作者monbydick (monbydick)
站内ChemEng
标题Re: [问题] 单元操作
时间Sun Oct 26 14:48:23 2008
※ 引述《ivy7132000 (ivy)》之铭言：
: 研究了很久 这题一直不知该怎麽写 小妹愚笨 麻烦大家帮忙
: A vertical cylindrical settler-separator is to be designed for separating a
: mixture flowing at 20.0 m3/h and containing equal volumes of a light
: petroleum liquid(密度b=875 kg/m3) and a dilute solution of wash water(密度
: a=1050
: kg/m3).
: Laboratory experiments indicate a settling time of 15 min is needed to
: adequately separate the two phases.
: For design purposes use a 25-min settling time and calculate the size of the
: vessel needed, the liquid levels of the light and heavy liquids in the
: vessel, and the height hA2 of the heavy-liquid overflow.
: Assume that the ends of the vessel are approximately flat, that the vessel
: diameter equals its height, and that one-third of the volume is vapor space
: vented to the atmosphere.
: 题目要求hA2的高度 有给答案 hA2=1.537m
: 是要先算质量流率吗??? 还是要算什麽??? 麻烦各为指点了 感激不尽!!
课本已经导出hA1跟密度B 密度A及hT(hA2+hB)的关系式
因此经过移项整理後 可以得到hA2跟密度B 密度A及hT(hA1+hB)的关系式
接着可以假设总体积流率 QT = QA + QB = 10 m^3/h + 10 m^3/h
= 1/6 m^3/min + 1/6 m^3/min
setlling time =25 min
=> VT = 25/3 m^3 = 25/6 m^3 + 25/6 m^3
然後依题目的假设
容器的直径 = 容器的高度 = X 因此容器的体积 = Pi/4X^3
由题目的叙述知道 Pi/4X^3 = VT + 1/3(Pi/4X^3)
由上式可解出 X = 2.5154 而由题意可以知道hA1 = hB = 1/3 X
接着带入hA2跟密度B 密度A及hT(hA1+hB)的关系式即可得到
hA2 = 1.537 m
--
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1F：推 ivy7132000:不好意思 可以请问那个X怎麽算的吗?? 10/27 20:48
2F：→ ivy7132000:我把X代入到Pi/4X^3 = VT + 1/3(Pi/4X^3) 里面 10/27 20:49
3F：→ ivy7132000:左右两边也无法相等 10/27 20:49
4F：推 ivy7132000:谢谢你唷 我自己研究出来了 那个应该是Pi*X^3/4 10/27 23:14
5F：→ ivy7132000:这样算出来答案就对了 谢谢你帮我 感激不尽^^ 10/27 23:14