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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

1-4

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

(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

, ,

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

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 

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.

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)

(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   

(

)

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

(

 = −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

(

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

 

2-10

−1

− 12

(

I1 2mx

I1 2ml 2

(a) Calculate Re =

) , q = −k  d d

)

dx x=l

V L

= , Pr = 

Cp



= , get

 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

= , 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)

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

 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):

 (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

(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

(2)

 = e−mLemx + emLe− mx = e

m(L− x)

+ e−m(L− x) 

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