华东理工大学化工原理习题答案(陈敏恒第三版).pdf

化工原理习题答案（上册） 第一章 流体流动 5 2 1 P （绝）= 1.28 ×10 N/m A 4 2 PA （表）= 2.66 ×10 N/m 2 W = 6.15 吨 4 3 F = 1.42 ×10 N 4 P = 7.77 ×10 Pa 4 H = 0.39m 5 △P = 2041×105N/m2 5 6 P = 1.028×10 Pa △h = 0.157m 7 P （绝）= 18kPa H = 8.36m 8 H = R PA PB 9 略 10 P = Paexp[-Mgh/RT] 2 11 u = 1 1.0m/s ; G = 266.7kg/m s qm = 2.28kg/s 12 R = 340mm 3 13 qv = 2284m /h 14 τ= 1463s 15 Hf = 0.26J/N 16 会汽化 2(P −P ) 17 u A 1 2 1 2 ρ(A 2 −A 2 ) 1 2 2(P −P ) u A 1 2 2 1 ρ(A 2 −A 2 ) 1 2 3 18 F = 4.02 ×10 N 19 略 20 u2 = 3.62m/s ; R = 0.41m 21 F = 151N -6 2 22 v = 5.5 ×10 m /s − u 23 =0.817 a = 1.06 u max 24 略 25 P （真）= 95kPa ; P （真）变大 26 Z = 12.4m 27 P （表）= 3.00 ×105N/m2 3 28 q = 3.39m /h P 变小 P 变大 v 1 2 3 29 qv = 1.81m /h 30 H = 43.8m 31 τ= 2104s 32 He = 38.1J/N 3 3 33 q =0.052m /s=186m /h v 3 3 34 qv1 = 9.7m /h ; qv2 = 4.31m /h 3 , 3 qv3 = 5.39m /h ; q v3 = 5.39m /h 35 qvB/qvC = 1.31 ; qvB/qvC =1.05 ;能量损失 5 36 P1 （绝）=5.35 ×10 Pa − 37 u = 13.0m/s 38 qv = 7.9m3/h 39 qVCO2 （上限）=3248l/h