化工原理习题解答 下载本文

??1076.kg/m3, ??1985.?10?5Pa?s

ms2?V?2?0.785d2u2?22n200?0.785?0.0332?12?1.076??1.104kg/s 22Q?ms2cp2?t2?t1??1.104?1.005??86?26??66.57kW

Re?du???0.033?12?1076.1985.?10?5?215.?104?104 (湍流)

1005.?103?1985.?10?5Pr???0697.

?0.02861cp???2?0.023?dRe0.8Pr0.4?0.023?0.02861?2.15?1040.033??0.8?0.6970.4?50.46W/(m2?K)

d1110.038 ??1?4?K1?1?2d21050.46?0.033解得:K1?43.63W/(m2?K)

?tm94?34??59℃

94ln34??120

26 94 120 86 34

Q66.57?103A1???25.86m2

K1?tm43.63?59?A1???d1?l?n

?l?A125.86??1.084m

??d1?n3.14?0.038?200解法2

??1???2 ?K2??2?50.46W/(m2?K)

Q66.57?103A2???22.36m2

K2?tm50.46?59?l?A222.36??1.079m

??d2??n3.14?0.033?200可见这种近似是允许的。

??0.05m (2) 改为?54?2mm列管,d2n??08.n?08.?200?160

l??108.m

49

? 令空气出口温度为t2热量衡算:

??26)t2??26Q?ms2cp2(t2 ??Qms2cp2(86?26)60?A2??tm??2?A2??tm?Q?K2 ??QK2A2?tm?2A2?tm (1)

速率方程: (2)

式中 ?tm?59℃

?tm??t2??60 94ln120?t2?120

26 94

120 t2?

120?t2?

???d2??l?n?d2??n?0.05?160A2????1.212 A2??d2?l?nd2?n0.033?200?d2?Re???2??????Re??2d21.80.8d?2?d20.8???d2???d???2?0.8?u??????u?1.80.8d?2?d20.8???d2???d???2?0.8??d?2n?2????d????n??????2??

0.8?d2??n????d?????n???2????0.033??1????????0.05??0.8??0.566将以上各值代入(2)后再与(1)式联立

t2??26?0566.?1212.?60 94ln?59120?t2???26??t2解得:t2??73℃

27、试计算一外径为50mm,长为10m的氧化钢管,其外壁温度为250℃时的辐射热损失。若将此管附设在:

(1)与管径相比很大的车间内,车间内为石灰粉刷的壁面,壁面温度为27℃,壁面黑度为0.91; (2)截面为200mm?200mm的红砖砌的通道,通道壁温为20℃。 解:由表3-8查的氧化钢管黑度为? 1=0.8,石灰粉刷壁面的黑度? 2=0.15 (1)由于炉门被极大的四壁包围,由表3-9知

?=1,A=A1=3.14?0.05?10=1.57m2,

C1-2=? 1C0=0. 8?5.669=4.535W/(m2?K4)

所以 Q1?2?C1?2?1?2A?(T?T14?)?(2)4?

100??10050