Fundamentals of heat and mass transfer [6th ed] 9780471457282, 0471457280, 9780471761150, 047176115X

Table of contents :
Node 4......Page 59
Node 9......Page 60
Coefficient matrix [A]......Page 79
Finite-Difference Method of Solution......Page 129
Untitled......Page 840

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PROBLEM 4.1 KNOWN: Method of separation of variables for two-dimensional, steady-state conduction. FIND: Show that negative or zero values of λ2, the separation constant, result in solutions which cannot satisfy the boundary conditions. SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties. ANALYSIS: From Section 4.2, identification of the separation constant λ2 leads to the two ordinary differential equations, 4.6 and 4.7, having the forms d2X d2Y 2 λ + X = 0 − λ 2Y = 0 (1,2) 2 2 dx dy and the temperature distribution is

θ ( x,y ) = X ( x ) ⋅ Y ( y ) .

(3)

Consider now the situation when λ2 = 0. From Eqs. (1), (2), and (3), find that X = C1 + C2 x, Y = C3 + C4 y and θ ( x,y ) = ( C1 + C2 x ) ( C3 + C4 y ) .

(4)

Evaluate the constants - C1, C2, C3 and C4 - by substitution of the boundary conditions: x = 0: θ ( 0,y ) = ( C1 + C2 ⋅ 0 ) ( C3 + C4 ⋅ y ) = 0 C1 = 0 y = 0: θ ( x,0 ) = ( 0 + C2 ⋅ x ) ( C3 + C4 ⋅ 0 ) = 0 C3 = 0 x = L: θ ( L,0 ) = ( 0 + C2 ⋅ L )( 0 + C4 ⋅ y ) = 0 C2 = 0 y = W: θ ( x,W ) = ( 0 + 0 ⋅ x )( 0 + C 4 ⋅ W ) = 1 0 ≠1 The last boundary condition leads to an impossibility (0 ≠ 1). We therefore conclude that a λ2 value of zero will not result in a form of the temperature distribution which will satisfy the boundary conditions. Consider now the situation when λ2 < 0. The solutions to Eqs. (1) and (2) will be (5,6) X = C5 e - λ x + C 6 e + λ x , Y = C7 cos λ y + C8sin λ y and

θ ( x,y ) = ⎡⎣ C5e-λ x + C6 e+λ x ⎤⎦ [ C7 cos λ y + C8sin λ y ].

Evaluate the constants for the boundary conditions. y = 0 : θ ( x,0 ) = ⎡ C5e-λ x + C6 e-λ x ⎤ [ C7 cos 0 + C8sin 0] = 0 ⎣ ⎦ x = 0 : θ ( 0,y ) = ⎡ C5e0 + C6 e0 ⎤ [ 0 + C8sin λ y ] = 0 ⎣ ⎦

(7)

C7 = 0 C8 = 0

If C8 = 0, a trivial solution results or C5 = -C6. x = L: θ ( L,y ) = C5 ⎡e-xL − e+xL ⎤ C8sin λ y = 0. ⎣ ⎦ From the last boundary condition, we require C5 or C8 is zero; either case leads to a trivial solution with either no x or y dependence.

PROBLEM 4.2 KNOWN: Two-dimensional rectangular plate subjected to prescribed uniform temperature boundary conditions. FIND: Temperature at the mid-point using the exact solution considering the first five non-zero terms; assess error resulting from using only first three terms. Plot the temperature distributions T(x,0.5) and T(1,y). SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties. ANALYSIS: From Section 4.2, the temperature distribution is n +1 + 1 ⎛ nπ x ⎞ sinh ( nπ y L ) T − T1 2 θ ( −1) θ ( x, y ) ≡ = sin ⎜ . (1,4.19) ⎟⋅ T2 − T1 π n L ⎠ sinh ( nπ W L ) ⎝ n =1 Considering now the point (x,y) = (1.0,0.5) and recognizing x/L = 1/2, y/L = 1/4 and W/L = 1/2, n +1 + 1 ⎛ nπ ⎞ sinh ( nπ 4 ) T − T1 2 θ ( −1) = θ (1, 0.5 ) ≡ sin ⎜ . ⎟⋅ T2 − T1 π n 2 ⎠ sinh ( nπ 2 ) ⎝ n =1 When n is even (2, 4, 6 ...), the corresponding term is zero; hence we need only consider n = 1, 3, 5, 7 and 9 as the first five non-zero terms.

∑

∑

2 ⎧⎪ ⎛ π ⎞ sinh (π 4 ) 2 ⎛ 3π ⎞ sinh ( 3π 4 ) + + sin ⎜ ⎟ ⎨ 2sin ⎜ ⎟ π ⎪⎩ ⎝ 2 ⎠ sinh (π 2 ) 3 ⎝ 2 ⎠ sinh ( 3π 2 ) 2 ⎛ 5π ⎞ sinh ( 5π 4 ) 2 ⎛ 7π ⎞ sinh ( 7π 4 ) 2 ⎛ 9π + sin ⎜ + sin ⎜ sin ⎜ ⎟ ⎟ 5 ⎝ 2 ⎠ sinh ( 5π 2 ) 7 ⎝ 2 ⎠ sinh ( 7π 2 ) 9 ⎝ 2

θ (1, 0.5 ) =

θ (1, 0.5 ) =

2

π

⎞ sinh ( 9π 4 ) ⎫⎪ ⎬ ⎟ ⎠ sinh ( 9π 2 ) ⎪⎭

[0.755 − 0.063 + 0.008 − 0.001 + 0.000] = 0.445

(2)

T (1, 0.5 ) = θ (1, 0.5 )( T2 − T1 ) + T1 = 0.445 (150 − 50 ) + 50 = 94.5D C .