求：?t 解： ml?h??l?160?2016.3?0.8?10?3?6.27
ttlch(ml)?t0f?ch(ml)?1?84ch(6.27)?60ch(6.27)?1?84.09℃
ch(6.27)?264.24
?t?tf?tlt?100%?84.09?84f84.09?100%?0.11%
27.已知：??3mm,l?16mm (1)??14W0m/?(kh)?,W802m/?( k ) (2)??40W/(m?k),h?125W/(m2?k) 求：?f 解：（1）ml?hU?Al?h2L?L?l?2h??l?16?10?3?2?80140?3?10?3?0.312 ?th(ml)f?ml?th(0.312)0.312?0.97 （2）ml?hUl?h2LL?l?2h??l?16?10?32?125?A??40?3?10?3?0.73 ?th(ml)thf?ml?(0.73)0.73?0.853 28.已知：d1?77mm,d2?140mm,??4mm,P?25mm,??50W/(m?k) h?60W/(m2?k),t0?320℃，tf?75℃ 求：ql 解： l?12(d2?d1)?31.5 lc?l??2?33.5
r2c?r1?lc?72
f??(r2c?r1)?4?10?3?(72?38.5)?10?3?1.34?10?4m2
16
P
lc?32?2h????33.5?10??f?123?32?2?60???0.821 ??4??50?1.34?10?12
r2c72??2.15 r133.5 查图得： ?f?0.78 每片肋片的散热量为Q1 Q1??fQ0??fhF(t0?tf) ?2?(r2c?r?fh(t0?tf) 1) ?2?(72?38.5)?10?0.78?60?(320?75)?266.7W 每米肋片管的散热量为：
ql?nQ1?(n?1)Q2 n? ?41?266.7?40?1.48?11kW Q2为两肋片间的表面的散热量 Q2??d1P(t0?tf)
???77?10?25?10?(320?75)?1.48W 29.略
30.已知：l1?l2?3?2.2m,??0.3m,??0.56W/(m?k),tw1?0℃，tw2?30℃
求：ql
2?3?322?6221000?1?41片/米 253?L?10L
??0.3Al?L2.2?L??7.33L S2?2?2??0.3解： S1?A1?l1?L?l2l117
1? 5Q(2S1?2S2?4S3)??t? ql? , ?t?tw1?tw2 LL S3?0.54L l1,l2? ?(2?10?2?7.33?4?0.54)?0.56?(30?0) ?618.6W/m
31.已知：d?165mm,tw1?90℃，H?1.5m,??1.05W/(m?k),tw2?6℃ h?20W/(m2?k) 求：ql
λtw2H 解： l?r,H?3r
2?l 2Hln()rQs??t2?? ql????t
2Hllln()r ∴ s? ?tw1d2??1.05?(90?6)?154.2W/m
2?1.5??ln???0.165/2?32.已知：l1?l2?0.52?0.52m2,H?0.42m,??0.023W/(m?k),tw1?30℃ tw2??14℃, Q?34W 求：? 解： S1?l2l1l1?l2?,S2?l1?H?,l1?l 2
H S3?0.54H,S4?0.54l1 Q?(S1?4S2?4S3?4S4)??t ??Q底=0l1l2?4l1HQ?4S3?4S4??t?2?0.52?0.52?4?0.52?0.4234?4?0.54?0.42?4?0.54?0.520.023?(30?14)
?3.62?10m?36.2mm
33.已知：??5mm,??2.54?m,P?2MPa,?t?80℃，??180W/(m?k)
18
求：?tc
解：由??2.54?m,P?2MPa，查表得，Rc?0.88?10?4(m2?k)/W Q??t???Rc??? ?t?t3?t1
t1t1At2Bt3xt1t2At2BRct3?t 再由 Q?c，?tc?t2A?t2B
Rc得
?tc?2??Rc?Rc?t?0.8?810?80?4℃9 ?35?1042??0.8?8?10180?4λλ
第三章 非稳态导热
1.略 2.略 3.略 4.略
5.已知：d?0.15mm,cp?420J/(kg?k),??8400kg/m,h1?58W/(m?k) h2?126W/(m2?k) 求：?01,?02
324?d??d??cp????cp???cpV3?2??2??8400?420?0.15?10?3?1.52(s) 解：?01???2h1F3h12?3?58?d?h14????2?3 同理：?02?6.略
?cp??2?d????8400?420?0.15?10?3?0.7(s) 3h22?3?12637.已知：d?0.5mm,??8930kg/m,cp?400J/(kg?k),t0?25℃，tf?120℃
h?95W/(m2?k), 求：?,t 解：由
??1%,??22W/(m?k)（康铜） ?0?t?tf??1% ?0t0?tf19
1t0(?tf?)? t?tf?0.0?h1?20?0.01?(25?120℃) 119.05RVh95?0.5?10?313F ?Biv????3.6?10?4?0.1M??0.1
??2?3?223 故满足集总参数法的求解条件，有：
??e?BiFo ?0VV1?38930?400??0.5?10?cpV?3ln??ln(1?10?2)?14.43s ????hF?095?28.已知：??3mm,F?1?1m,h?39W/(m?k),??48.5W/(m?k)，t0?300℃，
22tf?20℃，a?12.7?10?6m2/s,t?50℃
求：?
3330??1?012 解： ?BiV?2??0.98?10?3?0.1M??0.1
?48.53 ? 满足集总参数法的求解条件，故有：
h????cV?e ?0p?hF ?????cpVhFln??V???ln ?0haF?0348.5?1??10?350?202 ??ln?328s ?639?12.7?10?1300?209.略
10.已知：t0?80℃，d?20mm,tf?20℃，u?12m/s,??5min,t?34℃ ??8954kg/m,cp?383.1J/(kg?k),??386W/(m?k) 求：h
解：假设可使用集总参数法，故有：
3??e?0?hF??cpV
120895?438?3.1???cpV?22ln?? ?h???F?05?60?31034?20ln?83.2W/(m2?k) 80?20 20