ε = ε θ

ID number: Heat Transfer

Name: 1^st Semester, 2008

Problem 1 2 3 4 5 Total

Points

Mid Term Exam 1 (April 14)

1. Consider a very long fin with uniform cross-sectional area shown below. The temperature distribution for the long fin is given by

( ) exp

b c

T x T hP

T T kA x

∞

∞

⎛ ⎞

− = ⎜⎜− ⎟⎟

− ⎝ ⎠

When T∞ = 25ºC, h = 100 W/m^2.K, Tb = 100ºC, k = 400 W/m^.K, L = 10 cm, W = 10 cm, t = 0.2 cm, find

a) fin effectiveness b) fin resistance c) fin efficiency

a)

w/o f f

q

ε = q ,

( ) ( )

0

f c c b c b

x c

dT hP

q kA kA T T hPkA T T

dx ₌ ^∞ kA ^∞

= − = − = − ,

( )

w/o c b

q =hA T −T_∞

( )

( ) ( )

w/o

2

c b

f f

c b c

hPkA T T k W t

q kP

q hA T T hA hWt

ε ^∞

∞

− +

= = = =

−

( )

²

4

2 400 10 0.2 10 100 10 0.2 10 63.9

−

−

× + ×

= =

× × ×

T∞, h Tb

L

W t

x

2

, ,

63.9

t b c

f

t f c c

R hPkA kP

R hA hA

ε = = = =

c)

( )

( )

max

1

c b

f f

f b

hPkA T T

q hPkWt kWt

q hA T T hPL L hP

η ^∞

∞

= = − = =

−

( )

1 400 0.1 0.002

0.626 0.1 100 2 0.1 0.002

× ×

= =

× +

2. Consider a slab of thickness L = 0.15 m as shown below. One side (x = 0) is at T1 = 400 K and the other side (x = 0.15) at T2 = 300 K. The thermal conductivity of the slab depends on temperature in such a way that k(T) = k0e^aT, where k0 = 0.05 W/m^.K, a = 4.7×10^-3 K^-1. a) Find the steady-state temperature distribution T(x) in the slab. Hint: d ( )dT 0

dx k T dx

⎛ ⎞ =

⎜ ⎟

⎝ ⎠

b) What is the heat flux q′′[W/m²] across the slab?

a) ( )dT ₁

k T C

dx = , k e dT₀ ^aT =C dx₁ , ⁰

( )

1

k aT

d e C dx

a =

1 2

0

,

aT a

e C x C

= k + T(0)= →T₁ e^aT1 =C₂

( ) 2

T L =T → ² ₁ ¹

0

aT a aT

e C L e

=k + , 0

(

^{2 1}

)

1

aT aT

k e e

C aL

= −

(

^{2 1}

)

₁

0

1 2

0 0

1 1

( ) ln ln

aT aT

k e e aT

a a

T x C x C x e

a k a k aL

⎛ − ⎞

⎛ ⎞ ⎜ ⎟

= ⎜⎝ + ⎟⎠= ⎜⎝ + ⎟⎠ ^1ln

(

^{eaT2 eaT1}

)

^{x eaT1}

a L

⎛ ⎞

= ⎜⎝ − + ⎟⎠

b) 0

(

^{2 1}

)

0 1

aT aT

aT k e e

dT dT

q k k e C

dx dx aL

′′ = − = − = − = − −

(

^{4.7 103 400 4.7 103 300}

)

2

3

0.05 174.3 W/m

4.7 10 0.15 e ^{× −×} e ^{× −×}

−

= − =

× ×

L = 0.15 m x

T1

T2

4

3. A composite cylindrical wall is composed of two materials of thermal conductivity kA

and kB, which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature T∞,i

and provides a convection coefficient hi at the inner surface of the composite. The outer surface is exposed to ambient air, which is at T∞,o and provides a convection coefficient ho. Under steady-state conditions, a uniform heat flux of q′′_h is dissipated by the heater.

a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. Hint:

(

2 1

)

,cond ,conv

1 1

ln / 1

2 , 2

t t

r r

R R

Lk r Lh

π π

= =

b) Obtain an expression that may be used to determine the heater temperature, Th.

c) Obtain an expression for the ratio of heat flows to the outer and inner fluids,q_o′ /q_i′. How might the variables of the problem be adjusted to minimize this ratio?

a)

b)

(

²

) (

^,2 1

) (

^, 3 2

)

1 B 3 A

2 1 ln / 1 ln /

2 2 2 2

h i h o

h h i o

i o

T T T T

q q r q q

r r r r

r h k r h k

π

π π π π

∞ ∞

− −

′ = ′′ = +′ ′ = +

+ +

T∞,i T1 Th T2 T∞,o

(

2 2

)

h h

q′ =q′′ πr

q′i q′_o

1

1

2πr h_{i 3}

1 2πr h_o

(

2 1

)

B

ln / 2

r r πk

(

3 2

)

A

ln / 2

r r πk

c)

( )

( )

( ) ^{( )}

( ) ^{( )}

,

2 1

3 2

, 3 1

, 3 2

2 1 , 3

1

ln / 1

ln / 1

2 2

2 2

ln / 1

ln /

1 2 2

2 2

h o

h o

o o i

h i

i

h i

o i

T T

r r

r r T T

r h k

q r h k

T T

q r r

T T

r r r h k

r h k

π π

π π

π π

π π

∞

∞

∞

∞

−

⎛ ⎞

− +

+ ⎜ ⎟

′ = = ⎝ ⎠

′ − ⎛ ⎞

− ⎜ + ⎟

⎝ ⎠

+

To reduce q_o′ /q_i′, one could increase kB, hi, and r3/r2, while reducing kA, ho, and r2/r1.

6

4. A two dimensional rectangular plate is subjected to the boundary conditions shown.

Derive an expression for the steady-state temperature distribution T(x,y).

Hint: sin₂ cos ² sin 2

sin , sin

2 4

ax x ax x ax

x axdx axdx

a a a

= − = −

∫ ∫

2 2

2 2 0

T T

x y

∂ +∂ =

∂ ∂ b.c. x: T(0, ) 0, ( , ) 0y = T a y = y: ( ,0) 0, ( , )T x = T x b = Ax Assume ( , )T x y =X x Y y( ) ( ), then X Y ²

X^′′= − Y^′′= −λ X: X′′ +λ²X =0, (0) 0, ( ) 0X = X a =

1 2

( ) sin cos

X x =C λx C+ λx, x(0) 0= →C₂ =0, x a( ) 0= →C₁sinλa=0 To be non-trivial, _n n

a n

a

λ = π →λ = π . Thus _n( ) _nsinn x X x a

a

= π Y: Y y( )=C₃sinhλy C+ ₄coshλy, Y(0) 0=

(0) 0 4 0

Y = →C = Thus ( )_{n n}sinhn y Y x b

a

= π

1

( , ) _nsin sinh

n

n x n y T x y c

a a

π π

∞

=

=

∑

^,

1

( , ) _nsin sinh

n

n x n b T x b Ax c

a a

π π

∞

=

= =

∑

0

0

sin

sinh sin

a

n a

Ax n xdx c a

n b n x

a a dx π

π π

=

∫

∫

5. The finite difference method is used to determine the temperature distribution in the solid shown below and the heat transfer rate from the hot gas flow to the coolant through the solid. The thermal conductivity of the solid is k = 25 W/m^.K, and other conditions are; ho = 1000 W/m^2.K, T∞,o =1700 K, hi = 200 W/mK, and T∞,i = 400 K.

a) Write down the finite difference equations for nodal points 1, 3, 8, 12, 15, 16, 18, and 19 in terms the temperatures of neighboring nodes

b) Express the heat transfer rate per unit length q′to the channel in terms of the nodal temperatures.