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精品课程附件, 兰州大学
2. (16分)解:
7000???1000?(85?20)?126389W (2分)
3600?Tm?(108?20)?(108?85)= 48.44oC (2分)
108?20ln108?85u??7000?30.95kg?m2?s?1 (1分)
3600??4?0.022?200R0.02?30.95e?du???1.98?10?5?31263 ?.8空?0.023?dR0ep0.4r ?0.023?2.85?10?20.02?31260.83?0.70.4 ?112.1W?m?2?K?1 1d?d1dK?2??2?d??2?1 1d1m?2?1d1?2 ?25112.1?20?11?104?0.0113
K = 88.5W?m-2?K-1 A需??K?T?126398.5?48.44?29.48m2 m88A实 = n?d外l = 200 ? ? ? 0.025 ? 2 = 31.42m2 A实A?1.066? 1.1 ?能完成上述传热任务。 需
3. (14分)解:(1)
qn,c1?YA,20.02?0.004q?YA,n,BX?A,1?XA,20.008?0?2 N?1?mqn,BYA,1?mXA,2mqn,B?OG1?mqln??(1?)??q? n,Bn,CYA,2?mXA,2qn,C??qn,C5 (共7页)
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(2分)
(2分) (2分) ( ( ( ((( ( (
精品课程附件, 兰州大学
?11.50.02?01.5??ln?(1?)??2.77 (2分) 1.5?20.004?02??1?2或 ?YA,1 = YA,1 – mXA,1 = 0.02 – 1.5 ? 0.008 = 8?10-3 (1分) ?YA,2 = YA,2 – mXA,2 = 0.004 (1分) ?YA,m?YA,0.02?0.0041?YA,2?3 (2分) ??5.77?10?3lnYA,1?mXA,18?10YlnA,2?mXA,20.004?0
NYA,1?YA,2OG??Y?0.02?0.0045.77?10?3?2.77 A,mH10OG?2.77?3.61
(2) 其他条件不变,YA,2变,则NOG不变 N?11?mqln??(1?mqn,B)YA,1?mXA,2mqn,B?OG,B?q/?? nn,CYA?mXqn,C?q?,2A,2?n,C ?1ln?1?1.5??(1?1.52)0.02?01.5?0.002?0?2???4.66 2 H/ = HOG? NOG = 3.61? 4.66 = 16.82m ?H = 16.82 – 10 = 6.82 m
4. (19分)解:(1)0.75?RR?1 R = 3 0.2?xdR?1?xd3?1 xd = 0.8
qn,D?xw0.35?xwq?xfn,Fxd?x?w0.8?x?0.4 xw = 0.05 w ?
qn,Dq?40% ? qn,D = 0.4 ? 100 = 40kmol?h-1 n,F qn,W = qn,F - qn,D = 100 – 40 = 60 kmol?h-1 6 (共7页)
(2分) (2分) (2分)
(2分) (1分) (1分) 2分) 2分) 2分)
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((( ( (精品课程附件, 兰州大学
qn,V = ( R + 1) qn,D = (3 + 1 ) ? 40 = 160 kmol?h-1 (1分) ? = 0
qn,V/ = qn,V - qn,F = 160 – 100 = 60 kmol?h-1 (1分) qn,L/ = qn,L = qn,V - qn,D = 160 – 40 = 120 kmol?h-1 (1分) 提馏段操作线方程: ym?1?qn,Lqn,V//xm?qn,Wqn,V/xW = 2xm – 0.05 (2分)
*(2) y1?2.5x12.5?0.7 ??0.8573 (2分) 1?(2.5?1)?0.71?(2.5?1)x1 y2 = 0.75 ? 0.7 + 0.2 = 0.725 (1分) y1 = 0.8 (1分) Em,V,1?
二、(11分)选择题(每小题各一分)
1. (B) 2. (C) 3.(A) 4.(C) 5.(B) 6.(B) 7.(C) (B) (C) 8.(A) 9.(B)
三、(20分)填空题(每个空各一分)
1. ? ,A ; 2. 0 2. 饱和蒸汽,水 3. 降低,增大,增大,增大; 4. 0.045,0.0667; 5. 6 5779; 6. 0.35; 7. 吸收 8. 0.33 kg水/(kg绝干),0.27 kg水/(kg绝干),0.13 kg水/(kg绝干); 9.
y1?y20.8?0.725??0.566 9 (2分) *0.857?30.725y1?y22V ; 10. 减小; 11.
7 (共7页)
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