Distillation IV McCabe thiele method (2)

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Distillation IV McCabe thiele method (2). Mass Transfer for 4 th Year Chemical Engineering Department Faculty of Engineering Cairo University. Today isA. Special cases of binary systems distillation using McCabe Thiele method. The Simple Case. Only one feed Consisting of two sections - PowerPoint PPT Presentation

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Distillation IV McCabe thiele method (2)

Distillation IVMcCabe thiele method (2)Mass Transfer for 4th YearChemical Engineering DepartmentFaculty of EngineeringCairo UniversityToday isASpecial cases of binary systems distillation using McCabe Thiele method.

Only one feedConsisting of two sectionsCondenser and ReboilerNo side productsSteps:q-lineRminRop and Top section oper. LineBottom section oper. lineThe Simple CaseXDXFXW

(XD,XD)

(XW,XW)Other CasesEnriching SectionStripping SectionComplex Feed or Multiple feedsOpen SteamTop Side ProductBottom Side Product

NOTE: q-line NOT ONLY represents feed, but it represents any stream that changes the flow rates inside the column.1- Enriching Section

Used in case when its needed to recover light component from feed containing little amount of it.Feed composition is near that of the bottom product (xF is very small)Feed is usually saturated vapourNo reboiler is used. 1- Enriching SectionSteps:Feed is saturated vapourNo. of stages in bottom section =0Get point of intersection of bottom section andq-line directlyDraw the top section operating line.XDXW(XD,XD)XF2- Stripping Section

Used in case when its needed to recover heavy component from feed containing little amount of it.Feed composition is near that of the top product (xF is very big)Feed is usually saturated liquidNo reflux is needed.2- Stripping SectionSteps:Feed is saturated liqiudNo. of stages in top section =0Get point of intersection of top section andq-line directlyDraw the bottom section operating line.XDXWXF3- Multiple FeedsGenerally F1 is saturated liquid, and F2 is saturated vapour.The column can now be divided into THREE section: Enriching, Stripping and Middle section.Both operating lines of the top and bottom sections will not be changed.Only we will need to get the middle section line.

3- Multiple FeedsSince we assumed constant molar flow rates in the column, we can say that:L=L+F1V=V

AndL=LV=V+F2

To draw the middle section operating line we have to calculate its slope from the above relations

3- Multiple FeedsGiven information:F1, xf1, F2, xf2, xD, xW, RF1 is saturated liqiudF2 is saturated vapour

Steps:Top section operating line as it isXDXF1XW

(XD,XD)(XW,XW)XF23- Multiple FeedsMiddle line starts from the end of the top section and ends at the beginning of the bottom section.So point of intersection of top section and first q-line is on the middle section line.We need another point or a slope to draw the middle line.XDXF1XW

(XD,XD)(XW,XW)XF23- Multiple FeedsThe slope of any operating line is L/V in the section it represents.So in the middle section slope is L/V

As we knowL=F1+L and V=VWhere L=D*R and V=L+D=(R+1)DSo we can now get L/V and draw its lineXDXF1XW

(XD,XD)(XW,XW)XF2L/V4- Upper Side ProductAny side product is withdrawn as saturated liquid.As this side product will change the flow rates inside the column, there will be a q-line representing the side product.The column will be divided into 3 section: Top, middle and bottomStill the top section not affected, and we want to draw the middle section line.

4- Upper Side ProductTo get Operating line of the middle section do MB on the loop:V=L+S+DV.yn+1=L.xn+S.xS+D.xD(equation of st. line as S,D,xS,xD are constants)

This line starts at the intersection of the top section with the q-line of side product

4- Upper Side ProductSteps:Side product is saturated liquid.Feed q-line is drawn whatever its stateDraw the top section line

We need to use the equation of the middle section to draw its operating line.XDXSXW

(XD,XD)(XW,XW)XF4- Upper Side ProductBack to equations:V=L+S+DV-L=S+DV.yn+1=L.xn+S.xS+D.xDWe know a point on the line and need to get another point to draw itThe easy point is on 45 lineV.x=L.x+S.xS+D.xD(V-L).x=S.xS+D.xD(S+D).x=S.xS+D.xDXDXSXW

(XD,XD)(XW,XW)XF

x=y5- Bottom Side ProductThe column will be divided into 3 section: Top, middle and bottomStill the top section not affected, and we want to draw the middle section line.We will derive the middle section operating line equation (as in previous case)

5- Bottom Side ProductTo get Operating line of the middle section do MB on the loop:L=V+S+WL.xm+1=V.ym+S.xS+W.xW(equation of st. line as S,W,xS,xw are constants)

This line ends at the intersection of the bottom section with the q-line of side product5- Bottom Side ProductSteps:Side product is saturated liquid.Feed q-line is drawn whatever its stateDraw the top section line

We need to use the equation of the middle section to draw its operating line.XDXSXW

(XD,XD)(XW,XW)XF5- Bottom Side ProductBack to equations:L=V+S+WL-V=S+WL.xm+1=V.ym+S.xS+W.xwWe know a point on the line and need to get another point to draw itThe easy point is on 45 lineL.x=V.x+S.xS+W.xW(L-V).x=S.xS+W.xW(S+W).x=S.xS+W.xWXDXFXW

(XD,XD)(XW,XW)XS

x=y6- Open Steam

Used in cases when the feed contains water, so heat is added to column in the form of direct heating by steam instead of reboiler.The bottom section operating line will change as the equations will be changed.Still the intersection of the top section line and the q-line is on the bottom section line.6- Open Steam

Operating line of bottom section:L+S=V+WL-V=W-SL.xm+1+S.yS=V.ym+W.xW

To draw this line we need another point, try 45o lineL.x=V.x+W.xW(L-V)x=W.xW(W-S)x=W.xW

6- Open SteamSteps:Draw the top section lineDraw the q-lineLocate the point derived on the 45o lineDraw the bottom section line.Here the bottom section will end at a point (xw,0)

Usually S needed to be calculated.XDXFXW

(XD,XD)(XW,ys)

4-100 Kgmol/hr of saturated liquid containing 70 mol% benzene enters a stripping tower at 1 atm. The bottom product flow rate is 15 Kgmol/hr containing only 10% benzene. Saturated steam at 4 atmospheres is available for the reboiler duty. (Hv=2740 KJ/Kg, hL=610 KJ/Kg).Calculate:a- The overhead product flow rate and its composition.b- Number of theoretical plates required.c- Steam consumption in reboiler.Equilibrium data:

lBenzene=7360 Cal/gmol, lToluene=7960 Cal/gmolT, C80859095100105110PBo, mmHg76087710161068134415321800PTo, mmHg205345405475557645760Stripping towerF=100 Kgmol/hr(saturated liquid)xF=0.7P=1 atmW=15 Kgmol/hrxW=0.1Steam: Hv=2740 KJ/Kg, hL=610 KJ/Kglst=2130 KJ/Kg

a)D=F-W=100-15=85 Kmol/hrxD=(F.xF-W.xW)/D=(100*0.7-15*0.1)/85=0.806T, C80859095100105110PBo, mmHg76087710161068134415321800PTo, mmHg205345405475557645760X10.780.580.480.260.130Y10.900.780.6780.460.260b)NTS= Reboiler + 4.6c)Qr=mst.lst=V.lrlr is calculated at yr

lr =0.2*7360+0.8*7960lr =7840 Cal/gmol

V=???Slope of operating line is L/V=1.1667And W+V=LGet L and VV=90 kmol/hrL=105 Kmol/hrL/V

FinallyQr=90*7840=705600 Cal/hr= 2949.408 KJ/hrlst=2130 KJ/Kg2949.408 =2130*mstmst= 1384.7 Kg/hr