SOLUTIONS MANUAL for Analytical Heat Transfer 2nd Edition by Je-Chin Han & Lesley Wright. ISBN13: 97

Chapter 1 1-1

Consider a differential volume: dV = rd dr dz Consider heat conduction in r-direction, -direction, z-direction: T T q = −k A = −k rddz r r r r T drdz T q = −k A = −k   r r  T = −k rddr T qz = −k Az z z Consider energy balance and constant properties to get (1.18)

1-2

Consider a differential volume: dV = r sin d rd dr = r 2 sin dd dr Consider heat conduction in r-direction, -direction, z-direction: T T T q = −k A = −k r sin d rd = −kr 2 sin d d r r r r r T rdrd  T drd  T q = −k A = −k = −k   r sin  r sin  sin  T T T q = −k A = −k r sin ddr = −k sin ddr    r r  (a) see (1.17) d 2T (b) 0 2 = dx (c) T = c1 + c2 x at x = x1 , T1 = c1 + c2 x1 x = x2 , T2 = c1 + c2 x2 T ( x − x2 ) + T2 ( x1 − x) Solve for c1, c2, obtain T = 1 x1 − x2 dT " T1 − T2 = constant value (d) q = −k x dx = −k x − x

1-3

1

1-4

2

Refer to Example 1.1 (a), (b), (c)

1

(a) at

,

Solve for

, can be determined can be determined

(b) at

,

Solve for , Solve for and from above (a) and (b) can be determined can be determined (c) In the above (b), replace 1.5

.

Refer to Example 1.2 (a), (b), (c)

(a) at

Solve for (b) at

by

,

, ,

2

Solve for , and then solve for , from above (a) and (b) can be determined can be determined and

(c) In the above (b), replace

by

.

3

Chapter 2 2-1

Consider one-D conduction through plane walls: T,o − T,i (1) q= 1 L L 1   + + + Rt",c +   ho  k TBC  k blade hi T −T =  L s,o ,i 1 (2) + k h  blade i T −T = s,i ,i (3) 1 hi Let (1)=(2), solve for blade outer surface temperature Ts,o , Ts,o = 1,174 K , below 1,250K , O.K. Let (2)=(3), solve for blade inner surface temperature Ts,i , Ts,i = 1,104K .

If without TBC, Rt",c = 0 , Ts,o above 1,250K, No good. TBC thickness: too thick too brittle

2-2

Find steel pipe AISI 1010 conductivity k from Tables: k = 58.7W mK at 400K Consider one-D conduction through circular tube walls: T1 − Ts q = (1) L ln ( r2 r1 ) 1 + 2k 2 r1 hi = 2r 2h 2(T 2 − T 2 ) + 2r 2 (T s4 − T 4 2)

(2)

Let (1)=(2), solve for tube outer surface temperature Ts, Ts=502.4K q W From (1), = 1,837 m L 2-3

(a)

2 1 d  dT  q   r   o r  + 1−    = 0 r dr  dr  k   ro   4 dT = − qo  1 r 2 − 1 r  C r dr 2 + 1  k 2 4 ro  dT q 1 1 r3  C o =−  r− 2 + 1 dr k2 4 ro  r qo  1 2 1 r 4  T =− r − 2  + C1 ln r + C2  k 4 16 ro 

4

dT = 0 ,  C1 = 0 dr q 1 2 1 2 r = r , T =T = − o r − r +C o s  2 k 4 o 16 o    4   r  T = Ts + qo  3 ro2 − 1 r 2 + 1 2  k  16 4 16 ro   1 q From (2.27), h (T s − T  ) = −k dT = −k  − 4 ro  ko   dr r =ro q r where Ts = T + o o 1 d  dT  41 h r + a + b (T − T  ) = 0 (b) r dr  dr  k    dT  b 2  a   2 d   r dr r dr + r T − T −  = 0    k a   b  Let  = − T − , m 2= b T   b  k    d  d  2 2 r2  r function +m r  =0 dr dr Bessel   From (2.48), with  = 0 , x = r From (2.49),  = a0 J0 (mr ) + a1Y0 (mr ) d = −ma J (mr ) − ma Y (mr ) From A.3.1, 0 1 11 dr d at r = 0 , = 0 , but Y (0) = −,  a = 0 1 1 dr 0 at r = 0 ,

(

)

at r = ro ,  r = s = Ts − T − a  = a0 J0 (mro ) o  b  J mr 0 ( )   = s J 0 (mro ) dT d where h (T − T ) = −k = −k = km J1 (mro ) s s  J 0 (mro ) dr r =ro dr r =ro 2-4

For copper tube at 300K, k = 403 W m K h P m2 = 2  m = 22.46 m−2 k Ac M = h2 PK Ac = 0.1084W K From (2.35), qf = 0.2W (given)  b = 5.59 K From (2.34),  ( x) can be obtained since b, m, L, h2, and k are given now.

5

2-5

2-6

(a) Energy balance d  dT  (2.47): k A dr − P h (T − T ) dr = 0 dr  c dr     Ac = 2 rt , P = 2r d 2T dT h − r (T − T ) = 0 r 2 + dr dr k t h 2 Let  = T − T , m = kt d  d  2 2 r dr r dr − m r  = 0   (b) Modified Bessel Function (2.52):  = a0I0 (mr ) + a1K0 (mr ) d =0  a =0 K (0) →  at r = 0 , 1 1 dr at r = R ,  = TR − T = R = a0I0 (mR)   = I0 (mr ) R I0 (mR) (c) Hotter at the center d  d  2h (a) From (2.47): r  r − m2r 2 = 0 where m2 = ,  = T − T  dr dr kt   From (2.52):  = a0I0 (mr ) + a1K0 (mr ) at r = ri ,  = i = (Tb − T ) = a0I0 (mri ) + a1K0 (mri )  ( ro ) h at r = r , = −  (r ) o o r k h a mI (mr ) − a mK (mr ) = − a I (mr ) + a K (mr ) 0 1 o 1 0 o  1 1 o o k 0 0 Solve for a0 and a1 from (1) and (2) q  (b) Radiation gain, let  = T − T − r = a I (mr ) + a K (mr ) .  0 0 1 0 h q  Temperature will be lower by r . h

(1)



2-7

(a) Same as 2-5 (a), but P = 2 2r , m = 2

(b) Same as 2-5 (b) (c)

2h kt

qr (c) Radiation loss, let  = T − T + =  I0 (mr )  R h I 0 (mR)

6

(2)

qr . h

Temperature will be lower by 2-8

2-9

(a) Same as 2-5(a) (b) Same as 2-5(b) qr (c) Radiation loss, let  = T − T + =  I0 (mr )  R h I 0 (mR) q  Temperature will be lower by r . h From (2.46), Ac =

 x2 4 l2

d , P =

d  2 d  4hl dx  x dx  − kd x = 0 d  2 d  2 dx x dx − m x = 0   

(

1

)

1

x

d ,  = T − T , x=from pin tip.  l 4hl let m2 = kd

2

(

1

1

 = −a 0 x− 2 I 1 2mx 2 − a1 x− 2 K 1 2mx 2

) Modified Bessel

at x = 0 , K1 (0)   , at the tip of pin, a1=0 1

(

1

at x = l ,  = Tb − T = b , at the tip of pin, b = −a 0l − 2 I 1 2ml 2 

 

2-10

b

=

x

l

−1

2

− 12

(

1

(

1

I1 2mx

2

2

I1 2ml 2

(a) Calculate Re =

) , q = −k  d d

)

)

f

dx x=l

4

V L

= , Pr = 

Cp



= , get

hL

 k k Consider one-D conduction through plane walls: Ti − T q= =  t  +  t  +  t  + 1 k k k h  A  B  C T −T T −T T −T (b) q = i 2 = 2 3 = 3 4 =   t   t   t  k k k  A  B  C (c) C, B, A T −T T −T T −T (d) same as (b): q = i t 2 = 2 t 3 = 3 t 4 =        k k k  C  B  A L

7

= , get h = 

2-11

d 2T = 0 , T = C1 + C2 x dx2 (b) at x = 0 , T = T0 = C1 , x = L , T = TL = T0 + C2 L  T = T0 + (TL −T0 ) x L (a)

(c) Energy balance

  ( qx +  AVCT ) − (q x +  AVCT ) + ( q x +  AVCT ) dx = 0  dx   d  dT  d ( AVCT )  − −kA + dx = 0 d

dx

dx

dx

 k d  dT  V dT dT = , where  = , let  = dx  dx   dx C dx d V =  (d) dx V x dT  = e + C1 = dx V  x T = e + C1x + C2 V at x = 0 , T = T0 , x = L , T = TL ,Vsolve for C1, C2, V L x x  x      T = T0 − (T0 − TL ) + 1− e  − 1− e  L V L V   2-12

Radiation effect on a long thin fin: d 2 ( T − T )  D − (a)  (T 4 − T 4 ) = 0 2 dx k Ac (b) temperature order: (1) Aluminum  (3) Steel  (2) Aluminum with paint due to  effect  (4) steel with paint

2-13

(a) See equation (2.47) (b) See equation (2.53) (c) One-D, s.s., finite-difference energy balance in cylindrical coordinate

2-14

Calculate: P for circular cross-section fin =  Ac for circular cross-section fin =  P for triangular cross-section fin =  Ac for triangular cross-section fin =  qf

( PAc ) 2 ,  f ( P Ac ) 2 1

2-15

Same as problem (2-9).

2-16

From equation (2.52):

1

8

 (mr ) = a0I0 (mr ) + a1K0 (mr ) at r = r1 ,  = b = a0I0 (mr1 ) + a1K0 (mr1 ) (1)  ( r2 ) h = −  (r ) at r = r , 2 2 r k h  a mI (mr ) − a mK (mr ) = − a I (mr ) + a K (mr ) 0 0 2 1 0 2  0 1 2 1 1 2 k from (1) and (2), solve for a0, a1: b K0 (mr1 ) a = 0 I (mr ) − a1 I (mr ) 0 1  0 b 1 K0 (mr1 ) C1  a =   C  I (mr ) C 1 1 (mr ) C  1+ 1 2 1 2  0   I 0 where h C = mI (mr ) + I (mr ) 2 1 1 2 k 0 h C = mK (mr ) − K (mr ) 2 1 1 2 k 0 

2-17

From equation (2.52):  (mr ) = a0I0 (mr ) + a1K0 (mr ) at r = r1 ,  = b = a0I0 (mr1 ) + a1K0 (mr1 )

(1)

at r = r2 ,  = T2 − T = 2 = a0I0 (mr2 ) + a1K0 (mr2 ) b K0 (mr2 ) −  2 K 0 (mr1 ) a0 = I (mr ) K (mr ) − I (mr ) K (mr ) 0

1

0

2

0

2

0

1

0

1

0

2

0

2

0

1

(2)

b I0 (mr2 ) −  2 I 0 (mr1 ) a1 = K (mr ) I (mr ) − K (mr ) I (mr ) 2-18

From equation (2.33):  ( x) = C1emx + C 2e−mx (1) at x = 0 ,  = b = C1 + C2 d  ( L) = 0 = C memL − C me−mL at x = L , 1 2 dx from (1) and (2), solve for C1, C2: e − mL C1 = b mL − mL e +e emL C2 = b mL e + e−mL

9

(2)

 = e−mLemx + emLe− mx = e

m(L− x)

+ e−m(L− x) 

2

=

cosh m ( L − x )

cosh mL emL + e− mL ( emL + e− mL ) 2 d (0) −m sinh m ( L − x) hP sinh mL b q f = −kAc = −kAcb = kAc kAc cosh mL dx cosh mL x=0

b

= hPkAc b tanh mL 2-19

at x = 0 ,  = b = C1 + C2

(1)

(2) at x = L ,  = L = C1emL + C 2e−mL from (1), (2), solve for C1, C2: −mL C =   e − L  b 1 b mL e− mL − emL   −e C2 = b  L−mLb mL e −e  L b − e −mL mx emL − L b −mx (L b )sinh mx + sinh m ( L − x) = e + mL −mL e = e −e sinh mL mL −mL b  e −e cosh mL −L b q f = hPkAc  b sinh mL 2-20

at x = 0 ,  = b = C1 + C2 at x =  ,  = 0 = C1  + 0  C2=b  =  be−mx

 C1=0

q f = hPkAc b

10

Chapter 3 3-1

2-D, S.S., let  = T − To 2 2 + = x2 y2 0 B.C.s: y = 0 , T = To ,  = 0 y = b , T = To ,  = 0 T = 0 ,  = 0 x =0, x x T  x = a , −k = h (T − T ) , −k = h ( − )   x x Separation of variables  = X ( x ) Y ( y ) = C1 sinh  x + C2 cosh xC3 sin  y + C4 cos  y  where C4 = 0 , at y = 0 C1 = 0 , at x = 0 0 = C2C3 cosh  x sin b , at y = b  0 = sin b  nb = n , n = 0,1, 2

  = C2C3 cosh  x sin  y =  Cn sin n y cosh n x  at x = a , −k = −k  C sin  y  sinh  a n n n n x = h ( −  ) = h  Cn sin n y cosh n a −   b

h sin n y dy  Cn = ( h cosh  a + k 0 sinh  a ) b sin2  y dy n

n

b 

b

where sin n y dy =  0 b

n



n

n 0



1− (−1)  n 

sin2  y dy = 2 n

b

0

n = n 3-2

b

Assume a long rod with square cross-section floating as a diamond-shape, treat its lower-edge point as x = 0 , y = 0 .

Let  = T − T , B.C.s: y = 0 ,  =  f

11

y = L ,  = − h  y k x = 0 ,  =f x = L ,  = − h  x k Use superposition principle: Let  = 1 + 2 B.C.s for 1 : y = 0 , 1 = 0  1 h y = L, =−  y k 1 x = 0 , 1 =  f 1 h x = L, =−  x k 1 Separation of variables: 1 = X Y , X = C1 sinh n x + C2 cosh n x Y = C3 sin n y + C4 cos n y at y = 0 , C4 = 0 h at y = L ,  C cos  L =− C sin  L n n 3 n k 3 hL 1 cot  L = −   L= n n k n L h at x = L , C  cosh  L + C  sinh  L = − (C sinh  L + C cosh  L) 2 n n n 2 n 1 n n k 1 A  n sinh n L + h k cosh n L = −C2 C1 = −C2 B n cosh n L + h k sinh n L  1 =  Cn sin n y cosh n x − A B at x = 0 , 1 =  f , solve for Cn , 



(

)

L

C = n

 f sin n ydy 0

=

(1− A B ) sin  ydy ( L

2

n

0

L n   n   1− ( −1)  L  1− A B 2

f

)

Let x  y , y  x , solve for 2 Then  = 1 + 2 3-3

Let  = T − T = 1 + 2 1 = X Y Separation of variables

12

X = C1 sinh x + C2 cosh x

Y = C sin y + C cos y 3

4

at x = 0 , 1 = 0 to determine Cn 1 = 0 to get C ~ C at x = 2L , 1 2 x  at y = 0 , −k 1 = h to determine  1 n y  at y = L , −k 1 = 0 to get C ~ C 3 4 y

2 = X Y X = C1 sin x + C2 cos x

Y = C sinh y + C cosh y 3

4

at x = 0 , 2 = 0 to determine C2 = 0 2 = 0 to get n at x = 2L , x  at y = 0 , −k 2 = h to determine C ~ C 2 3 4 y  at y = L , −k 2 = −q to get C n y

Then  = 1 + 2 =  3-4

(1) Let  = T − To , refer equations (3.9) and (3.10) for solution.

(2) Follow same method for equations (3.9) and (3.10): n x X = C sin 1 a n y n y Y = C sinh + C cosh 3 4 a a n n at y = b ,  = 0 = C sinh b + C cosh b 3 4 a a n C = −C tanh b 4 3 a n x n y nb n y    = X Y = C sin  C sinh − tanh cosh 1 3  a a a  a  n x  n y nb n y  =  C sin sinh − tanh cosh n a a a  a  n x  nb  − tanh at y = 0 ,  =  =  C sin s n a  a  a n x −s sin a dx 0 Solve for Cn = = nb a n x dx tanh sin2 a 0 a

13

(3) X = C1 sinh n x + C2 cosh n x Y = C3 sin n y + C4 cos n y

at x = 0 , C2 = 0 at y = 0 , C4 = 0 at y = b , 0 = C3 sin nb , nb = n

 = X Y =  Cn sinh n x sin n y

at x = a ,  =  s =  Cn sinh n x sin n y b

Solve for Cn =

s sin n ydy

=

a

b

sinh na sin 2  n ydy

 a

(4) X = C1 sinh n x + C2 cosh n x Y = C3 sin n y + C4 cos n y

at x = a , C2 = −C1 tanh na at y = 0 , C4 = 0

at y = b , 0 = C3 sin nb , nb = n  = X Y = C1 (sinh n x − tanh na cosh nx) C3 sinh n y = Cn sin n y (sinh n x − tanh na cosh n x)

at x = 0 , s = Cn sin n y (− tanh na) b

Solve for Cn =

−s sin n y dy 0

b

=

tanh na sin  n y dy 2

0

3-5

(1) Let  = T − T1 , refer to equation (3.11) (2) Follow the same method for equation (3.11) n X = C sin x 1 a n n y Y = C3 sinh y + C4 cosh a a n at y = b ,  = 0 , C = −C tanh b 4 3 a n  n n n  x sinh y − tanh b cosh y  = XY = C sin n  a a a  a  at y = 0 ,  = − q = C sin n x  n   n a  a  y k a q n − sin x dx a Solve for Cn = k a 0 = n n 2 sin x dx a 0 a

14

(3) X = C1 sinh n x + C2 cosh n x Y = C3 sin n y + C4 cos n y

at x = 0 , C2 = 0 at y = 0 , C4 = 0 at y = b , 0 = C3 sin nb , nb = n

 = XY =  Cn sinh n x sin n y  q at x = a , =− = C  cosh  a sin  y n n n n x k b q s sin n y dy k 0 Solve for C = =

−

n

n

n

 b

n

 cosh  a sin  y dy 2

0

(4) X = C1 sinh n x + C2 cosh n x Y = C3 sin n y + C4 cos n y

at x = a , 0 = C1 sinh na + C2 cosh na  C2 = −C1 tanh na at y = 0 , C4 = 0 at y = b , 0 = C3 sin nb , nb = n

 = XY = Cn sin n y sinh n x − tanh na cosh nx at x = 0 ,  = − qs = C sin  y    n n n x k b q − s sin n y dy k 0 Solve for C = = n

b

n  sin  n y dy 2

0

3-6

(1) Let  = T − T1 , refer to equation (3.12) (2) Follow the same method for equation (3.12) n X = C sin x 1 a n n y Y = C3 sinh y + C4 cosh a a n n n at y = b ,  = 0 = C sinh b + C cosh b ,  C = −C tanh b 3 4 4 3 an a n  a n n  x sinh y − tanh b cosh y  = XY = C sin n a  a a a   at y = 0 , −k = h ( − ) y n  n   n  n   x = h  C n sin x  − tanh b − −k C n sin  a a a a      

15

h a 

sin

n

x dx a k 0 = Cn = h n n  a n  2 b+ x dx − tanh  sin a a  0 a  k (3) X = C1 sinh n x + C2 cosh n x at x = 0 , C2 = 0 Y = C3 sin n y + C4 cos n y at y = 0 , C4 = 0 at y = b , 0 = C3 sin nb , nb = n 

 = XY =  Cn sinh n x sin n y  at x = a , −k = h ( − ) x −k  Cn n cosh n a sin n y = h   Cn sinh n a sin n y −   h b  sin n y dy k 0 = Solve for Cn = n n b  h   2 n n   cosh  a + k sinh  a  0 sin  y dy (4) X = C1 sinh n x + C2 cosh n x at x = a , 0 = C1 sinh na + C2 cosh na  C2 = −C1 tanh na Y = C3 sin n y + C4 cos n y at y = 0 , C4 = 0 at y = b , 0 = C3 sin nb , nb = n

 = XY = Cn sin n y sinh n x − tanh na cosh nx  at x = 0 , −k = h ( − ) x −k  Cn sin n y ( n ) = h   Cn sin n y ( − tanh n a ) −   h b  sin n y dy k 0 = Solve for Cn = b n n  kh   n 2   − tanh  a  sin  y dy 0

3-7

From Figure 3.4,  = 1 + 2 + 3 For 1 : 1 = X Y = (C1 sin n x + C2 cos n x)(C3 sinh n y + C4 cosh n y ) at x = 0 , C2 = 0 , at x = a , 0 = C1 sin na , na = n 1 , at y = b , h1 = −k y h (C3 sinh nb + C4 cosh nb) = −k (C3n sinh nb + C4n cosh nb) C4 = −C 3 h sinh nb + kn cosh nb = −C 3 A h cosh  b + k sinh  b n

n

n

16

1 = Cn sin n xsinh n y − Acosh n y 1 q =− =  Cn sin n x (n ) at y = 0 , y ka q − sin n xdx k 0 solve for Cn = = a 2 n sin n xdx

 0

For 2 : X = C1 sinh n x + C2 cosh n x Y = C3 sin n y + C4 cos n y at y = 0 , C = 0 , at y = b , h C cos  b = k C sin  b  3

4

n

4

n

 n

h

= tan  b ,  = 

k n

at x = a , 0 = C1 sinh na + C2 cosh na , C2 = −C1 tanh na  = X Y = Cn cos n y (sinh n x − tanh na cosh n x) at x = 0 ,  =  ( y ) = Cn cos n y (− tanh na) b

Solve Cn =

− ( y )cos n y dy 0 b

for given  ( y )

=

tanh na cos2  n y dy  0

For 3 : X = C1 sinh n x + C2 cosh n x at x = 0 , C2 = 0 Y = C3 sin n y + C4 cos n y at y = 0 , C3 = 0 h = tan  b ,   =  at y = b , h C cos  b = kC  sin  b ,  4 n 4 n n n n k n

3 = XY = Cn sinh n x (cos n y )

at x = a , 3 = 0 =  Cn sinh n a cos n y b

0  cos n y dy

Solve for Cn =

0

b

=

sinh na cos  n y dy  2

0

3-8

(1) From Figure 3.6, equation (3.17) and (3.18): Let  = T − T1 X = C1 sin n x + C2 cos n x Y = C3 sin n y + C4 cos n y Z = C sinh  2 +  2 z + C cosh  2 +  2 z 5

n

n

6

n

n

at x = 0 , C2 = 0 , at x = a , 0 = C1 sin na , na = n at y = 0 , C4 = 0 , at y = b , 0 = C3 sin nb , nb = n

17

n

n

(  +   C ) + C cosh (  +   C )  C = −C tanh (  +  C )  = XY Z =  C sin  x sin  y sinh ( z ) − tanh ( C ) cos ( z )  +  +  + at z = C , 0 = C5 sinh 6

2 n

5

n

2 n

n

2 n

2 n

2 n

2 n



n

2 n

6

2 n

2 n

at z = 0 ,  =  =  C sin  x sin  y − tanh 0 n n n  

2 n

2 n

(  +  C ) 2 n

2 n



ba

Solve for Cn =

tanh

−o  sin n x sin n y dxdy 00

(

n2 +  n2

C

) 

b a

sin2  x sin2  y dxdy n

00

=

n

(2) same as (1), except: at z = 0 , C6 = 0

(  +  z) at z = C ,  =  =  C sin  x sin  y sinh (  +  C )  = X Y Z =  Cn sin n x sin n y sinh o

n

n

2 n

2 n

2 n

n

2 n

ba

Solve for Cn =

sinh

o  sin n x sin n y dxdy 00

(

n2 +  n2

C

) 

b a

sin2  x sin2  y dxdy n

0 0

3-9

=

n

(1) From equation (3.19): Let  = T − T0 X = C1 sinh n x + C2 cosh n x Y = C3 sin n y + C4 cos n y at y = 0 , C4 = 0 , at y = b , 0 = C3 sin nb ,  nb = n  at x = 0 , −k = 0 , C = 0 ,  = XY = C cosh  xsin y 1 n n n x  at x = a , −k = h ( − ) , x −k  Cn ( −n sinh n a sin n y ) = h   Cn cosh n a sin n y −   h b  sin n y dy k 0 Solve for C = = n cosh  a −  sinh  a b sin2  y dy n

n

n

 0

18

n

2 n



d

=−

q

x + C , at x = 0 , C = 0 1 1 dx k 1q 2   =− x + C , at x = a , −k = h ( −  ) 2  x  1 h2 k 2   2 k qa + qa + h   Solve for C2 =   k +h   = +  =  (2) Same as (1), just replace x → y , y → x , a → b , b → a .

From (3.30):

3-10

See Figure 3.3, the solution is shown in equation (3.12) with b = a .

3-11

(c) Let  = T − T X = C sinh  x + C cosh  x = C e−n x + C en x 1

n

2

n

Y = C3 sin n y + C4 cos n y

1

at x =  , C = 0 2

2

at y = 0 (centerline), C3 = 0  at y = L , −k = h , −k sin  L = h cos  L n n n y  n = 

  = X Y =  C ne−n x cos  n y

at x = 0 ,  = b =  Cn cos n y L

b  cos n ydy

Solve for Cn = L 0

=

 cos  ydy 2

n

0

3-12

Let  = T − T0 , a = w , b = 2w X = C1 sin n x + C2 cos n x

at x = 0 , C2 = 0 at x = a , 0 = C1 sin na , na = n at y = 0 , C4 = 0

Y = C3 sinh n y + C4 cosh n y

  = X Y =  Cn sin n x sinhn y

at y = b = 2w ,  = Tm sin ( x w) = Cn sin nx sinh nb a

Solve for C = n

Tm sin ( x w)sin n x dx 0

, where  = n

a

sinh nb  sin  n x dx 2

0

19

n 

a

=

n w

a

T m

=

(

) sin (n x )dx

sin  x



w

0

a

w

(

sinh  b sin2 n x



n

w

0

)dx

= 3-13

Let  = T − T1 Superposition:  = 1 + 2 For 1 : X = C1 sinh n x + C2 cosh n x Y = C3 sin n y + C4 cos n y

at x = 0 , 1 = 0 , C2 = 0 at y = 0 , 1 = 0 , C4 = 0 at y = b , 1 = 0 = C3 sin nb , nb = n

 1 = X Y =  Cn sinh n x sinhn y

at x = a , 1 = (T2 − T1 ) = Cn sinh na sinn y b

Solve for Cn =

(T2 − T1 ) sin n y dy 0

b

=

sinh na sin  n y dy  2

0

For 2 : at x = a , 2 = 0 = C1 sinh na + C2 cosh na ,  C2 = −C1 tanh na  2 = XY = Cn (sinh n x − tanh na cosh nx) sin n y at x = 0 , 2 = (T2 − T1 ) = Cn (− tanh na) sin n y b

Solve for Cn =

(T2 − T1 ) sin n y dy

=

0

b

− tanh na sin2  n y dy

 0

3-14

Same as problem (3-1)

3-15

Same as problem (3-11) with 2H = 2L .

20

Chapter 4 4-1

(a) Check Bi = hL k = 0.5  0.1 , one-D transit, let  = T − T 2 1  = , separation of variable – see equation (4.5) and (4.6) 2 x  t  = 0, C = 0 X = C sin  x + C cos  x at x = 0 , 1 1 n 2 n x  at x = L , −k = h , −k (−sin  L) = h cos  L  = C3 e−nt n n n x  nL = hL k cot nL =  2

  = X = C cos  xe −n t 2

n

n

at t = 0 ,  = Ti − T = i =  Cn cos n x L

Solve for Cn =

i  cos n x dx L

0

 cos2  x dx

i 2sin n = n + sin n cos n

=

n

(c)  4-2

e− t 2

−

0 h2t

e kC , if k  C

t

Same as problem (4-1), forced convection water, Bi = hL k  0.1 , one-D transient.

See equation (4.5) and (4.6).

4-3

(a) Natural convection air, Bi = hL k  0.1 .

Lumped capacitance method: From equation (4.4) − hAs t  = eCV , where  = T − T ,  i = Ti − T i (b) Forced convection water, Bi = hL k  0.1 , one-D transient, see equations (4.5) and (4.6). (c) T(final surface temperature) =T Water will heat wall faster due to higher h.

4-4

(a) Same as problem (4-3) (b), see equation (4.5) and (4.6). (b) Let  = T − T X = C1 sin n x + C2 cos n x  at x = 0 , = 0 , C1 = 0 x

21

(

)

at x = L ,  = Ts − T = 0 = C2 cos n L ,  nL = n + 1 2  , n = 0,1, 2,

  = X =  C

cos  x e− t 2

n

n

n

at t = 0 ,  = i = Ti − T =  Cn cos n x L

Solve for C = n

i  cos n x dx



0

L

2sin  ( −1)

=

+ sin  cos  = n

i

n

cos  x dx 2

n

0



n

n

n

(

2 ( −1) i

n

 n

)

n =  2 ( −1) cos x e−n t , where  = n + 12  , n = 0, 1, 2, L n n i n (c) Same t Figure 4.4 (a) (b) but with reversed profiles ( T  Ti ).



2

4-5

(a) See Figure 4.10 and equation (4.26). (b) (i) Same as Figure 4.10 (ii) Same as Figure 4.9 (iii) Same as Figure 4.11

4-6

(a) Same as example (4.3) except at x = 0 : T , T = T , let  = T − T q − h (T − T ) = −k  i i x   q − h = −k , take Laplace: x  q − s x − h = −k , where  = C e 2 x s s x − s x q  − hC e  = −kC2 − s −  e 2 s q  C2 = s h + k s

(

)

(

 =

q

(

)

)

− s x

, take inverse Laplace: s hk + s  h q  x h2   x = erfc − exp x +   h +  t erfc t      k h  k2  k  4 t 4 t   0 if q h 0.

(b) Ts

e

( k) ,  k , t t

1 2

to reach Ts .

Since Ts ~ t/k, therefore, higher conductivity requires longer time to reach Ts.

22

4-7

(a) See Figure 4.12, but in the ice part, temperature profiles are similar to those shown in Figure 4.10 for constant q at x = 0 but with revised profiles (heat removed from surface). You need to add Tm which is lower than Ti in the figure so that you can  t see ( ) moving in the x-direction during freezing period. (b) See Figure 4.12, temperature profiles are similar to those shown in Figure 4.9 for T = Ts at x = 0 but with reversed profiles (heat removed from surface). You need

 t to add Tm between Ti and Ts in the figure so that you can see ( ) moving in the xdirection during freezing period. (c) If consider natural convection takes place in liquid region with a constant h ,  (t ) moves faster which will affect temperature profiles in the ice and water regions.

This is because natural convection has better heat transfer than conduction.

4-8

(a) See Figure 4.15, but in the ice part, temperature profiles are similar to those shown in Figure 4.10 for constant q at x = 0 . (b) See Figure 4.15, but temperature profiles are similar to those shown in Figure 4.9 for T = Ts at x = 0 . (c) See Figure 4.15, but temperature profiles in liquid part are similar to those shown in Figure 4.11 (in fluid side).

4-9

(a) See equation (4.14) and (4.15): Solution for 2 x , 2y , 2 z can be obtained from equation (4.9) by using x, y, z respectively.

Solution for 1 can be obtained from equation (2.20) by using a for L . (b) See equation (4.14) and (4.15): Solution for 2 x , 2y , 2 z can be obtained from equation (4.6) by using x, y, z respectively.

Solution for 1 can be obtained from equations (2.20) and (2.23) by replacing a for L .

4-10

(a) Same as Figure 4.7, replacing Ti by Ts , and Ts by Tw . (b) Same as equation (4.18).

4-11

2

1  T − Ts = , let  = ,= 2 x  t T −T w

x 4 t

s

From equation (4.19): d 2 d =− 2 2 dd d 2  = C1e− d

23



 = C1  e− d + C 2 2

0

B.C.s: at x = 0 ,  = 0 ,  = 1, C2 = 1  at x =  (t = t ) ,  = = 1 1 4 t1 x  = 1− linear temperature 2 x  x 1− = C  e− d +1  C =−



1



1

 1

1

0

 e− d 2

0



  = 1−

x 0

 1



e− d 2

2

e− d

0

= 1−

x

where t = t1 , x =   linear temperature  x let x =  (t ) =  (t ) =  , t = t ,  ( x, t ) = 1− linear.

1 1 1  4-12

(a) For copper plate – Lumped Capacitance method From equation (4.4): −

hLc  t

Tb − T = e k Lc  t =  Ti − T 1 where Lc = b 2 (b) For plastic plate – one-D transient – separation of variable From equation (4.6), replacing L by b , let x = 0  t =  h can be estimated from Reynolds number (given).

2

4-13

Same as problem (4-3).

4-14

Undergraduate heat transfer problem.

4-15 Undergraduate heat transfer problem. 416 Undergraduate heat transfer problem.

4-17 Lumped Capacitance method, undergraduate heat transfer problem. 418 Same as example (4-1):

24

T − T Ti − T R = C1 J0 (r ) + C2 Y0 (r )

=

at r = 0 , C2 = 0 at r = r0 , 0 = C1 J0 (r0 ) ,  n r0 = 

at t = 0 , r0 = 1

 r r J  0  n r  dr  0  Cn = r0 0  r  =  r J 02  n r  dr  0  0  −  t r  =  C n e 2 n J 0  n r  r  0  o

4-19

2

T − T

Same as example (4-2):  =

, U = r Ti − T R = C1 sin r + C2 cos r at r = 0 , U = 0 , R = 0 , C2 = 0 at r = r0 , U = 0 , R = 0 , 0 = C1 sin r0 ,  n r0 =   r  −n Fo U =  Cn sin  n  e  r0  at t = 0 ,  = 1, U r0 = r  r r sin   n r  dr  0  Solve for C = 0 = n r0  r r sin2   n r  dr  0  0 2

4-20

Same as equation (4.9).

4-21

Refer to text 4.4.2, Figure 4.15, special case for slow melting. See Figure 4.15, but temperature profiles are similar to those shown in Figure 4.9 for T = Ts at x = 0 . And

, b can be determined, as shown in the text.

4-22 (a) Refer to text, Figure 4.4 (a) but temperature profiles are reversed due to heating from fluid instead of cooling to fluid. The final temperature of the plate is the same as the fluid temperature. (b) At x = L, the temperature rise of stainless steel is quicker than aluminum. But at x = 0, insulated surface with zero temperature gradient, the temperature rise of aluminum is quicker than the stainless steel. Since Ts ~ t/k, therefore, higher conductivity material (aluminum) requires longer time to reach Ts than low conductivity material (stainless steel).

25

Chapter 5 5-1

(1) See Example 2. Refer to Figure 5.2 for surface nodes, perform energy balance With various boundary conditions. Need to determine  A and C  .

(2) Same as (1).

(3) Same as (1).

5-2

From equation (1.18), one-D, hollow cylinder, S.S., q = 0 : 1   T  d 2T 1 dT r = + = 0  2 r r dr r dr   0 , Ti+1 − 2Tr 1 Ti+1 − Ti−1 i + Ti−1 + = 0 for interior nodes 2 ri 2r ( r ) where ri = i r , i = 1, 2, 3

N

(1) at r = r1 (i = 1) , T = T1 , at r = rN (i = N ) , T = TN get  A and C  .

(2) Interior nodes, same as (1), but at r = r1 , r = rN , use convection boundary conditions.

(3) Interior nodes, same as (1), but at r = r1 , use convection boundary condition, at r = rN , use heat flux boundary condition.

Or, perform energy balance directly at each note with the cylindrical coordinate.

Refer to the text 5.4.1, case 1 and case 2.

5-3

From Figure 5.8, and equation (5.16) by adding radiation out of surface node ( T1 ): T2 − T1 T1− T1 x h (T − T ) −  (T 4 − T 4 = C  1 1 sur ) + k x t 2 o where T1=surface note=400 C, T2=next to surface note=380oC, T1’=surface  t temperature h x  1after1 t=100 sec= For stability, x2  1+ k   2 = 2 , x =   

5-4

(a) Same as problem (3-11), for (a) and (c) (b) For a typical node on one of the surface of the fin, perform energy balance with the convection boundary condition.

5-5

(a) q

+q n−1

T (b) k

dT

=  A x

P

dt −T P

I  1

n+1

n

+ q = mC n+1

−T

P

n−1

x

P n

+

T k

x

c

2

dT dt

 T P+1 − T P n n + Cx  x + nx  = Ac  2 t 

26

I 2  1  1 + n = TnP+1 − T nP ) T + T − 2T (  n  (c) n−1 n+1 + k Ac  2  Fo 2  I  1  P P P P+1 + n + 1− 2Fo T ( ) n  Tn = Fo Tn−1 + Tn+1 + kA  2   c   t 1  .

For stability, 1− 2Fo  0 , or Fo = 2 x 2 P

P

P

5-6

(1) Same as equation (5.23) for interior nodes.

Refer to Figure 5.2 for surface nodes, perform energy balance with various boundary conditions.

(2) Same as (1) (3) Same as (1)

5-7

(1) Same as equation (5.25) for interior nodes.

Refer to Figure 5.2 for surface for surface nodes, perform energy balance with various boundary conditions.

(2) Same as (1) (3) Same as (1)

5-8

From equation (1.18), one-D, transient, q = 0 , hollow cylinder: d 2T 1 dT 1 T + = dr2 r dr  t ReferPto equation (5.13) for interior nodes: T − 2T P + T P 1 T P − T P 1 Ti P+1 − Ti P i+1 i i−1 i+1 i−1 + = 2  t ri 2r ( r ) (1) at r = r1 (i = 1, inner wall) , T = T1 (given) at r = rN (i = N, outer wall) , T = TN (given) ri = i r , i = 1, 2, 3

N.

(2) at r = r1 , r = rN , use convection boundary conditions. Interior nodes same as (1).

(3) at r = r1 , use convection boundary condition, at r = rN , use heat flux boundary condition. Interior nodes same as (1).

Or, perform energy balance directly at each note with the cylindrical coordinate.

Refer to the text 5.4.2, case 3.

5-9

Same as problem (5-8), but refer to equation (5.24) for implicit form, for interior nodes:

27

(1), (2), (3), surface nodes are similar to problem (5-8), but with explicit form.

Or, perform energy balance directly at each note with the cylindrical coordinate.

Refer to the text 5.4.2, case 3.

5-10

(a) Explicit format from (5.22) by using i, j, k: T P −T P T P −T P T P −T P T P −T P i, j +1,k i, j ,k i+1, j,k i, j ,k i, j ,k + i, j −1,k i, j ,k + i−1, j ,k y y x x + x y Ti,Pj ,k +1 − Ti,Pj ,k + Ti,Pj ,k −1 − Ti,Pj ,k z z P+1 P 1 T −T i, j ,k i, j ,k + = t z   t , similar to equation (5.23) as Let x = y = z , Fo = x2

P P P P P P P  Ti,Pj+,k1 = Fo (T i+1, j,k + Ti−1, j ,k + Ti, j +1,k + Ti, j −1,k + Ti, j ,k +1 + Ti, j ,k −1 ) + (1− 6 Fo)Ti, j ,k

For stability, 1− 6 Fo  0 , or Fo  16 . (b) Implicit format from (5.24) by using i, j, k: Same as part (a), but in left-hand-side of equation, replace P step by P+1 step.

Similar to equation (5.25) as P+1 + T P+1 + T P+1 + T P+1 + T P+1 + T P+1 P  (1+ 6 Fo)T i,P+j ,k1 − Fo (T i+1, j ,k i−1, j ,k i, j +1,k i, j −1,k i, j ,k +1 i, j ,k −1 ) = Ti, j ,k

28

Chapter 6 6-1

(a) Refer to Figure 6.4 for Pr=0.1<1 (gas), Ts  T (b) Refer to Figure 6.4 for h vs. x (c) Rex , Pr , Nux (d) Nu x = RemxPrn = a RemxPrn , depending on laminar or turbulent flow. k (e) q = h (T − T ) , where h = a Rem Prn x x x x s  x = negative value

6-2

Same as problem (6-1), except q x = positive value, Ts  T

6-3

Refer to text for detail

6-4

Refer to text for detail

6-5

Refer to text for detail

29

Chapter 7 7-1

(a) Refer to Figure 7.2 for both water and air (same one curve) Refer to Figure 7.4 for water (Pr>1) and air (Pr<1) (b) For u(y): at given x, , ,U , get Rex = u

From Figure 7.2, read

U x 

value at given  (i.e., change “y” value, or given

U

=

y Re x value) x

Then get u(y) from read

u U

, since U is known

For T(y): at given x, , ,U , get Rex =

U x 

T − Tw value at given  (i.e., change “y” value, or T − Tw U value,  =  ) and Pr given  = y x  T ( y) − Tw Then get T(y) from read , since T and T are known w  T − Tw From Figure 7.4, read

7-2

(a) U = U T = a + by at y = 0 , T = Tw , at y = T ,T = T T −T T = Tw − w  y T

From energyTwintegral − T equation: d T −k =k = c U (T − T )dy T 0  dxd 0 T p  y T y = c U (T − T )(1− )dy k

d

1

dx 0

p

(c U ) p  2 T T dx 1 d ( 2 ) 2k d =T T = T dx cpU 2 dx



=

30



w



T

T =

4kx ~ 4x cpU

T k y 0 (b) h = = = Tw − T  T −k

 Nux =

7.3

hx x = = k T

k 4kx cpU x = 1 Re  Pr x 2 4kx cpU

(a) For water, refer to Figure 6.4 for Pr>1,   T For liquid metal, refer to Figure 6.4 for Pr<1, T   (b) Refer to text fir detail (c) Refer to Figure 7.2 for water and liquid metal (same one curve) Refer to Figure 7.4 for water (Pr>1) and liquid metal (Pr<<1) (d) Same as problem (7-1)(b) T    −k y 0 −k (T − Tw )  qw  y 0 (e) h = = = Tw − T Tw − T Tw − T  =k  Rex , where  = Re x y x  0 x    get   0,liquid metal 0 from part (d), and 0,water   But, kwater  kliquid metal Therefore, hliquid metal  hwater

7.4

(a) Refer to text for detail (b) Same as Problem (7-2)(a) (c) Same as Problem (7-2)(b) q T = q = −kd , d = − w (d) T = c + dy at y = 0 , −k 0 k y at y =  ,T = T , c = T + qw  T   k T + q T = T kw (T y) − From Energy Integral equation:

31

q =

d

T

c U (T − T )dy =

p  dx 0 U d 1 2 1 = cp  ( T ) k dx 2 w

T =

2kx

cpU



d

T

dx 0

c U qw p



k

( − y)dy T

2x

But, from (7-2)(a), T 4x for Tw = constant T −k y 0 −k qw k k (e) h = = = Tw − T qw  T T k hx x x = 1 Re x  Pr Nux = k =  = T 2 2kx cpU But, from (7-2)(b), Nu = 1 Re  Pr for T = constant w x x 2  Nux qw Nux T w , T (x) qw  T (x) T w  7.5

(a)(b)(c)(d) same as problem (7-3) except change liquid metal to air. However, for   part (e), h  h since both k  k and  0,water 0,air water air water air  

7-6

(a) Refer to text for detail (b) Same as problem (7-4)(d) (c) Same as problem (7-4)(e) (d) See example 7.1 in the text book, but with the parabolic velocity and temperature profiles.

Thus, velocity profile is Solve for

as

Thus, temperature profile is

If , The above Integral equation becomes Solve for

as

32

Therefore, heat transfer coefficient and Nusselt number:

For parabolic velocity profile, boundary layer thickness is greater than slug velocity profile (zero boundary layer thickness, therefore, local Nux parabolic velocity  Nux slug velocity (e) If consider a uniform suction through the wall (i.e., V = −V0 ), boundary layer will be thinner, therefore, local Nux will be higher than without suction (V = −V0 = 0 ) 7-7

Refer to problem (7-3) for detail

7-8

For (a) and (b); refer to problem (7-4) (a)(b)(c) For (c); if consider a uniform blowing through the wall (i.e., V = V0 ), boundary layer thickness will be thicker, therefore, local Nux will be lower than without blowing (V = V0 = 0 )

7-9

For (a) (b) and (c); refer to problem (7-4)(a)(b)(c) For (d); refer to problem (7-6) (e)

7-10

For (a); refer to (7-4)(a) For (b); u = a + by at y = 0,u = 0, a = 0 y =  , u = U , b =

U



u y  = U 

From Momentum Integral equation: u d = − w =  u(u −U )dy 0 dx y U U y d  U  =  ( y −U )dy



dx d0 =



dx 0 d



2 2 y

 2 y

[U  ]dy 2 −U  U2 d 21  21  = ]=  [U  −U 2 6 dx 6 1 2  6 dx 3 1 U d  d = dx , integrate, [  ] = [ x]x  = 0 0  6 dx 2 U U 



33

12x or  = 3.464 U x Rex 

 =

T , b = − qw at y = 0, qw = −k 0 k y at y =  ,T = T , a = T + qw  T   k T

T = a + by

For (c);

qw (1− y )   k T T T Integral equation: From Energy d u qw d (T − T )dy u(T − T )dy = T = U     c p dx 0 dxy0 U  d T y qw (1− )dy qw   = c U dx 0  k T  p  T T y Let  = , = ,  T

T − T =

 (1−) d = d  2 3 1( − 2 )dy T T 0 c U dx  dx p   d 1 2 3  = (   ) U dx 6 6 2 3   = x+C at x = 0, = 0, C = 0 k

=

3 =

d

U 6x

1 0

1

1

, where  2 =

12x

from part (b)

U 

U  2 11 2 Pr − 1 1 1  = ( )3 (Pr) 3 , 2 =

 =  = T

1

1 3

−

1

( ) Pr 3 2

12x U

=

2.749x Re Pr

q qw , where Tw − T = w  T , from (c) at y=0 k Tw − T =k

For (d); h =

T

x hx kx  k =  k = Nu x =  T

T

34

1 − 2 x

−1 3

=

1 1 1 Re2x Pr 3 2.749 1

1

= 0.3637 Re Pr For (e); T = a + by 2 x

3

at y = 0,T = T w y = T ,T = T y y − Tw − T , or T − T = 1− T =T y, or T − Tw = w T T − Tw T Tw − T T From Energy Integral equation: d T c u(T − T )dy q = −k T = w 0 dp T y  y )dy  (T dy 0 k (T − Ty ) =  c U (1− −T ) w − w  0 p  T dx 0  T T k d  1 y2 y3  − = cpU  T dx 2  3 T 0 

= cpU −

d 1 2 ( T ) dx 6 

1 3

T =  Pr T 1 −k 0 3 k Re x Pr k k y h = = = 1 = − Tw − T T x 12  Pr 3 1 1 1 hx 1  Nux = =  Rex Pr 3 = 0.2887 Rex 2 Pr 3 k 12 Therefore, Nux qw  Nux Tw For (f); If consider a uniform suction through the wall (i.e., V = −V0 ), boundary Layer will be thinner, local Nux will be higher than without suction (V = 0 ) 7-11

(a) Refer to text for detail 1 5x at y =  where  = 3 Rex U = 1  5x  U = 5  = y  1.67  x 3 Rex  x 3 1 From Table 7.1 (or Figure 7.2): for  = 1.67 at any (x, y =  ) points 2

35

u f  =  0.52 U Similarly, From Figure 7.4 (or Table 7.1 with Pr=1) T − Tw

=

(=

u−0

= f ) = 0.52 at  =

T − Tw U − 0 (b) If U , f  and  w  at any given x:

f

−

1

5

= 1.67

3

3

U 2 ,  w

U 2 If x , h  and q  at any givenU : −

1

1

hx x , q = x 2 (a) Refer to problem (7-11): Pr = 1 , from Table 7.1 1 u T − Tw y =  , = 2.5, f  = (= f )  0.75  0.75, = T − Tw 2 U 1 u T − Tw y =  , = 1.25, f  =  0.4, =  0.4 T − Tw 4 U (b) Same as problem (7-11) 2

7-12

7-13

(a) Vx = a + by, y = 0,Vx = 0, y =  ,Vx = V y V = V x   Vy = −V0 (suction) d  Integral d  (with suction): From Momentum equation  = u2dy −U ( udy − V ) dU 0  0 dx 0 dy dx 0 d  2  =  d  y 2 y V 0 V 2 dy −V ( 0 V dy − V0 ) dx 0   dx  d 1 (− V 2 ) + (V V ) 0  dx 6   d 1 = (− V  + V0 )  dx 6  (b) T = a + by, y = 0,T = T1 = a, y = T ,T = T = T1 + bT y T − T1 = (T − T1 ) T T d Integral From Energy equation (with suction): −k = ( c u(T − T )dy + c V (T − T ) T p 0 1  p  y 0 dx 0

36

d

=



dx 0

−k (T − T1 ) kT d or =

T

p

−T )

= (T T

c u[(T − T ) − (T − T )]dy + c V (T − T ) 

1 T

d

1

p 0

V y

c V [1− exp( 0

−

1

y

)](1 dx 0 V yp  y T c V [1− exp( 0 −1)dy + c V  

1

p   dx 0 )( kT d T T If  T constant, dx 0 = 0, T= c V

p



)dy + c V (T − T ) p

0

1



0

p 0

Nu =

hx

=

k 7-14

k  x = x = xcpV0 k T k T

(a) at x = 0.05m,  =

5x

=

U x 

 = U f (0) U = 0.332U U = w   x x 1 U  h=k  (0) = 0.332 Pr3 k U = x x (b) at x = 0.05m y = 0.002m u U = from Table (7.1) to get f  = = , f  = ,  =  x U If Pr  1 (air Pr = 0.7 ) T − Tw = From Table (7.1) to get  = T − Tw same way for x = 0.05m, y = 0.04m Given: Air properties at 350K : K = 3010−3 W mK ,  = 20.9210−6 m2 s , Pr = 0.7

=y

7-15

1 y 3 u 3 (a) Refer to text: = ( y ) − ( ) U 2  2 

 x 4.64x = 1 U 2 Rex (b) For qw =constant, refer to problem (7-20)(a): qw T 2 y 1 y 3  (x) = 4.64

T −T = 

k

[ −( )+ ( ) ] 3  3  T

T

37

d T Integral equation: From Energy q = u(T − T )c dy w

k dx

c U



0

=



T

d

p

3 y

1 y 3 2

 dx  [ ( ) − ( ) ][ − ( 0

T

y

1 y 3 ) + 3 ( ) ]dy

T 2  2  3 T 1   = [ 2 ( T ) + 8  2 ( T )3 105 T  dx T 60    d 2 T 60 ] , if T  1, ( T )3 = [ T 0 U dx    T , follow similar procedure outlined in the text: To solve for  −1 T  0.77 Pr 3 , T  3.57x 1 1  3 2 Rex Pr qw h = qw 3 k x T − T = q 2 = 2 

p



d



w

=

w T

k 0.42

From text,

T

k

3

1

1

 for qw =constant

Rex 2 Pr 3

x T



−1

 Pr 3 for T = constant, and then w

1 1 k hx = 0.323 Rex 2 Pr 3 for Tw = constant x  hx qw hx T w 

7-16

(a) Refer to problem (7-14) for detail, u and T can be calculated at x = 0.03m , y = 0.002m (b) hx  with x , Nux  with x ,

7-17

−

hx

1

x 2

hx Nux = k

1

x2

(a) Refer to problem (7-14) for detail, u can be determined at x = 0.2m 5x (b) u( y) ,   = 3.9510−3 m , where u = 0.99U U x  1 1 − − 5x −3 T ( y) , T =  Pr 3 Pr 3 = 4.47 10 m , where Pr = 0.7 for air U x  T   if assume Pr  1

38

7-18

(a) Refer to problem (7-24) for air properties at 350K Calculate Rex = 239103 x, Pr = 0.7 From equation (7.39) at x0 = 0.1m , calculate Nux : 0.1 3 −1 hx 4 3 Nux  140 x[1− ( ) ] = x k 1 − 0.1 3 −1 h = 4.2x 2 [1− ( ) 4 ] 3 x 0.2 = q = h(Tw − T ) = 200h Heat loss for 0.1m  x  0.2m q = q dxw= 0.2 200hdx  412W



w

0.1



w

m

0.1

−

1

(b) If isothermal plate, x0 = 0, h = 4.2 x 2 , qw = 

0.2

200hdx  752W

0

7-19

m

(a)(b), refer to text for detail (c) for invicid fluid, assume u = U (plug flow model) assume T = a + by + cy2 + dy3 at T = Tw , y = 0 3 y 1 y From text: T − T = 1− ( ) + ( )3 Tw − T 2 T 2 T T 1 d equation: 3 y 1 y From Energy3Integral −k =k = c U [1− ( ) + ( )3 ]dy T 0 p  2 dx 0 2  2  y T T T k 3 1 = d (3  ) cpU 2 T dx 8 T 4k 8kx dx =  T dT T = c U c U p



p

T −k y 0 3 k h= = , Tw − T 2 T

Nu =

1

= 3 k 2 8

U x cp   k

1

 Nu = 0.53(Rex Pr) = 0.53Pe qw T = − = (a) y = 0, qw = −k 0 ,  a1 y 2 k T = = =  = = y 0, y2 0 2a2 , a2 0 a 2 , a = − 1 =  y =  , T = 0 = a + 3a  T  1 3 T 3 3T 2 y y = , T =T = a +a +a  3 2

7-20

hx



2

T

T



0

1 T

3 T

39

a  a  3 = − 1 T− 3 T q 2 y 1 y T − T = w T [ − ( ) + ( )3 ]  k 3  3 a0 = T

T

T

T T 0 at x1 = 0 at x2 , but T at x1  T at x2 y y (c) for frictionless flat plate,  = 0, u = U d T Integral equation: From Energy q = c U (T − T )dy (b)

p   dx 0 T (x) can be obtained by using ( T − T ) from part (a) qw 3 k by using T − T from part (a) h= = w  Tw − T 2 T hx 3 x Nu = = , where  (x) =  T x k 2 T w



7-21

(a)(b) For frictionless plate at Tw , u = U ,  = 0, t (x) =  2T (c) y = 0, T = Tw , =0 y2 T =0 y = T , T = T , y 3 y 1 y From text, T − Tw = ( ) − ( )3 T − Tw 2 T 2 T T −k y 0 3 k hx 3 x (d) h = = , Nu = = , see problem (7-19) Tw − T 2 T k 23 Ty 1 y 3 (e) −k T = 3 k = d c U [1− ( ) + ( ) ]dy 0 p  2 dx  2  2  y T T T Solve for T (x) =  see problem (7-19)

7-22

Refer to problem (7-2) 1 Re x  Pr Nux = 2

7-23

(a) Refer to problem (7-10)(e) 1 k h = , where T =  Pr −3 ,  = T (b) If Pr  1, hwater (Pr  1)  hair (Pr  1)

40

12x U 

hliquid metal (Pr  1)  hwater (Pr  1)  hair (Pr  1) 7-24

(a) Rex =

U x



= 4.78104

Laminar flow at x = 10cm 1

U x 1  = 0.332(  ) 2 (Pr)3 , h  with x  k  qw = h(Tw −T ) = 100h W m2 , where h = 19.34 at x = 10cm hx

= 1934W m2

7-25

(a) Refer to problem (7-10)(c)(d) for qw = constant: 1

1 hx Nux = = 0.3637 Rex 2 Pr 3 k (b) Refer to problem (7-10)(e) for Tw = constant: 1

1 hx Nux = = 0.2887 Rex 2 Pr 3 k  Nux qw Nux T w 

7-26

(a) Refer to equation (7.30). (b) Refer to equation (7.30) and Figure 7.11, heat transfer coefficient for m = 0.5 is higher than m = 0 due to higher temperature gradient shown in Figure 7.11. (c) Refer to equation (7.30) and Figure 7.12, heat transfer coefficient depends on both thermal conductivity and temperature gradient. Pr = 0.01 will provide higher heat transfer coefficient although Pr = 5 has greater temperature gradient as shown in Figure 7.12, but Pr = 0.01 (liquid metal) has much higher thermal conductivity than Pr = 5 (water).

7-277-27 Refer to text 7.4.2 High Velocity Flow over a Flat Plate.

The surface heat flux becomes: where

Here, for laminar boundary layer heat transfer, h can be calculated from:

For high-speed flow over a flat plate, heat transfer can be negative, zero, or positive, depending on if the wall temperature is less than, equal to, or greater than the adiabatic wall temperature, as shown in Figure 7.19.

If the coming flow Mach number is greater than 1, use shock wave equations, refer to problem 7-28, to obtain velocity, and temperature after the shock (velocity decreases and

41

temperature increases). Use these new velocity and temperature to determine adiabatic wall temperature and heat transfer coefficient (with lower Reynolds number). And heat flux can be calculated.

7.28 Region 1 = before the shock wave Region 2 = after the shock wave At altitude = 17500 m, Cp/Cv =

= 0.131 kg/m3

= 216.7 K,

= 1.4, R = 287 J/kg-K, Cp =

= 1.0045 103, Pr = 0.7

From compressible flow:

or

, outside the stagnation region.

Use reference temperature to compute thermal fluid properties,

From properties table, Pr = 0.7, k = 37.7 10-3 W/m-K, ReR =

= 25.75 10-6 m/s

= 1.35 106, based on nose radius R0 , for flow at stagnation region. , when R0 = nose radius

, it can be varied by using different properties from the table. This heat flux should be removed to maintain the nose at 80

42

7.29 (a)

At y = 0, u=a=0 at y =

,

at y =

,

Solve b, c for velocity profile Solve a, b, c for temperature profile (b) insert u and T from (a) into momentum and energy integral equation, respectively.

Solve for and .

except Pr << 1 (c) same as (a), except

is replaced by

,

.

, increases,

(d)

for blowing for suction

43

decreases.

for blowing for suction

For blowing case:

decreases,

increases.

decreases,

increases.

For suction case: the reverse is true.

7.30 Refer to problem (7-29). (a)

At y = 0, At y =

, ,

,

where (b)

44

Chapter 8 8-1

(a) 0 = −

1 P

 x

T , u

x

= c =

2T y2

T 0 =0 y y = l, u = u, T = Tw T T b qw = = (b) = constant x x cpul u T ( y2 − b2 ) T = Tw + 2 x T l u T h = −k y = −k 2 x 2l = 3k T −T 1 u T l w b l2 3  x 1 l 1 l u T 2 2  ( y − l ) dy = T 1 u T 2 Tdy = T + where T = − l B.C.s: y = 0, u = u,

w  l 0  w 2 x 3  x l 0 hDe h4l Nu = = = 12 , where De = 4l k k (c) Slug flow velocity has thinner boundary layer thickness (  ) than parabolic velocity profile, h is inversely proportional to boundary layer thickness (  ) 1 h h h slug flow parabolic velocity  b

8-2

(a) 0 = −

1 P

 x

T , u

x

=

2T

y2 T =0 B.C.s: y = 0, u = um , y 0 y = l, u = 0, T = Tw

y (b) u = um (1− ) l y T 2T = um (1− ) l x y2 Let um T = constant = c ( q = constant ) w  x 2 y T c(1− ) = 2 l y

45

T

1

y2 ) + c , where c = 0 due to T = 0 1 0 1 y 2l y 1 1 1 T = ( y2 − y3 ) + c 2 , where T = Tw at y = l, c2 = Tw − c l 2 2 6l 3 1 2 y3 1 2  − − l ) T = Tw + c( y 6l 3 2 l udy  = 1u u= 0 l 2m l  uTdy = ... = T − c 4 l 2 T = 0 w b ul T 1 15 −k − kc 2 l y 0 15 k  h= w b = =− c 4 l2 T −T 8l 15 hDe 15 k 4l Nu = =−  = 7.5 where D  4l h k 8 l k (c) Parabolic velocity profile had greater velocity gradient all wall, greater Nu too ( Nu = 8.24 from text) = c( y −

3

8-3

y2

From text, u = um (1− 2 ) , where um = centerline velocity 2 H From Energy equation: 2T u T 3 1 y2 T T = constant = = um (1− 2 ) , where x H x y2  x 2  3 um T 1 2 1 y4 T= [ y − ] + c y + c2 2 1 2  x 2 12 H T = 0 (insulated) B.C.s: y = −H , y y = H , T = Ts (unknown surface temperature)

Solve for c1, c2 , 3 u mH 2 T y4 6 y2 8 y T = [− + + −13] + T 4 2 s 24  x H H H H 2 T = − H uTdy = ... = − 52 u mH T + T m s um 2H 35  x

46

T −k y H −2k umH T qs x 70 k h = T − T = T − T = 52 u H 2 T = − 52 H s

m

s

m

m

35  x 70 Nu = = = = 5.38 , where D  4H h k 52 H k 13 hDh

8-4

70 4H k

(a) u = constant - fully developed T (x, r) - developing (b) For thermally fully developed, T

T

= constant, T −T

x

m

= constant

s,r1

= constant (c) u = constant, x T 1  T c u =k (r ) simplified energy equation p x r r r 1d dT 1 dT = constant = c (d) (r )= u r r r  x dT = c 1 r2 r + c1 r 2 1 T = c r 2 + c ln r + c 1 2 4 T 1 B.C.s: at r = r , = 0, c = − r 2 1 0 2 0 r 1 r = r , T = T , c = T − cr 2 − c ln r s 1 i i s 2 4 i 1 2 1 2 1 2 1 2 T = cr − r ln r + T − cr + r ln r s i 4 i 2 0 4 20 c 2 1 r = T + (r − r 2 ) − r 2 ln s i 4 2 0 ri c T −k( r − 1 2 1 −k i r0 ) r =r 2 2 ri i (e) h = qw r = = Ts − Tm ro Ts − Tm Tro s − 1 Tm2 r 1 2 [ (T − T )2rdr  r ln − c(r 2 − r 2 )]rdr ri 2 o i s r 4 T − T = ri i = s m  (ro2 − ri2 ) (ro2 − ri2 ) hDh 4 A 4 (ro2 − ri2 ) Nu = where Dh = c = = 2(r o− r i) k f 2 (r + r ) o

8-5

Refer to Example 8.1 in the text

47

i

2

3

y2

(a) ub = um , u = um (1− 2 ) 3 2 b 96 u D 4A (b) f = , it should be 24 instead of 96, where Dh = c = 4b, Re Dh = m h ReDh  P hDh (c) Nu = 8.235 = , where Dh = 4b k 8-6 8-7

8-8

Refer to problem (8-2) 2u

T 2T , u =  y2 x y2 (b) y = 0, (lower plate), u = 0 , y = 2H (upper plate), u = uT , u = c y + c , c = 0 , c = uT 1 2 2 1 2H y uT u = 2H T = constant due to qs = constant x y T 2T uT = 2 2H x y dT u T 1 2 = T y + c1 dy 2H x 2 u T 1 3 T= T y + c y + c2 1 2H x 6 B.C.s: y = 0, dT = − qs = c 1 dy k uT T 1 q  (2H )3 − s 2H + c y = 2H , T = T (unknown) = s 2 2H x 6 k (a) 0 =

(a) From problem (8-7) y u= V 2H V T 1 3 y +c y+c T= 2 2H x 6 1 B.C.s: at y = 0, T = T0 = c2 y = 2H , T = T , c = [T − T −

V

T 1

(2H )3]

1

2H x 6 2H (b) From Energy equation and parabolic velocity from example 8.1: 1

1

1

48

0

u T

=

2T

=

T 1 3

y2

 x y2 um (1− 2 ) 3 um T 1x 2 21 y4 H T= ( y − ) + c y + c2 2 1 2  x 2 12 H B.C.s: at y = 0, c2 = T = T0 3 um dT 1 1 (2H )4 1 2 at y = 2H , c = {T − T − [ (2H ) − ]} 2 1 0 2  dx 2 12 H 2 2H 8-9

(a) For the developing flow and temperature, refer to text for detail (b) For the fully developed region, refer to text for detail T (c) (1) No viscous dissipation, wall = 0 y  0 = − (out of channel) (2) With viscous dissipation, T wall y

8-10

(a) u = constant, T = fully developed region hDe h4l = Refer to problem (8-1), Nu = 12 = k k (b) u = constant, T = developing region = T (x, y) T 2T = 2 x y Separation of variable method: T (x, y) = T (x) T ( y) u

Nu (b)  Nu (a ) due to thinner thermal boundary layer hx  with x  (c) h  with x , but Nu = k ( Tw − Tb )  with x  Refer to text for detail 8-11

(a) Viscous dissipation = u

 u 2 ( ) , it should be included for high speed flow, cp y

large, or for high  fluid such as plastic flow y (b) and (c), refer to problem (8-1) i.e.,

8-12 This is a undergraduate level heat transfer problem: use log temperature difference method (LTDM): q = mcp (To −Ti ) = hAs Tlm Ti − To (Ti − Ts ) − (To − Ts ) = , As =  DL Ti Ti − Ts ln ln To To − Ts Since everything given, except To , so calculate To = , then, get q = 

where Tlm =

49

8.13 (a) flow between two parallel plates with a gap of 2H, at uniform wall temperature.

Energy equation becomes

B.C.s: x = 0, T = T0 y = 0, = 0 y = H, T = Tw separation of variables: Let

At y = 0, At y = H,

= 0, c1 = 0 = 0, 0 =

At x = 0,

where

where

50

= can be determined If consider uniform heat flux condition, ,

= given, but Tw(x)

= can be determined If wall temperature increasing linearly with the plate length, increases, h(x) in case (b) is higher than in case (a) after 1/2 of the plate length, but lower in the first 1/2 of the plate.

Similar trend is true if wall heat flux increasing linearly with the plate length.

8-14 From text: (1) (2) (3) From (2):

where

(4) From (1):

From integration table:

51

(5) Thus, where

= can be determined

8-15 = given

(1)

From text: (2) (3) From (1)

From (2)

(4) Then , energy balance

, from (4)

52

From (1)

can be determined

can be determined,

If

from (4):

,

53

Chapter 9 9-1

Refer to text: y G 1  = ( rx ) 4 , where G = g  qw x 4 rx 2 k x 4  (x,) T − T f () = Grx 14 ,  = Tw − T 4 ( ) 4 f  + 3 ff  − 2 f 2 +  = 0   + 3Pr f   = 0

9-2

(a) Refer to text, Figure 9.1 −1

(b) hx 1x 4L, i.e., h is 4 inversely proportional to x to the ¼ power (c) h = h dx = h from text L 0 x 3 L hx L 4 L 4Lk = h = 0.443(G Pr) 1 4 (d) Nu = rL k 3k L 3k L x

1

9-3

= 0.59(GrL Pr) 4    u= =  y  y G 1 1 G 1 = 4 ( rx 4  ( rx ) 4 ) f 4 x 4 =

2 x

V =−

1

2

(Grx )  x

=−

f



g (Tw − T ) 14 y Grx 14 ) = ( ) 2 4 x x 4 f () =  (x,) G 1 4 ( rx ) 4 4 T − T = Tw − T

where  = y(

    x

G 1 − 5 g (Tw − T ) 14 = −4 ( rx 14 ) ) f y(− x 4 )( 4 4 4 2 y G = f ( rx ) 1 4 ( Grx ) 14 x 4 4 y 1 2  = f 1 5 (Grx ) 4 4 4 x    1 y Grx ) 14 (− =  = ) ( x  x 4 x 4

54

   1 G 1 =  =   ( rx ) 4 y  y x 4 2     1 G 1 1 G 1 1 G 1 rx 4 rx  ( ) =  4 = (  ) =  ( )  ( rx ) 2 y2   y y x 4 x 4 x2 4 Substitute into equations (9.4) and (9.5), obtain equations (9.9) and (9.10) 9-4

From text, equation (9.15:) hx G 1  Nu = = −( rx ) 4  x k 4  0  where − 0 = 0.508 for Pr = 0.733   Nu 0.508( 1 ) 1 4 (G ) 14 = 0.539(G ) 14 x = rx rx L 44 h = h dx = h from text, 

x



3 L 1 4 L 4L k = h =   0.539(G ) 4  Nu = x rL k 3 L k 3k L 0

x

hx L

1

= 0.478(GrL ) 4 9-5

(a) From equation (9.19), 1  4 3 ( 2 ) 14  1   (Grx  Pr2 ) 4 Nux = 1 4 5 1+ 2 Pr 2 + 2 Pr  1

1

= 0.596(Grx  Pr2 ) 4 if Pr → 0 (b) From equation (9.19) 1   4  1 3 ( 2 ) 1 4  1  (Grx Pr) 4 Nux = 1 1 4 5  +2 + 2 1  Pr  Pr 2 11 1 2 4 if Pr → = 0.596( ) 4(Grx Pr ) 2 1 = 0.496 (Grx  Pr) 4 9-6

Refer to Figure 9.3: Perform force balance (refer to text) to get equation (9.22) Perform energy balance to get equation (9.23)

55

9-7

h =

T −k y 0 

=

2k (Tw − T )

=

2k

Tw − T (Tw − T ) (x)  (x) hx 2x 2x Nux = kx =  (x) = 1/ 4  20  x  240(1+ 21Pr)  Pr Grx     x

1

  4  Pr Grx  = 20  15(1+ ) 21Pr   1

  4 1   Pr 4 = (R )  ax 20 15(Pr+ )  21   1

 0.413Rax 4

for Pr = 0.733

9-8 Integral Method Assume velocity profile , B.C.s: , , , At

, from momentum equation

(1) Where Assume temperature profile

56

B.C.s: , , ,

(2) Put (1) into momentum integral equation:

Get (3) Put (2) into Energy Integral Equation: (4) Let

,

From (3): (5) From (4): (6) From (5), (6) are valid for any value of x:

Put

½, and

¼ into (3) and (4), solve for

:

From (3):

57

9-9 (a) Inclined plate at Heating fluid at If , , For upward forced, h will decrease For downward forced, h will increase Refer to Figure 9.4. Replace by (b) Refer to Reference [9].

9-10 If If , , For upward forced, h will increase For downward forced, h will decrease Refer to Figure 9.4. Replace by (b) Refer to Reference [9].

58

Chapter 10 10-1

Refer to text for detail +

10-2

T =

(a) 0  y  13.6 : T+ =

q

*

w

c pu*



*

, u =

w



+

,u =

u u*

+

, y =

yu*



+

y+

1− Ry + dy + = Pr y + = Tw − T

  +1 0

T*

H



Pr

13.6  y+  : +

T =

y

+

(1−

y+ +

)

y+

T

−T

+

R+ dy = 2.44 ln = 13.6T * y 13.6 (1− + ) R 2.44 y+ (b) at y = 0.05cm, u* = 10 m s , T * = 3C, Tw = 100C, 13.6

Pr = 0.7,  = 20.9210−6 m2 s yu* = 239  u+ = 18.36 m s , u = 183.6 m s T −T 100 − T13.6 T + = w 13.6 = = 9.52, T = 71.44C 13.6 * T 3 T − T 71.44 − T And T + = 13.6 * = = 6.994, T = 50.46C T 3 y+ =

10-3

(a) Refer to problem (10-2)(a) (b) Refer to the text: T −T R+ +13.6 Pr)T * + T − T = (2.5 ln b w  0.833 , where T − T = T − T c w c 13.6 13.6 w T −T 13.6 c w qw h= , Nu = hD , T * = qw * D c u k T −T w

 Nu =

b

p

ReD Pr

f 2 Re 0.833[13.6 Pr+ 2.5 ln( D 27.2 0.079 where f = Re1/D 4 D

59

f 2)]

For Re D = 104, Pr = 1, NuD = 35.5 0.4 Compared with NuD = 0.023Re0.8 = 36.5, ~ 3% difference D Pr

If V ,  , NuD  10.4

(a) Refer to text for detail, but for parallel plates with a gap of b : T * =

qw c pu*

1  7  7 b 2     T − T y u y 0 1 uTdy = 1  , T = b b u , T − Tw =  1  2 c  b c w  b 0 udy 2  2  Tw − Tb  0.89(Tw − Tc )  1 +  = 0.89T *[5 Pr+ 5ln(5 Pr+1) + 2.5ln 2 b  ]  30  1

1



hDe

h2b

q h = T −w T , Nu = k = k w



b

h2b

Nu =

 , where D  2b e

Re Pr

=

f 2

 1 +  0.89[5 Pr+ 5 ln(5 Pr+1) + 2.5 ln 2 b  30 ]     + bu* 1 0.079 V 2b * f, f = , b = where u = V 1 , Re = 4 2   Re (b) For Re = 2103, 2104, 2105, Pr = 0.7, b = ,  =, b+ = , Nu =  k



and compared with Nu = 0.023Re0.8 Pr0.4 =  10-5

(a) From equations (10.11) and (10.12):   u  u u u ( − uv) = (  +  ) u +v = M x y y  y y y assume −uv =  u ,  =  M y   k T T T ( − vT ) =  (  +  M ) T u +v = x y y cp y y Pr Prt y  k T  assume Pr = ,  = , −vT  =  , Pr = M H t  c y  





p

H

(b) For fully developed turbulent pipe flow: u = 0, from text, assume −uu = M u , − T  = H T , v = 0, r r x

60

d  du 1 dP 1 [r( − uv)] +  dx r dr  dr d 1 dP 1 [r( +  du ) ] =− + M  dx r dr dr

0=−

u

 k T T 1 [r( ( ) − vT )] = x r r cp r 1  = [r( +  ) T ] H r r r

10-6

(a) Refer to* text for three-region velocity profile, where yu 0.079 VD + * −6 2 ,  = 15.8910 m s y = , u = v f 2, f = , Re = 1   Re 4 u u+ = u*  You can plot u = u( y) , by varying y = 0 to 2.5cm 5 yu * + = (b) Let y 5= , you can calculate y = * =laminar sublayer u  4 − = 1.54210 m

10-7

(i) and (ii) Rex =

ux  where air properties at 450 K:  = 32.3910−6 m2 s , k = 0.0373W mK ,  = 0.774 Kg m3 , Pr = 0.686

Re x = 3.7 104

Laminar flow at x = 0.1m , universal profiles

Turbulent flow at x = 1m , three region profiles Re x = 3.7 105 (iii) For turbulent flow, refer to text −1

C fx = 0.026 Re x 7 = 4.16310−3 1  = C ( U 2 ) = 0.232 N m2 s fx  2 w u* = = 0.5475 m s  St x = 2.53510−3, from equation (10.55) Nux = Stx  Rex  Pr = 644.27 Nuxk h= = 24W m2k x x  qx = hx (Tw −T ) = 7209.4 w m2 10-8

(a) Refer to problem (10-5) (b):

61

1 dP

1 d

[r( +  ) du ] M  dx dr r dr where    M , assume M =  M (r) = given du =0 B.C.s: r = 0, u = 0, r = R, dr (b) Refer to text: u2rdr V = = 0.82U = constant =

 2rdr 0.079 (c) C f = Re0.25 D

max

10-9

(a) Refer to problem (10-5) (a) for detail (b) Refer to problem (10-1) for detail

10-10

(a) M

du

= −uv , turbulent stress, turbulent flow structure, dominated term

dy for turbulent flow M uv turbulent stress , = , relation between turbulent stress and Prt =  H vT  turbulent heat flux turbulent heat flux, Prt  0.9 for common turbulent flow, imply

vT   0.9uv , or  H = 0.9M (b) Refer to problem (10-1) for detail 10-11 (a) Refer to text Use where

from the text

Where

62

(b) Where

= can be calculated

can be calculated (c) Check

= can be calculated at x = 20 cm, y = 0.2 cm

can be calculated

(1) Where

can be calculated

can be calculated From (1): T = can be determined

63

10-12 Van Driest Mixing Length Theory

Let

,

,

Where , for

,

for You can plot vs. for And compare with Prandtl Mixing Length Theory,

64

10-13 Derive the turbulent pipe flow velocity profile (Figure 10.14) and friction factor (Figure 10.12) for sand grain roughness. See Chapter 16.

Note: How to solve the problem if this is a turbulent boundary layer flow?

10-14 Derive the turbulent pipe flow temperature (similar to Figure 10.14) profile and heat transfer coefficient (similar to Figure 10.12) for sand grain roughness.

See Chapter 16.

Note: How to solve the problem if this is a turbulent boundary layer flow?

10-15 Refer to Chapter 8, problem 8-14, and refer to Chapter 8.

where

can be determined

Re = turbulent flow.

Need to provide dimensions, thermal fluid properties, you can plot , , vs. and find the maximum at certain x location (near the end of the tube length).

65

,

Chapter 11 11-1

(a)  = 0.3F0−4 + 0.7 (1− F0−4 ) = 0.673 where F0−4 = 0.0667 at Ts = 2000m  K (From Table 11.1)

 = 0.3F0−4 + 0.7 (1− F0−4 ) = 0.405 where F0−4 = 0.7378 at Tsur = 6000m  K (From Table 11.1) (b) q = G − E = T 4 − T 4 "

net

sur

= 5.67 10

11-2

−8 



 =  = 1−  = 1−

0.405 (1500 ) − 0.673 ( 500 )  = 113.868103 W m2  4

Gref

G qn = qc + E + Gref − G "

s

= 1−

4

800 = 1− 0.8 = 0.2 1000

"

= h (Ts − T ) + T s4 + G ref − G = 42W m2 11-3

(a) From Figure 11.6, for black body, max = E = 0.4E ,b

for   3m

= 0.8E ,b

for   3m

2897.6 = 5.8m 500

(b)  = 0.4F0−3 + 0.8(1− F0−3 ) = 0.795 where F0−3 = 0.01376 at Ts = 1500m  K

 = 0.4F0−3 + 0.8(1− F0−3 )

= 0.574 where F0−3 = 0.564 at Ts = 4500m  K q = G − E = T 4 − T 4 "

net

sur

= 5.67 10

(c) q"

  with T 

net " net  =0.4

q

−8 

0.574 (1500 ) − 0.795 ( 500 )  = 161.946 103 W m2 

s " net  =0.795

q

s 4

4

"  qnet  =0.8

(d)   with Ts    constant for Ts = 500 ~ 1000K 11-4

Energy balance on the panel:  s Gs = T s4 Ts =  s Gs 1 4      Pannel A:s = 0.5F0−3 + 0.2 (1− F0−3 ) = 0.491, where F0−3 = 0.96998 at T = 3 5800 = 17400m  K

66

Pannel B:s = 0.1F0−3 + 0.01(1− F0−3 ) = 0.0973, where F0−3 = 0.96998 at T = 17400m  K Pannel A:  = 0.5F0−3 + 0.2(1− F0−3 )  0.2 if Ts 500K , T = 1500m  K , F0−3 0.002(negligible) Pannel B:  = 0.1F0−3 + 0.01(1− F0−3 )  0.01 14  0.491 1300 Ts =  0.2 5.67 10−8  = 487K for panel A   =  0.0973 1300 −8 1 4 0.01 5.67 10 = 687K for panel B 

11-5



(a) Refer to example 12-2, to get T,i greenhouse temperature. (b)For global warming, Gatm  due to CO2 emission.

11-6

(a)  = 0.9F0−0.3 + 0(F0−2 − F0−0.3 ) + 0.9(1− F0−2 )

0.9

(b) s = 0.9F0−0.3 + 0(F0−2 − F0−0.3 ) + 0.9(1− F0−2 ) = 0.9 0.0393 + 0.9(1− 0.945) (c) Refer to Example 11-2:  = 0.085, G = 900W m2 , G s

s

= T 4 ,  atm

sur

atm

= 22 C, T = 27 C ,  = 0.9

h = 5W m2K , T

,i

i

0.085

g

  0.9, G =  T ,   0.9, T = 22 C, T

= 273K

4

i

i

i

i

i

i

0.9,

sky

 h0 (T,0 − Tg ) =  = +, T,o  Tg −, T,o  Tg 11-7

(a)    1 for   1m , to transmit solar energy (b)   1 for   1m , to absorb solar energy (c) Refer to Chapter 4.1, dT VC = hA (T − T ) + A  (T 4 − T 4 ) s  i s w i dt

11-8

(a) Refer to Chapter 2 for the concept of convection and radiation resistances and flows and draw the thermal network diagram for the collector. The collector plate absorbed radiation from solar and sky, but lost heat to convection water flow from its lower surface, and lost heat to natural convection between the collector plate and the glass cover window, and emission from the collector plate from its upper surface. The collector plate temperature is higher than water, glass cover, and outside air. (b)

67

 q" +  q" = q" s s

s sky

+ q" + E

convection

0.8q" + 0.8T 4 = h s

sky

q

flow

(T

loss

collector

−T

collector

fluid

) + h (T gap

−T

collector

glass

) + 0.8T 4

collector

=

" convection

(c) If s = 0.95,  = 0.1, " qconvection =

11-9

Refer to problem (11-8).

11-10 Refer to problem (11-2).

11-11 (a) solar = 0.8F0−2 + 0.3(1− F0−2 ) 0.8

where F0−2  0.94 at T = 2m  5800K

sky  0.3 where F0−2 = 0 at T = 2m 233K  plate  0.3 where F0−2 = 0 at T = 2m333K =  T 4 + h (T −T ) −  600 − T 4 (b) q" 

heater

s

 97W m



s

solar

sky

sky

2

11-12 (a) Refer to examples 1 and 2: For the collector:

 q" +  q" = q" s s

s sky

convection

+ q" + E loss

collector

4 0.8q s + 0.8T sky = h flow (Tcollector − T fluid ) + h gap (Tcollector − Tglass ) + 0.8Tcollector "

4

For the glass: where αs = solar absorptivity for absorption of Gλ,s ~ Eλ,b (λ, 5800 K) (b) For the glass: The glass is totally transparent for λ < 1 μm, and opaque with an absorptivity α = 1 for λ > 1 μm.

For the collector plate: The collector plate has very high absorptivity α1 (for λ < 1 μm) and low absorptivity α2 (for λ > 1 μm).

68

Chapter 12 12-1

Refer to text for details.

12-2

Follow Hottel’s crossed-string method for 2-D geometry, refer to text for details.

12-3

Follow double area integration and contour integration for 3-D geometry, refer to text for details.

69

Chapter 13 13-1

(a) Refer to Figure 13.5 for detail. (b) F12 (top to bottom wall)  0.5 = F21, F1R (top to side wall) = 0.5 = F2R From equation (13.11), T1 = 923K , q = 20kW , T2  796K

13-2

(a) From Hottel’s crossed-string method, F12 (top to bottom wall) = 0.618 = F21, F1R (top to side wall) = 0.382 = F2R From equation (13.11), T1 = 923K , q = 20kW , T2  560K

13-3

(a) Refer to Figure 13.5 and equation (13.11): Need to assume the emissivity of bottom wall ~ 0.7, the emissivity of top wall ~ 0.9 (see problem 13.2): q12 (bottom to top)  25.5kW (b) q = q = Eb1 − J1 to get J =  12 1 1 1 − 1 A11 q = q = J2 − Eb 2 to get J =  12 2 2 1−  2 A22 J −J J − J2 Then from q1R = 1 R = qR 2 = R to get J R =  1 1 A1F1R AR FR 2  J = E = T 4 to get T  1225K R

b,R

R

(c) 1 (bottom) ,

13-4

R

q12 , TR 

2 (top) ,

q12 , TR 

R  or ,

No effect on q12 or TR

(a) q1 (bottom) = q12 (bottom to top) + q1s (bottom to surroundings) = A1 F12 ( Eb1 − Eb2 ) + A1 F1s ( Eb1 − Ebs ) =, where F12 = , F1s =  A1 ( Eb1 − Eb 2 ) = (b) q1 = −1  1 1 −1  F12 +  +   F F  1R 2R  

70

(c) Solve J1 and J 2 from Eb1 − J1 J − J )+ F (J − J ) = F12 ( 1 2 13 1 3 1 − 1 1 Eb 2 − J2 J −J +F J −J ) = F21 ( 2 1 ) 23 ( 2 3 1 − 2 2 Then q1 = A1F12 ( J1 − J2 ) + A1F13 (J1 − J3 ) , where J3 = 0, F12 =  , F13 =  (d) 13-5

From equation (13.11), q1 = , where F12 = , F1R =  (a) F11 = 0, F22 = 0, F33 = , F31 = F13 = , F32 = F23 =  F12 = , F13 = , F21 = F12 = , F23 =  Refer to text for 3-surface enclosure problem, to get J , J , J with q = 0 , i.e., J = E = T 4 1

2

3

3

3

b3

3

Then get q1 = q2 =  (b)Refer to Figure 13.5 for detail, to get q1 = q2 from equation (13.11) by using F1R = F13 , F2R = F23 13-6

(a) and (b), refer to problem (13-4)(c) (c) F12 = 1, F13  0, A1F12 = A2 F21, F21  0, F23 = 1 Refer to problem (13-4)(c), solve for J1 , J , Then q = A ( J − J ) =  A (T 4 − T 4 ) 1

1

1

2

1 1

1

2

(d)Energy balance: 1 = 1  G +  T 4 =  T 4 + h (T − T ) 1

s

1

sky

1

1

1



T1 =  if Gs is given, or Gs = 0

13-7

(a) J =  T 4 + 0 =  1 1 1 G =  T 4 + (1−  ) T 4 =  1

2

2

J2 = G1 =   T 4 −T 4 1 2 (b) q = 1−  2 1+

(

2

)

=

1

1

from equation (13.8) with 1 = 1, F12 = 1

2

13-8

(a) F12 = 1 A A F = 1 F = 1 = 21 A2 12 A2 

(b)From equation (13.8), to get q12 = 

71

13-9

(a) Refer to text for detail for a 3-surface enclosure problem with given temperature. (b) Refer to text for detail for a 3-surface enclosure problem with given heat flux. (c) Similar to (a) and (b). (d)Similar to (a) and (b). (e) Similar to (a) and (b).

13-10 (a)Refer to Figure 13.5 for detail.

F12 (top to bottom wall)  0.5 = F21, F1R (top to side wall) = 0.5 = F2R From equation (13.11), T1 = 923K , q = 20kW , T2  796K (b) The required heater power will be higher than (a) due to the energy absorbed by the side walls. To solve the problem, refer to Matrix Liner Equations for Case A for a three-surface enclosure problem as shown in Figure 13.7 and related equations.

72

Chapter 14 14-1

(a) Refer to Figure 14.7 (a), (b). (b) From equation (14.29), Ebg − J1 Ebg − J2 + qg = qg1 + qg 2 = 1 1 A1 g A2 g Where J1 and J 2 can be obtained from equation (14.27), J − Ebg E −J q = b1 1 = 1 + J1 − J2 1 1− 1 1 1 

q = 2

A11 Eb 2 − J2 1−  2

=

+

1

J2 − J1 1

A2g

A2 F21 (1−  g )



A22

A1F12 (1− g )

A1 g J2 − Ebg

Where A1F12 = A2 F21 , assume gray gas, g = g 14-2

T14 −  T 241−  , where F12 = 1 1 2 1 + + A F A22 1 12 A11 (b) Same as problem (14-1) (c) From Figure 14.10 and Equation (14.36), Ebg −  T 24 qg 2 = −1     A  + 1  + 1 − 2  2 g A1 1  A22  + A1F12 (1− g )  1 g  (a) q12 =

1− 

qg 2 (c)  qg 2 (b) 14-3

(a) Same as problem (14-2)(a). (b) Refer to Chapter 13.2, if T = T2 ( ) J =  T 4 + (1−  ) J dF 2 1

1

1



1

j

1j

Where  J jdF1 j =  J j ( )  K (1− j ) dAj j

Ji = function of ( ) =    q12 = 1 T1 4 − J j ( )  1− 1

73

 T g4 − J 1  T g4 − J 2 from equation (14.29) + g 1 1 A1 g A2g Where J1 , J 2 can be obtained from equation (14.27),  T14 − J 1 J 1 −  T g4 + J1 − J2 q1 = 1− 1 = 1 1

(c) q =

A11

A1 g

 T24 − J 2 J 2 −  T g4 + q2 = 1− 2 = 1 A2g

A22

A1F12 (1− g ) J2 − J1 1

A2 F21 (1− g )

Where A1F12 = A2 F21, gray gas,  g = g , constant.

14-4

Refer to problem (14-2)(c), from Figure 14.10 and equation (14.36), T 4 − T 4 g 2 qg 2 = −1 −1 1−  2  2 g  A1 1   A  + A  +  1 g + AF (1−  ) 1 12 g    2 2   4V L = 0.9  = 0.6, Pc = 0.5atm, from Figure 14.2,  g = g  0.18 As F12 = 1, A1, A2 , Tg , T2 , are given. qg 2  51.4kW

14-5

14-6

Same as problem (14-4), except use L = 0.6, Pc = 0.2atm, from Figure 14.2 to get lower value of  g = g 2 1 (a) Lq = L q dx = T −k (T )dT = 210−7 (T 2 − T 2 ) c



10−7 0



c

T1

2

2

4

2

2

1

2

( 5 −1 )10 = 120W m 0.2 For optically thin gas, from equation (14.42),  (T14 − T 24 ) 5.67 10−8 (54 −14 ) 108 = = 186.2W m2 qR = 1 1 1 1 + −1 + −1 1  2 0.1 0.1 q = q + q = 306.2W m2 qc =

c

R

(b)For optically thick gas, from equation (14.43),

74

q =  (T 4 − T 4 ) R

1

2

4 1 3 L

= 5.67 10−8 (54 −14 )108

4

1  = 78.624W m2 3 0.2 100

q = qc+ q R = 198W m2 q (b)  q (a) 14-7

(a) Refer to problem (14-6): qc = 0.024k W m2 = 24W m2 qR = 49.77W m2 for optically thin gas q = q + q = 73.77W m2 c

R

(b) qR = 11.94W m2 for optically thick gas q = q + q = 35.94W m2 c

R

14-8

Refer to problem (14-6).

14-9

From equation (14-33), q = A  ( T 4 −  T 4 ) net

s

g g

g s

Where  g = c + w − 

with   0.01

g = c +  w − 

with   0.01

From Figure 14.2, to determine  c ,  w at L = 0.9D = 0.36m = 1.15 ft Pc L = Pw L = 1.15 0.15 = 0.1725 atm-ft Tg = 1273K c

0.069, w

0.085, g

0.069 + 0.085 − 0.01 = 0.144

From equation (14.34) and (14.35), to determine w , c T T at P L s = P L s  0.6 0.1725 = 0.105 atm-ft c w Tg Tg Ts = 773K 0.45  1273    w  0.083, w = 1   0.083 = 0.104  773 0.65  1273  c  0.08, c = 1   0.08 = 0.111  773  g  0.111+ 0.104 − 0.01 = 0.205 q ' =  ( 0.4 ) 5.67 10−8 0.144 (1273) − 0.205 ( 773)  = 21.9W m2   4

14-10 Same as problem (14-4), except use

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4

L = 0.6m, Pc = 0.2atm, from Figure 14.2 to get lower value of  g , and  g = g for

gray gas assumption 14-11 Refer to problem (14-1) for detail.

14-12 Same as problem (14-11) by changing surface A1T11 to surface A2T22 .

14-13 Refer tp Figure 13.7, but with hot gray gas. (a) Follow procedure outlined in Chapter 14.2.

(1) Let equation (14.24) = equation (14.27) set i = 1 for surface 1, j = 1, 2, 3, to get a11 J1 + a12 J2 + a13 J3 = c1 set i = 2, j = 1, 2, 3, for surface 2, to get a21 J1 + a22 J2 + a23 J3 = c2 set i = 3, j = 1, 2, 3, for surface 3, to get a31 J1 + a32 J2 + a33 J3 = c3 From [A][J] = [C], solve for [J], i.e., J1, J2, J3, then from equation (14.24), to get qi, where i = 1, 2, 3.

Finally, from equation (14.29), to get qg =  (2) Alternately, from equation (14.28), apply for surface 1, 2, 3 (i = 1, 2, 3), at each surface, j = 1, 2, 3, to get [A][J]=[C], solve for [J], i.e., J1, J2, J3, then from equation (14.24), to get qi (i = 1, 2, 3), from equation (14.29), to get qg.

Note: you can give some numbers of T1, T2, T3, Tg, etc, to calculate q1, q2, q3, qg. (b) Similar to part (a).

Apply equation (14.27) for surface 1 (i = 1) and j = 1, 2, 3, where Ebg can be obtained from equation (14.29), i.e., E − J1 Ebg − J2 Ebg − J3 qg = bg + + 1 1 1 A1 g A2 g A3 g Ebg = , insert Ebg into equation (14.27), then get a11 J1 + a12 J2 + a13 J3 = c1 Do same for surface 2 (i = 2) and j = 1, 2, 3, and insert Ebg into Equation (14.27) to get a21 J1 + a22 J2 + a23 J3 = c2 Do same for surface 3 (i = 3) and j = 1, 2, 3, and insert Ebg into Equation (14.27) to get a31 J1 + a32 J2 + a33 J3 = c3 to get [J] from [A][J] = [C], then from equation (14.24) to get Ebi (i = 1, 2, 3), E = T 4 (i = 1, 2, 3), to get Ti bi

i

From equation (14.29) to get E = T 4 , to get Tg =  bg

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g

Note: you can give some numbers of q1, q2, q3, qg, etc, to calculate T1, T2, T3, Tg. (c) Combine part (a) and part (b).

Note: you can provide some numbers of T1, T2, q3, Tg, etc, to calculate q1, q2, T3, qg. (d) Similar to part (c). (e) Similar to part (c).

14-14 Refer to problem (14-13) for details, except now view factors (F12, F13, etc) are different between cylindrical furnace and cubic furnace.

14-15 (a) Refer to Figure 14.12 for special case 6. (b) If we consider an energy balance between gases and enclosure surfaces i, the heat transfer rate (energy releases) from the gases to the enclosure can be obtained from equation (14.29) for i = 1, 2, and 3. However, we still need Equations 14.24 and 14.27 or Equation 14.28 to solve Ji using the matrix method. Also refer to example 14.1. (c) Refer to Figure 14.10 for special case 4, and use equation (14.36) to obtain the heat transfer rate from the gases to the surface 1. The heat transfer rate in the (c) is higher than in (b) due to re-radiation from the top and side walls. Also refer to example 14.1.

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