2009年湖北工业大学化工原理(A)考研试题及答案(6)

五、解：(1) Q=qmCp =15×103×1.76×103×(50-20)/3600=2.20×105W

Δtm=[(130-20)+(130-50)]/2=95℃

以外表面为基准的传热系数：

Ko=(1/αo+bdo/λdm + do/αidi)-1

–1 =[1/10000 +0.00250.025/(450.0225) +0.025/(700×0.02)]

=513 W/（m2.K）

(2)qv=qm/ρ=15×103/(3600×858)=0.0049m3/s

A= qv/u =0.0049/0.5 =9.8×10-3 m2

又A= nπd2/4

∴ n=9.8×10-3/(0.785×0.022) =31 (根)

由于Ao=Q/(KΔtm) =2.2×105/(513×95) =4.5m2

Ao=nπdoL

∴L=4.5/(31×3.14×0.025) =1.85 m 取L=2m

(3)因为管程流速小，α也小，故应强化管程传热，如改为双管程。

六、解：(1) y1 = (100×10-3/32)/(1/22.4) = 0.07

Y1= y1/(1-y1) =0.07/(1-0.07) = 0.0753

Y2= Y1(1-η) = 0.0753(1-0.98) = 0.001505

X1e = Y1/m = 0.0753/1.15 = 0.0655

X1= 0.67X1e= 0.0439

G = 1000/22.4 = 44.64kmol.h-1

GB=44.64(1-0.07) = 41.52 kmol. h-1

LS=GB(Y1-Y2)/X1=41.52(0.0753-0.001505)/0.0439

=69.8 kmol.h-1

(2) VS= (π/4)D2u = 1000/3600 = 0.278 m3/s

D= (4VS/(πu))0.5= (4×0.278/(π×0.5))0.5= 0.841 m

(3) LS/GB=(Y1-Y2)/X1=(0.0753-0.001505)/0.0439 =1.681

S=m/(LS/GB) = 1.15/1.681 = 0.684

NOG =1/(1-S)Ln[(1-S)Y1/Y2+S] = 8.87

HOG = GB/Kya

=41.52/(0.5×180×(π/4)×0.8412) = 0.831m

∴H =HOG×NOG=7.37 m