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could you show clearer steps and reasoning behind working for this question. working and answers included

could you show clearer steps and reasoning behind working for this question. working and answers included

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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3 Tutorial 4 - Question Q1 Solution a) Define the light key (LK) and heavy key (HK) components. F = 2000 kmol/h XMET = 0.18 XETH = 0.32 X = 0.26 Xa=0.24 ! RETH = 95% + F = 2000 kmol/h XMET= 0.18 XETH = 0.32 XP = 0.26 X8 = 0.24 B Rop=98% Assumptions: F = 2000 kmol/h XMET = 0.18 XETH = 0.32 XP = 0.26 X₂8 = 0.24 100% recovery of methanol in D 100% recovery of n-butanol in B Tutorial 4 - Question Q1 Solution The composition of D and B is controlled by the separation of ethanol and n-propanol. VolatiliyETH > Volatiliy Light key = ethanol Heavy key = n-propanol b) Determine the outgoing compositions for all the components. DMET=RMET, D(F-XMET) = (F-XMET) DETH = RETH D(FXETH) Dop=(1-RPB)(F-XnP) Doa Rua D(FX) = 0 1 Mass balance of every component y compa applied to the whole column BMET=RMET, B(F-XMET) = 0 BETH=(1-RETH, D) (F-XETH) B₂P=R₂PB(F*XP) BBR-B, (F-X₂B) = (F-X₂B) Tutorial 4 - Question Q1 Solution b) Determine the outgoing compositions for all the components. DMET= 2000-0.18 = 360 kmol/h DETH = 0.95(2000-0.32) = 608 kmol/h Dp = 0.02(2000-0.26) 10.4 kmol/h Doa = 0 kmol/h University of Strathclyde BMET=0 kmol/h ВМЕТ BETH = 0.05(2000-0.32) = 32 kmol/h Bp=0.98(2000-0.26) = 509.6 kmol/h BB= 2000-0.24 = 480 kmol/h * University of Strathclyde University of Strathclyde XUET = 0.37 XETH = 0.62 X₁P = 0.01 X₁8 = 0 XMET = 0 XETH = 0.03 XP = 0.50 X-8 = 0.47
Transcribed Image Text:3 Tutorial 4 - Question Q1 Solution a) Define the light key (LK) and heavy key (HK) components. F = 2000 kmol/h XMET = 0.18 XETH = 0.32 X = 0.26 Xa=0.24 ! RETH = 95% + F = 2000 kmol/h XMET= 0.18 XETH = 0.32 XP = 0.26 X8 = 0.24 B Rop=98% Assumptions: F = 2000 kmol/h XMET = 0.18 XETH = 0.32 XP = 0.26 X₂8 = 0.24 100% recovery of methanol in D 100% recovery of n-butanol in B Tutorial 4 - Question Q1 Solution The composition of D and B is controlled by the separation of ethanol and n-propanol. VolatiliyETH > Volatiliy Light key = ethanol Heavy key = n-propanol b) Determine the outgoing compositions for all the components. DMET=RMET, D(F-XMET) = (F-XMET) DETH = RETH D(FXETH) Dop=(1-RPB)(F-XnP) Doa Rua D(FX) = 0 1 Mass balance of every component y compa applied to the whole column BMET=RMET, B(F-XMET) = 0 BETH=(1-RETH, D) (F-XETH) B₂P=R₂PB(F*XP) BBR-B, (F-X₂B) = (F-X₂B) Tutorial 4 - Question Q1 Solution b) Determine the outgoing compositions for all the components. DMET= 2000-0.18 = 360 kmol/h DETH = 0.95(2000-0.32) = 608 kmol/h Dp = 0.02(2000-0.26) 10.4 kmol/h Doa = 0 kmol/h University of Strathclyde BMET=0 kmol/h ВМЕТ BETH = 0.05(2000-0.32) = 32 kmol/h Bp=0.98(2000-0.26) = 509.6 kmol/h BB= 2000-0.24 = 480 kmol/h * University of Strathclyde University of Strathclyde XUET = 0.37 XETH = 0.62 X₁P = 0.01 X₁8 = 0 XMET = 0 XETH = 0.03 XP = 0.50 X-8 = 0.47
Multicomponent Flash Distillation A mixture of methanol (1), ethanol (2), n-propanol (3) and n-butanol (4) is fed into a distillation column. The feed flow rate is 2000 kmol/h and its mole composition is given in the table below. The goal is to achieve 95% recovery of ethanol in the distillate and 98% recovery of n-propanol in the bottoms. a) Define the light key (LK) and heavy key (HK) components. b) Determine the outgoing compositions for all the components. Assumption: methanol and n-butanol are only present in the distillate and bottom streams, respectively Component z methanol (1) 0.18 ethanol (2) 0.32 n-propanol (3) 0.26 n-butanol (4) 0.24
Transcribed Image Text:Multicomponent Flash Distillation A mixture of methanol (1), ethanol (2), n-propanol (3) and n-butanol (4) is fed into a distillation column. The feed flow rate is 2000 kmol/h and its mole composition is given in the table below. The goal is to achieve 95% recovery of ethanol in the distillate and 98% recovery of n-propanol in the bottoms. a) Define the light key (LK) and heavy key (HK) components. b) Determine the outgoing compositions for all the components. Assumption: methanol and n-butanol are only present in the distillate and bottom streams, respectively Component z methanol (1) 0.18 ethanol (2) 0.32 n-propanol (3) 0.26 n-butanol (4) 0.24
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