第四章 传热(习题及解答2007版)

1/K'=1/α1+1/α2+R 两式相减得 R=1/K'-1/K=1/256.2-1/504=1.92×10-3 m2·℃/W 4.11 解: (1)求逆流操作平均温差 △tm逆=(△t1-△t2)/ln(△t1/△t2)=[(243-162)-(155-128)]/ln[(243-162)/(155-128)]=49 ℃ (2)求并流操作时平均温差 T1=243 ℃ → T2'=? t1=128 ℃ → t2'=? 由热量衡算:

WhCph(T1-T'2) = KA[(T1-T'2)+(t'2-t1)]/ln[(T1-t1)/(T'2-t'2)] (1) 又: WhCph(T1-T'2) = WcCpc(t'2-t1) (2) 将(t'2-t1)从(2)解出代入(1)式,等式两边消去(T1-T'2) 得到下式:

ln[(T1-t1)/(T'2-t'2)]=[KA/(WhCph)][1+WhCph/(WcCpc)] (3) 对逆流操作,同理可推得下式:

ln[(T1-t2)/(T2-t1)]=[KA/(WhCph)][1-WhCph/(WcCpc)] (4) (3)/(4)得到:

ln[(T1-t1)/(T'2-t'2)]/ln[(T1-t2)/(T2-t1)] = [1+(WhCph)/(WcCpc)]/[1-(WhCph)/(WcCpc)] (5) (WhCph)/(WcCpc)=(t2-t1)/(T1-T2) = (162-128)/(243-155) = 0.386

ln[(T1-t2)/(T2-t1)]=ln[(243-162)/(155-128)] = 1.1 代入(5),得到: T'2-t'2=9.6 ℃ (6) 并流工况热衡算: WhCph(T1-T'2) = WcCpc(t2'-t1) 代入数值得到: t'2+0.386T'2 =221.8 ℃ (7) (6)、(7)联立求解得: T'2=167℃, t'2=157.4℃ 则并流平均温差: △tm并=[(T1- t1)-(T'2-t'2)]/ln[(T1-t1)/(T'2-t'2)]=[(243-128)-(167-157.4)]/ln(115/9.6)=42.5 ℃ 4.12 解: (1)热负荷 Q=WhCph(T1-T2)=(2000/3600)×1.84×103×(80-40)=4.09×104 W △t1=80-30=50 ℃, △t2=40-20=20℃ △tm=(△t1-△t2)/ln(△t1/△t2)=(50-20)/ln(50/20)=32.7 ℃ K=Q/(A△tm)=4.09×104/(2.8×32.7)=446.7 W/(m2·℃) (2)Q=WcCpc(t2-t1); Wc=Q/[Cpc(t2-t1)]= 4.09×104/[4.187×103×(30-20)]=0.977 kg/s=3517 kg/h 4.13 解: (1)主要仪器:水银温度计,热电偶温度计,流量计。 (2)温度:苯蒸汽温度T,[℃];水进、出口温度t1,t2,[℃];内管内、外表面平均温度,即在内管内侧与外侧分别测若干处的壁温,然后取其平均值twm及Twm,[℃] 。 流量:水量W1,[kg/h]; 尺寸:内管内、外直径及有效长度,[m]。 (3)计算步骤:

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Q=WcCpc(t2-t1) [W];Ao=πdoL [m2];Ai=πdiL [m2] α苯=Q/[Ao(T-Twm)] [W/(m2·℃)]; α水=Q/[Ai(t wm- tm)] [W/(m2·℃)]

4.14 解: (1)求水与管壁之间对流传热系数 tm 1= (20+40)/2=30 ℃ Re1= duρ1/μ1=0.02×1×995.7/(80.07×10-5)=2.487×104>10000 Pr1= Cp1μ1/λ1=4.174×103×80.07×10-5/0.617=5.42 (120>pr>0.7) L/d= 3/0.02=150>60 ∵ 流体被加热 ∴ α1= 0.023(λ1/d)Re10.8Pr10.4=0.023×(0.617/0.02)×3285.1×1.97=4592 W/(m2·℃) (2)求空气与管壁之间对流传热系数 tm2=(20+40)/2=30 ℃ Re2=duρ2/μ2=0.02×10×1.165/(1.86×10-5)=12527>10000 Pr2=Cp2μ2/λ2=1.005×103×1.86×10-5/0.02675=0.7 α2=0.023(λ2/d)Re20.8Pr20.4=0.023×(0.02675/0.02)×1898×0.867=50.6 W/(m2·℃) 4.15 解: (1)求管内水的对流传热系数 Re=duρ/μ=0.02×1.2×992.2/(0.656×10-3)=36300>10000 Pr =Cpμ/λ=4.174×103×0.656×10-3/0.634=4.3 (120>pr>0.7) L/d=2/0.02=100>60 Nu=0.023Re0.8Pr0.4=0.023×(36300)0.8×(4.3)0.4=183.2 α=(λ/d)Nu=(0.634/0.02)×183.2=5808 W/(m2·℃) (2)求总管数改为50根时管内对流传热系数 因为传热面积不变

nπdL=n'πdL',L'=(n/n')L=(60/50)×2=2.4 m 又:n0.785d2u=n'0.785d2u' u'=(n/n')u=(60/50)×1.2=1.44 m/s 湍流传热时:α'/α =(u'/u)0.8 ∴ α'=α(1.44/1.2)0.8 = 6763 W/(m2·℃) 4.16 解: 新的管内对流传热系数为α'1 α'1/α1 = (u'/u)0.8/ (d'i/di)0.2 =(di/d'i)2×0.8 (di/d'i)0.2=(di/d'i)1.8=(20/27)1.8=0.583 α'1 = 0.583α1 4.17 解: Re =duρ/μ=0.077×0.5×850/(26×10-3)=1259 <2300,为层流 Pr =Cpμ/λ=2×103×26×10-3/0.13=400 (6700>pr>0.6), RePrd/L =1259×500×0.077/6=6463>10 故选用公式:

Nu =1.86Re1/3Pr1/3(L/d)1/3(μ/μw)0.14

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α =1.86(λ/d)Re1/3Pr1/3(L/d)1/3(μ/μw)0.14 =1.86×0.13/0.077(1259)1/3(400)1/3(0.077/6)1/3(26/3)0.14 =79.2 W/(m2·℃) 为了考虑Gr的影响,计算Gr如下:

Gr=gd3ρ2β△t/μ2

=9.81×(0.077)3(850)2×0.001×(150-40)/[(26×10-3)2] =5.265×105 >25000 校正因数f为: f =0.8×(1+0.015Gr1/3)=0.8×[1+0.015×(5.265×105)1/3]=1.769

∴ α=79.2×1.769=140 W/(m2·℃) 4.18解: (1)求环隙流体的对流传热系数: 套管环隙的当量直径为: de=d2-d1=0.045-0.025=0.02 m Re=deuρ/μ=0.02×0.7×1836/(6.4×10-3)=4020,(2300

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α'=fα=0.805×1078 = 868 W/(m·℃) (2)对流传热的热通量:q = α'△t = 868×(70-60) = 8700 W/m2 4.19 解:(此题为流体在弯管道中强制流动的问题,先计算Re断定流型,以便确定计算公式。) 四组蛇管并联的横截面积:S=4×(π/4)d2=4×0.785×0.0382=0.00454 m2 管内流速:u=V/S=2.7/(3600×0.00454)=0.165 m/s

计算Re,判断流型: Re=duρ/μ=0.038×0.165×1200/(2.2×10-3)=3420 Re在2300与10000之间,故流型属过渡流,先按湍流计算,再用过渡流校正,然后再用弯管公式校正。 tm=(32+8)/2=20℃ Cp=0.9×4.183×103=3764.7 J/(kg·℃),λ=0.9×0.5989 =0.539 W/(m·℃) ∴ Pr=Cpμ/λ=3764.7×2.2×10-3/0.539=15.4 计算对流传热系数,铜氨液冷却。 Nu=0.023Re0.8Pr0.3

α=0.023(λ/d) Re0.8Pr0.3 =0.023×(0.539 /0.038)(3420)0.8(15.4)0.3=497.7 W/(m2·℃) 过渡流校正: α'=α[1-6×105/(Re1.8)]= 497.7×[1-6×105/(3420)1.8]=367.7 W/(m2·℃) 弯管校正:

α''=α'(1+1.77d/R)=367.7(1+1.77×0.038/0.285)=454.5 W/(m2·℃) 4.20 解:

A=(π/4)(d22-d12)=(π/4)(0.0512-0.0382)=9.08×10-4 m2 G=ms/A=2730/(3600×9.08×10-4)=835 kg/(m2·s) de=d2-d1=0.51-0.038=0.013 m

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Re=deG/μ=0.013×835/(0.38×10-3)=2.86×104, Re0.8=3674 Pr=Cpμ/λ=1840×0.38×10-3/0.128=5.46, Pr0.3=1.66 ∴ α=0.023(λ/de)Re0.8Pr0.3=0.023×(0.128/0.013)×3674×1.66=1381 W/(m2·℃) 4.21解:(此题为流体在非圆形管中的对流传热问题,故先求出当量直径,然后用圆形直管公式作近似计算。)

de=4[(π/4)D2-86(π/4)d2]/(πD+86πd)=(D2-86d2)/(D+86d) =(4002-86×252)/(400+86×25)=41.67 mm tm=(120+30)/2=75 ℃ 甲烷的密度:ρ=PM/(RT) P=1.013×102 kPa M=16 kg/kmol T=273+75=348 K 则 ρ=1.013×102×16/(8.314×348)=0.56 kg/m2

Re=deuρ/μ=0.04167×10×0.56/(0.018×10-3)=12964>10000 Pr=Cpμ/λ=2.43×103×0.018×10-3/0.0399=1.1 (120>pr>0.7) ∴ α=0.023(λ/de)Re0.8Pr0.3=0.023×(0.0399/0.04167)(12964)0.8(1.1)0.3 =0.023×(0.0399/0.04167)×1950.7×1.03=44.2 W/(m2·℃) 4.22 解:(此题为列管式换热器管外强制对流问题) 当量直径 de=4[(1/2)31/2t2-(π/4)d02]/(πd0) =4×[(1/2)×31/2×512-(π/4)×382]/(3.14×38)=37.4 mm 管外流体流过的截面积 S=hD(1-d0/t)=1.25×2.8(1-38/51)=1.035 m2 管外流体的流速 u=V/S=4×104/(3600×1.035)=10.7 m/s Re=deuρ/μ=0.0374×10.7×0.845/0.0239×10-3=14150>10000 Pr=Cpμ/λ=1.014×103×0.0239×10-3/0.03524=0.7 α=0.36(λ/de)Re0.55Pr1/3(μ/μw)0.14 近似取(μ/μw)0.14=1 则 α=0.36×(0.03524/0.0374)×(14150)0.35×(0.7)0.33=58 W/(m2·℃) 考虑到部分流体在挡板与壳体之间隙短路,取实际对流传热系数为计算值的0.8倍。 α=0.8×58=46.4 W/(m2·℃) 4.23 解:(由于海水在槽内流速很小,海水和冷却排管的传热可按大空间自然对流处理。)

tm= (56+42.5)/2=49.3 ℃ Gr=βg△td3ρ2/μ2 =4.45×10-4×9.81×(56-42.5)×0.1143×988.42/(0.5565×10-3)2 =2.75×108 Pr=Cpμ/λ=4.174×103×0.5565×10-3/0.6463=3.594 GrPr=2.75×108×3.594=9.88×108

对水平圆柱体,GrPr在104~109范围内,C=0.53,n=1/4 Nu=0.53(GrPr)1/4 α=0.53(λ/d)(GrPr)1/4=0.53×(0.6463/0.114)×(9.88×108)1/4=533 W/(m2·℃) 4.24 解:

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