Bottom operating line
the materials balance the blow number of plats (m) , Lm+1 = Lm
Lm = Vm +W , then Vm = Lm –W
ym Vm = Lm Xm+1 - W Xw ym = ( Lm / Vm) Xm+1 -( W / Vm ) Xw
q- line equation yq = (q / q-1) Xq – Xf / q-1
First point (Xf,Xf) , 2- second point ( Xf / q) , 0))
For finding the q-line , drawing the line contact between the top operating line and bottom operating line
How to limited q
If feed is sub cooled feed temperature (TF) less than boiling point (TB)
Q =( Cp(TB-TF ) + ? ) / ? q ? 1
The feed is liquid at it is boiling point ( X / Tf= TB ) q= ? / ? = 1
1 2
Lm = Ln + qF , Vm = Vn + F(q-1) or Vm= Lm –W
No. of stage -1 = No. of plates
q- line , q = heat required to vaporize 1 mole of feed / latent heat of feed , when the feed is partially vaporized , two phase ( vapour + liquid ) , for example the feed 20% liquid ad 80% vapour , and Xf=0.4 , then q=0.2 ? / ? =0.2
1st point ( 0.4,0.4), 2nd point (Xf / q , 0) = 0.4 / 0.2, 0 ) =( 2 ,0)
saturated vapour Tf= TB , q= 0 / ? =0
The feed is super-heated Tf ? TB , q= cp( TB-Tf ) / ? , q < 1
Calculation the minimum Reflux Ratio (Rm)
From equations [Rm=1/(?AB-1)[ xd/xf-?AB (1-xd)/(1-xf) ] , ?AB = ?A / ? B= PA? / PB?
?AV=?(?T ?F ?B)
– from drawing
1-Drawing line from Xf (q-line) and interception the curve at point E ,
2- Drawing the line from Xd to interception the curve and y- axes at point A
A( 0, Xd / Rm+1) , example A= 0.6 then 0.6 = Xd / Rm+1 , 0.6 = 0.97 / Rm+1
Rm = 0.62
Calculation of Minimum No. of plates (N min )
By equations Nmin= (log[(Xd/(1-Xd))((1-Xw )/Xw))/(log ?AB)
The feed at its boiling point q=1 ( saturated liquid )
By drawing
Example 1/ Abinary mixture of n-heptaine at its boiling point containg 70% mole percent of heptane is to be continuously distillated to give a top product of 98 % mole and bottom product of 1 mole% heptane. The reflux ratio is 3 and the vapour pressure of n- heptane is twice the vapour pressure of octane at the same temp.
Obtain the number of theoretical plate required .
Calculate the min-number of theoretical plate (Nm)
Calculate the min- reflux ratio (Rm)
Solution
?AB = ?A / ? B= PA? / PB? = 2 PB? / PB? =2 , A = n-heptane
yA = ?AB XA / ( 1+ XA (?AB -1)) = 2 XA / 1+(2-1) XA = 2 XA / 1+ XA
XA 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
yA 0 0.18 0.33 0.46 0.57 0.66 0.75 0.82 0.88 0.95 1
Overall material balance
F = D+W 100= D+W (1)
Material balance on the top
F Xf = DXd + WXw , 0.7 (100)= 0.98 D+ 0.1 W 70 = 0.98 D+0.1W (2)
W= 31.82 and D= 68.18
Top operating line
yn=( Ln)/( Vn) Xn+1+D/Vn Xd ,R= Ln/D , Ln = 3 *68.1=204.3 kmole , Vn=Ln+D = 204.3+68.1=272.4 kmole then yn=( 204.3)/( 272.4) Xn+1+68.1/272.4 0.98
yn= 0.75 Xn+1 + 0.245 operating line , can drawing this line between two points
(0,(D/ Vn)Xd)) ( Xd,Xd) (0,0.245) and (0.98,0.98)
Bottom operating line : ym=Lm/Vm Xm+1-(_Vm^W)Xw , Lm= Ln + qF , q=1 since the feed is at its boiling point
Lm= Ln+F Lm = 100+204.3= 304.3 Kmole
Vm = Lm-W = 304.3 -3189= 272.41 kmole
ym=304.3/272.4 Xm+1-(_272.41^31.89)Xw ym=1.117Xm+1-0.0117 Bottom operating line
First point ( Xw, Xw)= (0.1,0.1) , if X= 0.5 then ym = 1.117(0.5) – 0.0117 = 0.54
Second point (0.5, 0.54)
From the diagram , No.of stage =14 , No,of plate =13
Nmin=log??[(Xd/(1-Xd)) ((1-Xw)/Xw)]?/(log?AB) = log??[(0.98/(1-0.98)) ((1-0.1)/0.1)]?/log2 ? 9
Rm=1/(?AB-1)[ xd/xf-?AB (1-xd)/(1-xf) ] = 1/(2-1) [ 0.98/0.7-2 (1-0.98)/(1-0.7) ]=1.26
المادة المعروضة اعلاه هي مدخل الى المحاضرة المرفوعة بواسطة استاذ(ة) المادة . وقد تبدو لك غير متكاملة . حيث يضع استاذ المادة في بعض الاحيان فقط الجزء الاول من المحاضرة من اجل الاطلاع على ما ستقوم بتحميله لاحقا . في نظام التعليم الالكتروني نوفر هذه الخدمة لكي نبقيك على اطلاع حول محتوى الملف الذي ستقوم بتحميله .
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