RRA1A2B1B2C1C21012RRR12RRR12R333A3B3C310RRR
R1?R,3R2?AR2?BR2? CRA1?RB1?RC1?R?A1?B1?C135?10?3130?10?3 ?R1????2???R3?0.1307(m2?k)/W
?A1?B1?C11.531.53?A2?B2?C335?10?3130?10?3 R2????2???0.221(m2?k)/W
?A2?B2?C31.530.742 ?R??11??5.04?10?2(m2?k)/W
11112??2??R1R20.13070.22116.已知：d1?160mm,d2?170mm,?1?58W/(m?k)，?2?30mm,?2?0.093W/(m?k)
?3?40mm,?3?0.17W/(m?k),tw1?300℃，tw4?50℃
求：1）R?1,R?2,R?3; 2) ql: 3) tw2,tw3. 解：
32123tw11tw4 11
1）R?1?12??112??212??3lnd21170?ln?1.664?10?4(m?k)/W d12??58160d2?2?21170?60?ln?0.517(m?k)/W d22??0.093170d2?2?2?2?31170?60?80?ln?0.279(m?k)/W
d2?2?22??0.17170?60 R?2?ln R?3?ln ?R?1?R?3?R?2 2) ql??t?t300?50???314.1W/m
?R?iR?2?R?30.517?0.279tw1?tw2 得 R?1 3）由 ql? tw2?tw1?qlR?1?300?314.1?1.664?10?4?299.95℃ 同理：
tw3?tw4?qlR?3?50?314.1?0.279?137.63℃ 17.已知：?1??2,?2? 求：
1?1,d2m?2d1m 2ql 'qld?2?1d?2?1?2?21 R??ln0?ln02??1d02??2d0?2?11d?2?1d?2?1?2?21 R??ln0?ln02??2d02??1d0?2?1' 解：忽略管壁热阻
d2mtw1d1m22d0tw3121ql?
?t'?t,ql?' （管内外壁温tw1,tw3不变） R?R?d0?2?1d0?2?1?2?21?ln'qlR?2??2d02??1d0?2?1?'??
d0?2?1d0?2?1?2?211qlR?ln?ln2??1d02??2d0?2?11ln 12
d0?2?11d0?4?1?lnd02d0?2?1 ?
d0?4?11d0?2?1ln?ln2d0d0?2?1ln 由题意知： d1m? d2m1[d0?(d0?2?1)]?d0??1 21?[d1m?(d1m?2?2)]?d0?3?1 2 即：d2m?2d1m?d0?3?1?2(d0??1)?d0??1 （代入上式）
15ln3?lnqR23?1.277 ?l'???qlR?1ln3?ln523' 即： ql'?0.78q3l
ql'?ql ???21.7%即热损失比原来减小21.7%。
ql18.已知：d?1mm,Rl?2.22?10?3?/m,??0.15W/(m?k)
tw1max?65℃，tw2?40℃，??0.5mm,
求：Imax
2 解： ql?ImaRxl?tw1tw2tw1ma?xtw2
1d?2?ln2??d??????65?40??123.7(A) ????32.22?101?2?0.5???ln????2??0.151?1212ql ?Imax??t?t??w1maxw2?Rllnd?2?d?2??19.已知：d1?85mm,d2?100mm,?1?40W/(m?k),tw1?180℃ ?2?0.053W/(m?k),tw3?40℃，ql?52.3W/m 求：?2 解： ql?Rλ1tw1tw2lnd2d1lnRλ2tw3d2+2δd2?t?R?1?R?2tw1?tw3
d2d2?2?211ln?ln2??1d12??2d22πλ12πλ2 13
整理得：
2180?401100)2??0.053(?ln)d22??2(ql?2??1lnd110052.32??4085 ?2?(e?1)??(e?1)?72mm
22?t1d 或：R?2?R?1,故有 ql??t?R?2tw1?tw3
d2?2?21ln2??2d2d ? ?2?2(e22??2?tql ?1)?7m2m20.已知：d1?0.35mm,?1?3mm,?2?30mm,r?199.6kJ/kg,tw1?(?273.15?77.4)℃
tw3?25℃，?2?0.03W/(m?k),?1?16.3W/(m?k),??1h
求：m
t?t 解： Q?w3w1
R?F1?R?F2
tw1RλF1tw21-1RλF2tw31-1(r1r2)4πλ2(r2r3)4πλ1tw3?tw1
111111(?)?(?)11114??1d(d1?2?1)4??2(d1?2?1)(d1?2?1?2?2)122222?(25?273.15?77.4) ?
111111?(?)??(?)16.30.350.3560.030.3560.416 ?102.7W
? 或：
R?F1?R?F2,故有： Q?tw3?tw12?(25?273.15?77.4)?0.03??102.7W
11111(?)(?)4??2r2r30.3560.416Q?102.7?3.6??1.85kg/h r199.6 m?21.略 22.略
ld2??m2??0,??t?tf2dx23. 解： x?0,???1?t1?tf
x?l,???2?t2?tf
t1tf,h12ft ,t >tt214
解微分方程可得其通解： ??c?mx1emx?c2e 由此得温度分布（略）
24.已知：l?25mm,??3mm,??140W/(m?k),h?75W/(m2?k),t0?80℃
tf?30℃，qx?l?0 求：?,ql 解： ml?hU?Al?h2L??L?l2h???l0.0252?75140??31?30?0.4 725 m?18.9 ???ch[m(l?x)]ch[0.4725?0ch(ml)?(80?30)?18.9x]ch(0.4725)
?44.91ch(0.4725?18.9x) ?t?30?44.91ch(0.4725?18.9x)
qQl?L?hUmL?(ml)?2h0thm?0th(ml) ?2?7518.9?(80?30)th(0.4725)?174.7W/m 25.已知：??15mm,l?20mm,??48.5W/(m?k),tl?84℃，t0?40℃ h?20W/(m2?k)
求：?t 解： ml?hU?Al?h?d??d?l?h??l?0.12?2048.5?1.5?10?3?2
?0t0?tfl??ch(ml)?t?ch(ml)
l?tf ?ttlch(ml)?t0f?ch(ml)?1?84ch(2)?40ch(2)?1?99.93℃
ch(2)?3.7622
?t?tf?tl99.93?84t?100%?99.93?100%?15.9%
f26.已知：??0.8mm,l?160mm,t0?60℃，??16.3W/(m?k)，其他条件同25题
15