1 / 29

Distillation IV McCabe thiele method (2)

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

oksana
Download Presentation

Distillation IV McCabe thiele method (2)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Distillation IVMcCabe thiele method (2) Mass Transfer for 4th Year Chemical Engineering Department Faculty of Engineering Cairo University

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

  3. The Simple Case • Only one feed • Consisting of two sections • Condenser and Reboiler • No side products Steps: • q-line • Rmin • Ropand Top section oper. Line • Bottom section oper. line (XD,XD) (XW,XW) XW XF XD

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

  5. 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 (xF is very small) • Feed is usually saturated vapour • No reboiler is used.

  6. 1- Enriching Section Steps: • Feed is saturated vapour • No. of stages in bottom section =0 • Get point of intersection of bottom section andq-line directly • Draw the top section operating line. (XD,XD) XW XF XD

  7. 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 (xF is very big) • Feed is usually saturated liquid • No reflux is needed.

  8. 2- Stripping Section Steps: • Feed is saturated liqiud • No. of stages in top section =0 • Get point of intersection of top section andq-line directly • Draw the bottom section operating line. XW XF XD

  9. 3- Multiple Feeds Generally 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.

  10. 3- Multiple Feeds Since we assumed constant molar flow rates in the column, we can say that: L’=L+F1 V’=V And L”=L’ V’=V”+F2 To draw the middle section operating line we have to calculate its slope from the above relations

  11. 3- Multiple Feeds Given information: F1, xf1, F2, xf2, xD, xW, R F1 is saturated liqiud F2 is saturated vapour Steps: Top section operating line as it is (XD,XD) (XW,XW) XW XF1 XD XF2

  12. 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. (XD,XD) (XW,XW) XW XF1 XD XF2

  13. 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’=F1+Land V’=V Where L=D*R andV=L+D=(R+1)D So we can now get L’/V’ and draw its line (XD,XD) L’/V’ (XW,XW) XW XF1 XD XF2

  14. 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.

  15. 4- Upper Side Product To get Operating line of the middle section do MB on the loop: V=L+S+D V.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

  16. 4- Upper Side Product Steps: • Side product is saturated liquid. • Feed q-line is drawn whatever its state • Draw the top section line We need to use the equation of the middle section to draw its operating line. (XD,XD) (XW,XW) XW XS XD XF

  17. 4- Upper Side Product Back to equations: V=L+S+D V-L=S+D V.yn+1=L.xn+S.xS+D.xD 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.xS+D.xD (V-L).x=S.xS+D.xD (S+D).x=S.xS+D.xD (XD,XD) x=y (XW,XW) XW XS XD XF

  18. 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)

  19. 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.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 product

  20. 5- Bottom Side Product Steps: • Side product is saturated liquid. • Feed q-line is drawn whatever its state • Draw the top section line We need to use the equation of the middle section to draw its operating line. (XD,XD) (XW,XW) XW XD XS XF

  21. 5- Bottom Side Product Back to equations: L’=V’+S+W L’-V’=S+W L’.x’m+1=V’.y’m+S.xS+W.xw 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.xS+W.xW (L’-V’).x=S.xS+W.xW (S+W).x=S.xS+W.xW (XD,XD) x=y (XW,XW) XW XF XD XS

  22. 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.

  23. 6- Open Steam Operating line of bottom section: L’+S=V’+W L’-V’=W-S L’.x’m+1+S.yS=V’.y’m+W.xW To draw this line we need another point, try 45o line L’.x=V’.x+W.xW (L’-V’)x=W.xW (W-S)x=W.xW

  24. 6- Open Steam Steps: • Draw the top section line • Draw the q-line • Locate the point derived on the 45o line • Draw the bottom section line. Here the bottom section will end at a point (xw,0) Usually S needed to be calculated. (XD,XD) (XW,ys) XW XF XD

  25. 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/gmol

  26. Stripping tower F=100 Kgmol/hr (saturated liquid) xF=0.7 P=1 atm W=15 Kgmol/hr xW=0.1 Steam: Hv=2740 KJ/Kg, hL=610 KJ/Kg lst=2130 KJ/Kg a) D=F-W=100-15=85 Kmol/hr xD=(F.xF-W.xW)/D=(100*0.7-15*0.1)/85=0.806

  27. b) NTS= Reboiler + 4.6 c) Qr=mst.lst=V’.lr lris calculated at yr lr=0.2*7360+0.8*7960 lr=7840 Cal/gmol

  28. 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’

  29. Finally Qr=90*7840=705600 Cal/hr= 2949.408 KJ/hr lst=2130 KJ/Kg 2949.408 =2130*mst mst= 1384.7 Kg/hr

More Related