Chapter02.pdf

2. HEAT CONDUCTION CONCEPTS, THERMAL RESISTANCE, AND THE OVERALL HEAT TRANSFER COEFFICIENT

2.1

Prove that if k varies linearly with T in a slab, and if heat transfer is onedimensional and steady, then q may be evaluated precisely using k evaluated at the mean temperature in the slab.

Solution: Differential Equation, 2

2

∂ T

∂ T

2

2

2

∂ T

q& 1 ∂T 2 ⋅ ∂t ∂x + ∂y + ∂z + { k = 1 α2 3 14243 = 0 , since T ≠T ( y or z )

=0

= 0 , since steady

The ordinary differential equation is d 2T =0 dx 2 Integrating twice T = C1 x + C2 Two Boundary Condition, T(x = 0) = T1, and T(x = L) = T2, Solving for C1 and C2. T1 = C2 T2 = C1L + C2 T − T2 C1 = 1 L Then (T − T )  T −T  T =  1 2  x + T1 = T1 − 1 2 x L  L  T − T1 x = T2 − T1 L q = −k q=k

dT dx

= −k

(T − T ) d  T −T   T1 − 1 2 x  = k 1 2 dx  L L 

∆T

L Using k = k o (1 + aT ) dT = − ko (1 + aT ) dx dx qdx = − k o (1 + aT )dT q = −k

dT

Integrating: T qL = −k o [T + 12 aT 2 ]T

2

1

qL = −k o [(T2 − T1 ) + 12 a (T2 − T1 2

2

)] 1


(T2 − T1 ) + 12 a (T2 2 − T12 )

q = −ko

L

q = − k o [1 + 12 a(T2 + T1 )] q = k o [1 + 12 a (T1 + T2 )] q = k o [1 + 12 a(T1 + T2 )]

(T2 − T1 ) L

(T1 − T2 ) L ∆T

L since ko [1 + 12 a (T1 + T2 )] = kave, q may be evaluated precisely using k evaluated at the mean temperature in the slab.

2.2

Invent a numerical method for calculating the steady heat flux through a plane wall when k(T) is an arbitrary function. Use the method to predict q in an iron slab 1cm thick if the temperature varies from 100 C on the left to 400 C on the right. How far would you have erred if you had taken kave = ( kleft + kright) / 2 ?

Solution: Tabulating Thermal Conductivity of iron slab, Table A.1. T,

C -100

k

98

0 100 200 300 400

Then q = − k

, W/m.K

84 72 63 56 50

dT

dx Numerical Method, q = constant, i =5

q∆x = −

∑ k ∆T i

i

i =1

where i is the number of increment stages. ki = average at that stage, and ∆Τi = change of temperature at that stage.

The ∆x = 1 cm = 0.01 m Numerical Calculation: Stage

Range Temp., C

∆ Ti

2

ki

ki∆Ti


1 2 3 4 5

-100 to 0 0 to 100 100 to 200 200 to 300 300 to 400

-100 -100 -100 -100 -100

91 78 67.5 59.5 53

-9100 -7800 -6750 -5950 -5300

i =5

∑ k ∆T = -34,900 W/m i

i

i =1 i =5

q∆x = −

k ∆T

∑ i =1

i

i

q(0.01 m) = - (-34,900 W/m) q = 3,490,000 W/m2

Using average value: kave = (kleft + kright) / 2 = (91 + 53) / 2 = 72 W/m.K k ave Tleft − Tright q=− L (72 )( − 100 − )400 q=− = 3,600,000 W/m2 0.01 error = ∆q = 3,600,000 W/m2 – 3,490,000 W/m2 = 110,000 W/m2 or 3.2 % above the

numerical result. 2.3

The steady heat flux at one side of a slab is a known value qo. The thermal conductivity varies with temperature in the slab, and the variation can be expressed with a power series as i =n

k=

∑ AT

i

i

i =0

(a) Start with eqn. (2.10) and derive an equation that relates T to position in the slab, x. (b) Calculate the heat flux at any position in the wall from this expression using Fourier’s law. Is the resulting q a function of x ? Solution: (a)

∂T Eq. (2.10) ∇ ⋅ k ∇T + q& = ρ c t 12∂3 =0 {

= 0 since steady

∇ ⋅ k ∇T = 0

d (k ∇T ) =0 dx k ∇T = C

3


k

dT = C1 dx

i n i   ∑ AiT dT = C1dx i 0  i n   AiT i dT = C1 ∫ dx ∫  ∑ i 0  =

=

=

=

Boundary Conditions: x = 0, T = T1 and x = L, T = T2 i +1 i =n AiT1 C1 (0) + C 2 = = C2 i +1 i =0 i +1 i=n AiT1 C1 (L ) + C2 = C1 L + = i +1 i=0

∑

∑

i =n

∑ i=0

C1 =

i +1

AiT2 − i +1 L

i= n

∑ i= 0

i=n

∑ i=0

i +1

AiT2 i +1

i +1

AiT1 i +1

Then i=n

∑ i=0

AiT i +1  i = n AiT2 = − i + 1  i =0 i + 1 i +1

∑

i =n

∑ i =0

i +1

AiT1 i +1

 x  i n AiT1i 1   + ∑  L  i 0 i + 1 =

+

=

q = − k dT dx Differentiating above result,

(b)

i n i  dT  ∑ AiT  i 0  dx =

=

 i n AiT2 i 1 ∑  i 0 i +1 =

i 1 dT  i n AiT2 k = ∑ dx  i 0 i + 1 +

=

i=n

−

=

 1    i + 1  L  i 0 i 1 AiT1  1    i + 1  L  +

=

=

∑ i =0

i =n

−

∑

AiT1

i +1

=

+

Then,

i

=n

i

=0

q = 

∑

AiT2

i +1

i +1

i =n

−

∑ i=0

 1    i + 1  L 

AiT1

i +1

The resulting q is not a function of x. 2.4

Combine Fick’s law with the principle of conservation of mass (of the dilute species) in such a way as to eliminate ji, and obtain a second-order differential equation in m1. Discuss the importance and the use of the result.

Solution: Eq. (2.19), Fick’s law 4


r j1 = − ρD 12∇m1 Then ∂m1 j1 = − ρD 12 ∂x

 ∂m1

∆j1 = − ρD12 

 ∂x

  ∂x x 

∂m1

− x + δx

By law of conservation of masses, ∆j1 = 0. ∂m1 ∂x

x + δx

∂m1

− ∂m1 ∂x −

∂x

x + δx

=0 x

∂m1 ∂x

δx

2

x

=

∂ m1 ∂x

2

=0

2

∂ m1 ∂x

2

=0

The importance and the use of the result simplify the means to use the Ohm’s law for electrical resistance, or heat transfer equation for steady state analogy. 2.5

Solve for the temperature distribution in a thick-walled pipe if the bulk interior temperature and the exterior air temperature, T∞i and T∞o , are known. The interior and the exterior heat transfer coefficients are hi and ho , respectively. Follow the method in Example 2.6 and put your result in the dimensionless form. T − T∞i T∞i − T∞o

= fn (Bii , Bio , r ri , ro ri )

Solution: Follow Ex. 2.6 Eq. (2.23)

∂ (C1 ln r + C2 )  ∂r 

h [(C1 ln r + C2 ) − T∞ ] = − k  h [(C1 ln r + C2 ) − T∞ ] = −

To > T∞o

kC1 r

Then, r = ro , ho [(C1 ln ro + C 2 ) − T∞o ] = −

kC1 ro

5


ho [T∞o − C1 ln ro − C2 ] =

kC1 ro

at r = ri , in reverse. hi [T∞i − (C1 ln ri + C 2 )] = −

kC1 ri

kC1 hi [(C1 ln ri + C 2 ) − T∞i ] = ri Then: kC1 C T∞o − C1 ln ro − C 2 = = 1 ho ro Bio C1 ln ri + C2 − T∞i =

kC1

C1

=

hi ri

Bii

Adding:

 1

T∞o − T∞i − C1 (ln ro − ln ri ) = C1 

−

 Bio

 1  Bi  o

T∞o − T∞i = C1 

−

r 1 + ln o r Bii  i

T −T ∞o ∞i C1 =  1 r 1 − + ln o  r Bi Bi i  i  o

C2 = T∞i + C2 = T∞i +

)

  

    

   

C1 − C1 ln ri Bii T∞o − T∞i

 1 Bii   Bio

1 Bii

  

−

(T o − T i ) ln ri ∞

∞

 1  r  1 − + ln o    r  Bi Bi i  i   o  1  1  r   r  1 1 − + ln o   = T i Bii  − ) +( ln o   + T o − T i − Bii T o − T i ln ri C2 (Bii  r  r  Bi Bi Bi Bi i  i  i  i   o  o  1         T −T i  1 1 1 r r  − (T o − T i ) ln ri − + ln o   = T i  − + ln o   +  o C2       o i i o i i i  Bi Bi  r   Bi Bi  r   Bi   1    (T i − T o ) − lnr i   Bii  C2 = T i −  1  ro  1 − + ln    r   i   Bio Bii r 1 − + ln o r Bii  i

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

6

∞

∞


Then: T = C1 ln r + C2

    T o −T i  ln r + T T =  1  ro   1 + + ln    r   i    Bio Bii

   Bii  − i  1  ro   1 − + ln    r   i   Bio Bii  1  r  (T − T ) 1 (T o − T i ) ln r + T i  Bio + Bii + ln roi  − oBii i + (T i − T o )ln ri T=  1  r  1 + + ln o   r   i   Bio Bii  1 r  r  (T − T i ) 1 (T o − T i ) ln  + T i  + + ln o  − o r Bi Bi Bii i  i  ri   o T=  1  r  1 + + ln o    r   i   Bio Bii  r  (T − T i ) (T o − T i ) ln  − o Bii  ri  +T i T=  1  ro   1   + + ln    o i i  Bi Bi  r   r 1  (T o − T i )ln  +    ri  Bii  T −T i =  1  r  1 + + ln o   r  Bi Bi i  i   o r 1 ln  + T −T i  ri  Bii = T o −T i r  1 1 + + ln o  Bio Bii  ri  r 1 ln  + T −T i  ri  Bii =− ∞

 1

(T i − T o ) ∞

∞

− ln r i 

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

∞

T∞i − T∞o

1 + 1 + ln ro  r  Bio Bii  i

2.6 Put the boundary conditions from Problem 2.5 into a dimensionless form so that Biot number appear in them. Let the Biot numbers approach infinity. This should get you 7


back to the conditions for Example 2.5. Therefore, the solution that you obtain in Problem 2.5 should reduce to the solution of Example 2.5 when the Biot numbers approach infinity. Show that this is the case.

Solution: Result from Problem 2.5.

  T − T∞i

=−

T∞i − T∞o

ln r  + 1  ri  Bii

r  1 1 + + ln o  r  Bio Bii  i

if Bii → ∞, Bio → ∞ Then, Ti → T∞i , To → T∞o T − Ti

=−

Ti − To

T − Ti To − Ti = −

r ln  + 0  ri   ro    ri 

0 + 0 + ln

r ln   ri   ro    ri 

ln

The result is the same as Example 2.5.

2.7 Write an accurate explanation of the idea of critical radius of insulation that your kid brother or sister, who is still in grade school, could understand. (if you don not have an available kid, borrow one to see if your explanation really works.)

Solution: Step 1. Write down equation (2.25). Q=

Ti − T∞

r ln o  ri  1  + 2π kl h 2π ro l Step 2. Multiply numerator and denominator by ri.

8


ri (Ti − T∞ )

Q=

r ri ln o  ri  ri  + 2π kl h 2π ro l Step 3. Let x = ro/ri, then: ri (Ti − T∞ ) Q= 1 r ln( x ) + i h 2π xl 2π kl

Step 4. Let constant a = ri (Ti − T∞ ) , b=

1 r , c= i . 2π kl h 2π l

b  + c ln x  x  

a

Q=

−1

= a

b + c ln x x Step 5. Differentiate y with respect to x then equate to zero to find x for maximum or minimum Q. Q = a (bx −1 + c ln x )

−1

dQ

= −a

(bx

−1

dx

=

x x=

)  − bx −2

−2

+

c x

=

−2

+

c

=0

x



− bx

c

+ c ln x

0

b x2 b

c Step 6. To check if this is maximum or minimum. Solve for the second derivative, positive for minimum and negative for maximum. dQ −2 −1 −2 −1 = − a (bx + (c )ln x −) bx + cx dx dQ −2 −1 −2 −1 =

+

−

dx a (bx c( ln ) x )bx cx d 2Q −1 = −2a (bx + c(ln ) x −3 − bx )( − 2 + cx)−1( bx − 2 −(cx ) −1 + a bx)−1 + c ln x − 2 − 2bx −3 + cx −2 dx 2 d 2Q −1 = 2a (bx + c( ln ) x −3 − )bx( −2 + cx −(1) 2 − a bx)−1 + c ln x − 2 2bx −3 − cx − 2 dx 2 assume, 0.5 cm OD copper line, ri = 0.005 m / 2 = 0.0025 m, l = 1 m, k = 0.074 W/m.K, h = 20 W/m2.K

9


b= c=

1

=

h 2π l ri

2π kl

=

1

(20)( 2π)( 1)

= 0.00796

0.0025 = 0.00538 (2π)( 0.074)( ) 1

 1    h 2π l  b k = x= = c  ri  h ri    2π kl 

=

d 2Q = 2a [(0.00796 )( 1).48( dx 2

[

− a (0.00796 )(

1).48 (

−1

+

0.074

(20)( 0.0025 ) −1

+

= 1.48

) ln 1.48 ]( [ 0.00796 )( ) 1( .48 0).(00538

0).(00538 ) ln 1.48

−3

] ([2 0.00796 )( ) 1.(48 −2

−3

−2

)− 0.00538

−)(0.00538 )

1.48

−1

−2

]

2

]

d 2Q −12 −3 = 2a (2,382,200)(1.2 × 10 ) − a (17,837 )(2.4547 × 10 ) = −43.87a is negative, dx 2 therefore, it is maximum at x = 1.48 = ro/ri.

Step 7. Critical radius of insulation = rcrit = ro. rcrit k = ri h ri k rcrit = h

2.8 The slab shown in Fig. 2.22 is embedded on five sides in insulation materials. The sixth side is exposed to an ambient temperature through a heat transfer coefficient. Heat is generated in the slab at the rate of 1.0 kW/m 3. The thermal conductivity of the slab is 0.2 W/m.K. (a) Solve for the temperature distribution in the slab, noting any assumptions you must make. Be careful to clearly identify the boundary conditions. (b) Evaluate T at the front and back faces of the slab. (c) Show that your solution gives the expected heat fluxes at the back and front faces. Fig. 2.22

10


Solution: q& = 1.0 kW/m3 = 1000 W/m3 k = 0.2 W/m.K h = 20 W/m2.C T∞ = 25 C Assume no thermal radiation.

(a) First boundary condition, Eq. 2.11 2 2 1 ∂T T ∂ T ∂ T q& + + + = 2 2 2 ∂x ∂y ∂z k α t 12∂3 14243

∂

2

= 0 , since T ≠T ( y or z )

Then, d 2T q& =− 2 dx k q& 2 T =− x + C1 x + C 2 , 2k

= 0 , since steady

dT q& = − x + C1 dx k

Second boundary condition, qconvection = qconduction at the wall h (T − T∞ ) = −k ∂T ∂x L = thickness ∂   q& 2  q& 2    q&  x + C1 x + C2  − T  = − k  − x + C1 x + C2  = − k  − x + C1  ∂x  2k    k   2k 

h  −

∞

at x = 0 11


 q& )(0)(2 + C1 0 2 k 

h  −



 

 q& 0  k

+ C 2  − T∞)( = − k  −



 

+ C1 

h (C2 − T∞ ) = − kC1 C1 =

h k

(T

∞

− C2 )

at x = L

 q& 2    q&  L + C1 L + C 2  − T  = −k  − L + C1    k   2k 

h  −

∞

q& 2 h L + (T)∞ − C2 L + C 2 − T∞ =( 2k k q& 2 h L hL − L + T∞ − C2 + C 2 − T∞ 2k k k q&L q& k C2 = − − + T∞ 2h h 2 h  q&L q&k  − 2 + T∞   C1 = T∞ −  − k  2h h  −

q&

k h

h

h k

) L−

=

q& h

T∞ − C2

L − T∞ + C2

 q&L q&k  q&  L k  + 2 =  +    2h h  k  2 h  q&x 2 q&  L k  q& L q& k +  + x − − +T T =− 2k k  2 h  2h h 2  x2  L 1   L k  + x +  + 2  T − T = − q&  −   2k  2k h   2h h   C1 =

h k

∞

∞

(b) q& = 1.0 kW/m3 = 1000 W/m3 at the front: x = L = 0.10 m

 (0.10 )2

T f − 25 = −(1000 )

 (0.10)

−

 2(0.20)  2(0.20)

+

 (0.10 ) (0.20 )   1 (0.10 ) +  + 2   20   2(20) (20)  

T f - 25 = 2 T f = 27 C

at the back: x = 0



 (0.10 )



 2(0.20 )

Tb − 25 = −(1000 )0 − 

+

 (0.10 ) (0.20 )   1  (0) +  + 2  20   2(20) (20)  

Tb - 25 = -3 Tb = 22 C

12


(c ) Check of heat flux.

 x2

 L 1  L + x +   2k h   2h dT x L 1 = − q&  − −  dx  k 2k h 

T − T∞ = − q& 

 2k

−

+

k 



h 2 

at the back of the slab, x = 0 qb = − k

∂T

x L 1 − −   k 2k h  x

kq& 

=

∂x

x =0

=0

 L 1 +  qb = − kq&   2k h  qb = −

q& L q& k  q&L q&k  − = − +  2 h  2 h 

at the front of the slab, x = L

x L 1 − −   k 2k h  x x L L L 1 −  q f = kq&  −  k 2k h 

q f = −k

q& L

qf =

∂T

=

∂x

−

2

kq& 

=

=L

q& k h

Putting equation into neat dimension form.

 x2

k   L 1  L + x +  + 2   2k  2k h   2h h   q&  x 2  L k   kL k 2  T − T = −  −  + x +  + 2  k 2 2 h  2 h h  q& L2  x 2  1 k   k T − T∞ = − q& 

−

∞

−

T

=−

T

∞

T − T∞ = −

−

k  2 L

2

+

 2 L

+

h L  x  2h L 2

k 2 

+

h 2 L2 

2 2k  x   k 2k 2   q& L2  x   + 2 2    +    −  1 + 2k  L   h L  L   h L h L  

2k  x   x   k 2k 2   T − T∞ 1  = 1 + + 2 2    −   −  2 q&L 2  h L  L   L   h L h L   k 2

13


or 2 k  x   x   k  2k  T − T∞ 1  = 1 +   −   −  1 +  q&L2 2  h L  L   L   h L  h L  k 2



d  T − T∞  2 dx  q& L k 

 2  1  2k  1   x   1     − 2     = 0  = 1 +  2  h L  L   L   L   

x = 1 1 + 2k L 2  h L 

2(0.2 )  1 +  = 0.6 L 2  (20)( 0.)1  1  2(0.2 )  T −T ( 0).6( − ) 0.6 = 1 + 2 q&L 2  (20)( 0.)1  x

=

1

∞

2



0.2  (20)( 0.)1

− 

k x T − T∞ = 0.12 , = 0.6 q& L2 L k x x T − T∞ T − T∞ and at = 0, = -0.06 : = 1, = 0.04 q& L2 q& L2 L L k k

 2(0.2 ) 1 +  (20)( 0.)1

  

Plotting:

q& L q& k  q& L q& k  − − − −  = q&L 2 h  2 h  Heat energy generated at the back of the slab is equal to half of the total energy plus the film coefficient effect. Heat energy generated at the front of the slab is equal to the half of the total energy less the film coefficient effect. Thus giving the solution appears to be correct.

Therefore

Total

Energy

Generated

=

14

q f − qb =


2.9 Consider the composite wall shown in Fig. 2.23. The concrete and brick sections are of equal thickness. Determine T1, T2, q, and the percentage of q that flows through the brick. To do this, approximate the heat flow as one-dimensional. Draw the thermal circuit for the wall and identify all four resistances before you begin. Figure 2.23

Solution: Thermal Circuit:

∆T = Tw1 − Tw 2 = 370

C – 66 C = 304 C

15


1

Rtotal = Rt fir +

1 Rtcinder

Rt fir =

L fir

Rtbrick =

Rt pine =

Rtbrick

0.025 m

(0.11W / m ⋅)(C) A

Lcinder

Rtcinder =

Rtotal =

=

k fir A fir

=

kcinder Acinder Lbrick

=

k brick Abrick

L pine k pine Apine

0.2273 + A

+ Rt pine

1

+

=

=

0.2273 A

0.075 m

( 2)

(0.76 W / m ⋅ C ) A 0.075 m

(0.69 W / m ⋅ C )(A 2 ) 0.05 m

(0.147 W / m ⋅)(C ) A

=

=

=

0.1974 A

0.2174 A

0.3402 A

1 0.3402 + A A A + 0.1974 0.2174

0.671 A ∆T 304 Q= = = 453.06 A Rtotal  0.671     A  Rtotal =

q = Q = 453.06 W/m2 A Solving for T1 and T2.

Fir: Q=

∆T fir

Rt fir

453.06 A =

(370C − T1 )  0.2273     A 

T1 = 267 C

Yellow Pine: Q = ∆Tpine Rt pine

453.06 A =

(T2 − 66 )  0.3402     A 

T2 = 220 C

16


For concrete cinder block. (267 − 220) ∆Tcinder Q1 = = = 238.1A Rtcinder  0.1974     A  For building brick (267 − 220 ) ∆Tbrick Q2 = = = 216.2 A Rtbrick  0.2174     A  Q = Q1 + Q2 = 238.1A + 216.2A = 454.3A W q = 454.3A ~ 453.06A Percentage of q that flows through the brick 216.2 A Q = 2 (100 ) % = ( ) 100% = 47.6 % Q1 454.3 A

2.10 Compute Q and U for Example 2.11 if the wall is 0.3 m thick. Five (each) pine and sawdust layers are 5 and 8 cm thick, respectively; and the heat transfer coefficients are 10 on the left and 18 on the right. T∞l = 30 C and T∞r = 10 C. Solution:

Thermal Circuit

17


1

Rtotal = Rt1conv +

1 Rt pine

Q=

∆T

Rttotal

=

1

+

+ Rt 2 conv

Rt sawdust

T∞l − T∞r 1 1 1 + + hl A  k p Ap k s As  hr A

  L

+

 L 

L = 0.3 m hl = 10 W/m2.C hr = 18 W/m2.C Thermal conductivity of pine and sawdust, Appendix A. k p = 0.14 W/m.C k s = 0.06 W/m.C T∞l = 30 C T∞r = 10 C Assume 1 m length Ap = 5(0.05 m)(1 m) = 0.25 m2 As = 5(0.08 m)(1 m) = 0.40 m2 Q=

30 − 10 1 1 1 + + (10)( 0.65 ) ( 0.)(14 0) .25 (0.06)( 0.40 ) )  (18)( 0.65 +  (0.3) (0.3)  

Q = 3.7565 W

18


U=

Q

3.7565

=

)( 30 −)10 A∆T (0.65 2 U = 0.29 W/m .K

2.11 Compute U for the slab in Example 1.2.

Solution: Fig. 1.7

Ls = 2 mm = 0.002 m Lc = 3 mm = 0.003 m ks = 17 W/m.K kc = 372 W/m.K

Thermal Circuit

Q=

∆T

∑ Rt

=

∆T 2 Ls L + c k s A kc A

19


U=

U=

Q A∆T

=

1 2 Ls Lc + ks kc

1 = 4109 W/m2.C 2(0.002 ) 0.003 + 17 372

2.12

Consider the tea kettle in Example 2.10. Suppose that the kettle holds 1 kg of water (about 1 liter) and that the flame impinges on 0.02 m 2 of the bottom. (a) Find out how fast the water temperature is increasing when it reaches its boiling point, and calculate the temperature of the bottom of the kettle immediately below the water if the gases from the flame are at 500 C when they touch the bottom of the kettle. Assume that the heat capacity of the aluminum kettle is negligible. (b) There is an old parlor trick in which one puts a paper cup of water over an open flame and boils the water without burning the paper. Explain this using an electrical analogy.

Solution: (a)

U = 192.1 W/m2.K hb = 5000 W/m2.K cp = 4187 J/kg.K Q = mc p

dT

dt T = 100 C

= UA∆T = UA(T∞ − T )

 dT  2 2  = (192.1 W/m .K)(0.02 m )(500 C – 100 C)  dt 

(1 kg)(4187 J/kg.K) 

dT dt = 0.367 C/s Q = hb A(T)1 − T (= UA) T∞ − T

(5000)(0.02)(T1 – 100) = (192.1)(0.02)(500 – 100) T1 = 115.37 C

(b)

Assume negligible paper cup thickness. 20


Burning temperature of paper cup is at 235 C from various sources. Check for temperature of paper cup at water boiling temperature of 100 C. Use the same data as in (a) Tboiling water = 100 C T flame = 500 C q=

T flame − Tboiling water 1 L 1 h + k + hb

h = 200 W/m2.K hb = 5000 W/m2.K L

≈0 k Thermal Circuit

Then, 500 − 100 q= = 76,923 W/m2. 1 1 + 200 5000 Tc = paper cup temperature q 76,923 = 500 − h 200 Tc = 115.4 C << 235 C Tc = T flame −

Therefore paper cup will not burn. 2.13

Copper plates 2 mm and 3 mm in thickness are processed rather lightly together. Non-oil bearing condenses at Tsat at = 200 C on on the oneother side ( h =12,000 W/m2steam .K) and methanolunder boils pressure under pressure 130 C 2 ( h =9,000 W/m .K). Estimate U and q initially and after extended service. List the relevant thermal resistance in order of decreasing importance and suggest whether or not any of them can be ignored.

Solution:

21


Thermal conductivity of copper @ 150 C, k = 239 W/m.K Initially: Rtotal = Rt steam + Rt c + Rt c + Rtmethanol 1

2

After Extended Service: Rtotal = Rt steam + Rtc + Rt contact + Rtc + Rt methanol 1

2

List of thermal resistance in order of decreasing importance, L (0.002 m ) 8.3682 × 10 −6 Rtc1 = kA = (239 W / m.K )A = A Rtc = 2

Rtcontact

L

=

kA

(0.003 m ) (239 W / m.K ) A

=

1.2552 × 10 −5 A

1 = hc A

hc = 20,000 W/m2.K Rtcontact =

1 1 5.0 ×10 −5 = = hc A (20,000W / m 2 .K )A A

Rt steam =

1 1 8.333 × 10−5 = = 2 hs A (12,000 W / m .K )A A

Rt methanol =

1

1

=

hm A

=

2

1.111 × 10 −5

(9,000 W / m .K )A

A

These shows that not any of them can be ignored. Initially: Rtotal = Rt steam + Rt c + Rt c + Rtmethanol 1

2

8.3682 × 10 −6 1.2552 × 10 −5 8.333 × 10 −5 1.111× 10 −5 1.1536 × 10 −4 Rtotal = + + + = A A A A A ∆T Q= = UA∆T Rtotal 1 1 2 = 8669 W/m .K U= = −4 Rtotal A (1.1536 × 10 ) q=

Q

= U∆T

= (8669)(200 – 130) = 606,830 W/m2.

A

After Extended Service: Rtotal = Rt steam + Rtc + Rt contact + Rtc + Rt methanol 1

2

22


8.3682 × 10−6 1.2552 × 10 −5 5.0 × 10 −5 8.333 × 10 −5 1.111 × 10 −5 + + + + A A A A A 1.6536 × 10 −4 Rtotal = A ∆T = UA∆T Q= Rtotal 1 1 2 = 6048 W/m .K U= = −4 Rtotal =

×

Rtotal A (1.6536 10 ) Q q = = U∆T = (6048)(200 – 130) = 423,360 W/m2. A

2.14

0.5 kg/s of air at 20 C moves along a channel that is 1 m from wall to wall. One wall of the channel is a heat exchange surface ( U = 300 W/m2.K) with steam condensing at 120 C on its back. Determine (a) q at the entrance; (b) the rate of increase of temperature of the fluid with x at the entrance; (c) the temperature and heat flux 2 m downstream.

Solution: & = 0.5 kg/s m T1 = 20 C L=1m

2

U .K T∞== 300 120 W/m C

(a) At the entrance q = U (T∞ − T1 ) = (300 W/m2.K)(120 C – 20 C) = 30,000 W/m2

(b) At the entrance dQ = UdA(T∞ − T ) = m& c p dT

Let w = width dA = wdx

 dx    dr 

& = ρwL m

Then dQ = Uwdx (T∞ − T ) = ρwLc p dT dT

=

U (T∞ − T )

ρc p L L = 1.0 m dt

23


Density of air at 20 C, ρ = 1.205 kg/m3. Specific heat of air at 20 C, cp = 1006 J/kg.K dT

=

dt

(300 )( 120 −) 20 = 24.75 C/s (1.205)( 1006 )() 1

(c) For square channel, w = L = 1 m dT = U (T∞ − T ) dt ρc p L dT dx ⋅

U (T∞ − T )

=

ρc p L

dx dt

dT  m&  U (T∞ − T )  = ρc p L dx  ρwL  dT

=

dx

∫

Uw(T∞ − T ) & cp m

dT

=

T∞ − T

T

∞

− T1

T

∞

− T2

ln

∫

Uwdx & cp m

 Uwx  =  m& c p

at x = 2 m, solve for T2.

 120 − 20  (300)(1)( 2)  = )  120 − T2  (0.5)( 1006

ln

T2 = 89.7 C q2 = U (T∞ − T2 ) = (300 W/m2.K)(120 C – 89.7 C) = 9090 W/m2.

2.15

An isothermal sphere 3 cm in diameter is kept at 80 C in a large clay region. The temperature of the clay far from the sphere is kept at 10 C. How much heat must be supplied to the sphere to maintain its temperature if kclay = 1.28 W/m.K? (Hint: You must solve the boundary value problem not in the sphere but in the clay surrounding it.)?

Solution: Laplacian equation (2.12) 2

2

∇ T =

∂T ∂x

2

2

+

∂ T ∂y

2

2

+

∂ T ∂z

2

For clay from the source in any position: 24

2. HEAT CONDUCTION CONCEPTS, THERMAL RESISTANCE, AND THE OVERALL HEAT TRANSFER COEFFICIENT 2

∂ T ∂x

2

2

=

∂T ∂y

2

2

∂ T

=

∂z

2

2

∇ T =3

∂T ∂x

2

2 2

=3

∂ T ∂y

2

2

=3

∂ T ∂z

2

Heat diffusion equation: ∇ ⋅ k∇T + q& =

ρc

∂T

∂t 123

= 0 since steady

2

∇ T =−

q&

k Note: x, y, z are interchangeable, use the following step.

Boundary condition no.1 for x-coordinate. 2 ∂ T q& 3 2 =− ∂x k 2 ∂ T q& ∂x

2

=−

3k q& =− x + C1 ∂x 3k q& 2 T =− x + C1 x + C2 6k Other Boundary condition, y- and z- coordinates. q& 2 T =− y + C1 y + C 2 6k q& 2 T =− z + C1 z + C2 6k Boundaries: x1 = 3 cm / 2 = 1.5 cm = 0.015 m ∂T as y 2 → ∞, z 2 → ∞, x2 → x1 , = 0 at maximum ∂x ∂T q& =− x1 + C1 = 0 ∂x 3k q& C1 = x1 3k as x2 → ∞, z 2 → ∞, y2 → y1 , T1 = 80 C q& 2 T1 = − y1 + C1 y1 + C2 6k q& but is also C1 = y1 3k then ∂T

25


q& 2 q& 2 q& 2 y1 + y1 + C2 = y1 + C2 6k 3k 3k q& 2 C2 = T1 − y1 3k as x2 → ∞, y2 → ∞, z 2 → 0, T2 = 10 C q& 2 T2 = − z 2 + C1 z 2 + C 2 = 0 + 0 + C2 6k T2 = C2 q& 2 T2 = T1 − y1 3k q& 2 y1 = T1 − T2 3k x1 = y1 = z1 = 0.015 m q& 2 x1 = T1 − T2 3k 3k (T1 − T ) q& = 2 x1 T1 = −

4 3 but Q = q&V = q& πx1  3   4 3   3k (T1 − T2 )  Q =  πx1   2  x1 3   Q = 4π x1k (T1 − T2 ) Q = 4π (0.15)(m 1.28W /)(m.K 80 C ) − 10 C = 16.9 W

2.16

Is it possible to increase the heat transfer from a convectively cooled isothermal sphere by adding insulation? Explain fully.

Solution: Determine first the temperature distribution and heat flux. Step 1. T = T(r) Step 2. Laplacian equation 2 1 ∂ 2 (rT ) 1 ∂  ∂T  1 ∂T 2 ∇ ≡ + + 2 2 2 2 T r ∂r r sin θ ∂θ  sin θ ∂θ  r sin θ ∂φ 2 For isothermal, or uniform sphere surface temperature 1 ∂ 2 (rT ) 2 ∇ T ≡ r ∂r 2 Step 3. Heat diffusion equation ∂T ∇ ⋅ k∇T + q& = ρ c ∂t 26


∇ T+

q&

1 ∂T α t 12∂ 3

=

k

{

=0

= 0 since steady

2 1 ∂ (rT ) ∂ (rT ) = 0 , =0 2 r ∂r 2 ∂r 2

Step 4. Integrate once:

∂ (rT ) ∂r

= C1 ; integrate again: rT = C1r + C 2

T = C1 + C2 r Step 5. The first boundary condition is T(r = r1) = Ti. The second boundary condition must be expressed as an energy balance at the outer wall.

qconvection = q conduction at the wall or h (T − T∞ )r = ro = − k

∂T ∂t

r = ro

Step 6. From the first boundary conditon we obtain C T = C1 + 2 r   C  ∂  C  h  C1 + 2  − T∞  C1 + 2  = −k r  ∂t  r  r = ro   r = ro



h  T∞ − C1 −



C2 

C  = k 22 ro  ro

Eliminating C1: C kC2 T∞ − C1 − 2 = 2 ro h ro C kC2 C1 = T∞ − 2 − 2 ro h ro Then, C C kC 2 C2 Ti = C1 + 2 = T∞ − 2 − + 2 ri ro h ro ri Ti − T∞ = C2  1 − 1 − k 2   ri ro h ro  C2 =

(Ti − T∞ )ro Ti − T∞ = 1 1 k ro k − − −1 − 2 ri ro h ro ri h ro

27


(Ti − T )ro  1 

k   1 −   ro k   ro  h ro   − 1 −  h ro   ri (Ti − T )  k  1 −  C1 = T − h ro   ro  k   − 1 −  h ro   ri ∞

C1 = T∞ −

∞

∞

Step 7. T = C1 + T = T∞ −

C2 r

 (k ) Ti − T  ro  1 −  +    ro k   h ro   ro k  r   − 1 −   − 1 −  h ro  h ro   ri  ri

(Ti )− T

∞

∞

Rearranging: ro k ro − 1+ −1+ r h ro T − T∞ = = r k ro Ti − T∞ ro −1− −1− ri h ro ri

1 Bi 1 Bi

  ro   2   − r =− = k ∂r k  ro 1 (Ti T )  −1−  Bi   ri ∂T

Step 8. q radial

∞

Q = qradial (4π r 2 )

  Q = k  ro r  i   Q = k  ro r  i Q

=

  (T − T ) 1  i −1 − Bi    4π ro (T − T ) k  i −1− h ro  4π ro

∞

∞

Ti − T∞ Ti − T∞ = 1 1 1 1 1 1 1  − −  −  − 2 2 4π ri k 4π ro k h 4π ro 4π k  ri ro  h 4π ro

Step 9. Determine critical radius of insulation, setting

28

dQ dro

=0


dQ dro

 1  2 1   4π ro k 2  h 4π ro 

− (Ti − T∞ )

=

 1 1 1  −  −   4π k  ri ro 

2

+

 =0 h 4π ro  2

3

1 2 + cannot be equal to zero except for approaching infinity. As the 2 3 4π ro k h 4π ro outside radius ro increases the heat transfer always decreases, so it is not possible. Since

2.17 A wall consists of layers of metals and plastic with heat transfer coefficients on either side. U is 255 W/m2.K and the overall temperature difference is 200 C. One layer in the wall is stainless steel ( k = 18 W/m.K) 3 mm thick. What is ∆T across the stainless steel? Solution: q = U (T2 − T1 ) =

k∆T

L T2 − T1 = 200 C, L = 3 mm = 0.003 m (18)∆T q = (255)(200) = (0.003) ∆T = 8.5 C

2.17

A 1 % carbon-steel sphere 20 cm in diameter is kept at 250 C on the outside. It has 8 cm diameter cavity containing boiling water ( h inside is very high) which is vented to the atmosphere. What is Q through the shell?

Solution: Neglect h inside. Thermal conductivity of 1% carbon steel at 175 C. k = 42.25 W/m.K ro = (1/2)(20 cm) = 10 cm = 0.10 m ri = (1/2)(8 cm) = 4 cm = 0.04 m Derivation: 1 ∂ 2 (rT ) =0 r ∂r 2 ∂ 2 (rT ) = 0 2 ∂r ∂ (rT ) ∂r

= C1

rT = C1r + C 2 C T = C1 + 2 r

29


T (r = ri ) = Ti = 250 C T (r = ro ) = To = 100 C C2

Ti = C1 +

ri C2

To = C1 + −

ro

1 

=

To Ti C2 =

−

C 2  ro

1



ri 

To − Ti 1 1 − ro ri





1 (To − T )i  

 ro  1 1 − ro ri

To = C1 +





1 (To − T )i  

 ro 

C1 = To −

1 1 − ro ri





1 (To − Ti )  (To − Ti ) 1 

 ro  + 1 1 − ro ri

T = To −

1

(To − Ti )

r 1 1 − ro ri

T − To =

∂T

−

r

1 1 − ro ri

1  ro 

1 2  r 

− (To − Ti ) =

∂r

Q = − kA

1−1 ro ri ∂T ∂r

A = 4π r 2

30


Q = −k

Q=

2.18

(To − T)(i 4)π 1 1 − ro ri

− (42.25 )(

250 − 100 )( ) 4π = 5,309 W 1 1 − 0.10 0.04 A 1% carbon-steel sphere 20 cm in diameter is kept at 250 C on the outside. It has an 8 cm diameter cavity containing boiling water ( hinside is very high) which is vented to the atmosphere. What is Q through the shell.

Solution: To = 250 C Ti = 100 C Boiling ro = (1/2)(20 cm) = 10 cm = 0.10 m ri = (1/2)(8 cm) = 4 cm = 0.04 m

For 1 % carbon steel, k = 42 W/m.K at 20 0 C, Table A.1 dT Q = − kA dr Qdr = − k (4π r 2 )dT dr Q r 2 = −4π kdT r dr Q ∫rio 2 = −4π k (To − Ti ) r ro  1 Q −  = −4π k (To − Ti )  r  ri

1 1  Q  −  = −4π k (To − Ti )  ri ro  Q=

2.19

− 4π k (To −)Ti

1 1 − ri ro

=

( )(4π −

) − 100 42 250 = -5278 W 1 1 − 0.04 0.10

A slab is insulated on one side and exposed to a surrounding temperature , T∞ , through a heat transfer coefficient on the other. There is nonuniform heat generation in the slab such that q& = A(W / m 4 ) [x (m )] , where x = 0 at the insulated wall and x = L at the cooled wall. Derive the temperature distribution in the slab.

Solution: 2 ∂ T q =− 2 ∂x k 31


∂ T

Ax k Integrating once, ∂T Ax 2 =− + C1 ∂x 2k ∂x

2

=−

T =−

Ax 3

+ C1 x + C2 6k qconvection = qconduction ∂T h (T − T∞ )x = L = − ∂x x = L

 Ax 3

h−

  AL3 h−  6k



 Ax 2

+ C1 x + C2 − T∞ 

= −k  −

x L    AL2 + C1 L + C 2 − T  = − k  −   2k 2 3 k AL  AL − C1 L C2 = T −  C1 − + h 2k  6k 6k

=

2k

∞



+ C1 

x

=L

 + C1  

∞

at x = 0  Ax 3   Ax 2  + C1 x + C2 − T∞  = −k  − + C1  h−  6k  x=0  2k  x=0 h (0 + 0 + C 2 −) T(∞ )= − k 0 + C1 kC C2 = T∞ − 1 h Then, k AL2  AL3 C2 = T∞ −  C1 − − C1 L + h 2k  6k AL2

+

AL3

− C1 L = 0 6k AL AL2 C1 = + 2h 6k k  AL AL2  C2 = T∞ − 2  +  h  2h 6k 

2k

Ax 3  AL AL2  + +  x + T∞ − kAL2 − 6k  2h 6k  2h Ax 3  AL AL2  kAL + + T − T∞ = − x − 2 − 6k  2h 6k  2h T =−

AL2 6h AL2

6h

k  1 L  T − T∞ = − +  x −  +  6k 2  h  h 3k  Ax 3

AL 

32


2.20

800 W/m3 of heat is generated within a 10 cm diameter nickel-steel sphere for which k = 10 W/m.K. The environment is at 20 C and there is a natural convection heat transfer coefficient of 10 W/m2.K around the outside of the sphere. What is its center temperature at the steady state?

Solution: 1 ∂ 2 (rT ) q& =− r ∂r 2 k 3

q W/mcm) = 5 cm = 0.05 m ro == 800 (1/2)(10 k = 10 W/m.K T∞ = 20 C h = 10 W/m2.K 2 ∂ (rT ) q& =− r 2 ∂r k ∂ (rT ) q& 2 =− r + C1 ∂r 2k q& 3 rT = − r + C1 r + C2 6k q& 2 C T =− r + C1 + 2 6k r ∂T h (T − T∞ )r = ro = − k ∂r r = ro ∂T

=−

∂r

q& C r − 22 3k r

C  q& 2  r + C1 + 2 − T  6 k r  r

C   q& r − 22  3 k r r  ro q& 2 C k  q& C  − ro + C1 + 2 − T = −  − ro − 22  ro h  3k 6k ro 

h−

= −k  −

∞

=

∞

−

q& ro

2

6k C1 =

+ C1 +

kC2 h ro

2

−

C2 ro

C2 ro

− T∞ =

+

kq&

3h k

kq&

3h k

ro +

ro +

q&ro

kC2 h ro

2

2

6k

+ T∞

Let Ti center temperature, r = 0 q& 2 C T =− r + C1 + 2 6k r q& 3 rT = − r + C1 r + C2 6k at r = 0 , C2 = 0

33

= ro


C1 =

kq& ro

3h k

+

q& ro

2

6k

+ T∞

then, T =−

q&

r2 +

kq& ro

q& ro

2

+ T∞ 6k 3h k 6k center temperature , r = 0 2 kq&ro q& ro Ti = + + T∞ 3h k 6k (10)( 800 )( 0.05 ) (800)( 0.05 ) 2 Ti = + + 20 3(10)( 10) 6(10 ) Ti = 21.37 C

2.21

+

An outside pipe is insulated and we measure its temperature with a thermocouple. The pipe serves as an electrical resistance heater, and q& is known from resistance and current measurements. The inside of the pipe is cooled by the flow of liquid with a known temperature. Evaluate the heat transfer coefficient, h , in terms of known information. The pipe dimension and properties are known. [Hint : Remember that h is not known and we cannot use a boundary condition of the third kind at the inner wall to get T(r).]

Solution: T∞ = inside bulk temperature, C ro = outside pipe diameter, m r i = inside pipe diameter, m k = thermal conductivity, W/m.K q& = heater load, W/m3

Step 1. T = T(r) Step 2. 1 ∂  ∂T  1 ∂ 2T ∂ 2T q& 1 ∂T + + = r + 2 2 2 r ∂r  ∂r  1 r 4 ∂4 ∂ z k t θ2 α 44 3 12∂3 = 0 , since T =T (φ , z )

= 0 , since steady

1 ∂  ∂T  q& ∂  ∂T  q& r r  = − ; r =− r ∂r  ∂r  k ∂r  ∂r  k Step 3. Integrate once, ∂T q& r 2 r =− + C1 ∂r 2k ∂T q& r C1 =− + ∂r 2k r Step 4. Integrate again.

34


T =−

q&r 2

+ C1 ln r + C 2 4k Step 5. Second boundary condition.

qconvection = qconduction ∂T h (T − T∞ ) = −k ∂r  q&r 2    q&r C1  h  − 4k + C1 ln r + C2  − T∞  = − k  − 2k + r 



q& r 2



4k

h T∞ +

q&r



2

+

kC1 r

2

q& r kC1 − C1 ln r − C2 = − + 4k 2h h r Step 6. Substitute r = ri, r = ro then determine C1 and C2. Equation 1 2 q&r q&r kC T∞ + o − C1 ln ro − C 2 = − o + 1 4k 2h h ro Equation 2 2 q&r q& r kC T∞ + i − C1 ln ri − C 2 = − i + 1 4k 2h h ri Subtract Eq. 2 from 1 q& (ro 2 − ri 2 )− C1 ln ro  = − q&(ro − ri ) + kC1  1 − 1  4k 2h h  ro ri   ri  T∞ +

q&r



− C1 ln r − C 2  = −

 ro 2 − ri 2

   4k   C1 =  ro  k  1 1  ln  +  −   ri  h  ro ri   r 2 − ri 2 ro − ri    ln ro + k  + q&  o 2  4 2 k h h r & qr q& r o   + o C2 = T∞ + o −  4k 2h r  k  1 1 ln o  +  −  r h r r  i  o i q& 

C 2 = T∞ + q&

 ro 2  ro  ln  4k  ri

+

ro − ri 2h

 ro  +  4h

−

ro

2

4h ri

2

−

ro ln ro

4k

2

+

ri ln ro

4k

−

ro ln ro

+

2h

ri ln ro

−

2h

 ro  k  1  +   ri  h  ro

ln

35

ro

+

4h −

1  ri 

ri

2

4h ro

−

k

2h 2

+

kri

2h 2 ro

+

ro

2h

 ro  k  + 2  ri  2h

ln

−

kro 



2h 2 ri 


C 2 = T∞ + q&

 ro 2 ln) (ri −  4k

ro

−

2

4h ri

r

2

)+(( i ) ln ro

+

4k

r kri ro kro  ln ro +) ( i + − ln ri −  2 2h 4h ro 2h ro 2h 2h 2 ri  2

ri

 ro   +  ri 

k 1 1  −  h  ro ri 

ln

 ri 2  ln)(ro  4k C 2 = T∞ + q& 

ro

−)(

2

4k

ln ri +

ri

2

r

2

)(o

−

4h ro

4h ri

+)(

ri

ln ro −

2h

ro

2h

 ro  k  1 1   +  −   ri  h  ro ri  2 2 1 2  1  ri 2  − ro  + 1 (ri ln ro  (ri ln ro − ro ln ri ) +  ri  2h 4h  ro  4k C 2 = T∞ + q& r  k  1 1 ln o  +  −   ri  h  ro ri 

ln ri +

kri 2

kro 



−

2h 2 ri 

2h ro



ln

− ro

ln ri ) +

k 2h 2

 ri   ro

−

ro  

 ri  

Then:

T =−

q&r

+

4k

 r2

 ro  r 2  1   +  ri  4 h  ro

ln

q& −

 4k

T − T∞ =

 ro 2 − ri 2 ro − ri    4k + 2h  ln r   + T∞  ro  k  1 1     ln  +  +   ri  h  ro ri 

1

q& 

2

−

q& 

 4 k

+

(r

i

2

ln ro − ro ln ri ) + 2

 ro  k  1  +   ri  h  ro

T − T∞ =

 1  2  ro   − r ln  + ro 2 ln r − ro 2 ln ri r   4k   i

− ri

2

  

ln r + ri ln ro  + 2

 1  2  r  2  r  2  ro ro ln  − ri ln  − r ln  ri   ro   ri  4k 

q&  T − T∞ =

 ri 2 ro 2  1  + + (ri ln ro r ri  2h  o r  k  1 1  ln  o  +  +   ri  h  ro ri 

−

 ri   ro

+

ro 

 ri 

2 2 r  r  r2  1 1  1 1 r k r  − + (ro ln r − ri ln r + ri ln ro − ro ln ri ) +  i − o  + 2  i − o   4h  ro ri  2h 4h  ro ri  2h  ro ri   r  k  1 1 ln o  +  −   ri  h  ro ri 

 r 2  1  +   4h  ro

−

r 1 1  + ro ln ri  2h   ri

 ro  k  1  +   ri  h  ro

−

1  ri 

For a given r and T, solve for h ,

B=

k 2h 2

1  ri 

ln

 2  r  2  r  2  ro  ro ln  − ri ln  − r ln   ri   ro   ri   r  r  1  r 2 r2  1 1  1   −  +  ro ln  − ri ln  +  i  4  ro ri  2   ri   ro  4  ro

Let A =

− ro ln ri ) +

2   2  1 1 2 1 2 1 1 1 2 + ro ln r − ri ln r + ro ln r − ri ln r + (ri ln ro − ro 2 ln ri ) + 1  ri − ro  + 1 (ri ln ro − ro ln ri ) + k 2  ri − ro  4k 2h 2h 4k 4 h  ro 2h  ro ri   ri  4 k ri  2h

ln

q& 

1 4h

1 4k

36

−

2 ro 



ri 

  r  1  ri 2   − ri ln  +    ro  4h  ro

−

2 ro 

r + k2  i 2h  ro

ri 

−

ro 

 ri  


r   − o  2  ro ri  r  D = ln o   ri  1 1 E = k  −   ro ri 

C=

k  ri

−

F = T T∞ q& 1 x= h Then: ( A + Bx + Cx 2 ) F= (D + Ex) FD + EFx = A + Bx + Cx 2 Cx 2 + (B − EF ) ( x + ) A − FD = 0 Using, ax 2 + bx + c = 0 x=

b 2 − ac

−b±

a

, use positive value since x =

1 is always positive. h

a=C b = B − EF c = A − FD

For r = ro , T = To , thermocouple measurement A= B= B=

 2  ro   ro   ro   2 2 ro ln  − ri ln  − ro ln   = 0 r r  i  o  ri    2 r   r  1  r 2 ro  1 1  1   −  +  ro ln o  − ri ln o   +  i  4  ro ri  2   ri   ro   4  ro 2 r   r  1  r 2 ro  1 1  1   −  +  ro ln o  − ri ln o   +  i  4  ro ri  2   ri   ro   4  ro 1 4k

2

2

2

B = ro  1 − 1  + 1 ro ln ro  + 1  ri − ro   4  ro ri  2 ri   ri  4  ro Then, r  kr a = C =  i − o  < 0 2  ro ri 

37

−

−

2 ro 



ri 

2 ro 



ri 


ro  1 2

B=

 4  ro

−

r 1 1  + ro ln o ri  2  ri 1

2 b = B − EF = ro 

 ro

c = A − FD = −

−

 1  ri 2  +    4  ro

r 1 1  + ro ln o ri  2  ri

−

2 ro 



ri 

 1  ri 2  +    4  ro

−

2 ro 

  − k 1 r  o

ri 

−

1  (T − T∞ )  ri  q&

(To − T∞ )  ro  ln

 < 0  ri 

q&

b 2 − 4ac > 0 , then −b±

x=

b 2 − 4ac 1 = a h a

h= −b±

2.22

b 2 − 4ac

>0.

Consider the hot water in Problem 1.11. Suppose that it is insulated with 2 cm of a material for which k = 0.12 W/m.K, and suppose that h = 16 W/m2.K. Find (a) the time constant T for the tank, neglecting the casing and insulation; (b) the initial rate of cooling in C/h ; (c) the time required for the water to cool from its initial temperature of 75 C to 40 C ; (d) the percentage of additional heat loss that would result if an outer insulation were held on by eight steel rods, 1 cm in diameter, between the inner and outer casings.

Solution: (a)

T = time constant

ρ cV mc = hA hA m = 100 kg A = 1.3 m2 h = 16 W/m2.K Specific heat of water at 75 C, Table A.6 cp = 1009 J/kg.K T =

J /) kg.K T = (100)(kg 1009 (16W / m2 .)(K 1).3 m 2 T = 4851 s

(b) Q = UA(T − T∞ ) = − mc p

dT dt

T∞ = 20 C

38


T = 75 C to 40 C 1 1 L = + U h k L = 2 cm = 0.02 m k = 0.12 W/m.K 1 1 0.02 = + U 16 0.12 U = 4.364 W/m2.K

Initial rate of cooling: )( 1.3 )75 − 20 dT − UA(T − T)∞ ( − 4)(.364 = = = 0.0031 C/s = 11.16 C/hr dt mc p (100 ) ( 1009 ) (c)

∫

dT − UA(T − T∞ ) = dt mc p dT

(T − T )

=

− UAt

mc p

∞

 T2 − T  − UAt  =  T1 − T  mc p  40 − 20  − (4.364)( 1).3 t ln = (100 ) ( 1009 )  75 − 20  ln

∞

∞

t = 17,992 s = 5 hrs

(d) Derive and solve for the resistance of steel rods.

Equation of circle: ( x − a )2 + y 2 = r 2 2

y 2 = r 2 − (x − a )

A = π y 2 = π [r 2 − ( x − a ) dT Q = − kA dx Qdx = −π kdT 2 r 2 − (x − a )

2

Qdx

 x−a 1−    r 

]

2

2

= −π r kdT

but du −1 = tanh u + C ∫ 1− u2

39


x−a r 1 du = dx r then, 1 Qdx r = −π r kdT 2 x−a 1 −  r  u=

1 Qdx r Q ∫x1 = −π r k∆T 2  x−a 1−    r  x2



 x2 − a  1  x − a   − tanh  1  = −π r k∆T  r   r 

Q  tanh −1 



Q=−

−

∆T

Rr

 1  x2 − a  1  x1 − a   tanh  r  − tanh  r       −

Rr =

1 L = + U ′A kA

−

π rk 1 1 hA+ (8) Rr

x1 = 0.02 m x2 = 0.02 m + 0.01 m = 0.03 m a = 0.02 m + 0.005 m = 0.025 m r = 0.005 m

For steel rod, ks = 43 W/m2.K, Table A.1 at 1% carbon steel. [tanh −1 ()1 − tanh( )−1 − 1 ] Rr = ) π (0.005)( 43 -1 since tanh (1) → ∞ , tanh-1(-1) → −∞ use 1 ~ 0.999,999,999,999 −1

−1

Rr = [tanh ()1 − tanh () ) − 1 ] = 35 π (0.005)( 43

1 U ′A

=

L kA

+

1 1 hA+ (8) Rr

40


1 0.02 = + U ′(1).3 ( 0.)(12) 1.3

1

(16) ( 1.3)

+

1

(35)

(8)

U’ = 4.3766 W/m2.K 2 ∆U = 4.3766 – 4.364 = 0.0126 W/m .K ∆U 0.0126 (100) % = ( ) 100% = 0.29% Percentage increase = U 4.364

A slab of thickness L is subjected to a constant heat flux q1, on the left side. The right-hand side if cooled convectively by an environment at T∞ . (a) Develop a dimensionless equation for the temperature of the slab. (b) Present dimensionless equation for the wall temperatures as well. (c) If the wall is firebrick, 10 cm thick, q1 is 400 W/m2, h = 20 W/m2.K, and T∞ = 20 C, compute the left hand and right hand temperatures.

2.23

Solution: (a) q1 = − k dT

=−

dx

∂T ∂r

= −k

dT dx

q1 k

T =−

q1

x + C1

Then atk the outer wall ∂T h (T − T∞ )x = L = −k ∂r x = L

 q1 x + C1 − T  k  q h  − 1 L + C1 − T  k h−

∞

∞

  = q1 x L   = q1  =

q + 1 L + T∞ h k Then: q q q T = − 1 x + 1 + 1 L + T∞ k h k C1 =

q1

T − T∞ = q1  L + 1 − x  k h k

 L− x 1  +  h  k

T − T∞ = q1 

(b)

Dimensionless equation

41


 L− x 1  +  h  k q1 L  L − x k  +   k  L hL 

T − T∞ = q1  T − T∞ =

T − T∞ L − x k = + q1 L L hL k

(c)

firebrick, kFor = 0.1 W/m.KTable A.2, Appendix A q1 = 400 W/m2 h = 20 W/m2.K T∞ = 20 C L = 10 cm = 0.1 m At x = 0, left hand Tl − 20 0.1 − 0 0.1 = + 0.1 ( )(20 ) 0.1  (400)( 0) .1   0.1  Tl = 440 C At x = L = 0.10 m, right hand Tr − 20 0.1 − 0.1 0.1 =

 (400)( 0) .1   0.1 

+

0.1 ( )(20 ) 0.1

Tr = 40 C

2.24

Heat flows steadily through a stainless steel wall of thickness Lss = 0.06 m, with a variable thermal conductivity of kss = 1.67 + 0.0143 T (C). It is partially insulated on the right side with glass wool of thickness Lgw = 0.1 m, with a thermal conductivity of kgw = 0.04. The temperature on the left-hand side of the stainless steel is 400 C and on the right-hand side of the glass wool is 100 C. Evaluate q and Ti.

Solution: q = −k

∂T

∂x For stainless steel, T1 = 400 C, T2 = Ti but unknown, k = 1.67 + 0.0143 T = a + bT. dT q = −(a + bT ) dx 1 2  qx = − aT + bT  + C1 2   at the left hand side x = 0, Tl = T1 = 400 C

42


1 2 bTl 2 1 2 qx = a (Tl − T ) + b(Tl − T 2 ) 2 at the left hand side x = x2, For glass wool, kgw = 0.04 W/m.K 2 ∂ T q =− 2 ∂x kLgw C1 = aTl +

∂T

q

=−

∂x

T =−

kL gw

x + C1

q x 2 + C1 x + C 2 2kL gw

at x = Lgw + Lss = 0.1 m + 0.06 m = 0.16 m, T = Tr = 100 C q 1 2  Tr = − L2 +  aTl + bTl  L + C2 2kL gw 2   C2 = Tr + T =−

q 1  2  L2 −  aTl + bTl  L 2kL gw 2  

q 1 2 q 1 2    x 2 +  aTl + bTl  x + Tr + L2 −  aTl + bTl  L 2kL gw 2 2kL gw 2    

T − Tr =

q L2 − x 2 ) −  aTl + 1 bTl 2 (L − x ) 2kL gw ( 2  

Equating to Find T at x = Lss = 0.06 m, L – x = Lgw.  1 2   1 2 T − Tr +  aTl + 2 bTl  L gw  2kL gw a(Tl − T ) + b(Tl − T 2 )     2 L2 − Lss

=

2

Lss

 1   ) +( 0.0143 )( ) 400 ( )2()(0.10)( 2) 0.04 0.10 T − 100 + (1.67)( 400 2    

(0.16 ) (2 − )0.06 2

T + 81.2

(1.67 )( 400 −) T ( + =

2

668 − 1.67T + 1144 − 0.00715T 2.75 0.06 0.02182T + 1.772 = 1812 − 1.67T − 0.00715T 2 =

2

+

−

=

0.00715T 1.69182T 1810.228 0 T 2 + 236.62T − 253179 = 0 Ti = 398.6 C 1 2 2 qLss = a(Tl − Ti ) + b(Tl − Ti ) 2 1 q(0).06( = )(1.67 400 − ) 398 ( .6 (+) ) 0( .0143 ) [ 400 2 − 398.6 2 43

2

]

1 0)(.0143 ) [ 400 2 − T 2 ] 2 0.06


q = 172.3 W/m2.

Rework Problem 1.29 with a heat transfer coefficient ho = 40 W/m2.K on the

2.25

outside (i.e., on the cold side). Solution: k = k o (1 + aT 2 ) ko = 0.15 W/m.K a = 10-4 C-2 Ti = 100 C T∞ = 0 C L = 0.005 m 2 ∂ T ∂T = 0, = C1 2 ∂x ∂x T = C1 x + C2 at x = 0 , T = Ti Ti = C1 (0) + C 2 C2 = Ti ho (T − T∞ )x = L = − k

∂T ∂x

= −ko

(1 + aT ) ∂T 2

o

ho (C1 L + C2) −( T∞ = − k)o [1 + a C1 L + C2 k =( − )o [1 + a C1 L + C2 ho

(C1 L + C2) − T

∞

(C1 L + Ti)− T

∞

(C1)L + Ti ko aC1 ho ko aC1 ho ko aL2 ho ko aL2 ho

=(−

− T∞ = −

(C1 L +)T( i

(C

1

2

2

∂x

x =L

2

2

]C

1

]C

1

ko

) [1 + a C1 L + Ti 2 ]C1

ho

ko C1

ko aC1

ho

ho

( )−

+ )C1 L + Ti +

C1 L + Ti

k o C1 ho

− T∞ = 0

L2 + 2C1 LTi + Ti ) + C1 L + Ti + 2

2

ko C1 ho

− T∞ = 0

2

3

C1 +

2k o aLTi 2 ko aTi kC C1 + C1 + C1 L + o 1 + Ti − T∞ = 0 ho ho ho

 2ko aLTi 2  k o aTi ko C1 +  + + L C1 + (Ti − T∞ ) = 0  ho ho  ho  2

3

C1 +

This is a cubic form of equation, solve by finding the root. C 1 < 0 since T∞ < Ti Solving the cubic equation.

44


2Ti 2  Ti 1 h  h (T − T ) C1 +  2 + 2 + o C1 + o i 2 ∞ = 0 L aL ko aL  ko aL L 2

3

C1 +

Using Cardano’s Method: x 3 + a1 x 2 + b1 x + c1 = 0 2Ti T 1 h h (T − T ) ; b1 = i2 + 2 + o ; c1 = o i 2 ∞ L L aL k o aL k o aL 2

a1 = =

p1 p1 =

−

a1

b1

2

Ti

=

2

2

+

L2

3

ho 1 4Ti + − aL2 k o aL 3L2 2

ho

1 Ti − aL2 3L2

+

k o aL

3

q1 = c1 +

q1 =

q1 =

2a1 − 9a1b1 27

ho (Ti − T∞ ) k o aL2

 8Ti 3   2Ti  Ti 2  − 9  2 3   L   L  L

2 +

+

1 ho   + 2 aL k o aL 

27

ho (Ti − T∞ ) k o aL2

3

3

3

3

16Ti 18Ti 18Ti 18hoTi − − − L3 L3 aL3 k o aL2 + 27

16T3 i − 18T3 i − 18T3i − 18hoT2i ho (Ti − T∞ ) L L aL k o aL + q1 = 2 k o aL 27 q1 = q1 =

ho (Ti − T∞ ) 2

3

−

k o aL

ho (Ti − 3T∞ )

u=3

2

3

−

3k o aL q1

±

2

x = C1 =

q1

2Ti 2Ti 2hoTi − − 3 3 27 L 3aL 3k o aL2

2

+

4

2Ti 2Ti − 27 L3 3aL3 p1

3

27 a1

p1

−u − 3u 3 in order for C1 < 0, u > 0.

Then, u=3

q1

+

2

x = C1 =

q1

2

+

4 p1

3u

−u −

p1

3

27 a1 3

C2 = Ti

45


T = C1 x + C2

Substitute Values, ho = 40 W/m2.K a = 10-4 C-2 ko = 0.15 W/m.K Ti = 100 C T∞ = 0 C L = 0.005 m p1 = q1 =

ho

1 Ti 40 1 (100 ) = 8.0 x 10 8 − = + − 2 2 2 −4 −4 (0.15)(10 )(0.005) (10 )(0.005) 3(0.005)2 aL 3L 2

+

k o aL

ho (Ti − 3 ) T∞

3k o aL2

2

2Ti 2T)(i ) ( 40 [100 − 3 0 ] 2(100) 2(100) − = 2 − 3 − 3 3 3 −4 27 L 3aL 3(0.15)(10 )(0.005 ) 27(0.005 ) 3(10 −4 )(0.005) 3

−

3

q1 = −2.3704 × 1012 − 2.3704 × 10

(− 2.3704 ×10 )( ) + 12 2

12

u=3

+

2 2Ti 2(100) a1 = = = 40,000 L 0.005

4

8.0 × 108 27

3

= 14,930

8 C1 = 8.0 × 10 − 149300 − 40000 = -10,402 3(14930 ) 3 C 2 = 100 C

T = C1 x + C2 = − 10402 x + 100

Tabulation: x, m 0.000 0.001 0.002 0.003 0.004 0.005

T, C 100 89.6 79.2 68.8 58.4 48.0

Plot:

46


q = ho (To)(− T∞ = ho − 10402 ) L + 100 − T∞ q = (40 ) [− 10402 ( ) 0.005 + 100 − 0] = 1,920 W/m2.

2.26

A scientist proposes an experiment for the space shuttle in which he provides underwater illumination in a large tank of water at 20 C, using a 3 cm diameter spherical light bulb. What is the maximum wattage of the bulb in zero gravity that will not boil the water?

Solution: ro = (1/2)(3 cm) = 1.5 cm = 0.015 m T∞ = 20 C Ti = 100 C 2

∇ T+ 2

∂ T ∂x

+

2

∂y

2

=

2

∂y

∂ T ∂x

2

=−

2

∂ T ∂x ∂T

2

∂z

∂T

2

3

+

2

2

∂ T ∂x

q& = 0 k 2 2 ∂ T ∂ T

=− =−

2

2

=−

q& k

2

=

∂ T ∂z

2

q&

k q&

3k q&

x + C1 3k q& 2 T = − 6k x + C1 x + C2 ∂x

q& 2 y + C1 y + C 2 6k q& 2 T =− z + C1 z + C2 6k T = To , (x, y,)z( = r)o(,0,0 )=( 0, r)o ,0 = 0,0, ro T =−

47


To = −

q& 2 ro + C1 ro + C 2 6k

as z → ∞, y → ∞, x = ro ,

∂T ∂x

= 0,

maximum T.

q& r o + C1 3k q& C1 = r o 3k q& 2 q& 2 To = − ro + ro + C 2 6k 3k q& 2 To = C 2 + ro 6k as z → ∞, y → ∞, x = 0, T = T∞ ,

0=−

T∞ = −

q&

6k

(0) 2 + C( )1 0

+ C2

C2 = T∞ To = T∞ + To − T∞ =

q&

6k q& 6k

ro

2

ro

2

Q q& = V To − T∞ = V =

Q 2 ro 6kV

4 3 π ro 3

Q Q 2 ro = 4 8kπ ro  3  6k  π ro  3  Q = 8kπ ro (To − T∞ ) For water at 20 C, k = 0.0181 W/m.K Q = 8(0.0181 )( )(π 0.)(015 100) − 20 = 0.55 W To − T∞ =

2.27

A cylindrical shell is made of two layers – an inner one with inner radius = ri and outer radius = rc and an outer one with inner radius = rc and outer radius = ro. There is a contact resistance, hc, between the shells. The materials are different, and T1(r = ri) = Ti and T2(r = ro) = To. Derive an expression for the inner temperature of the outer shell ( T2c).

48


Solution:

r  r  ln c  ln o  ri  1   rc  + + ∑ Rt = hc Ac 2πlk1 2πlk 2 ∑R

∑ Rt

1

=

+

hc Ac

t

1

=

 rc    ri  Ac

ln

+

=

1

+

2πlk1 Ac 2πlk 2 Ac r  r  rc ln c  rc ln o  r  i  rc  +

hc Ac

∑ Rt Ac

 ro    rc  Ac

ln

k1 Ac

k 2 Ac

 rc  r   rc ln o   ri  +  rc 

rc ln +

hc

k1 Then. T − Ti T − Ti Q= o = 2c ∑ Rt Ac ∑ Rtc1 Ac

k2

 rc   r 

rc ln

∑ Rtc1 Ac

=

h1c +

k1

i



To − Ti

r  r  rc ln c  rc ln o  1  ri  +  rc  + hc

k1

T2 c = Ti +

k2

T2 c − Ti

=

1 + hc

 rc    ri 

rc ln

k1

  r  rc ln c    1  ri   (To − Ti ) +  hc  k1      rc   ro  rc ln  rc ln  1  ri  +  rc  + hc

k1

k2

or Q=

To − Ti

∑ Rt Ac

=

To − T2 c

∑ Rtc 2 Ac

49


To − Ti

r  r  rc ln c  rc ln o  r 1  i+  rc  + hc

k1

=

To − T2c

 ro    rc 

ro ln

k2

k2

  ro    ro ln   r (To − Ti )  c    k2     rc   ro  rc ln  rc ln  1  ri  +  rc  +

T2 c = To +

hc

2.28

k1

k2

A 1 kW commercial electrical heating rod, 8 mm in diameter and 0.3 m long is to be used in a highly corrosive gaseous environment. Therefore, it has to be provided with a cylindrical sheath of fireclay. The gas flows by at 120 C, and h = is 230 W/m2.K outside the sheath. The surface of the heating rod cannot exceed 800 C. Set the maximum sheath thickness and the outer temperature of the fireclay. [Hint: use heat flux and temperature distribution. Then use the additional convective boundary condition to obtain the sheath thickness.]

Solution: ri = inside radius of insulation = (1/2)(8 mm) = 4 mm = 0.004 m ro = outside radius of insulation T∞ = 120 C h = 230 W/m2 Ti = 800 C l = 0.3 m Q = 1 kW = 1000 W 2

∇ T+

q&

=

k

{

=0

∂

1 ∂T t α2∂3 1 = 0 , since steady

∂

1  r T  = 0 r ∂r  ∂r  ∂  ∂T  r =0 ∂r  ∂r  ∂T r = C1 ∂r

50

2. HEAT CONDUCTION CONCEPTS, THERMAL RESISTANCE, AND THE OVERALL HEAT TRANSFER COEFFICIENT ∂T

C1 r T = C1 ln r + C2 Then, ∂T qradial = − k ∂r Q = qA = q (2π rL ) =

∂r

Q  C1  q = 2π rL = − k  r  Q C1 = − 2π rkL

at ri = 0.004 m, Ti = 800 C Ti = C1 ln ri + C 2 Q ln ri + C 2 2π kL Q C2 = Ti + ln ri 2π kL Q Q T =− ln r + Ti + ln ri 2π kL 2π kL

Ti = −

T − Ti = −

Q ln r 2π kL  ri 

r Q ln 2π kL  ri

T = Ti −

  

and h (T − T∞ )r = ro = − k

∂T ∂r

r = ro

  r  Q C h Ti − ln o  − T  = − k 1 2π kL  ri  ro     r  Q Q h Ti − T − ln o  = 2 kL r 2 ro L π π  i   ∞

∞

Ti − T∞ −

Q ln ro  = Q 2π kL  ri  2π ro Lh

 r  Q  k + ln o   = Ti − T∞  r  2π kL  ro h  i  For fireclay, k = 1.0 W/m.K Substitute values: 51


1000  1  ro  + ln  = 800 − 120 = 680  2π (1)( 0.)3 ( ro 230 )  0.004  By Trial and error: ro = 0.0088 m Maximum thickness = ro – ri = 0.0088 m – 0.004 m = 0.0048 m = 4.8 mm To = Ti −

 ro  1000  0.0088   = 800 − ln  2π kL  ri  2π (0.3)  0.004  Q

ln

To = 382 C

2.29

A very small diameter, electrically insulated heating wire runs down the center of a 7.5 mm diameter rod of type 304 stainless steel. The outside is cooled by natural convection ( h = 6.7 W/m2.K) in room air at 22 C. if the wire releases 12 W/m, plot Trod vs. radial position in the rod. (Stop and consider carefully the boundary conditions for this problem.)

Solution: 2

∇ T+

q& k

=

1 ∂T t α 12∂3 = 0 , since steady

1 ∂  ∂T  q& r =− r ∂r  ∂r  k ∂  ∂T  q r =− r ∂r  ∂r  k ∂T q& 2 r =− r + C1 ∂r 2k ∂T q& C =− r+ 1 ∂r 2k r q& 2 T =− r + C1 ln r + C2 4k at r = ri = 0 ∂T q& 2 r =− r + C1 ∂r 2k 0 = 0 + C1 C1 = 0 q& 2 T =− r + C2 4k h (T − T∞ )r = ro = − k

∂T ∂r

r = ro

52


 q& 2  ro + C 2 − T   4k r

h−

C2 =

q&

q&

 q&  ro   2k 

= −k  −

∞

= ro

2

ro + ro + T∞ 2h 4k Then, q& 2 q& q& 2 T =− r + ro + ro + T∞ 4k 2h 4k q& 2 q& 2 T − T∞ = 4k (ro − r ) + 2h ro

Plot: ro = 0.0075 mm, h = 6.7 W/m2.K Q q& = L2 πro Q = 12 W/m L T∞ = 22 C For 304 stainless steel, say k = 14 W/m.K 12 q& = = 67,906 W/m3 2 π (0.0075) (67,906 ) T − 22 = [(0.0075)2 − r 2 ]+ (67,906 ) (0.0075) 4(14) 2(6.7 ) T = 60.075 − 1212.607r 2 r, m 0.0000 0.0015 0.0030 0.0045 0.0060 0.0075

r, mm 0.0 1.5 3.0 4.5 6.0 7.5

T, C 60.075 60.072 60.064 60.050 60.031 60.007

53


2

To = 60.075 − 1212.607(0.0075 ) = 60.007 C

2.30

A contact resistance experiment involves pressing two slabs of different materials together, putting a known heat flux through them, and measuring the outside temperature of each slab. Write the general expression for hc in terms of known quantities. Then calculate hc if the slabs are 2 cm thick copper and 1.5 cm thick 2

aluminum, if q is 30,000 W/m , and if the two temperatures are 15 C and 22.1 C. Solution: Known quantities: T1 = outside temperature of left slab T4 = outside temperature of right slab L1, L2 = thicknesses k1, k2 = thermal conductivities Let left slab has a higher temperature than right slab. T −T   T − T3   = hc (T2 − T3 ) q = − k1  2 1  = − k 2  4 L  1   L2  T2 = T1 − T3 = T4 +



qL1 k1 qL2 k2

q = hc  T1 −



qL   −  T4 + 2  k1   k 2 

qL1  

54




 L1



 k1

q = hc (T1 − T4 ) − q hc =

hc =

+

L2 

 k 2 

q

  L1 L2  (T1 − T4 ) − q +   k1 k 2    1

(T − T )  L 1

q

4

L 

−  k1 + k 2  1 2



Copper: L1 = 2 cm = 0.02 m, k1 = 398 W/m.K Aluminum: L2 = 1.5 cm = 0.015 m, k2 = 237 W/m.k T1 = 22.1 C, T2 = 15 C, q = 30,000 W/m2. 1 hc = = 8,122 W/m2.K (22.1 − 15)  0.02 0.015  − +  30,000  398 237 

2.31

A student working heat transfer problems late at night needs a cup of hot cocoa to stay awake. She puts milk in a pan on an electric stove and seeks to heat it as rapidly can,continuously. without burning the milk, turning stoveanonanalogous high and stirring as theshe milk Explain how by this workstheusing electric circuit. Is it possible to bring the entire bulk of the milk up to the burning temperature without burning part of it?

Solution: If you put the milk into a hot pan and leave it undisturbed the milk closest to the source of the heat will reach the burn temperature quickly and scorch. If you stir the milk constantly, the movement will keep redistributing the added heat throughout the milk so it heats up more uniformly. When the more uniformly heated milk reaches the same burn temperature, it will scorch. Milk get scorched at 185 F (85 C), which makes it taste unpleasant. Same data as in (a) = 100 C T flame = 500 C =

− +

1

55

2. HEAT CONDUCTION CONCEPTS, THERMAL RESISTANCE, AND THE OVERALL HEAT TRANSFER COEFFICIENT −

=

1

+

+

Where

1

≈0

h = 200 W/m2.K Same as in example 2.10. hb = 5000 W/m2.K assumed for stirred milk. Same as in example 2.10. L k ≈0 Thermal Circuit

Burning at the bottom or surface of the pan. Then, =

500 − 85 2 1 = 83,000 W/m . 200

= milk temperature =

−

1

+

+

83 000 =

1

500 − 1 1 + 200 5000

1   1 +  200 5000   = 68.4 C . Therefore the milk must be heated to approximately 65 to 70 C. Heating to approximately, it is possible to reach the burning temperature without burning =

−

= 500 − 83 000

part of it. 2.32

A small, spherical hot air balloon, 10 m in diameter, weighs 130 kg with a small gondola and one passenger. How much fuel be consumed (in kJ/h) if it is to hover at low altitude in still 27 C air? houtside = 215 W/m2.K, as the result of natural convection.)

Solution: 56


ro = (1/2)(10 m) = 5 m T∞ = 27 C h = 215 W/m.K

Derivation of equation: 1 ∂ 2 (rT ) q& =− r ∂r 2 k 2 ∂ (rT ) q& =− r 2 ∂r k ∂ (rT ) q& 2 =− r + C1 ∂r 2k q& 3 rT = − r + C1 r + C2 6k q& 2 C T =− r + C1 + 2 6k r ∂T q& C2 =− r− 2 ∂r 3k r ∂T h (T − T∞ )r = r o = − k ∂r r = ro









h  − q& ro 2 + C1 + C2 − T∞  = − k  − q& ro − C22  ro ro   6k  r =r o  3k



h  −

q&

 6k

C1 =

q&

3h

2

ro + C1 + ro +

k h ro

2

C2 ro

  r

C2 −

C2 ro

+

 q& C2   3k ro + r 2  o  

= k

− T∞ 

=ro

q& 2 ro + T∞ 6k

 q&

Q = h Ao (T − T∞ )r = r o = q& V = kAo 

 3k

ro +

1 ro

2



C2 



 & 1 2  q 4 3 q&  π ro  = k (4π ro ) ro + 2 C2  3 3 k r   o   C2 = 0 C1 = q& ro + q& ro 2 + T∞ 3h 6k q 2 q& q& 2 T =− r + ro + ro + T∞ 6k 3h 6k q& 2 T − T∞ = (ro − r 2 ) + q& ro 6k 3h

57


Density of air = ρ =

m V

4 3 π ro 3 3m 3(130 ) = ρ= = 0.2483 kg/m3. 3 3 4π ro 4π (5) From Appendix A, Table A.6, Temperature of air at this density is 1422 K = 1149 C. At uniform air temperature inside balloon. To = 1149 C V =

q& ro 3h Qro Q = T − T∞ = 2 3h V 4π ro h T − T∞ =

2

Q = 4π ro h (T − T∞ ) - equation derived. Q = 4π (5 )( 215 )( 1149)− 27 Q = 75,784,640 J/s = 272,824,703 kJ/h 2

2.33

A slab of mild steel, 4 cm thick, is held at 1000 C on the back side. The front side is approximately black and radiates to black surroundings at 100 C. What is the temperature of the front side?

Solution:

(To − Ti )

= σ (To − T∞ ) L Ti = 1000 C + 273 = 1273 K −k

4

4

T∞ = 100 C + 273 = 373 K L = 4 cm = 0.04 m σ = 5.67 x 10 -8 W/m2.K4

Thermal conductivity of mild steel at 1000 C, App. A k = 28 W/m.K (T − To ) 4 4 k i = σ (To − T∞ ) L (1273 − To ) 4 −8 4 (28) = (5.67 × 10 )(To − 373 ) (0.04)

(5.67 ×10 )T −8

o

4

+ 700To − 892,198 = 0

To = 1138.4 K To = 865.4 C

58


2.34

With reference to Fig. 2.3, develop an empirical equation for k(T) for ammonia vapor. Then imagine a hot surface at Tw parallel with a cool horizontal surface at a distance H below it. Develop equations for T(x) and q. Compute q if Tw = 350 C, Tcool = -5 C, H = 0.15 m.

Solution: From Fig. 2.3, ammonia vapor is nearly a straight line, use k = a + bT T = 0 C, k = 0.0234 W/m.K T = 100 C, k = 0.0498 W/m.K Then a = 0.0234, b = 2.64 x 10 -4 k = 0.0234 + 2.64 × 10 −4 T dT dT q = −k = −(a + bT ) dx dx

 

qL = −  aT +

T2

1 2 bT  2  T1

qL = − a(T2 − T1 ) −

1 2 2 b(T2 − T1 ) 2

T1 = Tw T2 = Tcool , L = H = 0.15 m

1 2 2 qL = a (T1 − T2 ) − 2 b(T1 − T2 ) 1 (2.64 ×10 − 4 )(T12 − T2 2 ) 2 q= L 1 2 2 0.234(Tw − Tcool ) − (2.64 × 10 − 4 )(Tw − Tcool ) 2 q= L 2 2 0.234(Tw − Tcool ) − (1.32 × 10 −4 )(Tw − Tcool ) q= L 0.234(T1 − T2 ) −

x  L

0.234(Tw − T ) − (1.32 × 10 − 4 )(Tw − T 2 ) = [0.234(Tw − Tcool ) − (1.32 ×10 − 4 )(Tw − Tcool )] 2

2

 x 

1773(Tw − T ) − (Tw − T 2 ) = [1773(Tw − Tcool ) − (Tw − Tcool )] 2

2

2

L x 2  − Tw − 1773Tw = 0 L  2 2 2  x  T 2 + 1773T − Tw + 1773Tw − [1773(Tw − Tcool ) + (Tw − Tcool )]  = 0  L   T 2 + 1773T + [1773(Tw − Tcool ) + (Tw − Tcool )] 2

2

59

2


− 1773 +



 x    L 

17732 + 4Tw + 1773Tw − [1773(Tw − Tcool ) + (Tw − Tcool )]

T=

2

2

2



2



 x    L 

T = −886.5 + 0.5 17732 + 4Tw + 1773Tw − [1773(Tw − Tcool ) + (Tw − Tcool )] 2

2

2



x 

T = −886.5 + 0.5 3,143,529 + 4Tw + 7092Tw − 4[1773(Tw − Tcool ) + (Tw − Tcool )] 2

2

2

L Compute q:

) (− ) − 5 0.234[350 − (− 5)] − (1.32 ×10−4 ) (350 0.15 2

q=

2.35

2

= 662 W/m2.

A type 316 stainless steel pipe has a 6 cm inside diameter an 8 cm outside diameter with a 2 mm layer of 85 % magnesia insulation around it. Liquid at 112 C flows inside, so hi = 346 W/m2.K. The air around the pipe is 20 C, and ho = 6 W/m2.K. Calculate U based on the inside area. Sketch the equivalent electrical circuit, showing all known temperatures. Discuss the results.

Solution:

r1 = (1/2)(6 cm) = 3 cm = 0.03 m r2 = (1/2)(8 cm) = 4 cm = 0.04 m r = 0.002 m + 0.040 m = 0.042 m 3

hi = 346 W/m2.K ho = 6 W/m2.K

Ti = 112 C To = 20 C

60


1 1 = + U i hi

 r2  r   r1 ln 3  r  1+  r2  + r1

r1 ln

2k ss

2k m

r3ho

For type 316 stainless steel,kss = 15 W/m.K from Appendix A For 85 % magnesia, km = 0.067 W/m.K 0.04   0.042  (0.03) ln  (0.03) ln  (0.03) 1 1 0 . 03 0 . 04 = + + +     (0.042)( )6 U i 346 2(15) 2(0.067) U i = 7.5104 W/m2.K U i (Ti − )To ( = hi) Ti − T1

7.5104(112 −)20 ( = 346 ) 112 − T1 T1 = 110 C 2k (T − T ) U i (Ti − To ) = ss 1 2 r  r1 ln 2   r1  7.5104(112 − 20 ) =

(2)(15 110) − T2  0.04    0.03 

0.03 ln T2 = 109.8 C U i (Ti − To ) =

2k m (T2 − T3 )

 r3    r2 

r1 ln

( 2) 0.(067 109) .8 − T3

7.5104(112 − 20 ) =

 0.042    0.04 

0.03 ln T3 = 102.25 C

Electrical circuit:

61


•

2.36

There is a negligible drop in temperature in stainless steel pipe, insulation and the inside film coefficient. The dominant factor is the outside film coefficient . Two highly reflecting, horizontal plates are passed 0.0005 m apart. The upper one is kept at 1000 C and the lower one at 200 C. There is air in between. Neglect radiation and compute the heat flux and the midpoint temperature in the air. Use a power-law fit of the formk = a (T 0C)b to represent the air data in Table A.6.

Solution:

Table A.6 T,

o

C

k

0.03656 0.03971 0.04277 0.04573 0.04863 0.05146 0.05425 0.05699 0.05969 0.06237 0.06501

727 827 927

0.06763 0.07281 0.07792

By curvefitting k = aTb a = 0.003225 b = 0.4608 q = −k

dT dx

q = − aT b =−

qL

dT dx

 a  b + 1 T

b +1



(W/m.K)

177 227 277 327 377 427 477 527 577 627 677

T2

 T1

T2 = 200C T1 = 1000 C L = 0.0005 m a (T2b +1 − T1b +1 ) q=− L(b + 1)

62


q= q=

a (T1b+1 − T2b +1 ) L(b + 1) 0.003225

()[ 1000 0.(4608) 1 − 200 0.4608 1 ] (0.0005 )( 0.4608 ) +1 +

+

= 96,358.3 W/m 2.

q x=

q=

L

2 b+ 1

2a T1 L(b + 1) (

− Tm

b+

1

)

2(0.003225) [(1000)0.4608+1 − Tm 0.4608+1 ] 96,358.3 = (0.0005 )( 0.4608 ) +1 Tm = 662.2 C 2.37

A 0.1 m thick slab with k = 3.4 W/m.K is held at 100 C on the left side. The right side is cooled with air at 20 C through a heat transfer coefficient, and 5 h = 5.1 W / m 2 (K ) (− 4  Twall) − T∞



1 4



. Find q and Twall on the right.

Solution: L = 0.10 m k = 3.4 W/m.K Tl = 100 C T∞ = 20 C 5 h = 5.1 W / m 2 (K ) (− 4  Twall) − T∞



q=k k



(Tl − Twall ) L

1 4

= h (Twall − T∞ )

5 (Tl − Twall )  2 = 5.1W / m ( K ) ( 4  Twall) − T   −



L



∞

5 4

5 (100 − Twall ) (3.4 ) = 5.1(Twall − 20 )4 (0.10) 5

5.1(Twall −) 20( 4 − 34) 100 − Twall = 0 Twall = 76.67 C q = k (Tl − Twall ) = (3.4 ) (100 − 76.67) = 793.22 W/m2. (0.10) L

2.38

Heat is generated at 54,000 W/m3 in a 0.16 m diameter sphere. The sphere is cooled by natural convection with fluid at 0 C, and

63


h = 2 + 6(Tsurface − T∞ )4  W/m2.K, ksphere = 9 W/m.K. Find the surface temperature 1





and center temperature of the sphere. Solution: q& = 54,000 W/m3. T∞ = 0 C k sphere = 9 W/m.K 1 ∂ 2 (rT ) q& + =0 r ∂r 2 k 2 q& 1 ∂ (rT ) =− 2 r ∂r k 2 ∂ (rT ) q& =− r 2 ∂r k ∂ (rT ) q& 2 =− r + C1 ∂r 2k q& 3 rT = − r + C1 r + C2 6k q& 2 C T =− r + C1 + 2 6k r q& C2 ∂T =− r− 2 ∂r 3k r

at the surface r = ro h (T − T∞ )r = ro = −k

∂T ∂r

r = ro

 q&  q& 2 C C  ro + C1 + 2 − T∞  = −k  − ro − 22  6 k r 3 k ro  o    q& 2 C2 q& kC2 − ro + C1 + − T∞ = ro + 2 6k ro 3h h ro but q& 2 C To = − 6k ro + C1 + r 2 o q& 2 C2 C1 = To + ro − 6k ro 

h  −

−

q& 2  q& 2 C2 ro +  To + ro − k ro 6k 6 

 C2  + − T  ro

∞

=

q& kC 2 ro + 2 3h h ro

64


q& kC2 ro + 2 h ro 3h

To − T∞ =

 h 2  q&  C2 =  ro  To − T∞ − ro  k 3 h     C1 = To +

q& 2  h  q&  ro −  ro  To − T∞ − ro  k h  6k 3  

 & 2  & 2 C1 = To + q ro −  h ro (To − T∞ ) + q ro 6k 3k k  T =−

q&

r 2 + To +

6k

T = To +

h  q& 2  h ro  q& 2  To − T∞ − ro  ro −  ro (To − T∞ ) + ro +  6k k 3 k k r 3 h       2

q&

q& 2 (ro − r 2 )+  h ro  To − T∞ − q& ro  ro − 1 6k 3h  r   k r 

Solve for surface temperatureT = To q& 2 (ro − ro 2 ) h ro  To − T∞ − q& ro  ro − 1 To = To + 6k 3h  ro  k ro   To = T∞ +

q& ro 3h

q& ro

To = T∞ +

1

3 2 + 6(To − T∞ )4 





3(To −)T∞ ( 2 + 6) To − T∞  = q&ro 1 4





T∞ = 0 C

3To 2 + 6To 4  = (54,000)(0.16) = 8,640 1





To 2 + 6To  = 2880



1 4



To = 129.5 C

Center Temperature, r = 0 q& ro = 0 Note that To − T∞ − 3h

Then: q&

(ro − r 2 ) 6k 2 (54000 )( 0) .16 2 q& r Tc = To + o = 129.5 + = 155.1 C 6k 6(9 ) T = To +

2

65


2.39

Layers of equal thickness of spruce and pitch pine are laminated to make an insulating material. How should the laminations be oriented in a temperature gradient to achieve the best effect?

Solution: Thermal conductivity of spruce and pitch pine, Appendix A kspruce = 0.11 W/m.K kpine = 0.14 W/m.K ∆T q= Rt In series: 1 1 1 Rt = ∑ = + = 16.234 k 0.11 0.14 ∆T q= = 0.0616∆T 16.234 In parallel: 1 1 = =4 Rt = ∑ k 0.11 + 0.14 ∆T q= = 0.0616∆T 16.234 q = ∆T = 0.25∆T 4 Therefore use series orientation with lower heat loss. 2.40

The resistance of a thick cylindrical layer of insulation must be increased. Will Q be lowered more by a small increase of the outside diameter or by the same decrease in the inside diameter?

Solution: 2π kl∆T ∆T = Q= ln(ro / ri ) Rt ln(ro / ri ) Rt = 2π kl For small increase in diameter let x be the ratio of that increase to diameter. Outside diameter increase: ln(ro / ri ) Rt = 2π kl

66


 ro + xro  r   − ln o   ri   ri 

ln ∆Rto =

2π kl ln(1 + x ) ∆Rto = 2π kl



 r   − ln o   ri − xri   ri 

ln

ro

∆Rti =

2π kl

 1   1− x 

ln ∆Rti =

2π kl

Since 1 for 1 > x > 0 1− x ∆Rti > ∆Rto and ∆qo > ∆qi 1+ x <

Therefore Q will be lowered more by a small increase of the outside diameter. 2.41

You are in charge of energy conservation at your plant. There is a 300 m run of 6 in. O.D. pipe carrying steam at 250 C. The company requires that any insulation must pay for itself in one year. The thermal resistance are such that the surface of the pipe will stay close to 250 C in air at 25 C when h = 10 W/m2.K. Calculate the annual energy savings in kW.h that will result if a1 in layer of 85 % magnesia insulation is added. If energy is worth 6 cents per kW.h and insulation costs $75 per installed linear meter, will the insulation pay for itself in one year?

Solution: Thermal conductivity of 85 % magnesia insulation k = 0.08 W/m.K ∆T Q= Rttotal Without insulation, OD = 6 in = 0.1524 m Q = h A∆T )( ) 300 ( 250)− 25 Q = (10)( )( π 0.1524 Q = 323,176 W With insulation 67


 ro    ri  +

ln

1 2π lk 2π ro lh ri = (1/2)(0.1524 m) = 0.0762 m ro = ri + 1 in = ri + 0.0254 m = 0.0762 m + 0.0254 m = 0.1016 m  0.1016  ln  1  0.0762  + Rttotal = = 2.43 x 10-3 K/W 2π (300 )250 ( 0−.080 ) ( 2π 0.)(1016)( )300 10 ∆T 25 Q= = = 92,593 W Rttotal 2.43 ×10−3 Rttotal =

Annual Energy Savings

Payback Period =

= (323,176 W – 92,593 W)(1 kW / 1000 W)(8760 h/yr) = 2,019,907 kW.h

($75.00 /)(m 300 ) m = 0.1857 year << 1 year (2,019,907 kW )( .h $0.06 ) / kW .h

Therefore insulation pays for itself in less than a year. 2.42

An exterior wall of a wood-frame house is typically composed from outside to inside, of a layer of wooden siding, a layer of glass fiber insulation, a layer of gypsum wall board. Standard glass fiber insulation has a thickness of 3.5 inch and a conductivity of 0.038 W/m.K. Gypsum wall board is normally 0.50 inch thick with a conductivity of 0.17 W/m.K, and the siding can be assumed to be 1.0 inch thick with a conductivity of 0.10 W/m.K. a. Find the overall thermal resistance of such a wall (in K/W) if it has an area of 400 ft2. b. Convection and radiation processes on the inside and outside of the wall introduce more thermal resistance. Assuming that the effective outside heat transfer coefficient (accounting for both convection and radiation) isho = 20 W/m2.K and that for the inside is hi = 10 W/m2/K, determine the total thermal resistance for the heat loss from the indoors to the outdoors. Also obtain an overall heat transfer coefficient,U, in W/m2.K. c. If the interior temperature is 20 C and the outdoor temperature is –5 C, find the heat loss through the wall inwatts and the heat flux inW/m2. d. Which of the five thermal resistance is dominant?

Solution: k1 = 0.10 W/m.K L1 = 1.0 inch = 0.0254 m k2 = 0.038 W/m.K L2 = 3.5 inch = 0.0889 m

68


k3 = 0.17 W/m.K L3 = 0.50 inch = 0.0127 m

 L L L  1  Rttotal =  1 + 2 + 3    k1 k 2 k3  A   0.0254 0.0889 0.0127  1  -3 = + +   = 6.67 x 10 K/W 0.038 0.17  400   0.10

(a) Rttotal

 1 L L L 1  1  Rttotal =  + 1 + 2 + 3 +    ho k1 k 2 k 3 hi  A 

(b)

 1 0.0254 0.0889 0.0127 1  1  -3 + + + +   = 7.04545 x 10 K/W 0.038 0.17 10  400   20 0.10

Rttotal = 

UA =

1 Rttotal

UA =

1 Rttotal

U=

(c)

1

(7.04545 ×10 )(400) −3

= 0.355 W/m2.K

Q = UA∆T

)( )400 (20 ( )− − 5 ) = 3550 W Q = (0.355 Q q = = U∆T = (0.355 ) (20() − 5 ) = 8.875 W/m2. A (d)

2.43

L2 K2

is the dominant one (glass fiber insulation).

 ln ro We found that the thermal resistance of a cylinder wasRtcyl =  1    2π kl   ri

.  

If ro = ri + δ , show that the thermal resistance of a thin-walled cylinder

(δ

<< ri ) can

Rtthin = δ

be approximated by that for a slab of thickness δ . Thus,

(kAi ) , where Ai

= 2π ri l

is the inside surface area of the cylinder. How

mych error is introduced by this approximation ifδ series.] Solution: 69

ri

= 0.2 ? [Hint: Use a Taylor


 1   ro  ln Rtcyl =   2π kl   ri

  

 1   ri + δ  ln Rtcyl =   2π kl   ri For small δ

  

ri

 ri + δ   δ  =  +  r ln i  ln1 ri  Using Taylor series: n ∞ (− 1) n+ ln(1 + x ) = ∑ x 1 for |x| < 1 n=0 n + 1 1 1 1 ln(1 + x ) = x − x 2 + x 3 − x 4 ........ 2 3 4 ln(1 + x ) ≈ x



ln1 +



δ ri

 δ  ≈  ri

 1  δ  Rtthin =   2π kl  ri

 δ  =  kAi

Error:

Error =

δ  δ − ln1 + ri  ri



ln1 +



2.44

δ ri

  

  0.2 − ln (1 + 0.2) = = 0.097 or 9.7 % ln(1 + 0.2)

A Gardon gage measures a radiation heat flux by detecting a temperature difference. The gage consists of circular constantan membrane of radiusR, thickness t, and thermal conductivity kct which is joined to a heavy copper heat sink at its edges. When a radiant heat fluxqrad is absorbed by the membrane, heat flows from the interior of the membrane to copper heat sink at the edge, creating a radial temperature gradient. Copper leads are welded to the center of the membrane and to the copper heat sink, making two copper-constantan thermocouple junctions. These junctions measure the difference∆T between the edge of the membrane, T(r = 0), and the edge of the membrane,T(r = R).

• •

The following approximations can be made: The membrane surface has been blackened so that it absorbs all radiation that falls on it. The radiant heat flux is much larger than the heat lost from the membrane by convection pr re-radiation. Thus, all absorbed radiant heat is removed from the

70


• •

membrane by conduction to the copper heat sink, and the other loses can be ignored. The gage operates in steady state. The membrane is thin enough ( t << R) that the temperature in it varies only with r, i.e. , T = T(r) only.

Answer the following questions. a. For a fixed copper heat sink temperature, T(r = R), sketch the shape of the temperature distribution in the membrane,T = T(r), for two arbitrary heat radiant fluxes qrad1 and qrad2, where qrad1 > qrad2. b. Find the relationship between the radiant heat flux, qrad, and the temperature difference obtained from the thermocouples,∆T. Hint: Treat the absorbed radiant heat flux as if it were a volumetric heat source of magnitudeqrad / t (W / m3). Solution: 1 ∂  ∂T  1 ∂ 2T ∂ 2T q& 1 ∂T + + = r + r ∂r  ∂r  r 2 ∂φ 2 ∂z 2 k t α 12∂3 14 4244 3 = = 0 , since T ≠T (φ , z )

0 , since steady

1 ∂  ∂T  q& q r  = − = − rad r ∂r  ∂r  k kt ∂  ∂T  qrad r r =− kt ∂r  ∂r  q ∂T 2 r = − rad r + C1 ∂r 2kt q ∂T C = − rad r + 1 ∂r 2kt r qrad 2 T =− r + C1 ln r + C2 4kt If r = 0, C1 = 0 q T = − rad r 2 + C2 4kt at r = R, T = T(r = R) T (r = R ) = −

qrad

R 2 + C2 4kt q C2 = T (r = R ) + rad R 2 4kt qrad 2 q (a) T =− r + T (r = R ) + rad R 2 4k ct t 4k ct t qrad 2 2 T= (R − r ) + T (r = R ) 4kct t

71


qrad (R 2 − r 2 ) = qrad R 2 − qrad r 2 4k ct t 4k ct t 4kct t Equation of parabola. y − k = 4ax 2 Here, q qrad − 4a = − rad , a = 4k ct t 16k ct t T − T (r = R ) =

x = r2

−

T

(b)

=

T (r

=

R)

qrad 2 2 − 4k ct t (R r )

at r = 0 T (r = 0 ) =

qrad

4kct t

R2

∆T = T (r = )R (− T) r (=

qrad =

2.45

0 )= T r = R −

qrad 2 R 4kct t

4kct t [T (r = R ) − ∆T ] R2 You have a 12 oz. (375 mL) can of soda at room temperature (70 F) that you would like to cool to 45 F before drinking. You rest the can on its side on the plastic rods of the refrigerator shelf. The can is 2.5 inches in diameter and 5 inches long. The can’s emissivity is ε = 0.4 and the natural convection heat transfer coefficient around it is a function of the temperature difference between 1

the can and the air: h = 2∆T 4 for ∆T in Kelvin. Assume that thermal interactions with the refrigerator shelf are negligible and that buoyancy currents inside the can will keep the soda well mixed.

72


a. Estimate how long it will take to cool the can in the refrigerator compartment, which is at 40 F. b. Estimate how long it will take to cool the can in the freezer compartment, which is at 5 F. c. Are your answers for part 1 and 2 the same? If not, what is the main reason that they are different? Solution: 1

h = 2∆T 4 ε = 0.4 D = 2.5 in = 0.0635 m L = 5.0 in = 0.127 m m = 12 oz = 375 mL = 0.375 kg

(a)

T∞ = 40 F, T1 = 70 F, T2 = 45 F

hrad = 4σεTm

3

1 1 (T + T∞ ) = (45 + 40) = 42.5 F 2 2 3 −8 = 4(5.67 × 10 )(0)( .4 42.5 + )273 = 2.849 W/m2.K

Tm = hrad

1

hc = 2(45 − 40)4 = 2.99 W/m2.K hT = 2.99 + 2.849 = 5.839 W/m 2.K hT A(T − T∞ ) = mc p

∆T

∆t assume cp = 4187 J/kg.K

π 2  π  ) 2(+ π 0)(.0635) 0.127 = 0.03167 m 2 D  + πDL =  (0.0635 4   2

A = 2

hT A(T − T∞ ) = mc p

∆T ∆t

(5.839 ) ( 0.03167 )( 45) − 40 ∆t

(b)

=

∆t

= 42454 s = 11.79 hrs T∞ = 5 F, T1 = 70 F, T2 = 45 F

hrad = 4σεTm Tm =

( 0.375 )( )(4187 70 ) − 45

3

1 1 (T + T∞ ) = (45 + 5) = 25 F 2 2 73


hrad = 4(5.67 × 10 −8 )(0.)(4 25 + 273 ) = 2.4 W/m2.K 3

1

hc = 2(45 − 5)4 = 5.03 W/m2.K hT = 2.4 + 5.03 = 7.43 W/m 2.K hT A(T − T∞ ) = mc p

∆T

∆t assume cp = 4187 J/kg.K

π 2  π  ) 2(+ π 0)(.0635) 0.127 = 0.03167 m 2 D  + πDL =  (0.0635 4   2

A = 2

hT A(T − T∞ ) = mc p

∆T ∆t

(7.43 )( 0.03167 )( 45) − 5 ∆t

(c)

( 0.375 )( )(4187 70 ) − 45 =

∆t

= 4182 s = 1.16 hrs The answer is not the same as the room temperature will still dominate the cooling time.

74

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