Rectification Process ( Thermal Separation Techniques)

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Chemicaleng Participant Join Date: Mar 2013 Posts: 2	Rectification Process ( Thermal Separation Techniques) 07/24/2014 9:43 AM
	Greetings everyone! I have one question concerning rectification process (countercurrent distillation). For example; if you have a process in which a feed of water-methanol solution (F) (22%mol) with a flow of 6.5 kg/h enters a column. Inside a column a feed of vapours (V) has value of 7 kg/h with 95%mol. If reflux ratio is 2.5, what is the value of reflux going back into the column (L) and fraction of distillation residue (xw)? If my knowledge is correct then you get reflux (L) from reflux ratio and mass balance of vapours: R=L/D 2.5=L/D V=L+D 7=L+D From this you get: D=2 kg/h and L=5 kg/h Then from the mass balance of feed you would eventually get a distillation residue: F=W+D 6.5=W+2 W=4.5kg/h Fxf=Wxw+Dxd 6.50.22=4.5xw+20.95 xw=-0.104 (?????) Question I want to ask is, why do I get negative value of distillation residue? Is my procedure correct? Do you in this case consider absolute value of xw? Could someone please help me? Regards, Filip
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Crabtree Guru Join Date: Jul 2007 Location: Commissariat de Police, Nouvions, occupied France, 1942. Posts: 2599 Good Answers: 77	#1 Re: Rectification Process ( Thermal Separation Techniques) 07/24/2014 11:38 AM
	You have assumed that the concentration in the reboil vapours and that in the distillate are the same. Is that so? __________________ Good moaning!
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bigg Guru Join Date: Jun 2010 Posts: 1296 Good Answers: 104	#2 Re: Rectification Process ( Thermal Separation Techniques) 07/24/2014 6:35 PM
	You have an obvious material imbalance. You have 1.43 kg/h of methanol going in (6.5 * 0.22) and 1.8 kg/h coming out overhead (2 * 0.95), not including any losses in the bottoms product. That's why you get a negative value of bottoms fraction.
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ADHitechVN Participant Join Date: Jul 2014 Posts: 1	#3 Re: Rectification Process ( Thermal Separation Techniques) 07/26/2014 7:21 AM
	Hi, You have mistake about the unit of measure in the material balance equation. Need to transfer measurement unit of flows (F, D, W) from Kg/h to (Gf, Gd, Gw) in Kmol/h. Because concentrations xf, xd, xw calculated in units of mol %. Gf=F/Mf Kmol/h; Gd=D/Md Kmol/h. There are Mf, Md, Mw: Molar Mass of Feed, Distillate and Bottom fraction. Molar Mass of Methanol CH4O: M1=46.07; Molar Mass of H2O: M2=18.02; Mf=(xfM1 + (100-xf)M2)/100= (2246.07+(100-22)18.02)/100=24.188; Md=(xdM1 + (100-xd)M2)/100= (9546.07+(100-95)18.02)/100=44.667; Gf=6.5 / 24.188 = 0.269 Kmol/h; Gd=2 / 44.667 = 0.045 Kmol/h. Material balance equation for liquid: Gw = Gf - Gd = 0.269 - 0.045 = 0.224 Kmol/h. Material balance equation for absolute methanol: Gwxw = Gfxf - Gdxd Kmol/h; Gfxf = 0,269 * 22% = 0.0591 Kmol/h; Gdxd = 0.045 95% = 0.0425 Kmol/h . Gwxw = Gfxf - Gdxd = 0.0591 - 0.0425 = 0.0166 Kmol/h RESULTS: xw = (Gwxw)/Gw = 0.0166 / 0.224 = 7.4 % mol. Best Regards ADHitechVN
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bigg Guru Join Date: Jun 2010 Posts: 1296 Good Answers: 104	#4 In reply to #3 Re: Rectification Process ( Thermal Separation Techniques) 07/26/2014 9:28 AM
	Good catch - I missed the mol%.
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ADHitechVN (1); bigg (2); Crabtree (1)

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