0934 mg/mL

4–2

Solutions Chapter 4

4.1.5

Step 5:

Basis: 1 day

Steps 2, 3, 4: W ton 1.00 H2O F=2 ton

P (ton) NaOH

mass fr NaOH 0.03 H2O 0.97 1.00

H2O

Step 6: Unknowns, 2: P, W Step 7: Balances 2: NaOH, H2O, total (2 of the 3)

Steps 8 & 9: Total: 2 = W + P NaOH: 2 (.03) = P(.18)

P = 1/3 ton (666 lb)

4.1.6

W = 1 2/3 ton (3334 lb)

Steps 2, 3, 4: 500 lb 10% solution

A 20% solution

3000 lb 13% polymer

pure solvent S

Step 5:

Basis: 3000 lb of final product

Steps 6 and 7: Unknowns: A and S; balances: total and polymer

4–3

Mass fr. 0.18 0.82 1.00

Solutions Chapter 4

Steps 8 and 9: Total wt

A + S + 500 = 3000 A + S = 2500

Polymer

0.2A + 50 = 390 A = 5 (340) =

1700

S = 2500 – 1700 = 800

4.1.7

Steps 1, 2, 3 and 4: N = nitrocellulose S – solvent

Treat the problem as a steady state flow problem, or as an unsteady state batch system. As a flow system:

M (lb) N 100% F (lb) N S

0.055 0.945 1.000

Step 5:

Plant

P 1000 lb

Mass fr. N 0.08 S 0.92 1.00

Basis: 1000 lb P

Steps 6 and 7: Two unknowns, F and M. Two balances can be made, N and S. Steps 8 and 9:

Solve to get

N:

F ( 0. 055) + M (1. 00) = 1000 (0. 008)

S:

F (9. 045) + M ( 0) = 1000 ( 0. 92)

4–4

⎫ F = 973. 5 lb ⎪ ⎬ M = 26. 5 lb ⎪ ⎭ P = 1000.0 lb

Solutions Chapter 4

4.1.8 Steps 1, 2, 3 and 4:

F = 100 kg mol/min CH4 He

Mol fr. 0.80 0.20 1.00

W (kg mol) CH 4 He

System P (kg mol)

CH4 He

Mol. n CH 4 n He

Mol fr. 0.50 0.50 1.00

Step 5:

Basis

1 min → 100 kg mol F

Step 4:

To get P, calculate first the average mol. wt. of F and P, or transform the mole fractions to mass fractions.

Step 5:

Basis: 1.00 mol F Mol

CH4 He

0.80 0.20 1.00 Basis : 1.00 mol P Mol

CH4 He

0.50 0.50 1.00

100 kg mol F

MW

kg

16.03 4

12.8 0.8 13.6

MW

kg

16.03 4

8.01 2.0 10.01

13.6 kgF 0.20 kgP 1kgmol P 1.00kg mol F 1kgF 10.01kgP

P = 27.2 kg mol Steps 6, 7, 8 and 9:

100(0.80) = 27.2(0.50) + n CH 4 ⎫ ⎬ 100(0.20) = 27.2(0.50) + nHe ⎭

n CH 4 nHe

Mol

66.4 6.4

72.8

4–5

In W

Mol.fr.

0.912 0.088

1.00

Solutions Chapter 4

4.1.9

Fermenter 40% void 60% cell volume

F1

Separator

discharge

W1

F2 wet cell 75% H2 O 25 % dry cell

100% H2 O

Basis: 1000 cm3 F1

Drycells =

1000cm 3 in fermenter

3

60cm cell 1.1gcell 25gsolids 3 100 cm in fermenter 1cm 3 cell 100g cell

= 16.5gDrycell / 1000cm 3 in fermenter

4.1.10

a. Basis: 1 kg mole of mixture Three components exist: (CH4)x, (C2H6)x, (C3H8)x Let A, B, C respectively represent kg mol of each mixture; these are the unknowns. Equations:

0.25A + 0.35B + 0.55C = 0.30

(CH4)x balance

0.35A + 0.20B + 0.40C = 0.30

(C2H6)x balance

0.40A + 0.45B + 0.05C = 0.40

(C3H8)x balance

There is a unique solution to the set of equations (in kg mol) The solution is

A = 0.600

B = 0.350

C = 0.05

b. It is proposed to prepare the final mixture by blending four different compounds (A, B, C, D); there will still be three equations, but now there will be four unknowns. Since the rank is now less than n, there will be an infinite number of possible blends of the four mixtures. Not required: (An optimization of a revenue function subject to the equations is needed.)

4–6

Solutions Chapter 4

4.1.11

a. A 100% CO2 100 lb = 2.27 lb mol

CH4 100% (MW = 16)

P (lb) = ? mol fr. CH4 0.9796 CO2 0.0204 (average) 1.000

F=? (lb)

100 lb 1 lb mol = 2.27 lb mol 44 lb

Steps 2, 3, 4: Step 5: Basis 1 min

Step 6: Unknowns: F, P Step 7: Balances: CH4, CO2 Steps 8 and 9:

Balances in moles CH4: F (1.00) + A(0) = P (0.9796) CO2: F (0) + 2.27 = P (0.0204)

F= b.

P = 111.41 lb mol

111.27 lb mol 0.9796 lb mol CH4 16 lb CH4 = 1746 lb/min 1 lb mol P 1 lb mol CH 4

(a)

Redo the problem with a new composition for F: CH4 CO2

0.99 0.01 1.00

CH 4 : F (0.99) + A(0) = P (0.9796) ⎫ ⎬ F = 3520 lb/min CO2 : F (0.01) + 2.72 = P (0.0204) ⎭

error

4–7

3520-1744 100 = 50% (b) 3520

Solutions Chapter 4

4.1.12

Steps 1, 2, 3, and 4: NH 3 F(kg)

Pipeline

72.3 kg/min for 12 min P (kg)

Mass fr. NH 3 0 gas 1.00 1.00

NH 3 gas

Mass fr. 0.00382 0.99618 1.00

Step 5: Basis: 1 min Steps 6 and 7:

Two unknown, F and P. Two balances can be made, NH3 and gas. You can use the total balance as a substitute.

Steps 8 and 9: Total

F + 72.3 = P

NH3

F (0) + 72.3 = P (0.00382) P =18,900 kg / min

Step 10:

F = 18,900kg/min or 1.14 × 106 kg/hr

Check using the gas balance

4.1.13

Steps 1, 2, 3, and 4: 10 lb/hr Mass fr. H2 O 1- 0.00012 0.00012 P Br 1.00

F Mass fr. H 2 O 1.00 Br 0 1.00

Step 5: Basis: 1 hr Step 6: Unknowns:

F, P 4–8

Solutions Chapter 4 Step 7: Balances: Steps 8 and 9: Total: Br: H2O:

H2O, Br F + 10 = F (0) + 10 F (1.00) + 0

= =

P

P (0.00012) ; P = 8.33 × 104 lb/hr P (1-.00012) 4 4 F = 8.33 × 10 − 10 = 8.33 ×10 lb / hr

4.1.14

Step 5: Basis: 1 year Steps 2, 3, 4:

P = 15 x106 lb

F = ? lb PET PVC

mass fr. 0.98 0.02 1.00

W PVC 100%

Steps 6 and 7: Two unknowns: F, W

Two balances: PET, PVC

Steps 8 and 9: PET balance: PVC balance:

F (.98) = P (.999990) ≅ 15 × 106 F (.02) = W (1.00) + P (0.000010) = W + (15 × 106 (10-5) Total balance could be used in lieu of one of the above F = 15 × 106 + W

From PET: F ≅ 15.3 × 106 lb W = (15.3 × 106) (0.02) – 15 × 106 (10-5) = 0.31 × 106 lb

4–9

mass fr. PET 0.999990 PVC 0.000010 (10 ppm)

Solutions Chapter 4

4.1.15 Basis. 300 g initial solution (a)

In Na2SO4 H 2O

g 100 200

MW Na 2 SO4 ⋅10 H2O is 322.2 MW 142.05 18.016

g mol 0.704 11.10

Out

Na 2SO4 ⋅10H2O

Material balances in g mol (in = out):

g mol

Na2SO4: 0.704 = 0.310 + n Na 2SO4

n Na 2SO4 = 0.394

H2O:

n H2O

11.10 = 0.310 (10) + n H 2O

= 8.0

Total Mother liquor:

(b)

g 100

8.394

g mol 0.310

g

%

55.97

28

144.1

72

200.07

100

Na 2SO4 is 28% and H2 O is 72%

100 g crystals 100 g solution = 33.3 g crystals/100 g initial soln. 300 g solution

4.1.16

Assume steady state flow problem (alternate is unsteady state batch problem). Steps 1-4: 200 g H2O 1.00

100 g

Na2B4O7 1.00

No rxn. No accum. P

(g) Na2B4O7 10H2O

F (g) Final solution mol fr. Na2B4O7 0.124 H2O 0.876 1.000

Calculate composition of P: Basis: 100 mol Na2B4O7.10 H2O

4–10

Solutions Chapter 4

Na2B407 H2 O

mol 1 10

Step 5:

Basis: 200 g H2O + 100 g Na2 B4O7

Step 6:

Unknowns: F, B

Step 7:

balances: Na2B4O7, H2O

Step 8:

Na 2 B4O7 :

Step 9:

MW 201.27 18

g 201.27 180.0 381.27

mol fr. 0.528 0.472 1.00

H 2O:

100 = P (0.528) + F (0.124) ⎫ ⎬ 2 indept. 200 = P (0.472) + F (0.876) ⎭

Total:

100 + 200 = P + F

Use H2O and Total to get P = 155.5g and F = 144.5 g ratio:

155.5 (100) = 51.8 g Na 2 B4O7 /100 g H 2O 300

Check use Na2B4O7 100.0 = 100.0 ok

Step 10:

4.1.17 Assume the process is a steady state one without reactions. Steps 1, 2, 3, 4:

F1

.

F2

.

FeC13 H2O

MW

Fe C13 Fe C13.6H2O Fe C13.H2O Fe C13.2.5 H2O H 2O

.

FeC13 6H2O 1000 kg

P

162.22 270.32 180.24 207.26 18.02

4–11

FeC13 2.5 H2O

Solutions Chapter 4

Calculate the compositions for just one iron compound. FeC13 is the simplest to use. For F1:

Basis: 1 kg mol

For F2:

FeC13 ⋅ 6H 2 O

FeC13 ⋅1H 2 O

kg mol 1 6

FeC13 H 2O

Basis: 1 kg mol

MW 162.22 18.03

kg 162.22 108.18 270.40

FeC13 H 2O

kg mol 1 1

MW 162.22 18.03

For P Basis: 1 kg mol kg mol 1 2.5

FeC13 H 2O

MW 162.22 18.03

Step 5: Basis: 1000 kg FeC13.6H2O

kg 162.22 45.08 207.30

Step 6: Unknowns P, F Step 7, 8, 9: Balances (kg) IN

OUT

⎛ 162.22 kg FeC13 ⎞ ⎛ 162.22 ⎞ ⎛ 162.22 ⎞ ⎟ + F2 ⎜ ⎟ = P ⎜ ⎟ kg tot ⎠ ⎝ 270.40 ⎠ ⎝ 180.25 ⎠ ⎝ 207.30

FeC13: 1000 ⎜

Total: 1000 + F2 = P

soln Step 10:

F2 = 1555.7 kg P = 2555.7 kg

Check: F2 + F1 = 1554.5 + 1000 = 2554.5

4–12

ok

kg 162.22 18.03 180.25

Solutions Chapter 4

4.1.18

Steps 2, 3, 4:

The process can be viewed as an unsteady process without reaction, or as a flow process. Step 5:

Take as a basis 100 g of Ba(NO3)2.

Step 4: The maximum solubility of Ba(NO3)2 in H2O at 100°C is a saturated solution, 34 g/100 g of H2O. Thus the amount of water required at 100°C is 100 g H2 O

100 g Ba(NO3 )2

34 g Ba(NO3 )2

= 294.1 g H2 O

If the 100°C solution is cooled to 0°C, the Ba(NO3)2 solution will still be saturated so that the composition of the final solution is mass fr. Ba(NO3)2:

5 = 0.0476 100 + 5

H2O:

100 = 0.9524 100 + 5

The composition of the crystals is Ba(NO3)2:

100 = 0.9615 100 + 4 4–13

Solutions Chapter 4

4 = 0.0385 100 + 4

H2O:

The composition of the original solution is BaNO3 H2 O

mass fr. 0.254 0.746

100g 294.1g

Steps 6 and 7: We have two unknowns, F and C, and can make two independent mass balances so that the problem has a unique solution. Steps 8 and 9: Balance

Final solution

Ba(NO3)2: H2O: Total:

F(0.0476) F(0.9524) F

Transport through boundary (out)

Initial solution 100 294.1 - (100 + 294.1)

= = =

-C(0.9615) -C(0.0385) -C

Solve the Ba(NO3)2 and total balance to get F = 305.2 g Step 10:

C = 88.89 g

Check using the water balance 305.2(0.9524) – 294.1 ? -88.89(0.0385) - 3.42 = -3.42

The Ba(NO3)2 that precipitates out on a dry basis is

88.89 g C 0.9615 g Ba(NO3 ) 2 = 85.5 g Ba(NO3 ) 2 1gC

4–14

Solutions Chapter 4

4.1.19

M.W. Na2 S2 O2 = 142 M.W. Na 2 S2 O2⋅ 5H2 0 = 232 feed = F 60% Na2S2 O2 1% impurity 39% H 2O

H2 O

W 0

H2 O

Wi S

I 10°C

C

Na 2 S2 O2⋅ 5H2 O crystals 0.06 lb satd. soln/lb crystals

II Na 2 S2 O2⋅ 5H2 O 0.1 % impurity 0% H 2O as H 2 O

D Process II Compositions: a)

satd soln 1.4 lb Na 2 S2 O2⋅ 5H2 O 1 lb H 2O ? lb impurity

dried cyrstals

Basis: 1 lb mol Na2S2O2 ⋅ 5 H2O, impurity free

Stream D Comp. lb mol Na2S2O2 H2O Total

1 5 6

mol wt

lb

142 18

wt fr 142 90 232

0.612 0.388 1.000

Basis: 100 lb D (Na2S2O2⋅ 5 H2O = 99.9 lb) Comp. Na2S2O2 H2 O impurity Total Stream C

99(0.612) 99(0.388)

lb = =

61.1 38.8 0.1 100.0

Let y = lb impurity/100 lb free water in saturated solution

Basis: 100 lb dry crystals, impurity free lb from dry Comp.

crystals

+

lb from adhering soln

from calculation below 4–15

= Total

Solutions Chapter 4

Na2S2O2

61.2

⎛ 85.7 ⎞ ⎟ (6)⎜ ⎝ 240 + y ⎠

⎛ 85.7 ⎞ ⎟ 61.2 + 6 ⎜ ⎝ 240 + y ⎠

H2O

38.8

⎛ 154.3 ⎞ ⎟ (6)⎜ ⎝ 240 + y ⎠

⎛ 154.3 ⎞ ⎟ 38.8 + ⎜ ⎝ 240 + y ⎠

impurity

-----

⎛ y ⎞ ⎟ (6)⎜ ⎝ 240 + y ⎠

Totals

100.0 Stream S

Comp.

⎛⎜ y ⎞ ⎟ ⎝ 240 + y ⎠

6

106

Basis: 100 lb free water in satd. soln., impurity free

lb from salt lb from free water

Na2S2O2

1.4(61.2) = 85.68

H2O

1.4(38.8) = 54.32

Total

100

wt. fr.

wt.fr.total 85.7

0.357

85.7 (240 + y)

154.3

0.643

154.3 (240 + y)

__ 140.00

100

240.0

1.000

y 240 + y

impurity = Total Basis: 100 lb F

(Rather than use the impurity balance, use the total balance instead if you want) Water in, Wi, comes from balances on Unit I. Total:

100 + Wi = S + C

4–16

Solutions Chapter 4

⎛ 85.7 ⎞ ⎟S+ 60= ⎜ ⎝ 240 + y ⎠

Na 2 S2 O2

H 2O

⎛ 154.3 ⎞ ⎟S + 39 + W i = ⎜ ⎝ 240 + y ⎠

⎛ 85.7 ⎞ ⎟ 61.2 + 6⎜ ⎝ 240 + y ⎠ 106 ⎛ 154.3 ⎞ ⎟ 38.8+ 6⎜ ⎝ 240 + y ⎠ 106

⎫ ⎪ Unknowns: W ,S,C, y i C⎪ ⎬ Total:4 ⎪ C⎪ ⎭

Balances on Unit II needed as well because 4 unknowns exist. Total:

C = W0 + D

Na 2 S2 O2

H 2O

⎛ 85.7 ⎞ ⎫ ⎟C 61.2 + 6⎜ ⎪ Unknowns:W o ,D ⎝ 240 + y ⎠ = 0.611D ⎪ 106 ⎬ Total:2 ⎛⎜ 154.3 ⎞ ⎟C ⎪ 38.8+ 6 ⎝ 240 + y ⎠ = W o + 0.388D⎪ ⎭ 106

6 equations and 6 unknowns Note that the complex term involving y can be eliminated to solve for D, W0, Wi, and S, i.e., in effect making total Na2S2O2 and H2O balances overall the process. The solution is

4.1.20

(a)Wi = 23.34 lb

(b)66.5%

Basis: 100 lb pulp as received Comp H2 O

lb = % 22

Pulp

78

Total

100

Freight cost =

$1.00 2000 lb 100 lb 1 ton

Assume air dried pulp means the 12% moisture pulp. Allowed water =

78 lb pulp 12 lb H 2 O 88 lb pulp

= 10.65 lb H 2 O

4–17

= $20.00/ton

Solutions Chapter 4

Pulp on contract basis = 78 + 10.65 = 88.65 lb Basis: 1 ton pulp as received Cost =

4.1.21

88.65 lb 12% pulp rec'd 1 ton shipped $60.00 100 lb shipped

1 ton

$53.19/ton

You pay for soap plus transportation. Basis: 100 kg soap with 30% water

Convert soap in wet soap to soap in dry soap. Data: W (wet soap) H2O 30 kg soap

D(dry soap) 5%

70 kg

95%

100 kg

100

Soap balance: 0.70W = 0.95D

or 0.70(100) = 0.95D

D = 73.68 kg of which 70 kg is soap

$6.05 = $36.05 100

Cost of W at your site (containing 70 kg soap):

100 ($0.30 +

Cost of D at your site (containing 70 kg soap):

⎛ x$ $6.05 ⎞ 73.68 ⎜ + ⎟ = $36.05 ⎝ kg 100 ⎠

x = $ 0.43/kg

4–18

Solutions Chapter 4

4.1.22

The problem here is to decide on the balances to make. Not all will be indept balances. Steps 1, 2, 3, and 4: F = 100 lb H2 O VM C Ash

H(lb)

Mixer

Mass fr. 0.124 0.166 0.575 0.135 1.000

(lb)

P

Mass fr. H2 O VM C Ash

0.10 ω VM ωC 0.10 1.00

Mass fr. VM 0.082 C 0.887 H2O 0.031 1.000

Mass mH O 2 m VM mC m Ash P

Step 5:

Basis: F = 100 lb

Step 6:

Unknowns:

m H 2 O , mVM, mC, mAsh, P, H (6)

Step 7:

Balances:

H2O, VM, C, Ash, Σmi = P, m H 2 O /P = 0.10 mAsh/P = 0.10: (7), and presumably 6 are independent

Steps 8 and 9: H2O (lb):

In 100 (0.124) + H (0.082)

=

Out m H 2O

VM :(lb)

100 (0.166) + H (0.082)

=

mVM

C (lb):

100 (0.575) + H (0.887)

=

mC

Ash (lb)

100 (0.135) + H (0)

=

mAsh

=

P

Total

100 + H

4–19

Solutions Chapter 4

Solution: mAsh = 13.5 lb

m H 2 O = 13.51

Not all of the equations have to be written down as above. Note that ash is a tie element to P, so that the ash balance gives 100(0.135) + H(0) = P(0.10)

P = 135 lb

and mAsh = 13.5 lb

From a total balance H + 100 = P

H = 35 lb

Check via H2O balance

Step 10:

12.4 + 35(0.031) = 13.5 ⎫ ⎬ ok 13.49 =13.5⎭ 4.1.23

Step 5:

Basis: 1 day

Steps 2, 3, and 4:

% C3 C3 i − C4 n − C4 C5 +

% 1.9 51.6 46.0 0.5 100.0

D F=5804

i − C4 n − C4

kg mol/day B

Steps 6 and 7:

i−C n- C 4 C 5+

3.4 95.7 0.9 100.0

% 1.1 97.6 1.3 100.0

There are two unknowns, and we have 4 independent equations, but the precision of measurement is not the same in each equation, hence different results are obtained depending on the equations used. Choose the most accurate to use. 4–20

Solutions Chapter 4

C3 (0.019)(5804) = i-C4 n-C4 C5+ Total:

In 110 2992 2667 37 5804

= = = = =

0.034D 0.057D + 0.011B 0.009D + 0.976B 0.013B D+B

Steps 8 and 9: Use total + i-C4 balances:

kg mol/day B = 2563 D = 3240

2992 = 0.975(5804-B) + 0.011 B Use total + n-C4 balances:

B = 2704 D = 3100

2667 = 0.009 (5804 - B) + 0.976B The other two balances give

C3 as a tie component: kgmol D 5804 kgmol F 0.019kg mol C3 1kg mol D = 3243 day 1.00 kgmol F 0.034 kg mol C3 1day C5+ as a tie element: 5804 kgmol F 0.006kg mol C5 + 1kgmol B = 2679 kgmol B / day 1.00 kg mol F 1day 0.013 kgmol C5 +

4–21

Out

Solutions Chapter 4

4.1.24

Steps 1, 2, 3, and 4: Assume the other components have the same density as water in all flows. 3.785 × 106 L/day 150 mg/L BOD

a.

B 5 mg/L BOD 0.3 m 3 /s

A

C

BOD = ? = x

Step 5: Basis: 1 day

0.3m 3 3600s 24hr 1000 L 7 = 2.592 ×10 L / day 3 s hr day m

A= Steps 6 and 7: Unknowns: C, x Balances:

H2O, BOD

Steps 8 and 9: Total mass balance A + B = C

1day A L 1kg 1day B L 1kgB 1day C L 1kg C + = day L day L day L BOD mass balance −3

−3

1day A L 5×10 gBOD 1day B L 150 ×10 gBOD 1day C L x g BOD + = day L day day day L 2.592 × 107 kg + 3.785 × 106 kg = C = 2.9705 × 107 kg 5 × 10-3 A + 150 × 10-3 B = xC 5× 10 −3A +150 ×10 −3 B x= C 4–22

Solutions Chapter 4

5 × 10−3(2.592 × 10 7 ) + 150 × 10 −3(3.785 × 106 ) = 2.9705 × 107

−3

= 23.475 × 10 g / L

b. g/L

N BOD = ? B 530 x 106 L/day 3x10-3 g/L BOD

Total mass balance

= 15.8x106 L/day C

A

5x10-3 g/L BOD

A+B=C

1day A L 1kg 1day B L 1kg 1day C L 1kg + = day L day L day L 530 × 106kg + 15.8 × 106 kg = C = 5.458 × 108 kg BOD mass Balance −3

−3

−3

1day A L 3× 10 gBOD 1day B L N ×10 g 1day C L 5×10 g + = day L day L day L 3 × 10-3A + N × 10-3 B = 5 × 10-3C 5 × 10 −3 C − 3 × 10 −3 A N= B (5 × 10−3 )(5.458 × 10 8 ) − (3 × 10−3 )(5.30 × 108 ) ≅ 72.1x10 −3 g / L = 7 1.58 × 10

4–23

ok

Solutions Chapter 4

4.1.25

The extractor is assumed to be a steady state process. Steps 1, 2, 3, and 4:

All the known data have been placed in the figure. 9.7 L/min, Solvent (organic stream)

Aqueous Stream F = 100 L/min

Waste (aqueous stream) W(L/min) Enzyme C W E Water, etc. P (L/min) Enzyme C PE Solvent

P

CE =18.5 CW E

Step 5: Basis: 1 min (F = 100L) Step 6: Unknowns are P and W and their concentrations. Steps 7 and 8: We can make enzyme, solvent, and aqueous balances, plus we have

C PE =18.5C W E The solvent and aqueous phases are tie components. Balances

In Out P C g (W L) C W 100 L 10.2 g (P L) E E g + = L L L

Enzyme (g) Phase relation (g/L) (D = 18.5) Balances

P

Out

9.7L ρ Sg (P L ) ρ P g = L L 100 L ρ F g W L ρ W g = ρ F ≅ ρ W and ρ= g ρ L L

Solvent Aqueous P = 9.7L 1020 = 97 C

W

C E = 18.5C E In

W = 100 L P E

+ 100 C

W E

= 9.7 (18.5) C PE + 100 C W E

g CW E = 3.65 L

C PE = 67.53 4–24

P

Solutions Chapter 4

Fraction recovery =

67.53 g 9.7 L L 10.2g 100L L

= 0.642(1020) = 655 g/min

4.1.26

Assume the process is in the steady state, and no reaction occurs. Solution assumes no S is in stream A and no N2 in stream B. Steps 1, 2, 3 and 4:

mass fr. A ω NH 3

NH 3

ω HA2 1.00

N2

mass fr. 0.0633 0.9367 1.00

mol fr. A x NH 3

NH 3

x NA2

S

1.00

mass fr. ω BNH 3 ω SB 1.00

Gas 1000 kg mol fr. NH 3 0.10 N2

0.90 1.00

Step 5: Basis 1.0 hr A , ω NA2 , ω BNH 3 , ω SB , a total of 6. Step 6: The unknowns are A, B, ω NH 3

Steps 7 and 8: Three compound balances can be made, N2, NH3, and S. One relation is given: A B ωNH 3 = 2 ωNH 3

4–25

Solutions Chapter 4 Two summations exist:

∑ω

A i

= 1 and

∑ω

B i

=1

hence, we have an adequate number of balances (unless some are not independent) to solve the problem. Material balance equations in grams Total:

F+S= A+B 1000 + 1000 = A + B

NH3:

A 1000(0.0633) + 0 = Aω NH + Bω BNH 3 3

N2 :

A 1000(0.9367) + 0 = Aω NH +0 2

S:

0 + 1000 = 0 + Bω SB Constraints A ω NH + ω NA2 = 1 3

ω BNH 3 + ω SB = 1 A ω NH = 2ω BNH 3 3

Solve in Polymath Note: The equations can be converted to linear equations by letting A A m NH = ω NH A, m NA2 = ω NA2 A, etc. The set can be solved in 3 3 A . Polymath or reduced to a quadratic equation in m BNH 3 or m NH 3

4.1.27

Basis: 12 hours The process is illustrated in the figure

MTBE

H2O mass fr. = 1.00

MTBE H2O

4–26

H2O MTBE

Solutions Chapter 4 The MTBE entering the pond in 12 hours is

25 boats 0.5 L gasoline 1000 cm3 0.72 g 0.10 g MTBE = 900 g MTBE 1 boat IL 1 cm3 1 g gasoline The pond holds (ignoring the MTBE in the pond which is negligible)

3000 m 1,000 m 3m 1000 kg H 2O = 9 × 109 kg H 2O 1 m 3 H 2O The increase in the concentration of MTBE is 900 g MTBE 1 kg = 1× 10-10 g MTBE/g H 2O 9 1000 g 9 × 10 kg H 2O

4–27

Solutions Chapter 5

BaCl2 + Na2SO4 → BaSO4 + 2NaCl

5.1.1

Mol. wt.: 208.3

142.05

233.4

58.45

a. Basis: 5.0 g Na2SO4

Example:

5 g Na 2 SO 4 1 g mol Na 2 SO 4 1 g mol BaCl 2 208.3 g BaCl 2 = 142.05 g Na 2 SO 4 1 g mol Na 2 SO 4 1 g mol BaCl 2

7.33 g BaCl 2 Problem b. c. d. e. f. g. h. i.

Basis 5 g BaSo4 5 g NaCl 5 g BaCl2 5 g BaSO4 5 lb NaCl 5 lb BaCl2 5 lb Na2SO4 5 lb NaCl

Answer 4.47 g BaCl2 8.91 g BaCl2 3.41 g Na2SO4 3.04 g Na2SO4 6.08 lb Na2SO4 5.59 lb BaSO4 8.21 lb BaSO4 9.98 lb BaSO4

AgNO3 + NaCl → AgCl + NaNO3

5.1.2

Mol. wt.: 169.89

58.45

143.3

85.01

a. Basis: 5.0 g NaCl

1 g mol NaCl 1 g mol AgNO3 169.89 g AgNO3 Example: 5.0 NaCl = 58.45 g NaCl 1 g mol NaCl 1 g mol AgNO3

14.53 g AgNO3 Problem b. c. d. e. f. g. h. i.

Basis 5 g AgCl 5 g NaNO3 5 g AgNO3 5 g AgCl 5 lb NaNO3 5 lb AgNO3 5 lb NaCl 5 lb AgNO3 5–1

Answer 5.92 g AgNO3 10.00 g AgNO3 1.721 g NaCl 2.04 g NaCl 3.44 lb NaCl 4.22 lb AgCl 12.27 lb AgCl 4.22 lb AgCl

Solutions Chapter 5

5.1.3

a.

The balanced equation is:

3 As2S3 + 4 H2O + 28 HNO3 = 28NO + 6 H3AsO4 + 9 H2SO4 and the reaction is unique within relative proportions. b.

2 KC1O3 + 4 H C1 = 2 KC1 + 2 C1O2 + C12 + 2 H2O

or

KC1O3

or

3 KC1O3 + 10 HC1 = 3 KC1 + 2 C1O2 + 4 C12 + 5 H2O

+ 6 H C1 = KC1

+ 3 C12

+ 3 H2 O

We classify these reactions as non-unique since they are not simply proportional equations, but are linearly independent, and infinitely many equations can be obtained by linear combination.

5.1.4

Basis: 2 g mol C: 6 × 12 = 72 H: 8 × 1 = 8 mol. wt. = 72 + 8 + 96 = 176

2 g mol 176 g 1 lb = 0.775 lb 1 g mol 454 g

5–2

O: 6 × 16 = 96

Solutions Chapter 5

5.1.5

a) C H Zn O

LHS

RHS

4 10 1 14

4 10 1 14

Yes, the equation is balanced. b) Basis: 1.5 kg ZnO

1.5 kg ZnO 1000 g ZnO 1 gmol ZnO 1 gmol DEZ 123.4 gDEZ 1.0 kg ZnO 81.40 g ZnO 1 gmol ZnO 1 gmol DEZ

×

c)

1.0 kg DEZ = 2.27 kg DEZ 1000 g DEZ

20 cm 3 H2 O 1.0 g H2 O 1.0 gmol H2 O 1 gmol DEZ 123.4 g DEZ 1 cm 3 H2 O 18.0 g H2 O 5 gmol H 2O 1 gmol DEZ = 27.4 g DEZ

5.1.6

Basis: Data given in problem statement 55.847 gFe 1 gmol Fe 1 gmol X2 O3 2 gmol X = 1 gmol X 1 55.847 gFe 2 gmol Fe 1 gmol X2 O3 55.847 gFe 1 gmol Fe 1 gmol X 2 O3 3 gmol O 16 g O = 240 g O 1 55.847 gFe 2 gmol Fe 1 gmol X2 O3 1 gmol O

50.982 g X 2 O3 − 24.0 g O = 26.982 g X 26.982 g X g = 26.982 hence X =A1 1 gmol X gmol

5–3

Solutions Chapter 5

5.1.7

Basis: Data given in problem statement 1.603g CO2 1 gmol CO2 1 gmol C 12.00g C = 0.437 g C 1 44.01g CO2 1 gmol CO2 1 gmol C 0.2810gH2 O 1 gmol H2 O 2 gmol H 1.008g H = 0.03144 g H 1 18.02g H2 O 1 gmol H2 O 1 gmol H

0.6349 g CxHyOz – 0.437g C – 0.03144g H = 0.1661g O Based on O: 0.1661g O 1 gmol O 0.01038 gmol O = = 1 gmol O 1 16.00g O 0.01038 0.437 g C 1 gmol C 0.03642 gmol C = = 3.5 gmol C 1 12.00 g C 0.01038 0.03144 g H 1 gmol H 0.03119 gmol H = = 3 gmol H 1 1.008 g H 0.01038

The formula in integers is C 7 H6 O2 5.1.8

Basis: Data given in the problem Mass of hydrate Mass of dry sample Mass of water

10.407 g -9.520 g 0.887 g

9.520 g BaI2 1 g mol BaI2 = 0.0243 g mol BaI2 391 g BaI2 0.887 g H2 O 1 g mol H2 O = 0.0493 g mol H2 O 18.02 g H2 O 0.0243 = 0.49, or 2 H 2 O for 1 BaI2 . Thus the hydrate is BaI2 ⋅ 2H 2 O. 0.0493

5–4

Solutions Chapter 5

5.1.9

Basis: 2000 tons 93.2% H2SO4 (1 day)

a.

2000 tons soln 0.932 ton H 2SO4 1 ton mole H 2SO4 1 mol S 32 ton S 1 ton soln 98 ton H 2SO4 1 mol H 2SO4 1 ton mol S = 609 ton S

1.5 mol O2 32 ton O2 b. 609 ton S 1 ton mol S = 913.5 ton O2 32 ton S 1 mol S 1 ton mol O2 1 mole H 2 O 18 ton H 2 O = 342.6 ton H2O c. 609 ton S 1 ton mole S 1 mol S 1 ton mol H 2 O 32 ton S 5.1.10

Basis: 1 lb Br2

2Br − → 1Br2

MW:

2Br– + Cl2 → 2Cl– + Br2 70.9 159.8

MW:

Br2 + C2H4 → C2H4Br2 28 187.9 6

a.

0.27 lb acid 1×10 lb seawater = 2.08 lb 98% H SO / lb Br 2 4 2 2000 lb seawater 65 lb Br 2

b.

1 lb Br 2 mole Br 2 1 mole Cl 2 70.9 lb Cl 2 = 0.445 lb Cl 2 159.9 lb Br 2 1 mole Br 2 mole Cl 2

1 lb Br 2 1×10 6 lb sea water c. = 15,400 lb seawater 65 lb Br 2

d.

1 lb Br2

mol Br2 1 mol C2 H 4 Br2 187.9 lb C2H 4Br2 = 1.176 lb C2 H 4 Br2 159.9 lb Br2 1 mol Br2 1 mol C2 H 4 Br2

5–5

Solutions Chapter 5

5.1.11

FeSO4 ⋅7H2O

MW 277.9

FeSO4 ⋅H 2O

MW 169.9

FeSO4 ⋅4H2O

223.9

FeSO4

151.9

It is best to evaluate the three types of ferrous sulfate in terms of the cost/ton of FeSO4 a.

FeSO4 ⋅7H2O:

Basis: 475 ton material

$83,766.25 = $176.35/ton FeSO4 ⋅ 7H2 O , which is equivalent to 475 ton FeSO4 ⋅ 7H 2 O

$176.35 277.9 ton FeSO4 ⋅ 7H2 O 1 ton mol FeSO4 ⋅7H2 O 1 ton FeSO4 ⋅ 7H2 O 1 ton mol FeSO4 ⋅ 7H2 O 1 ton mol FeSO4

1 ton mol FeSO4 = $323/ton FeSO4 151.9 ton FeSO4 b.

FeSO4 ⋅4H2O: $323 151.9 ton FeSO4 1 ton mol FeSO4 1 ton mol FeSO4 ⋅ 4H2 O ton FeSO 4 1 ton mol FeSO4 1 ton mol FeSO4 ⋅ 4H2 O 223.9 ton FeSO 4 ⋅ 4H 2O = $219 / ton FeSO4 ⋅ 4H2O

c.

FeSO4 ⋅H 2O: $323 151.9 ton FeSO4 1 ton mol FeSO4 1 ton mol FeSO4 ⋅H 2O ton FeSO 4 1 ton mol FeSO4 1 ton mol FeSO4 ⋅ H2 O 169.9 ton FeSO4 ⋅ H2 O = $289 / ton FeSO4 ⋅ H2O

5–6

Solutions Chapter 5

5.1.12

The reaction is C 4H10 + 6 12 O2 → 4CO2 + 5H2O Basis: 1 mol C4H10

⎛ mol O2 for complete combustion ⎞ Minimum O2 ≅ (LFL) ⎜ ⎟ mol C4 H10 ⎝ ⎠ ≅ (1.9%) (6.5/1) = 12.4%

5.1.13

Na2CO3 + Ca(OH2 ) → 2NaOH + CaCO3 Basis: 1 ton of soda ash (Na 2 CO3 )

1 ton Na 2 CO3 1 ton mol Na2 CO3 2 ton mol NaOH 40.0 ton NaOH = 106 ton Na 2 CO3 1 ton mol Na 2 CO3 1 ton mol NaOH

0.755 ton NaOH $130 1 ton Na 2 CO3 = $172 / ton Na 2CO3 1 ton Na 2CO3 0.755 NaOH The above result is for the case of free Ca(OH2). Otherwise, the value must be reduced to compensate for the cost of the CA(OH2)—which might come from the CaCO3 produced, and possibly be cheaper than Ca(OH2) purchased directly.

5–7

Solutions Chapter 5

5.1.14

Basis: 1 ton dolomite Assume complete reactions

CaCO3 + H2SO4 → CaSO4 + H2O + CO2 MgCO3 + H2SO4 → MgSO4 + H2O + CO2 Comp.

%

lb

CaCO3 MgCO3 SiO2

68 30 2

1360 600 40

100

2000

Total

a.

mol.wt. 100.0 84.3 60.0

lb mol that react 13.60 7.13 20.73

20.73 lb mol react 1 mol CO2 produced 44 lb 1 mol react ton dolomite 1 lb mol CO2 = 912 lb CO2

1 mol H 2 SO4 reqd 98 lb H 2 SO4 100 lb acid = 2160 lb acid b. 20.73 lb mol react 1 mole H 2 SO4 94 lb H 2 SO4 1 mol react ton dolomite

5–8

Solutions Chapter 5

5.1.15

Basis: 1 kg dichlorobenzene (DCB) The balanced stoichiometric reaction for dichlorobenzene is shown below: C6 H 4 C12 + 6.5 O2 → 6 CO2 + H 2 O + 2 HC1

Therefore, for 1 kg of dichlorobenzene, the following moles of HC1 are produced:

1 kg DCB 1 kg mol DCB 2 kg mol HC1 = 0.0136 kg mol HC1 147 kg DCB 1 kg mol DCB The balanced stoichiometric reaction for tetrachlorobiphenyl is as shown below: C12 H6 C14 + 12.5 O 2 → 12 CO 2 + H 2O + 4 HC1

Therefore, for 1 lb of tetrachlorobiphenyl (TCB), the following moles of HC1 are produced: Basis: 1 kg TCB 1 kg TCB 1 kg mol TCB 4 kg mol HC1 = 0.0138 kg mol HC1 290 kg TCB 1 kg mol TCB

Thus, the amount of acid produced does not change significantly (≈ 1.5%) , and neither will the amount of base required for neutralization.

5.2.1

Basis: 1 L solution = 1000 g H2O

50 g H2 S 1 g mol H2 S 1 g mol HOCl 52.45g HOCl 100 g HOCl soln 10 6 g soln 34 g H2 S 1 g mol H2 S 1 g mol HOCl 5 g HOCl

1000 g soln 2 = 3.08 g HOCl solution 1 L soln You can use the density of H2O for the density of the solution as the H2 S content has negligible effect on the density.

5–9

Solutions Chapter 5

5.2.2

Basis: 10 lb moles phosgene For phosgene: ξ=

5.2.3

n f − n i 10 − 0 = = 10 reacting moles 1 1

n CO,i = n CO,f -ν COξ = 7 − (−1)(10) =

17 lb mol

n C12 ,i = n C12 ,f -νC12 ,iξ = 3 − (−1)10 =

13 lb mol

Basis: 135 mol CH4 ξ=

0 − 135 = 22.5 reacting mols −6

5.2.4

2FeS + 3O2 → 2FeO + 2SO2 MW

87.85

32

71.85

64

Basis: Use 100 g slag Component FeO Fes For the FeO: ξ=

g=% 80 20 100

MW 71.85 87.85

g mol 1.113 0.228

nf − ni 1.113-0 = = 0.557 reacting moles ν 2

For the FeS: ξ=

0.228 − n i nf − ni = = 0.557 ν -2

ninitial FeS = 1.341 g mol

5–10

Solutions Chapter 5

5.2.5

Basis: 1080 lb bauxite A12 O3 +3H 2SO4 → A12 (SO4 )3 + 3H 2 O 1080 lb bauxite 55.4 lb A12 O3 1 lb mol A12 O3 = 5.87 lb mol A12 O3 100 lb bauxite 101.9 lb A12 O3

ξ=

0 − 5.87 = 5.87 reacting moles −1

Basis: 2510 lb H2SO4 (77%) 2510 lb H2SO4 (77%) 0.77 lb H 2SO4 1 lb mol H 2SO 4 = 19.70 lb mol H 2SO4 1 lb H2SO4 (77%) 98.1 lb H 2SO4

ξ=

0 − 19.70 = 6.57 reacting moles −3

(a)

A12O3 is the limiting reactant and H2SO4 is the excess reactant.

(b)

%xs H2SO4:

(19.70)(1/ 3) − 5.87 6.57 − 5.87 (100) = (100) = 11.9% 5.87 5.87

or H2SO4 (77%) required: %xs =

(5.87)(3)(98.1) = 2244 lb 0.77

2510 − 2244 = 12% 2244

Basis: 2000 lb A12(SO4)3 2000 lb A12 (SO4 )3 1 lb mol A12 (SO4 )3 1 lb mol A12 O3 342.1 lb A12 (SO4 )3 1 lb mol A12 (SO4 )3 101.9 lb A12 O3 1 lb bauxite =1075 lb bauxite 1 lb mol A12 O3 0.554 lb A12 O3 1075 × 100 = 99.5% 1080

5–11

Solutions Chapter 5

5.2.6

Basis: 100 kg of fusion mass BaSO 4

+

4C → BaS

+

4CO

Comp.

Wt.% = kg

MW

kg mol

BaSO4 BaS C Ash

11.1 72.8 13.9 2.2 100.0

233.3 169.3 12.0 ____

0.0476 0.430 1.16 ____

BaSO4 orig. = 0.0476 + 0.430 = 0.4776 mol Corig. = 1.16 + 4 (0.430) = 2.88 mol

Carbon is the limiting reactant.

Needed C for complete reaction = 4 (0.4776) = 1.9104 mol % excess =

2.88 − 1.91 × 100 = 50.8% excess C 1.91

Degree of completion =

0.4776 − 0.0476 = 0.900 0.4776

You could calculate the extent of reaction from the BaSO4, and use it to get the original amounts of reactants.

5–12

Solutions Chapter 5

5.2.7

CO + C12 → COC12

Basis: 20 kg products Comp. C12 COC12 CO

kg 3.00 10.00 7.00 20.00

kg mol 0.0423 0.1011 0.250 0.393

mol.wt. 70.91 98.91 28.01

kg mol C12 in 0.0423 0.1011 0.143

kg mol CO in 0.1011 0.250 0.351

The extent of reaction is CO:

0 − 0.351 = 0.351 −1

C12:

0 − .143 = 0.143 (smallest ξ ) −1

C12 is the limiting reactant; CO is the excess reactant a.

0.351 − 0.143 (100) = 145% excess CO 0.143

b.

0.1011 (100) = 70.7% C12 converted 0.143

c.

0.1011 kg mol COC12 = 0.205 (0.143 + 0.351) kg mol reactants

5.2.8

Amount of o-n hydrolized:

Total protein:

0.10 m mol 1000 µ mol = 100 µ mol 1 m mol

1.00 mg 0.10 mL = 0.100 mg 1 mL

Specific activity =

100 µ mol = 200 µ mol/(min)(mg protein) (5 min)(0.10 mg protein)

5–13

Solutions Chapter 5

5.2.9

Basis: 100 kg mol A 100 kg mol C 6H 5NO2 86.9 kg mol C 6H 8NOCl 0.95 kg mol C 6H 7NO 3 kg mol C 8H 9NO2 100 kg mol C 6H 5NO2 kg mol C 6H 8NOCl 4 kg mol C 6H 7NO

→ 0.619 fraction overall conversion.

5–14

Solutions Chapter 5

5.2.10

C5 H7 O2 N C5 H7 O2 N

60 7 32 14 113 kg/k mol

NH +4 N 14 4 H4 18

20,000 kg C 5 kg NH +4 = 1000 kg NH +4 100 kg C 1000 kg NH +4 kg mol NH 4+ = 55.556 k mol NH +4 + 18 kg NH 4 55.556 kg mol NH +4 1 kg mol C5H 2O2 N 95 kg mol NH +4 113 kg C5H 2O2 N 55 kg mol NH +4 100 kg mol fed kg mol C5H 7O 2 N = 108.43 kg C5H7O2N 5.2.11

2C2 H4 + O2 → 2C2 H4O For 100% conversion of C2H4 according to reaction (a):

For reaction b3:

2 mol C2 H 4O / time =1 2 mol C2 H 4 / time 1 mol C2 H 4 O/time = 0.75 1.33 mol C2 H 4 /time 5–15

Solutions Chapter 5

5.2.12

Step 4: The molecular weights needed to solve the problem and the gram moles forming the basis are: Component Sb2S3 Fe Sb FeS

kg 0.600 0.250 0.200

Mol. wt. 339.7 55.85 121.8 87.91

g mol 1.766 4.476 1.642

The process is illustrated in Fig. P9.19 1.766 g mol Sb2S3 Reactor

4.476 g mol Fe

Fe S

1.642 g mol Sb

Figure P9.19 Step 5:

Basis: Data given in problem statement

Steps 6-9: (a)

To find the limiting reactant, we examine the chemical reaction equation. The ratio of Sb2S3 to Fe in the equation is 1/3 = 0.33. In the actual reaction the corresponding ratio is 1.766/4.476 = 0.395, hence Sb2S3 is the excess reactant and Fe is the limiting reactant. The Sb2S3 required to react with the limiting reactant is 4.476/3 = 1.492 g mol.

(b)

The percentage of excess reactant is % excess =

(c)

1.766 − 1.492 (100) = 18.4% excess Sb 2S3 1.492

Although Fe is the limiting reactant, not all the limiting reactant reacts. We can compute from the 1.64 g mol of Sb how much Fe actually does react:

5–16

Solutions Chapter 5 1.64 g mol Sb 3 g mol Fe = 2.46 g mol Fe 2 g mol Sb

If by the fractional degree of completion is meant the fraction conversion of Fe to products, then Fractional degree of completion = (d)

2.46 = 0.55 4.48

Let us assume that the percent conversion refers to the Sb2S3 since the reference compound is not specified in the question posed. 1.64 g mol Sb 1g mol Sb 2S3 = 0.82 g mol Sb 2S3 2 g mol Sb

% conversion of Sb 2S3 to Sb = (e)

0.82 (100) = 46.3% 1.77

The yield will be stated as kilograms of Sb formed per kilogram of Sb2S3 that was fed to the reaction Yield =

0.200 kg Sb 0.33 kg Sb = 0.600 kg Sb2S3 1 kg Sb2S3

5–17

Solutions Chapter 5

5.2.13

Reaction on which to base excess is Fe2O3 + 3C → 2Fe + 3CO Basis: 1 ton Fe2O3 a. Theoretical C =

1 ton Fe 2O3 2000 lb Fe 2O3 1 lb mol Fe 2O3 3 lb mol C 12 lb C = 451 lb 1 ton Fe 2O3 159.7 lb Fe 2O3 1 lb mol Fe 2O3 lb mol C

600 œ451 × 100 = 33% excess C 451 b.

1 ton Fe 2O3 1200 lb Fe lb mol Fe 1 lb mol Fe 2O3 159.7 lb Fe 2O3 1 ton Fe 2O3 55.85 lb Fe 2 lb mol Fe 1 lb mol Fe 2O3

= 1720 lb Fe2O3 reacts

1720 × 100 = 86.0% Fe O 2 3 2000 c. (i)

1 ton Fe 2O3 183 lb FeO lb mol FeO 1 lb mol CO 28 lb CO = 35.62 lb CO 1 ton Fe 2O3 71.85 lb FeO 2 lb mol FeO 1 lb mol CO

1 ton Fe 2O3 1200 lb Fe 1 lb mol Fe 3 lb mol CO 28 lb CO = 902.4 lb CO 1 ton Fe 2O3 55.85 lb Fe 2 lb mol Fe 1 lb mol CO Total (35.62 lb CO + 902.4 lb CO) = 938 lb CO

(ii)

1 ton Fe 2O3 1200 lb Fe lb mol Fe 3 lb mol C 12 lb C = 387.1 lb C 1 ton Fe 2O3 55.85 lb Fe 2 lb mol Fe lb mol C

1 ton Fe 2O3 183 lb FeO lb mol FeO 1 lb mol C 12 lb C = 15.3 lb C 1 ton Fe 2O3 71.85 lb FeO 2 lb mol FeO lb mol C 387.1 + 15.3 = 402.4 lb C used / 1ton Fe2O3

5–18

Solutions Chapter 5

or use

938 lb CO 1 lb mol CO 1 lb mol C 12 lb C 28 lb CO 1 lb mol CO 1 lb mol C Selectivity: Mass basis:

d.

1200 lb Fe formed = 6.56 183 lb FeO formed

1200 lb Fe lb mol Fe = 21.49 lb mol Fe 55.85 lb Fe 183 lb FeO × Mole Basis:

lb mol FeO = 2.55 lb mol FeO 71.85 lb FeO

21.49 lb mol Fe = 8.44 2.55 lb mol FeO

5.2.14

Cl2 + 2NaOH → NaCl + NaOCl + H2O MW: 71.0 40.01 58.5 74.5 18.0 Basis: 1145 lbm NaOH + 851 lb Cl2 (→ 618 lbm NaOCl) a. Determine limiting reactant Assume all Cl2 reacts, calculate lb of NaOH required

NaOH required =

2 lb mol NaOH lb mol Cl 2 851 lb Cl 2 40.01 lb NaOH 71 lb Cl 2 lb mol NaOH lb mol Cl 2

= 959.11 lb NaOH required

C12 is limiting reactant

or use extent of reaction NaOH: C12:

1145 = 28.63 lb mol 40 851 = 11.99 lb mol 71.0

(0-28.63)/-2 = 14.3

(0-11.99)/ − 1 = 12.0 (minimum)

5–19

Solutions Chapter 5

b. % excess NaOH =

=

lb mol NaOH in excess (100) lb mol NaOH for rxn

1145 lb m/NaOH

lb mol NaOH 1 œ 959.11 40.01 lb NaOH 40.01 (100) 1 959.11 40.01

= 19.4% excess NaOH 618 lb mol C12 that reacted c. Degree of completion = = = 74.5 = 0.692 lb mol C12 available to react 851 71.0 ξ 8.30 or = =0.69 ξ min 12.8

lb d. Yield of NaOCl = 618 lb NaOCl = 0.726 851 lb Cl 2 lb e.

618 = 8.30 lb mol NaOC1 745

ξ=

8.30 − 0 = 8.30 reacting moles 1

5–20

Solutions Chapter 5

5.2.15

4HC1+ O 2 → 2C12 + 2H 2O

MW

35.45 32

71.0

18

Basis: 100 mol gas Exit comp.

Entering moles HC1 O2

% mol

HC1

4.4

C12

19.8

H2 O

19.8

4.4

-

(2)(19.8)

39.6

-

(1/2)(19.8)

-

9.9

O2

4.0

-

4.0

N2

52.0

-

-

44.0

13.9

100.0 (The N2 balance gives the O2 as 13.8).

Calculate the extent of reaction for complete reaction For HC1:

0 − 44.0 = 11.0 (the minimum) −4

For O2:

0 − 13.9 = 13.9 −1

(a)

HC1 is the limiting reactant 44.0 4 100 = 26.36% 44.0 4

139 −

(b)

% excess =

(c)

degree of completion =

44 − 4.4 = 0.90 44

(or 9.9 from (d)/11.0) = 0.90 (d)

extent of reaction ξ =

19.8 − 0 = reacting moles 2

5–21

Solutions Chapter 5

5.2.16

Basis: 30 mol CO, 12 mol CO2, 35 mol H2O feed = 1 hr CO CO2 H2O

mol 30 12 35

mol F

CO + H2O

MW: a)

CO CO2 H2 18 H2O

P

28

18

CO2 + H2

fed in eqn CO 1 CO 30 = = 0.86 < = =1 H2 O 1 H2 O 35

44

2

CO is limiting reactant

b)

H 2O is the excess reactant

c)

18 = 0.514 35

d)

18 = 0.60 30

e)

18 kg mol H2 2.016 kg H 2 1 kg mol H 2O 1 kg mol H2 35 kg mol H2 O 18 kg H2 O

(for each mol of H2 produced, 1 mol CO reacts)

= 0.0576 kg H2 f)

1 mol of CO2 is produced for each mol of H 2 mol CO 2 produced 18 = 0.60 30 mol CO fed

(g)

ξ=

18 − 0 = 18 reacting moles using H 2 1

5–22

Solutions Chapter 5

5.2.17 Basis: 100 lb mesitylene (C9H12) The reactions are

C9H12

MW 120.1

C9 H12 + H 2 → C8 H+10 CH 4

C8H10 106.08

C8 H10 + H 2 → C7 H+8 CH 4

C7 H8

92.06

The initial selectivity is: C8 H10 /C7 H8 = 0.7 and 0.8 The number of moles of C9H12 in the basis is

100 lb C9 H12 1 lb mol C9 H12 = 0.833 lb mol 120.1 lb C9 H12 For the selectivity of 0.7 (the equations are in lb mol) (1) C8 H10 + (1)C7 H8 = 0.833

C8 H10 / C7 H8 = 0.7 with the results:

(lb mol)

(in lb)

C7 H8 = 0.490 C8 H10 = 0.343

45.1 36.4

For the selectivity of 0.8 the results are

C7 H8 = 0.463 C8 H10 = 0.370

42.6 39.3

The catalyst used in the respective cases is Selectivity of 0.7

Selectivity of 0.8:

0.490 lb mol C7 H8 92.06 lb C7 H8

1 lb catalyst 500 lb C7 H8

1 lb mol C7 H8 0.463 lb mol C7 H8 92.06 lb C7 H8

1 lb catalyst 500 lb C7 H8

The economic analysis of the change is

5–23

1 lb mol C7 H8

= 0.090 lb

= 0.085 lb

Solutions Chapter 5

Income change Increase in C8H10: Decrease in C7H8:

$ (39.3 − 36.4)($0.65) = +1.89 (45.1 – 42.6) ($22) =

− 0.55

Expense change (decrease) Change in catalyst used: (0.090-0.085) ($25) = + 0.13 Net change

5.3.1

$1.47

Basis: 1 mol A A → 3B

50 percent conversion gives as the product: A B

mol 0.5 1.5 2.0

(a)

Mole fraction A = 0.5/2.0 = 0.25

(b)

Extent of reaction ξ =

0.5 − 1.0 = 0.5 using A −1

5–24

Solutions Chapter 5

5.3.2

Step 5:

2 FeS2 + 121 O2 → Fe2 O3 + 4 SO2

Basis: 100 lb pyrites in

Steps 2, 3, and 4: SO3

Y

Feed 100 lb pyrite 32% S 10 lb S

Z

X O2 N2

mol fr 0.21 0.79

Gas

mole %

SO 2

13.4

O2

2.7

N2

83.9 100.0

Air Cinder : Fe2 O3 , inert

1.00

Let X = mol air in, Y = mol SO3 out, Z = mol flue gas out Steps 6 and 7: Using a reduced set of variables: Unknowns (4): X1 Y1 Z, cinder Balances (4): O, N, S, cinder Steps 8 and 9: 1.

Sulfur Balance:

(32 + 10)/32 = Y + 0.134Z

2.

2O Balance:

0.21X = 1.5Y + (0.134 + 0.027) Z

+ 3.

2N Balance:

1mol S mol Fe 1mol Fe 2 O3 1.5mol O2 2mol S 2 mol Fe 1mol Fe 2 O3

0.79 X = 0.839Z

Solve (1), (2), (3): Y = 0.118 mol, Z = 8.9 mol Step 10: Check: mol S = 0.118 + 0.134(8.9) = 1.31 % conversion of S to SO3 = (11.8)/(1.315) = 8.98% 5–25

Solutions Chapter 5

5.3.3

Steps 1, 2, 3 and 4: mole % Product

Methanol F CH3 OH 100%

Reacts P A

Air

O2 N2

mol frac 0.21 0.79 1.00

Step 5:

Basis: 100 mol P

Step 6: Unknown

F, A

Step 7: Balances

N, H, O, C

H2 O

5.9

O2

13.4

N2

62.6

CH2 O

4.6

CH3 OH

12.3

HCOOH

1.2 100.0

More balances than unknowns. Are they independent? Consistent? Steps 8 and 9:

Balances in moles

Check by first solving C, N (tie elements) C:

F (1) = 4.6 + 12.3 + 1.2 = 18.1

F = 18.1 mol

2N:

A (0.79) = 62.6

A = 79.24 mol

F A Check via O: (2) (79.24)(.21) + 18.1

51.38

? =

13.4(2) + 5.9+4.6+12.3+1.2(2)

? =

52.0

Close but not exactly the same, more oxygen out than in. Might be caused by round off. Check via H:

4(18.1)

? =

5.9(2) + 4.6(2) + 12.3(4) + 1.2(2)

72.40

? =

72.6

Not exactly the same; more hydrogen out than in, but again may be caused by round off. The conclusion is that probably no error exists in the measurements. 5–26

Solutions Chapter 5

5.3.4

a.

Additional information was the molecular weights of the compounds and the chemical reaction equations. MW CaCO3 = 100.0, MW MgCO3 = 84.3, MW CO2 = 44.

b.

Assume a batch process Final

Initial F

CaCO3

CaO

MgCO3

MgO

P

G CO2

c.

Basis: 1 g sample

d.

F = 1.000 g; P is not explicitly given but can be calculated: P = 0.550 g; G is not explicitly given and can be calculated, G = 0.450 g

e.

Mass fraction of CaO in P and F, mass fraction MgO in P and F (or mass of each)

f.

Species balances

g.

It appears to be a carbon balance

h.

0 degrees of freedom

i.

Yes

5–27

Solutions Chapter 5

5.3.5

Assume a steady state flow process with reaction. Step 5: Basis: 1 hr (6.22 kg mol = F) Steps 1, 2, 3 and 4:

H2 (g)+SiHC13 (g) → Si(s)+3HC1(g)

F = 6.22 kg mol Mol fr. 0.580 H 2

Reactor

Mol fr. mol W(kg mol) HCl x HCl nHCl SiHCl3 x SiHCl 3 n SiHCl 3 H2

P

0.420 SiHCl 3 1.00

Si

0.223 nH 2

Mol fr. 1.00 MW Si = 28.086 n H 2 /W = 0.223

Step 6:

Unknowns:

W, P, 3 n i

Step 7:

Balances:

element balances are H, Si, Cl, ∑ ni = W, n H 2 /W = 0.223

Or, using the extent of reaction Step 6:

Unknowns = 6:

W, P, 3ni ,ξ Step 7:

Balances = 6 Species balances: H2, SiHC13, HC1, Si ∑ ni = W

n H2 /W = 0.223 Steps 8 and 9:

Balances are in kg mol.

5–28

Solutions Chapter 5

Cl (kg mol):

Out In = 6. 22(0. 420)(3) nHCl + n SiHCl3 (3)

Si (kg mol):

6.22(04.20)(1) = P(1) + n SiHCl3

H (kg mol):

6.22[( 0.580 )( 2) + 0.420(1)] = n HCl + nSiHCl3 + nH 2 (2 )

Balances using ξ :

H2 : ⎡⎣n H2 − (0.580)(6.22) ⎤⎦ = ( 1)(ξ ) SiHC13 : ⎡⎣nSiHC13 − (0.420)(6.22) ⎤⎦ = ( 1)(ξ ) HC1:

− −

[n HC1 − 0] = (+3)(ξ ) [nSi − 0] = (+1)(ξ )

Si : ξ = 1.83 Solution:

W = 8.05

P=1.83

n H 2 = 1.78

n HCl = 5.49

n SiHCl3 = 0.7824

1.83kgmol Si 1hr 20 min 28.086 kgSi = 17.13kgSi 1hr 60min 1kgmol Si 1.46 kg Si initial 18.59 kgfinal Si

5–29

Solutions Chapter 5

5.3.6

Steps 1, 2, 3, and 4: G(kg mol)

Ore Cus Cus* inert

SO3

P(kg mol)

Reactor n CuS (kg mol)

Y (kg mol) CuO A(kg mol) Air O2 N2

*

100%

SO2

mol fr. 0.072

O2

0.081

N2

0.847 1.000

unreacted CuS inert

mol fr. 0.21 0.79 1.00

This CuS doesn’t react.

This is a steady state open process with reaction. Step 5: Basis: 100 kg mol P Steps 6 and 7: Unknowns: n CuS ,A, G, Y Element balances: S, O, N, Cu Steps 8 and 9: Balances are in kg mol on the elements S:

n CuS (1) = G(1) + 0.072 (100)

2O:

A(0.21) = G(1.5) + 100(0.072) + 100(0.081) + Y( )

2N: Cu:

A(0.79) = 100(0.847) n Cus (1) = Y(1)

1

2

5–30

Solutions Chapter 5

Solution yields (in kg mol): G = 1.81, nCuS = Y = 9.01

1.81 mol SO3 1 mol S 1 mol SO3 = 0.20 of CuS that reacts goes to SO3 20% 9.01 mol CuS that reacts 1 mol S 1 mol CuS If you use the extent of reaction, you need to write down the balanced reactions

CuS + O2 → CuO + SO2 SO2 +

1 2

(1) (2)

O2 → SO3

Steps 6 and 7: The unknowns: 4 above plus ξ1 and ξ2 Balance: 6 species balances Steps 8 and 9: CuO: CuS

Y 0

= =

0 + (1) (ξ1 ) n CuS + (−1)(ξ1 )

O2 :

P(0.081)

=

A(0.21) + (−1)( +ξ1−) (

N2 SO2 SO3:

P(0.847) P(0.072) G

ξ1 = 9.01

ξ2 = 1.81

= = =

A(0.79)

0 + (1)(ξ1 ) + ( 1)(ξ2−) 0 + (1)(ξ2 )

5–31

1 2

)(ξ 2 )

Solutions Chapter 5

5.3.7

(1)

6HF(g) + SiO2 (s) → H2SiF6 (l ) + 2H2O(l )

(2)

H2SiF6 (g) → SiF4 (g) + 2HF(g) The net reaction is

4HF + SiO2 → SiF4 + 2H2O Steps 2, 3, and 4: SYSTEM mol fr. 0.50 HF 0.50 N2 1.00

F

SiO2 on wafers

P

The process is an unsteady state, open process Step 5:

Basis: 1 mol F entering

Steps 6 and 7: Unknowns: 4 ni , P1 ξ, loss SiO2 from wafer (L) Balances: HF, N2 , SiF4 , H2O, SiO2 (5 species) Other equations: ∑ ni = P; 10% HF reacts Steps 8 and 9:

SiO2:

HF:

n HF = 0.50 + (−4)ξ = 0.50(.90) = 0.45 ξ = 1.25 ×10−2 reacting moles

SiF4:

n SiF4 = 0 + (1)(1.25×10-2 ) = 1.25 × 10-2 mol

H2O: N2 :

n H2O = 0 + (2)(1.25 × 10-2 ) = 2.50 × 10-2 mol

n N2 = 0.50 mol

n SiO2 ,f − n SiO2 ,i =(−1)(ξ ) = −1.25 × 10-2 mol

5–32

nHF nN2

nSiFH nH2O

Solutions Chapter 5

Composition HF SiF4 H2 O N2

mol 0.45 1.25 10-2 2.50 10-2 0.50 0.9875

mol fr. 0.456 0.013 0.025 0.506 1.000

If you use element balances: Step 6 and 7: The unknowns: 4ni, P, L The balances: H, F, Si, O, N (presumably 4 are independent) Other equations: ∑ n i = P, 10% HF reacts Steps 8 and 9: H: F: Si: O: N:

0.50(1) = n HF + 2n H O 0.50(1) = n HF + 4nSiF (nSiO ,f -nSiO ,i ) = nSiF 2(nSiO ,f -nSiO ,i ) = n H O 2(0.50) = 2n N 2

4

2

2

2

4

2

2

2

n HF = 0.90(0.50)=0.45

5–33

Solutions Chapter 5

5.3.8 We will view this process as a steady state process in an open system with flow in and out, and a change in the material in the vessel from the initial to the final states. For component i, Equation (10.1) becomes Equation (10.6) R

n iout = n ini + ∑ νijξ j j=1

The bioorganisms do not have to be included in the solution of the problem. Steps 1, 2, 3, and 4 Figure P10.9 is a sketch of the process.

CO2 (g)

H2O C6H12O6

H2O (88%)

F C6H12O6 (12%)

C2H5OH C2H3CO2H Figure P10.9

Step 5 Basis: 3500 kg F Step 4

Convert the 3500 kg into moles of H2O and C6H12O6.

n initial = H O

3500(0.88)

2

n initial = C6 H12 O6

18.02

= 170.9

3500(0.12) 180.1

= 2.332

Step 6 and 7 The degree of freedom analysis is as follows: Number of variables: 9

5–34

Solutions Chapter 5

Out Out Out Out n InH2O , n CIn6H12O6 , n Out H2O , n C6 H12O6 , n C2 H5OH , n C2 H2CO2 H , n CO2 , ξ1 , ξ2

Number of equations: 9 Basis: F = 3500 kg of initial solution (equivalent to initial moles of H2O plus moles of sucrose) Species material balances: 5

H2O, C2 H12O6 , C2 H5OH, C2 H3CO2 H, CO2 Specifications: 4 (3 independent)

n InH2O = 170.4 or n CIn6H12O6 = 2.332 (one is independent, the sum is F in mol) n Out C6 H12 O6 =

90 = 0.500 180.1

n Out CO2 =

120 = 2.727 44

The degrees of freedom are zero. Step 8

The material balance equations, after introducing the known values for the variables are:

H 2O :

n Out H2O = 170.9 + (0)ξ1 + (2)

C2 H12O6 :

0.500 = 2.332 + (−1)ξ1 + (−1)ξ2

(b)

C2 H5OH :

n Out ξ C2 H5OH 0 + 2ξ1 + (0)

(c)

2

C2 H3CO2 H : n Out ξ C2 H3CO2 H + 0 + (0)ξ1 + (2) CO2

2

2.727 = 0 + (2)ξ1 + (0)ξ

2

2

ξ

(a)

(d) (e)

If you do not use a computer to solve the equations, the sequence you should use to solve them would be (e), plus (b), (a), (c), (d) Step 9

The solution of Equations (a) – (e) is

ξ1 = 1.364 kg moles reacting

ξ 2 = 0.469 kg moles reacting

5–35

Solutions Chapter 5

Results

Conversion to mass percent

Species

kg moles

MW

kg

mass %

H 2O

171.8

18.01

3094.1

89.0

C2H5OH

2.727

46.05

125.6

3.6

C2H3CO2H

0.939

72.03

67.5

1.9

CO2

2.272

44.0

100.0

2.9

C2H12O5

0.500

180.1

90.1 3477.3

2.6 1.00

Step 10 The total mass of 3477 kg is close enough to 3500 kg of feed to validate the results of the calculations.

5.3.9

Steps 1, 2, 3 and 4:

SiH4 +O2 → SiO2 +2H2 4PH3 + 5O2 → 2P2O+5 6H2 100% O2 100% SiH4 100% PH

B

A D

C

3

SiO2

H2 100%

E

P2 O5 Step 5: Basis:

1 g mol PH3

Step 6: No. Unknowns:

E A, B, D, E, n P 2 O 5

5

Step 7: Element balances:

Si, H, P, O, ωEP2O5

5

Note: n EP 2 O 5 + n ESiO2 = E Steps 8 and 9:

5–36

Solutions Chapter 5 P Bal (in grams):

in

Out

⎫ 1 g mol PH3 2 g mol P2O5 2 g mol P 31gP = 31g P ⎪ 4 g mol PH3 1 g mol P2O5 1 mol P ⎪ ⎬ in P2O5 1 g mol PH3 2 g mol P2O5 5 g mol O 16 g O ⎪ = 40 g O ⎪ 4 g mol PH3 1 g mol P2O5 1 g mol O ⎭ Total: or since all the P is in stream E: Si Bal:

________ 71 g P2O5

1 g mol PH3 2 g mol P2O5 142 g P2O5 = 71g P2O5 4 g mol PH3 1 g mol P2O5

in Out B g mol SiH4 1g mol SiO2 1g molSi 28.1g Si = 28.1B g Si 1g mol SiH4 1g mol SiO2 1g mol Si B g mol SiH4 1g mol SiO2 2 g mol O 16 g O = 32 B g O 1g mol SiH4 1g mol SiO2 1g mol O

Total: 71 = 0. 05 71 + 60.1B

1 g mol PH 3

3.55 + 3.0058B = 71

34 g PH 3 = 34 g PH 3 1g mol PH 3

23.5 . g mol SiO2 32.1g SiH 4 = 7.54 g SiO2 O2 1g mol SiH4

754 34

= 22.2

gSiO 2 gPH

5–37

60.1 B g SiO2 B = 23.5 g mol

Solutions Chapter 5

5.3.10

Steps 1, 2, 3 and 4:

Cu

Cu(NH 3 ) 4 Cl 2 NH 4 OH

A

B C

D E

Cu(NH 3 ) 4 Cl H2 O

F Cu Step 5:

Basis: 1 board 3

4in 8in 0.03in (2.54cm) 8.96g l gmol Cu = 2.219g mol Cu in 3 cm 3 63.55g Cu

A:

Cu foil

F:

Cu remaining

1 − 0.75 2.219 gmol Cu

Steps 6 and 7: Unknowns:

= 0.555g mol Cu

B, C, D, E

Balances: Cu, N, Cl, H, O should be enough if one is redundant Steps 8 and 9: Cu Balance:

2.219g mol Cu+

(Balances to be solved:)

Bg mol Cu(NH3 )4 Cl 2

= 1.664 gmol Cu +

D gmol Cu (NH3 ) 4 Cl

1gmol Cu 1g mol Cu(NH3 )4 Cl 2 1gmol Cu 1gmol Cu (NH3 )4 Cl

5–38

Solutions Chapter 5

Cl Balance:

Bg mol Cu(NH3 )4 Cl 2

D g mol Cu(NH3 )4 Cl 2 gmol Cl 1gmol Cl = gmol Cu(NH3 )4 Cl 2 gmol Cu(NH3 ) 4 Cl

N Balance:

Bg mol Cu(NH3 )4 Cl 2

=

D gmol Cu(NH3 ) 4 Cl

Solution: Cu overall:

Cg mol NH4 OH 1gmol N 4 gmol N + gmol Cu(NH3 )4 Cl 2 gmol NH4 OH 4gmol N g mol Cu(NH4 )Cl

A + B = F + D →A + B = F + (2B) 2.219 + B = 0.555 + (2B) B = 2.219 - 0.555 = 1.664 g mol Cu(NH3)4Cl2

Cl:

D = 2B= 2(1.664) = 3.328 g mol Cu(NH3)4Cl N:

4B + C = 4D

C = 4(D-B) = 4(3.328 - 1.664) = 6.656 g mol NH4OH MW of NH4OH N 14 H4 4 O 16 H 1 35

Cu(NH3)4CL2 Cu N H Cl2

35gNH4 OH 6.656gmol = 232.0 g NH4 OH gmol

63.55 56.00 12.00 70.90 202.45

202.45 (g/gmol) (1.664 g mol) = 336.88 g Ca ( NH 3 ) 4 Cl 2 5–39

Solutions Chapter 5

5.3.11

F

Feed Mixer

contaminated sand ( 16,000 lb )

F'

Incinerator

60% combustible

sand plus combustion products

H

O2 enriched air x O 2 = 0.4 100% excess Basis: 8 tons of contaminated sand containing 30% PCB hexane

a)

4, 4' dichlorobiphenyl ⇒

Cl

Cl

C12H8Cl2, MW = 222

8tons sand + PCB 0.3ton PCB 2000lb PCB = 4800 lbPCB tonsand + PCB ton PCB If we want the net feed to be 60% combustible, we see that the sand is a tie component: A sand balance around the feed mixer: 16000(0.7) = F'(0.4) so F' =28,000 lb. A total balance around the feed mixer.: F + H = F' ⇒ H = 12, 000lb m hexane b)

C1 2H8Cl 2 + 13.5O2 → 12CO2 + 2HCl + 3H 2 O C 6 H1 4 + 9.5O2 → 6CO2 + 7H2 O

Even with total combustion, we will produce undesirable HCl, which will adversely affect the environment if it is simply vented. Therefore, it would be wise to condense the HCl and H2O by cooling. Perhaps the acid solution might be sold to help pay for the process. (c) 21.62 moles PCB

P

139.5 moles nC 6

16.54% CO2 12.20% O2 H2 O = ? HCl = ? N2 = ?

A enriched air (40% O2, 60% N 2 )

5–40

Solutions Chapter 5

By a tie component for CO2 , we know: 12(21.62) + 139.5 (6) = 0.1654 P hence P = 6629 mol By an O tie: 0.40(2)A = 0.1654(2)(6629) + 0.1220(2)(6629) + x H 2 O (1) (6629)

x H 2 O (6629) =

21.62 PCBReacted 3H2 O 139.6 C6 Reacted 7 H2 O = = 1042 1 PCB 1 C6

so x H 2 O = 0.1572 From O tie balance: A = 6065.4 mol of 40% O2 air Thus: (N2) in = (N2) out = 0.6(6065.4)=3639.2 mol N2

xN2 =

3639.2 = 0.549 6629.0

We can get HCl by a Cl balance:

21.6PCBReacted 2 HCl = 43.2mol H Clin product 1PCB

xHCl= 0.0065 Component CO2 O2 H 2O HCl N2

xi 0.1654 0.1220 0.1572 0.0065 0.5490

We know % excess O2 used is : % excess =

100 [O2 fed − O2 req' d ] (O2 req' d)

O2 fed = 0.4 (air fed) = 0.4 (6065.4) = 2426.2 mol O2 Also, since we have total combustion, (O2 req'd) = (O2 consumed) = (O2 fed - O2 out) = 2426.2 - 0.1220(6629) = 1617.5 moles req'd ⎛ 2426.2 − 1617.5⎞ 50%excessO 2 used Thus: % O2 = 100 ⎝ ⎠ = 1617.5 5–41

Solutions Chapter 5

5.3.12 & 5.4.1

Steps 1, 2, 3, and 4: To solve this problem you must recall that CaCO3 reacts with H2SO4 to form CaSO4 and that CaCO3 when heated yields CO2 and CaO whereas when CaSO4 is heated at the same time it does not decompose. Although the problem seems to be underspecified, when you draw a diagram of the process and place the known data on it, the situation becomes clearer. Assume the process is a steady state open one with reaction. Then the element balances are just in = out. If you use the extent of reaction, you make species balances. Note that we have designated the composition of P and A in terms of the mass of each component, m Pi , rather than the mass fraction ω Pi , because this choice makes the element or species balances linear (avoids products such as ωP). The CO2 liberated from the sludge is equivalent to 1lbCO2 1lb mol CO2 1lb mol CaCO 3 100.09lb CaCO3 10 lbP 44lb CO2 1lb mol CO2 1lb mol CaCO3 = 0.227 lb CaCO3/lb P A (lb) waste acid H2 SO4

H 2O

lb m H SO A

2

m

Y (lb mol)

4

A H 2O

CO2 gas 1.000 = mol fr.

∑ =A lb Feed F (lb)

Reaction Vessel

P (lb) dry sludge

mass fr. CaCO 3 inert

0.95 0.05 1.00

Water W (lb) H2 O mass fr. = 1.00

Step 5:

lb CaCO 3

m PCaCO

CaSO 4

m CaSO

inert

m inert

3

P

4

P

∑ = P (lb) also

m PCaCO P

3

= 0.227

Basis: F = 100 lb A

Step 6: The variables whose values are unknown are A, m AH 2 SO 4 ,m H 2 O ,m PCaCO3 m PCaSO 4 , m Pinert , P, W, and Y, a total of 9. 5–42

Solutions Chapter 5

Step 7: The element balances that can be made are Ca, C, O, S, H, inert (6 total) and we know Σ m Ai = A , Σ m Pi = P, and m PCaCO3 / P = 0.227 for a total of 9. If the equations are independent, we can find a unique solution.

(

)

Steps 8 and 9: The equations (in = out) are (except for the inert balance which is in lb) in moles 1 lb mo lCa 95 lb CaCO3 1 lb mol CaCO3 100.09 lb CaCO3 1lb mol CaCO3

Ca:

m PCaCO3 lbCaCO3 1lb mol CaCO3 1lb mol Ca = 100.09lb CaCO3 1lb mol CaCO3 m PCaCO3 lbCaSO4 1lb mol CaSO4 1lb mol Ca + 136.15lb CaSO4 1lb mo l CaSO4 1 lb mol C 0.95 lb CaCO 3 1 lb mol CaCO3 Y lb mol CO2 1 lb mol C = 100.09 lb CaCO3 1lb mol CaCO 3 1 lb mol CO2

C:

m PCaCO3 lbCaCO3 1 lb mol CaCO3 1 lb mol C + 100.09 lb CaCO3 1 lb mol CaCO 3 A

m H 2 SO 4 lb H2 SO4 1 lb mol H2 SO4 1 lb molS S: 98.08 lb H2 SO4 1 lb mol H2 SO4 m PCaSO4 lb CaSO4 1lb mol CaSO4 1lb molS = 136.25lbCaSO4 1lb mol CaSO4

m AH 2 SO 4 lb H2 SO 4 1 lb mol H2 SO 4 1 lb mol H 2 2H : 98. 08 lb H 2 SO4 1 lb mol H2 SO4 +

m AH 2 O lb H2 O 1 lb mol H2 O 1 lb mol H 2 18. 016 lb H2 O 1 lb mol H 2 O

=

W lb H2 O 1 lb mo l H2 O 1 lb mol H2 18. 016 lb H2 O 1 lb mol H2 O

O: +

95 lb mol CaCO 3 1 lb mo l CaCO3 3 lb mol O 100.09 lb CaCO3 1 lb mol CaCO 3 m AH 2 SO4 lb H2 SO4 1 lb mol H2 SO4 4 lb mol O 98.08 lb H2 SO4 1 lb mol H2 SO4 5–43

Solutions Chapter 5 A

m lb H2 O 1lb mol H2 O 1lb mol O + H 2O 18.016lb H2 O 1lb mol H 2 O =

Y lb mol CO2 2lb mol O 1mol CO2

+

m PCaCO3 lbCaCO3 1lb mol CaCO3 3lb mol O 100.09lb CaCO3 1lb mol CaCO 3

m PCaSO4 lb CaSO4 1lb mol CaCO 3 4lb mol O + 136.15CaCO 4 1lb mol CaSO4 +

W lb H2 O lb mol H 2 O 1lb mol O 18.016lb H2 O l lb mol H 2O

Inert: 5 lb inert = m Pinert lb inert The solution of these equations using a computer code would give all of the values of the unknown quantities. However, a review of the set of equations, after having gone into great detail about each equation, indicates that only 4 of the equations have to be solved to get the desired answer, namely the Ca balance, the Σm Pi = P, and the inert balance plus m PCaCO3 = 0.227 : Σm Pi = P:

P m PCaCO3 + m PCaSO4 + 5 = P = m CaCO / 0.227 3

Ca:

P P 0.949 = 0.010 m CaCO3 + 0.00734 m CaSO4

m PCaCO3 = 28.55 lb and 100

28.55 = 30%unreacted . 95

5–44

Solutions Chapter 5

5.5.1

Basis: 100 kg mol

synthesis gas

Burner

flue gas

air One composition is unknown, the flue gas. After selection of a basis, the air flow is known. Hence, this problem can be solved by direct addition and subtraction Unknowns:

(CO2, H2O, N2,O2) = 4

Balances:

(C, H, O, N) = 4

Comp.

%=mol

CO2 O2 CO H2 N2

6.4 -------------0.2 ----------(0.2) 40.0 CO+1/2 O2→CO2 20.0 50.8 H2+1/2 O2→H2O 25.4 2.6 -------------XS O2: (0.40)(45.2) = 18.08 Total O2 in 63.28 N2 in with O2 (63.28)(79/21) = Totals

Comp.

mol

Reaction

%

CO2

46.4

13.0

H 2O

50.8

14.3

240.65

67.6

N2 O2 Totals

reqd.O2

18.1

5.1

356.0

100.0

5–45

CO2

H2 O

N2

O2

6.4 40.0 50.8

46.4

50.8

2.6 238.05 240.65

18.08 18.1

Solutions Chapter 5

5.5.2

Steps 1, 2, 3, and 4:

H2 O W

C H

mass fr. 0.80 0.20

F

30 lb

mol n CO n CO2 nO2

coal A

1.00

O2 N2

nN2 600 lb mol fr. 0.21 0.79

∑ =F

1.00

Steps 5 and 6: With a basis of 30 lb coal, only n N 2 ,n O2 ,n CO2 ,n CO ,F and W are unknown. Total 6 Step 7: Can make C, H, O, N element balances. plus

n CO2 nCO

=

3 and ∑ n i = F 2

Do calculations in moles, but you don't need to make all of the element balances Comp. C 2H

lb mol wt lb mol 24 12 2 6 2 3 30 3.5lb mol O2 1.00lb mole air = 16.67lb mol air reqd. 0.21lb mol O2

or

4.35− 3.5 (100) 3.5

= 24.3%

5–46

Rxn C+O2→CO2 H2+1/2O2→H2O

reqd O2 2.0 1.5 3.5

Solutions Chapter 5

5.5.3

Steps 1,2 3 and 4:

H2 O 100 % mol %

W P

mol frac F (mol) = ? CH 4 x N2 y A

1.00 O2 N2

CO

1.0

C O2

8.7

O2

3.8

N2

86.5

air

100.0

mol frac 0.21 0.79 1.00

Step 5: Basis: 100 mol P Step 6: Unknowns:

x, y, A, W, F Total 5

Step 7: Element Balances: Equations:

C, H, N, O

a)

If x and y are moles, not mole fractions, add: x + y = F

b)

If x, y are mole fr., add x + y = 1.00

Steps 8 and 9: Use x and y as mol: C balance:

x = 8.7 + 1.0 = 9.7

2N balance:

y + .79A = 86.5

2H balance:

2x = W = 2(9.7) = 19.4

2O balance:

0.21A = (1/2) (W + 8.7) + 1.0 + 3.8

Use 2O bal:

A=

A and P will be mol

9.7 + 8.7 + 0.5 + 3.8 = 108 0.21 5–47

Total 5

Solutions Chapter 5

Use 2N bal:

y = 1.10

required O2: CH4 +2O2→ CO2+2H2O

(1)

(9.7)(2) = 19.4 mol total O2 in: (0.21) (108) = 22.7 % xs air = % xs O2 =

3.3 (100) 19.4

= 17%

(a) (b)

excess O2= 22.7 - 19.4 = 3.3 mol

Percentage of CH4, N2: CH4 =

10.2%

9.7 (100) = 89.8% 9.7 + 1.1

N2 =

Alternate solution: Use extent of reaction and species balances. Another equation needed for CO: (2) CH 4 + 1.5O2 → CO + 2H 2 O Species balances to get ξ1 and ξ 2 : F CH 4 : O = n CH + ( −1)+ξ1 − ( 1)ξ 2 4

CO2 : 8.7 = 0 + (1)ξ1 CO: 1.0 = 0 + (1)ξ 2

ξ1 = 8.7 reacting moles ξ 2 = 1.0 reacting moles

F Thus n CH = 8.7 + 1.0 = 9.7 mol and other compounds follow. 4

5–48

1.1 (100)= 9.7+1.1

Solutions Chapter 5

5.5.4

Steps 1, 2, 3 and 4: H2 O

100% W

CH4 100%

P

F

A

CO2 20.2 O2 4.1 N2 75.7 100.0

enriched O2 40 N2 60 air 100

Step 5:

Basis 100 mol P

Step 6:

Unknowns: F, A, W (all compositions known)

Step 7:

Element balances: C, H, O, N (1 extra balance)

Steps 8 and 9: C balance: 20.2 = F 2N balance: 75.7 = 0.6A, A = 126.2 2H balance: 2F = W, W = 40.4 2O balance: 0.4A = 1/2 W + 20.2 + 4.1, 50.5 ≠ 20.2 + 20.2 + 4.1 = 44.5 Analysis isnot correct.

5–49

Solutions Chapter 5

5.5.5

Steps 2, 3, and 4:

H2 O 100% W F

mass frac C 0.90 H 0.06 inert 0.04

P

Furnace A

1.00 O2 N2

mole %

R

mol frac 0.21 0.79

mass frac 0.10 C inert 0.90

1.00 Step 5: Basis: 100 kg mol P

13.9

C 13.9

CO

0.8

0.8

O2

4.3

N2

81.0

CO2

100.0

14.7

1.00

Steps 6 and 7: Unknowns: F, R, A, W

4 Probably ok but check for redundancy

Balances:

C, H, O, inert, N2

5

Steps 8 and 9: Balances (in moles):

C: H:

F(0.90) R(0.10) = P(0.139+0.008) + + W(0) 12 12 F (0.06) = P (0) + R(0) + W (2) 1.005

2N:

A (0.79) = P(0.81) + R (0) + W (0)

2O:

A (0.21) = P (0.139 +

0.008 W + 0.043) + 2 2

Inert (mass): F (0.04) = R (0.90) 5–50

Solutions Chapter 5

From 2O,

100(.81) = 102.53kgmol 0.79 W = 5.863 kg mol

From H,

F = 196.98 kg

From inert,

R = 8.7548 kg

A=

From 2N,

196.98(0.06) = 5.94 2 5.94 O2 in: = 2.96 2 196.98(0.90) C in = = 14.77 12

H in:

Step 10:

Check via C: 196.98 (0.90) /12 = 100 (0.147) + (8.7548 (0.10)/12)

ok

Required O2: 14.7 + (5.863/2) = 17.63 kg mol Total O2in = 102.53 (0.21) = 21.53 Excess (21.53 - 17.63)/17.63 = 0.221 or 22% Alternate solution using extent of reaction: C+O2 → CO2 1

C+ O2 → CO

(1) (2)

2

1

2H+ O 2 → H 2 O

(3)

2

CO 2 :

P F n CO =13.9 = n CO +(1) ξ1 2 2

ξ1 = 13.9

CO:

P F nCO =0.8 = nCO +(1) ξ2

ξ 2 = 0.8

N2 :

n PN2 = 81.0 = n AN2

A = 102.5 mol

A O2

n = 0.21(102.5) = 21.5 mol

O2:

1

1

2

2

4.3 = 21.5 + (-1) ( ξ1 ) + (- )( ξ 2 ) + (- ) ( ξ 3 )

C:

0 = nCF + (-1)(13.9) + (-1)(0.8)

H:

0 = nFH + (-2) (2.90)

nCF = 14.7 mol

nHF = 5.80 mol (some round off)

5–51

ξ 3 = 2.90

Solutions Chapter 5

5.5.6

Basis: 100 mol F P

F = 100 mol Moles = % CH4 : 60 C 2 H6 : 20 CO: 5 O2 : 5 N 2:

A 0.21 O2 0.79 N 1.00 2

10 100

50% xs air: CH4 + 2O2

n CO2

Moles 281.25 1058

or

x CO2

n H 2O

x H 2O

n

O2

xO2

nN2

x n2

∑ =P

∑ =1.00

Calculate O2 and N2 in with the air in reqd O2 (mols) 60 (2) = 120 → CO2 + 2H2 O

C 2 H6 + 312 O2 → 2CO2 + 3H 2O

20(3.5) = 70 2.5 ⎛1 5⎝ ⎞⎠ = 192.5 2

1

CO + O 2 → CO 2 2

Less O2 in gas

5.00 Net required O2 xs O2: 187.5 (.5) = Total N2 in with O2

187.5 93.75 281.25

281.25 0.79 = 1058 0.21

Material balances (elemental): in (F)

+

in (A)

=

out (P)

C:

60 + 2 (20) +5 +

0

=

n CO2

H:

60 (4) + 20 (6) +

0

=

n H 2 O (2)

5 +5 2

+

281.25 =

O as O2:

5–52

n CO2 +

n H 2O 2

+ nO 2

Solutions Chapter 5

N as N2:

10

+

n O 2 = 288.75 - 105 -

nN2

=

180 = 93.75 2

moles

composition of stack in % moles

n CO2 = 105

7.26 12.44 73.82 6.48

n H 2 O = 180 n N 2 = 1068 n O 2 = 93.75 Total moles = out

1058

1446.75

100.00

5.5.7

Steps 1, 2, 3, 4: F = 100 (basis) mol = % CO 13.54 CO2 15.22 H2 15.01 CH4 3.20 N2 53.03 100.00

P CO2 O2 N2 H2O

O2 reqd 6.77 -7.505 6.40 --20.675

A

mol fr.

mol fr.

x y z 1-x-y-z 1.00

0.143 0.037 0.724 0.096 1.00

0.21 O2 0.79 N2 1.00

Step 5: Basis: 100 mol F Step 6: Unknowns: P, x, y, z Step 7: Balances

C, H, O, N

ok

Steps 8 and 9: Air in is based on complete combustion CO +

1 2

O2 → CO2 and CH 4 + 2O2 → CO2 +2H 2 O

5–53

mol calculated 31.96 8.27 161.92 21.41

Solutions Chapter 5

xs O2 = 0.40 (20.675) = 8.27 reqd O2 = 20.675 total O2 28.945 N2 in with air

28.945

28.945 ⎛ .79 ⎞ ⎜ ⎟ = 108.89 mol ⎝ .21 ⎠

Air = 137.84 mol

Balance 2N C 2H

N2 out = 108.89 + 53.03 = 161.92 mol CO2 out = 13.54 + 15.22 + 3.20 = 31.96 mol H2O out = 15.01 + 2(3.20) = 21.41 mol

2O

O2 out = 28.945 + (as O2)

13.54 21.41 + 15.22 − 31.96 − = 8.27 mol 2 2

Alternate solution: Use the extent of reaction and species balances

5.5.8

Steps 2, 3, and 4: 7.9%N2 21%O2

Step 5: Step 4:

A

F = 100 mol

P

CS2 40.0 SO2 10.0 H2O 50.0

xCO 2 xSO 2 xO 2 xN

W 100% H2O

Basis: 100 mols feed; assume complete combustion Equations: CS2 + 3O2 → CO2 + 2SO2

O2 required: % excess:

40 mol CS2 3 mol O2 = 120.0 mol O2 1 1 mol CS2 O 2 excess × 100 =% excess O 2 required

Steps 6, 7, 8 and 9: 5 unknowns and 5 equations → unique solution 5–54

2

Solutions Chapter 5

Element balances Component C

Mol in 40.0

Mol out Px CO 2

2S

1

2O

10.0 + (50.0) + .21A

2

1

(10.0) + 40.0 1

2N 2H

.79A 50.0

(0.02)P

2 1

(x C O2 + 0.02 + x O2 )P+ W

2

2

(1-x O2 − 0.02 − x CO2 )P W

2(45.0) = 4500 .02 x CO2 = 0.0089

P=

% Excess =

0.1829(4500 molO2 ) (100) = 686% 120 mol

Alternate solution: Use extent of reaction and species balances.

5.5.9

Steps 2, 3 and 4: W

Fuel F mass fr. C 0.74 H 0.14 ash 0.12 1.00

Flue gas

A air O2 0.21 N2 0.79 1.00

Step 5:

H2O 1.00

R ash 1.00

Basis: P = 100 mol

Step 6: Unknowns:

F,A,R,W ⎫⎪ a discrepancy unless one ⎬ Step 7: Balances: C,H,O,N,ash ⎪⎭ balance is not independent

Steps 8 and 9:

all balances in moles

5–55

P CO2 CO O2 N2

mol 12.4 1.2 5.7 80.7 100.

Solutions Chapter 5

C balance:

F(0.74) = 12.4 + 1.2 = 13.6 12 F=

(13.6)(12) = 220.5 lb 0.74

2N balance:

A(0.79) = 80.7

A = 102.2 mol

2H balance:

F(0.14) = W; 2

W = 15.44 mol

2O balance:

A (.21) = W

(1.00) 1.2 + 12.4 + + 5.7 2 2

21.46 = 26.42

not equal hence

Some discrepancy exists

A total balance can be used, but must be in mass, and the mass of P calculated. 5.5.10

Basis: 100 moles exhaust gas Comp CO2 O2 N2 Total

exhaust gas mol 16.2

tie element

exit gas % 13.1

4.8 ⎫ ⎬ 79.0⎭ 83.8 100.0 16.2 mol CO2 100.0 mol exit gas = 123.6mol exit gas 100.0 mol exhaust gas 13.1mol CO2

exhaust gas = 100.0 mol air leaked in = 23.6 mol

23.6 = 0.236 mol air / mol exhaust gas 100.0

5–56

mol 16. 2

Solutions Chapter 5

5.5.11

Steps 2, 3, and 4: H2 O 100% W mol

Sludge S mass

mass frac C S H2

O2

0.40 0.32 0.04 0.24

mol

1.00

O2 N2

mole % P mol

Furnace A

F mass

mol frac

Fuel oil lb

0.21 0.79

mC mH

1.00

Step 5: Basis: 100 mol P

CO

2.02

CO2

10.14

O2

4.65

N2

81.67

SO 2

1.52 100.00

∑ F (lb)

Steps 6 and 7: Unknowns:

S, W, F (or mH),A, mC

5 Element Balances: S, C, H, O, N

Element Balances (moles) S: 2N:

C: 2O:

0.32S = 32

1.52

A(.79) = 81.67

S(0.40) m C + =10.14+2.02 12 12

S = 152 lb A= 103.38 lb mol (2998.01 lb)

mC = 85.12 lb

W 2.02 0.24(152) + A (0.21) = + 1.52 + 10.14 + 4.65 + 2 2 32 W = 11.04 lb mol (198.71 lb)

5–57

Solutions Chapter 5

0.04(152) m H + =W 2.03 2.03

2H:

mH = 16.32 lb

Check via a total mass balance 152 + 16.32 + 85.12 + 2998.01 3251.45

? = ? =

198.71 + 3035.56 3234.27 close enough

lb 85.12 16.32 101.44

C H

152 lb S = =1.50 F 101.44 lb

5.5.12

mass fr 0.84 0.16 1.00

B

Basis: 1 min

[CaO]

= theor.

ft 3

1000 gal min

1.05×62.4 lb soln. 0.02lb H 2SO4 mol H 2SO4

7.48 gal

56.1 lb CaO mol CaO

ft

3

lb soln.

mol CaO

98 lb H 2SO4 mol H 2SO4

=100.3 lb CaO/min

Feed rate of CaO = 1.2 [CaO ]theor . = 120.4 lb CaO/min CaSO4 production rate =

×

100.3 lb CaO min

60 min 24 hr 365 d hr

d

yr

mol CaO

mol CaSO4 136.1 lb CaSO4

56.1 lb CaO mol CaO

= 63,947 ton/yr

5–58

mol CaSO4

ton 2000 lb

Solutions Chapter 5

5.5.13

Basis: 100 mol of product gas

NH3 + 2O2 → HNO3 + H2O The product gas from such a reactor has the following composition (water free basis): 0.8% NH3 9.5% HNO3 3.8% O2 85.9% N2 Determine the percent conversion of NH3 and the percent excess air used. in n out NH3 = n NH3 +

NH3

(9.5) υ

0.8 = n inNH3 + ( 1)(9.5) − F = ninNH3 = 0.8 + 9.5 = 10.3 mol n out − n in =ξ υ For HNO3 :

9.5 − 0 = 9.5 = 1

ξ

Percent conversion f = (100%) ξ /F = 92.2% Calculate the entering oxygen using a N2 balance: N2: 0.79A = 85.9; therefore, A = 108.73 mol Calculate the theoretical O2 needed for the reaction (O2)theor = 2 × 10.3 = 20.6 mol Feed rate of O2 = 0.21A = 22.83 mol % Excess O2 = [(22.83-20.6)/20.6](100) = 10.84%

5–59

Solutions Chapter 5

5.5.14

Basis: 100 mol of product gas (20.5%, C2H4O; 72.7 N2; 2.3 O2; and 4.5%CO2

C2 H 4 +

1 O2 → C2 H 4O 2

C2H4 + 3O2 → 2CO2 + 2H2O Calculate ξ1 :

n iout − n ini ξ1 = υi

Use C2H4O:

Calculate ξ2 :

Use CO2:

ξ2 =

ξ1 =

20.5 − 0 = 20.5 1

4.5 − 0 = 2.25 2

Calculate the entering air using a N2 balance: N2: 0.79A = 72.7; therefore, A = 92.03 mol Calculate the moles of C2H4 entering (F) in n out CH4 − n CH4 = υ1ξ1 + υ2 ξ2

0 − F = −ξ1 − ξ2 hence F = 22.75 O2 feed rate = 0.21A = 19.33 mol (O2)theor = F/2 = 11.375 mol % Excess O2 = [(19.33-11.375)/11.375](100) = 70.0% To get the ethylene feed:

22.75 C2 H4 100,000 ton C2 H4O 2000 lb C2 H4O mol C2 H 4O 20.5 mol C2 H4O yr ton C2 H4O 44 lb C2 H 4O ×

yr d 28 lb C2 H 4 = 16,123 lb C2 H 4 /h 365 d 24 h mol C2 H 4

5–60

Solutions Chapter 5

5.5.15

mole fr

mol fr

CH4

0.70

C3 H8

0.05

CO

0.15

O2

0.05

N2

0.05

Air

1.00

mol frs 0.21 0.79

Fuel

Flue gas

Flare

F mol

0.0773

H2 O

0.1235

O2

P mol

O2 N2

CO2

N2

A mol

xO2 xN2 1.00

1.00 Unknowns A, F, P, x O 2 , x N 2 , Pick F = 100 as basis

4

Balances: C, H, O, N (is one redundant ?) plus 0.0773+0.1235+ x O 2 + x N 2 = 1 Balances: C: 100 (0.70) + 100 (0.05) (3) + 100 (0.15) N2:

100 (0.05) + 0.79A

H2:

100 (0.70) (2) + 100 (0.05) 4

O2:

100 (0.15)

1 2

= P (0.0773) = P xN2

= P (0.1235) (redundant with C)

+ 100 (.05) + A (0.21)

Σxi Solve C (or H2) balance for P; Solve N2, O2 and Σxi = 1

5

0.7992 P = 1294

0.1235 ⎡ ⎤ = P ⎢(0.0773) + + x O2 ⎥ 2 ⎣ ⎦ = xO2 + xN2

simultaneously for A, x O 2 , x N 2

A = 1203 moles

x O 2 = 0.0648 and x N 2 = 0.7344 (not needed)

Calculation of the required O2

Required O2

C H4 + 2O 2 → CO2 + 2H2 O

70(2)

5–61

= 140

Solutions Chapter 5

C3 H8 + 5O2 → CO 2 + 4H 2 O CO+ 12 O2

5(5)

→CO2

15(

1 2

O2 present in F A(0.21) = less

= 25 ) = 7.5 -5.0 167.5 reqd

252.54 O2 in

167.5 reqd O2 85.04 xs O2 85.04 (100) = 51 % 167.5

5.5.16 F = 100 kg mol

CO2 N2

C 1.00 O2 0.21 N2 0.79

C + O2 → CO2 Basis: 100 kg mol coke Step 5: a.

O2 in = 100 kg mol

100 kg mol C 1 mol O 2 79 mol N 2 = 376 kg mol N 2 1 mol C 21 mol O2 Steps 6 and 7: Unknowns: moles (ni) CO2, O2, and N2 in P, and P Balances: elements C, O, N Equations:

P i

∑n

=P

or add ξ and have C, CO2, N2, O2 species balances

5–62

in

Solutions Chapter 5

Steps 8 and 9: Element balances C: 2O: 2N:

in 100 100 376

out n CO n CO +n O nN

= = =

2

2

n O2 = 0

2

2

Component in P N2 O2 CO2

kg mol 376 0 100 476

mol % 79 0 21 100

With 50% xs air, xsO2, is 50 mol The total O2 in is 150 kg mol The N2 in is 150 0.79 = 564 0.21

The N and C balance are the same but the 2O balance is In 150

out −

n O2

reacts =

100

Component in P N2 O2 CO2

n O2 = 50

kg mol 564 50 100 714

mol % 79 7 14 100

c. Same 150 mol O2 enter and 564 mol N2 exit but O2 is different in Oxygen balance:

150

mol CO2 used –

90

5–63

–

Used CO 5 = 55 kg mol O2 lb fg

Solutions Chapter 5

Component CO2 CO O2 N2 Total

kg mol 90

mol% 12.51

10

1.39

55 564 719

7.65 78.45 100.00

5.5.17

Steps 1, 2, 3 and 4: B CH3CHO 1.00 P = 100 mol (Basis) F CH3CH2OH 1.00

CO2 O2 CO H2 CH4 N2

A mol. wt. CH3CHO 44 CH3CH2OH 46

W H2O 1.00

Air .21 O2 .79 N2 1.00

Step 5: Basis: 100 mol P Step 6: Unknowns: F, A, B, W Step 7: Balances:

C, O, H, N

ok

Steps 8 and 9: Via algebra 2N balance: A(.79) = 85.2 A = 107.85 mol C balance: F(2) = B(2) + 5.6 2H balance: F(3) = B(2) + W(1) + 12.3 2O balance:

1

1

1

2

2

2

F( ) + (.21)A = B ( ) + W ( ) + 3.95

5–64

% 0.7 2.1 2.3 7.1 2.6 85.2 100.0

C 0.7 2.3

O2 0.7 2.1 2.3/2

7.1 5.2

2.6 5.6

H2

3.95

12.3

Solutions Chapter 5

Solution:

B = 44.1 mol F = 46.9 mol W = 40.2 mol

44.1 mol acet. 44 kg acet. 100 mol P 1 kg mol EtOH = 0.899 100 mol P 1 kg mol acet. 46.9 mol EtOH 46 kg

kg acetaldehyde kg ethanol

Alternate solution: Use extent of reaction and species balances, but it is more complicated.

5.5.18

Basis: F = 100 lb given in Example 10.9

This basis gives P = 50 lb moles. Note: If the problem calculations were redone because of the stated additional reactions, the value of P = 50 for two significant figures would not change. Similarly, the other values of the moles in P would not change. Thus, the SO2 + CO2 is 0.154 (50) = 7.7 lb mol. The values of the pertinent components (for NOx use NO1.5) for this problem are lb mol MW kg ELU/kg ELU NOx: (0.0024) (80.6) = 0.193 2.2 4.25 0.22 0.93 CO:

(0.0018) (7.7)

=

0.090

28

2.52

0.27

0.68

SO2:

(0.014) (7.7)

=

0.002

64

0.128

0.09

0.012

CO2:

(0.986) (7.7)

=

7.59

48

364

0.10

36.4

Total the total lb mol of nitrogen (N) entering is 2(50) (0.806) + 0.001 = 80.6

5–65

38.0

Solutions Chapter 5

5.5.19 Assume ammonia and glucose (MW = 180.1) are fed in stoichiometric proportions. The reaction equation has to be set up and balanced to use the value given of 60% (mole assumed) conversion of glucose.

a C6 H12O6 + b NH3 → c CH1.8O0.5 N0.2 + d C2 H6O + e CO2 + f H2O Pick a basis of 4.6 kg of EtOH (C2H6O, MW = 46.07)

4600 g EtOH 1 g mol EtOH = 100 g mol EtOH produced 46.07 g EtOH To balance the chemical reaction equation use element balances. Take a basis of a = 1. Assume 60% conversion of the glucose means that 1 mole of glucose produces 0.6 mol of ethanol, not that one mole of glucose produces 3 mol of ethanol. C: H: O: N:

6 = c + 2d + e 12 + 3b = 1.8c + 6d + 2f 6 = 0.5c + d + 2e + f b = 0.2c

Specifications 0.60 = d Results

a = 1 (the basis) b = 0.8 c=4

d = 0.6 e = 0.8 f = 1.8

167 g mol C6 H12O .08 g mol NH 3 = 133.6 g mol NH 3 1 g mol C6 H12O 167 g mol C6 H12O6 180.1 g mol C6 H12O6 = 30,076 g or 30.1 kg C6 H12O6 1 g mol C6 H12O6 133.6 g mol NH 3 17.03 g NH 3 = 2,275 g or 2.28 kg NH 3 1 g mol NH 3

5–66

Solutions Chapter 6

6.1.1

6.1.2

2 (The two overall balances on A and B sum up to the overall balance on the system)

4 (such as 2 component balances for each unit)

6.1.3

Unit I involves A, B, C Unit II involves D, E Unit III involves A, B, C, D

Components 3 2 4

Number of independent balances

9

If A and B are always combined in the same ratio, then you have to reduce the independent balances by 1 for Unit I and 1 for Unit III, a total of 2. 6.1.4

Six independent eqautions if all compositions are known. The sum of the components (in moles) would equal the total flow, hence not all of the equations that could be written would be independent, only 2 per subsystem. Total balances: Condenser F2 = F 4 + F5

(F3 does not mix, hence cancels out or can be omitted)

Column F1 + F5 + F9 = F 2 + F8 Reboiler F8 =F9 + F7 (F6 does not mix, hence cancels out or can be omitted) Component balances: In addition, (2) component balances could be written for each component for each subsystem cited above. Multiply Fi by the composition xij.

6–1

Solutions Chapter 6

6.1.5

Unknowns:

stream flows including F: Mass fractions (7 plus 3 in F) 10

Independent material balances: Species in 1 : 3 2:2 3:3

6 16

8

Other independent equations: ∑ ω i = 1 in each stream including F 8

6.1.6

Basis: F = 100 kg mass fr. A 0.60 B 0.20 C 0.20 1.00

m = mass fraction mass fr. A 0.10 B 0.85 P2 C 0.05 1.00

P1

(1) % = kg A 15 B 30 C 55 100

(2)

F = 100 kg

(a) SEPARATE ANALYSIS

16

W2=20 kg W1 mass fr. A m1A B m1B C m1C 1.00

Column (1) balances ⎤ = W1 (m1A ) + P1 (0.60) ⎤ ⎥ any 3 ⎥ 1 (2) B: 0.30(100) = W1 (m B ) + P1 (0.20) ⎥ ⎥ 4 indept. eqns. are indept.⎥ ⎥ 1 (3) C: 0.55(100) = W1 (m C ) + P1 (0.20) ⎥ eqns. ⎥ ⎥ ⎥⎦ 100 = W1 + P1 ⎥ (4) m1A + m1B + m1C = 1 ⎥ ⎦ (1) A: 015(100)

Unknowns: W1 ,P1 ,m1A ,m1B ,m1C 5 – 4 = 1 degree of freedom

5 unknowns

6–2

A 0.002 B m2B C m2C 1.00

Solutions Chapter 6

Column (2) balances (5) A: m1A W1

⎤ = 0.10 P2 + 0.002 (20) ⎤ ⎥ any 3 ⎥ 2 (6) B: m W1 = 0.85P2 + m B (20) ⎥ are indept.⎥ 5 indept. eqns. ⎥ ⎥ 1 2 (7) C: m C W1 = 0.05P2 + m C (20) ⎥ eqns. ⎥ ⎥ ⎥⎦ W1 = P2 + 20 ⎥ m1A + m1B + m1C = 1 ⎥ ⎦ 2 2 0.002 + m B + mC = 1 1 B

Unknowns: W1 , P2 , m1A , m1B , m1C , m 2B , m C2 > unknowns 7 – 5 = 2 degrees of freedom JOINT ANALYSIS Adding the 2 units to make a joint system No. of Unknowns • From the separate calculations start with: • Delete the 4 common variables W1 , m1A , m1B , m1C that are treated as unknowns in both subsystems leaving: • Delete the equation m1A + m1B + m1C that will no longer be independent in the joint system (it appears in both subsystems) once leaving:

No. of Degrees indept. Eqns. of freedom

12

9

3

8

9

-1

8

8

0

ANALYSIS OF OVERALL SYSTEM System overall balances

(1) A: 100 (0.15)

= P1 (0.60) + P2 (0.10) + 20 (0.002) ⎤ ⎥ (2) B: 100 (0.30) = P1 (0.20) + P2 (0.85) + 20 (m 2B ) ⎥ 4 indept. (3) C: 100 (0.55) = P1 (0.20) + P2 (0.05) + 20 (mC2 ) ⎥ ⎥ eqns. ⎥ ⎥ 0.002 + m2B + mC2 = 1 ⎥⎦ 2 2 Unknowns: 4 unknowns P1 , P2 , m B , m C 4 - 4 = 0 degrees of freedom as expected

6–3

Solutions Chapter 6

6.2.1

The results show that the system of equations is very sensitive to small perturbations in the coefficients (measurements).

6–4

Solutions Chapter 6

6.2.2

Basis: 1 hr (1000 kg)

a. Overall balances: Total: 1000 = W + D Salt: 1000 (0.0345) = 0 + 0.069D

W= 500 kg

D = 500 kg

b. Salt balance on the freezer: 1000 (0.0345) = 0 + 0.048B

B = 718.75 kg

Total balance on the freezer: 1000 = A + B

A = 281.25 kg

Total balance around filter: B = 718.75 = C + D

6.2.3

C =218.75 kg

Basis: 220g IgG from the reactor Fraction recovered =

140 = 0.64 220

6–5

Solutions Chapter 6

6.2.4

Two subsystems exist, hence 4 component balances can be written. No reaction occurs and the process is assumed to be in the steady state. Steps are omitted here. The balances are System : Splitter

System : Stock Chest

Total: R = E + P Fiber : 2.34 = xE+xP Water: 7452 = 4161+3291

Total: P +N = L Fiber: xP+xN =103.26 Water: 3291 + 18 = 3309

(a)

Also N(0.15) = 18 N = 120 0.85 (N) =xN = 0.85 (120)

Overall balances Total: R+ N = E + L Fiber: 2.34 + xN = xE + 103.26 Water: 7452 + 18 = 4161 + 3309

(b)

(c)

Substitute (b) into (c), solve (c) for xE , and then solve (a) for xP. x N =102

x P =1.26

x E =1.08

all in kg

6.2.5

Step 5:

Basis: 100 kg mol = F

Steps 1, 2, 3, and 4:

H2 O 100% W

F (kg mol)

E (kg mol)

P (kg mol)

Furnace CH4 H2 C2 H 6

mol frac 0.70 0.20 0.10 1.00

Duct mole frac.

mol

-

n ECO

CO 2 A (kg mol) O2 N2

mol frac 0.21 0.79

O2 H2 O N2

0.02

n

mole frac. Air ?

2

E O2

-

n HE O

-

nN

1.00

E

B (kg mol)

2

E

1.00

2

O2 N2

mol frac 0.21 0.79

CO 2

-

mol n PCO

O2

0.06

n

N2

-

nPN

1.00

P

2

P O2

2

1.00

Assume an air leak occurs first and use material balances to calculate the amount. This is a steady state flow process without reaction. 6–6

Solutions Chapter 6

Overall system Step 6:

Unknowns are:

A, B, W, P, x N 2 (or n N2 ), x CO2 (or n CO2 )

Step 7:

Balances are:

C,H, O, N, Σ x Pi = 1 (o r ∑ n Pi = P)

Steps 8 and 9: In C(kg mol):

Out

0.70(100) + 0.10 (100) 2

=

n CO2

=

W (2)

n CO2 = 90 kg mol H (kg mol):

0.70 (100)(4) + 0.20 (100)(2) + 0.10 (100)(6) W

=

190 kg mol

190(1.00) + 90 + 0.06P 2

2O (kg mol): 0.21A + 0.21B

=

2N (kg mol): 0.79A + 0.79B

=

n PN 2

=

P

90 + 0.06 P + n PN 2

A balance about the furnace or about the duct is needed; or these two alone would have been sufficient, omitting the overall balance. System: Furnace (or Duct) E

Step 6: Unknowns:

A, n ECO2 ,n EO 2 , nH 2 O , n EN2

Step 7: Balances:

C, H, O, N2, ∑ nEi = E

Steps 8 and 9: C (kg mol):

H (kg mol):

0.70 (100) + 0.10 (100)(2) n ECO2 = 90 kg mol

=n ECO2

E

0.70 (100)(4) + 0.20 (100)(2) + 0.10 (100)(6) = n H 2 O E n H 2 O = 190 190 2O (kg mol): 0.21A = n EO 2 + 90 + 2 6–7

Solutions Chapter 6

E = nN2

2N (kg mol): 0.79A

0.02

= n EO 2 / E

90 + 190 + n EO 2 + n EN 2 = E

Solution:

kg mol in E n ECO2

kg mol

90

A

982

n H 2O

E

190

E

1077

n EO 2

21.5

n EN 2

776 1077.5

From the overall N2 balance: O2 balance:

0.79 (982) + 0.79B = n PN 2

(1)

0.21 (982) + 0.21B = 185 + 0.06P

(2)

(1)

22.2 + 0.21B = 0.06P

B = 207 kg mol

(3)

90 + (0.79)(982) + 0.79B = 0.94P

P = 1095 kg mol

It is an air leak. 207 kgmol / 100 kgmol F

6–8

Solutions Chapter 6

6.2.6

Steps: 1,2,3, and 4: This is a steady state process with no reaction taking place. All the compositions are known except for stream C. We can pick the overall system first to get D, and then make balances on the first (or second) units to get C and its composition. Step 5:

Basis:

F = 290 kg (equivalent to 1 second)

Step 6:

Unknowns:

A, B, C, D and E and the composition of C

Step 7:

Balances:

NaCl, HCl, H2SO4, H2O, inert solid

Step 8:

For the overall system there are 4 unknowns (C is excluded) and 5 species balances:

Overall balances (kg)

In

Out

NaCl: HCl: H2SO4: H2O:

A(0.040) A(0.050) A(0.040) A(0.870)+B(0.910)

= = = =

290(0.0138) 290(0.0255)+D(0.020)+E(0.015) 290(0.0221)+D(0.020)+E(0.015) 290(0.9232)+D(0.960)+E(0.970)

Inerts: Totals:

B(0.09) A+B

= =

290(0.0155) D + E +290

Steps 6 and 7: Two of the equations are not independent: HC1 and H2SO4. HC1: D(0.020) + E (0.015) = 7.40 – 5.00 = 2.40 H2SO4: D(0.020) + E (0.015) = 6.41 – 4.00 = 2.41 Thus, overall the degrees of freedom are 0. Steps 8 and 9: The solution of the equations is (in kg/s)

A = 100 ,

B = 50 ,

D = 60 and E = 80

from which C = 150 via a total balance about the first unit.

6–9

Solutions Chapter 6

6.2.7

Notation: Subscripts Benzene = bz, Xylene = xy, Solids = s W = mass fraction, Fi = liquid stream, Si = solids stream 0

ωS bz = 1.0 0

ωS xy = 0

2

1

ωS xy

1

ωS xy

0

2

2

ωS bz = (1-ωSxy)

1

1

ωS bz = (1-ωSxy)

2

ωS s = 0 1.0

ωS s = 0 1.0

ωS s = 0 1.0

S0

S1

S2

1

2 F1

F2 2

2

1

1

ωF bz = (1-ωFxy)

F0

ωF bz = (1-ωFxy)

2

1

ωF xy

ωF xy

2

ωF s = 0.9 1.0

0

ωF bz = 0 0

ωF xy=0.1

1

ωF s = 0.9 1.0

0

ωF s = 0.9 1.0

Step 5: Basis is 1 hr (F0 = 2000 kg and S0 = 1000 kg) Steps 6 and 7: 2

1

1

Unknowns:

Unit 1: ωFxy , ωSxy , ωFxy , S1 , F1 , F2 ⎫⎪ ⎬ Net = 8 S1 F1 S2 1 1 2 Unit 2: ωxy , xy ω, xy ω, S , F , S ⎪⎭

Material Balances:

Unit 1: bz, xy, s Unit 2: bz, xy, s

Additional Equations: ωFxy = ωSxy and ωFxy = Sxy ω (the related equations for benzene are redundant). Total equations = 8. 2

1

1

2

The solution is simplified if two balances are made first (not an essential step): All material balances are in kg. Solids balance on Unit 2 and also Unit 1: 2000 (0.9) = F1 (0.9) 2000 (0.9) = F2 (0.9)

F1 = 2000 F2 = 2000

6–10

Solutions Chapter 6 Total balance on Unit 1 and also Unit 2: 2000 + S1 = 2000 + S2 1000 + 2000 = 2000 + S1

S1 = S 2 S1 = 1000 S2 = 1000

The other balances are Unit 2: 1

1

2

Benzene:

1000(1 − ωSxy ) = 2000(1 − ω Fxy ) + 1000(1 − ωSxy )

Xylene:

2000(0.1) + 1000ωSxy = 2000 Fxy ω + 1000 Sxy ω

1

1

1

2

2

10ωFxy = Sxy ω Unit 1: Benzene:

1.0(1000) + 2000(1 − ωF ) = 2000(1 − ωFxy ) + 1000(1 −ωSxy )

Xylene:

0 + 2000ωFxy = 2000ωFxy + 1000(ωSxy

1

1

2

2

2

1

1

)

1

10ωFxy = Sxy ω Solve the equations to get the compositions of the streams:

b)

Stream

Component wt fr

Stream

Component wt fr

S0

Bz 1.0 Xy 0.0

F0

Bz 0.9 Xy 0.1

S1

Bz 0.97 Xy 0.03

F1

Bz 0.082 Xy 0.018

S2

Bz 0.82 Xy 0.18

F2

Bz 0.097 Xy 0.003

Xylene in Feed

= 2000 × 0.1 = 200 kg

Xylene in Product

= 1000 × 0.18 = 180 kg

% Recovery =

180 ×100 = 90% 200

6–11

Solutions Chapter 6

6.2.8

Step 5: Basis: F = 100 kg and D3 = 10 kg Steps 2, 3, and 4: A

0.50

B D1 C

0.23

D2

0.27

A

0.17

B

0.10

C

0.73

D 3

1.00

1.00

D3

A

mA

B

mB

C

0

D3

10 kg

F = 100 kg P1

1 A

0.50

B

0.20

C

0.30

P2

2

3

P3

A

A B C

B 0

C

A

0.70

B

0.30

C

0

1.00 1.00

1.00 E

A B C

0 1.00

Additional information:

P3 = 3D3 = 30 kg P2 = D 2

m PA2 = 4m PB2 Begin with overall balances. Steps 6 and 7:

Unknowns 4: D1 , D 2 , m AD3 , m BD3 Balances (3) : A, B, C (plus total) Implicit equation (1): m DA3 + m DB3 = 10 kg

⎫ ⎪ ⎬ Degrees of freedom = 0 ⎪ ⎭

Steps 8 and 9: Overall balances Total :

100

= D1 + D2 + 10 + 30

6–12

or 60 =D1 + D2

Solutions Chapter 6 C:

100 (.30) = 0.27D1 + 0.73 D2 +0

Solve these two equations to get D1 = 30 and D2 = 30 Get m DA 3 and m DB3 from A balance and m DA 3 + m DB 3 = 1 A:

100(.50) = 30(.50) + 30(.17) + 10 m DA 3 + 30 (.70) so, m DA 3 = 0.89

m DA 3 + m DB 3 = 1 so

D

m B 3 = 0.11

Step10: Note - Can check using the B balance (not independent) Balances on Unit No. 3 P

A

m A2

B

m PB2

C

0 1.00

A

m

E A

B

m

E B

C

0

D 3

A

0.89

B

0.11

C

0 1.00

P2 = 30 E

1.00

P3

3

A

0.70

B

0.30

C

0 1.00

(D3 = 10; P3 = 30)

Steps 6 and 7: Unknowns (5): E, mPA2 , mPB2 , mAE , mBE Balances (2): A, B Implicit equations (2): mPA2 + mPB2 = 30, mAE + mBE = E Other equations: mPA2 = 4mBP2 Steps 8 and 9: Total:30 + E = D3 + P3 = 10 + 30 = 40

so

One component balance:

6–13

E = 10 kg

Solutions Chapter 6 A: 30 m PA2 + 10 m EA = 10 (.89 ) + 30 (. 70) = 29. 9⎫ ⎬ not independent equations P E B: 30 m B2 + 10 m B = 10 (.11) + 30 (. 30) = 10.1 ⎭

∑ m i 2 = m A2 + m B2 =1⎫ m PA2 = 0.80 P ⎬ m B2 = 0.20 m PA2 = 4m PB2 ⎭ 30 (.80) + 10 m EA = 29.9 so that m EA = 0.59 P

P

P

E

E

so that or 59%

E

m A + m B =1

or 41%

m B = 0.41 _____ 1.00

6.2.9

Steps 1, 2, 3 and 4: mol % 64.29H2 14.29SiCl 4 21.42 H2 SiCl 2

HCl (g) A 100% Si B

Step 5: Basis:

C

Reactor

D Separator

H 2 ,SiCl 4 ,HSiCl 3 H 2 SiCl 2

100% E

HSiCl 3

100 kg B

100 kg Si 1kg mol Si = 3.560 kg mol Si 28.09 kgSi Steps 6 and 7: Unknowns:

A, D, E

Balances:

Steps 8 and 9: Balances to be solved: System: overall process Si overall mole balance kg mol Si = 3.560

6–14

H, Cl, Si

Solutions Chapter 6

= D kg mol gas

+

⎡ 0.1429 kg mol SiCl 4 1 kg mol Si 0. 2142 kg mol H 2 SiCl2 1 kg mol Si ⎤ + ⎢⎣ kg mol gas kg mol SiCl 4 kg mol gas kg mol H 2Si Cl 2 ⎥⎦

E kg mol HSiCl3

1 kg mol Si = 0. 3571D + E kg mol HSiCl3

Cl overall mol balance ⎡ 0.1429kg mol SiCl 4 4 kg mol Cl A kgmol HCl 1kg mol Cl = D kg mol gas ⎢⎣ kgmol SiCl 4 kg mol gas kg mol HCl

+

0.2142 kgmol H2 SiCl 2 2 kgmol Cl ⎤ E kg mol HSiCl3 3kg mol Cl + kg mol gas kgmol H2 SiCl2 ⎥⎦ kg mol HSiCl 3

A=

D + 3E

H overall mol balance

⎡ 0.6429kg mol H2 2 kgmol H A kgmol HCl 1kgmol H = D kg mol gas ⎢⎣ kg mol H2 kgmol gas kg mol HCl

+

0.2142 kgmol H2 SiCl 2 2 kgmol H ⎤ E kg mol HSiCl3 1kg mol H + kg mol gas kgmol H2 SiCl2 ⎥⎦ kg mol HSiCl 3

A= Solution:

2

1

3

1.71421D + E 3.560 = 0.3571 1D + E

1

A

= D + 3E

2

A

= 1.7142D + E

3

2E

= 0.7142 1D

E

= 0.3571 1D

4 → 3.56

4

= 0.3571D + 0.3571 D 6–15

Solutions Chapter 6

H Si Cl3

D

= 4.98

E

= 1.78

A

= 10.32

MW 1.01 28.09 106.35 135.45 kg/mol

135.45 kg HSiCl 3 1.78 kg mol = 241 kg HSiC13 kg mol HSiCl 3

6.2.10

Step 5: Basis: 100 kg mol = F Steps 1, 2, 3, 4: W

F (kg mol)

E (kg mol)

Furnace

mol fr. CH4 0.70 H2 0.20 C2H6 0.10 1.00 A (kg mol) mol fr. 0.21 O2 0.79 N2 1.00

100% H2O

Duct

mol fr. mol CO2 nECO2 O2 0.02 nEO2 H2O nEH 2O N2 nN2 1.00

E

P (kg mol) mol fr. mol

B (kg mol) Air ? mol fr. 0.21 O2 0.79 H2 1.00

nPCO2 CO2 O2 0.06 nPO 2 nPN2 N2 1.00

P

Assume first an air leak occurs and use material balances to calculate the amount. This is a steady state flow process without reaction. Overall system Step 6: Unknowns are: A, B, W, P, x N2 (or n N2 ), x CO2 (or n CO2 )

6–16

Solutions Chapter 6 Step 7: Balances are: Step 8:

C, H, O, N,

∑x

P i

=1 (or ∑ niP = P)

In 0.70(100) + 0.10(100)2

C (kg mol):

Out = n CO2

n CO2 = 90 kg mol 0.70(100)(4) + 0.20(100)(2) + 0.10(100)(6) = W(2) W = 190 kg mol 190(1.00) 0.21A + 0.21 B = + 90 + 0.06P 2 0.79A + 0.79 B = n PN2

H (kg mol): 2O (kg mol): 2N (kg mol):

90 + 0.06P + n PN2

=P

A balance about the furnace or about the duct is needed, or those two alone would have been sufficient, omitting the overall balance. System: Duct E Step 6: Unknowns: A, E, n CO , n OE 2 , n EH2O 2

Step 7: Balances:

C, H, O, N2 , ∑ niE = E

Steps 8 and 9: C (kg mol):

0.70(100) + 0.10(100) (2) =

E n CO 2

E = 90 kg mol n CO 2

H (kg mol):

0.70(100)(4) + 0.20 (100) (2) + 0.10(100) (6) = n EH2O

n EH2O = 190 2O (kg mol):

0.21A = n OE 2 + 90 +

2N (kg mol):

0.79A = n EN2

190 2

90 + 190 + n OE 2 + n EN2 = E 0.02 = Solution

n

E CO2

n EH2O n OE 2 n EN2

n O2 E kg mol in E 90 190

A

kg mol 982

E

1077

21.5 776 6–17

Solutions Chapter 6 1077.5 From the overall 2N balance

(1)

0.79(982) + 0.79B = n PN2

2O balance (1') 22.2 + 0.21B = 0.06P

0.21(982) + 0.21B = 185 + 0.06P (2) ⎫ B = 207 kg mol ⎬ (3) 90 + (0.79)(982) + 0.79B = 0.94P ⎭ P = 1095 kg mol 207 kg mol / 100 kg mol F

6.2.11

Assume steady state process P = 100 mol

100% H2 O W gas

F2

F

1

mol fr. C 0.357 H 0.643 1.00

oil Aair

A1

A2

Basis: 100 mol natural gas Comp.

mol=% atoms C

atoms O

atoms H

CH4 C2H2 CO2

96 2 2

96 4 2

--4

384 4 --

Total

100

102

4

388

Basis: 100 mol oil Comp.

atoms C

atoms H

(CH1.8)n

100

180

Basis: 100 mol flue gas (dry)

6–18

Solutions Chapter 6 Comp. mol%

mol C mol O

atoms H

mol N2

CO2 CO O2 N2 Total

10.0 10.00 20.00 -0.63 0.63 0.63 -4.55 -9.10 84.82 --100.00 10.63 29.73

Note:

If have separate air streams, we have 5 unknowns and can't solve.

----

84.82 84.82

Basis: 100 mol natural gas (or use 100 mol dry gas product.) Let F2

= mol oil

F1 or P

= mol dry flue gas

A

= mol air to natural gas fed boiler plus oil fed boiler

W

= mol water associated with the dry flue gas

Four balances: In = Out C balance 102 + F2

= 0.1063 P

(1)

= 0.8482 P

(2)

N2 balance 0.79A H balance 388 + 1.80 F2 = 2 W

(3)

O balance 4 + 0.42A

= 0.2973P +W

(4)

P

= 1729 mol dry flue gas

W

= 268 mol H2O

F2

= 82 mol oil and 1 mol C = 1 mol oil so

C in oil = F2

= 82 mol

6–19

Solutions Chapter 6 Total C = 102 + 82

= 184 mol

% C burned from oil = (0.82)(100) = 44.5% 184 6.2.12

Steady-state, reaction take place mA=. 30 mB = .70 1.00

NaC1 A mANaC1 salt mAH2O H2O B

H2O

C12 H2

D

E

G F

C

30%

Overall balances Steps 6 and 7: 5 unknowns: A, B, D, E, G 4 balances: Na, C1, H, O other equations: A/B = 30/70 Steps 8 and 9: (a)

Percent conversion of salt to sodium hydroxide. Basis: 1 lb product = H 1b mol of NaOH:

1 lb 0.5 lb NaOH lb mol NaOH = 1.25 ×10-2lb mol NaOH lb 40 lb NaOH lb mol NaC1:

1 lb 0.07 lb NaC1 lb mol NaC1 = 1.20 ×10-3 lb mol NaC1 lb 58.45 lb NaC1

1.25 ×10−2 Conversion = (100) = 91.2% 1.25 ×10−2 + 1.20 ×10−3

6–20

H

50 % NaOH 7% NaC1 43% H2O

Solutions Chapter 6 (b)

How much chlorine gas is produced per lb of product?

1.25×10-2 lb mol NaOH 0.5 lb mol C12 70.9 lb C12 = 0.44 lb C12 /lb product lb mol NaOH lb mol C12 (c)

⎛ 58.45 lb ⎞ A = (1.25×10-2 lb mol + 1.20 × 10-3 lb mol ) ⎜ ⎟ ⎝ lb mol ⎠ = 0.80 lb Using salt as a tie component:

C=

A 1 = 0.8 lb = 2.67 lb 0.30 0.30

B = C – A = 2.67 – 0.8 = 1.87 lb Balance on oxygen:

⎛ 1.87 lb mol 0.43 lb mol 0.50 lb mol ⎞ ⎛ 18 lb ⎞ G = ⎜ − − ⎟⎟ ⋅ ⎜ ⎟ = 1.22 lb ⎜ 18 lb 18 lb 40 lb mol ⎝ ⎠ ⎝ ⎠ 6.2.13

Step 5: Basis = 100 lb A Steps 2, 3, and 4: Fe added: 36(100) = 3600 lb MW TiO2 = 79.9 lb/lb mol MW Fe = 55.8 lb/lb mol

MW H2SO4 = 98.1 lb/lb mol MW TiOSO4 = 159.7 lb/lb mol

6–21

Solutions Chapter 6

For Part (a), the system boundary has been drawn on a light solid line. Steps 6 and 7: F Unknowns (9): B, C, J, K, H, F, ξ1 , ξ 2 , m Fe

Species balances (7): TiO2 , Fe, H2SO4 , H2O, TiOSO4 , O2 , inert Other equations (2): Reactions 1 and 2 are complete Degrees of freedom = 0 Steps 8 and 9 a. lb H2O removed in evaporator (J): Mol TiO2 in slag:

Mol Fe in Slag:

(0.70)(100 lb) lb mol = 0.876 lb mol 79.9 lb

(0.80)(100 lb) lb mol = 0.143 lb mol 58.8 lb

Amount of H2SO4 based on theoretical requirements of Equations (1) and (2) 0.876 + 0.143 = 1.02 lb mol H2SO4 6–22

Solutions Chapter 6

1.02 lb mol H 2SO 4 98.1 lb = 100 lb H 2SO 4 lb mol 100 lb B= = 149.2 lb Water in stream B = 49.2 lb 0.67

O2 in:

0.143 lb mol Fe 0.5 lb mol O2 32 lb O2 = 2.29 lb O2 =C (0.0715 lb mol) 1 lb mol Fe 1 lb mol O2 Using TiO2:

ξ1 =

0 − 0.876 = 0.876 moles reacting ( − 1)

Using O2:

ξ2 =

0 − 0.0715 = 0.143 moles reacting (0.5)

TiOSO4 in stream K K n TiOSO = 0 + (1) (0.876) = 0.876 lb mol TiOSO4 4

K=

0.876 lb mol 159.7 lb 1 = 170.6 lb 1 lb mol 0.82

H2O in K = 170.6 (0.18) = 30.7 lb Water formed in the reactions (use ξ1 and ξ2 ) :

n out,rxn = (1)(0.876) + (1)(0.143) = 1.019 lb mol (or 8.36 lb H2O) H2 O Water exiting from the evaporator J (lb): 49.2 + 18.36 – 30.7 = 36.9 lb b.

Exit H2O from dryer: inlet air =

18 lb mol 29 lb = 522 lb dry air lb mol

inlet water =

0.036 mol H 2 O 18 mol air 18 lb = 11.7 lb H 2O mol air lb mol

water added to air = (0.18) 170.9 lb – (0.876 lb mol) 6–23

18 lb = 15.0 lb lb mol

Solutions Chapter 6

Humidity = (15.0 lb H2O + 11.7 lb H2O)/522 lb air = 0.051 lb H2O/lb air c.

The pounds of TiO2 produced (P): By reaction (3), 0.876 lb mol of TiOSO4 goes to TiO2. (0.876) (79.9) = 70 lb TiO2

6.3.1

a. 1; b. 3; c. 0; d. 7 a.

b.

6–24

Solutions Chapter 6

c.

d.

6–25

Solutions Chapter 6

6.3.2

Step 5: Basis: 60 kg W Pick the overall system Steps 6 and 7: Unknowns: F, P Balances: A, B Steps 8 and 9: Total F= 60 + P ⎫ F = 380 kg ⎬ B: 0.80F = 0 + 0.95P ⎭ P = 320 kg

Pick mixing point as the system Steps 6 and 7: Unknowns: G, R Balances: A, B (or total as alternate) Steps 8 and 9: Total: 380 + R = G ⎫ ⎬ R = 126.7 B: 0.80(380) + R(0) = 0.60G ⎭

R 125.7 = = 0.33 kg R/kg F F 380

6.3.3

Basis: 100 kg of fresh feed Overall balance around junction 100 + R = F KC1 balance around mixing point

6–26

Solutions Chapter 6 20 + 0.6 R = 0.4 F = 0.4 (100 + R) 20 + 0.6 R = 40 + 0.4 R R = 100 kg R/100 kg fresh feed

6.3.4

Steps 2, 3, and 4: B1 0.75 B2 0.25 S 0.00 1.00 B1 = butene B2 = butadiene S = solvent

A

5,000 lb/hr

I C (lb) B1 B2 S

E (lb) = 10,000

II

B1 0.0 B2 0.01 S 0.99

B (lb) B1 1.00 B2 0 S 0

D (lb) B1 0.05 B2 0.95 S 0

Step 5: Basic: 1 hr (A = 5000 lb) Select whole process as the system. Steps 6 and 7: Unknowns: B, D Balances: Total, B1, B2, S (not all independent) Steps 8 and 9: Total balance: 5000 = B + D B1 balance: 5000 (0.75) = B(1) + D(0.05) D = 1316 lb

B = 3684

Apparently the calculated values are not correct. (The value for C can be obtained from balances on Unit I or II).

6–27

Solutions Chapter 6

6.3.5

Basis: 1 hour

An overall balance shows that Rout from the adsorber must equal Rin to the adsorber if a steady state exists. Let Ri = R. Adsorber balance of U (units are U): 600 mL 1.37U R mL 5.2U 600 mL 0.08U R mL 19.3U + = + 1 mL 1 mL 1 mL 1 mL

600 (1.37-0.08) = R(19.3-5.2) R = 55 mL/hr

6.3.6

Step 5: Basis: 1000 kg = W ≡ 1 hour Steps 1-4:

Revised compositions are in mass fractions. Note: compositions identical

W (mass) = 1000 kg

E (kg)

Wet cereal

Exit air H2O air

mass fr. H2O 0.200 cereal 0.800 1.00

mol fr. 0.263 0.737 1.000

Recycle R (kg)

D (mass) Dried cereal

mass fr. 0.050 H2O Cereal 0.950 1.00

G (kg) mol fr. H2O 0.066 air 0.934 1.00

mass fr. 0.042 0.958 1.000

mol fr. H2O 0.0132 0.9868 air 1.000

6–28

F (kg) mass fr. 0.0082 0.992 1.000

Fresh Air

Solutions Chapter 6

Conversion of mole fractions to mass fractions is not required, but since both mass fractions and mol fractions are listed as data, convert all to mass fractions (can’t convert cereal to mol) to avoid confusion in making the balances. Fresh air, Basis 100 mol mol MW mass(kg) air 98.68 29 2861.7 H2 O 1.32 18 23.76 100.00 2885.48

Exit air, Basis 1.00 mol mol MW mass mass fr. 0.737 29 21.37 0.819 0.263 18 4.73 0.181 1.00 26.11 1.000

mass fr. 0.992 0.0082 1.000

Air entering drier, basis 1.00 mol mol MW mass mass fr. air 0.934 29 27.09 0.958 H2O 0.066 18 1.19 0.042 28.27 1.000 Overall balances

(unknowns D, E, F; balances H2O, cereal, air)

Total: can be made in mass: 1000 + F = E + D in cereal (dry): 1000 (0.80) + F (0) = E (0) + D (0.950) D = 842 kg 1000 (0.20) + F (0.0082) = E (0.181) + D (0.050) H2O: air: 1000 (0) + F (0.992) = E (0.819) + D (0) 3 are independent eqns. check by 4th eqn. a.

D = 842 kg E = 906 kg

F = 748 kg/hr

Balance on mixing point (to get R) Total: R + F = G air : R (0.819) + F (0.992) = G (0.958) H2O: R (0.181) + F (0.0082) = G (0.042) 2 equations are independent, check via 3d equation b.

R = 183 kg/hr

6–29

Solutions Chapter 6

6.3.7

Steps 1, 2, 3 and 4: H 2 O 100% W

KNO3 Solution P

Evaporator mass frac. H2 O

0.50

KNO3

0.50 1.00

mass frac.

10,000 kg/hr 20%

mH O m KNO

F

H2 O

0.80

KNO3

0.20

Feed

1.00

2

300°F M 50% KNO3

3

∑P

R Recycle 100

°F

Saturated Solution Crystallizer

0.6 kg KNO 3

mass frac. H2O

0.625

HNO3

0.375

kg H 2O

1.000

Boundary line for overall balance C mass frac. H2 O 0.04 KNO3

Boundary line for balance around crystallizer

0.96 1.00

Step 5: Basis: 1 hr = 10,000 kg KNO3 solution Step 4 cont’d: Compute the weight fraction composition of R. On the basis of 1 kg of water, the saturated recycle steam contains (1.0 kg of H2O + 0.6 kg of KNO3) = 1.6 kg total. The recycle steam composition is 0.6 kg KNO3 1 kg H2 O = 0.375 kg KNO3 / kg solution 1 kg H2 O 1.6 kg solution or 37.5% KNO3 and 62.5% H2O (which has been added to the figure). Analysis of complete process ( 6 streams): Unknowns:

W, M, C, R, m PH 2 O m PKNO 3

=6

Balances:

(2 units + 1 mix point) × 2 components

=6

6–30

Solutions Chapter 6

Other compositions are (in %) M H2O 50 KNO3 50

F 80 20

W 100 0

C 4 96

R 62.5 37.5

Start with overall balances as substitute for unit balances Steps 6 and 7:

Unknowns:

W, C,

Balances:

2 components

Overall balance to calculate C:

Total:

10,000

=W+C

KNO3: 10,000 (0.20) = W (0) + C (0.96)

10, 000 kgF 0.20 kgKNO3 1kgcrystals = 2083 kg / hr crystals = C 1kg F l hr 0.96 kgKNO3 W = 10,000 - 2083 = 7917 kg To determine the recycle stream R, we need to make a balance that involves the stream R. Either (a) a balance around the evaporator or (b) a balance around the crystallizer will do. The latter is easier since only three rather than four (unknown) streams are involved. Total balances on crystallizer: M=C+R

a

M = 2083 + R

Component (KNO3) balance on crystallizer: M ω M = Cω C + Rω R 0.5M = 0.96C + R (0.375) Solving equations a and

b

b we obtain

0.5 (2083 + R) = 0.375R + 2000 R = 7670 kg/hr

6–31

Solutions Chapter 6

6.3.8

Single pass conversion is: a.

fsp =

−υLR ξ feed n reactor LR

Single pass conversion based on H2 as the limiting reactant: 1.979 − 3.96 = 0.99 −2 0 − 3.96 ξ max = = 1.98 −2

ξ=

SP conversion =

−(−2)(0.99) = 0.50 3.96

Use Eq. (12.1) and Eq. 12.2 with H2 the limiting reactant b.

0.99 =

fsp =

c.

− υLR ξ −( − 2) ξ = ξ = 0.980 fresh feed n LR 1.98

−(−2)(0.98) = 0.986 1.989

Overall conversion of H2 f overall of H 2 =

d.

−(−2)(0.99) = 1.0 1.98

Overall conversion of CO f overall of CO =

−(−1)(.99) = 0.99 1

6–32

Solutions Chapter 6

6.3.9

Steps 2, 3, and 4: Recycle R(mol) CO, H2O F(mol)

Reactor

mol % 52 CO nCO 48 H2O nH2O

L(mol)

P(mol)

H2: 3 mol %

Step 5:

CO2 H2 CO

Basis: P = 100 mol

Pick the overall process as the system. Step 6:

F Unknowns: n CO , n FH O

Step 7:

Balances:

2

C, H, O

Steps 8 and 9: Element balances C (mol): n CO = 48 + 4 = 52

Step 10:

⎫⎪ ⎬ total 100 H (mol): n H2O (2) = 48 (2) n H2O = 48⎪⎭ O (mol): n H2O (1) + n CO (1) = 48(2) + 4 check is ok.

a.

Composition of fresh feed

40% H 2 O and 52% CO

Alternate solution: Use extent of reaction Based on Then

48 − 0 = 48 1 F nCO = 4 + ξ = 4 + 48 = 52 mol F n H2O = 0 + ξ = 48 mol

CO2:

ξ=

To get the recycle, make a balance about the separator (no reaction occurs)

6–33

mol % = mol

Separator 48 48 4 100

Solutions Chapter 6

H 2 balance (mol): Total (mol):

L(0.03) = 48⎫ L=1600 mol ⎬ L=100 + R ⎭ R=1500 mol

1500 mol R = 31.3 48 mol H 2

b. 6.3.10

Steps 2, 3, and 4: MW Ca (Ac)2 = 158.1

MW HAc = 60

100 kg Ca(Ac)2 1 kg mol Ca(Ac)2 = 0.633 kg mol Ca(Ac)2 158.1 kg Ca(Ac)2

Step 5:

Basis: 1000 kg of Ca(Ac)2 feed

Inspection of the diagram shows that the overall conversion of Ca(Ac)2 is 100% (none leaves the process). Thus, fOA = 1. You are given fSP = 0.90. Then R = 0.703 kg mol

a.

fSP 0.90 6.33 = = f OA 1 6.33 + R

b.

The single pass conversion for Ca(AC)2 is 0.90. 0.90 =

− υ LR ξ −( − 1)ξ = feed n LR 6.33

or 111 kg

ξ = 5.697

Overall HAc balance:

6.33 kg mol Ca (Ac) 2

2 kg mol HAc = 12.66 kg mol HAc or 760kg 1 kg mol Ca(Ac) 2

6–34

Solutions Chapter 6

6.3.11

Steps 2, 3, 4: MW C2 H2 = 26.02 MW Zn = 65.37

MW C2H2Br4 = 346

a. C2 H5 produced per hour Step 5: Basis: 1 hr ≡ 1000 kg C2 H 2 Br4 (2.890 kg mol) Make overall balances: Get the extent of reaction using C2 H 2 Br4 n out − n in = υξ 0 – 2.890 = (-1) ( ξ ) ξ = 2.890 reacting moles C2 H2 balance: n C2H2 = 0 + (1)(2.890) = 2.890 kg mol

2.890 (26.02) = 75.2 kg ZnBr2 balance: n ZnBr2 = 0 + 2(2.890) =5 .78 kg mol

Alternate solution 100 kg C2 H2 Br4 1 kg mol C2 H2 Br4 1 kg mol C2 H2 26 kg C2 H2 = 75.2 1 346 kg C2 H2 Br4 1 kg mol C2 H2 Br4 1 kg mol C2 H2

b.

Recycle Make balance on the mixing point. First, get the feed of C2H2Br4 to the reactor 0.80 =

−υξ -(-1)(2.896) = n feed n feed LR LR

C2 H 2 Br4 balance:

n feed LR = 3.61 kg mol

2.890 + R = 3.61

R = 0.72 kg mol

0.72 (346) = 249 kg Alternate solution (in kg). Balance over separator. (1000 kg + R) (0.20) = R R = 250 kg

6–35

Solutions Chapter 6

c.

Feed rate required for 20% excess Zn in reactor:

1000 kg C2 H 2 Br4 1 kg mol C2 H 2 Br4 2 kg mol Zn 65.37 kg Zn 1.2 kg Zn in feed = 454 kg 1 346 kg C2 H 2 Br4 1 mol C2 H 2 Br4 1 kg mol Zn 1 kg Zn required

d.

Mole ratio of ZnBr2 to C2H2 in final products: 5.78 = 2 2.89

2:1 (as per reaction)

6.3.12

Step 5:

Basis: 100 kg F1 ≡ 1 hr

Overall balance Step 6 and 7: Unknowns: F2, P1, P2 Element balances: Ca, Na, C1, CO3 (enough, not all independent) CO3 balance: 900 kg CaCO3 1 kg mol CaCO3 1 kg mol CO3 100 kg CaCO3 1 kg mol CaCO3

= 9.00 kg mol CO3

100 kg Na 2 CO3 1 kg mol Na 2 CO 3 1 kg mol CO 3 0.94 kg mol CO3 = 106 kg Na 2 CO3 1 kg mol Na 2 CO 3 9.94 kg mol CO3

a.

9.94 kg mol CO3 1 kg mol Na 2 CO3 106 kg Na 2 CO3 = 1054 kg Na 2 CO3 1 kg CO3 1 kg mol Na 2 CO3

6–36

Solutions Chapter 6 Species balance about the reactor plus the separator on CaCO3: IN   OUT [1000(0.90) + R] (.24) = R    UNREACTED

R = 284 kg

Alternate solution: Use the extent of reaction to get the same results.

6.3.13

This is steady state process with reaction and recycle. Pick the overall process as the system. Step 5: Basis: 100 kg mol CH4 Steps 1, 2, 3, and 4:

The system is as shown in the diagram with recycle added. 0.21 0.79

Air

kgmol O2 230 N2 865.24

CH4

CO2 O2

100 kg mol

H 2O NO N2

R

Calculate the moles of each component entering. C H4 (g) + 2O2 (g)→C O2 (g)+ 2 H2 O (g)

Calculation of the required O2 and accompanying N2 in moles req'd O2

100(2) = 200

xs

200 (0.15) Total O2

= 30 230

6–37

Solutions Chapter 6

⎛ 230 .79 ⎞ ⎟ = N2 in ⎜ .21 ⎠ ⎝

865.24

The exit concentrations (via stoichiometry) are CO2 O2 H2 O NO N2

Mol 100 30 200 415 × 10-6 865.24 1195.24

Next, consider adding recycle as shown in the diagram. Recycle is not involved in the overall balance, hence the concentration of NO will not be affected because the extent of reaction with and without recycle remains the same. The recycle does reduce the combustion temperature, which in turn will reduce the exit concentration of NO.

6.3.14

Basis: 1 hr ≡ 1L = F

all concentrations are g/L. W

D

F=1L

Zn 0.10 Ni 1.00 H2O

Zn 100 Ni 10.0

P0

Zn 190.1 Ni 17.02

Figure P12.9 Pick the overall process as the system. Steps 6 and 7: Balances:

3

H2O, Zn, Ni

Unknowns:

3

W, D, P

6–38

(all in L)

Solutions Chapter 6 Steps 8 and 9:

D(L) 0.10g + P 0 (190.1g / L) L

Zn:

100 (1) + 0

=

Ni:

10(1) +0

= D (1.00) + P0 (17.02)

Solve to get P0 = 0.525L

D = 1.056L

Pick Unit 3 as the system: Steps 6 and 7:

Steps 8 and 9:

Balances: Unknowns:

2 2

Zn, Ni R32, P2 (all in L)

Zn:

P2 (3.50) + 0 = 0.10(1.056) + R32 (4.35)

Ni:

P2 (2.19) + 0 = 1.00(1.056) + R32 (2.36)

Solve to get R 3 2 =2.75L/hr

6.3.15

H

Mass frac. P oil + dirt ω o+d P H2 O ω HO

ω

H

ωH O 2

2

1.00 P

H o+d

D

mass fr. 0.229 0.771

H2 O

0.9927

0.771 1.000

Filter R

G o+d

ω ω HGO

Mass frac. oil + dirt 0.229 H2 O

0.771 1.000

2

oil + dirt 0.0073

oil + dirt 0.229 H2 O

2910 gal/day

mass fr.

mass fr.

1.000

G

90 gal/day

F 3000 gal/day

1.000

6–39

Solutions Chapter 6 Pick the total process as the system Step 5:

Basis: 1 day (equivalent to D = 90 gal, F = 3000 gal, P = 2910 gal)

Steps 6 and 7: Unknowns:

P ωo+d ,

Balances:

oil + dirt, H2O

Steps 8 and 9: Overall oil + dirt balance:

ω

P H2 O

In Out P 3,000(0.0073) = 2,910(ωo+d ) + 90(0.229) P ωo+d =

21.9 − 20.61 = 4.43 10−4 × 2,910

(a)

To solve for R, make balances on the mixing point, the filter, and the splitter. Not all of the balances are independent: Total

oil + dirt

Splitter:

H = 90 + R

0.229H = 0.229(90) + 0.229(R)

Filter:

G = 2910 + H

ωGo+d G = 2910 (4.43 × 10-4) + H(0.229)

Mixing point: 3000 + R = G

0.0073 (3000) + 0.229R = ωGo+d G

Unknowns: H, R, G, ω Go+d Balances:

6 (4 independent)

Steps 8 and 9: The solution is

R = 57.2 gal/day

6–40

ωGo+d = 0.01145

Solutions Chapter 6

6.3.16

Start with the overall system

mass fr. 0.95 rice

mol fr. H2 O

0.0473

gas

0.9527 1.0000

0.05

H2 O

1.00

P (lb) S (lb mol)

Rice Product Dryer

mass fr.

W (lb mol)

F (lb)

mol fr.

rice

0.75

x

H2 O

0.25

x

w

0.0931

H2 O w

0.9069

gas

1.00 Steps 1, 2, 3 and 4:

1.0000 In the diagram use mol and mass percent for compositions and mol and mass for flow as specified.

Step 5:

Basis

P = 100 lb

Step 6:

Unknowns:

F, S, W

Step 7:

Can make total and 3 component balances: rice, H2O, dry gas (G)

Steps 8 and 9: lb

Rice:

F(0.75) = P (0.95) = 100 (0.95)

lb mol

Dry gas:

S (0.9527) = W (0.9069)

lb mol H2O

H2O:

S (0.0473) +

P (0.05) F (0.25) = W (0.0931) + 18.02 18.02

F = 126.67 lb S = 27.35 lb mol W = 28.75 lb mol

6–41

Solutions Chapter 6

R S

Make balances on the mixing point (easiest) lb mol gas:

Dryer

F1

(27.35)(0.9527) + R (0.9069) = F1 x FDG1

F

lb mol H2O:

(27.35)(0.0473) + R (0.0931) = F1 x H12 O

lb mol total:

27.35 + R = F1

x FH12O = 0.0520 1 x FDG + x HF12O =1

R = 3.12 lb mol / 100 lb P

0.0931 H 2O R

or, make balances on separation point H2O: total (or G):

P 1 0.0931 H 2O

P1 (0.0931) = R (0.0931) + 30.768 (0.0931) (only 1 independent equation) 0.0473 S + 0.0931 R = F1 (0.052) 27.35 + R = F1 1.385 + 0.0931 R = 0.052 (27.35 + R) = 1.523 + 0.052R 0.041 R = 0.138 R = 3.37lb mol

6–42

30.78 lb mol 0.0931 H2 O

Solutions Chapter 6

6.3.17

F° Organic Solvent 100% I

Aqueous Phase FA = 10 L/min A C0 =

F°

FA = 10 L/min

100 g/L

A

O

C1 = 0.01g / L

Fermentation product balance:

100

g 10L / min 10 L / min 0.1g / L = F°(0.01) + L

F° =

1000 − 1 = 99, 900 L min 0.01

B:

F2 °

F1 ° 10L/min I

II

A

C1

FA = 10 L/min CA = 100 g/L o

C1 = 0.1 g/L

O A C1 = 0.1C1

Fermentation balance:

F3 ° 10L/min

III

A C2

O

A

C 2 = 0.1C 2

A

C3 =0.1 g/L O

A

C3 = 0.1C 3

Assuming F1° = F2° = F3° = F°/3

I:

100

A

II:

C1

g L F° A L 10 min = C A ⋅10 L 1 mn + 3 × 0.1C (1) g L F° A L 10 min = C A ⋅10 L 2 min + 3 × 0.1C (2)

6–43

F =10L A

Solutions Chapter 6

A

III:

C2

g L F° A L 10 min = C A ⋅10 L 3 min + 3 × 0.1C3 (3)

Solving (1), (2) and (3) Simultaneously

C 1 = 10.0 g L

C 2 = 1.0 g L

A

C 3 = 0.1 g L

A

A

F = 2702 L min C: Fermentation Product Balance: I:

100 g L 10 L min A A A + 0.1 C 2 ⋅ F° = 0.1 C1 ⋅ F° + C1 ⋅ 10

II:

C1 ⋅10 + 0.1(0.1)F° = 0.1C 2 ⋅ F° + C2 ⋅10

III:

C 2 ⋅10 + 0 = 0.1(0.1) ⋅ F° + 0.1 (10)

A

A

A

A

Solving (1), (2) and (3) Simultaneously A

C1 = 10.36g / L F = 960 L min

A

C 2 = 1.06 g / L

A

C3 = 0.10 g / L

This one is the least

6.3.18

Steps 1, 2, 3, and 4:

Mol. Wt: Na2S, 78; CaCO3, 100; Na2 CO3, 106; CaS, 72 Step 5: Basis 1 hr 6–44

Solutions Chapter 6

Steps 6, 7, 8, and 9: System: Overall (System A) Na Balance:

= b.

(1000)(0.45)lb Na 2S 1 lb mol Na 2S 2 lb mol Na 78 lb Na 2S 1 lb mol Na 2S m Na 2CO3 lb 1 lb mol Na 2CO3 2 lb mol Na 100 lb Na 2CO3 soln 106 lb Na 2CO3 1 lb mol Na 2CO3 80 lb Na 2CO3

mNa2CO3 = 764 lb/hr

System: Reactor plus separator (System B) in

out

gen. Consumption accum

Na 2S balance: ⎡100(0.45) R ⎤ R ⎡100(.45) R ⎤ + − + 0 − ⎢ + ⎥ 0.90=0 ⎢ ⎥ (lb mol) ⎣ 78 78 ⎦ 78 78 ⎦ ⎣ 78

[1000(0.45) + R] (0.10) = R a.

R = 50 lb/hr

Alternate solution reacts.

Na2S is the limiting reactant (LR). Mole Na2S = (0.45)(1000)/78 = 5.77. All of it

1.00 =

( − 1)( − 1)ξ , 5.77

ξ = 5.77

Overall fraction conversion of Na2S is 100%; fOA = 1. fSP n freshfeed 0.90 5.77 LR = = freshfeed = recycle f OA 1.00 n LR +n LR 5.77+R

5.19 + 0.9R = 5.77

R = 0.64 lb mol (or 50 lb)

6–45

Solutions Chapter 6

6.3.19

A basis of 1 hr requires the listing and solution of many simultaneous equations. A basis of 100 mol of toluene in G is more convenient Makeup M

H2 F Feed

1

total

2

3450 lb/hr C7 H 8

Gas Recycle

P Pure

RG

mol 1

CH4

Mixer

100% toluene

100% H2

n wH

only

2

n

W

w CH 4

One system

Gross Feed G 4H 2 4CH 4 1toluene

Reactor

B

Separator

Benzene 100% 1.

7.3.14

Basis: Data on gases cited in problem.

a. The cubic feet associated with “165 ft3” and “240 ft3” probably means that the values are the respective volumes of an ideal gas at standard conditions, because the volumes calculated for the gases in part b. below are quite different than the stated values. b. To find the amount of gas in the cylinder, first find the approximate volume in the cylinder.

Vcyl =

π d2h 4

(3.14)(9)2 (52) = = 1.914 ft 3 (4)(1728)

Use the gas law as ratios:

p 2 V2 z RT = 2 2 p1V1 z1RT1 7–57

Solutions Chapter 7

⎛ p ⎞⎛ z T ⎞ V2 = V1 ⎜ 1 ⎟⎜ 2 2 ⎟ ⎝ p2 ⎠⎝ z1T1 ⎠ For C2H4: Tc = 283 K, pc = 50.5 atm Assume T1 = 80°F = 540°R = 300°K

300 ⎫ = 1.06 ⎪⎪ 283 ⎬ z1 = 0.35 1515 p r1 = = 2.05⎪ (14.7)(50.5) ⎪⎭

Tr =

(At standard conditions: z2 = 1.00)

⎛ 1515 ⎞⎛ 1 ⎞⎛ 492 ⎞ 3 V2 = (1.914) ⎜ ⎟⎜ ⎟⎜ ⎟ = 514 ft 14.7 0.35 540 ⎝ ⎠⎝ ⎠⎝ ⎠ For CH4: Tc = 191 K, pc = 45.8 atm

Tr =

300 = 1.57 191

pr2 =

2015 = 2.98 (14.7)(45.8)

z2 = 0.84

⎛ 2015 ⎞⎛ 1 ⎞⎛ 492 ⎞ 3 V2 = (1914) ⎜ ⎟⎜ ⎟⎜ ⎟ = 285 ft 14.7 0.84 540 ⎝ ⎠⎝ ⎠⎝ ⎠ MW of C2H4 = 28

MW of CH4 = 16

wt of C2H4 =

(514)

wt of CH4 =

(285)

(359) 16 (359)

Therefore c.

28

= 40.1

= 12.7

wt C2 H4 > wt CH 4

For C2H4:

pV = znRT

= z

m RT MW

7–58

Solutions Chapter 7

m=

(1515)(1.91)(28) pV(MW) = (0.35)(10.73)(590) zRT

= 40 lb C2 H4 For CH4:

m=

(2015)(1.91)(16) = 12.7 lb CH 4 (0.84)(10.73)(540)

Check with cylinder: 163 – 90 = 123 lb cylinder

105 – 9.5 = 135 lb cylinder

7.3.15

a)

p1V1 = z,n,RT

R = constant

pa V1 = z 2 n 2 RT1

V1 = constant, the vol. of cylinder

zn p1 = 1 1 p2 z 2 n 2

Tc = 133.0 K

Tr = T T =

24.44 + 273 = 2.24 133

c

pr1 =

pc = 34.5 atm

2000 psig + 14.7 = 3.97 14.7 psig/atm 34.5 atm

pr2 = 3.80 z1 = 1.0 ⎫ ⎬ Ideal gas behavior z 2 = 1.0 ⎭

p1 p2

⎛ n ⎞ ⎛ 1.0 ⎞ n = ⎜ 1 ⎟⎜ ⎟ = 1 ⎝ n 2 ⎠ ⎝ 1.0 ⎠ n 2

7–59

Solutions Chapter 7

n CO =

pV 14.7 psia = RT 536°R

Vcylinder =

10.73

175 ft 3 (psia ) ft 3

( )

= 0.447 lb mol

( lb mol)(°R)

nCO RT1 p1

( )

3 0.447 lb mol 10.73(psia ) ft 536°R Vc = = 1.276 ft 3 2014.7psia (lb mol)(°R)

n final,CO =

p2 Vc = RT

1924.7 psia 1.276 ft 3 = 0.427 lb mol ( psia ) ft 3 536°R 10.73 (lb mol )(°R )

( )

n CO, lost = (0.447 – 0.427)lb mol ≅ 0.020 lb mol rate of loss =

b)

n air

0.020 lb mol 48 hr

pV 14.7 psia = = RT 536.0°R

(

100 ppm = 100 10

–6

= 4.167 ×10−4

lb mol hr

1600 ft 3 = 4.09 lb mol air ( psia ) ft 3 10.73 (lb mol)(°R)

( )

)(4.09 lb mol) = 4.09 × 10 –4 lb mol

4.09 × 10-4 t= = 0.982 hr -4 lb mol 4.167 × 10 hr c)

t = 67 hr

CCO = d)

n CO =

4.167 × 10-4 67

= 0.0279 lb mol

0.0279 lb mol lb mol = 1.75×10-5 3 1600 ft ft 3

The CO alarms would go on to alert people in the building.

7–60

Solutions Chapter 7

7.3.16 3 The density of the gas must be 2.0 g/ m . MW is Si is 28.086 so ˆ = (2/28.086) = 0.0712 g mol/cm3. Let T - 298K. Also z = 1 at Tr = 1.98 V and Vri very large.

a)

zRT 1 8.314(Pa)(m3 ) = p= ˆ (g mol)(K) V

3

⎛ 100 cm ⎞ 298 K = 2.48 ×109 Pa ⎜ ⎟ ⎝ 1 m ⎠

b)

For a lower pressure, use a less dense gas (lower MW).

c)

At too high a temperature, you cannot get adequate density.

7.3.17 3

Basis: 240 ft methane at 80°F and 1964.7 psia. Assume the number of moles does not change. p1V1 nRT1z1 = p2 V2 nRT2 z 2 p1 T1 z1 = p2 T2 z2 T2 =

T1z1 p2 z 2 p1

Calculation of z1 Methane ⇒ Tc = 190.7K

pc = 45.8 atm

T1 = 80°F => 299.82K

Tr1 =

299.82K = 1.572 190.7K

pr1 =

133.7 atm = 2.919 45.8 atm

∴z1 = 0.83

7–61

Solutions Chapter 7

Calculation of z 2

( ) = (3000 psig + 14.7 psia )(1 atm 14.7 psia ) = 205.08 atm

p1 = (1950 psig + 14.7 psia ) 1 atm 14.7 psia = 133.7 atm p2

pr =

205.08 atm = 4.478 45.8 atm

(

)(

V = 240 ft 3 2.832 × 10–2 m 3 /1 ft3 100 cm3 /1 m3 6

= 6.797 × 10 cm

)

3

)

( )

3 RTc 82.06 cm (atm) /( g mol)( K) (190.7K ) ˆ Vc i = = pc 45.8 atm

= 341.68

cm3 g mol

(

(45.8 atm) 6.797 × 10 6 cm3 p1 V1 n= = (299.82K)(82.06)(0.83) z1 RT1

)

= 15244.5 g mol

V 6.797 × 10 6 cm3 cm 3 ˆ ∴V = = = 445.87 n 15244.5 g mol g mol ˆ V 445.87 ∴Vri = ˆ = = 1.305 Vc i 341.68 ∴z 2 = 1.06

T2 =

T1z1 p2 540°R( 0.83)(3014.7 psia ) = z 2 p1 1.06(1964.7 psia)

T2 = 649o R

(189°F)

7–62

Solutions Chapter 7

7.4.1

T = 180°F (640°R)P = 2400 + 14.7 = 2415 psia

Basis: 33.6 lb gas

pV = znRT Basis: 100 mol gas mol

mol.wt.

lb

CO2

10

44

440

CH4

40

16

640

C 2 H4 100

50

28 2560

n=

avg. mol wt. =

1400

2480 = 24.8 100

33.6 = 1.35 lb mol 24.8

Tcʹ′ = 0.10 (304.2) + 0.40 (190.7) + 0.50 (283.1) = 248.25 K pcʹ′ = 0.10 (72.9) + 0.40 (45.8) + 0.50 (50.5) = 50.86 atm p′r =

p 2415 = 3.23 ′ = p c (50.86 )(14.7)

Tr′ =

T 640 = 1.43 ′ = Tc (248.25)(1.8)

From Fig. in text, z ≅ .75

( 3)

znRT 0.75 1.35 lb mol 10.73(psia ) ft 640°R V= = (lb mol)(°R) p 2415 psia 3

= 2.88 ft at 180 ° F, 2415 psia of gas is cylinder volume

7–63

Solutions Chapter 7

7.4.2

0.20 EtOH 0.80 CO2 1.00

T = 500K V = 180 cm3 /g mol p=?

Basis: gas as above pcʹ′ = (0.20) (63.0) + (0.80) (72.9) = 70.92 atm Tcʹ′ = (0.20) (516.3) + (0.80) (304.2) = 346.6 K

p r′ = ? = T r′ =

500 = 1.44 346.6

Vci′ = ′

Vri =

p p = p c 70.92

RTc′ p′c

=

(1 atm )( 22.4 L) 1000 cm3 346.6 K = 401 cm3 /g mol (273 K )(1 g mol) 1 L 70.92 atm

180 = 0.45 401

From the compressibility chart, pr ≅ 2.5 , p ≅ 177 atm

7–64

Solutions Chapter 7

7.4.3

Basis: 540 kg gas @ 393 K, 3500 kPa absolute Component

kg

kg mol

mole fraction

pc, atm

Tc , K

CH4 C2 H6 C3H8 N2

100 240 150 50

6.25 8.00 3.41 1.79

0.321 0.412 0.175 0.092

45.8 48.2 42.1 33.5

191 305 370 126

Total

540

19.45

1.000

Tcʹ′ = (0.32l) (191) + (0.412) (305) + (0.175) (370) + (0.092) (126) = 262.8 K pcʹ′ = (0.321) (45.8)+(0.412) (48.2)+(0.175) (42.1)+(0.092) (33.5) = 44.96 atm T r′ =

3.93 = 1.50 262.8

p r′ =

3500 1 atm = 0.77 44.96 101.3

From Nelson and Obert Chart, zʹ′ = 0.935 MW (average) = 540/19.45 = 27.8 kg/kg mol

ρ=

=

3500 kPa

27.8 kg 393 K 0.935

(K) (kg mol) 3

kg mol 8.31 (kPa) (m )

31.9 kg m at 393K and 3500 kPa 3

7–65

Solutions Chapter 7

7.4.4

Use Kay’s Method pcʹ′ = Σnipci ; Tcʹ′ = ΣniTci Component C2H4 A He

n

pc

0.57 0.40 0.03

50.5 48.0 10.26

npc 28.8 19.2 0.308

Σnipci= 48.31 p r′ =

120 = 2.49 48.3

V=

( 0.70)(82.05)(298) 3 = 142.8 cm /g mol (120)

nTc

283 150.7 13.19

61.2 60.2 0.395

∑ n i Tci =

298 = 1.342 222

T r′ =

Tc

∴

221.8

z = 0.70

⎛ 142.8 –140 ⎞ 2.00% % diff = ⎝ ⎠ 100 = 140 7.4.5

Basis: 30ft3 mixture of 100 atm and 300°F Comp

Mol. frac

Tc

pc

Tcʹ′

pʹ′c

C2H4

0.60

283

50.5

170

30.3

A

0.40

150.7

48

60.2

19.2

230.2

49.5

Trʹ′ = T/Tcʹ′ = pʹ′r =

760 = 1.835 (230.2)(1.8)

p 100 = = 2.02 pʹ′c 49.5

z = 0.95

7–66

Solutions Chapter 7

Avg. Mol. wt. = (0.60)(28) + (0.40)(40) = 32.8 lb/lb mol pV = znRT

n=

pv (100)(14.7)(300) = = 57.0 lb mol zRT (0.95)(10.71)(760)

Assume capacities are given at storage conditions. Then type IA is selected because it is the cheapest. Number of tanks required =

(57.0)(32.8) = 30 62

7.4.6

Steps 2, 3, and 4: Assume a continuous process although a batch process would be acceptable and give the same results.

F=? mol fr. 0.20 C2H4 0.80 N2 1.00

Step 5: Basis:

P=?

Mixer

G 100% C2H4 @ 1450 psig and 70oF

G at given conditions

Steps 6 and 7:

Unknowns: Balances:

F, P

⎫ ⎪ C2 H 4 , N 2 , (or total) ⎬ degrees of freedom = 0 ⎪ ⎭

Steps 8 and 9: Calculate G:

Tr =

mol fr. C2H4 0.50 N2 0.50 1.00

70 + 460 = 1.044 508.3

7–67

MW 28 28

Solutions Chapter 7

1450 + 14.7 = 1.992 735

pr =

z = 0.34

V=

(359)(n)(14.7)(530)(0.34) (2640) = (1465)(492) (1728)

n=

(2640)(1465)(492) = 1.157 lb mol (1728)(359)(14.7)(530)(0.34)

Balances: C2H4

F(0.20) + 1.157 = P(0.50)

Total

F + 1.157 = P F = 1.928 lb mol

7.4.7

P = 3.085 lb mol

Basis: Gas at 50 atm and 600°R

From the compressibility charts: pr =

50 = 3.5 14.3

Tr =

600 = 1.2 40.0 + 460

z = 0.58 100,000 std ft 3 1 lb mol 359(0.58) act ft 3 ⎛ 1 ⎞ 600 3 Q= = 1415 actual ft /hr 3 ⎝ ⎠ hr 1 lb mol 359 std ft 50 492

or use pV = znRT twice p2 V2 z 2 n2 RT2 = p1V1 z1n1 RT1 V2

⎛ T ⎞⎛ p ⎞⎛ z ⎞ = V1 ⎜ 2 ⎟ ⎜ 1 ⎟ ⎜ 2 ⎟ ⎝ T1 ⎠ ⎝ p 2 ⎠ ⎝ z1 ⎠

7–68

Solutions Chapter 7 ⎛ 600 ⎞ ⎛ 1 ⎞ ⎛ 0.58⎞ = 100,000⎝ 492 ⎠ ⎝ 50 ⎠ ⎝ 1.00 ⎠ = 1415 actual ft 3 /hr 7.4.8

Basis: initial gas at 200 psig and final gas at 1000 psig MW C2H4 = 28 initial 214.7/14.7 pr1= = 0.290 50.5

final 1015/14.7 pr2= = 1.37 50.5

Tr1 = 1.05

Tr2 = 1.05

z1 = 0.90

z2

= 0.35

Use pV = znRT n 2 p 2 z2 = n 1 p 1 z1

–1

1015 0.35 = 215 0.90

–1

= 12.14

Alternate way to get z is to use the Pitzer ascentric factor. z = z0 + z1 ω z1 = 0.910 + (-0.023)(0.085) = 0.91 z2 = 0.37 + (0.1059)(0.085) = 0.38 Material balance W1 + Wc = 222 W2 + Wc = 250

W2 – W1 = 28 lb

Solve (I) and (II) to yield n1 = 0.089 and n2 = 1.089

7–69

n2 – n1 = 28/28 = 1

(II)

Solutions Chapter 7

a.

Mass of gas initially = (0.089) (28) = 2.49 lb (28 lb) ($0.41/lb) = $11.48 billing

b.

Wt. of cylinder = (222 – 2.49) lb = 219.5 lb

c.

V=

n1z1 RT 0.089 0.9 10.73 298(1.8) 3 = = 2.17 ft p1 214.7

7.4.9

Step 5:

Basis: 1 hr.

Steps 2, 3, and 4: Steady state, open system liquid 30,000 = F(kg) mass fr. Bz 0.50 Tol 0.30 Xy 0.20 1.00

mole fr. Bz 0.912 Tol 0.072 Xy 0.016 1.000

Separator

vapor P1(kg) MW 78.11 92.13 106.16

kg 71.24 6.63 1.70 79.57

liquid P3(kg)

mass fr. Bz 0.06 Tol 0.09 Xy 0.85 1.00

liquid P2 = 9,800 (kg) mass fr. 0.895 0.084 0.021 1.00

mBz mTol mXy Σ = 9,800

mass fr.

Convert F to moles and then to mass: Basis: 100 kg = F Component

kg

MW

kg mol

mol fr.

Tc(K)

pc(atm)

Bz

50

78.11

0.640

0.592

562.6

48.6

Tol

30

92.13

0.324

0.299

593.9

40.3

Xy

20

106.16

0.118

0.109

619

34.6

1.082

1.000

100

7–70

Solutions Chapter 7

Calculate the kg of F; get z:

Tcʹ′ = 562.6 (0.592) + 593.9 (0.299) + 619 (0.109) = 578.1 pʹ′c = 48.6 (0.592) + 40.3 (0.299) + 34.6 (0.109) = 44.6 Tr =

607K =1.05 578.1K

pr =

26.8 atm = 0.60 44.6 atm

z = 0.80 Together mass of feed:

pV 26.8 atm 483 m3 101.3 kPa (kg mol) (K) n= = zRT 0.80 1 atm 8.314(kPa)(m3 ) 607 K = 324.7 kg mol F = (324.7) (100)/(1.082) = 30,017 kg, say 30,000 kg The material balances are in kg. Step 6:

Unknowns: P1 , P3 , m Bz , m Tol , m Xy

Step 7:

Balances: Bz Tol

mBz 3 = = 1.5 mxy 2

Xy

∑ mi = 9,800 Step 8:

The balances in kg are (only 3 are independent mass balances)

(1)

Bz

30,000 (0.50) = P1 (0.895) + m Bz + P3 (0.06)

(2)

Tol:

30,000 (0.30) = P1 (0.084) + mTol + P3 (0.09)

(3)

Xy:

30,000 (0.20) = P1 (0.021) + mXy + P3 (0.85)

(4)

Total 30,000

= P1

+ 9,800 + P3 7–71

Solutions Chapter 7

(5)

m Bz + mTol + m Xy = 9800

(6)

m Bz = 1.5m Xy

Step 9: Solution: Using (1), (2), (3), (5), and (6) via Polymath kg

mass fr.

P3 = 5500 kg/hr

mBz = 1518

0.156

P1 = 14,700 kg/hr

mtol = 7270

0.745 0.103

mXy = 1012 9800

7–72

1.00

Solutions Chapter 8

8.2.1

a. (1); b. (2); c. (3); d. (4); e. (3); f. (1)

8.2.2

8.2.3

F=2–P+C=2–2+2= 2 8.2.4

F=2–P+C (a) (b)

F=2–2+1= 1 F=2–3+2= 1

8.2.5

C=1 P=2 F=2–2+1=1

8.2.6

C = 3 (O2, N2, H2O) P=2 F=2–2+3= 3 8–1

Solutions Chapter 8

8.2.7

The number of components is 3, but one independent reaction exists among them so that C=2 NH 4 C1(s) Ä NH3 (g) + HCl(g)

P=2 F=2–2+2= 2

8.2.8

(1) F = 2 – P + C The rank of Ca C O

CaCO3(s) 1 1 3

CaO(s) 1 0 1

⎡1 0 9 ⎤ ⎢1 1 0⎥ is only 2 so C = 2 ⎢ ⎥ ⎢⎣3 2 0 ⎥⎦

P = 3 hence F = 2 – 3 + 2 = 1

8.2.9

(a) 40ºC; (b) 190ºC; (c) 60ºC; (d) Compound B

8.2.10

(e)

8–2

CO2(g) 0 1 2

Solutions Chapter 8

8.2.11

a.

F=2–P+C C=3 P = 2 (assume that p is not unreasonably high)

F=2–2+3= 3 b.

Possible variables are: pressure mol or mass fraction of H2O

⎫ ⎪ acetic acid ⎬ only 2 are independent ethyl alcohol⎪⎭

8.2.12

(a)

F = 2 –P + C = 2 – P + 2 if F = 0 P=4–F Max P = 4

(b)

F=2–P+C 3 = 2 – 2 + C C=3

8.3.1

All true

8.3.2

(a) higher; (b) lower; (c) higher; (d) lower; (e) no change; (f) lower; (g) higher; (h) lower; (i) higher; (j) lower; (k) lower; (l) higher

8.3.3

From the p-T diagram for water you can see that raising the pressure by a large amount on ice at constant temperature causes the water to go from the solid phase to the liquid phase.

8–3

Solutions Chapter 8

8.3.4

The vapor pressure of methanol is less than that of gasoline so that at lower temperatures the fuel-to-air mixture is insufficient for combustion. Automotive parts can be made of alcohol tolerant materials such as stainless steel or other corrosion resistant materials and additives to the methanol, such as 15% gasoline, can help solve the problem of difficult cold-weather engine starts.

8.3.5

Appendix G (T is in K) (a)

CD (T is in ºC)

Acetone at 0ºC ln (p*) = 16.6513 −

2940.46 T − 35.93

log10 (p*) = 7.2316 −

p* = 70.51 mm Hg (b)

p* = 70.55 mm Hg

Benzene at 80ºF ln (p*) = 15.9008 −

2788.51 T − 52.36

log10 (p*) = 6.90565 −

p* = 102.73 mm Hg (c)

1277.03 T + 237.23

1211.033 T + 220.79

p* = 102.74 mm Hg

Carbon tetrachloride at 300 K ln (p*) = 15.8742 −

2808.19 T − 45.99

log10 (p*) = 6.8941 −

p* = 123.81

1219.58 T + 227.17

p* = 122.89 mm Hg

8–4

Solutions Chapter 8

8.3.6

Fit the Antoine Equation using the three data points to get A, B, and C. In mm Hg kPa 2.53 15.0 58.9 1n(18.98) = A −

mm Hg 18.98 112.54 441.90 B = 2.9434 C + (−40+ 273)

1n(112.54) = A −

B = 4.7233 C + (−10+ 273)

1n(441.90) = A −

B = 6.0911 C + (−20+ 273)

The solution from Polymath is A = 16.5386

B = 2707.43

At 40ºC, ln (p*) = 16.538 −

C = -33.8541

2707.43 −33.85 + (40 + 273)

p* = 939 mm Hg (125 kPa) 8.3.7

At the triple point, all the phases (solid, liquid and vapor) exist in equilibrium. Therefore, the vapor pressure of liquid ammonia = the vapor pressure of solid ammonia 15.16 – 3063/T = 18.7 – 3754/T T = 195 K

8.3.8

No. It is a typo. Probably the number was 98.8ºC because if you equate the pressures from the two equations, T = 371.75K, or 98.7oC.

8–5

Solutions Chapter 8

8.3.9

For benzene (p* is in mm Hg and T in K) 1n(p*) = 15.9008 −

2788.51 T − 52.36

p* = 760 mm Hg

ln(p*) = 6.6331

T = 353K agrees with data base on CD For toluene 1n(p*) = 16.0137 −

3096.52 T − 53.67

p* = 760 mm Hg

T = 383.7 agrees with data base on CD

8.3.10

Merge the two equations (A1 MW = 27) 1/2

W = 10-4 = 5.83 × 10-2

(p v )(27) T1/2

where p v =

[10

1.594×104 ) T

(8.79 −

]

1/ 2

T

The solution from Polymath is

T = 1492K

8.3.11

ˆ =x V ˆ ˆ All of the values of specific volume are obtained using V with vapor vapor + (1-x vapor )Vliquid ˆ data from the steam tables. V

(a)

ˆ = 0.5 (1.673) + 0.5 (0.001043) = 0.837 m3/kg V

(b)

ˆ = 0.5 (0.6058) + 0.5 (0.001073) = 0.303 m3/kg V

(c)

ˆ = 0.3 (350.8) + 0.7 (0.01613) = 105.25 ft3/lb V

(d)

ˆ = 0.7 (1.8431) + 0.3 (0.0187) = 1.296 ft3/lb V

8–6

Solutions Chapter 8

8.3.12

(a) T; (b) F; (c) T; (d) T; (e) F; (f) T

8.3.13

Basis: 2.10 lb water Specific volume:

ˆ = (1 − x) V ˆ +xV ˆ V l g

0.35 Specific volume of water = Vˆl = = 0.175m3 /kg 2

From the steam tables: Specific volume of saturated liquid, Vˆl = 0.001 088 m3 /kg

specific volume of saturated vapor = Vˆg = 0.414 m3 /kg

Then, 0.175 = (1 – x) 0.001 088 + x (0.414) The quality = x = 0.42

8.3.14

Since the two phases coexist in equilibrium, we have a saturated mixture, and the pressure must be the saturation pressure at the given temperature: p = p* @ 90ºC = 70.14 kPa ˆ and V ˆ values are V = 0.001036 m3 /kg and V ˆ = 2.361 m3 /kg At 90ºC, V l l g g

Add the volume occupied by each phase: ˆ +m V ˆ V= Vl + Vg = mV l g g

= (8 kg)(0.001 m3/kg) + (2 kg)(2.361 m3/kg) = 4.73 m3

8–7

Solutions Chapter 8

8.3.15

v=

ˆ Q mV = A A

where v = velocity, ft/s ˆ = specific volume of the fluid, ft3/lb V A = pipe cross sectional area, ft2 ˆ can be obtained from the steam tables: V ˆ = 0.9633 ft3/lb V 2

⎛ 2.9 ⎞ 2 The area is A = ∏ d / 4 = ⎜ ⎟ = 0.046 ft 4 ⎝ 12 ⎠ 2

v=

∏

25,000 lb 0.9633 ft 3 ft = 5.24 ×105 or 145 ft/s 2 hr lb 0.046 ft hr

8.3.16

The hot oil vaporized the water, and the increase in volume in the system caused the damage.

8.3.17

Use data from the steam tables at the initial condition of p = 101.33 kPa and saturated water (i.e. two phases): ˆ = 1.044 cm3/g and V ˆ = 1673.0 cm3/g. V liq vap

Then the mass of both phases in the container can be calculated: m liq = (0.03 m3) (100 cm/m)3 / 1.044 cm3/g = 28,376g m vap = 2.97 m3 (100 cm/m)3 / 1673.0 cm3/g = 1,775 g

mt otal

= 30,511 g

8–8

Solutions Chapter 8

At the final conditions all of the liquid has been evaporated, and only saturated vapor will exist having a specific volume of: Vv ap = (3.00 m3) (100 cm/m)3 / 30,511 g = 98.3 cm3/g

Interpolating in the steam tables (212-T) ºC/(212-214) ºC = (99.09-98.3) cm3/g/(99.09-95.28) cm3/g T = 212oC

(1985.2-p) kPa/(1985.2-2065.1) kPa = (99.09-98.3) cm3/g/(99.09-95.28) cm3/g p = 2002 kPa

8.3.18

Basis: 10 lb of water Specific volume:

ˆ = 10.0 = 4.975 ft 3 /lb V 2.01

From the paper steam tables (in ft3/lb) at 80 psia: ˆ = 0.0176 V liq

ˆ = 5.476 V vapor

Mass balance:

m L + m V = 2.01 Volume balance:

10.0 = m L (0.0176) + m V (5.476) Mass of liquid is 0.184 lb Mass of vapor is 1.826 lb The volume of liquid is ˆ = (0.184)(0.0176) = 3.28 × 10−3 ft 3 Vliq = m liq V liq

The volume of vapor is

8–9

Solutions Chapter 8

ˆ = (10.0 − 3.28 × 10−3 ) = 9.997 ft 3 Vvapor = m vap V vapor

Quality is 1.826 / 2.01 = 0.908

8.3.19

Log (T)

(log (T))

5.6101 5.7038 5.7991 5.8861 5.9402 5.9915 6.0403 6.0868 6.1527 6.2146

31.4727 32.5331 33.6295 34.6462 35.2856 35.8976 36.4847 37.0488 37.8561 38.6214

DATA: (log (T))3

2

vp

176.5329 185.5619 195.0204 203.9313 209.6027 215.0795 220.3767 225.5079 232.9186 240.0166

0.61130 3.53600 17.21000 62.15000 128.80000 245.60000 437.00000 733.20000 1454.00000 2637.00000 1n p* = a+b ln T + c(ln T)2 + d(ln T)3

THE COEFFICIENTS ARE -1199.2548 532.2715

a b

-79.0261 3.9638

c d

8.3.20

Based on data from the steam tables: State 1: liquid

State 3: solid

State 5: liquid (saturated)

State 2: superheated vapor

State 4: saturated vapor

State 6: vapor (saturated)

You can calculate the properties of a mixture of vapor and liquid in equilibrium (for a single component) from the individual properties of the saturated vapor and saturated liquid by averaging the properties of the two saturated phases. The weights are the respective amounts of each phase.

8–10

Solutions Chapter 8

8.3.21

Based on data from the steam tables: State 1: saturated vapor

State 3: two phase (saturated liquid and vapor)

State 2: saturated liquid

State 4: superheated vapor

8.3.22

Prepare a Cox chart as described in the text. Use semi-log paper with the horizontal axis log10. Obtain vertical scale rules by use of steam properties at even integers for the temperature. Or you can use Antoine Equation, Appendix G, to get 2 points and draw a line between them.

Tcritical (o F) (a) Acetone (b) Heptane (c) Ammonia (d) Ethane

454 512 270 89.7

p critical cox chart (psia) 670 700 1500 750

8–11

pcritical (psia) 691 397 1636 708

Solutions Chapter 8

8.3.23

Steps for preparing a Cox Chart (using water as the reference substance): 1.

On logarithmic paper, set up the pressure scale on one axis. The other axis will be a nonlinear temperature scale, which will be set up by the following method.

2.

Draw a diagonal line from the origin to the opposite corner of the graph.

3.

From the steam tables, obtain values for the vapor pressure of water at evenly spaced temperature increments.

4.

Plot these vapor pressures along the diagonal line.

5.

Now, beginning with the first point on the diagonal, draw a line extending from that point to the temperature axis and label the intersection as the corresponding temperature for that vapor pressure.

6.

Repeat step five for every point on the diagonal line. This process will establish the temperature scale.

7.

Using the established temperature scale, plot the data for benzene and draw a line through the points.

8.

From this line, obtain the vapor pressure for benzene at 125ºF.

The vapor pressure for benzene at 125ºC (257oF) is estimated to be 3.3 atm.

8–12

Solutions Chapter 8

8.3.24

The procedure is the same as outlined for P16.28. The result from the Cox chart is that the vapor pressure for aniline at 350o is 22 atm .

8.3.25 From the CD that accompanies the text the respective vapor pressures at 300K are in increasing value

Ethanol Methanol MTBE

p* in mm Hg

PEL (ppm)

65.8 139.7 294

1000 200 100

The relative values of p* are about the same as the inverse of the relative values of the PEL.

8.4.1

(a) Volume increases

(b) Pressure increases

8–13

Solutions Chapter 8

8.4.2

Basis: Data given in problem statement

dry N2

H 2O



27 C p* = 3.536kPa p = 101.3kPa

27 C

(a) pT = p N 2 + pH 2 O = (101.3 + 3.536) kPa = 104.8kPa

(b)

n H 2O n N2

=

p H 2O p N2

=

3.536 kPa = 0.0349 101.3 kPa

8.4.3

T= − 20oC p* =14.1 mm Hg

O2

p t =760 mm Hg

N2 C 6 H1 4

pair =760 − 14.1= 745.9 mm Hg pC6H14 = pC6H14 =14.1 mm Hg

Basis: 760 mol saturated gas

(or use 100 mol)

p ⇒ moles O2 N2 C6H14

mol = 156.6 = 589.3 = 14.1 760

.21 (745.9) .79 (745.9)

For complete combustion C6H14 + 9 1/2 O2 → 6 CO2 + 7 H2O 14.1 mol C6H14 required 14.1 (9.5) = 134 mol O2 required 156.6 – 134 = 22.60 mol O2 excess

22.60 × 100 = 17 % 134

8–14

Solutions Chapter 8

8.4.4

mol. wt. CCl3NO2 = 164.5 Basis: 1 mol saurated gas mol fr. 0.02 0.98 1.00

CCl3NO2 air

100 kPa 0.02 mol CCl 3NO3 760 mm Hg = 15.0 mm Hg 1 mol gas 101.3 kPa (a)

Using linear interpolation, which may introduce a slight error, 15.0 mm Hg corresponds to

⎛ 15.0 − 13.8 ⎞ o ⎜ ⎟ (5) = 1.3 + 15 = 16 C ⎝ 18.3 − 13.8 ⎠

(289.5K)

Basis: 100 m3 at 100 kPa and 16.5°C

(a)

3 100 m air 100 kPa 273 K 101.3 kPa 289.5 K

1 kg mol air 3

22.4 m air at SC

0.02 mol CCl3 NO2 164.5kgCCl 3 NO2 = 13.95 kg 0.98mol air 1kgmol CCl 3 NO2

8.4.5

Basis: 1 mol air y = 0.12

pH2O = pT (0.12) =101.3 kPa (0.12)

= 12.2 kPa The dewpoint is the temperature at which p*H O =12.2 kPa , or from the steam tables 323K 2

or 50 C o

8–15

Solutions Chapter 8

8.4.6

Assume the tank with air fills with the hazardous liquid vapor at 80ºF. The pressure in the tank with air will become after the transfer (10 + 14.7) + 13 = 37.7 psia hence the seal will possibly rupture during the transfer

8.4.7

At 60ºF, p*H O =0.52 in Hg; at 75ºF, p* = 0.87 in Hg 2

Basis: 12,000 ft3 air at 75ºF and 29.7 in Hg absolute 12,000 ft 3

0.52 in Hg H 2 O 492oR 18 lb H 2 O 1 lb mol = 9.62 lb H 2 O 359 ft 3 at SC 29.92 in Hg total 535oR lb mol

8.4.8

Basis: 1 gal benzene (sp. gr. 0.879, MW 78.1) at 750 mm Hg, 70ºF Moles of air =

(3600 ft 3 ) (750) (14.7) = 9.18 lb mol (10.73) (760) (530)

Moles of benzene =

(1) (1ft 3 )(0.879)(62.4) = 0.094 mol (7.48)(78)

nTotal = 9.18+0.094 = 9.274 mol yBz =

0.094 = 0.0101 9.274

mol % = 1.01% therefore beneath explosive limit

8–16

Solutions Chapter 8

8.4.9

Basis: 350 ft3 C2H2 – O2 mixture at 25ºC, 745 mm Hg n1 =

pV (745)(14.7)(350) = = 0.874 lb mol RT (760)(10.73)(298)(1.8)

⎛ 745 ⎞⎛ 14.7 ⎞⎛ 300 ⎞⎛ 1 ⎞ n 2 = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 0.671 lb mol ⎝ 760 ⎠⎝ 10.73 ⎠⎝ 333 ⎠⎝ 1.8 ⎠

p*H2O at 60ºC = 149.4 mm Hg 149.4 y H2O at2 = = 0.201 745 n H2O = (0.201)(0.671) = 0.135 lb mol C2 H 2 +

5 O 2 → 2CO 2 + H 2 O 2

You get 1 mol H2O/mol C2H2 on reaction Therefore moles C2H2 = 0.135 mol O2 = 0.874 – 0.135 = 0.739 lb mol y O2 at 1 =

0.739 = 0.845 0.874

VO2 =(0.845)(350) = 296 ft 3 at 745 mm and 25ºC VC2H2 = 350 − 276 = 54 ft 3 at 745 mm and 25ºC

8.4.10

Elephant seals have an average body temperature several degrees higher than ours. That means their breath can hold a bit more moisture than ours, and on that day, that extra moisture could have been just enough for condensation to occur. But there are other reasons the elephant seals’ breath may have been easier to see. It could be that it was a few degrees colder on the beach than where the observer was sited. And, as you know if you have ever experimented with seeing your breath, the bigger the volume of air, the more moisture available to condense out and so the condensed moisture was more visible. 8–17

Solutions Chapter 8

8.4.11

(a) p*Bz exists if saturation occurs. Then, with ptot = 100 kPa pBz = 100 (0.014) = 1.4 kPa, which corresponds to

−11oC from Perry or use Antoine Eqn. and solve for T ( p*Bz = 10.5 mm Hg) ln (10.5) = 15.9008 -

b)

(-15oC from the CD)

2788.51 T − 52.36

Repeat for p*Bz = 8.0 kPa (60 mm Hg) corresponding to 15.4oC

8.4.12

Assume equilibrium. At 75˚F, from the Antoine equation, the vapor pressure of benzene is 90 mmHg. Assume the barometer is 760 mm Hg abs.

p∗B 90 mol Bz mol Bz = = = 0.13 ∗ mol air mol air pt − pB 760 − 90 a) The OSHA limit is 1 ppm over 8 hours. The above greatly exceeds this limit. b) No, because the garage is probably not well ventilated, and also if a water heater is in the garage, the LEL (Lower Explosive Limit) may be exceeded.

8.4.13

Basis: 50 ft3 air at 29.92 in. Hg and 70ºF p*H2O at 50ºF = 0.36 in Hg

The amount of water per hour is 1.68 lb H 2O 50 ft 3 60 min 1 lb mol 429oR 0.36 in. Hg of H 2O 18 lb = 3 o min 1 hr 359 ft 530 R 29.92 in. Hg total lb mol hr

Alternate solution: make air and water balances. 8–18

Solutions Chapter 8

8.4.14

Basis: 1 lb air If the air is saturated, p*C8 = 2.36 in Hg and pair = 29.66 - 2.36 = 27.30 in Hg.

p*C8 pair

=

2.36 mol C8 = 0.0864 27.30 mol air

(a)

lb air lb air 27.30 mol Air 29 lb air 1 lb mol C 8 = 2.94 = l bC 8 lb C8 2.36 mol C 8 1 lb mol air 114.2 lbC 8

(b)

nc 2.36 mol C 8 = 8 = 0.080 which is also the fraction of the volume. (27.30 + 2.36) mol total n tot 2.94lb air 1 lb mol air 359 ft 3 at SC 580 R 29.92in Hg lb C8 29 lb air 1lb mol air 492 R 29.66in Hg

(c)

= 43.3 ft 3 at 120 ˚F and 29.6 in. Hg/lb octane

8.4.15

At equilibrium, the pressure of Hg is its vapor pressure −4 p* H g 1.729× x10 g mol Hg = = = pt nt pair 99.5g mol air

pHg

nHg

Note: pt = pair + pHg = pair for all purposes so

1.729×10 −4 gmol Hg 99.5gmol total gas

Basis: 1.729 × 10-4 g mol Hg −4

1.729×10 gmol Hg 200.59g 1000mg = 38.14 mgHg 1gmol Hg 1g nRT = p 3 99.5gmol air 293.15K 8.314(kPa)(m ) 1kg 3  = 2.44m at 20 C and 99.5kPa 99.5 kPa (kg mol)(K) 1000g

V=

8–19

Solutions Chapter 8

15.65

mg at 20˚C and 99.5 kPa. m3

Level is not acceptable. A mercury spill can be cleaned up reasonably effectively by dusting with sulfur, then vacuuming with a vacuum cleaner specifically designed for mercury pick-up, which prevents the escape of mercury vapor. The instructor should point out to the student that the assumption of the “no ventilation” is not a very good approximation unless the surface of the mercury is rather large or there is truly no ventilation in the storeroom.

8.4.16

Step 5: Basis: 1 min Steps 2, 3, and 4: System is loading plus water seal, steady state, open C4 300cm3 20oC 100 kPa

Comp.

P

Water seal

Loading

Air = ?

0.015

H2O(sat) Total

p*H2O at 20oC = 2.34 kPa 3

pV 100.0 kPa 300 cm3 ⎛ 1 m ⎞ (kg mol)(K) n c4 = = ⎜ ⎟ RT 293.15 K ⎝ 100 cm ⎠ 8.314(kPa)m3 )

= 1.231 × 10-5 kg mol = 1.231 × 10-2 g mol C4 Steps 6, 7, 8, and 9: At P, the mole fraction of water is 2.34/120 = x H O = 0.0195 2

The moles of P come from a C4 balance (C4 is a tie element) P = 0.821 g mol

8–20

kPa

air C4

1.231 × 10-2 = 0.015P

mol fr.

2.34 1.00

120

Solutions Chapter 8

Then the air in is 0.821 (1-0.0195 – 0.015) = 0.793 g mol V=

nRT 0.793×10-3 kg mol 293.15 K 8.314 (kPa)(m3 ) = p 100 kPa (kg mol)(K)

V= 1.93 ×10-2 m3 at 100.0 kPa and 20ºC per min

8.4.17

Basis: 1 hr The question really is: does the Hg condense at 150ºF, or is it still a vapor (and thus not collected). The moles of Hg are: 40,000 (0.023 × 10-2)

lb Hg 1 lb mol Hg = 0.046 lb mol 200 lb Hg

The moles of gas are (ignoring the Hg): 40,000 lb gas 1 lb mol gas = 1250 lb mol 32 lb gas

y Hg =

0.046 = 3.68 10−5 0.046 + 1250

×

The partial pressure of the Hg in the gas is yp total (3.68 × 10-5)(14.7) = 5.4 × 10-4 psia The Hg will remain a vapor and not condense.

8–21

Solutions Chapter 8

8.4.18

Basis: 20,000 ft3 moist air at entering conditions in 1 day

20,000 ft 3 273 800 1 lb mol = 57.18 lb mol total 280 760 359 ft 3 at SC

(0.532 lb mol H2O) System: Refrigerator is system Steps 6 and 7:

Unknowns: W2 and P; balance: H2O and air

Steps 8 and 9: ⎫ ⎛ 0.99.12 ⎞ ⎛ 0.13 ⎞ H 2 O: 57.18 ⎜ ⎟ = W2 (1) + P ⎜ ⎟ ⎪ lb mol ⎝ 106.63 ⎠ ⎝ 106.63 ⎠ ⎪ ⎬ P = 56.72 ⎛ 105.64 ⎞ ⎛ 106.5 ⎞ ⎪ Air: 57.18 ⎜ ⎟ = W2 (0) + P ⎜ ⎟ not accurate ⎪ W2 = 0.4613 ⎝ 106.63 ⎠ ⎝ 106.63 ⎠ ⎭

Total: 57.18

= W2

+P

Basis: 30 days 30 (0.4613) (18) = 249 lb H 2O/30 days

8–22

Solutions Chapter 8

8.4.19

Steps 1-4: air F(mol) H2O 25oC 100 kPa

remove 50% H2O

(saturated) air H2O

P

100 kPa W (mol) 100% H2O

dew point = 16oC

mol fr. 100-p*P/100 p*P/100 1.00

at exit p* = ?

At 16ºC

0.57 in Hg 0.28 psia * From steam tables p* = 1.92 kPa p H2O ≅ 13.6 mm Hg =1.81 kPa pair = 100 – 1.92 = 98.08 Thus pair = 100 – 1.81 = 98.18 kPa

Step 5: Basis is 100 Mol F Step 6: Unknowns: W, P Step 7: Balances: H2O, air Steps 8 & 9: Balances around process in mol Air

⎛ 100 − p*H2O ⎞ ⎛ 98.08 ⎞ 100 ⎜ = W(0) + P ⎜ ⎟ in P ⎟ 100 ⎝ 100 ⎠ ⎝ ⎠

H2O

⎛ p*H2O in P ⎞ ⎛ 1.92 ⎞ 100 ⎜ ⎟ ⎟ = W(1) + P ⎜ ⎝ 100 ⎠ ⎝ 100 ⎠

Total

100

Also, p* = f(T)

=W+P

vapor pressure relation

•

The exit gas has one-half the entering H2O, or ½ (1.92) = 0.96 mol thus W = ½ (1.92) = 0.96 mol H2O.

•

The exit air has 98.08 mol dry air.

•

At the exit p H O = p*H O = y H O (100) = 2

2

2

0.96 100 ≅ 0.97 kPa 0.96 + 98.08

8–23

Solutions Chapter 8

•

From the steam tables, this corresponds to about

280 K (7oC)

, or use Antoine Eq.

Note: Taking ½ of (1.92) = 0.96 and dividing by 100 instead of (pH O +pair ) is wrong. The answer is close to correct because of the very small amount of water. 2

8.4.20

Basis: 1000m3 sat. air at 99 kPa and 30ºC p*H2O at 30ºC = 0.6155 psia = 31.8 mm Hg = 4.24 kPa 1000 m3

or

1 kg mol 99 kPa 273 K = 39.3 kg mol 3 22.415 m a tSC 101.3 kPa 303 K

n=

99 kPa 1000 m3 (kg mol)(K) =39.3 kg mol 303 K 8.314 (kPa)(m3 )

Initial: 4.24 ⎞ Initial mol H2O = 39.3 ⎛⎜ ⎟ = 1.68 kg mol H 2 O ⎝ 99 ⎠

99 − 4.24 ⎞ Initial mol air = 39.3 ⎛⎜ ⎟ = 39.3 − 1.68 = 37.62 kg mol air ⎝

99

⎠

Final: At 14ºC and 133 kPa, the air is still saturated, and nair = 37.62 kg mol p*14oC H2O = 0.2302 psia = 11.9 mm Hg = 1.59 kPa = p H 2O @14oC p H2 O pair

=

n H2 O n air

so

n H O ⎫ There is the same as 1.59 = 2 ⎬ (133 − 1.59) 37.62 ⎭ an H2O material balance

n H2O = 0.46 kg mol

1.68 – 0.46 = 1.22 kg mol or 22 kg

8–24

Solutions Chapter 8

8.4.21

Steps 2, 3, and 4: C2H6

CO2 H2O O2 N2

Combustion chamber

AIR

100 kPa

(20% excess)

Assume combustion is complete and the gases are ideal. The process is open, steady state with reaction. The first objective is to make material balances. C2 H6 +3.5O2 → 2CO2 + 3H2 O

Step 5:

Basis: 100 mol C2H6

Step 4:

The entering O2 is The excess O2 is (0.20) (3.50) = Total O2 is

350 mol 70 mol 420 mol

0.79 ⎞ The N2 is 420 ⎛⎜ ⎟ =

1580 mol

⎝ 0.21 ⎠

Steps 6 and 7: Unknowns: 4 (The components of the flue gas) Balances: C, H, O, N, total of 4 Degrees of freedom = 0 Steps 8 and 9 (balances in moles, in = out): C: 2N: H: O:

2(100) = n CO 1580 = n N 6 (100) = 2n H O 420(2) = 2n CO +n H O +2n O Total

n CO 2 = 200 mol

2

n N2 = 1580 mol

2

n H2O = 300 mol

2

2

2

n O2 = 700 mol 2150 mol

2

The dewpoint is the temperature at which the flue gas is saturated p*H2O = y H2O pTotal =

300 (100 kPa) = 14.0 kPa 2150

This pressure of H2O corresponds (from the steam tables) to 325.7 K (52.6o C) 8–25

Solutions Chapter 8

8.4.22 mol % 4.5 CO2 26.0 CO 13.0 H2O 0.5 CH4 56.0 N2 100.0

F

P Pt = 98 kPa

nCO 2 nH O 2 nN 2 nO 2

Air 10% excess O2 22.55 mol N2 84.83

}

Step 4:

Calculate the xs air and add total air to the diagram

Step 5:

Basis: 100 mol synthesis gas

Step 6:

Unknowns: n O ,NCO ,n H O ,n N ,P

(5 total)

Step7:

Balances

(5 independent equations)

Steps 8, 9:

Use element balances

2

2

2

2

C, O, H, N, ∑ n i = P

In

out

C:

4.5

+ 26.0 + 0.5

=

n CO 2

n CO 2 = 31.0

H:

0.5 (4)

+ 13 (2)

=

n H2O (2)

n H2O =

O2:

4.5

+

N2:

56.0

+ 84.83

y H2O =

26.0 + 22.55 = 2

=

n CO2 +

n N2

n H2O 2

28 2

n O2 = 2.05 n N2 =

140.8 187.9

28 / 2 14 = 187.9 187.9

⎛ 14 ⎞ p*H2 O = p H2 O = 98 ⎜ ⎟ =7.3 kPa (or 1.06 psia) ⎝ 187.9 ⎠

This is equivalent to a temperature of 104oF from the steam tables (40ºC)

8–26

Solutions Chapter 8

8.4.23

Basis: 100 mol gases leaving absorber CH4 + 2O2 →CO2 + 2 H2 O

0.8335mol N2 0.21mol O 2 1mol CO2 = 0.1108 mol CO2 0.79 mol N2 2 mol O2 1 - 0.8335 - 0.1108 = 0.0557 mol H2O; 0.00557 (20) = 1.114 psia 

Dew point ≈ 105 F at 1.114 psia b) At constant temperature of 130 ˚F, the gas must be compressed until the partial pressure increases and becomes equal to the vapor pressure in order to reach the point of condensation. From Steam tables, the vapor pressure of water at 130 ˚F is 2.223 psia. The total pressure at which the partial pressure of water (mol fraction = 0.0557) will be 2.223 psia is pH 2 O = p t y H 2 O 2.223 = 40.1psia 0.0557 8.4.24

Basis: 1 lb mol of benzene free air Barometric Pressure

=

mm Hg 742

p* benzene at 40ºC

=

181

Partial pressure of air

=

561

Partial pressure of benzene

=

181

At 25 psig total pressure or p* benzene at 10ºC Partial pressure of air

⇒ 0.323 lb mol benzene

2052 =

45.4 2006.6

Final partial pressure of benzene in air

=

45.4 ⇒ 0.023 lb mol benzene 8–27

Solutions Chapter 8

0.323 – 0.023

= 0.3 lb mol C6H6 recovered

a.

(0.3)(100) = 92.9% recovery (0.323)

b.

2 psig is 103 + 742 = 845 mm Hg (0.023)(845) = 18.9 mm Hg partial pressure of benzene in the recycled air. (1.023)

8.4.25

Steps 2, 3, and 4:

The dewpoint of 140ºF gives p*H2O = 149.3 mm Hg Step 5: Basis: 100 lb wet solids in (BDS is bone dry solids) Steps 8 and 9: BDS balance: 60 lb BDS = 0.90 (wet solids out) wet solids out = 66.7 lb From a total balance 100 – 66.7 = 33.3 lb H2O evaporated = 1.85 lb mol H2O evaporated For the air, Assume the air exits at 1.0 atm Composition in:

H2O is

10 (100) 800

8–28

=

1.25%

Solutions Chapter 8

790 (100) 800 149.3 H2O is (100) 800

=

Air

=

Air is

Composition out:

98.75% =

18.7%

81.3%

Overall balance (mol) on the gas phase Ain + 1.85 = Aout Air balance (mol) on the gas phase 0.9875 Ain = 0.813 Aout Ain = 8.619 lb moles wet air in 8.619 lb mol 359 ft 3 660o R 760 = 3940 ft 3 of wet air @ 800 mm Hg and 200ºF lb mol 492o R 800

8.4.26 Basis: 1 hr Calculate the amounts of the various components in the gas phase: Water

H 2O in = 0 (by specification) H 2O out (saturated): p* at 300 K = 3.51 kPa = p H 2O

y H2O =

3.51 kPa = 0.0319 110 kPa

CO2 CO2 in:

y CO2 = 0.00055

CO2 out: y CO2 = 0.125 O2 O2 in: yO2 = 0.21 8–29

Solutions Chapter 8

O2 out: yO2 = 0.0804 Assume the other gas in and out is N2. The entering gas is

600 m3 120 kPa

(kg mol)(K) = 28.87 kg mol 300 K 8.314(kPa)(m3 )

The exit gas can be obtained from a N2 balance

28.87(1 − 0 − 0.00055 − 0.21) = n out (1 − 0.0319 − 0.125 − 0.0804) n out = 29.89 kg mol In one hour in the steady state the moles of CO2 produced were

29.89(0.125) − 28.87(0.00055) = 3.72 kg mol The mole of O2 consumed were

28.87(0.21) − 30.14(0.0804) = 3.64 kg mol RQ =

3.72 = 1.02 3.64

8.4.27

Step 5:

Basis: 100 lb mol flue gas

Steps 2, 3, and 4: 2.5 lb 97.5 lbF

W

H2O Coal (dry) F % C 80 H 6 O 8 Ash 6 100

P (flue gas)

Air

H2O CO2 CO N2 O2

mols fr. 0.140 0.004 0.800 0.056 1.000

N2 0.79 O2 0.21 1.00 H2O, dew point 50oF; p*H2O = 0.3624 in Hg

8–30

Solutions Chapter 8

Steps 6 and 7: Unknowns: F, A, W Equations: Material balances: C, H, O (the ash is ignored) Steps 8 and 9 (balances are in moles; in = out) To get F: C: (0.80F)(1/12) = (0.140 + 0.004) 100 To get A: N2: 80 = A(0.79)

F = 216 lb or 18.0 lb mol

A = 101.27 lb mol

H: Hydrogen from the dry coal plus the water in the coal is (MW H = 1.008):

⎡ 216(0.06) 2.5 216 1 2 ⎤ 1 + ⎢ ⎥ = 1 97.5 18.016 ⎦ 2 ⎣ 1.008

6.74 lb mol H2O

In A: p*H2O = 0.3624 = yAH2O ptotal = yAH2O (29.90) yAH2O = 0.0121 The water in A is (0.0121) (101.27) =

1.23 lb mol

Total H2O in P: W =

7.97 lb mol

The dew point is 7.97 (29.92) = 2.21 in. Hg equivalent to 105oF 107.97

8.5.1

(a) F; (b) F;(c) F; (d) T; (e) T; (f) T

8.5.2

For multiple components, the vapor pressure varies with composition as well as with temperature, and to be precise the composition should be stated.

8–31

Solutions Chapter 8

8.5.3

Henry’s Law is p = Hx, a straight line with no intercept. A plot of the data indicates Henry’s law fits the data well.

8.5.4

Use Henry’s law p = Hx. The MW of CHC13 is 119.4. The total pressure is 1 atm. x=

p 0.024(1) = = 1.41 × 10−4 mol fraction H 170

The concentration in mg/L is Water:

1000 = 55.6 g mol/L 18

The number of moles of CHC13 is (55.6)(1.41× 10−4 ) = 7.84 ×10−3 g mol/L −4 1 − 1.41×10 Concentration =

119.4 g 7.84×10-3g mol 1000 mg = 940 mg/L g mol 1L 1g

8–32

Solutions Chapter 8

8.5.5

p = Hx where p is in kPa and x is mol fraction The mole fraction of O2 in the gas phase is y O p O2

y O2 =

pTotal

=

2

Hx O2 pTotal

The moles of O2 and H2O are: 6 mg 1g 103 L 1 g mol O 2 = 0.1875 g mol/m3 in water L 1000 mg 1 m3 32 g O 2

O2 :

H 2 O:

x O2 =

3 10×10 18

3

= 55.6 × 103g mol/m 3

0.1875 = 3.37 10−6 55.6 × 103 + 0.1875

pTotal = 1 atm

y O2 =

×

4.02×106 kPa 1 3.37×10-6 mol fr. = 0.134 mol fr. 101.3 kPa

Basis: 1 L gas nO = 2

pO

2

RT

V=

(101.3 kPa)(0.134)

1 atm 101.3 kPa

1L

(g mol)(K)

18 g

0.08206(L)(atm) 290 K 1 g mol

8–33

1000 mg 1g

=

0.101 mg per L

Solutions Chapter 8

8.5.6

Steps 2, 3, and 4 pTotal = 1 atm

and at 60ºF the vapor pressures are:

p*Toluene = 16 mm Hg

p*Benzene = 60 mm Hg

Assume the mixture to be ideal so that Raoult’s Law applies Step 5:

Basis: 1 mol liquid

Steps 6-9: yi =

pi p Total

=

p*i x i p Total

For toluene: yTol =

For benzene: (16)(0.60) = 0.0126 760

y Bz =

(60)(0.40) = 0.0316 760

The balance is air. 0.0126 0.0316 0.0442

Toluene Benzene Total

The vapor is flamable.

8.5.7

Use Raoult’s Law.

p Total = p*P (1 − x B ) +p*B x B

From Perry the vapor pressures are: p*Propane = 16.8 atm

p*Butane = 4.8 atm

pTotal = (100/14.7) atm = 6.80 atm 6.80 = 16.8(1 − x B ) + 4.8x B x Butane = 0.83 assuming the liquid phase is essentially all of the mixture.

8–34

Solutions Chapter 8

8.5.8

Mole fraction in the liquid phase is given by solving the following equations for x Bz : and yTol =p*Tol xTol /pTotal to get xBz = (pTotal − p*Tol )/(p*Bz − p*Tol ) yBz = p*Bz x Bz /pTotal The corresponding equilibrium mole fraction in vapor phase (y) is, yi = p*i x i /pTotal

Using these equations for the various temperatures given, the following data for x and y are obtained for the total pressure of 1 atm T(ºC) 80 92 100 110.4

xBz 1.000 0.508 0.256 0.000

yBz 1.000 0.720 0.453 0.000

From the above-calculated data, the following T-x diagram can be drawn.

8–35

Solutions Chapter 8

8.5.9

8.5.10

The mole fractions of each component are needed to apply Raoult’s law. Assuming a basis of 100 g of solution, we can construct the following:

Water Methanol

g

Molecular weight

25 75

18 32

Mol

Mol fraction

1.39 2.34 3.73

0.37 0.63 1.00

Raoult’s law is used to compute the vapor pressure (p*) of pure methanol based on the partial pressure required to flash: p = xp* p* = p/x = 62/0.63 = 98.4 mm Hg. Use the Antoine equation to get T ln(p* ) = 18.5875 −

3626.55 -34.29 + T

T = 293 K (20o C)

8–36

Solutions Chapter 8

8.5.11

p

Basis:

*

Nap

@ 180 ˚F = 460 mm Hg

p* H2 O @ 180 ˚F = 388 mm Hg = 460 + 388 = 848mm Hg

pMax = pH 2 O + pNaptha

(a)

p* Naptha @ 160 ˚F = 318 mm Hg p* H2 O @ 160 ˚F = 245 mm Hg Basis:

6000 lb feed

(5000) (0.928) = 4640 lb pure Naptha

lb mol H 2 O 245 = lb mol Nap 318 W1 n1 M1 = ; W1 W 2 n 2 M2

1000 - 600

= 0.77

⎛ 18 ⎞ = 4640 ⎝ (0.77) = 600 lb H2O Distilled 107 ⎠ = 400 lb H2 O left

(b)

8.5.12

Steps 2, 3, and 4: LH O = ? 2

100% H2O SO2 Air

y1 = 0.03 G = 84.9 m3 293 K pTotal = 1 atm

y2 = 0.003 SO 2 Air 290K pTotal = 1 atm LH O 2

nH O=? 2

nSO2= ? saturated

Step 5:

Basis: 1 min (85 m3 entering gas)

8–37

Solutions Chapter 8

Steps 6 and 7: L n LH2O , n SO , Air 2

Unknowns:

Equations. Material balances:

H2O, SO2, Air yi > Hxi

Steps 8 and 9: The liquid and vapor are assumed to be in equilibrium at the exit, and the water is saturated with SO2 so that y1 = Hx1

0.03 = (43) x SO = so that x SO2 = 0.00070 mol fraction 2

Material balance SO2 (moles): G (0.03 – 0.003) = L (0.00070 – 0) b.

g mol H 2 O L = 38.6 G g mol air

The respective flow rates are G=

85 m3 1 g mol gas = 3540 g mol gas 0.024 m3

L = (3540) (38.4) = 137 ×103 g mol water In terms of kg a.

L = (136) (18) = 2450 kg/min

8–38

Solutions Chapter 8

8.5.13

Use the Antoine equation or the physical property package on the CD to get the vapor pressure of pentane (P) and heptane (H). Assume ideal liquid and vapor. Basis: Data given in problem statement ln p*P (mm) = 15.8333 −

2477.07 T-39.94

p*P = 283.6 mm Hg (5.48 psia)

ln p*H (mm) = 15.8737 −

2911.32 T − 56.51

p*H = 20.6 mm Hg (0.40 psia)

Convert mass fractions to mole fractions: Basis: 100 g liquid P H

g 20 80 100

MW 72.15 86.17

g mol 0.28 0.93 1.21

mol. fr. 0.23 0.77 1.00

Equations to use, i = P and H ∑ yi = 1

a.

yi =

p*i x i pTotal

p*P p*H xP + xH = 1 pTotal pTotal

pTotal = (283.6)(0.23) + 20.6(0.77) = 80.95 mm Hg

3.19 in Hg 10.8 kPa 1.57 psia 16 (0.23) = 0.90 4.07 0.5 yH = (0.77) = 0.10 4.07 1.00 yP =

b.

yi =

p*i x i pTotal

8–39

Solutions Chapter 8

8.5.14

Data: MW Benzene (B) 78.1 Toluene (T) 92.1

p* at 60ºC 51.3kPa 18.5 kPa

Assume ideal gas and liquid. Apply Raoult’s Law. Calculate mole fractions in the liquid. b.

xB =

1 78.1

= 0.541

1 1 + 78.1 92.1 xT =

1 92.1

= 0.459

1 1 + 78.1 92.1

a.

The equations to use are x B p tot = x B p*B yT p tot = x T p*T yB + yT = 1 yB + x T = 1

(1 − y B )p tot = x T p*T ⎛ x B p*B ⎞ * ⎜1 − ⎟ p tot = x T p T p tot ⎠ ⎝ tot

p tot = x B p*B + x T p*T = (0.541)(51.3) + (0.459)(18.5) = 36.2 kPa

(b)

What will be the composition of this first bubble? yB =

x B p* B (0.541)(51.3) = = 0.766 p tot 36.2

y T = 1 − y=B

0.234

8–40

Solutions Chapter 8

8.5.15

Data: p* at –31.2ºC (kPa) Propane (P) 160.0 n-butane (B) 27.6 Assume ideal solution and vapor. Use Raoult’s law (a)

yP ptotal = x P p*P

yP +yB = 1

yBptotal = x Bp*B

yP +yB = 1

x p p*p

yP =

(1 − y P )p total = (1 − x P )p*B

p total

More manipulations give xP =

p total − p*B 101.3 − 26.7 = = 0.56 p*P − p*B 160.0 − 26.7

x B = 1 − 0.56 = 0.44 (0.56)(160.0) = 0.884 101.3

(b)

yP =

(c)

For propane TºC -0.5 -16.3 -31.2 -42.1

x 0.0 0.196 0.560 1.0

yB = 1 − 0.884 =

y 0.0 0.577 0.884 1.0

8–41

0.116

Solutions Chapter 8

8.5.16

Basis: 1.0 lb mol of mixture @ 200ºC Data: p*C7 @ 200o F = 14.7 psia p*C8 @ 200o F = 5.5 psia

Mole fraction C7 Partial Press C7 Partial Press C8

0 0 5.5 5.5

0.2 2.94 4.40 7.34

0.4 5.88 3.30 9.18

0.6 0.8 8.82 11.75 2.20 1.10 11.02 12.75

8–42

1.0 14.7 0 14.7

Solutions Chapter 8

C7 = 0.47: pTot = 11.7 psia

For

pC7 = 10.0 psia pC8 = 1.7 psia

8.5.17

Assume ideal solution. For the bubblepoint pTotal = p*1 (T)x1 + p*2 (T)x 2

(1)

pTotal = 200 psia or 10,342 mm Hg

xi (mol fr.) 0.20

n-pentane

⎡ 2477.07 ⎤ p*1 = exp ⎢15.8333 − (−39.94 + T) ⎥⎦ ⎣

(2)

0.80

n-hexane

⎡ 2697.55 ⎤ p*2 = exp ⎢15.8366 − (−48.78 + T) ⎥⎦ ⎣

(3)

p*i is mm Hg and T is in K

Solve Equation (1) to get from Polymath T = 447 K

8–43

Solutions Chapter 8

8.5.18

Assume ideal solution For the dewpoint 1 p Total

=

y1 y 2 + p*1 p*2

(1)

p Total = 100 psia or 5, 171 mm Hg yi (mol fr.) 0.20

n-pentane

⎡ 2477.07 ⎤ p*1 = exp ⎢15.8333 − (−39.94 + T) ⎥⎦ ⎣

(2)

0.80

n-hexane

⎡ 2697.55 ⎤ p*2 = exp ⎢15.8366 − (−48.78 + T) ⎥⎦ ⎣

(3)

p*i is mm Hg and T is in K

Solve Equation (1) to get from Polymath T = 413 K 8.5.19

Calculate the bubble point and dew point temperatures at 39.36 in. Hg = 1000 mm Hg. Use the Antoine equations to get p* based on the assumption the solution is ideal. ln p* = A −

B C+T

1 = Benzene 2 = Toluene

K1 =

p*1 pt

ln(K1 ) = 1n(p*1 ) − 1n(pt ) = 1n(p*1 )− 6.9078

Calculate 1nK1 and lnK2. Solve the bubble point equation and the dew point equation by a computer code (or Newton’s method) starting with T = 365K (slightly above p* for benzene). a. Bubble point temperature ∑ yi = 1 =

K ∑i x i

8–44

Solutions Chapter 8

If solution is ideal

Ki xi =

p*i x i p* p* and 0.5 1 + 0.5 2 − 1 = 0 pt pt pt

(1)

Use the Antoine equation to get p*i Benzene p*1 = 15.9008 −

2788.51 −52.36 + T

Toluene p*2 = 16.0137 −

3096.52 −53.67 + T

Solution of Equation (1) using Polymath: Bubble point temperature is 375 K b. Dew point temperature ∑ xi = 1 =

y 0.5 0.5 ∑i = + Ki Ki K2

(2)

Solution of Equation (2): Dewpoint temperature is 381.5 K

8–45

Solutions Chapter 8

8.5.20 V V

F

L

p = 80 psia (5.45 atm) T = 250oF (121.1oC)

L

Assume ideal liquid and vapor exist. Vapor pressure data at 121ºC from the Antoine equation (refer to problem P19.23 for the equations) p* (atm) 9.07 4.03

Pentane (P) Hexane (H) p P = p*P x P = 9.07x P

pH = p*H (1 − x P ) = 4.03(1 − x P ) pTotal = 5.45 = pP + pH = 9.07x P + 4.03(1 − x P )

b.

x P = 0.282 yP =

a.

(0.718 hexane)

pP p* x (9.07)(0.282) = P P = = 0.469 (0.531 hexane) pTotal pTotal 5.45

Basis: F = 1 mol F=L+V

L=1–V

F (xF) = Lx + Vy

1(0.40) = (1-V) (0.282) + V(0.469)

L = 0.37 mol

V = 0.63 mol

8–46

Solutions Chapter 8

8.5.21

Assume pentane is the major contributor to the gas-phase composition. yi =

pi p tot

For Raoult's law, *

pi = p i x i

*

pi =

so

yi p tot (0.018)(100) = = 36kPa xi 0.05

Using the Antoine equation, ln (36) =15.8333−

2477.07 −39.94 + T

T = 242 K

8.5.22

Assume that the pressure at the bottom of the lake = pCO2 p = pgh varies with height Avg. depth = 225 m

pCO2 =

1000 kg 9.8 m 225 m = 2.21 × 103 kPa + 1.01 × 102 kPa m3 s2 = 23 Atm

pCO2 = HxCO 2 ⇒

x CO2

=

23atm = 0.0134 1.7×10 3 atm mol fr

If the entire 200,000 tons were saturated at pCO but not supersaturated

200, 000 ton H2 O 1000 kg 1kgmol = 1.11 × 107 kg mol ton 18kg 0.0134 =

n CO2

nCO 2 is mol of CO2

1.11×107 + nCO 2 8–47

Solutions Chapter 8

n CO2 = 1.51 × 105 kg mol (1.5×10 5 )(8.314)(273) nRT 6 3 = VCO2 = = 3.39×10 m p 101.3kPa

Above would be the worst case. The CO2 would be less; p CO2 average might be a better choice to use in which case VCO 2 would be, say ½ the calculated amount.

8.5.23

a.

Apply Henry’s law to solve the problem.

p = Hx or x =

p H

From the internet, H = 43,600 where x is the mole fraction in the liquid and p is in bars. Enriched gas

(110)(0.397) kPa 1 bar 100 kPa x o2 = = 1.00×10-5 mol fr. 43,600 bar mol fr. b.

To compare enriched gas with nonenriched gas

x enriched x nonenriched

percent excess is:

⎛ p ⎞ ⎜ ⎟ ⎝ H ⎠enriched = ⎛ p ⎞ ⎜ ⎟ ⎝ H ⎠ nonenriched

=

(110)(0.397) = 1.89 (110)(0.21)

1.89 − 1 (100) = 89% 1

8–48

Solutions Chapter 8

8.5.24 VI y1I FI z1I

VII yII 2

I II

LI x1I = z1II

LII x1II

1 designates component A 2 designates component B For each of the separation units we can write a mass balance and an equilibrium relationship, e.g., Raoult’s Law. Material balance on Comp. 1 V y1 F z1

Fz1 = Vy1 + Lx1

(1)

Raoult’s Law

Sep. Unit L x1

y1pTotal = x1p*1

(2)

(1-y1 )pTotal = (1-x1 )p*2

(3)

Solve for y1 in (1): ⎛ Fz ⎞ ⎛ L ⎞ y1 = ⎜ 1 ⎟ − ⎜ ⎟ x1 ⎝ V ⎠ ⎝ V ⎠

(4)

Introduce in (2): ⎡⎛ Fz1 ⎞ ⎛ L ⎞ ⎤ * ⎢⎜ V ⎟ − ⎜ V ⎟ x1 ⎥ pTotal = x1p 1 ⎝ ⎠ ⎠ ⎣⎝ ⎦

Basis: 1 minute Introduce the given values p*1 = 10 kPa

p*2 = 100 kPa

FI = 100 mol

LI = 50 mol 8–49

Solutions Chapter 8

V I = 50 mol

VII= 25 mol

LII = 25 mol z1 = 0.50 The solution is p*1 =

10kPa

p*1 =

10kPa

p*2 =

100kPa

p*2 =

100kPa

Unit I FI = LI = VI = z1I =

Unit II FII = LII = VII = z1II =

100 mol/s 50 mol/s 50 mol/s 0.5

I

x1 = I

x2 =

50 mol/s 25 mol/s 25 mol/s 0.75975

II

x1 =

II

0.93391

II

II

0.06609

0.75975 = z1

0.24025 = z 2

x2 =

I 1

y =

0.24025

y

=

0.58559

I

0.75975

y2 =

0.41441

y2 =

II 1 II

I

II

p Total = 31.62 kPa

pTotal = 15.95 kPa

8.6.1

Apply K values to solve the problem. Get the mole fraction of the compound in the liquid phase. Basis: 100 g water (5.555 g mol) liquid

g

MW

g mol

g mol

x (mol fr.)

vapor K

y (mol fr.)

H 2O Glycerol

5.5

92.08

0.0597

5.555

0.0106

1.2 × 10-7

1.27 × 10-9

MEK

1.1

72.10

0.0153

5.555

0.00275

3.065

0.00843

Phenol

2.1

94.11

0.0223

5.555

0.00400

0.00485

1.94 × 10-5

The mole fractions in the gas phase are in the far right hand column. Glycerol and phenol have the lowest concentrations. If volatization is proportional to the vapor phase concentration, only MEK might be a problem. 8–50

Solutions Chapter 9

9.1.1

Basis: 1 lbm

252 cal 1 lb m 4 = 2.5 × 10 cal / kg a. 45.0 Btu 1 Btu 0.454 kg lb m 3 1.055 × 10 J 1 lb m 5 45.0 Btu = 1.048 × 10 J / kg b. 1 Btu 0.454 kg lb m –4 2.930 × 10 (kW)(hr) 1 lb m – 2 ( kW)(hr ) 45.0 Btu = 2.91× 10 c. 1 Btu 0.454 kg lb m kg 2

7.7816 × 10 (ft) (lb f) 4 d. 45.0 Btu = 3.5 × 10 ( ft )(lb f ) / lb m 1 Btu lb m 9.1.2

9.484 × 10 a. 4.184 J J (g) (°C)

–4

–41.6 J 9.484 × 10 b. J kg

–4

c.

0.59 (kg (m) 3

(s ) (K) = 0.341

Btu

1g 2.2 × 10

Btu

–3

1°C lb m 1.8°F

=

1.00 (lb m )(° F )

Btu kg 1g = –0.0179 1000 g 2.20 × 10 – 3 lb lb m m

1N –2

(kg) (m) (s )

–1 (J) (m ) 9.484 × 10 J 1N

–4

Btu

Btu (ft )(hr )(° F)

9.1.3 2

cal 252 cal hr ft a. 6000 Btu = 0.4521 2 2 2 (s) cm 2 (hr) (ft ) Btu 3600 s 30.48 cm

( )

cal (Δ) 1.8°F b. 2.3 Btu 252 cal lb = 2.3 ( g)(°C ) (lb) (°F) Btu 453.6 g (Δ) ° C

9–1

1 m 3600 s 1 K 3.2808 ft hr 1.8°F

Solutions Chapter 9

c.

J 200 Btu 1, 055 J hr (Δ )1.8°F ft = 3.46 (s)(cm )(° C) (hr )( ft )(°F ) Btu 3600 30.45 cm ( Δ )°C 3

d.

10.73 (lb f) (ft ) 144 in. 2

(in. 2 ) (lb mol) (°R) = 8.31

ft

2

3

Btu lb mol 1.055×10 J (Δ) 1.8°R Btu 778 (ft) (lb f) 453.6 g mol (Δ) ° K

J (g mol )(° K)

9.1.4

Basis: 1011 watts

108 W 8.6057 ×105 cal 1 hr 1 (min)(cm2 ) 1 m2 103 (W)(hr) 60 min 0.10 32.0 cal (100 cm) 2 = 4.5 × 104 m2 larg e

9.1.5

(a) T; (b) T; (c) F; (d) F; (e) T 9.1.6

(a) T; (b) T; (c) T; (d) T; (e) T; (f) F; (g) F 9.1.7

partial pressure volume specific gravity potential energy relative saturation specific volume surface tension refractive index

intensive extensive intensive extensive intensive intensive intensive intensive 9–2

Solutions Chapter 9

9.1.8

a. b. c. d.

Intensive Intensive Intensive Extensive

9.1.9

To convert to J/(min) (cm2) (°C), we set up the dimensional equation as follows 2

h=

2

0.026 G 0.6 Btu 1055 J 1 hr ⎛ 1 ft ⎞ ⎛ 1 in. ⎞ 1.8o F ⎜ ⎟ ⎜ ⎟ D0.4 (hr) (ft 2 ) (o F) 1 Btu 60 min ⎝ 12 in. ⎠ ⎝ 2.54 cm ⎠ 1o C = 8.86 × 10-4

G 0.6 J D0.4 (min)(cm2 )(o C)

9.1.10

a.

1 lb fat 1 kg 7,700 k cal day = 7 days 2.2 lb kg 500 k cal

b.

1 lb fat 1 kg 7,700 k cal 4.184 k J km 0.6214 mi = 22.75 mi 1 k cal 400 k J 1 km 2.2 lb kg

c. Total energy consumed is the same. The higher power consumption is compensated for in the shorter travel time.

9–3

Solutions Chapter 9

9.1.11

975 W A m 2 320 days 24 hr 5 0.21 3.6 × 106 J m2 1 day 25 1( kW)( hr )

a.

20

= 3 × 10 J

A = 2.5 × 101 1m 2

( 2.5×10 km ) 5

2

The capital cost is probably too great and the maintenance too high relative to other power sources. Also, the location would have to be in the far western united so power would have to be transmitted over long distances. b.

3×1020 J =

10,000 Btu 2,000 lb T tons 1055 J 0.70 lb ton 1 Btu

T = 2.0 ×1010 tons 2.0 ×1010 = 0.012 1.7 ×1012

9.1.12

First substitute for T

o

F

ToF = To C (1.8) + 32

Next change units k=

{a + b ⎡⎣T

}

(1.8) + 32⎤⎦ Btu 1055 J 1 ft 1 in 1 hr 1.80o F (hr)(ft)(o F) 1 Btu 12 in 2.54 cm 60 min 10 K o

C

= 1.05 ⎡⎣a + b(To C (1.8) + 32 ⎤⎦

J (min)(cm)(K)

or k = 1.05 a + 1.8 bT C + 33.2 o

9–4

Solutions Chapter 9

9.1.13

900,000 J (s)(10-3 ) kW = 105 kW or 100 MW -3 9×10 s 1J

The answer is yes .

9.1.14

(a)

Some examples are: How far would you have to run to lose 1 kg of fat? 32,000 kJ 1 km = 80 km 400 kJ

about the distance of 2 marathons! (b) How many days would you have to reduce your diet from 2400 calories to 1400 calories to lose one kg of fat? 7700 k cal 1 day = 8 days 1 kg fat 1000 k cal

9.1.15

A main meal is equivalent to about 4000 kJ. A ¼ hour is 900 seconds, so that (700 J/s) (900 s) / 1000 = 630 kJ, an insufficient amount of time. 9.1.16

A closed system because no mass exchange occurs with the surroundings.

9.1.17

Yes. Watts are power, and if the energy is transferred for a very short period to time, the power can be quite large. For example, 1J (a very small amount of energy) transferred in 10–6s is 1 MW!

9–5

Solutions Chapter 9

9.1.18

The trend is to advertise in a way that makes people select a product. And sometimes that just creates confusion. What the manufacturer is doing is basing the advertisement on the weight of the hot dog. A hot dog weights about 43 grams and has 8 grams of fat in it. They divided 8 by 43 and get 19 percent fat. They can round it off to 20 and say it’s 80 percent fat free. But you are concerned with the percentage of fat in the calories. You want 30 percent or less to come from fat. Your figures are right. 9.2.1

Basis: 40.0 Newtons × 6.00 m = 240(N)(m) (a)

(b)

(c)

(d)

240(N)(m)

1J = 240 J 1(N)(m)

240(N)(m) 0.738(ft)(lb f ) = 177.0(ft)(lb f ) 1(N)(m) 240(N)(m)

1 J 0.239 cal = 57.4 cal 1(N)(m) 1J

240(N)(m)

1 J 9.478 ×10-4 Btu = 0.226 Btu 1(N)(m) 1J

9.2.2

The work done by the cylinder-piston system is W = -p(V2-V1) =−

350 kPa 0.15 − 0.02 m3

1 kJ = 45.5 kJ (1 kPa)(1 m3 )

9–6

−

Solutions Chapter 9

9.2.3

No work is done because the boundary of the system remains fixed.

9.2.4

Basis: 2 kg mass PE = mgh =

12 kg 9.80 m 25 m 1(J)(s2 ) = 2940 J s2 1(kg)(m2 )

9.2.5

Any correct examples will be satisfactory for (b), (c), and (d). But for (a), liquid to solid, there are probably no examples. If any, they would be pathological cases. 9.2.6

(1) Solid melting. 9.2.7

Basis: 10 lb water (5 lb vaporized)

The work done by the piston is caused by the water vapor expanding at constant temperature and pressure. Ignore the volume of the liquid water at the initial conditions, and assume the initial amount of vapor is also negligible. From state to state 2 3 ˆ ΔV vapor = 4.897 ft / lb from the steam tables. For the 5 lb V2 = (5 lb)(4.897 ft 3 /lb) = 24.49 ft 3 W=−

89.65 lbf 144 in 2 (24.49 ft 3 ) = −3.16 × 105 (lbf )(ft) 2 2 in 1 ft

9.2.8

(a) T; (b) F; (c) T; (d) F; (e) F; (f) F; (g) T

9–7

Solutions Chapter 9

9.2.9

The comment is not a correct use of the term heat, which is a transfer of energy.

9.2.10

Conservation is confused with balancing. Heat is energy transferred from one system to another, and from the viewpoint of either system (for which an energy balance is made), heat is not conserved. Thus (a) is wrong, heat is part of a balance; (b) same as (a); (c) heat is not conserved—it is transferred and is part of the energy balance; (d) the answer should be no with the explanation offered.

9.2.11

(a)

Wrong. Heat is energy, not material.

(b)

Wrong. Heat is energy by definition.

(c)

Wrong. Heat is energy; cold is related to temperature.

(d)

Wrong. See (c).

(e)

Wrong. Heat is a transfer of energy.

(f)

Heat can be measured, but indirectly via temperature, mass, etc.

(g)

Wrong. See (a).

(h)

OK except for “or can be stored,” which is wrong.

(i)

Burning produces a change in the state of the reactants and products, but the change is not heat. Heat transfer can result from the change in state.

(j)

Heat cannot be stored and hence destroyed. Heat transfer can be terminated.

9.2.12

A better use of words than “absorbing heat” would be “to transfer heat.” Heat is really not “concentrated.” The internal energy of the medium increases.

9–8

Solutions Chapter 9

9.2.13

(a) F; (b) F; (c) T; (d) F; (e) T; (f) T but debatable; (g) F; (h) F

9.2.14

 Heat transfer rate is Q=hAΔT=h( π DL)ΔT = Q

5J 5 cm 1 m 100π m (120 − 20)oC 7854J/s 100 cm (s)(m 2 )( oC)

9.2.15

Basis: 1 minute KE = (1/2 ) mv2

v=

500,000 g 1 cm3 4 = 22,143 cm/min min 1.15g π(5)2cm2

1 500 kg KE = 2 min

2

2 ⎛ 22,143 cm ⎞ 1(J)(s ) ⎜ ⎟ 2 ⎝ min ⎠ 1(kg)(m )

2

2

⎛ 1 min ⎞ ⎛ 1 m ⎞ ⎜ ⎟ ⎜ ⎟ = 6810 J ⎝ 60s ⎠ ⎝ 100 cm ⎠

9.2.16

Basis: 1 lbm H2O

KE = 1/2 mv2

2 1 1 lbm ⎛ 10 ft ⎞ (lbf )(s2 ) KE = = 1.55 (ft) (lbf ) ⎜ ⎟ 2 ⎝ s ⎠ 32.2 (lbm )(ft)

9–9

Solutions Chapter 9

9.2.17

Relative to the reference values in the steam tables and the CD: (a)

U = 3528 kJ/kg (from the steam tables)

(b)

The quality is 0.13% and U = 1213 kJ/kg from the steam tables

(c)

At 100oC and 1000 kPa, U of the liquid is about U ≅ 532 kJ/kg from the steam table

9.2.18

Basis: 1 kg steam

1

2

250.5°C 4000 kPa

650°C 10,000 kPa

ΔU = ΔH – ΔpV or use U values directly from SI tables or computer code. Data here taken from the steam tables. State 1 (assumed saturated) State 2 (superheated) 250.5°C (523.5 K)

650°C (923 K)

482°F

1202°F

4000 kPa abs.

10,000 kPa abs.

p

580 psia

1450 psia

Vˆ

0.498 m3/kg

0.04117 m3/kg

ΔU

2602 kJ/kg

3338 kJ/kg

ΔH (ref. satd liquid at 0°C)

2801 kJ/kg

3742 kJ/kg

T

9–10

Solutions Chapter 9

9.2.19

Internal energy is the energy contained in the material of a system because of its molecular arrangement. Heat is energy that flows between the system and the surroundings because of a temperature gradient. Therefore they are not the same class of energy. 9.2.20

ΔH = ΔU + Δ (pV) When a liquid vaporizes Δ (pV) adds to ΔU.

9.2.21

∆H and ∆U are zero because ∆H and ∆U are point (state) functions (variables).

9.2.22

None. The initial and final states for U and H are the same.

9.2.23

The energy balance is ΔΕ = ΔU + ΔPE + ΔKE = Q + W Q >0 0 0 0

(a) (b1) (b2) (c)

ΔE >0 >0 >0 >0

W 0 >0 >0 >0

9.2.24

(a) (b) (c) (d) (e)

T F F T F

(f) (g) (h) (i) (j)

F F F T T

9–11

ΔU >0 >0 >0 >0

Solutions Chapter 9

9.2.25

(a) (b) (c) (d) (e)

F F F T T

(f) (g) (h) (i) (j)

T F T F F

(k) (l) (m) (n)

T F T T

9.2.26

(a) 1; (b) 3; (c) 5; (d) 2; (e) 4 The boiling temperature is at 4 reading to the left scale (100ºC) The freezing temperature is at 1 reading to the left scale (0ºC) 9.2.27

Tc = 540.2 K

pc = 27 atm

⎛ T ⎞ 0.0331⎜ b ⎟ − 0.0327 + 0.0297 log (p c ) kJ ⎝ Tc ⎠ ΔH v = Tb (g mol) (K) ⎛ T ⎞ 1.07 − ⎜ b ⎟ ⎝ Tc ⎠

ΔH v = 31.679

kJ g mol

The percent error is negligible in view of the precision of the data.

9.2.28

Below log is log10 . The critical temperature = 408.1 K

C p = A + Blog Tr Basis: 1 g mol

B=

Cp1 − Cp2 log Tr1 − log Tr2

;

A = Cp1

Blog Tr1 9–12

−

Solutions Chapter 9

Cp1 = 97.3

Tr1 = 300 / 408.1 = 0.735

log Tr1 = 0.134

C p2 = 149.0

Tr2 = 500 / 408.1 = 1.225

log Tr2 = 0.088

A = 128.5

B = 232.6

at 1000 K: Tr1000 = 1000 / 408.1 = 2.45

−

log Tr1000 = 0.389

Cp = 128.5 + 232.6(0.389) = 219.0 J /(g mol)(K) (100)

227.6 − 219.0 = 3.8% 227.6

9.2.29

ΔH = 2∫

250 + 273

50 + 273

(27.32 + 0.6226 ×10−2 T − 0.0950 ×10−5 T 2 )dT

ΔH = 11,980 J (The CD gives 11,784 J)

9.2.30

Integrate the heat capacity equation, use Table D3 in the Appendix, or use CD From the CD ΔH = 2580 J/g mol

9.2.31

2 ⎛ πD 2 ⎞ ⎟ 5.5 = π (0.25) ( 5.5) = 0.27 cm 3 . The density of Al is The volume of the wire is ⎜ ⎝ 4 ⎠ 4 3 19.35 g/cm , so 19.35(0.27) = 5.22g or 0.194 g mol Al. From Perry, Cp = 20.0 + 0.0135T (T in K, Cp in J/(g mol)(°C) and ∆Hfusion = 10,670 J/g mol at 660°C.

⎡660 +273 ⎤ ΔH = 0.194 ⎢ ∫ (20.0 + 0.0135T)dT + 10,670 ⎥ = 5560 J ⎣ 25+ 273 ⎦

(

)

5560 2.773 × 10 –7 = 1.54 × 10 –3 kWh

9–13

Solutions Chapter 9

9.2.32

Basis: 5 ft3 vessel

Vapor = 4 ft3, Liquid = 1 ft3

Steam tables: Sat'd steam at 1000 psia 3

3

ft Vv = 0.44596 lb

ft Vl = 0.02159 lb

lb

ml =

mv =

1 ft

3

lb 0.02159 ft

4 ft

3

3

lb 0.44596 ft

3

mass fr.

= 46.32

0.838

8.97

0.162

55.29

1.000

=

mt = Quality = 0.162

9.2.33

We will use the steam tables for this problem. Basis: 1 lb of H2O at 60°F From steam tables (ref. temp. = 32°F):

ˆ = 28.07 Btu/lb H

at 60°F

ˆ = 1604.5 Btu/lb H

at 1150°F and 240 psig (254.7 psia)

ˆ = (1604.5 – 28.07) = 1576.4 Btu/lb ΔH ΔH = 1576(8.345) = 13.150 Btu/gal Note: The enthalpy value that has been used for liquid was taken from the steam tables for the saturated liquid under its own vapor pressure. Since the enthalpy of liquid water changes negligibly with pressure, no loss of accuracy is encountered for engineering purposes if the initial pressure on the water is not stated. 9–14

Solutions Chapter 9

9.2.34

Basis: 3 kg H2O

The initial conditions are obtained from the SI steam tables for saturated liquid. Enthalpy is a function of the temperature and pressure, but the effect of pressure on liquid water under these conditions is negligible. Therefore the enthalpy of water at 300K and 101.3 kPa can be approximated by the enthalpy of saturated water at the same temperature but at the vapor pressure of 3.536 kPa. The final conditions are presumably a state in which water is all vapor. A check of the steam tables shows this assumption to be true. At T(K)

p(kPa)

300 800

3.536 1500

ˆ ΔH(kJ/kg) 111.7 3384.3

The enthalpy change is

ΔH =

3 kg (3384.3-111.7) kJ = 9817.8 kJ kg

9.2.35

The amount of energy that the water gives up in cooling to 0ºC is not enough to melt all the ice, since each gram of water will give up 209 J in cooling, while heating 1 g of ice to 0º and melting it requires that 2 × 40 + 335 = 415 J be expended (the heat capacity of ice is 2 J/(g) (ºC)). 9.2.36

Yes. 40ºF and saturated liquid is an arbitrary reference state for enthalpy that can be assigned a value of zero. Only enthalpy changes can be computed using the chart; the enthalpy itself cannot be determined.

9–15

Solutions Chapter 9

9.2.37

Basis: 1.2 ft3 gas at 7.3 atm p of 1 atm = 15,450 lbf/ft2 pV1.3 = const. V1 = (1.2)1.3 = 1.27 1.3

p1V1 = (15450)(1.27) = 19,620 1.3

19,620 W = − ∫=pdV − ∫= − 1.3 dV V

19,620 ∫

dV ⎡ 19,620 ⎤ V2 = − ⎢ 1.3 0.3 ⎥ V ⎣ −0.3V ⎦ V1

1.3

1.3 2

V

(V )(p1 ) (1.27)(7.3) = 1 = = 9.27 p2 (1.0)

V2 = 5.54 ft 3 V1 = (1.2)0.3 = 1.06 0.3

V2

0.3

W=

= (5.54)0.3 = 1.67 19,620 ⎡ 1 1 ⎤ (19,620)(0.344) − =− = −22,500(ft)(lbf ) ⎢ ⎥ 0.3 ⎣1.67 1.06 ⎦ 0.3

9.2.38

The density of air at 27ºC and 1 atm is ρ =

ρ=

101.3 kPa

Power =

(p)(MW) (RT)

29 kg air (kg mol)(K) = 1.18 kg / m3 1 kg mol air 8.314(kPa)(m3 ) 300 K

1 1 1  2 = (ρAv) v 2 = ρAv 3 mv 2 2 2

9–16

Solutions Chapter 9

Assume air flow through the windmill is equivalent to the average flow in a pipe of diameter 15m.

v=

20 mi 1.61×103m 1 hr = 8.94 m/s hr 1 mi 3600 s 2

⎛ 15 ⎞ π ⎜ m ⎟ 3 2 1 1.18 kg ⎛ ⎞ ⎝ 2 ⎠ ⎛ 8.94 m ⎞ 1(s )J (s)(W) 1 kW = 74.46 kW Power = ⎜ ⎟ ⎜ ⎟ 3 2 ⎝ s ⎠ 1(kg)(m ) 1 J 1000W ⎝ 2 ⎠ m

Electrical energy = (74.46)(0.30) = 22.3 kW

9.2.39

a.

Basis: 1 lbm of vehicle

25, 000 mi 1 hr 5280 ft = 36, 600 ft/s hr 3600 s 1 mi 9 2 1 1.34 × 10 ft 1 lb m 1 7 (ft) (lb f) K.E. = = 2.08 × 10 lb m 2 2 gc s 7

2.08 × 10 (ft) (lb f) 1 Btu = K.E. = 778 (ft) (lb f) 2.68 × 10 4 Btu / lb lb m m

b.

Heat capacity of, say, 1 Btu/(lb) (°F) gives 1 Btu 20°F = 20 Btu/lb m . (lb) (°F) Essentially all the energy must be transferred to the surroundings

9–17

Solutions Chapter 9

9.2.40

Assume 7 meters is a constant level difference hence h = 7m

Filled reservoir

7m

For one-half of the complete cycle (6 hours):

m = ρAh =

Δ(PE) =

3

2

10 kg 23 km 7 m m

3

3

10 m 1 km

2

= 1.61 × 10

11

kg

1.61×1011 kg 9.80 m 7 m 1J = 1.10×1013 J 2 (1 kg)(m 2 ) s s2

For a complete cycle:

2 1.10×1013J 0.85

1 min = 4.23×108 J/s 370 min 60s

8

or 5.06 × 10 W

9–18

Solutions Chapter 9

9.2.41

The heat capacity of a monoatomic ideal gas is volume is:

5 R , and the heat capacity at constant 2

5 3 R −R = R 2 2

You can show that C p = C Vˆ + R

ˆ ⎞ ⎡ ∂U ˆ + ∂ (pV) ˆ ⎤ ⎛ ∂U ⎞ ⎛ ∂H ⎛ ∂V ⎞ Cp = ⎜ ⎟ = ⎢ ⎥ = ⎜ ⎟ + p ⎜ ⎟ ∂T ⎝ ∂T ⎠p ⎝ ∂T ⎠p ⎣ ⎦ p ⎝ ∂T ⎠p ˆ is a function of T only (not of p or V ˆ ) so that For the ideal gas U

⎛ ∂U ⎞ ⎛ ∂U ⎞ ⎜ ⎟ = ⎜ ⎟ = Cv ⎝ ∂T ⎠p ⎝ ∂T ⎠Vˆ ˆ ⎞ ⎛ ∂V ⎛ R ⎞ p ⎜ ⎟ = p ⎜ ⎟ = R ⎝ p ⎠ ⎝ ∂T ⎠ p

ΔU = ∫ Cvˆ dT = (3/2 R)(50) = 624 J 9.2.42

Use the same reference state, so that

ΔH = ΔU + Δ (pV) For one mole of an ideal gas: pV = RT

ΔU = ΔH – ΔRT

= 6.05 ×105

J 8.314 J 103g mol [(100 + 273) − (273)]K – kg mol ( g mol )( K ) 1 kg mol

= 6.05 × 105 – 8.314 × 105 = −2.26 ×105 J/kg mol 9–19

Solutions Chapter 9

9.2.43

Basis: 100 mol gas Comp. CO H2 CO2

mol 68 30 2 100

ˆ (J/g mol) relative to 0°C and 1 atm ΔH CO H2 CO2 25°C = 1400°C =

298 K 1500 K 1673 K 1750 K

728 39,576 45,723 48,459

718 36,994 42,724 45,275

912 62,676 72,896 77,445

Basis: 1000 m3 at 101 kPa and 1673 K

1000 m3 273 K 1 kg mol = 7.28 kg mol 1673 K 22.4 m3 Calculation of ΔH:

CO H2 CO2

kg mol

ˆ (kJ/kg mol) ΔH

7.28 (.68) 7.28 (.30) 7.28 (.02)

-44,995 -42,010 -71,984

9–20

ΔH(kJ)

-222,743 -91,750 -10,481

Solutions Chapter 9

9.2.44 K 423

°C 150.0

°F 302

353.3

80.1

176

278.7

5.5

42

253.0

-20.0

-4

vapor

↓

→ satd

satd vapor

liquid

↓

Satd liquid

→ satd solid

↓

solid

Basis: 1 g mol From Table D.1 in the Appendix. Benzene properties: Mol. wt.

78.11

Boiling point

353.26 K

Melting point

278.69 K

ΔHvaporization

30.76 kJ/g mol

ΔH fusion

984 kJ/g mol K 423 353.26 278.69 253

°C 150 80.11 5.54 -20

Heat Capacity Equation Coefficients: Benzene(g) Benzene(1)

a 74.06 62.55

b 32.95 × 10-2 23.9 × 10-2

c -25.20 × 10-5 -

d 77.57 × 10-9 -

The Cp equation is in °C hence the limits on integration should be in °C. ΔH1 =

∫

353

423

(74.06 + 32.95 ×10−2 T − 25.20 ×10−5 T 2 + 77.57 × 10−9 T3 ) dT = -11,790 J/g mol

9–21

T °C K

Solutions Chapter 9

Condensation

ΔH2 = −3.076 ×104 J/g mol Liquid

ΔH3 = ∫

278.69

353.26

(62.55 + 23.4 ×10−2 T)dT = 10,180J/g − mol

Fusion

ΔH4 = −9.84 ×103 J/g mol Solid From Perry, 5th edition:

Cp (cal)/(g)(oC) 0oC -50oC

0.375 0.299

12.3 97.7

If a linear relationship exists between Cp and T, -11,790

Cp,av = 119 J/(g mol) (o C) ΔHs = ∫

−20.0

5.5

(119)dT = 3034 − J/g mol

-30,760 -10,180

5

Overall, ΔH = ∑ ΔHi = 65, − 604 J/g mol i =1

-9,840 -3,034 -65,604

−65,604 g mol 1000 g = −8.40×105 J/kg g mol 78.11 kg

9–22

Cp (J/(g mol)(oC))

Solutions Chapter 9

9.2.45

Basis: liquid water at 32°F and 1 atm

ΔH = 3 ΔH 300°F,1

a. b.

atm

– ΔH 32°F,1

atm

= 3 (1192 – 0) = 3576 Btu

Basis: liquid water at 40°F and 60 psia Δ H 32 ° F,60 psia = 0

c.

Δ H = 3 (11192 – 0) =

3576 Btu

Basis: liquid water 40°F and 60 psia

ΔH = ΔH 300°F,60 psia – ΔH 40°F,60 psia = (1181.4 – 8.05) = 1173.4 Btu d.

Basis: water-steam mixture of 60% quality

ΔH, Btu/lb

State 1

60% steam - 40% water at 300°F; sat'd steam 1179.7

2

80% steam - 20% water at 300°F; water ΔH = ΔH 2 – ΔH 1 ΔH 2 = (0.80) (1179.7) + (0.20) (269.6) = 998 Btu/lb ΔH 1 = (0.60) (1179.7) + (0.40) (269.6) = 816 Btu/lb

Enthalpy change needed = ΔH ΔH = ΔH 2 – ΔH 1 = 182 Btu / lb e.

Basis: steam at 500°F and 120 psia ΔH = ΔH 2 – ΔH 1 = 312.5 – 1276.7 = –964.2 Btu / lb

f.

ΔH = ΔH 2 – ΔH 1 = 487.8 – 1276.7 = 788.9 Btu / lb

9–23

269.6

Solutions Chapter 9

Data from text ΔHA = 1205.0 ΔHB use tables in pocket

210o F 5 psia

200°F 1148.3

250°F 1171.1

7 psia

1146.7

1170.2

78.17

84.24

38.88

41.96

a. Enthalpy decrease

Vˆ A =1.0451 ft3 /lb

VB

(h)

40 psia and 267.24°F sat steam/sat liquid

70 psia/302°F liquid

b. volume goes up

70 psia/304°F superheated steam

depending on heat supply (i) 2.5 ft3 tank of water @ 160 psia and 363.5°F. @ 160 psia 6363.5 total volume is occupied by sat steam

2.5 Vˆ steam = 2.834 ft 3 / lb ⇒ m steam = = 0.88 lb 2.834

(

 water = 1-m steam Vˆ Liq/ water = 0.0182 ft 3 / lb ⇒ m V water =

)

2.5 3 × 0.0182 = 0.016 ft 2.834 9–24

Solutions Chapter 9 Vsteam = 2.5 – 0.016 = 2.49 ft 3 Of 5 lb H2 O assume x lb is sat steam (5 – x )lb is sat. water

x( 2.834) + ( 5 – x)(0.0182)

= 2.5

⇒ x(2.834 – 0.0182)

= 2.5 – 5(0.0182)

⇒

x

=

2.5 − .0910 = 0.86 lb 2.816 Yes, it is possible

(k)

Enthalpy of 10 lb steam @ 100 psia = 9000 Btu Enthalpy/lb steam @ 100 psia = 900 Btu ˆ ΔH steam @ 100 psia = 1187.3 Btu ˆ ΔH water @ 100 psia = 298.43 Btu Let x be fraction of steam (1–x) is fraction of water (1187.3)x + (1–x) 298.43 = 900 x(1187.3–298.43) = 900 – 298.43

x=

or 68%

601.57 = 0.68 888.87

9.2.46

a.

State 1

State 2

T = 400 K

T = 900 K

p = 100 kPa

p = 100 kPa

From the steam tables in SI units (interpolation required): (1) ΔH 1 = 2729.8 kJ/kg

(2) ΔH 2 = 3763.6 kJ/kg

9–25

Solutions Chapter 9

ΔH = 3763.6 – 2729.8 = 1033.8 kJ/kg 1033.8 kJ 18 kg 2 kg mol 4 = 3.7217 × 10 kJ/2kg mol kg kg mol b.

Using the table for the enthalpies of combustion gases:

(1) ΔH = 4284 J/g mol

(2) ΔH = 22, 760 J/g mol

ΔH = 22760 – 4284 = 1.848 ∞ 104 J/g mol = 1.848 ∞ 104 J/kg mol

1.1848 kJ 2 kg mol 4 = 3.695 × 10 kJ/2 kg mol kg mol c.

Using the heat capacity equation for steam: 627°C + 273

ΔH =

900

(33.46 + 0.6880 × 10

C p dT = 127° + 273

–2

T + 0.7604 × 10

–5

T

2

400

- 3.593 ∞10–9 T3) dT = 33.46 (900 – 400) +

+

0.7604 × 10 3

0.6880 × 10 2

–5

(900 3 – 400 3 ) –

–2

2 2 ( 900 – 400 )

3.593 × 10 4

–9

4

4

(900 – 400 )

= 20,085 J/g mol = 20,085 kJ/kg mol 4

20,085 × 10 kJ 2 kg mol = kg mol 4.017 ×104 kJ/2 kg mol The steam tables are the most accurate value.

9–26

Solutions Chapter 9

9.2.47

a-1:

ˆ ≅ 0.23 ft 3/lb V

a-2:

gas

b-1:

42°F

b-2:

mixture of liquid and vapor

9.2.48

Basis: 1 g mol CO2

ΔH = m∫

100o C

50o C

(a + bT + cT

2

+ dT3 )dT

b c d ⎡ ⎤ ΔH = m ⎢a(100 − 50) + (1002 − 502 ) + (1003 − 503 ) + (100 − 4 50 4 ) ⎥ 2 3 4 ⎣ ⎦ ⎡ 4.233 ×10−2 2.887 ×10−5 2 2 (100 − 50 ) − (1003 − 503 )+ ⎢36.11(100 − 50) + 2 3 ΔH = 1 g mol ⎢ −9 ⎢ 7.464 ×10 (1004 − 504 ) ⎢⎣ 4 ΔH = 1956 J

⎤ ⎥ ⎥ ⎥ ⎥⎦

From the CO2 chart ΔH ≅ 198-178 = 20 Btu/lb or 2040 J/g mol. From the enthalpy tables (interpolating) ΔH = 1957 J/g mol. The CD gives 2038 J/g mol. 9–27

Solutions Chapter 9

9.2.49

ˆ − ΔH ˆ ) = 10(86 − 288) = − ΔH = 10(ΔH 2020 Btu . The CD gives –1962 J. 2 1 9.2.50

Assume the temperature is 80ºF, and the propane is saturated liquid and vapor. The corresponding pressure is the saturation pressure, namely about 2.3 atm. Another temperature would correspond to another pressure. After 80% of the propane is used, the conditions are still saturated liquid-vapor mixture at 80ºF and 2.3 atm.

9.2.51

Basis: 10 lb mol ideal gas

(a)

Use pV = nRT ˆ = Δ(RT) Δ(pV)

3 359 (ft) 1 atm 500°R 3 (1) 10 lb mol = 36.5( ft ) @ 100 atm, 40°F lb mol 100 atm 492°R

900°R (2) 100 atm = 180 atm 500°R

(3)

ΔH = ( Hˆ 2 – Hˆ 1) 10 = 10 [300 + 8.0 (440) – (300 + 8.0 (40))] = 32, 000 Btu

(b)

ˆ = 0. The equation for H is based on the reference conditions where H

ˆ = 0 when 300 = -8T or Tref = -37.5 ºF = -38.5ºC. U should have the same H

reference temperature.

Δ Uˆ = Δ Hˆ – Δ(p Vˆ ) = Δ Hˆ – ΔRT for an ideal gas ˆ Change units for H:

First term:

300 Btu 1 lb mol 1055 J 697 J = lb mol 454 g mol Btu g mol

9–28

To F =1.8To C + 32

Solutions Chapter 9

Second term:

o 8.0 Btu 1055 J lb mol (1.8To C + 32) F J = (33.5To C +595) o (lb mol)( F) Btu 454 g mol g mol

ˆ H ref T = 0 ΔH = H T − H ref T = H T

ΔRT = (RT)T − (RT)refT

ΔU = 697 + 595 + 33.5 To C − R(TABS − TABS

(T is absolute)

) REF T

TABS =To C + 273 TASS1REF.T = − 38.5 + 273 ΔU = 1292 + 33.5 To C − 1.987 (To C + 38.5) = 1215 + 31.5 To C 9.2.52

Steady state flow process, no reaction

700 kPa Dry steam, 100 kPa and 125°C ˆ = 2725.5 kJ / kg ΔH

wet steam ˆ =? ΔH

The energy balances reduces to ∆H=0 for this process. At 700 kPa saturated (438.1K) ˆ = 696.7 kJ / kg ΔH L ˆ = 2763.1 kJ / kg ΔH v Let x = vapor fraction (1 – x)(696.7) + x( 2763.1) = 2725.4 x = 0.98 9.3.1

All of them.

9–29

Solutions Chapter 9

9.3.2

Closed unsteady state system Q + W = ΔU 60 + W = 220

W = 160 Btu (work done on the system)

9.3.3

(a)

pump pump

system is the pump. Open, steady state. Q Very little

W Yes

ΔU No

ΔH Yes(flow)

ΔPE No

ΔKE No

(b)

system is the pump and motor. Open, steady state A little

Yes

No

Yes

No

No

(c)

System is the ice. Closed, unsteady state. Yes

No

Yes

9–30

Yes

No

No

Solutions Chapter 9

(d)

∞

system is the mixer and solution. The process is closed, unsteady state.

Possibly Yes

Yes

Yes

Yes

No

No

9.3.4

a. System:

can and liquid

Q = 0, W ≠ 0

b. System:

motor

Q = 0, W ≠ 0 But motor may get hot.

c. System:

pipe and water

Q = 0, W ≠ 0

9.3.5

a.

boundary

Steam out Energy out

Water in Energy in

Q

9–31

Solutions Chapter 9

b.

Q=0

Steam in Energy in

Turbine

Steam out Energy out

Work c.

Work

Battery

Q=0 W ≠0

9.3.6

No. The energy balance is Q + W = ΔU, and for Q = 0 and since ΔU = 0 for an isothermal process W = 0 .

9.3.7

Closed unsteady state system Q + W = ΔU In kJ: (30-5) + 0.5 = Uf – 10 Uf = 35.5 kJ

9–32

Solutions Chapter 9

9.3.8

Closed unsteady state system Q + W = ΔU

Q=0

W = ΔU = CvΔT W=

(100 W)(5 hr) 1 kW 3.6 × 103 kJ = 1.8 ×103 kJ 1000 W 1 kW

mols of air = n =

pV (100 kPa)(100 m3 ) (kg mol)(K) = = 3.97 kg mol RT 303 K (8.314)(kPa)(m3 )

1.8 ×103 = 30(3.97)(T − 30)

T = 45o C

9.3.9

Closed unsteady state process Q + W = ΔU + Δ(PE) Ignore change in center of mass of the air Q = 0 assumed ΔU = 0 assumed (to get maximum elevation) W = 100 J 100 = Δ(PE) = mgΔh = (0.990 kg) (9.80 m/s2) Δh (m) Δh = 10.3 m

9–33

Solutions Chapter 9

9.3.10

Closed, unsteady state process

Initial conditions

Final conditions

V = 1ft saturated dry satd. steam (the basis)

V = 1ft (know) + V2

Look up

At p = 1 psia T = 200ºF. The water is vapor

ˆ = 33.610 ft3/lb V ˆ = 1145.72 Btu/lb H ˆ = 1073.96 Btu/lb U

From the steam tables: ˆ = 1150.15 Btu/lb H ˆ = 1077.48 Btu/lb U ˆ = 392.713 ft3/lb V

p = 11.548 psia Calculate the lb of steam m=

1 lb 1 ft 3 = 0.02975 lb 33.610 ft 3

Q + W = ΔU

W = 0 (fixed boundary)

a.

Q = ΔU b&c. d.

ΔU = U2 – U1 = (1077.48) (0.02975) – (1073.96) (0.02975) = 0.105 Btu = Q The volume of the second tank is Vfinal = (392.713) (0.02975) = 11.68 V2 = 11.68 − 1 10.68 = ft 3

9.3.11

No. For each path Q + W = ΔU, and for the cycle 1 to 2 and back ΔU = 0. Addition gives (QA + WA ) + (QB + WB ) = ΔU1−2 + ΔU 2−1 =0 (QA + QB ) = − (WA + WB )

Note: The equations in the problem used a different sign for W. A similar analysis applies to the second equation in the problem. 9–34

Solutions Chapter 9

9.3.12

A closed steady-state system. The cooling at the maximum efficiency is 0.695 kW 0.948 Btu 3600s = 2372 Btu/hr (1 kW)(s) 1 hr

not 5500 Btu/hr. The main question with the advertisement is: where does the energy go that is removed from the room if the unit does not require outside venting?

9.3.13

A closed steady-state system  +W  = ΔU  =0 Q  =−W  Q is negative (heat loss) Q 3 -3  = − 2050 × 10 J 1 hr 10 kW = − 0.57 kW Q hr 3600 s 1(J)(s)

 = 0.57 kW W

9.3.14

A closed steady-state system  +W  = ΔU  =0 Q

 =−W  Q

 = 0.25 hp 0.7068 ⎛ Btu ⎞ 60 s = 10.6 Btu/min W 1 hp ⎜⎝ s ⎟⎠ 1 min

 = − 10.6 Btu/min Q

9–35

Solutions Chapter 9

9.3.15

Open system, unsteady-state. The system is the volume containing 1 lb of H2O at 27ºC (liquid). At the end of the process the system contains nothing. The water leaving is at 100ºC because the atmospheric pressure is 760 mm Hg. Q + W = ΔH

W=0

ˆ (1) − H ˆ (0) Q = ΔH = H out in

The reference temperature for enthalpy is 0ºC but the water is at ˆ =H ˆ ˆ 27ºC, H −H out 100 C 27 C o

o

Q = (1 kg) (2675.6 – 111.7) kJ/kg = 2558 kJ

9.3.16

Steps 1, 2, 3, and 4: Figure P22.27 shows the known quantities. No reaction occurs. The process is clearly a flow process (open system). Assume that the entering velocity of the air is zero. Step 5: Basis: 100 kg of air = 1 hr Steps 6 and 7: Simplify the energy balance (only one component exists):

ˆ + KE ˆ + PE)m] ˆ ΔE = Q + W − Δ[(H (1) The process is in the steady state, hence ΔE = 0.

9–36

Solutions Chapter 9

(2) m1 = m2 = m.

ˆ (3) Δ(PE)(m) = 0. (4) Q = 0 by assumption (Q would be small even if the system were not insulated). (5) v1 = 0 (value is not known but would be small). The result is

 )m] = ΔH + ΔKE ˆ + KE W = Δ[(H We have one equation and one unknown, W. ΔKE and ΔH can be calculated, hence the problem has a unique solution. Steps 7, 8, and 9

ΔH =

(509 − 489)kJ 100 kg = 2000 kJ kg

ΔKE =

1 2

m(v22 − v12 )

3 1 kJ ⎛ 1 ⎞ 100 kg (60 m ) = ⎜ ⎟ = 180 kJ 2 1000(kg)(m 2 ) s ⎝ 2 ⎠ (s) 2

W = (2000 + 180) = 2180 kJ (Note: The positive sign indicates work is done on the air.) To convert to power (work/time),

kW=

2180 kJ 1 kW 1 hr = 0.61 kW 1 hr 1 kJ 3600 s s

9–37

Solutions Chapter 9

9.3.17

(a)

Steady state process assumed.

 + PE  )m ⎤ + Q + W ˆ + KE ΔE = − Δ ⎡(H ⎣ ⎦  ; ΔPE  =0 1. Ignore PE  is small so ΔKE  =0 2. ΔKE

3. No reaction 4. W = 0 5. (E2 − E1 ) = 0, steady state

ΔH = Q  + PE  )m] + Q + W ˆ + KE ΔE = − [(H

(b) 1. 2. 3. 4. 5.

 ; ΔPE  =0 Ignore PE No reaction

(E2 − E1 ) = 0; steady state

Q = 0 – assumed W=0  =0 6. ΔKE

ΔH = 0

9.3.18

System: gas

(

)

 + PE  m⎤+ Q + W ˆ + KE Et − Et = − Δ ⎡ H ⎥⎦ 2 1 ⎣⎢ a.

no reaction  + PE  )m ⎤ = 0, no flow ˆ + KE -Δ ⎡(H ⎣ ⎦

ΔKE and ΔPE of gas = 0 Ut 2 − Ut1 = Q + W

9–38

Solutions Chapter 9 b.

Here W = 0 Ut 2 − Ut1 = Q

c.

Here Q = 0, Ut 2 − Ut1 = W

d.

Ut 2 − Ut1 = Q + W

e.

Here, the system is the gas

Q = 0, W = 0

Ut 2 − Ut1 = 0

9.3.19

(a) System is the tank (b) Closed, unsteady state process

(c) 2 ft3 8 lb H2O 100oF

W

The 8 lb H 2O occupy ! # 0.129 ft 3 hence the rest # ## of space is vapor. " (d1) We will ignore the # initial vapor (see note # # #$ at end of solution)

Basis: 8 lb H2O

" Initial data for H O are (satd liquid) 2 $ o ˆ = 0.01613 ft 3 /lb (d) # V T = 100 F $ ˆ =67.97 Btu/lb if neglect vapor p = 0.9487 psia ΔH % Basis: 1 hr #The energy balance is % (e) $ ΔE = [ΔU + ΔPE + ΔKE]inside = Q + W − ΔH − ΔPE − ΔKE % no change in PE or KE inside no mass transfer & ΔU = W = ΔH − Δ(pV) = ΔH − VΔp (f) W =

0.25 hp 0.7068 Btu 3600s = + 636 Btu 1(hp) (s) 1 hr

9–39

Solutions Chapter 9

This proves to be negligible quantity ↓ 3 2 1 Btu ˆ (8) − 2 ft (pfinal − 0.9487) lbf 144 in (g1) 636 = ΔH final 2 2 in 1 ft 748(ft)(lbf )

# % % (h) $ % % & (g2)

ˆ , and ΔH ˆ Since pfinal ,V are in the tables, you final final ˆ , and V ˆ vapor given an assumed p have to get a ΔH final final final ˆ that satisfy the equation. ΔH will be composed of saturated final

ˆ (8 − m) + ΔH ˆ (m) = ΔH ˆ . m = lb vapor. (g3) liquid and vapor. ΔH L v final

ˆ (8 − m) V 2 ft 3 = V +ˆ V (m). L ˆ ,V ˆ , ΔH ˆ , ΔH ˆ , and p final are all related in the steam tables. V L v L v

_________________ ˆ of saturated vapor = 350.8 ft3/lb so that 1.87 ft3 represents 0.0053 lb, a negligible * V amount relative to 2 lb.

9.3.20

(a)

The system is the mixing point of the 1000°F “air” and water at 70°F

(b)

Open system

(c) and (d) H2O at 70oF H = 38.05 Btu/lb Air at 1000oF H = 6984 Btu/lb mol

Products (100 lb)

System H2O @ 400oF, H = 1201.2 Btu/lb air @ 400oF, H = 2576 Btu/lb mol

The enthalpies of water come from the steam tables. The properties of air come from the tables in Appendix D7 (with interpolation). The reference temperature for the enthalpy is 32°F.

9–40

Solutions Chapter 9

ΔE = [ΔU + ΔPE + ΔKE]inside = Q + W − ΔH − ΔPE − ΔKE   

(e)

• • • • Thus

0 because steady state Q is assumed to be 0 because time is short in transit W = 0 (none specified) ΔKE = ΔPE = 0 none with mass flow

ΔH = 0

or

ΔHout = ΔHin

(f) The units in the balance will be Btu and lb

( n air ) ΔHˆ air

ˆ (g) = ( n ) ΔH ˆ + ( n ) ΔH ˆ + ( n H 2O ) ΔH H 2O air air H 2O H2O

@ 400o F

@ 400o F

@ 1000o F

(e)

(a) 70o F

(100)(2576) (6984) + (mH2O )(1238.9) = (100) + m H2O (38.05) 29 29 9.3.21

 + PE  )m] + Q + W for all parts of this problem ˆ + KE a. (E 2 − E1 )= − Δ[H

 ; ΔPE  =0 1. Ignore PE  is small so ΔKE  = 0 (or it can be included) 2. ΔKE 3.

No reaction

4.

W=0

5.

(E2 – E1) = 0; steady state ΔH = Q

b.

 ; ΔPE  =0 1. Ignore PE 2.

No reaction

3.

W=0

4.

(E2 – E1) = 0; steady state

5.

Q=0 9–41

Solutions Chapter 9

 )m] = 0 ˆ + KE -Δ[H

c.

 ; ΔPE  =0 1. Ignore PE  =0 2. ΔKE

3.

No reaction

4.

(E2 – E1) = 0; steady state

5.

Q = 0 (assume) ΔH = W

d.

1.

 ΔPE  =0 Ignore PE;

2.

No reaction

3.

) (E2 – E1) = 0; steady state (may be not the KE

4.

Q = 0 (assumed)

 )m] = W ˆ + KE Δ[H

9–42

Solutions Chapter 9

9.3.22

General balance is

 + PE  )m] + Q + W ˆ + KE ΔE = [ΔU + ΔKE + ΔPE]inside = − Δ[(H (1)

(2)

(3)

(4) (5)

(6)

(7) (8)

Assume all reactants are in bomb at start of reaction. a. Q

1 2 3

4 5 6

ΔU retain because changes ΔKE delete - no change ΔPE delete - no change

7 8

Q retain W delete (fixed boundary)

⎫ ΔH ⎪ ⎬ ΔKE delete as no mass flow in or out ⎪ ΔPE ⎭ Result: ΔU = Q

b. Gases Fuel

W Q ΔH = Q - W

1 ⎫⎪ 2 ⎬ΔE = 0 because steady state 3 ⎪⎭ 4

ΔH

keep

5

delete as negligible

6

delete not a factor

7

Q

keep

8

W

keep (electric work)

Result: ΔH = Q + W 9–43

Solutions Chapter 9

c.

1 ⎫⎪ 2 ⎬ΔU = 0 3 ⎪⎭ 4

steady state

ΔH is kept Result: ΔH = 0

5

ΔKE delete negligible

6

ΔPE delete not applicable

7

Q = 0 insulated applicable

8

W=0

no work done

9.3.23

Unsteady state, closed, isothermal process at 25oC Basis: 1 kg CO2

p1 = 550 kPa T1 = 25°C

p2 = 3500 kPa T2 = 25°C

ˆ = 0.0125 m3 /kg V 2 ˆ =170 kJ/kg ΔH

ˆ = 0.10 m 3 /kg From CO 2 ⎧⎪V 1 ⎨ ˆ = 205 kJ/kg chart ⎪⎩ΔH

2

(converted to SI units) Energy balance

ΔE = Δ[(U + PE + KE)]inside = Q + W − Δ(H + PE + KE)

ΔU = Q + W = ΔH − Δ(pV) so Q = ΔH − Δ(pV) − W ΔH = (170 − 205)(1) = 35kJ − W = 4.106 kJ

⎡ 3500 kPa 0.0125 m 3 Δ(pV) = ⎢ − kg ⎣⎢

Q = (−35) − (−11.25) − 4.106 =

550 kPa 0.10 m ⎤ ⎥ kg ⎦⎥ 3

−27.86 kJ / kg CO2 (removed) 9–44

⎡ 3 N ⎤ ⎢10 m 2 1J ⎥ ⎢ ⎥ = −11.25 kJ ⎢ 1 kPa 1(N)(m) ⎥ ⎣ ⎦

Solutions Chapter 9

9.3.24

Pick the room as the system. The simplified form of the energy balance is (for no mass flow): ΔE = ΔU = Q + W = ΔU where Q = 0 (room is insulated) W = the electrical energy provided freezer through the system boundary and is positive. (No volume change occurs). Hence ΔU = CvΔT is positive and ΔT is positive; i.e. the temperature increases.

9.3.25

This is an unsteady state closed process Basis: 1 lb water Data from the CD Initial Conditions

Final Conditions

T = 327.8ºF

T = 327.8ºF

p = 100 psia ˆ = 0.018 ft 3 /lb for liquid and 4.435 for vapor V

9–45

ˆ = 4.435 V

Solutions Chapter 9

ˆ = 298.475 Btu/lb H

ˆ = 298.146 Btu/lb U

V = 4.435 ft3

V = 4.435 ft3

At the final conditions the water is saturated vapor H = 1187.48 Btu/lb U = 1105.46 Btu/lb T = 327.75 ºF p = 100 psia still ΔH = 1187.48 – 298.475 = 889.0 Btu/lb ΔU = 1105.46 – 298.146 = 807.3 Btu/lb The energy balance is ΔU = Q + W W = 0 (fixed container) Q = ΔU = 807.3 Btu/lb

9.3.26

Closed unsteady state system Use the CD to get data Basis: 4 kg steam Initial Conditions

Final Conditions

Given:

T = 500K p = 700 kPa

Given: Known:

Look up:

ˆ = 0.320m3/kg V ˆ = 2903.94 kJ/kg H ˆ = 2680.51 kJ/kg U

Look up (2 phase system) the quality (trial and error on the CD): x = 0.437

T = 400K ˆ = 0.320m3/kg V

ˆ = 1486.81 kJ/kg H

9–46

Solutions Chapter 9

Calculate V = 0.320(4) = 1.280 m3/kg ˆ +W ˆ = ΔU ˆ Q

ˆ = 1408.23 kJ/kg U

p = 245.77 kPa

ˆ =0 W

Q = (1408.23 – 2680.51)(4) = (-1272.3)(4) = −5,089 kJ ˆ = ΔH ˆ − (ΔpV) ˆ . If you do not use the CD, ΔU

9.3.27

A closed, unsteady state system comprised of 2 tanks. Get the steam properties from the CD. Basis: 1 lb steam. State 1

State 2

m = 1 lb p = 600 psia ˆ = 0.795 ft3/lb V T = 500oF (960oR) ˆ = 1128.19 Btu/lb U ˆ = 1215.97 Btu/lb H V1 = 0.795 ft3

m = 1 lb* p = 330.45 psia ˆ = 1.590 ft3/lb V T = 500oF (960oR)* ˆ = 1157.03 Btu/lb U ˆ = 1253.98 Btu/lb H V2 = 2V1 = 1590 ft3*

* used to get state 2 properties The closed system is Q + W = ΔU ΔU = 1157.03 – 1128.19 = 28.84 Btu/lb W = 0 (fixed boundary) Q = ΔU = 28.84 Btu/lb ΔH = 1253.98 – 1215.97 = 38.01 Btu/lb

9–47

Solutions Chapter 9

9.3.28

Basis: 1 lb H2O at 212ºF, 1 atm Data from the CD ˆ U ˆ H ˆ V

Initial (liquid) 180.182 Btu/lb 180.226 Btu/lb 0.017 ft3/lb

Final (vapor) 1077.40 Btu/lb 1150.29 Btu/lb 26.773 ft3/lb

p = 1.00 atm Non-flow process (unsteady-state) Q + W = ΔU ΔU = (1 lb)(1077.40 – 180.182) Btu/lb = 897.22 Btu W = − ∫ pdV = (1 atm)(26.773 – 0.017) ft3/lb (2.72) Btu/(atm)(ft3) − = -72.78 Btu Q = 897.22 – (-72.78) = 970.0 Btu

ΔH = 1150.29 − 180.226 =

970.0 Btu

Flow process (unsteady-state) ΔU = Q + W –ΔH Assume the volume of the system is fixed at V = 0.017 ft3 and that W = 0. Also assume that nothing is left in the system after the water evaporates. ˆ (0) − U ˆ ΔU = U final initial (1) = − 180.182 Btu

ˆ ˆ (0) = 1150.29 Btu ΔH = H (1) − H in out 1212o F Q = ΔU + ΔH = − 180.182 + 1150.29 = 970.1 Btu

9–48

Solutions Chapter 9

9.3.29

Treat the system (the cylinder) as an unsteady state flow system. ΔU = Q + W - ΔH W=0 Q=0 U t − U t = −(Hout − Hin ) 2

1

n f = final moles in cylinder

ˆ =U ˆ U t2 T 1 , 40 atm ˆ =U ˆ U t1

n i = initial moles in cylinder

298K,1 atm

ˆ =0 H out ˆ ˆ H =H in

298 K, 50 atm

ˆ = 0 . Ignore the effect of pressure on Let the reference temperature be TR that makes H R ˆ ˆ the values of U and H.

ˆ =H ˆ + C (T − T ) H in R p in R

ˆ =0 H R

in

ΔU

out

ˆ −n U ˆ ˆ Ut 2 − Ut1 = n f U t2 i t1 = (n f − n i )Hin − 0 n f [Cv (T − TR ) − RTR ] − n i [Cv (Tin − TR ) − RTR ] = (n f − n i ) ⎣⎡Cp (Tin − TR ) ⎦⎤ = (n f − n i ) [(Cv + R)(Tin − TR )]

Multiply all of the terms out to get (note terms with TR cancel):

CV = C P – R R 2 = for diatomic gas CV 5

n f Cv T = n f Cv Tin + n f RTin − n i RTin

Insert pV = nRT to get ni and nf The equations reduce to n f Cv T = n f Cv Tin + n f RTin − n i RTin T ⎡ R p T R ⎤ = ⎢1 + − i ⎥ Tin ⎣ Cv pf Ti Cv ⎦

9–49

Solutions Chapter 9

⎡ 2 T = ⎢1 + 298 ⎣ 5

⎛ 1 ⎞⎛ T ⎞⎛ 2 ⎞ ⎤ − ⎜ ⎟⎜ ⎟⎜ ⎟ ⎥ ⎝ 40 ⎠⎝ 298 ⎠⎝ 5 ⎠ ⎦

T = 417 K

9.3.30

The tank is an open unsteady state system. f stands for final, i for initial, in for in, m for mass in lb. Data:

at 291°F and 50 psia 1179.39 1099.60 8.653

ΔU = Q + W – ΔH

at 14.7 psia saturated (212°F)

ˆ Btu/lb H ˆ Btu/lb U ˆ ft3/lb V W=0

1150.26 1077.37 26.818

Q=0

ΔU = -ΔH

ˆ −m U ˆ ˆ nf U f i i = (m f − m i )H in − 0 V = 50 ft 3 so mf =

mi =

50 = 5.78 lb 8.653 50 = 1.86 lb 26.818

ˆ ) − 1.86(1077.37) = (5.78 − 1.86) 1179.39 5.78(U f ˆ = 1146.56 Btu/lb at 50 psia U f Tf = 411o F

(superheated)

9–50

Hout = 0

Solutions Chapter 9

9.3.31

The general energy balance is

ˆ + KE ˆ + PE) ˆ m] ΔE = Q + W − Δ[(H (1) (a)

(2) (3)

(4)

(5) (6)

(1)

ΔE = 0

steady state flow process, one overall system (You can also pick two system, one the H2O and one the benzene).

(2)

Q is the net overall heat transfer to and from the heat exchanger with the surroundings. It is essentially 0 if exchanger is well insulated. If you pick 2 systems, Q is the heat transfer between the benzene and the H2O, and is not 0.

(b)

(3)

W=0

no mechanical work in the process

(4)

ΔH ≠ 0

for all streams

(5)

ΔKE = 0 negligible KE change

(6) (1)

ΔPE = 0 ΔE = 0

level exchanger assumed. steady state process

(2)

Q

not zero as isothermal

9–51

Q = ΔH

Solutions Chapter 9

(3) (4)

W ΔH ≠ 0

retain the pressure changes even if T is constant

(5)

ΔKE = 0 pipe diameter remains the same, and the flow rate is the same

(6)

ΔPE = 0

level pump lines

Q + W = ΔH 9.3.32

Unknowns

Relationships (R)

F P

(1) F = P + V (2) 0.05F = xPP ˆ + SΔH = VH + PH ˆ (3) FH F S V P  (4) SΔH = Q

 V

S  Q

S

 = UA(T − T ) (5) Q S V

TV XP Yes, two measurements must be made to balance the unknowns and the relationships. If x P and TV are measured: solve (R5) for Q . (R4) for S substitute (R2) and (R1) in (R3)  . Note that measuring and solve for F. Use (R2) to calculate P and (R1) to calculate V TV and S is not satisfactory because this combination eliminates either (R4) or (R5) as an independent relationship. Also note that if you neglect the vapor pressure of the organic, you can set Tv = 212ºF.

9–52

Solutions Chapter 9

9.3.33

Basis: 1 hour

Input conditions: 500o F ⎫ ⎬ h1 = 1264.7 Btu/lb 250 psi ⎭

Exit conditions: ⎫ h SV = 1150.4 Btu/lb ⎬ 15% liquid ⎭ h SL =180.07 Btu/lb 14.7 psi

ˆ = 0.15(180.07) + 0.85(1150.4) H 2

= 1004 Btu/lb The simplified energy balance is: ˆ =Q+W mΔH

ˆ = 1004 − 1265 = − 261 Btu/lb ΔH

ˆ = −261,000 Btu mΔH W=−

86.5 hp 1 hr

1 Btu = −2.20 × 105 Btu 2.93×10-4 (hp)(hr)

Thus, Q = mΔHˆ − W = − 2.61×105 + 2.2×105 = − 4.1×104 Btu hence, not adiabatic.

9–53

Solutions Chapter 9

9.3.34

Basis: 1 lb air

The energy balance is in the steady state. The system is the value.  + PE  )m ⎤ = Q + W ˆ + KE Δ ⎡(H ⎣ ⎦

if ΔKE ≅ 0

W=0

ΔPE = 0

Q ; 0

Hence ΔH = 0 ΔH = ∫ CpdT = 0 (a)

ΔT = 0 or

T2 = 250 K

Since ΔH = 0 (the kinetic energy term was negligible), the steady-state constraint requires that the mass flow rate be constant:  = ΔvS = Constant m

where: ρ = density v = velocity of the gas S = cross-sectional area of pipe

Since the diameter of the pipe does not change and density is a function of temperature and pressure, the gas velocity is also a function temperature and pressure. vSTP =

100 lb Air 1 lb mol 359 ft 3 1 hr 1 = 7.0 ft/s 1 hr 29 lb Air 1 lb mol 3600 s π(1.5/12 ft)2

Therefore, the velocity downstream of the valve is: v Outlet =

7.0 ft 250 K 1 atm = 3.2 ft/s s 273 K 2 atm

9–54

Solutions Chapter 9

9.3.35

Steady-state process. Basis: 1 second ˆ )=Q  +W ˆ −H   (H m out in

ˆ use the steam tables or a Cp equation. Using constant Cp is not the most To get H in accurate method, but for liquids for small temperature changes, it is a good approximation.  +W   )(Tout − Tin ) = Q (m)(C p

Because the tank is well-mixed, the outlet water temperature is the same as the temperature inside the tank T. T = Tout  (C p ) (T − Tin ) = 100 (20 − T) + 300 m

 = m

1 kg 1 min 1 kg = min 60 s 60 s

Cp = 4180 J/(kg)(K) 4180 (T − 40) = 100 (20 − T) + 300 60

T = 30.0o C

9–55

Solutions Chapter 9

9.3.36

Assumptions 1.

This is a steady-flow process since there is no change with time at any point, hence ΔE = 0.

2.

Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point.

3.

The kinetic and potential energy changes are negligible ΔKE ≅ ΔPE ≅ 0.

4.

Constant specific heats at room temperature can be used for air.

The simplified energy balance is ˆ )  +W ˆ −H  = ΔH  = m(H Q out in  = − 200W Q  = 15 kW W

Assume that with little error ΔH = CpΔT

with Cp = 1.00 kJ/(kg)(ºC)

Otherwise you will need tables or a data base for the properties of air. The ideal gas law gives the specific volume of air at the inlet of the duct 3 ˆ = RTm = [(0.287 kPa)(m )/(kg)(K)](290 K) = 0.832 m 3 /kg V 1 pin 100 kPa

9–56

Solutions Chapter 9

The mass flow rate of the air through the duct is determined from  = m

 V 150 m 3 /min ⎛ 1 min ⎞ 1 = = 3.0 kg/s ˆ 0.832 m 3 /kg ⎜⎝ 60 s ⎟⎠ V 1

Substituting the known quantities, the exit temperature of the air is determined to be (15 kJ/s) − (0.2 kJ/s) = (3 kg/s)[1.00 kJ/(kg)(oC)](Tout − 17)oC Tout = 21.9o C

9.3.37

The procedure is correct only for an open, steady state system. (a)

If the evaporation occurs from an open, unsteady-state system of fixed volume: ΔU = Q + W –ΔH

W=0

ΔU = Q – ΔH Data: ˆ = 419.5 kJ/kg U liquid

ˆ = 2506.0 kJ/kg U vapor

ˆ = 419.5 kJ/kg H liquid

ˆ = 2675.6 kJ/kg H vapor

Assume only 1 kg of water is in the system at t = initial. At t = final, 0 kg of water are in the system. U final = 0 Uinitial = (1 kg) (419.5 kJ/kg) = 419.5 kJ Hout = (1 kg)(2675.6 kJ/kg) = 2675.6 kJ H in = 0 -419.5 = Q – (2675.6-0) Q = 2256.1 kJ (b) If the evaporation occurs in a steady-state open system, m in is in kg and min − 1 = mout . The process is isothermal.

9–57

Solutions Chapter 9

ΔU = 0

W=0

OUT

IN

ˆ ˆ ˆ ⎤ Q = ΔH = ⎡⎣(m1 − 1)H liquid + (1)H vapor ⎦ − m1Hin ˆ =H ˆ H in liquid

Q = (1)(2675.6 − 419.5) 2256.1 = kJ The answer in the solution is ok, but note that the equation ΔU = Q − pΔV refers to a closed system, not an open one, in which the vessel expands against the atmosphere. This viewpoint is ok, but an unlikely procedure.

9.3.38

Open, unsteady state system – the adiabatic turbine  =0 Q

 =0 ΔU

 =0 ΔPE

 = ΔH  + ΔKE  W

Data from interpolating in the SI steam tables At 600ºC and 100 kPa 3697.9 0.4011

At 400ºC and 100 kPa 3278.2 3.103

ˆ H(kJ/kg) ˆ 3 /kg) V(m

The velocities at the inlet and outlet to the turbine are calculated from the volumetric ˆ flow rate = m V=

d2 v where d is the pipe diameter. Therefore, 4

⎛ kg ⎞ ⎛ m3 ⎞ (4) ⎜ 2.5 ⎟ ⎜ 0.4011 ˆ s ⎠⎝ kg ⎟⎠ ⎝  1V 4m m 1 v1 = = = 127.7 , 2 2 s (3.14159)(0.1 m) d in

9–58

Solutions Chapter 9

and

v2 =

ˆ  2V 4m m 2 = 158.0 2 s d out

⎡ ⎛ 1⎞ ⎛ 1 ⎞⎤ ΔKE = ⎜ ⎟ (2.5) ⎢158.02 − 127.7 2 ) ⎜ ⎥ kJ/s = 10.82 kJ/s ⎝ 2⎠ ⎝ 100 ⎟⎠ ⎦ ⎣

 = (2.5)(3278.2 − 3697.9) kJ/s = − 1049.25 kJ/s ΔH

 = − 1049.25 + 10.82 = 1038.4 kJ/s W (1038.4 kW)

9.3.39

This is an unsteady-state closed process. The energy balance reduces to

ΔE = ΔU + ΔPE + ΔKE = Q + W Q = ΔU = U2 − U =1

ΔPE = ΔKE = 0

ΔH − Δ(pV) = ΔH − (p2V2 − p1V2 )

where 2 (327.8ºF) is the final state and 1 (406ºF) is the initial state Get the data from the CD in the back of the book. The final state is saturated ˆ and H ˆ are in Btu/lb and V ˆ is ft3/lb. conditions. U

ˆ = 381.65 H 1,L ˆ U1,L = 380.72 ˆ = 0.019 V 1,L

ˆ = 1202.2 H 1,V ˆ U1,V = 1116.7 ˆ = 1.743 V

ˆ = 298.5 H 2,L ˆ U 2,L = 298.2 ˆ = 0.0177 V

1,V

2,L

ˆ = 1187.5 H 2,V ˆ U 2,V = 1105.5 ˆ = 4.433 V 2,V

From the steam tables, get the initial amount of steam and use as the basis: ⎛ 1 ⎞ 100 ⎜ ⎟ = 57.4 lb ignoring the liquid volume. (If you do not, you have to get the ⎝ 1.743 ⎠ average volume) and get the final state. Let y = the fraction of liquid at 100 psia 3 3 ˆ ˆ V liquid = 0.0177 ft / lb, Vvapor = 4.433 ft / lb

Basis: 1 lb water

y(0.0177) + (1 − y)(4.433) = 1.74 y = 0.6095 lb (liquid) x = 0.3905 lb(vapor) 9–59

Solutions Chapter 9

ˆ from the CD to calculate Q. Otherwise you have to calculate Δ(pV). Use values of U

Q = [(0.6095)(298.2) + (0.3905)(1187.5)]− [(1116.7)(1)] Q = −471 =

Btu or Q −2.71×104 Btu lb

9.3.40

This is an unsteady state process in a closed system, or a closed system with air and surroundings with water. From the latter viewpoint the system is the air, and ΔU = Q + W where Q is the heat transfer to the water from the air.

Q

5 lb H2 O

air

W (positive) For the system of the air

Q + W = ΔE = ΔU + ΔP + ΔK = ΔHΔ − (pV) For the water:

Q = ΔH – Δ( pV ) but Δ ( pV ) is negligible for liquid water so that T +2.3

Q = ΔH = 5

∫ C pdT

= 5 Cp (T + 2.3 – T ) for constant C p .

T

(or use the steam tables) C p = 8.0

Btu (lb mol )(° F)

Btu 2.3°F 1 lb mol H2 O 5 lb H2 O = 5.11 Btu (lb mol ) 18 lb H2 O Q is removed from the air, so for the system comprised of the air, Q = –5.11 Btu. Q = 8.0

9–60

Solutions Chapter 9

For the system of the air W=

12, 500(ft )(lb f ) 0.0012854 Btu = 16.07 Btu 1(ft )(lb f )

W is positive when the surroundings do work on the system of the air. 3

Basis: 3 ft air (not essential) ∆U = –5.11 Btu + 16.07 Btu = 10.96 Btu

9.3.41

Steps 1, 2, 3 and 4 Get the needed properties from the CO2 chart or tables (Be sure to use the same reference ˆ is the specific value). Let m CO2 be the mass of CO2 in the cylinder at the final state, V

ˆ ˆ volume, U is the specific internal energy of the CO2 at the end of the filling, and H final m is the specific enthalpy of the CO2 in the pipeline.

ˆ ≅ 153 Btu / lb and V ≅ 0.57 ft 3 / lb At 200 psia, 40°F, H Step 5 Basis: 3 ft3 of CO2 @ 200 psia (final conditions) System: The cylinder (open system, unsteady state) Steps 6 and 7:

ˆ ,m ,T Unknowns: V CO2 CO2 CO2 Equations: energy balance, enthalpy chart

ˆ chart to fix the conditions in the cylinder. The energy We need two points on the p − H balance reduces to (W = 0, Q assumed to be 0) ΔU = −ΔH

which gives one point:

9–61

Solutions Chapter 9

ˆ ) − m (U ˆ ˆ ˆ ) mCO2 (U −(H final in itial initial ) = min (Hin ) mout out

minitial = 0

mout = 0

m CO2 = m in

ˆ ˆ U final = Hin = 153 Btu / lb ˆ and The other point is the pressure p = 200 psia. Unfortunately the CO2 chart is p − H ˆ , and calculate ˆ , thus requiring a trial and error solution. Assume a T, lookup V not p − U ˆ =H ˆ − pV ˆ (ignore pV ˆ at the reference sate of –40oF). U ˆ in a ˆ = 153Btu / lb , then you have determined T. If you used tables of p vs U When U handbook, the calculations would be easier. ˆ ˆ ˆ H p V U Assume T = 160oF. T = 140 F. o

ˆ − pV ˆ = Then H ˆ − pV ˆ = Then H

The calculates are quite approximate but

T = 140 F o

mCO2

3 ft 3 = = 4.3 lb 0.70 ft 3 / lb

9–62

182-(200)(0.68)(0.185) = 157 178-(200)(0.70)(0.185) = 153

Solutions Chapter 9

9.3.42

Basis: 1 lb steam Closed system; hence Q + W = ∆E = ΔU

A

expand

130 psia, 600°F

75 psia, 600°F

const. T

B cool, const. V

adiabatic compression

60 psia

Q=0

C Basis: 1 lb steam

ˆ. Note: The CD will result in slightly different values for ΔH A

B

ˆ = 1329.6 – 1326.1 = 3.5 Btu / lb : ΔH

Use the steam tables or the CD directly for U, or calculate

ˆ = ΔHΔ ˆ − (pV)= ˆ ΔU

= 3.5 –

75 lb f 8.319 ft 3 130 lb f 4.760 ft 3 – 2 lb 2 in in lb = 3.5 – .95 = 2.5 Btu / lb

B

C :

ˆ =? Q

144 in ft

2

2

1 Btu 778 (ft) (lb f)

ˆ =? W

ˆ – pdV but pV relation is not known. W= ∫

W = 0 (at constant volume)

ˆ = 1232.7 – 1329.6 = –96.9 Btu / lb ΔH 2

144 in 1 Btu ˆ = – 96.9 –60 8.319 – 75 8.319 = –73.8 Btu/lb ΔU 2 778 (ft) (lb ) f ft ˆ = –73.8 Btu / lb ˆ = ΔU Q 9–63

Solutions Chapter 9

C

A :

( )

ˆ = 0 so W ˆ = ΔU=ΔH–Δ ˆ ˆ ˆ Q pV

ˆ = 1326.1 – 1232.7 = 93.4 Btu/lb ΔH Δ pV = 130 4.760 – 60 8.319 144

ˆ = 93.4 – 22.15 = 71.3 Btu/lb W

778

= 22.15 Btu/lb

ΔU = 93.4 – 22.15 = 71.3 Btu/lb

9.3.43

To make the process a steady state open process from (1) to (3) assume the system is the coil plus the compressor. Then

0

0 0 ˆ )m + Q + W] ΔE = –Δ[(Hˆ + Pˆ + K

or

W = ∆H – Q

Basis: 1 lb water evaporated ˆ @ (1): (0.98) (1143.3) + (0.02) (161.17) = 1123.6 Btu/lb Calculate Δ H

From (1) to (2): Balance on the compressor

ˆ – ΔH ˆ = 1235.2 – 1126.6 = 111.6 Btu/lb ΔH 2 1 9–64

Qˆ = –6 Btu / lb

Solutions Chapter 9

W = 111.6 –(–6) = 117.6 Btu/lb From (2) to (3): Balance on the coil ˆ – ΔH ˆ = 168 – 1235.2 = – 1067.2 Btu/lb ΔH 3 2

a.

Q

= 1067.2 = 9.1 W 117.6

ˆ = 0 so Q ˆ = ΔH ˆ = -1067.2 Btu/lb W

(Note: the problem asks for the heat transfer out as a + value)

⎡ 38.46 ft 3 0.98 (0.016)ft 3 0.02 ⎤ lb steam V1 = ⎢ + = 37.69 ft 3/lb ⎥ lb lb 1067.2 Btu ⎣ ⎦ Basis: 1,000,000 Btu/hr b.

106 Btu 1hr 1 lb steam 37.69 ft 3 = 589 ft 3 / min hr 60 min 1067.2 Btu lb steam

9–65

Solutions Chapter 9

9.3.44

Basis: 25 kg exit fluid

Q + W = ΔH

W=0

Q = ΔH = ΔH25 –ΔH15 –ΔH10 =ΔH25 – ( 2784.4 )(15) – (82.2 )(10 ) –50 J 25 × 103 g = –1250 kJ g exit ⎧ ΔHˆ L = 798.5 kJ / kg For the exit fluid ⎨ ˆ ⎩ ΔH = 2784.4 kJ / kg

ˆ ~ 10 J ) ( note ΔKE

Q=

–4

V

Let x = kg vapor, 25 – x = kg liquid –1250 =

x kg 2784.4 kJ (25-x) kg 798.5 kJ 15 kg 2784.4 kJ 10 kg 82.2 kJ – – + kg kg kg kg

– 1250 = 2784.4 x + 19,963 – 798.5 x – 41,766 – 822 x = 10.76

fraction vapor =

10.76 = 0.43 25.0

9–66

Solutions Chapter 9

9.3.45

Closed unsteady state process for each stage. Q + W = ΔU overall Calculate the moles of gas: a.

n=

pV (1 atm)(0.387 ft 3 )(lb mol)(o R) = = 1×10-3 lb mol RT (530o R)(0.7302)(ft 3 )(atm)

Known values p (atm) T (ºF) n (lb mol) V

A 1 170 (630ºR) 1 × 10-3 ?

B

C

1 70 (530ºR) 1 × 10-3 0.387

10 70 (530ºR) 1 × 10-3 ?

(Basis) D 10 823 (1283ºR) 1 × 10-3 ?

For an ideal gas, C v = 5/2 R (has to be looked up or calculated) and U and H are functions of T only Also, Δ(pV) = Δ (nRT) Calculate some of the missing values

b.

VA =

(1×10−3 )(0.7302)(630) = 0.460 ft 3 1

VC =

(1×10−3 )(0.7302)(530) = 0.0387 ft 3 10

VD =

(1×10−3 )(0.7302)(1283) = 0.0937ft 3 10

The work for each stage will be assumed to be ∫ pdV (ideal process) WAB (constant pressure process)

c.

W = − ∫ pdV = − p AB (VB − VA ) =−

1 atm (0.387 − 0.460)ft 3 2.72 Btu = 0.199 Btu (atm)(ft 3 )

9–67

Solutions Chapter 9

WBC (isothermal process)

d.

0.0387

W = − ∫=pdV − ∫=

0.387

0.0387 dV nRT dV −nRT =− ∫ 0.387 V V

= −(1×10 = −3 )(1.987(530)(ln 0.1)

nRT [ln V ]0.387

0.0387

2.42 Btu

WCD (constant pressure process)

e.

W = − ∫ pdV = − pCD (VD − V = c ) −(10)(0.0937 − 0.0387) = − 0.550 Btu

WDA (ideal adiabatic process) W = − ∫ pdV = − ∫

nRT dV However, both T and V vary. V

Instead use Q + W = ΔU

Q=O

R = 1.987 Btu/(lb mol) (ºR)

f.

⎛ 5R ⎞ W = ΔU = nC v ΔT = (10−3 ) ⎜ ⎟ ( 630 − 1283) = − 3.243 Btu ⎝ 2 ⎠

g.

Woverall = 0.199 + 0.891 – 0.550 – 3.243 = – 1.74 Btu

h. i.

ΔH = 0 (a cycle)

For the overall process Q + W = ΔU. Because of the cycle, ΔU = 0, and thus Q = -W = 1.74 Btu

9–68

Solutions Chapter 9

9.3.46

exit stock T2 = ?

Cooling water out 60oC W1

C 27.5 kg vapor 100 kg C 9.1% H2O 90.9% Toluene 150oC

Condenser

F

F liquid 90oC

P

Cooler

C 50,000 kg/day 20oC Cp = 2.1 kJ/(kg)(oC)

W Cooling water in 20oC

Basis: 1 day

27.5 kg F 50,000 kg C 1 day = 13,750 kg F 100.0 kg C day 13,750 kg F 9.1 kg H 2O = 1,251.25 kg H 2O 100 kg F 13,750 kg F – 1251.25 kg H2O = 12,498.75 kg toluene. Energy balance at condenser: The energy balance reduces to ΔH = 0 ΔHC + ΔHF = 0 T2

100

m C ∫ C p dT + m H 2 O ⎡ ∫ C p dT – ΔHvap 20 ⎣ 150 111

+ m tol ⎡ ∫ C p dT – ΔH vap ⎣ 150 50, 000 kg

2. 1 kJ

at 111°C

+∫

90

111

liquid 40oC

at 100°C

+∫

C p dT⎤ = 0 ⎦

T 2 – 20°C

(kg) (°C)

9–69

90

100

Cp dT ⎤ ⎦

Solutions Chapter 9

+ 1, 251.25 kg H 2O

+ 12, 498.75 kg tol

2.1 kJ (100 – 150)°C (kg) (°C) 1.3 kJ (111 – 150) °C (kg)(°C)

–

–

2260 kJ 4.2 kJ (90 – 100) °C + kg (kg) (°C)

230 kJ 1.7 kJ (90 – 111)°C =0 + kg (kg) (°C)

105,000 T2 – 2,100,000 – 3,011,758.75 – 3,954,604.5 = 0 (a)

T2 =

9,066,363.25 = 86.3o C 105,000

Enthalpy balance at cooler: ΔH = 0 out in (ΔHΔ W − H W ) − (ΔHΔ p − HF ) = 0

mW

+

60 40 40 m W ∫ C p dT – ⎡ m H 2 O ∫ C pdT – m td ∫ C dT⎤ = 0 ⎣ 20 86.34 86.34 p ⎦ 4.2 kJ (60 – 20 )°C 1,251.25 kg H2 O 4.2 kJ (40 – 90)°C = (kg)(°C) (kg)(°C)

12498.75 kg tol 1.7 kJ ( 40 – 90 )°C ( kg)(°C) m W 168 kJ – 262,762 kJ – 1,062,394 kJ kg

mW =

1,325,156 kJ = 7,887 kg H 2O 168 kJ/kg

9–70

Solutions Chapter 9

9.3.47

Basis: 1 lb water/steam Steady-state, open system for each unit

The data: ˆ H ˆ and quality or superheats at the numbered points: Tabulation of p, T, V, p (psia) T (oF) ˆ 3 / lb) V(ft qual./sup.heat

1 14.7 65 0.026 0%

2 250 65 0.016 0%

3 250 401 1.844 100%

4 250 550 2.202 149º

5 40 268 10.14 96.5%

6 0.306 65 0.016 0%

ˆ H(Btu / lb)

33.09

33.79

1201.1

1291

1137

33.05

Calculate the specific volume at point 5:

ˆ = 0.965(10.506) + 0.035(0.01715) = 10.14 V (a)

Heat to boiler:

Q = ΔH = H3 − H2 = 1201.1 − 33.8 = 1167.3 Btu / lb (b)

Heat to superheater:

Q = ΔH = H4 − H3 = 1291 − 1201.1 = 90 Btu / lb (c)

Heat removed in condenser = H6 − H5 = 33 −1137 = -1104 Btu/lb

(d)

Work delivered by turbine: Q + W = ΔH

W = ΔH = H5 − H4 = 1137 − 1291 =−

Q=0 (e)

154Btu / lb

Work required by the pump between 1 and 2: Q + W = ΔH

Q=0

hence W = ΔH

Because we do not have values of ΔH for compressed water at 65ºF, we will use W = ΔU + Δ(pV)

ΔU = ∫

65o F

65o F

Cv dT = 0

9–71

Solutions Chapter 9

Δ(pV) = p 2 V2 − p1V1 = (p 2 − p1 )V = Efficiency =

(250 − 14.7) 0.016 144 = 0.70 Btu / lb 778

Net work delivered 154 − 0.70 = Total heat Supplied 1167.3 + 90

= 0.121 For a water rate of 2000 lb/hr: hp =

154 Btu 2000 lb steam 1 hr 1.415 hp = 121 lb hr 3600s 1 Btu/s

You can improve the efficiency by: (a) Exhausting the turbine at a lower pressure. (b) Use higher boiler pressure. (c) Use higher superheat. 9.3.48

Basis: 100 lb/day liquid C12 feed at 8ºF Assume (1) Adiabatic operation of all units (2) Flow process, steady-state

R 2.5 T/day, 0oF

R' ) P D ) 100 T/day, -30oF

F 100 T/day, 8oF

First find the enthalpy lost or gained by the C12 in the heat exchanger from which we can get the work done by compressor A on the Freon. Overall C12 balance through the heat exchanger F+R=P 100 + 2.5 = 102.5

9–72

Solutions Chapter 9

Energy balance for C12 through the heat exchanger For a flow process:

Q + W = ΔH W=0 Q = ΔH

Choose as the reference state liquid C12 at -30ºF

ΔH = H Pout − H Fin − H R in =

8.1 Btu lb mol 102.5 lb ⎡⎣ −30 − (−30)o F⎤⎦ o (lb mol)( F) 71 lb 100 lb F

−

8.1 Btu lb mol 100 lb ⎡⎣8 − (−30)o F⎤⎦ o (lb mol)( F) 71 lb 100 lb F

−

8.1 Btu lb mol 2.5 lb Btu ⎡0 − (−30)o F⎤⎦ = 0 − 433.5 − 8.55 = −442.1 o ⎣ (lb mol)( F) 71 lb 100 lb F 100 lb F

Consequently

Q = −442.1 Btu /100 lb F This means heat is lost by the C12 and gained by the Freon Energy balance for the Freon flowing through compressor A: Q + W = ΔH But since the flow of Freon is cyclical ΔH = 0

Q Freon = −Q =C12

−(−442.1) = 442.1 Btu /100 lb F

WA = QFreon = 442.1

Btu 100 lb F

Energy balance for the C12 flowing through the compressor B: 2.5 lb/day @ -30ºF → 2.5 lb/day at 0ºF Q + W = ΔH Q = 0 9–73

Solutions Chapter 9 ΔH = Enthalpy change across compressor = H R − H R '

Other units W = 539 Btu/mole = 1,520,000 Btu/day ΔH =

+

H R = CpΔT + HΔvap

2.5 lb 8.1 [0 − (−30)] 100 lb F 71

4878 cal 1.8 Btu / lb 1 lb mol 2.5 lb g mol cal / g mol 71 lb 100 lb F

= 8.55 + 309 = 317.5

Btu 100 lb F

Total work input needed to make the process operational:

hp =

317.5 + 442.1 Btu 2000 lb 100 ton F 778(ft)(lbf ) 100 lb F ton day Btu

1 day (hp)(min) (24) × (60) min 33,000(ft)(lb f ) 0.30

= 82.9 horsepower (actual power input)

9–74

Solutions Chapter 9

9.3.49

Basis: 1 day; Reference temp = 80ºF

a. Material and Energy Balances: Overall Material: In = 1,000,000 lb Out : residue: 500,000 lb naptha: 200,000 lb gasoline: 300,000 lb Total out 1,000,000 lb Tower 1: Material feed reflux total

In 1,000,000 1,500,000 2,500,000

Out 2,000,000 500,000 2,500,000

vapor residue total

Energy: In: feed: ( 1×106 )(0.53)(480 − 90) 106 (100) + +(106 )(0.45)(500 − 480) = 6 reflux: (1.5 ×10 )(0.59)(180 − 90) = steam: (by difference) total:

315,000,000 Btu 79,700,000 Btu 114,500,000 Btu 509,700,000 Btu

Out: residue: (5 ×105 )(0.51)(480 − 90) vapor: (2 ×106 )(111) liquid: (2 ×106 )(0.59)(250 − 90) total (Btu):

99,200,000 222,000,000 188,500,000 509,700,000

= = = =

Furnace: Material

In 1,000,000 lb Energy In 6 feed: (10 )(0.53)(200 − 90) = 58,300,000 furnace ht: 257,200,000 total: (Btu) 315,500,000 Condenser I: Material In: vapor H2O: 304,500,000/(1)(120-70) 9–75

Out 1,000,000 lb Out 315,500,000

= =

2.0 ×106 lb 6.1×106 lb

Solutions Chapter 9

Out: liquid:

2.0 ×106 lb

Energy H2 O In: vapor: 222 ×106 + 188.5 × 106 Out: liquid: (2 ×106 )(0.59)(180 − 90) Removed by water Total out: Preheater I: Material In: liquid = 500 ×106 lb ; Out: vapor Energy In: liquid: (5 ×105 )(0.59)(180 − 90) steam: Total Out: vapor:

(5 ×105 )(0.59)(250 − 90) +(5 ×105 )(111) + (5 ×105 )(0.51)(200 − 250)

Tower II Material In: feed = 500,000 lb Out: reflux = 600,000 lb Total = 1,100,000 lb

= = = =

6.1×106 lb 410.5 ×106 Btu 106 ×106 Btu 304.5 ×106 Btu

=

410.5 ×106 Btu

=

500 ×106 lb

= =

26.5 ×106 Btu 78.65 ×106 Btu

=

105.15 ×106 Btu

=

105.15 ×106 Btu

vapor = residue = total =

Energy In: feed: reflux: (6 ×105 )(0.63)(120 − 90) steam: (by difference) Total: (Btu):

= = = =

900,000 200,000 1,100,000 105,150,000 11,300,000 42,700,000 159,150,000

Out: vapor: (9 ×105 )(0.63)(150 − 90) + 9 105 (118) =×140.05 ×106 residue: (2 ×105 )(0.58)(255 − 90) = 19.1× 106 Total (Btu): Condenser II Material In: vapor: H2O: Out:

liquid

= 159.15 ×106

= 0.9 ×106 lb 123.0.5 ×106 (1)(110 − 70)

= 3.08 ×106 lb = 0.9 ×106 lb 9–76

Solutions Chapter 9

H2 O

= 3.08 ×106 lb

Energy In: (Btu) Out: liquid: (9 ×105 )(0.63)(120 − 90) removed by H2O: Total (Btu):

= = = =

140.05 ×106 17 ×106 123.05 ×106 140.05 ×106

Heat Exchanger II (Assume Tout = 90ºF) Material In liquid: 300,000 lb H2O: 5,670,000/(1)(80-70) = 567,000 lb Energy In: (300,000)(0.63)(120-90) Out: liquid: removed by H2O

liquid: H2O:

= 5.67 ×106 Btu = 0.00 = 5.67 ×106 Btu

Heat Exchanger III Material In = Out: Bottoms 200,000 lb Charge 1,000,000 lb Energy In: (200,000)(0.58)(255-90) (1 × 106)(0.53)(90-90)

= =

19.15 ×106 0.0

Total (Btu): Out: (2 × 105)(0.58)(26.3-90) (1 × 106)(0.53)(140-90) Total (Btu):

= = = =

19.15 ×106 −7.35 ×106 26.5 ×106 19.15 × 106

Tout of 26.3ºF does not seem reasonable Heat Exchanger IV Material In = Out Bottoms 0.5 ×106 lb Charge 1.0 ×106 lb

9–77

Out 300,000 567,000

Solutions Chapter 9

Energy In: (5 × 105)(0.51)(480-90) (1 × 106)(0.53)(140-90) Total (Btu): Out: (5 × 105)(0.51)(257-90) (1 × 106)(0.53)(200-90) Total (Btu):

= =

99.3 ×106 26.5 ×106

=

125.8 ×106

= = =

67.5 ×106 58.3 ×106 125.8 × 106

Heat load of furnace = 315.5 × 106 - 58.3 × 106 = 257.2 ×106 Btu b.

Additional heat if charge to furnace is at 90ºF Total heat load = 315.5 × 106 Btu Additional heat = 315.5 × 106 – 257.2 × 106 = 58.3 ×106 Btu

9.3.50

Basis: 1 hour

Open steady-state process

Deaerator Balance

Material Balance

F1 + F2 + 50,000 = 105,000

F2 = 105,000 − 50,000 − F1

Energy Balance

F1 (1164.1) + F2 (48.0) + 50,000(218.9) = 105,000(218.9) F1 (1164.1) + (55,000 − F1 )(48.0)= 55,000(218.9)

a.

F2 = 8, 422 lb / hr steam

b.

F2 = 46,578 lb / hr make-up feedwater

Boiler Feedwater Pump 3 2 ⎛ ⎞ ⎛ ⎞ ⎛ (hp)(s) ⎞ ⎛ hr ⎞ ˆ = ⎛ 105,000 lb ⎞ 0.017 ft ⎛ 800 − 30 lbf ⎞ 144 in.  VΔp m 2 2 ⎜⎝ ⎟ ⎟⎠ ⎜⎝ 550(ft)(lb ) ⎟⎠ ⎜⎝ 3600s ⎟⎠ lb ⎟⎠ ⎜⎝ hr ⎠ ⎜⎝ in ⎟⎠ ⎜⎝ ft f = 100 hp 9–78

Solutions Chapter 9

c.

For 55% efficiency: 181.7 hp

d.

⎛ kW ⎞ (181.7 hp) ⎜ 0.7457 ⎟ = 135.5 kW hp ⎝ ⎠

e.

(135.5kW)(8760

f.

⎛ 1 ⎞⎛ 1 ⎞⎛ 0.7457 ⎞ Savings = (105, 000)(0.017)(200)(144) ⎜ ⎟⎜ ⎟⎜ ⎟ (8760 )( (0.05) ) ⎝ 550 ⎠⎝ 3600 ⎠⎝ 0.55 ⎠

hr )($0.05 / kWhr) = $59,360 / year yr

= $15, 420 / year Steam Drum

(100,000)(1230.7) + (5000)(471.7) − 105,000(218.9) = 102.4 106 Btu×/ hr g.

Steam Drum Heat Input = 102.4 ×106 Btu / hr

h.

Superheater = (100,000)(1351.8-1230.7) = 12.1×106 Btu / hr

Flash Drum

x = vapor flowrate

(100,000 − x)(218.93) + x(1164.1) = 100,000(330.65) x=

11,820 lb/hr

=

vapor flowrate

and

88,180 lb/hr

=

liquid flowrate

i.

Steam Lost =11,820 – 8,422 = 3,398 lb / hr

j.

Condensate Lost = 88,180 – 50,000 = 38,180 lb / hr 9–79

Solutions Chapter 9

9.3.51

Basis: one hour operation open, steady-state process. A material balance initially is necessary to determine amounts and compositions of streams. A.

Feed Component C2H6 C2H4

B.

Wt.% 65 35

lb 650 350

Product Have 97% recovery of ethylene lb C2H4 = (0.97) (350) = 340 Now, since the product is 98% C2H4

340 lb C2 H 4 2 lb C2 H6 = 6.95 lb C2 H6 98 lb C2 H 4 Component C2H6

Wt.% 2

lb 6.95 9–80

Solutions Chapter 9 C2H4 C.

98

340

Bottoms lb lb

C2H4 = 350 – 340 = 10 C2H6 = 650 – 6.95 = 643.05

Component

Wt.%

lb

C2H6 C2H4

98.47 1.53

643.05 10

We are now ready to determine, by a series of energy balances, the quantities desired. Energy Balance on System “A” (see Fig.)

⎧Energy out with Product + ⎪ Energy in with Feed + energy in with steam = ⎨Energy out with bottoms + ⎪Energy out with refrigerant ⎩ We will assign an arbitrary enthalpy of zero to the liquid feed as it enters the Pre-heater at –100ºF. Actually, any reference condition could be chosen since we are only concerned with enthalpy changes in the balances. ------------------------------------------------------------The feed undergoes at 20ºF rise before it enters the still. T = -80ºF and Cp feed = (0.65) (0.65) + (0.35) (0.55) = 0.62 Btu/(lb) (ºF) Enthalpy in with feed = (lb feed) (CpΔT) = (1000) (0.62) (-80-(-100)) = 12,400 Btu/hr ------------------------------------------------------------Enthalpy in with steam = (lb steam) (ΔHv) at 30 psig, ΔHv = 945 Btu/lb Enthalpy in with steam = (S) (945) btu/hr ------------------------------------------------------------From the data given, the temperature of the saturated liquid product is =30ºF since it is essentially pure ethylene. Cp = 0.55 Btu/(lb) (ºF) 9–81

Solutions Chapter 9

Enthalpy out with product = (346.95) (0.55) [-30-(-100)] = 13,360 Btu/hr The bottoms are practically pure ethane so, T = 10ºF, Cp = 0.65 Btu/(lb) (ºF) Enthalpy out with bottoms = (653.05) (0.65) [10-(-100)] = 46,700 Btu/hr ------------------------------------------------------------The refrigerant experiences a temperature rise of 25°F as it passes through the condenser. The heat capacity may be assumed to be 1.0 Btu/(lb) (°F) Enthalpy out with refrigerant = (lb refrig.) (1.0) (25) m Btu/hr Now, rewriting the energy balance “A” with a substitution of known terms: 12,400 + 945S = 13,360 + 46,700 + 25m

(1)

It is evident from this first balance that a second balance will be necessary to determine the unknown quantities. Energy Balance “B” Energy in with Overhead =

Energy out with refrigerant + Energy out with Product + Energy out with Reflux

Since we are using a reflux ration of 6.1 lb reflux = (6.1) (346.95) = 2,120 lb/hr. Now, the specific enthalpy of the reflux = the specific enthalpy of the product, since they are from the same stream, and the differences in enthalpy between the overhead and the product, or overhead and reflux, is merely the heat of vaporization of ethylene (neglecting the small amount of C2H6 present), therefore, we can rewrite Energy Balance “B” as: (lb Overhead) (ΔHv) = 25 m (2,120 + 346.95) (135) = 25 m or m = 13,320 lb/hr. Now substituting this value into equation (1) we have 12,400 + 945 S = 13,360 + 46,700 + (25) (13,320) 9–82

Solutions Chapter 9 so that S = 403 lb / hr a.

403 lb steam hr = 0.403 lb steam / lb feed hr 1000 lb feed

b.

13,320 lb refrigerant 1 ft 3 refrig. 7.48 gallons = 1,993 gallons of refrigerant/hr hr 50 lb refrig. ft 3

The bottoms leave the still and enter the Pre-heater at approximately 10ºF, the saturation temperature of pure C2H6. The final temperature of the bottoms as it leaves the Preheater can be calculated from a simple energy balance around the Pre-heater. c.

Enthalpy Balance “C” (lb bottoms) [0.65 Btu/(lb) (°F)] (10 – Tf) = (lb feed)

×[0.62 Btu /(lb)(o F)][−80 − (−100)] (653.05) (0.54) (10 – Tf) = (1000) (0.62) (20)

Tf = 19.2o F

9–83

Solutions Chapter 10

10.1.1

(b), (d), (f)

10.1.2

(a)

10.1.3

(c)

10.1.4

The ΔH o f is 0 by definition.

10.1.5

ˆ ° of Fe O is calculated as follows. ΔH f 2 3 (1)

ˆ ° – Σn ΔH ˆ° ΔHRxn = Σni ΔH fi i fi products

reactants

° ˆ ° (2 ) – ΔHˆ ° –822.200 kJ = ΔHˆ fFe O – Δ H fFe fO 2 3

2

( 32 )

ˆ° = ΔH fFe O – 0 – 0 2 3

(2)

ΔHRxn = –284.100 kJ

ˆ° ˆ° ˆ° = ΔH fFe O – Δ H fFeO (2) – ΔH f O 2 3

2

( 12 )

ˆ ° (2) – 0( 1 ) –284.100 = –822.200 – Δ H fFeO 2

ˆ ΔH fFeO = °

–538.10 kJ = –269.05 kJ 2 vs – 267 in Appendix F

10–1

Solutions Chapter 10

10.1.6

ΔH°(kJ) +123.8 +2220.0

C3H8 (g) → C3H6 (g) + H 2 (g) 3CO2 (g) + 4H2O(1) → C3H8 (g) + 5O2 (g) 4 × (−285.8) 4[H 2 (g) + 1/ 2O2 (g) → H 2O(1)] 3 × (−393.4) 3[C(s) + O2 (g) → CO2 (g)] ________________________________________________ +20.1 kJ/g mol

3C(s) + 3H2 (g) → C3H6 (g)

ΔHof of C3H6 = 20.1 kJ/g mol

10.1.7

Basis: 1 g mol C5H2 ΔH (kJ/g mol)

5 CO2 (g) + H 2O(l )

→

add : 5[C(s) + O 2 (g)] → 1

1

C5 H 2 (s)+ 5 O 2 (g)

2110.5

5[CO 2 (g)]

5[-394.1]

2

add: H 2 (g)+ O 2 (g)

→

H 2O(g)

-241.826

add: H 2O(g)

→

H 2 O(l )

-43.911

net: 5C(s)+H 2 (g)

→

C5 H 2 (s)

2

ΔH o f = -143.737 kJ/g mol

10.1.8

(a) (b) (c)

NH4 (l ) : -67.20 kJ/g mol from Appendix F Formaldehyde gas (H2CO): -115.89 kJ/g mol from Appendix F Acetaldehyde liquid (CH3CHO): For gas ΔH o f = 166.4 − kJ / g mol from Appendix F The heat of vaporization at 293.2 K is 25.732 kJ/g mol from the CD. The value is close enough to ΔHvap (298 K) to use. ΔHof = -166.4 + 25.732 = -140.7 kJ/g mol

10–2

Solutions Chapter 10

10.2.1

a.

Basis: 1 g mol NH3 ( g)

NH3 ( g) ˆ ° (kJ / g mol ) : ΔH f ° ΔHrxn =

b.

HCl( g)

+

–46.191

ˆ – ∑ ni ΔH i

products

→

–92.311

–315.4

ˆ ∑ ni ΔH i

reac tants

=

(1)( –315.4) – [(1)( –46.191) + (1)( –92.311)]

=

–176.9 kJ / g mol NH3 ( g)

Basis: 1 g mol CH4 (g )

CH4 (g ) + 2 O2 (g ) → CO2 (g) ˆ (kJ / g mol ) : ΔH f ° ΔHrxn =

c.

NH4 Cl (s)

–74.84

ˆ – ∑ ni ΔH i

products

0

+ 2 H 2 O ( )

–393.51

–285.840

ˆ ∑ ni ΔH i

reac tants

=

⎣⎡( 2)( –285.840) + (1)( –393.51)⎦⎤ – ⎣⎡(1)( –74.84) + 0⎤⎦

=

–890.4 kJ / g mol CH4 ( g )

Basis: 1 g mol C6 H 2 (g )

C6 H1 2(g ) ˆ ° (kJ / g mol ) : ΔH f

–123.1

→

C6 H 6 ( ) +48.66

+

3H2 (g) 0

ΔH°rxn = [ 0 + (1)(+48.66)] – [(1)(–123.1)] = 171.76 kJ / g mol C 6 H12 (g)

10–3

Solutions Chapter 10

10.2.2

a. Basis: 1 g mol CO2 (g) CO2 (g) + H2 (g) → CO (g) + H2O (1) ΔH°rxn = (– 110.52 – 285.84) – (– 393.51) = –2.849 kJ b.

Basis: 1 g mol CaO (s) ΔH°rxn = (–986.59 – 924.66) = [(–635.55 – 601.83) - 2 (285.84)] =

−102.19 kJ per mol CaO(s) Multiply by 2 for the reaction listed −204.38 kJ c.

Basis: 1 g mol Na2SO4 (s) ΔH°rxn = (–1090.35 – 110.54) – (–1384.49) = 183.59 kJ

d.

Basis: 1 g mol NaCl (s) ΔH°rxn = (–92.31 – 1126.33) – (–811.32 – 411.01) = 3.67 kJ

e.

Basis: 1 g mol O2 ΔH°rxn = [2 (–1384.5) + 4 (–92.31)] – [(–411.00) + 2 (–296.90) + 2 (–285.84)] = –1561.76

f.

Basis: 1 g mol SO2 (g) ΔH°rxn = (–811.32) – (–285.84 – 296.90) = –228.58 kJ

g.

Basis: 1 g mol N2 ΔHºrxn = 2 (90.374) = 180.75 kJ

h.

Basis: 1 g mol Na2CO3 (s) ΔH°rxn = [3 (–1117.13) + (–393.51)] – [–1130.94 + 2 (–373.21) + 4 (–296.90)] = –1867.54 kJ 10–4

Solutions Chapter 10

i.

Basis: 1 g mol CS2 (1) ΔH°rxn = (–1394.69 – 60.25) – (+87.86) = –287.61 kJ

j.

Basis: 1 g mol C2H2 (g) ΔH°rxn = (+105.02) – (52.28 – 92.31) = –145.05 kJ

k.

Basis: 1 g mol CH3OH (g) ΔH°rxn = (–115.90 – 241.82) – (–201.25) = –156.48 kJ

l.

Basis: 1 g mol C2H2 (g) ΔH°rxn = (–140.7) – (226.75 – 285.84) = −81.61 kJ

m.

Basis: 1 g mol C4H10 (g) ΔH°rxn = (52.28 – 84.67) – (–124.73) = 92.34 kJ

10.2.3

You want to get ΔH for 1

1

2

2

C3H8 (g) + Br2 (g) → C3H 7 Br(g) +

ΔH

H 2 (g)

(J)

(–b)

C3 H8 ( g) → C3 H6 ( g) + H2 (g )

+126,000

(+a)

HBr( g) + C3 H6 (g) → C3 H7 Br

–84,441

(+c)

H2 (g ) + Br( g) → 2HBr( g)

–36,233

(–d)

Br (g) → Br (l )

–15,355 Total

10–5

−10, 029 J / g mol C3H8

Solutions Chapter 10

10.2.4

Basis: 1 g mol Fe S2

ΔH o rxn = 567.4 − kJ / g mol FeS2 ( − 567.4) kJ 1 g mol FeS2 1000 g kJ = −4728.3 g mol FeS2 120 g FeS2 1 kg kg FeS2 Note: The information about conversion does not affect the value of the standard heat of reaction. It will affect the values in the energy balance.

10.2.5

G+E → ← E+F

ˆ ° is at 25°C, 1 atm ΔH Rxn

(

)

ˆ ° = ΣΔ H ˆ ° – ΣΔH ˆ ° = 1.040 × 10 9 – 0.990 × 109 J / g mol G ΔH Rxn f f prod.

react.

6

= 50 × 10 J / g mol G converted Since 0.48 reacts, ΔH°Rxn = 0.48( 50) × 10 6 J / g mol G

= 24 × 106 J / g mol G fed (not asked for)

10.2.6

No. The value reported is the heat transfer for constant volume, Qv. The heating value at constant pressure should be reported for the standard heat of reaction (an enthalpy change). Assume the reaction is at 25ºC.

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l ) Basis: 1 g mol CH4 Q = ΔU = ΔH – Δ(pV) If the gases can be treated as ideal Δ(pV) = Δ(nRT) = Δn (RT). Δn = 1 – 2 – 1 = -2 10–6

Solutions Chapter 10

The correction is ΔnRΤ = -2 (8.314) (298) = -4958 J/g mol In SI units of 273.15 K and 101.3 kPa, the volume is 22.415 m3 g mol. The correction is (ignoring the pressure change from 1000 kPa to p2)

−4958 = 221.2J / m3 or 0.22 kJ / m3 22.415 Otherwise you have to calculate p2 V − p1V where p2 = 1000 kPa and p1 = p*H2O + pCO2 . The value to be reported should be 39.97 – 0.22 = 39.75 kJ / m3

10.2.7

Yes. Calculate the sensible heats using the tables and using a common reference temperature.

10.2.8

C63H132O15 tristearin

3 C18 H36O2 + stearic and

3 C3H8O3 glycerol

ˆ o = −[(3)(−964.3) + (3)(−159.16)] − (−3820)(1) = − 449.6 kJ/g mol ΔH rxn

10.2.9

a.

If the exit temperature is > 25oC, and the dilutent is at the entering temperature of the reactants, then less heat has to be removed from the process.

b.

Remains the same as the presumption is in calculating ΔHºrxn that stoichiometric quantities react to completion.

c.

ˆ o = ΔH ˆ o with H O(g) = 241.826 At 25ºC, ΔH − kJ / g mol H2O. rxn f 2

At 500 K, the sensible heats have to be considered, and ½ O2 and H2 together have a larger ΔH than does H2O so that ΔHrxn at 500 K is less than ΔHrxn at 25ºC.

10–7

Solutions Chapter 10

10.2.10

S (s) + O2 (g)

SO2 (g)

ΔHRxn 600K (327oC) ΔHReact.

ΔHProd. 298K (25oC) ΔHoRxn

Basis: 1 g mol S(s) =1 g mol SO2 (g)

ΔHrxn, 600 K = ΔHorxn + ∑ ΔHprod, 600 K − ∑ ΔHreact, 600 K ΔHo rxn, 298 = ∑ ΔHo f − ∑ ΔHo f prod

react

= [−296.90) − [0 + 0] = −296.90 kJ / g mol S = − 296,900 J/g mol S

∑ ΔH

prod,600 K

ΔHSO2 ,600 K = ∫

327 25

(38.91 + 3.904 ×10−2 T − 3.104 ×10−5 T 2 + 8.606 10×−9 T3 )dT

= 13,490 J/g mol

∑ ΔH

(13,530 from Enthalpy Tables)

react,600 K

This assumes s is really a liquid at 600ºC, not a solid as in the equation

ΔHS,600K = ΔH 298−386 + ΔH fusion + H 386−600 Δ solid

=∫

386 298

liquid

(15.2 + 2.68 ×10−2 T)dT + 10, 000 + 1/ 2∫

600 386

(35.90 + 1.26 10−3×T)dT ½ (S from Perry) 2

= 2144 + 10,000 + 3975

10–8

Solutions Chapter 10

= 16,119 J/g mol

ΔHO2 ,600K = ∫

327 25

(29.10 + 1.158 ×10−2 T − 0.6076 ×10−5 T 2 + 1.311 10 × −9 T3 )dT

= 9,340 J/g mol (8899 from the enthalpy tables)

ΔH rxn,600K = −296,900 + (13, 490 − 25, 429) = 308,840 J / g mol S

10.2.11

C3H8 (g) → C2 H2 (g) + CH4 (g) + H 2 (g) Basis: 1 g mol C3H8

ΔH500 = ΔH25 + ∑ ΔHprod − ∑ ΔHreact ΔH o 25 = 226.75 + (−74.84) + 0 − (−103.85) = 255.76 kJ / g mol

ΔH = ∫ CpdT = aT + b / 2 T2 + c / 3 T3 + d / 4 T4 ΔHC3H8 = 68.032T + 1.130 ×10−1 T2 − 4.37 ×10−5 T3 + 7.928× 10−9 T4

500 25

= 55.53 kJ / g mol

ΔHC2H2 = 42.43 T + 3.027 ×10−2 T2 − 1.678 ×10−5 T3 + 4.55 ×10−9 T4

500 25

= 25.89 kJ / g mol

ΔHCH4 = 34.31 T + 2.735 ×10−2 T2 + 1.220 ×10−6 T3 + 2.75 ×10−9 T4

500 25

= 23.10 kJ / g mol

ΔHH2 = 28.84 T + 3.825 ×10−5 T2 + 1.096 ×10−6 T3 + 2.175 × 10−10 T4

500 25

= 13.83 kJ / g mol

ΔH500 = (255.76 + 25.89 + 23.10 + 13.83) − 55.53= 10–9

263.05 kJ / g mol

Solutions Chapter 10

10.2.12

All data is from Tables E.1 and F.1 of the Appendix. a.

o CH 3OH(g) + 1/ 2O 2 (g) 200 C H 2CO(g) + H 2O(g)

Basis: 1 g mol CH3OH

Comp.

Cp (J /(g mol)(K or oC))

T

ˆ o (kJ / g mol) ΔH f

CH3OH(g)

42.39 + 8.301× 10−2 T − 1.87 × 10−5 T 2 − 8.03 × 10−9 T 3

ºC

-201.25

O2(g)

29.10 + 1.158 × 10−2 T − 0.6076 × 10−5 T 2 + 1.311 ×10−9 T 3 ºC

H2CO(g)

34.28 + 4.268 × 10−2 T − 8.694 ×10−9 T 3

H2O(g)

33.46 + 0.6880 × 10−2 T +0.7604 × 10−5 T 2 − 3.593 × 10−9 T 3 ºC

ºC

0 -115.89 -241.826

25ºC = 298 K 200ºC = 473 K

ΔHo rxn ,25o C = [−115.89 + (−241.826)] − (−201.25) = 156.47 −kJ / g mol

= −156, 470 J / g mol Products Add the Cp equations for the two products and integrate (each involves 1 mole).

∑ ΔH prod,200o C = ∫ =

200 25

(67.74 + 4.95 ×10−2 T + 0.7604 ×10−5 T 2 − 1.228 ×10−8 T 3 )dT

12,850 J/g mol

Reactants

ΔHCH3OH = ∫

200

25

=

(42.39 + 8.301×10−2 T − 1.87 ×10−5 T 2 − 8.03 ×10−9 T3 )dT

9000 J/g mol

10–10

Solutions Chapter 10

ΔHO2 =

1 200 (29.10 + 1.158 ×10−2 T − 0.6076 ×10 −5 T 2 + 1.311× 10 −9 T 3 )dT ∫ 25 2 = 2,650

ΔHrxn,200o C = −156, 470 + 12,850 − 2,650

146, = 270 J / g mol

o SO 2 (g) + 1/ 2 O 2 (g)300 C SO3 (g)

b.

Basis: 1 g mol SO2 DATA: Comp.

Cp (J /(g mol)(K or oC))

O2(g)

29.10 + 1.158 × 10−2 T − 0.6076 × 10−5 T 2 + 1.311 ×10−9 T 3 ºC

0

SO2(g)

38.91 + 3.904 × 10−2 T − 3.104 × 10−5 T 2 + 8.606 × 10−9 T 3 ºC

-296.90

SO3

48.50 + 9.188 × 10−2 T − 8.540 ×10 −5 T 2 + 32.40 ×10 −9 T 3 ºC

-395.18

T

ˆ o (kJ / g mol) ΔH f

ΔHo rxn ,25o C = −395.18 + 296.90 = −98.28 kJ / g mol = 98, 280 −J / g mol Products

∑ ΔH prod,300o C = ∫

300 25

(48.50 + 9.188 ×10−2 T − 8.540 ×10−5 T 2 + × 32.40 10−9 T3 )dT

= 16, 740 J/g mol Reactants

ΔHSO2 = ∫

300 25

(38.91 + 3.904 ×10−2 T − 3.104 ×10−5 T 2 + 8.606× 10−9 T3 )dT

= 12, 180 J/g mol

ΔH O2 = 8, 470 J / g mol 1 ΔH rxn,300o C = −98, 280 + 16, 740 − 12,180 − (8, 470) = 97,960J − / g mol SO2 2

10–11

Solutions Chapter 10

10.2.13

Basis: 1 g mol SnO

SnO + 1/ 2O2 → SnO2 ΔHrxn25 = −577.8 − (−283.3 + 0) = 294.5 − kJ / g mol ΔH90 = ΔH rxn, 25 + ∑ ΔH prod − ∑ ΔH react C ΔH = ∫ 90 C dT 25o C p o

ΔHSnO = 39.33T + 7.575 ×10−3 T=2

363

2.88 kJ

298

ΔHSnO2 = 73.89 T + 5.02 ×10−3 T 2 +

2.16 ×104 T

363 298

= 2.89 kJ

ΔHO2 = 29.1 T + 5.79 ×10−3 T 2 − 2.025 ×10−6 T3 + 3.278 ×10−10=T 4

90 25

1.93 kJ

ΔH90 = −294.5 + 2.89 − (2.88 + 0.5(1.93)) = 295.46 − kJ / g mol

10.2.14

Basis: 100 g mol CO Assume pressure is 1 atm, and an open, steady state process.

CO +

1 2

O 2 → CO 2

The energy balance reduces to Q = ΔH. The reference temperature is 25°C.

100 g mol

CO

CO2 400oC

300oC 100 g mol O2 100oC

10–12

Solutions Chapter 10

Data: Sensible heats ˆ from CD : ΔH(kJ / g mol)* ΔH(kJ)

mol

o

T( C)

ˆ o (kJ / g mol) ΔH f

O2 CO

100 100

100 300

-110.541

2.235 8.294

223.5 -10,225

Out O2 CO2

50 100

400 400

-393.505

11.715 16.245

585.8 -37,726

Comp. In

ˆ using the tables in the appendix, but the process requires *You can calculate ΔH sensible interpolation. Q = ΔH = (−37,726 + 585.8) − (−10, 225 + 223.5) = − 27,139 kJ

10.2.15 Basis: 1 g mol yeast The heat of formation is

3.92 C + 6.5 H + 1.94 O2 → C3.92 H6.5 O1.94 The heat of combustion is for the reaction

C3.92 H6.5 O1.94 + 5.60 O2 → 3.92 CO2 (g) + 3.25 H2 O(l ) ˆ 0 − ∑ n ΔH ˆ 0 ⎤ −1,518 kJ = − ⎡⎣ ∑ n i ΔH f ,i i f ,i ⎦ product

reactants

ˆ0 ⎤ −1,518 = − ⎣⎡(3.92)(−393.51) + (3.25)(−285.840) − 5.60(0) − (1)ΔH f ,yeast ⎦

ˆ0 ΔH − kJ/g mol f ,yeast = 963.54 The molecular weight of C3.92 H 6.5O1.94 is 84.63. Per 100 g yeast

ˆ0 ΔH 1139−kJ/100 g yeast f ,yeast =

10–13

Solutions Chapter 10

10.2.16

The problem requires more information to solve. To make the solution easier but approximate, assume the reaction takes place at 25°C and 1 atm. Assume that the energy balance reduces to Q = ∆H, and that Q represents the desired kJ. Data: ˆ° ˆ° ΔH f , H 2 O = –285.840 ΔH f, CO2 = –393.51 , both in kJ/g mol. Ref. temp. = 25°C. Basis: 1 g mol glucose. MW of glucose = 180.

(C6 H12O6 )glucose + 6O2 (g) → 6H 2O(l ) + 6CO2 (g) ΔH° =ΔH°products –ΔH°reactants = [6( –285.840) + 6(–393.51)] – [1( –1260) + 6( 0)] = –2816 kJ

0.90 ( –2816 ) = 14.0 kJ/g glucose 180 At 37°C (body temperature), assume only the O2 is used in a reaction

6 g mol O2 22.4L at SC 310K 1 g mol suc. = 0.85 L/g at 37o C and 1 atm 1 g mol suc. g mol O2 273K 180g suc.

10.2.17

The reaction is

C3H8 (l ) + 5O2 (g) → 3CO2 (g)+ 4H 2O(l ) Need to calculate the heat of formation of C3H8 (l ) and H2O (l ) first

C3H8 (l ) ˆ of ,C3H8 ( l ) = ΔH ˆ ofC H8(g) − ΔH ˆ o vap(25o C) ΔH 3

= −24.820 − 3.820 =−

28.643 k cal / g mol

H2O(l ) ˆ o f ,H2O( l ) = ΔH ˆ o f ,H 2O(g) − ΔH ˆ o vap(25o C) ΔH 10–14

Solutions Chapter 10

= −57.798 − 10.519 = 68.317 − k cal / g mol Basis: 1 g mol C3H8 (l )

ΔHRxn,C3H8 ( l ) = 4(−68.317) + 3(−94.052) − (−28.643)

-526.781 = k cal/g mol

10.2.18

a. Presumably H2, O2, C1, HC1, and H2 are gases. (1)

6 g mol FeC13(-403.34 kJ/g mol FeC13) + 9 g mol H2O (-241.826 kJ/g mol H2O) -3 g mol Fe2O3 (-822.156 kJ/g mol Fe2O3)-18 g mol HC1(-92.312 kJ/g mol HC1) = −468.39 kJ

(2)

6 g mol FeC12(-342.67 kJ/g mol FeC12) + 3 g mol C12 (0.0 kJ/g mol C12) -6 g mol FeC13 (-403.34 kJ/g mol FeC13) = 364.02 kJ

(3)

2 g mol Fe3O4 (-1116.7 kJ/g mol Fe3O4) + 12 g mol HC1 (-92.312 kJ/g mol HC1) + 2 g mol H2 (0.0 kJ/g mol H2) – 6 g mol FeC12 (-342.67 kJ/g mol FeC12) -8 g mol H2O (-241.826 kJ/g mol H2O) = 649.48 kJ

(4)

3 g mol Fe2O3 (-822.156 kJ/g mol Fe2O3) – 2 g mol Fe3O4 (-1116.7 kJ/g mol Fe3O4) – ½ g mol O2 (0.0 kJ/g mol O2) = −233.07 kJ

(5)

6 g mol HC1 (-92.312 kJ/g mol HC1) + 3/2 g mol O2 (0.0 kJ/g mol O2) -3 g mol H2O (-241.826 kJ/g mol H2O) – 3 C12 (0.0 kJ/g mol C12) = 171.61 kJ

b. 2H2O

2H2 + O2

∑ ΔH

o n,i

= 483.65 kJ

10–15

Solutions Chapter 10

10.2.19

Basis: 1 g gasoline

ΔH =

40 kJ 0.84 = 33.6 kJ / cm3 g cm3

Methanol: Assume the reaction is:

ˆ o (kJ / g mol) ΔH f

1

CH3OH(l ) + 1

2

O2 (g) → CO2 (g) 2H + 2O(g) 0

− 238.64

-393.51

-241.826

ΔHRxn = 638.52 − kJ / g mol 638.52 kJ 1 g mol 0.792 g = 15.79 kJ / cm3 3 g mol 32.04 g cm

33.6 kJ cm3 methanol cm3 methanol = 2.13 cm3 gas 15.79 kJ cm3 gas

a. (2.13-1) 100 = 113% larg er tan k

Ethanol:

C2 H5OH(l ) + O2 (g) → 2CO2 (g) + 3H2O(g) ˆ o (kJ / g mol) ΔH f

− 277.63

0

-393.51 -241.828

ΔHRxn = 1235 − kJ / g mol 1235 1 g mol 0.789 g = 21.15 kJ / cm3 3 g mol EtOH 46.07 g cm

33.6 kJ cm3 EtOH = 1.59 cm3 gas 21.15 kJ

b. (1.59) 100 -100 = 59% larg er

10–16

Solutions Chapter 10

10.2.20

Basis: 1 g mol C2H4(g)

C2 H4 (g) ˆ o ⎛ kJ ⎞ ΔH f ⎜ ⎟ ⎝ g mol ⎠

+

52.283

2H 2 (g)

→

2CH 4 (g)

0

25o C

-74.84

ΔH o rxn = 2(−74.84) − 52.283 = 202.0 − kJ / g mol The reaction gives Q = ΔU = ΔH – Δ(pV). Assume ideal gases so that Δ(pV) = Δ(nRT) = Δn(RT) Q = -202.0 – (-1) (8.31 × 10-3) (298)

Q = −199.5 kJ / g mol

10.2.21

Basis: 1 g mol Cu and 1 g mol H2SO4 Data from Perry

Cu(s) + H2SO4 (l ) → H2 (g) CuSO + 4 (s) ΔH o rxn = (0 − 772.78) − (0 − 810.40) = 37.61 kJ For a constant volume process: Q = ΔU = ΔH – Δ(pV) = ΔH –Δn(RT)

Δn = 1

Q = 37.61 – (1) (8.314 × 10-3) (298) = 35.13 kJ/g mol Basis: 1 lb mol of each reactant

(37.13)(1000)J 454 g 1 Btu = 15,978 Btu / lb mol g mol 1 lb 1055 J

10–17

Solutions Chapter 10

10.2.22

Assume steady state flow process with reaction.

N2(g)

H2(g)

C6H12(g)

C6H6(g) Q Ref. temp. = 25°C

C6 H12 (g) → C6 H6 (g) + 3H2 (g) The standard heat of reactions is

ˆ o = [1(82,927) + 3(0)] − [1(−123,100)] = 2.06 ×105 J / g mol C H ΔH Rxn 6 12 Material balance: Basis: 1 g mol C6H12 (g) in

Step 5: Steps 6 to 9:

Make balances for each species so that the unknowns H2 and C6H6 can be calculated. Results:

n H2

=

0.70(1)(3)

=

out

=

0.70(1)(1)

= 0.70

n C6H12

out

=

0.30(1)

= 0.30

out

=

0.50

ˆ o f (J/gmol) ΔH

∫

T

25

Out:

In:

H2(g) C6H6(g) C6H12(g) N2(g)

0 + 82.927 × 103 - 123.1 × 103 0

C6H12(g) N2 (g)

-123.1 × 103 0

(a)

2.10

n C6 H 6 n N2 Energy balance Component

out

n C6H12 in = 1.0

ni

ΔH 300 (kJ) 25

0 0 0 0

2.10 0.70 0.30 0.50

0 0

1.00 0.50

16.865 71.226 -47.552 4.016 44.556 -123.1 0 -123.1

C p dT (or use tables)*; T = 300°C

Q = 1.44 × 105 J / gmol

8031 18,824 35,408 8,031

or

0 0

or

(b)

= 0.50

n N2 in

Q = + 1.68 × 105 J / gmol

*at 25°C exit temp. and entrance temp. these terms are 0.

10–18

Solutions Chapter 10

10.2.23

Basis: 1 gal of each fuel MW density (g/cm3) (1) Ethanol : C2 H5OH(l ) + 3O2 (g) → 2CO2 (g) + 3H2O(l ) (2) Benzene : C6 H6 (l ) + 7.5O2 (g) → 6CO2 (g) + 3H 2O(l ) (3) Isooctane : C8 H18 (l ) + 12.5O2 (g) → 8CO2 (g) +9H2O(l )

46 78 114

0.789 0.879 0.692

Example calculation

1 gal Bz 0.879 g Bz 3.785 L 1000cm3 1 g mol Bz 6 g mol CO2 Benzene : cm3Bz 1 gal 1L 78 g Bz 1 g mol Bz ×

44 g CO2 1 lb = 24.8 lb CO2 / gal Bz 1 g mol CO2 454 g

Other results: Ethanol: 12.5 lb CO2 / gal EtOH

Isooctane:

17.8 lb CO2 / gal iso

Carry out the standard combustion material balance for stoichiometric air. The ratios are at SC or the same temperature and pressure. Example of the calculation for benzene (Reaction 2) Use data from the CD, heat capacity equations, or tables to get the sensible heats. The data below are from the CD. Reference T = 25ºC. IN:

T(ºC)

mol

ˆ o f (kJ / g mol) ΔH

ˆ ΔH(kJ) ΔH sensible (kJ / g mol)

ΔH(kJ) C6 H 6 ( l ) O2 (g) N 2 (g)

25 100 100

1 7.5 28.21

+ 44.66 -

2.235 2.183

CO 2 (g) H 2O(l ) N 2 (g)

25 25 25

6 3 28.21

-393.505 -285.840 -

-

46.66 16.763 61.553

OUT:

10–19

2361.030 857.520 -

Solutions Chapter 10

ΔHRxn = (−2361.030 − 857.520) − (46.66 + 16.763 + 61.553) = 3093.6 − kJ / g mol Bz −3093.6 kJ 1 gal Bz 0.879 g Bz 3.785 L 1000 cm3 1 g mol Bz 1 Btu g mol Bz cm3 Bz 1 gal 1L 78 g Bz 1.055 kJ = −1.251× 105 Btu per gal Bz

(exothermic)

Ethanol:

−7.55 ×104 Btu per gal EtOH

Isooctane:

−1.21×105 Btu per gal isooctane

Ethanol Benzene Isooctane

lb CO2 /106 Btu 166 198 147

10.2.24

At 25°C and 1 atm H2O is a liquid. 1

H 2 (g) + O2 (g) → H 2 O(l ) 2

o

ΔH c (kJ/g mol): − 285.84

0

0 ΔHRxn25oC

ΔHReact

ΔHProd ΔHRxn0oC

Basis: 1 g mol H2 (g)

ΔHo rxn = −[∑ ΔHo c,Prod − ∑ ΔHo c,React ] 1

= −[(1)(0) − (1)(−285.84) − (= − )(0)] 2

285.84 kJ / g mol H 2

H2O at 25ºC is a liquid

10–20

Solutions Chapter 10

ΔH Pr od

4.184 J 18 g (−25o C) = ∫ o Cp dT = = 1.883 − kJ 25 C (g)(o C) 1 g mol 0o C

ΔH React = (1) ∫

0o C 25o C

1

0o C

2

25

CpH2 (g) dT + ( ) ∫

1

CpO2 (g) dT = −(718)(1) − (727)( ) = 1.081 − kJ 2

ΔHrxn (0o C) = (−1.883) − (−1.081) + (−285.84) = 286.64 kJ − / g mol H2

10.2.25 Basis: 1 g mol dry cells The heat of reaction is

ˆ 0 − ∑ n ΔH ˆ 0 ⎤ ΔH orxn = − ⎣⎡ ∑ n i ΔH i,c i i,c ⎦ products

reactants

= −[(1)(−1,517) + 2.75(−393.51) + 3.42( − 285.840) −6.67(−2,817) − 2.10(0)] = 15, − 213 kJ/g mol cells For 100 g dry cells

-15,213 kJ 1g mol 100 g dry cells = -1.800×104 kJ g mol dry cells 84.58 g dry cells

10.3.1

Steady State, open process. Ref T = 25ºC Basis: 100 g mol feed The moles out come from the material balance, and the other data from the CD. Q = ΔH ΔH data from the CD. Comp.

% = g mol

T(ºC)

ˆ o (kJ / g mol) ΔH f

IN with gas 10–21

ˆ ΔH sensible (kJ)

ΔH(kJ)

Solutions Chapter 10

CO2

6.4

500

-393.505

20.996

-2384.06

O2

0.2

500

-

15.034

3.04

CO

40.0

500

-110.541

14.690

-3834.04

H2

50.8

500

-

13.834

702.77

N2

2.6

500

-

14.242

37.03

100.0

-5,475.29

In with air O2

63.28

25

-

0

0

N2

240.65

25

-

0

0

303.93 Out with fg CO2

46.4

720

-393.505

32.006

-16,773.55

H2O(g)

50.8

720

-241.835

26.133

-10,957.66

N2

240.65

720

-

21.242

5,111.89

O2

18.1

720

-

22.555

408.25

356.0

-22,211.08

Q = ΔH = (−22, 211.08) − (−5, 475.29) = − 16,736 kJ (Heat leaving)

10–22

Solutions Chapter 10

10.3.2

Steady state, open process. Ref T = 25°C Basis: 100 g mol P W H2O coke F

40oC

Furnace

P 1100oC

CO2 CO O2 N2

mol % 16.1 0.8 4.3 78.8 1000

R Refuse 200oC

mass fr. C 0.90 H 0.06 inert 0.04 1.00

A

Air 40oC mol fr. O2 0.21 N2 0.79 1.00

C inert

mass fr. 0.10 0.90 1.00

Data come from the material balances and the CD.

C p (J / g mol)( o C) H 20.8 C 14.3 Comp. Percent (g)

g mol

ˆ o (kJ / g mol) T(ºC) ΔH f

ˆ ΔH sensible (kJ / gmol) ΔH(kJ)

In F (214.8 g) solid C

90.0

16.11

40

-

0.214

3.456

H

6.0

12.79

40

-

0.312

3.991

Inert

4.0

-

40

-

ignore

-

100.0

28.90

7.447

In air (99.7 g mol) O2

20.94

40

-

0.442

9.255

N2

78.76

40

-

0.436

34.339

99.70

43.595

10–23

Solutions Chapter 10

Out in P (g) CO2

16.1

1100

-393.505

24.744

-5937.05

CO

0.8

1100

-110.541

25.032

-68.41

O2

4.3

1100

-

26.209

112.70

N2

78.8

1100

-

38.891

3064.59

Water out (g)

100.0 6.80 1100

30.174

–2828.17 205.18

Refuse

10 g

-241.826

200

-

1.488 per g

14.88

Q = ΔH = (−2828.17 + 205.18 + 14.88) − (7.447 + 43.595) = − 2659.2 kJ (heat exiting)

10.3.3

Basis: 1 g mol CaC12 ⋅ 6H 2O(s) Unsteady state, open process. Reference T = 25ºC.

68o F → 20o C and 86o F → 30o C ΔU = Q − ΔH H2Oexiting

Assume ΔU ; ΔH inside the system

ˆ o (kJ / g mol) ΔH f

Data

Cp (J /(g)(o C))

MW

CaC12 ⋅ 6H 2O(s)

-2607.89

1.34

219

CaC12 ⋅ 2H 2O(s)

-1402.90

0.97

147

ˆ Calculate ΔU Comp. Initial CaC12 ⋅ 6H 2O(s)

g mol T(ºC)

1

20

MW

219.1

ˆ o (kJ / g mol) ΔH f

ΔHsensible (J)

ΔH(kJ)

-2607.89 (1)(1.34)(219.1)(25-20) -2606.422 10–24

Solutions Chapter 10

Final CaC12 ⋅ 2H 2O(s) Out H2O(g) 965.798

1

30

129.0

-1402.90 (1)(0.97)(129.0)(30-25) -1402.274

4

30

18.0

-241.826 (1)(4.18)(18)(30-25)

-

CaC12 ⋅ 6H2O(s) → CaC12 ⋅ 2H2O 4H + 2O(g) Per g mol CaC12 ⋅ 6H2O(s) :

Q = ΔU + ΔH exiting = [(−1402.274) − (−2606.422)] + (− 965.798) = 238 kJ

or 226 Btu/g mol CaC12 ⋅ 6H 2O(s)

200, 000 Btu 1 g mol CaC12 ⋅ 6H 2O(s) 219 g 1 kg = 194 kg 226 Btu 1 g mol 1000 g (427 lb) 10.4.1

(a)

0.005⎞ ⎛ + 4050(0.006) HHV = 14,544 (.80) + 62,028 ⎝ 0.0003 – 8 ⎠ = 11,640 Btu/lb LHV = 11,800 – 91.23(0.3) = 11,770 Btu/lb

(b)

HHV = 17,887 + 57.5(30) – 102.2 (0.5) = 19,560 Btu/lb LHV = 19,560 – 91.23 (12.05) = 18,460 Btu/lb

10.4.2

If the fuel cannot produce H 2O as a product, HHV = LHV, but the concept really refers to cases in which water is produced so that HHV ≠ LHV.

10–25

Solutions Chapter 10

10.4.3

Both H2 and O2 enter at 0°C, and the products leave at 0°C. The reference temperature is 25°C. Stoichiometric quantities react according to the reaction equation. H2 (g) + 1/2 O2 (g) → H2O (l ) ° ΔHf (kJ/g mol):

0

0

–285.83

Basis: 1 g mol H2 (g)

ΔHRxn( 0°C) =

ΣΔHPr oducts

ΣProducts

( –285.83) + ∫20°C 5°C

=

–

ΣΔHRe ac tan ts 4.184 J 18g H2 O dT = –287.71 kJ ( g)(°C ) g mol

Alternately you can use the steam tables or the CD to get the value. The reactants are gases, hence ΔH can be obtained from tables. ΣReactants

=

28.84 + 0.00765 × 10 –2 T )dT ( 25°C

(0 + 0) + (1)∫ + ( 12 )∫

0°C 25°

0°C

(29.10 + 1.158 × 10 –2 T )dT = –1.45 kJ

ΔHRxn( 0°C) = –287.71 – ( –1.45) = –286.26 kJ

10–26

Solutions Chapter 10

10.4.4

Basis: 100 lb mol gas The reaction for each compound is the stoichiometric amount of O2 going to CO2 gas and H2 O ( l ) ˆ Comp. lb mol ΔH (Btu) ΔH(Btu/lb mol) CO2 C2 H4 CO H2 CH4 N2

9.2 0.4 20.9 15.6 1.9 52.0 100.0

0 598,000 122,000 123,000 383,000 0

0 239,000 2,544,000 1,918,000 728,000 0 5,429,000

Note: NGI standard state is 60°F

Btu 5,429,000 Btu 1 lb mol 520°R 3 = 100 lb mol 359 ft 3 492°R ft 3

= 143 Btu / ft at std. conditions of NGI

10.4.5

Basis: 1 m3 gas at 25ºC and 1 atm with 40% relative saturation Calculate the moles of C9H12 gas. Its pressure at 25ºC is calculated as follows: ( p*H2O = 3.2 kPa) (3.2) (0.40) = 1.28 kPa, and thus the pressure of the C9H12 (g) is (101.3-1.28) = 100 kPa.

n=

pV 100 kPa 1 m3 1(kg mol)(K) = = 0.0404 kg mol RT 1 atm 298 K 8.314(kPa)(m3 ) Basis: 1 g mol n-propyl benzene (C6H5

⋅

CH2

⋅ C2H5)

The higher heating value is with H2O as a liquid. Also, at 25°C, C9H16 is a liquid; its normal boiling point is 432 K. C9H12 (l ) + 12 O2 (g) → 9 CO2 (g) + 6 H2O (l )

10–27

Solutions Chapter 10

ΔH° f

kJ : g mol

–38.40

0

–393.51

–285.840

ΔH° rxn = [6 (–285.840 + 9 (–393.51)] – [(1) (–38.40)]

= −5218.2 kJ/g mol C9 H12

HHV= −

−5218.2 kJ 40.4 g mol C9 H12 = 2.11×105 kJ/m3 3 g mol C9 H12 1 m gas

10.4.6

Basis: 100 mol of gas (a1) Comp.

mol.wt.

CH4 C2 H6 C3H8 C4H10 Total

16 30 44 58

Mol%= Vol.%

lb

88 6 4 2 100

1408 180 176 116 1880

–ΔH°c1 kJ/g mol

–ΔH°c2 kJ/g mol

890.35 1559.88 2220.05 2878.52

802.32 1427.84 2044.00 2658.45

mol. wt. mixture = 18.80 lb/lb mol

−ΔHo c1 =H2O(1) to CO2 (g) products;

− ΔHo c2 = H 2O (g) + CO2 (g) products

ΔHº1 rxn = – [ΔHc prod – ΔHºc react] = 1022.90 kJ/g mol (a1)

ΔH o rxn =

−1022.90 454 g mol Btu lb mol Btu = − 23,600 g mol lb mol 1.055 kJ 18.80 lb lb

HHV = 23,600 Btu/lb (a2)

ΔHºrxn = (–23,600) (18.80) = -439,000 Btu / lb mol

HHV = 439,000 Btu/lb mol (a3)

ΔHo rxn =

−439,000 Btu 1 lb mol = − 1160 3 lb mol 379.2 ft at 60o F, 760 mm Hg 10–28

Solutions Chapter 10

HHV

3

= 1160 Btu / ft at 60°F, 760 mm Hg

(b1) ΔH°2 rxn = –925.71 kJ/g mol ΔH° 2 rxn =

–925.71 454 = –21,200 Btu/lb 1.055 18.80

LHV = 21,200 Btu/lb (b2) ΔH°2 rxn = (–21,200) (18.80) = -398,000 Btu/lb mol

LHV = 398, 000 Btu / lb mol (b3) ΔH°2 rxn = (–398,000)/(379.2) = -1051

LHV = 1051 Btu/ft 3 at 60°F and 760 mm Hg

10.4.7

Write the equation for the combustion for 1 gram as: 1 g of Fuel or food + Oxygen → CO2 (g) + H2O(1) + N2 (g) Both the reactants and the products are at 25ºC and 1 atm. The chemical energy released by the complete combustion of 1 g of a food or fuel according to the above equation at standard conditions is the heat of combustion. Assume the ΔHc are additive. Then

ΔHC = (17.1)(28) + (39.5)(10) + (14)(4) = 930 kJ The High Energy bar has (capital C for calorie)

220 Calories 1000 calories 4.184 J 1 kJ = 921 kJ 1 Calorie 1 calorie 1000 J The values are close enough.

10–29

Solutions Chapter 10

10.4.8

When the combustion results in a single component.

10.4.9

Basis: 1 lb mixture Get the data for the ΔHc in Btu/lb, not Btu/mol. Component

Mass fr., ωi

Hexachloroethane, C2C16 Tetrachloroethene, C2C14 Chlorobenzene, C6H5C1 Toluene, C7H8

0.0487 0.0503 0.2952 0.6058

ΔHc (Btu / lb) 828 2141 11,876 18,246

ΔH(Btu)

Basis 1200 lb

40 108 3,506 11,053

58.44 60.36 354,24 726.96

Total 1.000 14,707 1200.00 _______________________________________________________________________ The Btu/lb are 14, 707 from the heat of combustion data, a deviation of 3.2% from 15,200. For DuLong’s equation, the component elements are needed. Basis: 1200 lb _______________________________________________________________________ C H C1 O MW Total (lb) _______________________________________________________________________ C2C16 24 0 213 0 237.00

ωi

0.10

0.00

0.90

0.00

lb

5.84

0.00

52.60

0.00

24

0

142

0

ωi

0.14

0.00

0.86

0.00

lb

8.45

0.00

51.91

0.00

72

5

35.5

0

C2C14

C6H5C1

10–30

58.44 166.00

60.36 112.50

Solutions Chapter 10

ωi

0.64

0.04

0.32

0.00

lb

226.71

14.17

113.36

0.00

84

8

ωi

0.91

0.09

0.00

0.00

lb

661.53

65.43

0.00

0.00

Total

902.53

79.60

217.87

0

1200

0.75

0.07

0.18

0

1.00

C7H8

ωi

0

0

354.24 9200

726.91

HHV = 14,455 C + 62,028 H – 7753.5O2 + 4050 S = (14,455)(0.75) + (62,028)(0.07)

= 15,183 Btu/lb (the higher heating value) a deviation of 0.11%

10.4.10

The reactions are:

C2 H5OH(1) + 3 O2 (g) → 2 CO2 (g) + 3 H 2O(1)

(1)

C8H18 (1) + 12.5 O2 (g) → 8 CO2 (g) +9 H2O(1)

(2)

⎡ ˆ o − ∑ n ΔH ˆ o ⎤⎥ ΔH rxn = − ⎢ ∑ n i ΔH c,i i c,i reac tan ts ⎣ products ⎦ For (1): −[(2)(0) + (3)(0) − (3)(0) − (1)(−1366.91) = 1366.91 − kJ / g mol C2 H5OH For (2): −[(8)(0) + (9)(0) − (12.5)(0) − (−1307.53)(4.184)] = 5470.71 − kJ / g mol C8H18

ˆ = 1307.53 k cal / g mol for n-octane liquid is from Perry. Note: the ΔH c

10–31

Solutions Chapter 10

Basis: 1 kg gasahol Component

mass.fr. = kg

C2H5OH

0.10

C8H18

0.90

Total

1.00

g mol

ˆ ΔH(kJ)

46.05

2.17

-2,966

114.14

7.89

-43,137

MW

−46,103

If the fuel were all octane

1 kg oc tan e 1 kg mol octane 1000 g −5470.71 kJ = −47,930 kJ 114.14 kg octane 1 kg g mol octane

−47,930 − (−46,103) (100) = 3.8% −47,930

10–32

Solutions Chapter 11

11.1.1

Basis: gas as given in problem statement

(a) 0.030 =

p H2O pTot − pH2O

=

p H 2O 101.6 − p H2O

p* H2 O at 60 °C % humidity

so, pH 2 O = 2.96 kPa

= 19.9 kPa

⎛ 2.96 ⎞ ⎡ 101.6 −19.9 ⎤ = 100 ⎝ = 12.2% 19.9 ⎠ ⎢⎣ 101.6 − 2.96 ⎥⎦

(b) Relative humidity (100)

2.96 19.9

14.9%

= *

(c) Dewpoint occurs where pH 2 O = 2.96 kPa, or T = 24oC from the steam tables.

11.1.2 *

At 27 °C, pH 2 O = 3.52 kPa. The actual pressure of the water vapor is obtained from

p=

nRT ⎛ 0.636 ⎞ ⎛⎜ 8.314 ⎞⎟ ⎛⎜ 300 K ⎞⎟ ⎛ ⎞ = 3.15 kPa =⎝ ⎠ ⎝ ⎠⎝ ⎠ 28.0 ⎠ V 18.01 ⎝

⎛ 3.15 ⎞ 100 ⎝ 3.52 ⎠

= 89%RH

11–1

Solutions Chapter 11

11.1.3

.5067 ---------p

.1217 -----

Initial state

40 80 T °F

*

At 80 °F, pH 2 O = 0.5067 psia from the steam tables *

At 40 °F, pH 2 O = 0.1217 psia * At 58 °F, pH 2 O = 0.2384 psia (a) 100

0.1217 psia = 0.5067psia

24%

(b) The mol fraction of H2O vapor is constant on compression unless saturation is reached. At the initial conditions

y H 2O =

0.1217 −3 = 8.28 × 10 14.696

⎛ 0.1217⎞ = 0.2437 psia At 2 atm, pH 2 O = p Tot y H2 O = (14.692)(2) ⎝ 14.696 ⎠

0.2437 > 1 hence water condenses and the relative humidity is 100% 0.2384

11–2

Solutions Chapter 11

11.1.4

P

T = 140°F

*

H 2O

= 5.878 in Hg

p = 30 in Hg H = 0.03 mol H2O/mol BDA ⇒ Basis: 1 mol BDA mol H 2O 0.03 = = 0.0291 = y H 2O mol H 2O + BDA 1 + .03

p H 2O = 30 (.0291) = 0.874 in Hg pair = 30 (1 – 0.0291) = 29.13 in Hg (a)

% rel. humidity = 100

p H2O p*H2O

⎛ 0.874 ⎞ = 100 ⎜ ⎟ = 14.9% ⎝ 5.878 ⎠

(b) Dew point is the temperature at which vapor first condenses on cooling at constant humidity, hence p* = 0.874 in Hg ~ 75°F.

11–3

Solutions Chapter 11

11.1.5

Steps 1, 2, 3, and 4: ptot = 275 kPa Air 30oC F (mol) 100 kPa RH = 75.0%

275 kPa

Compressor

Cooler

P(mol)

Air Saturated 20oC RH = 100%

0.341 kg H2O *

pH 2 O

*

= 0.6153 psia = 4.241 kPa

pH 2 O

pH 2 O = 4.241 (.75) = 3.18 kPa System: overall Step 5: Basis: 0.341 kg H2O Step 6 and 7: Unknowns:

(0.341/18) kg mol

F, P

Balances: air, H 2O

Steps 8 and 9: Balances are in kg mol

Air:

3.18 ⎞ H 2O: F ⎛ ⎝ 100 ⎠

V=

⎫ ⎪ ⎪ Solve anytwo toget ⎬ ⎪ F = 0.804 kg mol 0.341 ⎛ 2.34 ⎞ ⎪ = +P ⎝ 275 ⎠ ⎭ 18

0.341 18 ⎛ 100 − 3.18 ⎞ ⎛ 275− 2.34 ⎞ =P ⎝ F⎝ ⎠ 275 ⎠ 100

Total: F

= P+

0.804 kgmol 8.314 303 3 = 20.3 m at 30°C and 100 kPa 100

11–4

= 0.3388 psia = 2.34 kPa

Solutions Chapter 11

11.1.6

Basis: Air at T = 66 °F and 21.2 psia; assume V is constant. (a) Initial

pair ,i V = n air RT 6 6 F Final Assume all of the water evaporates, and check later on to see if the assumption is true.

pair , f + pH 2 O = p t so pair , f = pt − p H 2 O pair , f V = n air RT 1 8 0  F The material balance is simple: all the initial air = final air.

pair , i pt − p H 2 O

=

66 + 460 526 21.2 = = 180+ 460 640 33.0− pH 2 O

pH 2 O = 7.20 psia Since p* = 7.51 psia, all of the water can evaporate and the air will not be saturated. (b) Basis: 1 lb H 2O (0.0555 lb mol H 2O ) At the final condition

pH 2 O pt

=

7.20 n H 2 O 0.0555 = = 33.0 nt nt

so n = 0.254 lb mol

nRT ⎛⎜ 0.254 ⎞⎟ ⎛ 10.73 ⎞ ⎛⎜ 640 ⎞⎟ 3 = 53.0 ft = ⎝ ⎠ ⎝ ⎠ 33.0 ⎝ ⎠ p 0.17 lb H2 O 1lbH 2O = 0.254 - 0.0555 = 0.1985 lb mol = lb air 0.1985(29)

V= (c) n air

11–5

Solutions Chapter 11

11.1.7

Steps 1, 2, 3, and 4:

pt = 103.0 kPa F (kg mol)

P (kg mol)

Air: 22oC RH = 50%

Air: 72oC RH = 80%

W = 200 kg mol H2O 18

Data for p* from the steam tables:

T (°C)

p* (kPa)

22°C (295K) 72°C (345K)

2.622 33.77

These calculations provide the composition of F and P. In

kPa

Out

kPa

pH 2 O = 0.50 (2.60)

=

1.31

pH 2 O

pair = 103.0 - 1.31

=

101.69 pair

= 102.0 - 27.0

=

103.0

pt Step 5: Basis:

= 0.80 (33.77) = 27.02 = 76.0 pt

=103.0

W = 200 kg H 2O (1 hour)

Steps 6 and 7: Unknowns are: F, P

Balances are: H2O, air

Steps 8 and 9: This is a steady state process without a reaction Balances

In

Out

H 2O

F (1.311 0 3) + 2 0 01 8(1.00) = P ( 2 7. 01 0 3)

air:

F (1 0 1. 71 0 3)

F = 32.86 kg mol Bone dry air: Step 10:

= P( 7 61 0 3) P = 43.95 kg mol

F (1.071 0 3) = 32.45 kg mol Check using total balance

Weight in kg of dry air used = 32.45 (29 = 941 kg 11–6

P ( 7 61 0 3) = 32.43kgmol

Solutions Chapter 11

11.1.8

(a)

Entering

Leaving

80°F = 26.8°C

70°F = 21.0°C

p* at 26.8°C = 26 mm Hg

p* at 21.0°C = 19 mm Hg

Rel. Humidity =

(b)

p 5 mm Hg 100 = p* 26 mm Hg

= 94.8%

Entering

Leaving

V p = Vtot p tot

Vol fr. =

5 mm Hg 100 = 0.676% 740 mm Hg

H 2O:

Air:100 - 0.0676 = 99.324% (c)

p 18 mm Hg 100 = p* 19 mm Hg

= 19.2%

Vol fr. = H 2O:

Rel. Humidity =

p p tot

18 mm Hg 100 = 2.43% 740 mm Hg

Air: 100 - 2.43 = 97.57%

Entering

Leaving

Basis: 1 lb mol at 80°F, 740 mm Hg Vol % = mol %

Basis: 1 lb mol at 70°F, 740 mm Hg

lb

wt %

H 2O: (0.00676)(18)

0.12

0.4

H 2O: (0.0243)(18)

0.44

1.53

Air: (99.324)(29)

28.80 28.92

99.6 100.0

Air: (0.9757)(29)

28.30 28.74

98.47 100.00

(d)

lb

wt %

Entering

Leaving

Basis: 1 lb mol at 80°F, 740 mm Hg

Basis: 1 lb mol at 70°F, 740 mm Hg

Actual: pvap = 5 mm Hg

Actual: pvap = 18 mm Hg

pvapor-free gas = 740 – 5 = 735 mm Hg

pvapor-free air = 740 – 18 = 722 mm Hg

11–7

Solutions Chapter 11

Humidity =

n H 2O n dry air

=

g mol H 2O 18 g 1 g mol dry air g mol dry air 1 g mol H 2O 29 g

p H 2O p dry air Entering

Humidity = (e)

Leaving

5 18 g H 2O = 4.22×10-3 735 29 g dry air

18 18 g H 2O = 0.0155 722 29 g dry air

Basis: 1000 ft3 of mixture at 740 mm Hg and T Entering

Leaving

Vol % H2O = 0.676 %

Vol % H2O = 2.43 %

Vol H2O = (0.00676)(1000 ft3) = 6.76 ft3

Vol H2O = (0.0243)(1000 ft3) = 24.3 ft3

@ 26.8 °C and 740 mm Hg

@ 21.0°C and 740 mm HG

3

3

6.76 ft 1 lb mol 492°R 740 mm Hg 18 lb 3 540°R 760 mm Hg lb mol 359 ft

24.3 ft 1 lb mol 740 mm Hg 492°R 18 lb 3 760 mm Hg 530°R lb mol 359 ft

= 0.30 lbH 2 O / 1000 ft 3

(f)

at 80 F and 740 mm Hg

= 1.10 lb H2 O / 1000 ft 3 at 70°F and 740 mm Hg

Entering

Leaving

0.30 0.99324

1.10 0.9757 3

3

= 0.302 lbH 2 O / 1000 ft vapor free air (g)

= 1.127lb H2 O / 1000 ft vapor freeair

Basis: 800,000 ft3/day at 80°F, 740 mm Hg Entering

Leaving

Vol dry air = 794, 600 (

Vol % air = 99.324 %

530 ) 540

= 779,900 ft3 at 70°F

Vol dry air = (800,000)(0.99324) 11–9

Solutions Chapter 11 = 794,600 ft3

Vol % = 97.57 % Total vol =

Vol H2O vapor = (800,000)(0.00676) = 5400 ft3 at 80°F, 740 mm Hg 5400 ft

3

779, 900 3 = 799, 300 ft 0.9757

Vol H2O vap = 799,300 – 779,900

740 492 760 540

= 19,400 ft3 at 70°F, 740 mm Hg

= 4790 ft3 at SC

19, 400

492 740 = 17, 535 at SC 530 760

Water evaporated = 17,535 – 4,790 = 12,745 ft3 at SC 3

12, 745 ft 1 lb mol 18 lb = 639 lb H 2O/day 3 lb mol 359 ft Part g could also be worked using part f and the volume of dry air: Entering

Leaving

794,600 ft3 dry air

779,900 ft3 dry air

0.302 lb H2O/1000 ft3 dry air (see f)

1.127 lb H2O/1000 ft3 dry air (see f)

794, 600 0.302 = 240 lb H 2O 1000

779, 900 1.127 = 879 lb H 2O 1000

Water evaporated = 879 – 240 = 639lb H2 O / day

11.2.1

a.

Dew point = 10ºC

b.

%RH =

c.

H =

p H2O

p H2O pair

psat =

100 =

p10o C p27o C

100 =

1.27 kPa 100 = 38% 3.356 kPa

1.27 18 = 0.79 kg H 2O / kg air 101.3 − 1.27 29

11–10

Solutions Chapter 11

11.2.2

When the air is saturated.

11.2.3

Draw a horizontal line until it intersects with the saturation curve. The temperature at the intersection is the dew-point temperature. 11.2.4

When the air is saturated (100% relative humidity).

11.2.5

The two temperatures are approximately equal at atmospheric temperatures and pressure. 11.2.6

If by “air” is meant wet air, H =

0.02 lb H 2O = 0.204 0.98 lb dry air

From the SI humidity chart a.

Dew point = 25.2ºC

b.

%RH ≅ 59%

11.2.7

From the SI chart at the intersection of TDB = 30o C and RH = 65% H = 0.0174 kg H2O / kg dry air

11–11

Solutions Chapter 11

11.2.8

Let A = alcohol and C = CO2. MW of A = 46; MW of C = 44. At 40ºC, pA* = 134.26 mm Hg or 17.90 kPa a.

pA = 0.10 (100) = 10 kPa

H =

(46)(10) = 0.116 g A/g C (44)(100 − 10) p 10 (100) = (100) = 55.9% p* 17.90

b.

%RS =

c.

Cs = 1.00 + 1.88(H) = 1.00 + 1.88 (0.116) = 1.22 J/(K) (g C)

d.

ˆ = 2.83 ×10−3 T + 4.56 ×10−3 (H ) V' K = (2.80)(10-3)(313.15) + (4.56)(10-3)(0.116) = 0.877 m3 / kg C

c.

At saturation at 40ºC

H =

(46)(17.90) = 0.228 g A / g C (44)(100 − 17.90)

ˆ = (2.80)(10-3) (313.15) + 4.56 × 10-3 (0.228) = 0.878 m 3 / kg C V'

11.2.9

They are almost parallel to each other.

11–12

Solutions Chapter 11

11.2.10

Condenses water from the air in humid climates. 11.2.11

a.

From Humidity Chart, where tDB = 90ºC (194ºF) and tWB = 46ºC (115ºF) H = 0.049 kg H2O/kg air. Upon cooling to 43ºC (109ºF), no condensation occurs, therefore H is constant.

0.049 kg H 2O 29 kg air 1 kg mol H 2O kg mol H 2O = 0.079 1 kg air 1 kg mol air 18 kg H 2O kg mol air b.

c.

⎛ 273 + 43 ⎞ Final pressure = 100 ⎝ = 87.1 kPa 273 + 90 ⎠ At saturation: 0.079 =

p*H2O *

87.1 − p H2O

; solving p*H2O = 6.38 kPa

At the dew point, the vapor pressure of pure water is equal to 6.38 kPa. Dew point = 37°C (99°F) The same answer can be obtained by proceeding to the dew point at constant H on a humidity chart for the correct pressure.

11–13

Solutions Chapter 11

11.2.12

Basis: 1 lb dry air Data from the humidity chart.

Initial state: H = 0.0637 lb H2O/lb dry air

TDB = 180o F and TWB = 120o F

Hsaturated ≅ 120 Btu / lb dry air δHdeviation ≅ −1.5 Btu / lb dry air H = 118.5 Btu/lb dry air Final state

TDB = 115o F,

TWB = ?, H = 0.0657 lb H 2O / lb dry air

Hsaturated ≅ 101 Btu / lb dry air ΔH = 101 – 118.5 = −17.5 Btu / lb dry air

11.2.13

From the SI psychometric chart at 29ºC and 40% relative humidity read TWB = 19.3ºC Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wet-bulb temperature, which is the lowest temperature that can be obtained on an evaporative cooler. That is,

Tmin = TWB = 19.3o C

11–14

Solutions Chapter 11

11.2.14

Basis: 1 lb bone dry air (BDA)

Basis: 100 m3 Data from psychometric chart (BDA = bone dry air). a.

@ 1, Dew point = 23°C

b.

@ 1, humidity = 0.018 kg H 2O / kg dry air

@1, Relative humidity = c. d.

ΔH = Q + W

H 0.018 = (100 = ) 54.5% H 2 0.033

W=0

Q = ΔH 2 – ΔH 1 = (141.5 – 3.6)

kJ kJ kJ – (79.0 – 0.3) = 59.2 kg BDA kg BDA kg BDA

V 100 m 3 entering air kg BDA m= ˆ = = 112.36 kg BDA V 0.89 m3

Q= e.

59.2 kJ 112.36 kg BDA = 6, 651.7 kJ kg BDA

Adiabatic cooling by evaporation yields a saturated humidity of 0.041 kg H2O/kg BDA:

(0.041 − 0.018) kg H 2O 112.36 kg BDA 2.58 kg H2 O = 3 100 m kg BDA 110 m3 air f.

Texit = 37°C

11–15

Solutions Chapter 11

11.3.1

a.

The humidity of the entering air at 225ºF DB and 110ºF WB is obtained from the humidity chart.

Humidity = 0.031

lb H 2O lb dry air

Assume the exit air to be saturated at 125ºF. Humidity = 0.0955 b.

lb H 2 O lb dry air

Basis: 1 hr 10 tons 1 day 2000 lb = 835 lb/hr day 24 hr ton

Water in = (0.1) (835) = 83.5 lb/hr Water out =

(0.9) (835) lb dry grain

1 lb H 2O = 7.59 lb/hr 99 lb dry grain

lb H2O removed/hr = water in – water out = 83.5 – 7.59 = 75.9 lb H 2 O / hr

c. Product output = (0.9) (835) d.

lb H 2 O lb dry grain + 7.59 hr hr

24 hr = 18, 200 lb / day 1 day

7.59 lb H 2O/hr lb BDA = 1175 lb BDA/hr = (0.0955 – 0.0310) lb H 2O/lb BDA hr Q = ΔH = Enthalpy out – Enthalpy in = ΔHair + ΔHdry grain + ΔHwater = 1175

+

lb BDA Btu Btu 136.5 – (92.25 – 2.25) hr lb BDA lb BDA

(0.9) (835) lb dry grain 0.18 Btu (110 – 70) °F hr (lb) (°F)

11–16

Solutions Chapter 11

+

–

7.59 lb H 2O 1.0 Btu (110 – 32) °F hr (lb) (°F)

83.5 lb H 2O 1.0 Btu 2 (70 – 32) °F hr (lb) (°F) 4

= 5.46 × 104 + 0.54 × 104 + 0.059 × 104 – 0.32 × 104 = 5.74 × 10 Btu / hr 11.3.2

Assume in this problem 1. 2. 3. 4.

Steady operating conditions Dry air and water vapor are ideal gases ΔKE = ΔPE = W = 0 The mixing is adiabatic (Q = 0)

Data from the humidity chart: Stream 1: H1 = 110.3 kJ/kg dry air H = 0.0272 kg H2O/kg dry air Stream 2: H2 = 50.9 kJ/kg dry air H 2 = 0.0130 kg H2O/kg dry air The specific humidity and the enthalpy of the mixture can be determined from mass and energy balances for the adiabatic mixing of the two streams: Basis: 1 kg dry air Total Mass balance: 8 + 6 = 14 kg total Water mass balance: 8(0.0272) + 6(0.0130) = 14 (Hmixture) b.

H = 0.0211 kg H2O/kg dry air

11–17

Solutions Chapter 11

Energy balance (ΔH = 0) 8(110.3) + 6(50.9) = 14 (Hmixture) H = 84.8 kJ/kg dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychometric chart. a.

T = 30.7o C

c.

RH = 75.1%

11.3.3

Data from Humidity Chart

a.

Hair in = 0.01813 lb H 2 O / lb air

b.

Hair out = 0.031 lb H2 O / lb air H2O picked up = 0.031 – 0.01813 = 0.0129 lb H2O/lb air

Energy Balance: ΔHair out – ΔHair in = ΔHwater in – ΔHwater out Basis: 1 lb dry air

Ref. temp = 85ºF

ΔHH 2 O out ΔHair out     90–85 ) + 0.45 Btu 0.031 lb (90–85) 0.24 Btu 1 lb ( ( lb)(°F) ( lb)(°F) 11–18

Solutions Chapter 11

ΔHH2O out

ΔHair + H2O in

ΔHH2O in – ΔHH2O out

 +1040 Btu 0.0129 lb lb

   –0

 1 Btu m lb (102–89 ) = ( lb)(°F)

m = 1.136 lb H2O/lb air or 0.915 lb air / lb H2 O

c. % H 2 O vaporized = 0.0129 100 = 1.14% 1.136 11.3.4

Steps 1, 2, 3, and 4 The process will be assumed to be a steady state, open, continuous one with 5L flowing in (in the material balance the air is invariant). All of the data for the stream flows and compositions have been placed on the figure. The lungs will be the system.

P L air, 37oC

5 L air, 25oC p

H2O

pair ttot

Lungs 0.95 kPa 96.05 kPa 97.0 kPa

p

H2O

pair ttot

= p*

H2O

6.27 kPa 90.73 kPa 97.0 kPa

p*25o C = 3.17 kPa p H2O = 3.17(0.3) = 0.95 kPa Step 5:

Basis: 5 L air → 1 min

Steps 6, 7, 8, and 9: Material balances In: n air =

96.05 kPa

1 atm 5L 1(g mol)(K) = 0.194 g mol 101.3 kPa 298 K 0.08206(L)(atm)

11–19

Solutions Chapter 11

⎛ 0.95 ⎞ −3 n H2O ⎜ ⎟ (0.194) = 1.92 ×10 g mol 95.05 ⎝ ⎠ Out:

n air = 0.194 g mol ⎛ 6.27 ⎞ n H2O = 0.194 ⎜ ⎟ = 0.0134 g mol ⎝ 90.73 ⎠ Energy balance The energy balance reduces to Q = ΔH. The reference temperature will be 25ºC. Assume the increase in water vapor comes from water vaporized at 37ºC ˆ = 2414.3 kJ / kg). (ΔH vap ΔHin: Both the air and water enter at 25ºC so ΔH = 0 for both the input streams. ΔHout:

ΔHair = 0.194∫

310

298

(27.2 + 0.0041 TK )dT = 66.2 J 310

ΔH H2O = 1.92 ×10−3 ∫+ (34.4 0.0063 TK )dT 298

+ (0.0134 – 1.92 × 10-3) (2414.3)(18) = 499 J Q = ΔH = 499 + 66.2 = 565 J/min or 33.9 kJ / hr

11–20

Solutions Chapter 11

11.3.5

Basis: 180 kg/hr of product Material Balance

Dry product = a.

92 kg dry P 180 kg P = 166 kg dry P/hr 100 kg P 1 hr

Water removed from solid: In =

1.25 kg H 2O 166 kg/hr dry P – (0.08) (180) kg/hr 1.00 kg dry P

= 207.5 – 14.4 = 193 kg H 2 O / hr Basis: 1 BDA Water picked up in dryer out

in

0.0571 kg H 2O kg BDA

out, 53°

–

0.0083 kg H 2O kg BDA

in, 21°

= 0.0488

kg H 2O kg BDA

193 kg H 2 O kg BDA = = 3955 kg BDA / hr hr 0.0488 kg H 2O / kg BDA b.

Energy balance Air:

ΔH (kJ/kg BDA)

air in (21ºC, 52% RH ):

58.2

air out (53ºC, 60% RH):

219.7

ΔH =

161.5 kJ/kg BDA

Basis: 1 hr ΔH = (161.6) (3955) = 6.387 × 105 kJ Solid (ref. 21ºC) 11–21

Solutions Chapter 11

solid out: ΔH =

43 21

3

0.18 cal 4.184 J 10 (43 – 21) °C C pdT = = = 16.57 kJ/kg P (g) (°C) cal 10 3

solid in: ΔH = 0 (because of reference temperature) Basis: 1 hr ΔH = (16.57) (180) = 2983 kJ If the dryer and reheater are insulated, then for the system Q – W = ΔH and

W=0 5

Qreheater = 2983 + 6.387 × 105 = 6.42 × 10 kJ / hr

11.3.6

Initial air: TDB = 38ºC, TWB = 27ºC, H1 = 0.0175 kg/kg dry air Air from scrubber: T = 24ºC, RH = 100%, RH2 = 0.0188 kg/kg dry air Heated to 93ºC:

H3 = 0.0188

From drier: TDB = 49ºC,

H4 = 0.0377

(1000 kg/hr) (0.05) = 50.0 kg H2O to be evaporated 50.0/(0.0377 – 0.0188) = 2650 kg dry air/hr 2650 (0.0377) = 100 kg H2O/hr

⎛ 2650 100 ⎞ 22.4 273 + 49 V4 ⎜ + = 2560 m3 at 49ºC and 1 atm ⎟ 18 ⎠ 273 ⎝ 29 Heat supplied: Q = [(2650) (1.00) + 100 (0.200] (93 – 24) = 1.84 × 105 kJ/hr Water at 93ºC has p* = 79.4 kPa 11–22

Solutions Chapter 11

H3 =

(18) ( 79.4) = 2.25 kg H O/kg dry air 2 ( 29) (101.3–79.4 )

[(0.0188/2.25)] (100) = 0.83%

RH air from heater

Answers: a.

1. 2. 3. 4.

H = 0.0175 H = 0.0188 H = 0.0188 H = 0.0377

b. 1. 2. 3. 4.

42% 100% 0.83% 47%

c. 2650 kg dry air/hr d. 2560 m3/hr e. 1.84 × 105 kJ/hr

11.3.7

Step 5:

Basis: 1 hr

Assume: 1. 2. 3. 4.

ΔPE = ΔKE = W = 0, No reaction occurs, open, steady state process, ideal gas behavior

Data: Entrance air (in)

ˆ (kJ/kg dry air) H H (kg H2O/kg dry air) ˆ 3 / kg dry air) V(m

50.5 0.00587 0.88

Assume the properties of the wet penicillin are the same as those of water. ΔHvap at 34ºC = 2420.25 kJ/kg water. Let P = kg dry penicillin per hour, and F = kg dry air/hr. Steps 3 and 4

11–23

Solutions Chapter 11 Moist air TDB= 34oC TWB = ?

Penicillin

Penicillin

Dryer

34oC

34oC

P in

ω = 0.80

ωPout = 0.50

Moist air 3300 m3/hr TDB=34oC TWB = 17oC

air in =

4500 m3 1 kg dry air = 5114 kg dry air 0.88 m3

Steps 6 and 7: The exit conditions for the air are not known, but the H and H are related on the humidity chart hence only one is unknown. P (dry penicillin is unknown). The balances are water, dry air, and dry penicillin. Steps 8 and 9

TWB ( o C) Assume:

20 21 22

⎛ kg H 2O ⎞ H ⎜ ⎟ ⎝ kg dry air ⎠ 0.009 0.010 0.0115

ˆ H(kJ / kg dry air) 57.0 60.6 64.0

Water balance:

5114 (0.009 – 0.00587) = water evaporated = 16.0 kg 5114 (0.010 – 0.00587) = 21.1 kg 5114 (0.0115 – 0.00587) = 28.8 kg

Energy balance:

5114 (57.0 – 50.5) / (2420.25) = 13.73 kg 5114 (60.5 – 50.5) / (2420.25) = 21.1 kg 5114 (64.0 – 50.5) / (2420.25) = 28.5 kg

The solution is very sensitive to the values read from the psychometric chart. Assume the final TWB ; 21 or 22o C . a.

Water evaporated = 28.5 kg 11–24

Solutions Chapter 11 b. c.

ΔH (21ºC) = 51,140 kJ/kg dry air ⇒ 51,140 kJ / hr

Steps 6 and 7 Unknowns

P, exit TWB

Equations

air, water

Steps 8 and 9 Water balance on air gives water evaporated (5114) (0.010-0.00587) = 21.1 kg H2O Energy balance: (60.5 – 50.5) (5114) + 21.1 (2420.25) = 0 0.80 → 0.60

Use 21ºC

⎛ 0.80 0.50 ⎞ 21.1 kg H2O evap = P ⎜ − 3P ⎟ = ⎝ 0.20 0.50 ⎠

P=

21.1 = 7 kg / hr 3

11.3.8

R 80oF 120oF Hin=0.0031 lb H2O/lb DA

Hout=0.0292 lb H2O/lb DA

H = 0.0075 lb H2O/lb DA

Basis = 2000 lb dry waste (DW) Comp.

in %

out

waste in =

H2 O

63.4

22.7

2000 0.773 100 = 4, 220 lb 36.6

DW

36.6

77.3

DW out = 2,000 lb

→

11–25

Solutions Chapter 11 100.0

100.0

H2O evap.

= 2,220 lb

Basis: 1 lb dry air Air in, 80ºF, 29.92 in. Hg, and a wet bulb temperature = 54ºF:

ˆ = 22.62 − 0.18 =22.54 Btu / lb dry air Hin = 0.0031 lb H2O/lb dry air H' in Air out, 120ºF, 29.92 Hg, and a wet bulb temperature = 94ºF: Hout = 0.0292 lb H2O/lb dry air

ˆ ' = 61.77-0.4 = 61.37 Btu/lb dry air H

Overall

b.

Hout

=

0.0292

Hin

=

0.0031

H2O evaporated =

0.0261

air used per ton dust =

lb H2O/lb dry air lb H2O/lb dry air

(1 + 0.0031)

2220 0.0261

= 85, 400 lb moist inlet air per ton DW

Air out =

1.00

2220 0.0261

0.0292

(29)

= 2,930 lb mol dry air out

2220 0.0261

(18)

= 138 mol H2O out 3068 total mol out

c.

3068 359 580 = 1,300, 000 ft3 wet air exiting/ton DW 492

Air Recirculated R = lb air added at mixing point before kiln Water balance:

(0.0031) (1) + (0.0292) (W) = (0.0075) (1+W) 0.0292 W

R= a.

=

0.0075

-0.0075 W

-0.0031

0.0217 W

0.0044

0.0044 = 0.203 lb 0.0217

Recirculation =

0.203 100 = 16.8% of gas out is recirculated 1.203

11–26

Solutions Chapter 11

11.3.9

Adiabatic operation removes 425,000 Btu/hr from the process. Basis: 1 hr The energy balance is ΔH = 0 overall. a.

For the process:

TWB = 100o F⎫⎪ ⎬ RH = 1.3% TDB = 234o F ⎪⎭ b. c.

Temperature leaving the cooler is 61ºF (dew point constant H)

d.

ΔH = (72 − 27)Btu = / lb BDA 45 425, 000 Btu 1 lb BDA = 9450 BDA / hr hr 45 Btu

11–27