Fundamentals of Heat and Mass Transfer [Frank P.Incropera - David P.DeWitt] Solution Manual - CH0.pdf

PROBLEM
KNOWN: Method of separation of variables (Section ) for two-dimensional, steady-state conduction.
2
FIND: Show that negative or zero values of l , the separation constant, result in solutions which
cannot satisfy the boundary conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.
2
ANALYSIS: From Section , identification of the separation constant l leads to the two ordinary
differential equations, and , having the forms
d22XdY
+λλ22X=0 ­=Y0 (1,2)
dx22dy
and the temperature distribution is θ(x,y) =×X(x) Yy.( ) (3)
2
Consider now the situation when l = 0. From Eqs. (1), (2), and (3), find that
XC=1+C2x, Y=C3+C4y and θ(x,y) =(C1++Cx2) (C34Cy.) (4)
Evaluate the constants - C1, C2, C3 and C4 - by substitution of the boundary conditions:
x=0: θ(0,y) =(CC1+2×0)(C3+C41y) ==0 C0
y=0: θ(x,0) =(0+C2X)(C3+C43×0) ==0 C0
x=L: θ(L,0) =(0+C2L0)( +C42y) ==0 C0
y=W: θ(x,W) =(0+0×x0)( +=C4W) 1 01¹
2
The last boundary condition leads to an impossibility (0 ¹ 1). We therefore conclude that a l value
of zero will not result in a form of the temperature distribution which will satisfy the boundary
2
conditions. Consider now the situation when l < 0. The solutions to Eqs. (1) and (2) will be
-λλx+x
X=C5e+C6e, Y=+C78cos λλyCsin y (5,6)
and θx,y=éùCe-λλx++Ce+os λλyCsin y. (7)
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Evaluate the constants for the boundary conditions.
y=0: θ x,0=éùCe-λλx+Ce-os 0+Csin 0==0 C 0
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x=0: θλ(0,y) =éùCe00++Ce 0Csin y==0 C0
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If C8 = 0, a trivial solution results or C5 = -C6.
x=L: θλL,y=Céùee-xL­=+xL Csin y0.
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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
KNOWN: Two-dimensional rectangular plate subjected to prescribed uniform temperature boundary
conditions.
FIND: Temperature at the mid-point using the exact solut