作者monbydick (monbydick)
站內ChemEng
標題Re: [問題] 單元操作
時間Sun Oct 26 14:48:23 2008
※ 引述《ivy7132000 (ivy)》之銘言：
: 研究了很久 這題一直不知該怎麼寫 小妹愚笨 麻煩大家幫忙
: 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.
: 題目要求hA2的高度 有給答案 hA2=1.537m
: 是要先算質量流率嗎??? 還是要算什麼??? 麻煩各為指點了 感激不盡!!
課本已經導出hA1跟密度B 密度A及hT(hA2+hB)的關係式
因此經過移項整理後 可以得到hA2跟密度B 密度A及hT(hA1+hB)的關係式
接著可以假設總體積流率 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
然後依題目的假設
容器的直徑 = 容器的高度 = X 因此容器的體積 = Pi/4X^3
由題目的敘述知道 Pi/4X^3 = VT + 1/3(Pi/4X^3)
由上式可解出 X = 2.5154 而由題意可以知道hA1 = hB = 1/3 X
接著帶入hA2跟密度B 密度A及hT(hA1+hB)的關係式即可得到
hA2 = 1.537 m
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1F：推 ivy7132000:不好意思 可以請問那個X怎麼算的嗎?? 10/27 20:48
2F：→ ivy7132000:我把X代入到Pi/4X^3 = VT + 1/3(Pi/4X^3) 裡面 10/27 20:49
3F：→ ivy7132000:左右兩邊也無法相等 10/27 20:49
4F：推 ivy7132000:謝謝你唷 我自己研究出來了 那個應該是Pi*X^3/4 10/27 23:14
5F：→ ivy7132000:這樣算出來答案就對了 謝謝你幫我 感激不盡^^ 10/27 23:14