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Mass Transfer for 4 th Year Chemical Engineering Department Faculty of Engineering Cairo University

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Today isA Special cases of binary systems distillation using McCabe Thiele method.

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Only one feed Consisting of two sections Condenser and Reboiler No side products Steps: 1.q-line 2.R min 3.R op and Top section oper. Line 4.Bottom section oper. line The Simple Case XDXD XFXF XWXW (X D,X D ) (X W,X W )

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Other Cases 1.Enriching Section 2.Stripping Section 3.Complex Feed or Multiple feeds 4.Open Steam 5.Top Side Product 6.Bottom Side Product NOTE: q-line NOT ONLY represents feed, but it represents any stream that changes the flow rates inside the column.

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1- Enriching Section Used in case when it’s needed to recover light component from feed containing little amount of it. Feed composition is near that of the bottom product (x F is very small) Feed is usually saturated vapour No reboiler is used.

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1- Enriching Section Steps: Feed is saturated vapour No. of stages in bottom section =0 Get point of intersection of bottom section and q-line directly Draw the top section operating line. XDXD XWXW (X D,X D ) XFXF

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2- Stripping Section Used in case when it’s needed to recover heavy component from feed containing little amount of it. Feed composition is near that of the top product (x F is very big) Feed is usually saturated liquid No reflux is needed.

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2- Stripping Section Steps: Feed is saturated liqiud No. of stages in top section =0 Get point of intersection of top section and q-line directly Draw the bottom section operating line. XDXD XWXW XFXF

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3- Multiple Feeds Generally F 1 is saturated liquid, and F 2 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.

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3- Multiple Feeds Since we assumed constant molar flow rates in the column, we can say that: L’=L+F 1 V’=V And L”=L’ V’=V”+F 2 To draw the middle section operating line we have to calculate its slope from the above relations

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3- Multiple Feeds Given information: F 1, x f1, F 2, x f2, x D, x W, R F 1 is saturated liqiud F 2 is saturated vapour Steps: Top section operating line as it is XDXD X F1 XWXW (X D,X D ) (X W,X W ) X F2

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3- Multiple Feeds Middle 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. XDXD X F1 XWXW (X D,X D ) (X W,X W ) X F2

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3- Multiple Feeds The slope of any operating line is L/V in the section it represents. So in the middle section slope is L’/V’ As we know L’=F 1 +L and V’=V Where L=D*R and V=L+D=(R+1)D So we can now get L’/V’ and draw its line XDXD X F1 XWXW (X D,X D ) (X W,X W ) X F2 L’/V’

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4- Upper Side Product Any 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 bottom Still the top section not affected, and we want to draw the middle section line.

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4- Upper Side Product To get Operating line of the middle section do MB on the loop: V=L+S+D V.y n+1 =L.x n +S.x S +D.x D (equation of st. line as S,D,x S,x D are constants) This line starts at the intersection of the top section with the q-line of side product

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4- Upper Side Product Steps: 1.Side product is saturated liquid. 2.Feed q-line is drawn whatever its state 3.Draw the top section line We need to use the equation of the middle section to draw its operating line. XDXD XSXS XWXW (X D,X D ) (X W,X W ) XFXF

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4- Upper Side Product Back to equations: V=L+S+DV-L=S+D V.y n+1 =L.x n +S.x S +D.x D We know a point on the line and need to get another point to draw it The easy point is on 45 line V.x=L.x+S.x S +D.x D (V-L).x=S.x S +D.x D (S+D).x=S.x S +D.x D XDXD XSXS XWXW (X D,X D ) (X W,X W ) XFXF x=y

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5- Bottom Side Product The column will be divided into 3 section: Top, middle and bottom Still 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)

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5- Bottom Side Product To get Operating line of the middle section do MB on the loop: L’=V’+S+W L’.x’ m+1 =V’.y’ m +S.x S +W.x W (equation of st. line as S,W,x S,x w are constants) This line ends at the intersection of the bottom section with the q-line of side product

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5- Bottom Side Product Steps: 1.Side product is saturated liquid. 2.Feed q-line is drawn whatever its state 3.Draw the top section line We need to use the equation of the middle section to draw its operating line. XDXD XSXS XWXW (X D,X D ) (X W,X W ) XFXF

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5- Bottom Side Product Back to equations: L’=V’+S+WL’-V’=S+W L’.x’ m+1 =V’.y’ m +S.x S +W.x w We know a point on the line and need to get another point to draw it The easy point is on 45 line L’.x=V’.x+S.x S +W.x W (L’-V’).x=S.x S +W.x W (S+W).x=S.x S +W.x W XDXD XFXF XWXW (X D,X D ) (X W,X W ) XSXS x=y

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6- 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.

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6- Open Steam Operating line of bottom section: L’+S=V’+WL’-V’=W-S L’.x’ m+1 +S.y S =V’.y’ m +W.x W To draw this line we need another point, try 45 o line L’.x=V’.x+W.x W (L’-V’)x=W.x W (W-S)x=W.x W

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6- Open Steam Steps: 1.Draw the top section line 2.Draw the q-line 3.Locate the point derived on the 45 o line 4.Draw the bottom section line. Here the bottom section will end at a point (x w,0) Usually S needed to be calculated. XDXD XFXF XWXW (X D,X D ) (X W,y s )

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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. (H v =2740 KJ/Kg, h L =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: Benzene =7360 Cal/gmol, Toluene =7960 Cal/gmol

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Stripping tower F=100 Kgmol/hr(saturated liquid) x F =0.7P=1 atm W=15 Kgmol/hrx W =0.1 Steam: H v =2740 KJ/Kg, h L =610 KJ/Kg st =2130 KJ/Kg a) D=F-W=100-15=85 Kmol/hr x D =(F.x F -W.x W )/D=(100*0.7-15*0.1)/85=0.806

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b) NTS= Reboiler + 4.6 c) Q r =m st. st =V’. r r is calculated at y r r =0.2*7360+0.8*7960 r =7840 Cal/gmol

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V’=??? Slope of operating line is L’/V’=1.1667 And W+V’=L’ Get L’ and V’ V’=90 kmol/hr L’=105 Kmol/hr L’/V’

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Finally Q r =90*7840=705600 Cal/hr= 2949.408 KJ/hr st =2130 KJ/Kg 2949.408 =2130*m st m st = 1384.7 Kg/hr

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