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 rddz r r r r T drdz T q = −k A = −k r r T = −k rddr 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 dd 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 ddr = −k sin ddr 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: T1 − Ts q = (1) L ln ( r2 r1 ) 1 + 2k 2 r1 hi = 2r 2h 2(T 2 − T 2 ) + 2r 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 k2 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 = 2r 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 2r , 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)