Theoretical Losses in Gear Pumps

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THEORETICAL LOSSES

GEAR

TUMFS

THESIS Submitted in Partial Fulfilment of the requirements for the degree of MASTER

OF MECHANICAL

ENGINEERINC

AT THE POLITECHHIC

INSTITUTE

OF

BROOKLIN

W Joseph F. Pizzirusso May 1950

Approved!

Head of Department and Thq^s Advisor

ProQuest Number: 27591447

All rights reserved INFORMATION TO ALL USERS The quality of this reproduction is d e p e n d e n t upon the quality of the copy subm itted. In the unlikely e v e n t that the a u thor did not send a c o m p le te m anuscript and there are missing pages, these will be noted. Also, if m aterial had to be rem oved, a n o te will ind ica te the deletion.

uest ProQuest 27591447 Published by ProQuest LLO (2019). C opyright of the Dissertation is held by the Author. All rights reserved. This work is protected against unauthorized copying under Title 17, United States C o d e M icroform Edition © ProQuest LLO. ProQuest LLO. 789 East Eisenhower Parkway P.Q. Box 1346 Ann Arbor, Ml 4 8 1 0 6 - 1346

VITA ]^rn

-

June 4> 1926

Birthplace B.M.E.

-

- Brooklyn# New York 1947

-

Polytechnic Institute of Brooklyn

Bachelors? Thesis - ’* The Design# Construction# and Calibration of a Position Indicator for a Carburetor Needle Valve **

This thesis was begun in September 1949 and concluded in May 1950. The actual experimental work was conducted in the Mechanical Laboratory of the Polytechnic Institute of Brooklyn.

ACKNOWLEDGEMENTS The author wishes to express his sincere appreciation to Professor E, L. Midgette for his suggestions and advice. Appreciation is also expressed to Mr. C. Groth# Mr. W. Peterson, and Mr. H« Peterson of the Mechanical Laboratory for their aid in completing this thesis.

Abstract The purpose of this thesis was two fold, To determine how the clearance was situated between the gear faces and casing and to develop theoretical formulas that could accurately pre­ dict the discharge of a gear pump under a given set of condi­ tions. It was proposed to develop these formulas using basic hydraulic: equations and then to compare their results with actual test data. To accomplish the practical part of this pro* gram it was necessary to design and construct test apparatus. The results of this work proves that clearance is equally divided on each side of the gear face and that the discharge could be determined by a simple equation with surprising accuracy.

Table of Contents Page History

1

Use and Description of the Gear

h

Scope

7

Homenclatmre

S

Theory

9

Physical Characteristics of Oil

19

Desi^ and Construction

23

Theoretical Delivery

28

Experimental Procedure

33

Experimental Data and Results

35

Sunmary and Conclusions of Results

64

Bibliography

66

Table of Illustrations Page Typical Gear Pumps

5

Clearance and Principle Dimensions of a Gear Pump

11

Oil COiaracteristics

22

Test Apparatus

24

Diagrammatic Flow of Fluid

27

Possible Theoretical Deliveries

32

Theoretical Delivery Compared toActualDelivery

5^

Efficiency of Pump

63

1.

BISTORT Pumps are among the oldest of men’s aids, Their origin is lost in antiquity.lt is known however,that they were used to provide water for ancient Egypt,Chinq, India,Greece,and Rome, The history of nearly all machines used by en­ gineers today is a story of success,They tell of a steady development from a rudimentary beginning and record a series of successful adventures as greater and greater developments were attempted. The story of the positive action rotary pumps is an exception to this rule.Forover one hundred years,after its first design entered the me­ chanical world,engineers and designers struggled to bring into successful use this tool whose principle had undoubt­ ed advantages but with defects of a serious nature. Chief advantage is the ability to operate at high speeds,Since at this time there were not any mechanisms capable of producing high speeds,this advantage could not be realized, Th# disadvantages were overburdening. The length of the leakage line was much larger than that of the recipro­ cating pumps,The art of machining was not as good as it is today,To prevent external leakage packing had to be used,This in turn absorbed much of the power delivered to the pump. Rotary pumps were knovm and used during the six­ teenth century. Three distinct designs were developed

2.

by the end of that century,Two of the basic principles then ap­ plied are still in use,One of these was a typical gear pump invented by Pappenheim in 1636 ,This was the forerunner of the modern gear pump,It was constructed of two meshing gears having six teeth each. It was not until the nineteenth century that the gear pump began to make strides. This progress was mostly due to Watt’s development of the steam engine.More power and higher speeds,the food ofthe gear pump, were supplied. Still the leakage posed a serious problem. Again this type of pump seemedto slumber until the twentieth century,With modern advances in the science of machining, a gear pump could be made that didn’t leak. Then and only then did this pump assume its rightful place with the giants of the machine age. It has been found that the best gear pumps are not made merely by placing two ordinary gears in a suit­ ably shaped case but are carefully designed to suit the particular pntpose for which they are intended.The reason for this intricacy in design lies in the fact that the hydraulics and gear requirements conflict.Many modern pumps are made in which the wheels have teeth of involute form.With this form of tooth,if serious undercutting is to be avoided,there should be at least twelve teeth.On the other hand,a considerable increase indelivery can be obtained without altering the ®ize ofthe casing were it

3.

possible to reduce the number of teeth. While the involute shape is not the best

for gear

pumps,it has been adopted for a number of reasons,The most important being a comparative ease of accuracy in machining. Since the gear pump offers a wide range of designs, and the principle of positive acting rotary displacement, it still retains an attraction for engineers.

4.

Use and Description of the Gear Pump Bor many industrial applications, oil,water,chem­ icals,and other fluids must he handled under operating conditions for which reciprocating and centrifugal types of pumps are not suited.Rotary pumps have construction features that make them adaptable to many of these ap­ plications where the other pumps cannot give satisfactory service. In spite of their external resemblance to centrif­ ugal pumps,rotary pumps are more like reciprocating pumps because both are positive-displacement types,The rotary type combines the constant-discharge characteristic of the centrifu^l type with the positive-discharge feature of the reciprocating type.The flow from a reciprocating type is pulsating,whereas that from the rotary types is relatively constant. Figure I illustrates the gear-type pump.One gear is keyed to the driving shaft and rotates with it while driving the other gear mounted on a trunnion,Care has to be taken in designing the gears so that they will form a tight seal where they mesh and at the same time will not trap oil in the root of the teeth to build up high pressures.The gears rotate in the direction indicated by the arrows and have very close clearances with the cas­ ing.When the pump is operating,liquid flows in between the gear teeth on the low pressure of inlet side in very

5.

OUTLET

~ ■'V

T

l£^

ypical

Gear P u m p

F ig. I

b.

mach the same manner that it flows into the cylinder of a reciprocating pump when the piston is on the suction stroke.As the gears rotate, the fluid is trapped between their teeth and the casing.It is carried around to the high pressure of discharge side.Since the liquid cannot escape back to the low pressure side, a pressure builds up sufficient to force it out the discharge opening.lt should be noted that each gear tooth acts like the plunger in a reciprocating pump to forcé the liquid out of the discharge opening. Gear pumps are intended to handle fuel oil, lubri­ cating oil,or other thick and viscous liquids that have lubricating qualities.Therefore further lubrication of moving parts is not necessary.

7.

Scope Ihis paper concerns itself basically with the mathematical determination of the delivery of a gear pump so that at any known set of conditions: pump characteristics fluid properties speeds pressure differential the discharge can he accurately computed. A secondary problem was to determine whether the clearance between the casing and gear faces is located equally on both sides of the gear face.This problem arises because the gear has freedom of axial movement. To accomplish these objectives, the theory of the flow in a gear pump was investigated and actual tests were conducted to verify the theory.The latter required the design and construction of test apparatus to complete the experimental part of the project.

B.

Homenolattire actual delivery, cu.in./min. 01 “ ideal delivery,cu,in./min.

Os - loss in delivery due to slip,cu.in./min. Or - loss in delivery due to canitation,cu.in./min. Od = gain in delivery due to viscous drag,cu.in,/min. n,ni,U2 ,= revolutions of gears, R.P.M. a,ai,a2 = addendum of gears,inches. b,b^,T)2 = face width of gears,inches. B,Dil,D2 - pitch diameter of gears,inches. ® outlet pressure,p.8 .i. ?2 - inlet pressure,p/s.i.

P = P2 -E2 - pressure differential,p.s.i. 1

®

1

- clearance between gear and

casing,inches. X,X^,l2 ,X^,lj^ “ length of flow path,indhes. M - viscosity of fluid,lbs.-min.per inch square. d^fd^ - outside diameter of gears,inches. - diameter of shafts,inches, V - velocity, inches/min. - » - specific gravity.

S r Ti^cosity,Saybolt Universal Seconds. Z - viscosity,poises.

Theory As prèviously stated,the action of a gear pump is the trapping of fluid between thegeai? teeth and transpor­ ting it to the outlet where it

is

forced out .Promthis

conception of the function of this pump,the delivery can be readily determined.For n revolutions of the gear, the delivery will be the product of the volume of fluid trap­ ped between the teeth and the number of revolutions.This volume delivered by one spur gear can be approximately found as the product of the addendum,the pitch,diameter, and the face width of the geardivided by two.The deliv­ ery for one gear will be;

QX “ 2yn

aPb

Since most gear pumps are constructed with the meshing of two gears, a driver and driven,the delivery will be: qi = 2 ir m

ZTOg ■t

or

Qi =

=2^2% tgj

This simple trapping and transporting action of the gear teeth does not tell the entire story: concerning the delivery of the pump,since the pressure difference between inlet and outlet results in a secondary flow that reduces delivery.This flow is commonly known as the slip.

10.

A second reduction in ideal delivery may be due to cav­ itation in the neighborhood of the inlet and air entrap­ ped in the fluid stream,Due to the viscous property of the fluids,an increase in delivery is also realized,This is caused by the dragging of the fluid contained in the clear­ ance passages to the outlet by the rotation of the gears. This type of flow is commonly called viscous drag.Thus, the resulting delivery will be: Q. — Qi—Qs—Qr-|" Qd The problem then is to determine these secondary flov/s before the actual pump delivery can be computed. Since OX has already been defined as a function of speed and physical bharacteristics of the pump ,Qs may also be defined in a similar manner. To derive this expression,à study must be made to determine how this flow acts and the factors that in­ fluence its magnitude.This flow occurs in the passage caused by the necessary clearances and is caused by a pressure dillerential between inlet and outlet.Figure II illustrates existing clearances and principal dimensions in a gear pump. To simplify this work,several assumptions must be made: 1.

The viscosity of the fluid is such that the flow

in all parts of the unit is governed by viscous forces, that is,the flow will be laminar and not turbulent.

//

û. s (S. cc < w e u.

0 l/i z g (n Z

1

Û w

CL

O

H 2 CL w

V ;s < i r < w

U

w ir

12.

2. The passage is similar to two parallel plates through which the flow passes. Therefore the flow diagram will he:

Where

is greater than

and b is the width

of the passage. From basic hydraulics, the laminar flow of a fluid having a viscosity Af between two plates a distance c apart caused by a differential pressure P,having a path

of length L,and width b,will be directly proportional

to the pressure difference across the plate, the width of the passage,and the cube of the distance separating the plates, and be inversely proportional to the viscosity of the fluid and to the length of the flow path. /. %8 Pbc3 /y If For this set of conditions: Qs -

^

Pbc.3

If this formula governs the slip flow in a gear pump, the Qs for a pump can be computed.Slip occurs between the faces of the gears,teeth ends,and the casing of the p^p^igure II illustrates the passages for slip flow.

13.

The gears are assumed to be solid disks as the fluid car­ ried between the teeth will have the same characteristics as the gear. The slip flow between the gear faces and casing. For gear I : With

clearances equal to c^ and

width equal

to (&i-&2 );and length equal to L^.Where L^is an average length; IÎ-X,

°3 L

(ai-Ag) 1.3

and 04 (&!-&) ---- E5---

For gear II With clearances equal to c^ and c^,vfidth equal to (d^-djj^) ,and length equal to

Where

is an average

length: =5

3

(*3 -4 4 )

and = -I— 12 M For slip flow over the teeth. Gear I With clearance equal to c^,width equal to b^,and length equal to I^:

14.

Gear II With clearances equal to 0 3 ,width equal to bp,and length equal to

Summing these losses: Qs =

P TTTf

°5

®6

%

— I

Simplifying: -

P —

. °2 ^î>2 + — [g-

(*3 ^ + 04 ^) 4

r 3 3, {05 + 06 )

s-\

To further simplyfy this analysis,it will be assumed that the fluid is solid,therefore the cavitation loss will be zero. Or = 0

In analyzing the viscous drag flow, it is neces­ sary to study this flow which also occurs in the clearance passages of the pump. Again these passages are assumed to be ti?o parallel plates,one stationary,the other having a velocity v.

15.

If the liquid adjacent to the gear has the same velocity as the gear,and the fluid adjacent to the casing has the same velocity as the casing which is zero ^ d laminar flow ezists;the flow diagram will be:

V width of the passage is equal to b. The flow from this system will be proportional to the plate velocity,the distance between plates,and the width of the passage. qd

Therefore:

¥cb

V/here the average

velocity for this system is one

half the plate velocity: qd =

¥ cb ?

Appljring this law to the pump characteristics we can determine the pump qd. The resultant drag across the gear faces will be zero due to the fact that while one half of the gear de­ livers fluid to the outlet,the second half is returning a like amount to the inlet.Therefore they mulify each other,

16,

Viscous drag across the teeth: For gear X Vdiere clearance equals c^,width equals

velocity

equals 2TTnj^p^,where n^ equals revolutions per minute and r^ equals radius of the gear; qd ÏÏ---

For gear II-; Where clearance equals Cg,width equals bg,velocity equals 2TT

,where n^ equals revolutions per minute and

T2 equals radius of the gear: qd = 2lTiL^T^Q2'b2

Z Summing these flows: qd = IT

B2?2G2^2)

Therefore the quantity of fluid delivered by the pump will be:

q = TT (n^a^D^b^f n2a2D2b2) + T ^1 -^2 ) (=3 ^+ ^

\~W~

^

1.3

If both gears are similar,that is : Li= L2>^ “

&2"^4' ^1" ^2

then the equation simplifies to q = 27T(naDb) + TT nrb (0^+ 02) —

^^^292^) > + ^ 3-^4 (*5^+ °6^)| lyV

17.

The face clearance of the gears can act in either of two ways. 1. It can he divided equally on each side of the gear,therefore c^ = or 2. It can all be on one side,therefore c. = 0, 1 c^ s 0 , and 0^3 = 2 0 3 , c = 2 c^. In case I: q - 271 naDb4 TT nrb (c^-#-

“

Case II q = 27T naDb^iT nrb(c^-l-c^)

—

r 5

(0 ] 3 + C 2 ^

3^7/

' >+ h

-^ 2

3

[(203)^+ (20g)^'

=-

-JJ

Further sin^lification can be accomplished if the respective clearances are equal. ®1 * ®2 »

" 0&

Case I q = 2TTnaDb 4gTTnrbc^ —

IS.

Case II q = 2 TTnaDb 'fZTT nrbc^

or q - 2 7T naDb isn nrbo^-

19.

Physical Characteristics of Oil The Kinematic viscosity of the oil was determined at three temperatures: 100

130

and 210

These were converted to viscosity in Saybolt Universal Seconds,From these three points a straight line curve was constructed,Therefore the viscosity of the oil at any temperature included in these limits can he de­ termined, Mexfcjthis viscosity was converted to poises hy utilizing the formula found in "Design of Machine Members^, by Yallance and Boughtie. This formula is : Z = ^

(0.22 S- 180)

Where: S = Saybolt seconds f = Specific gravity Z = poises From this the viscosity in lb,-min, per square inch was computed. The specific gravity of the oil was determined at 66 ^F and using the correction factors for oil found in the "Hew and Revised Tag Manuel for Inspectors of Petroleum",26ï^ Edition by R,!!,Wilhelm,the specific gravity was determined at any desired temperature,These charac-

20.

teristics of the oil were plotted and shown in Figure III. The actual values are shown in Table Ho. I. Sample calculation: i - density

Z - i

Z

= poises

S

= Viscosity Saybolt Seconds

{0.22 S - 180) at 140

F

Z = .886 TÜÜ

(0.22(68) - 180 “55

Z = .886 TCC

( 1^0 )

Z = .1090 1 lbs.-sec. ft/!

= 478.3 poises

Viscosity in Iba.-min. ■ .1090 lbs.-sec. % 1 ft.^ _ z 1 min. la 4 478.8 ft. 144 in.^ qOsac.

À4 = viscosity = 26I.5 z 10“1®

Ibs.-Mn. in.^

21. Table Hç. 1 Oil Pl^sical Characteristics

•f

f

8

120 125

.894 .892

.84 .80

130

,890

135

Folses .1460

.3520

.1368

.3300

.76

.1276

.3075

.888

.72

.1183

;%855

140

,886

;68

.1090

.2615

145

.885

%1020

.2460

150

.883

.65 :62

.0930

.2240

155

.881

.0874

22105

.0800

.1930

160

:56

Conversion from Kinematic Viscosity to Saybolt universal Viscosity.

Kinematic Viscosity

Saybolt Universal Viscosity

100

270

9B

130

160

76

210

45

23

22.

M■

r rc

43C

23.

Design and Construction Ü5iis part of the project was to design the test apparatus which is composed of three essential components. 1. A gear pump# 2. A driving motor. 3. Circulating and collecting equipment.% i s test equipment is shown in Figure J2. First a gear pump of unknown mechanical properties was selected.This pump ims investigated and its physical dimensions were measured so that they could he used in connection with the theoretical part of this paper.These are noted here and are used later in the theoretical cal­ culations. casing width

.433 inches

gear width

.429 inches

clearance

.004 inches

diameter of casing

1.602 inches

diameter of gear

1.600 inches

clearance

.002 inches

clearance on the circumference

.001 inches

diameter of shaft

.546 inches

number of teeth

14

average length of path across gear face

1.26 inches

length of path from inlet to outlet across gear teeth 4,11 inches

24.

%

25,

A motor was selected capable of producing enough power to drive this pump.The characteristics of this motor are : Direct Current Five horsepower 115 volts 40 amperes 1800 R.P,M. The circulating and collecting equipment was de­ signed and constructed to fulfill the following conditions: 1. A complete circulation of fluid must be obtained to result in stable operating conditions. 2. Collection of discharged oil should be obtained for a given time. 3. Temperature of the fluid should be maintained within a given range so that a comparison can be made between theoretical and actual discharge. This set of conditions was satisfied by the circu­ lation system illustrated in Figure III.The storage fluid was surrounded by cool water to keep the temperature fair­ ly constant.The fluid flows from this storage tank through a system of pipes,passes a check valve and inlet pressure guage to the gear pump. It leaves the gear pump and is [measured by an outlet pressure guage,The fluid then flows through a pressure adjusting valve which is a manually operated needle valve that regulates the outlet pressure.

26.

From here the fluid passes to a three-way valve that can direct the flow back to the storage tank of divert it to the measuring tanlc.The flow cyble is illustrated diagrammat ically in Figure V.

id

s

Ui S a: g

to

:I!::; mV '

a

=) li u.

o

o Sri

I

i

Û

N o

LZ

28.

Theoretical Delivery Characteristics for pump used: gears are identical a » .100 inches D = 1.400 inches b - .429 inches @1 " @2 = ,001 inches r = .800 inches c^+C6 = ®3 ‘*’®4 = .004 inches di - d^ s 1,600 inches = &4 * .546 inches di _d2 = &3 -

= 1.054 inches

n = 1800 revolutions per minute « 4,11 inches ^3 = 1.26 inches 1. Qi = 2TTaDbn

QX = 2TT(.100) { 1,400) (.429) (1800) Qi = 683.89 ou, in. 2. For case I where clearance is equal on both sides of the gear face: Qd = Trnrb(Gl+ 0%) Qd=TT(l800) (.800) (.429) (,002) Qd =

3,89 cu. in.

3. Loss due to slip flow. An average value of viscosity was used so that a

29.

clearer comparison could be made between actual and the­ oretical values.This average does not cause a noticeable error in the final computations. 4 =

262 z 10-10

ibsv-mln, in.z L3

%8 = ;:

p

I2 (È62zio-10 )

2(.&29)(ljilO-9 I4(1.054)|gl0-9

% . igfa;Æc-i°i qs =

P________ (256 .0 9 ) 12(i6S)

qs =

P {.0813)

' 1”-=)

The discharge for each pressure is shown in Table Ho.2. For ease H with clearance on only one side of gear 0 C».

=

=

203

2 cc = 2 (.002 ) inches L-iu, e-»» -10-10.

i' ,-10) 03 qs =

^

‘ 10-10,10|1,%Ü»I10-0

P (2.09 z 10-10 + 10.16 z 10 - 8 ) i?(552 z id-Iü)

P

12(2b2)

(1018.09)

qs » P (.3365) The discharge for each pressure is shown in Table No. 3.

30.

%ble Ho. 2 Theoretical Discharge for Equal Clearance on Each Side of the dear Face. Qi^Qr Cu.in./2ij[n.

Qs ou. in./min.

683.89

0

683.89

4.07

679.82

100

8.13

675.76

150

12.20

671.69

200

16.26

667.63

250

20.33

663,56

300

24.39

659.50

400

32.52

651.37

500

40.65

643.14

600

43.78

700

56.91

635.11 .-• 626;98

800

65)04

618.85

900

73^17

610.72

Pressure PSI 0 50

Q ou.in

31,

Table No. 3 Theoretical Discharge with Clearance on only One Side of the dear Face.

Pressure PSI

gi:Qr: ou,ln.ymln.

-

ou.ln.yoia.

%

000

• /min 633.89

50

1 6 .8 8

66 7 .0 1

100

33.65

650.2 4

150

50.53

633.36

200

67.30

616.5 9

250

84^18

599.71

300

582.94

400

100i