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PROBLEM 4.3

KNOWN:  Temperature distribution in the two-dimensional rectangular plate of Problem 4.2.

FIND:  Expression for the heat rate per unit thickness from the lower surface (0 ≤ x ≤ 2, 0) and result

 based on first five non-zero terms of the infinite series.

SCHEMATIC: 

ASSUMPTIONS:  (1) Two-dimensional, steady-state conduction, (2) Constant properties.

ANALYSIS:  The heat rate per unit thickness from the plate along the lower surface is

( ) ( )

x 2 x 2 x 2

out y 2 1y 0x 0 x 0 x 0 y 0

T

q dq x,0 k dx k T T dxy y

∂ ∂θ 

∂ ∂  

= = =

== = =   =′ ′= − = − − = −∫ ∫ ∫   (1)

where from the solution to Problem 4.2,

( ) ( )( )

n 11

2 1 n 1

1 1 sinh n y LT T 2 n xsin

T T n L sinh n W L

π π θ 

π π 

+∞

=

− +−   ⎛ ⎞≡ =   ⎜ ⎟−   ⎝ ⎠

∑ . (2)

Evaluate the gradient of θ from Eq. (2) and substitute into Eq. (1) to obtain

( )  ( ) ( ) ( )

( )

n 1x 2

out 2 1

n 1 y 0x 0

1 1 n L cosh n y L2 n xq k T T sin dx

n L sinh n W L

π π π 

π π 

+=   ∞

=   ==

− +   ⎛ ⎞′   = −   ⎜ ⎟

⎝ ⎠

∑∫  

( )  ( )

( )

n 1 2

out 2 1x 0n 1

1 12 1 n xq k T T cos

n sinh n W L L

π 

π π 

+∞

==

⎡ ⎤− +   ⎛ ⎞⎢ ⎥′   = − −   ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∑  

( )  ( )

( )  ( )

n 1

out 2 1n 1

1 12 1q k T T 1 cos n

n sinh n Lπ 

π π 

+∞

=

− +′   ⎡ ⎤= − −⎣ ⎦∑   < 

To evaluate the first five, non-zero terms, recognize that since cos(nπ) = 1 for n = 2, 4, 6 ..., only the n-

odd terms will be non-zero. Hence,

Continued …

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PROBLEM 4.3 (Cont.)

( )  ( )

( ) ( )

  ( )

( )  ( )

2 4

out1 1 1 12 1 1

q 50 W m K 150 50 C 2 21 sinh 2 3 sinh 3 2π π π 

⎧ − + − +⎪′   = ⋅ − ⋅ + ⋅ ⋅⎨

⎪⎩

o

 

( )

( ) ( )  ( )

( ) ( )  ( )

( ) ( )

6 8 101 1 1 1 1 11 1 1

2 2 25 sinh 5 2 7 sinh 7 2 9 sinh 9 2π π π 

− + − + − +

+ ⋅ + ⋅ + ⋅

⎫⎪⎬⎪⎭

 

[ ]outq 3.183kW m 1.738 0.024 0.00062 (...) 5.611kW m′   = + + + =   < 

COMMENTS:  If the foregoing procedure were used to evaluate the heat rate into the upper surface,

( )x 2

in y

x 0

q dq x, W

=

=

′ ′= − ∫ , it would follow that

( )   ( ) ( ) ( )n 1

in 2 1n 1

1 12q k T T coth n 2 1 cos nn

π π π 

+∞

=

− +′   ⎡ ⎤= − −⎣ ⎦∑  

However, with coth(nπ/2) ≥ 1, irrespective of the value of n, and with ( )n 1

n 1

1 1 n

∞+

=

− +⎡ ⎤⎢ ⎥⎣ ⎦∑  being a

divergent series, the complete series does not converge and inq ′   → ∞ . This physically untenable

condition results from the temperature discontinuities imposed at the upper left and right corners.