化工基础作业答案

整理文档很辛苦,赏杯茶钱您下走!

免费阅读已结束,点击下载阅读编辑剩下 ...

阅读已结束,您可以下载文档离线阅读编辑

资源描述

第一章5.答:在水池水面和高位档水面间列伯努利方程:10.答:11.答:13.答:第二章8.答:12.答:24Q=*0.836*60-20=222.9/360030-10t==18.230ln10Js()Δ14.答:211212211122122=qC(t-t)=1.25*1.9*50=118.75/t-t30-10tm===18.2t30lnln10t2tm=1111+ln++1.70*0.012525-2*2.5==0.02=0.01252=118.752=1074A=Kgslrrrrrmrmlm苯苯℃παα其中;代入,得得π传热面积(按管外面积计)13.43mΦΔΔΔΔΔΔΦλΦ第三章2.答:1225*166**11*112min123Vmin0.20Y==0.25Y=0;1-0.20X=0P3.04*10*0.20===0.0561.08*101.084*10X==0.05941-Y-YL==4.21VX-X1*10*10.2q==35.7/22.4L=V*4.21*220=33.1/xxxmolskgs;()3.答:1111*121*2*22L12min*V12L12V121112OGmOG1y==0.0099101.3-1yY==0.011-y0.01X==0.01320.70Y=Y1-=0.00040.0004X==0.0005260.70X=0Y=0qY-Y==0.727qX-XqY-Y=1.5*0.727=qX-XX=0.0088Y=0.00669Y-YN==7.0YH=0.13();()ΦΔv20.45q==0.0155/29H=0.923mKmolsm4.答:VGY0VYG0V23220-42Yqh=KAqK=hAq=0.10/=150/A=r=2.54K=1.75*10/(*)KmolsmmmKmolms由πααα5.答:3V12*1*22*21*1m12OGmOG1000*10273q=**(1-0.05)=10.98/3600*22.42930.05Y==0.05261-0.05Y=0.0079X=0.0263X=0.0040X=0Y=0X=0.0175Y=0.035Y=0.0121Y-YN==3.7Y10.98H==0.7418.87*0.785H=2.74molsΔΔAA.OAA.OA'A.OARA1VA.OA11.xckt=1-xc=1.0k=0.025*60t=1hx=0.6C=2.0x=0.754.kt1kt=ln=2.40791-x1kt=lntx=70.7%1-x7.Vxt==qkc1-x第七章,代入,,得当时,由于操作条件和反应时间相同时,恒定在活塞流反应器中,在全混流反应器中,,代入,得()RVA1A.O3-1-1''RA2A1VA.OA2A2A1'3RV=0.5q=0.02x=0.68c=1.6k=0.166mkmolminVx-xt==qkc1-xxxkV=0.47m,代入,,,得()代入,,,得

1 / 10
下载文档,编辑使用

©2015-2020 m.777doc.com 三七文档.

备案号:鲁ICP备2024069028号-1 客服联系 QQ:2149211541

×
保存成功