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SOLUTION MANUAL for
SEPARATION PROCESS ENGINEERING. Includes Mass Transfer Analysis 3rd Edition (Formerly published as Equilibrium-Staged Separations) by Phillip C. Wankat
SPE 3rd Edition Solution Manual Chapter 1 New Problems and new solutions are listed as new immediately after the solution number. These new problems in chapter 1 are: 1A3, 1A4, 1B2-1B4, 1D1. A2.
Answers are in the text.
New problem for 3rd edition. Answer is d.
Everything except some food products has undergone some separation operations. Even the water in bottles has been purified (either by reverse osmosis or by distillation).
New problem for 3rd edition. Many homes have a water softener (ion exchange), or a filter, or a carbon water “filter” (actually adsorption), or a reverse osmosis system.
New problem for 3rd edition. For example: the lungs are a gas permeation system, the intestines and kidney are liquid permeation or dialysis systems.
New problem for 3rd edition. You probably used some of the following: chromatography, crystallization, distillation, extraction, filtration and ultrafiltration.
New problem for 3rd edition. Basis 1kmol feed.
Weight fraction ethanol = 18.4/29.2 = 0.630 Flow rate = (1500 kmol/hr)[(29.2kg)/(1 kmol)] = 43,800 kg/hr.
Chapter 2 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 2A6, 2A9 to 2A16, 2C4, 2C8, 2C9, 2D1.g, 2.D4, 2D10, 2D13, 2D24 to 2D30, 2E1, 2F4, 2G4 to 2G6, 2H1 to 2H3. 2.A1.
Feed to flash drum is a liquid at high pressure. At this pressure its enthalpy can be calculated as a liquid. eg. h TF,Phigh
Tref . When pressure is dropped the mixture is above
its bubble point and is a two-phase mixture (It “flashes”). In the flash mixture enthalpy is unchanged but temperature changes. Feed location cannot be found from T F and z on the graph because equilibrium data is at a lower pressure on the graph used for this calculation. 2.A2.
2.A4. 1.0 Equilibrium (pure water)
zw = 0.965 Flash operating line
2.A6. New Problem. In a flash drum separating a multicomponent mixture, raising the pressure will: i. Decrease the drum diameter and decrease the relative volatilities. Answer is i. 2.A8.
a. K increases as T increases b. K decreases as P increases c. K stays same as mole fraction changes (T, p constant) -Assumption is no concentration effect in DePriester charts d. K decreases as molecular weight increases
New Problem. The answer is 0.22
New Problem. The answer is b.
New Problem. The answer is c.
New Problem. The answer is b.
New Problem. The answer is c.
New Problem. The answer is a.
New Problem. a. b. The answer is
The answer is 36ºC
2.A16. New Problem. The liquid is superheated when the pressure drops, and the energy comes from the amount of superheat. 2.B1.
Must be sure you don’t violate Gibbs phase rule for intensive variables in equilibrium. Examples: F, z, Tdrum , Pdrum F, h F , z, p F, TF , z, p
F, TF , z, Tdrum , p drum
Drum dimensions, z, Fdrum , p drum
Drum dimensions, z, y, p drum
F, TF , x, Tdrum F, TF , y, x
This is essentially the same problem (disguised) as problem 2-D1c and e but with an existing (larger) drum and a higher flow rate. With y = 0.58, x = 0.20, and V/F = 0.25 which corresponds to 2-D1c. lb mole If F 1000 , D .98 and L 2.95 ft from Problem 2-D1e . hr Since D α V and for constant V/F, V α F, we have D α With F = 25,000:
Fnew Fold = 5, Dnew = 5 Dold = 4.90, and Lnew = 3 Dnew = 14.7 . Existing drum is too small. 2 2 Fexisting D exist 4 2 Feed rate drum can handle: F α D. gives 1000 .98 .98 Fexisting 16,660 lbmol/h Alternatives a) Do drums in parallel. Add a second drum which can handle remaining 8340 lbmol/h. b) Bypass with liquid mixing
y = .58, V = .25 (16660) = 4150 16,660
Since x is not specified, use bypass. This produces less vapor. c) Look at Eq. (2-62), which becomes
Bypass reduces V c1)
Kdrum is already 0.35. Perhaps small improvements can be made with a better demister → Talk to the manufacturers. c2) ρv can be increased by increasing pressure. Thus operate at higher pressure. Note this will change the equilibrium data and raise temperature. Thus a complete new calculation needs to be done. d) Try bypass with vapor mixing. e) Other alternatives are possible. 2.C2.
K i zi L 1 Ki F
From data in Example 2-2 obtain:
2.C8. New Problem.
Solve for L & V Or use lever arm-rule
New Problem. Derivation of Eqs. (2-62) and (2-63). Overall and component mass balances are,
Substituting in equilibrium Eqs. (2-60b) and 2-60c)
Dividing numerator and denominator by F and collecting terms.
K i,V L 2 zi L 1 1 K i,V F
Slope op. line See graph. y b.
K i,L1 L 2 z i L K i,L1 L 2 1 1 K i,V F K i,L1 L 2 1 z i
L V 3 2, y x 0.77 and x 0.48
x i,liq 2 , we have x i,liq1
900 . Rest same as part a.
0.2 on equil. Diagram and y
V F z 1.2 0.25 . From equil y d. Plot x 0.45 on equilibrium curve.
Slope Plot operating line, y e. Find Liquid Density.
.2 32.04 32.04 .7914
4 V V F .2 0.51 . From mass balance F 37.5 kmol/h.
Find Vapor Density.
1 atm 26.15 g/mol
20.82 22.51 0.925 g/ml
(Need temperature of the drum)
MW v y m MW m y w MW w .58 32.04 .42 18.01 Find Temperature of the Drum T: From Table 3-3 find T corresponding to y .58, x 20, T=81.7 C 354.7K v
Find Permissible velocity:
u perm 3600 Thus, D
0.0744, and n Flv
14.19 3600 8.98 10
250 26.15 454 g/lb 4
750 lbmol/h, and WL
Flv Then K drum
g/ml 28316.85 ml/ft
1.705 ft. Use 2 ft diameter.
L ranges from 3 D 6 ft to 5 D=10 ft Note that this design is conservative if a demister is used. f. Plot T vs x from Table 3-3. When T 77 C, x 0.34, y 0.69. This problem is now very similar to 3-D1c. Can calculate V/F from mass balance, Fz Lx Vy. This is V z y 0.4 0.34 Fz F V x Vy or 0.17 F y x 0.69 0.34 g. Part g is a new problem. V = 16.18 mol/h, L = 33.82, y= 0.892, x = 0.756. 2-D2.
Work backwards. Starting with x 2, find y2 = 0.62 from equilibrium. From equilibrium point V plot op. line of slope This gives L V 2 1 V F 2 3 7. F 2
0.51 x1 (see Figure). Then from equilibrium, y1
V F 0.6 V L y x z V F V Op. eq. 2 y x 2 3 3 See graph: y 0.55 x M 0.18
T ~ 82.8 C linear interpretation on Table 2-7 .
V 2.985 and V F Can also calculate V/F from slope.
0.545 @ equil 0.3 0.8
0.545 0.6215. 3 7 Can also draw line of slope through equil point. 3
2.D4. New problem in 3rd edition. Highest temperature is dew point Set
K ref TNew If pick C4 as reference: First guess
K ref TOld K bu tan e
4.0145 T too low K i 3.1 1.0 .125 Guess for reference K C4 4.014 T 118 C : K C3 8.8,
Use 90.5º → Avg last two T
T ~ 87 88º C Note: hexane probably better choice as reference. a) v1 = F2 v2
z Plot 1st Op line.
V1 = 687.5 kmol/h = F2
y = x = z = 0.55 to x1 = 0.3 on eq. curve (see graph) 0.55 0.80 .25 0.454545 .55 0 .55
3, y x V 0.25F 0.66 2.64. At y 0, x 2 0.25
0.66. Plot op line
0.63 . 171.875 kmol/h
V F = 1.0 kmol/min
T = 50ºC P = 200 kPa
zc4 = 0.45 zc5 = 0.35 Zc6 = 0.20
0 , First guess V/F = 0.6
1 1.4 .6 1 0.2 0.6 Use Newtonian Convergence
Kc4 = 2.4 Kc5 = 0.80
0.00028 1 1.4 0.6377 1 0.2 0.6377 1 0.7 0.6377 Which is close enough. yi K i x i zi 0.45 x c4 0.2377, V 1 1.4 .6377 yc4 2.4 0.2377 1 Ki 1 F 0.35 x c5 0.4012, y c5 0.8 0.4012 0.3210 1 0.2 0.6377
0.1466 V 1 4.6 0.2276 1 KM 1 F 5.6 0.1466 0.8208 0.8534 , y M K M x M
Use Rachford-Rice eqn: f
Converge on V F
Find K i from DePriester Chart: K1
0 . Note that 2 atm = 203 kPa.
we obtain x1 .0077, x 2 .0809, x 3 .9113 V 1 Ki 1 F From yi K i x i , we obtain y1 .5621, y 2 .3649, y3 .1048 Need hF to plot on diagram. Since pressure is high, feed remains a liquid h F CPL TF Tref , Tref 0 from chart From x i
Where x EtOH and x w are mole fractions. Convert weight to mole fractions. Basis: 100 kg mixture 30 30 kg EtOH 0.651 kmol 46.07 70 kg water 70 18.016 3.885 Total = 4.536 kmol 0.6512 100 Avg. MW 0.1435, x w 0.8565 . 22.046 Mole fracs: x E 4.536 4.536 Use CPL at 100 C as an average CP value. EtOH
C PL Per kg this is
h F 0.946 2000 189.2 kcal/kg which can now be plotted on the enthalpy composition diagram. Obtain Tdrum 88.2 C, x E 0.146, and y E 0.617 . For F 1000 find L and V from F = L + V and Fz Lx Vy which gives V = 326.9, and L = 673.1
Note: If use wt. fracs. CPL
23.99 & CPL MWavg
217.6 . All wrong.
2.D.10 New Problem. Solution 400 kPa, 70ºC From DePriester chart Know y i
35 Mole % n-butane
K i 1 zi 0 z C3 1 z C6 z C 4 .65 z C6 V 1 Ki 1 F zC6 zC6 V For C6 0.7 z C 6 0.7 1 0.7 V V F 1 K C6 1 1 0.7 F F V z C6 0.7 0.49 F 4 .65 z C 6 0.9 .35 0.7z C 6 RR Eq: 0 V V V 1 4 1 0.9 1 0.7 F F F 2 equations & 2 unknowns. Substitute in for z C6 . Do in Spreadsheet. R.R.
Use Goal – Seek to find V F. V 0.594 when R.R. equation 0.000881 . F V z C6 0.7 0.49 0.7 (0.49)(0.594) 0.40894 F L F 0.6 V F 0.4 & L V 1.5 Operating line: Slope 1.5, through y x z 0.4
For problem 2.D1c, plot x = 0.2 on equilibrium diagram with feed composition of 0.3. The resulting operating line has a y intercept z V / F 1.2 . Thus V F 0.25 (see figure in Solution to 2.D1) Vapor mole fraction is y = 0.58. Find Liquid Density.
Find Vapor Density.
.8 18.01 18.01 1.00
20.82 22.51 0.925 g/ml
(Need temperature of the drum)
MW v y m MW m y w MW w .58 32.04 .42 18.01 26.15 g/mol Find Temperature of the Drum T: From Table 2-7 find T corresponding to y .58, x 20, T=81.7 C 354.7K
1 atm 26.15 g/mol
Find Permissible velocity:
1.25 K drum,vertical
250 26.15 lb lbmol
Raoult’s Law: K C 4
g/ml 28316.85 ml/ft 3
3.41 ft and L 13.6 ft
0.0744, and n Flv
17.74 3600 8.98 10
250 26.15 454 g/lbm
2.D13. New Problem. The answer is ycresol = 0.19582 xp Since x c 0.3, x p 0.7, y p 1 1 xp
0.442, and K drum ,horiz
K drum ,vertical
750 lbmol/h, and WL
0.3 0.7 1 11121 1844.36 1 1 0.4 1 1 0.4 P P Solve for Pdrum = 3260 mmHg zi xi V 1 Ki 1 F .3 11121 x C4 0.1527, y C 4 K C 4 x C 4 0.1527 11121 3260 1 1 .4 3260 1844.36 x C6 1 x C 4 0.84715, y C6 0.84715 0.47928 3260 Check 1.00019 2.D15.
This is an unusual way of stating problem. However, if we count specified variables we see that problem is not over or under specified. Usually V/F would be the variable, but here it isn’t. We can still write R-R eqn. Will have three variables: zC2, ziC4, znC4. Need two other eqns: z iC4 z nC4 constant, and z C2 z iC4 z nC4 1.0 Thus, solve three equations and three unknowns simultaneously. Do It. Rachford-Rice equation is, K C2 1 zC2 K iC 4 1 z iC 4 K nC 4 1 z nC 2 0 V V V 1 K C2 1 1 K iC 4 1 1 K nC 4 1 F F F Can solve for zC2 = 1 – ziC4 and ziC4 = (.8) znC4. Thus zC2 = 1 – 1.8 znC4 Substitute for ziC4 and zC2 into R-R eqn. K C2 1 .8 K iC 4 1 K nC 4 1 1.8 z nC4 z nC 4 z nC 4 V V 1 K C2 1 1 K iC 4 1 1 K nC 4 F F K C2 1 V 1 K C2 1 F Thus, z nC 4 K C2 1 .8 K iC 4 1 K nC 4 1.8 V V 1 K C2 1 1 K iC 4 1 1 K nC 4 F F Can now find K values and plug away. KC2 = 2.92, KiC4 = .375, KnC4 = .26. Solution is znC4 = 0.2957, ziC4 = .8 (.2957) = 0.2366, and zC2 = 0.4677
z C1 0.5, z C4 0.1, z C5 0.15, z C6 0.25, K C1 50, K C4 .6, K C5 .17, K C6 1st guess. Can assume all C1 in vapor, ~ 1/3 C4 in vapor, C5 & C6 in bottom V / F 1 .5 .1 / 3 .53 This first guess is not critical.
49 .5 1 49 .53 Eq. 3.33
87.6 kmol/h and L 150 87.6 z C1
y C1 K C1 x C1 50 0.016883 Similar for other components. 2-D17.
L F 1.5 V 0.4F 400, L 600 Slope Intercepts y = x = z = 0.70. Plot line and find xA = 0.65, yA = 0.77 (see graph) b. V = 2000, L = 3000. Rest identical to part a. c. Lowest xA is horizontal op line (L = 0). xA = 0.12 Highest yA is vertical op line (V = 0). yA = 0.52. See graph a.
V = 600, L = 400, -L/V = -0.667. Find xA = 0.40 on equilibrium curve. Plot op line & find intersection point with y = x line. zA = 0.52 zh 1 zi V xh 2.D18. From x i , we obtain V F Kh 1 1 Ki 1 F Guess Tdrum , calculate K h , K b and K p , and then determine V F .
Check: Initial guess: If x h
.85 then Tdrum must be less than temperature to boil pure hexane
94 C . On this basis 85° to 90° would be reasonable. Try 85°C.
0.6 1 V 0.85 1.471 . Not possible. Must have K h F 0.8 1 73 C where K h 0.6 . Then K b 3.8, K p 9.9 . 0.6 1 .85 .6 1
Converge on T ~ 65.6 C and V F ~ 0.57 . 2.D19.
90% recovery n-hexane means 0.9 Fz C6 Substitute in L
F V to obtain z C6 .9
C 8 balance: z C6 F
L x C6 1 V F x C6
Two equations and two unknowns. Remove x C6 and solve
z C6 Solve for V F.
. Trial and error scheme.
Pick T, Calc K C6 , Calc V F, and Check f V F If not K ref new
K ref Told 1 df T
F .9 .37 .1 Rachford Rice equation 2.1 .4 f 1 2.1 .231 1
K ref Tnew Converge on TNew
0.28745 use .28 1 0.28719 ~ 57 C. Then K C4 2.50, K C8 .67, and V F
2.D20. New Problem. The K values are: K E 8.7 , K B 0.54 , K P 0.14 Can use Eq. (2-40), (2-41) or (2-42). If we use (2-42) the R – R eqn
K i 1 zi 0 V 1 Ki 1 F 7.7 .2 .46 z B
Soln to Binary R.R. eq.
V 1 3.8 .5309 1 K C2 1 F x C5 0.8177 , y C5 0.1251 Need to convert F to kmol. Avg MW 0.55 30.07 0.45 72.15
0.5265 0.51977 z B
MW V 0.8749 30.07 For liquid assume ideal mixture: V1
RT atm g 700 kPa 35.33 101.3 kPa mol v ml atm 82.0575 303.16K mol K WL K drum : Use Eq. (2-60) with FlV WV
V MWV u Perm 3600
0.621 0.009814 ft
0.0145229 u Perm
For vapor: ideal gas:
kg kmol kg 1000g
K iP 1 z iP V 1 K iP 1 F
K NP 1 z NP V 1 K NP 1 F
3.011679 20 219.61 1027.256 mm Hg , K iP 1027.256 760 1.35165
5. 0°C, 2500 kPa Fig 2.12: K M
0.4095 V 1 F 0.5905; yip K iP x iP
0.6 (equal split ethylene and ethane)
0.5 in both wt & mol frac., as does z NP .
2.75943 90 204.64 574.68 mm Hg , K NP 574.68 p tot 0.75616
0.104, x ethylene
2.D24. New Problem. p = 300 kPa At any T. K C3
Substitute 1st equation into 2nd
At 300 kPa pure propane K C3
1.0 boils at -14°C
At 300 kPa pure n-hexane K C6 at -14°C
K C3 1 K C6 K C3
1.0 boils at 110°C 1 K C6 1 K C6 0 K C3
1 K C6 K C3 0 K C3
Pick intermediate temperatures, find K C3 & K C6 , calculate x C3 & y C3 .
T 0ºC 10ºC 20ºC 30ºC 40ºC 50ºC 60ºC 70ºC
1.45 2.1 2.6 3.3 3.9 4.7 5.5 6.4
0.027 0.044 0.069 0.105 0.15 0.21 0.29 0.38
x C3 1- 0.027 = 0.684 1.45 - 0.027 0.465 0.368 0.280 0.227 0.176 0.136 0.103
0.9915 0.976 0.956 0.924 0.884 0.827 0.75 0.659
L V 0.6 0.4 1.5 x C3 0.3 , V F 0.4, Operating line intersects y x 0.3, Slope 1.5 L F y x z V V F 0.3 at x 0, y z 0.75 V 0.4 Find yc3 = 0.63 and xC3 = 0.062 Check with operating line: 0.63 1.5 .062 0.75 0.657 OK within accuracy of the graph.
0.63 0.062 10.2 , DePriester Chart T = 109ºC
2.D25. New Problem. 20% Methane and 80% n-butane V Tdrum .50 ºC , 0.40 , Find p drum F K A 1 zA K B 1 zB V 0 f V V F 1 KB 1 1 KA 1 F F Pick p drum
(Any pressure with K C1 Trial 1
Need lower p drum
0.541 0.055769 K C1 17.4
0.0159 , OK. Drum pressure = 1100 kPa
yC1 2.D26. New Problem. a)
b) Stage is equil.
V 1 16.4 .4 1 F 17.4 0.02645 0.4603
Can solve for L and V from M.B. 100 = F = V + L 45 Fz 0.8V 0.2162L Find: L = 59.95 and V = 40.05 y C3 0.8 K C3 3.700 x C3 0.2162
.2552 0.7838 These K values are at same T, P. Find these 2 K values on DePriester chart. Draw straight line between them. Extend to Tdrum , p drum . Find 10ºC, 160 kPa. 2.D27. New Problem. a.)
VP 191.97 mmHg b.)
log10 VP 6.853 1064.8 / T 233.01 Solve for T = 71.65ºC c.) Ptot 191.97 mm Hg [at boiling for pure component Ptot d.)
30 233.01 637.51 mm Hg
637.51 500 1.2750 1171.17
30 224.41 187.29 mm Hg
If K A & K B are known, two eqns. with 2 unknowns K A & y A
1 0.3746 1.2750 0.3746
y C5 K C5 x C5 1.2750 0.6946 0.8856 f.) Overall, M.B., F = L + V or 1 = L + V C5 : Fx F Lx Vy .75 0.6946 L + 0.8856 V Solve for L & V: L = 0.7099 & V = 0.2901 mol g.) Same as part f, except units are mol/min.
2.D28. New Problem.
F L From example 2-4, x H With h D
V MWv u perm 3600
C=4, MWv = 97.39 lbm/lbmole (Example 2-4)
From Example 2-4, K vertical
0.00314 lbmol 0.51 3000 hr
8.231 ft s [densities from Example 2-4]
lbmol lbm 97.39 h lbmol ft s lbm 8.231 3600 0.1958 3 s h ft 1530
20.27 ft 1 Use 5 20 or 5 22 ft drum. 2 2.D29. New Problem. The stream tables in Aspen Plus include a line stating the fraction vapor in a given stream. Change the feed pressure until the feed stream is all liquid (fraction vapor = 0). For the PengRobinson correlation the appropriate pressure is 74 atm. The feed mole fractions are: methane = 0.4569, propane = 0.3087, n-butane = 0.1441, i-butane = 0.0661, and n-pentane = 0.0242. b. At 74 atm, the Aspen Plus results are; L = 10169.84 kg/h = 201.636 kmol/h, V = 4830.16 kg/h = 228.098 kmol/h, and Tdrum = -40.22 oC.
Graph for problem 2.D30.
2.E1. New Problem. From Aspen Plus run with 1000 kmol/h at 1 bar, L = V = 500 kmol/h, WL = 9212.78 kg/h, WV = 13010.57 kg/h, liquid density = 916.14 kg/m3 , liquid avg MW = 18.43, vapor density = 0.85 kg/m3 , and vapor avg MW = 26.02, Tdrum = 94.1 oC, and Q = 6240.85 kW. The diameter of the vertical drum in meters (with u perm in ft/s) is D = <[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]>0.5 = 0.5 Flv = (WL/WV)[ρV/ ρL]0.5 = (9212.78/13010.57)[0.85/916.14]0.5 = 0.02157 Resulting Kvertical = 0.404299, and uperm = 13.2699 ft/s, and D = 1.16 m. Appropriate standard size would be used. Mole fractions isopropanol: liquid = 0.00975, vapor = 0.1903 b. Ran with feed at 9 bar and pdrum at 8.9 bar with V/F = 0.5. Obtain WL = 9155.07 kg/h, WV = 13068.27, density liquid = 836.89, density vapor = 6.37 kg/m3 D = <[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]>0.5 = 0.5 Flv = (WL/WV)[ρV/ ρL]0.5 = (9155.07/13068.27)[6.37/836.89]0.5 = 0.06112 Resulting Kvertical = .446199, uperm = 5.094885 ft/s, and D = 0.684 m. Thus, the method is feasible. c. Finding a pressure to match the diameter of the existing drum is trial and error. If we do a linear interpolation between the two simulations to find a pressure that will give us D = 1.0 m (if linear), we find p = 3.66. Running this simulation we obtain, WL = 9173.91 kg/h, WV = 13049.43, density liquid = 874.58, density vapor = 2.83 kg/m3, MWv = 26.10 D = <[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]>0.5 = 0.5 Flv = (WL/WV)[ρV/ ρL]0.5 = (9173.91/13049.43)[2.83/874.58]0.5 = 0.0400 Resulting Kvertical = .441162, uperm = 7.742851 ft/s, and D = 0.831 m. Plotting the curve of D versus pdrum and setting D = 1.0, we interpolate pdrum = 2.1 bar At pdrum = 2.1 bar simulation gives, WL = 9188.82 kg/h, WV = 13034.53, density liquid = 893.99 , density vapor = 1.69 kg/m3, MWv = 26.07. D = <[4(MWV) V]/[3600 π ρV uperm (1 m/3.281 ft)]>0.5 = 0.5 Flv = (WL/WV)[ρV/ ρL]0.5 = (9188.82/13034.53)[1.69/893.99]0.5 = 0.0307 Resulting Kvertical = .42933, uperm = 9.865175ft/s, and D = 0.953 m. This is reasonably close and will work OK. T drum = 115.42 oC, Q = 6630.39 kW,
Mole fractions isopropanol: liquid = 0.00861, vapor = 0.1914 In this case there is an advantage operating at a somewhat elevated pressure.
This problem was 2.D13 in the 2nd edition of SPE. a. Will show graphical solution as a binary flash distillation. Can also use R-R equation. To generate equil. data can use x C6 x C8 1.0, and yC6 yC8 K C6 x C6 K C8 x C8 1.0 Substitute for xC6
K C6 K C8 Pick T, find KC6 and KC8 (e.g. from DePriester charts), solve for xC6. Then yC6 = KC6xC6 T°C
125 120 110 100 90 80 66.5 Op Line Slope
4 3.7 3.0 2.37 1.8 1.4 1.0
1.0 .90 .68 .52 .37 .26 .17
0 .0357 .1379 .2595 .4406 .650 1.0
yC6 = KC6 xC6 0 .321 .141 .615 .793 .909 1.0
1.5 , Intersection y = x = z = 0.65.
See Figure. yC6 = 0.85 and xC6 = 0.52. Thus KC6 = .85/.52 = 1.63. This corresponds to T = 86°C = 359K
b. Follows Example 2-4.
1.0 90.38 g/mol 0.00307 g/ml ml atm RT 82.0575 359K mol K Now we can determine flow rates V V F .4 10, 000 4000 lbmol/h F pMW v
42.57 19135 .19135
10.3 ft. Use 10.5 ft.
L ranges from 3 × 10.5 = 31.5 ft to 5 × 10.5 = 52.5 ft. Note: This uPerm is at 85% of flood. If we want to operate at lower % flood (say 75%) have u Perm75% 0.75 0.85 u Perm85% 0.75 0.85 .63 5.56 Then at 75% of flood, ACs = 94.44 which is D = 10.96 or 11.0 ft. 2.F1
Benzene-toluene equilibrium is plotted in Figure 13-8 of Perry’s Chemical Engineers Handbook, 6th ed. 2.F2.
See Graph. Data is from Perry’s Chemical Engineers Handbook, 6th ed., p. 13-12.
Intercept Stage 2)
Intercept Stage 3)
Dew Pt. Calc. Want Try
Bubble Pt. At P = 250 kPa. Want Guess
Converge to T 124 C . This is a wide boiling feed. Tdrum must be lower than 95°C since that is feed temperature. First Trial: Guess Td,1 70 C : K1 7.8, K 2 1.07, K 3 Guess V F
0.5 . Rachford Rice Eq. 7.8 1 .517
V F .6 gives f .6
By linear interpolation
.0016 which is close enough for first
25 C . (Perry’s 6 ed; p. 3-127), and (Perry’s 6th ed; p. 3-138)
0.576 cal / (g C) 44
20 to 123 C, CpL3
25.34 kcal/(kmol C) .
65.89 kcal/(kmol C)
39.66 kcal/(kmol C) . (Himmelblau/Appendix E-7)
Cpv a bT cT 2 propane a = 16.26 b = 5.398 × 10-2 c = -3.134 × 10-5 -2 n-pentane a = 27.45 b = 8.148 × 10 c = -4.538 × 10-5 -3 **n-octane a = 8.163 b = 140.217 × 10 c = -44.127 × 10-6 ** Smith & Van Ness p. 106 Energy Balance: E(Td) = VHv + LhL – FhF = 0 Fh F 100 .577 25.34 .091 39.66 .392 65.89 95.25 297, 773 kcal/h Lh L
3597.4 16.26 5.398 10
6302.88 27.45 8.148 10
0.67 9895.3 8.163 140.217 10
E Tdrum Converge on For
60,101 Thus, Tdrum is too high. Tdrum
Fh F 297, 773; Lh L 90, 459; VH v Thus Tdrum must be very close to 57.3°C. x1 .136, x 2 .101, x 3 .762
y1 .328, y 2 .081, y 3 .041 V 51.3 kmol/h, L 48.7 kmol/h Note: With different data T drum may vary significantly. 2.F4.
z 2.5 .25 V V = 4 kmol/h, L = 6 kmol/h.
From the graph, x = 0.19 y = 0.34 Equilibrium is from NRTL on Aspen Plus.
2.G1. Used Peng-Robinson for hydrocarbons. Find Tdrum 33.13 C, L 34.82 and V 65.18 kmol/h In order ethylene, ethane, propane, propylene, n-butane, xi (yi) are: 0.0122 0.0748 , 0.0866 0.3005 , 0.3318 0.3781 , 0.0306 0.0404 , 0.5388 0.2062.
Used Peng-Robinson. Find Tdrum 30.11 C, L 31.348, V 68.66 kmol/h. In same order as 2.G1, are: 0.0189 0.1123 , 0.0906 0.3023 , 0.3255 0.3495 , 0.0402 0.0501 , 0.5248 0.1858 .
Compares to graph with
0.55 . Different equilibrium data.
2.G4. New Problem. COMP METHANE BUTANE PENTANE HEXANE V/F = 0.58354 2.G5.
x(I) 0.12053E-01 0.12978 0.29304 0.56513
y(I) 0.84824 0.78744E-01 0.47918E-01 0.25101E-01
New Problem. Used NRTL. T = 368.07, Q = 14889 kW, 1st liquid/total liquid = 0.4221, Comp Furfural Water Ethanol
Liquid 1, x1 0.630 0.346 0.0241
Liquid 2, x2 0.0226 0.965 0.0125
Vapor, y 0.0815 0.820 0.0989
2.G6. New Problem. Used Peng Robinson. Feed pressure = 10.6216 atm, Feed temperature = 81.14oC, V/F = 0.40001, Qdrum =0. Note there are very small differences in feed temperature with different versions of Aspen Plus. COMP METHANE BUTANE PENTANE HEXANE V/F = 0.40001
x(I) 0.000273 0.18015 0.51681 0.30276
y(I) 0.04959 0.47976 0.39979 0.07086
2.H1. New Problem. The spreadsheet with equations for problem 2.D16 is shown in Appendix B of Chapter 2. The spreadsheet with numbers for i-butane replacing n-butane is below. MC flash, HW 2.G.b., MC flash with ibutane K const. aT1 aT2 aT6 ap1 ap2 ap3 M -292860 0 8.2445 -0.8951 59.8465 0 iB -1166846 0 7.72668 0.92213 0 0 nPentane -1524891 0 7.33129 0.89143 0 0 nHex -1778901 0 6.96783 0.84634 0 0 p T deg R 509.688 psia 36.258 F 150 zM 0.5 z iB 0.1 z np 0.15 znhex V/F 0.602698586 0.25
guess KM KiB KnPen KnHex xM xib xnPen xnHex Sum RR M RR nB RRnP RRnHex sum RR
51.86751896 0.926804057 0.175621816 0.05400053 Use goal seek for cell B24 to = 1.0 change B9 0.015793905 0.104615105 0.29812276 0.581601672 1.000133443 0.803396766 -0.007657401 -0.2457659 -0.550194874 -0.000221409
2.H3. New Problem. Use the same spreadsheet as for problem 2H1, but with methane feed mole fraction = 0. Answer: V/F = 0.8625, xib 0.08596648 xnPen 0.203540261 xnHex 0.710481125 KiB 3.886544834 KnPen 1.264637936 KnHex 0.574940847 yib = xib Kib = .33411 and so forth
Chapter 3 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 3A7, 3A10, 3A11, 3C3, 3C4, 3D4, 3D8, 3G2. 3.A7.
Simultaneous solution is likely when one of the key variables can be found only from the energy balances. For example, if only 1 of x D , x B , D, B, FR A dist are given energy balances will be required. This is case for most of the simulation problems and for a few design problems. In some simulation problems the internal equations have to be solved also. a.
x D , x B , opt feed, Q Re b
x D , x B , opt feed, Q C x D , x B , opt feed, S (open steam), sat’d vapor steam All of above with fractional recoveries set instead of x D , x B D, x B , opt feed, L/D b. N, N F , col diameter, frac. recoveries both comp. N, N F , col diameter, FR A dist, Lo D N, N F , col diameter, FR A dist, QR N, N F , col diameter, FR A dist, QC N, N F , col diameter, x D , QC
N, N F , col diameter, S (sat’d steam), sat’d vapor steam, x D or x B Many other situations are possible [e.g., 2 feeds, side streams, intermediate condensers or reboilers etc.] 3.C1.
See solution to problem 3-D2.
See solution to problem 3-D3.
New Problem in 3rd Edition. Fmix
F1z1 F2 z 2 Fmix
Bx B (Mole frac. MVC)
Now solve like 1 Feed Column Fmix & z mix . From Eq. (3-3),
z mix x B Fmix kmol/h. xD xB B Fmix D kmol/h.
New Problem in 3rd Edition. See solution to 3D4, Part b.
Ftotal xD xB F2 1500 kg/h
1000 .60 500 0.10 1500
0.43333 0.0001 1500 0.85 0.0001
F1z1 F2 z 2 Ftotal
D 1500 764.62 735.38
Mass balance calculation is valid for parts a & b for problem 3G1. a)
Lo 3, Eq 3-14 QC 1 L0 / D D h D H1 D h D is a saturated liquid at x D 0.85 wt. frac. From Fig. 2-4, h D ~ 45 kcal/kg H1 is saturated vapor at x D
0.85, H1 ~ 310 kcal/kg
1 3 764.62 45 310
EB around column.
810, 497 kcal/hour
h F1 81 C, 60 wt% ethanol ~ 190kcal / kg; h F2 20 C, 10 wt% ethanol ~ 10kcal / kg h B (sat’d liquid – leaves equil contact, ~ 0 wt% ethanol) ~ 100 kcal/kg, Qcol = 0 (adiabatic)
B, x B =0.0001 Sat’d liqd Approximately x B ~ yV ~ x L . Thus h L
1838.45 640 Dh D
h B 100 . HV 735.38 100
640 kcal kg 2573.83 100
992, 763 kcal / h
34407.9 7353.8 190, 000 5000 992, 763
Column: mass bal: F + S = D + B (1) MVC: Fz + SyS Dx D Bx B (2) Note: yS
energy bal: Fh f
Condenser: mass bal. : V1
energy bal.: V1H1
Solve Eqs. (1) and (2) to get:
Fz Fx B Sx B xD xB
100 .05 100 .05 .6 .05
Note: Not Eq. (3-3). Solve Eqs. (4) and (5) to get:
Substitute Q C into Eq. (3):
From Figure 2-4: h F
F S D h B Fh F SH S 1 D h D H1 8, h D
163.6 92 100 8 36.4 65 408
External balances: F + C = B + D
F = 2000, C = 1000, z = .4, x C
h C sat 'd liquid
50, h B sat 'd liquid
L V B Lx N Vy Re b
For a total reboiler: x N
92, H D sat 'd vapor
617 (saturated vapor at y N 1 0.05 ) Fz Cx C Fy D Cy D Solve Eqs. (1) and (2) for B: B x B yD 800 1000 1600 800 Thus B 800 and D 2200 .05 .8 From Eq. (3),
804500 1532.4 kg/h 617 92
New Problem in 3rd Edition. F B D MVC Fz Bx B Dy D But given recoveries. Thus, use: Fz F,M (Frac Rec Methanol in distillate)
Fz F,W (Frac Rec water in bottom)
yD,M unknown, x B,W unknown.
Methanol 29.7 100 0.3 .99
If 99% methanol recovered in distillate, 1% is in bottoms 0.3 100 0.3 0.01
1.4 100 0.7 0.02 xi
2% water in distillate
29.7 1.4 31.1 kmol h 68.9 kmol h OK
Bx B,M B 0.3 68.9 0.00435
Reflux liquid is in equilibrium with vapor y D,M
From equilibrium data (Table 2-7) x M,0 ~ 0.893 (linear interpolation) E.B. Partial condenser: V1H1
35, 270 J [email protected] C 35, 270 kJ / kmol
choose MeOH reference 64.5ºC.
40, 656 J mol@100 C 40, 656 kJ/kmol
choose water reference 100ºC. The condenser is at 66.1ºC (linear interpolation Table 2-7).
y1,M CP,V,M T1 64.5
V1 is at T1 in equilibrium with y1,M
y1,W CP,V,W T1 100
0.914. From Table 2-7 T1 ~ 67.6 C
Assuming only constant & linear T term are important in CP,V eqs., C P,V methanol Tavg
42.93 0.08301 66.05
33.46 0.00688 83.8
67.6 100 83.8 C . 2 J 1000 mol kJ kJ 48.41 48.41 mol kmol 1000J kmol C
66.65 C . For water, Tavg
J kJ 34.04 o mol C kmol C 67.6 100 32.4 34.04
yD,M CP,V,M TD 64.5
CP,V,M TM,avg . Tavg,M
42.93 0.08301 65.3
CP,V,W TW,avg . TW,avg
Note λ terms dominate.
33.46 0.00688 83.05 66.1 100
yD,W CP,V,W TD 100
100 66.1 83.05 2 34.03
1829.5 Reflux liquid at 66.1ºC and x M,0
73.88 51.91 35534.3 0.893, x W,0 0.107
Reference MeOH 64.5ºC, water reference 100 ºC h 0 CPL,M x M,0 TM CPL,W x W,0 TW
Overall EB Fh F or QR
CPL,M 0.00435 99.2 64.5 99.2 64.5 2 13.53 60.06
From Table 2-7, TF
CPL,W 0.99565 99.2 100 75.86 0.1683 81.86
46.5 kJ kmo l 0.3, z W
CP,LM 0.03 78 64.5
78 64.5 2 71.25 and CP,L,M
805.4 kJ kmol Then
31.1 35534.3 1,105,116
Feed is saturated liquid at z M
0.00435 and x B,W
Interpolating in Table 2.7 TBot
kJ kmol C 149 kJ kmol
h B is saturated liquid with x B,M
75.86 0.1683 71.25
805.4 3, 424, 479 kJ h
Mass Balances: F = D + S + B, Fz Dx D Sx S Bx B Solving simultaneously, B = 76.4 kg/min, D = 13.6 kg/min. Condenser: QC V1 h 0 H1
From Figure 2-4, h 0
7.7 kcal/kg (x = .9, T = 20°C),
290 kcal/kg (y = .9, sat’d vapor).
Thus, Q C = 54.4 (7.7 – 290) = -15,357 kcal/min Overall Energy Balance: Fh F
Dh D ShS Bh B Fh F QC
.7, sat'd Liq'd ; h F
.01, sat'd Liq'd , h D
From Eq. (3-3), D = F Then B = F = 1502. Lo Condenser: V
76.4 99 z xB xD xB
h D H V D Lo D 1
With 99.9% nC5 have essentially pure nC5 . Thus, it is at its boiling point.
h D HV QC Overall:
11,369 998 4 Dh D
Distillate is at boiling point of pure nC5 is at boiling point of nC6
K C5 1.0 on DePriester Chart) = 35°C. Bottoms
Converting to °F: 35°C = 95°F, 67°C = 152.6°F, 30°C = 86°F. Note feed is obviously a subcooled liquid. Arbitrarily, pick a liquid at 0°F as reference. (This will not affect the result and other values can be used.) CPF x C5CPLC5 z C6 CPLC6
Distillate is almost pure nC5 . Liquid at 95°F
Bottoms is almost pure liquid nC6 at 152.6°F.
0.7 0.001 0.999 0.001
B F D 299.6 kmol/h Lo D D 2.8 700.4 Condenser: Lo
Only this reflux is condensed since product is a vapor. QC Lo where λ is for essentially pure n-pentane.
Btu 2.20462 lbmol lbmol 1 kmol Btu 1J J 49,154, 204.85 5.18176 1010 -4 h 9.486 10 Btu h 1966.1
From overall balance QR
Distillate is vapor at b.p. of pure n-pentane (35°C from DePriester chart, K C5 1.0 ) Bottoms is boiling n-hexane (67°C) Conversions: 35°C = 95°F - distillate & Feed and 67°C = 152.6°F - bottoms As reference, arbitrarily choose liquid at 0°F. Feed is subcooled liquid.
CPF hF Distillate H D
z C5CPLC5 z C6CPLC6 CPF TF 0 C5
43.3Btu lbmolo F
15,140.5 Btu lbmol
Bottoms is pure C6 @152.6 F
CPLC6 Tbot 0 kmol h
7889.4 Btu lbmol kmol h
2.20462 lbmol kmol
Btu 1J h 9.486 10-4 Btu
New Problem in 3rd Edition.
98% rec. E in distillate, 81% rec water in bot.
.98 90 0.6885 128.1 B Bottoms .02 90 .81 210
Vapor H1 y1 V1 L0 D
x 0 in equilibrium with y 0 , thus from equation data x 0 Entering vapor y1 (from graph) 0.61
L0 D 256.2 128.1 384.3 kmol h.
E.B. on PC. V1H1 Qc DHdist L0 h 0 . Can use Figure 2-4 by converting mole fracs to mass fracs. Basis 1 kmole. .6885 mol E MW 46 31.671 kgE Distillate
5.607 kgW 37.28 kg total Mass frac. E = 31.671 37.28 0.8496
.3115 mole W MW 18
7.02 kgW 35.08 kg total Mass frac E = 28.06 35.08 0.7999 .39 mole W 18
Liquid reflux L 0
7.65 kgE 34.1 total Mass frac E = 26.45 34.1 0.7757 ~ 310 kcal kg, H1 ~ 330 kcal kg, h 0 ~ 65 kcal kg 0.425 mole W 18
From Figure 2-4, Hdist
L 0 h 0 V1H1 128.1 364.3
2, 400,517 kcal h
kmol 37.28 kg kcal 310 256.2 34.1 65 hr kg kg
35.08 kg 330 2, 400,517 kcal hr kmol Qc DHdist Bh B
DHdist 1, 480, 426 kcal h.
To find Fh F and Bh B , need to convert mole frac to wt frac. Basis 1 kmol 30 mole % E: .3 mole 46 13.8 Feed
12.6 total 26.4 kg kmol 13.8 26.4 0.5227
70% W : .7 18 Mass frac E Bottoms
17.811 total 18.293 kg kmol 0.48162 18.28 0.0263
0.98953 mole 18 Mass frac E From Figure 2-4
kmol 18.29316 kg 97 kcal 26.4 kg kcal 300 70 h kmol kg kmol kg 1, 480, 426 305,525 554, 408 2, 400,517 3,632,069kcal h
3D9. New Problem 3rd Edition. B = (xD – z)/(xD – xB)F = [(0.9999 - 0.76)/(0.9999 – 0.00002)](500) = 120
Bottoms is almost pure water.
9.72 kcal mol 9720 kcal kmol 1.750 106 kcal h
2 atm × 101.3 kPa/atm = 202.6 kPa. Pentane Recovery: 0.995 Fz P
0.995 1000 0.55 0.9993
B = 1000 – 547.6333 = 452.3667 Since
Pentane Recovery Bot F z p , 1 .995 1000 0.55 452.3667
mol frac pentane
Distillate is essentially pure Pentane. Bottoms Pure in Hexane. From DePriester Chart K P 1@ p 202.6 kPa when Tdist 59.5 C
202.6 kPa when Tbot
For Total Condenser, Eq. (3-14)
h D pure pentane hD
kcal 59.5 25 kmol C
Lo D h D H1 D Tref choose Tref 25 C
assuming λ is independent of temperature
Btu 1 lbmol 0.252 kcal kcal 7680.196 lbmol 0.454 kmol Btu kmol kmol kcal Eq. (3-14) is QC 1 2.8 547.6333 6310.5 13,132, 288 h h kcal kcal h B pure hexane CPLC6 Tbot Tref 51.7 94 25 3567.3 kmol C kmol H1 1369.65 11369
Feed is a liquid at 65°C
CPC5 zC5 CPC6 zC6 TF Tref hF
547.6333 1369.65 QR
13, 692, 081 kcal/h.
Note that QC and Q R are relatively close. 3.E1.
Was 3.D8 in 2nd Edition.
0.990 M.B. F + S = D + B
100 kmol h z M 0.6
Since steam is pure, yS,M
Unknowns S, D, B, Need E.B.
Fh F QC SHS Pick as basis liquid at 0°C, h W Assume ideal mixtures.
0 (essentially steam table choice)
CPavg TP Tref where CPavg
Felder & Rouseau p. 637 CPW
0.07586 16.83 10 5 T kJ/mol
Know F, z M , x D , x B
3.1078 kJ/mol feed = 3107.8 kJ/kmol
Can also use steam table for water
H S is sat’d vapor steam 1 atm,
kJ 18.0 kg kg kmol
Steam Table F&R, p. 645 H S =48,168 kJ/kmol
h D is sat’d liquid at x D
0.99 . From Table 2-7, T = 64.6°C
CPavg 64.6 0 where CPavg
5.2479 kJ mol 5247.9 kJ kmol
h B : Since leaving an equilibrium stage, sat’d liqd. 2% MeOH Table 2-7, T = 96.4°C
CPavg 96.4 0 where CP avg
7.28509 kJ mol 7285.09 kJ mol
do EB around condenser
40.656 kJ mol kJ 0.99 35.27 0.01 40.656 35.324 35,323,86 kJ kmol mol 35,323.86 2.3 1 D 116,568.7D kJ h
Plug Q C & numbers into E.B.
116,568.7D 48,168S 5247.9D 7285.09B
or 310,780 + 48,168S = 121,816.6D + 7285.09B Solve simultaneously with 2 MB. 100 + S = D + B 60 + 0 = 0.99D + 0.02B One can use algebra or various computer packages. Obtain:
D = 56.33 kmol/h, B = 211.71 kmol/h
S 168.04 kmol/h, QC E2.
Was 3.D9 in 2nd Edition.
2 eq. 3 unknowns 1 L0 / D D h o Condenser: QC
Note Eq (3-14) not valid. For enthalpy pick reference pure liquid water 0°C and pure liquid methanol 0°C. Felder & Rouseau: CPMeOH
75.86 0.01683T at Tavg
Assuming distillate pure methanol, boils at 64.5°C
CPMeoh ,liq T Tref
4928 35270 J/mol
Overall Energy balance: F h F
4928.0 J mol 40,198 J/mol
141, 080D J/h where D is mol/h
Bottoms is essentially pure H 2 O at 100°C
7540 40656 J/mol
For feed. 60 mole % Methanol boils at 71.2°C (Table 2-7).
Now, Eqs are (1) F + S = D + B or 500,000 + S = D + B (2) Fz Dx D Bx B or (500,000) (.6) = 0.998D + 0.0013B (3) QC (4) Fh F
Lo D h D H1 or QC 141, 080D D Sh S QC Dh D Bh B or (500,000) (54137) + S (48196) 1
+ Q C = D(4928) + B (7540) Solve simultaneously: D = 298.98, B = 1245.5, S = 1044.2 kmol/h Q C = - 4.218 × 10+7 kJ/h 3.F1.
An enthalpy composition diagram is available on p. 272 of Perry’s Chemical Engineer’s Handbook, 3rd ed., 1950. Eq. (3-3)
0.79 0.004 25, 000 0.997 0.004
Note that N 2 mole fractions were used since N 2 is more volatile. B = F – D = 5211.5 From enthalpy comp. diag. h D
1 Lo D D h D H1 QR
0, H1 1350 kcal/kmol, h B 160, h F
5 19788.5 0 1350
133,572, 000 kcal/h
95, 030, 000 kcal/h
We will use the enthalpy composition diagram on p. 3-171 of Perry’s 6th edition or p. 3-158 of Perry’s 5th ed.Do for 1 kmol of feed: Conversion of feed from kg to moles. Basis 100 kg 30 kg NH 3 = 1.765 kmol 70 kg H 2O 3.888 Total 5.653 kmol Thus 1 kmol is 100/5.653 = 17.69 kg
Will work problem in weight fractions since data is presented that way. 95% recovery: (0.95) Fz = Dx D or, D = (.95) Fz / x D = (.95) (17.69) (.3)/(.98) = 5.15 kg. B = F – D = 12.54 kg
From diagram: h D
1 Lo D D h D H1
G1. a.) Using NRTL. QC b.) QC G2.
5562 kcal/kmol feed
7815 kcal/kmol of feed
New Problem in 3rd Edition. ASPENPlus. D = 988, L/D = 3, Peng Robinson, N
778,863 kcal/h, QR
1, 064,820 kcal/h, QR
20 (arbitrary values in Radfrac)
4.4426 107 Btu h, 4.9852 107 Btu h
SPE 3rd Ed. Solution Manual Chapter 4 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 4A6, 4A13, 4C10, 4C16, 4D6, 4D9, 4D13, 4D15, 4D18, 4E4, 4E5, 4H1 to 4H3. 4A1.
Point A: streams leaving stage 2 (L2, V2) Point B: vapor stream leaving stage 5 (V5) liquid stream leaving stage 4 (L4) Temp. of stage 2: know K y 2 / x 2 , can get T from temperature-composition graph or DePriester chart of K = f(T,p). Temp. in reboiler: same as above (reboiler is an equilibrium stage.)
a. Feed tray = .6, z = 0.51 (draw y = x line), yF =0.52, xF = 0.29. b. Two-phase feed. c. Higher
New Problem in 3rd Edition. Answer is a. See Table 11-3 and 11-4 for a partial list. New Problem in 3rd Edition. A. Answer is b B. Answer is a C. Answer is a D. Answer is a E. Answer is b F. Answer is a G. Answer is b If feed stage is non-optimum, the feed conditions can be changed to have an optimum feed location.
a. Use columns in parallel. Lower F to each column allows for higher L/D and may be sufficient for product specifications. b. Add a reboiler instead of steam injection. Slightly less stages required and adds 1 stage. c. Make the condenser a partial instead of a total condenser. Adds a stage. d. Stop removing side stream. Fewer stages are now required for the same separation. e. Remove the intermediate reboiler or condenser and use it at bottom (or top) of column. Fewer stages, but all energy at highest T (reboilers) or lowest T (condenser) for same separation. Many other ideas will be useful in certain cases. 4C7. Easiest proof is for a saturated liquid feed. Show point z, y D satisfies operating equation. Solution: Op. Eq.
y D D Fz Bx B Which is external mass balance.
Can do similar for enriching column for a saturated vapor feed. 4.C10. New Problem in 3rd Edition. If we consider λ, the latent heat per mole to be a positive quantity, then QR V . With CMO and a saturated liquid feed V V (1 L / D) D , and then
New Problem in 3rd Edition. Define a fictitious total feed FT , z T , h T
FT FT Intersection of top & bottom operating lines must occur at feed line for fictitious feed F T. (Draw a column with a single mixed feed to prove this.) This feed line goes through y x z T
with slope where
and H mix , h mix are saturated
vapor and liquid enthalpies at feed stage of column with mixed feed.
Given p, L/D, saturated liquid reflux, x D , x B
opt feed locations, z1 , z 2 , F1 , F2 , h F1 , h F2
zT z1 Plot top op line. Plot all 3 feed lines. Draw
line from point A to y = x =
op. line. Connect pts B & C to get q 2 Fbot. 2
where H mix & h mix are vapor and liquid enthalpies on feed stage of mixed column
FT Usual CMO assumption is λ >> latent heat effects in either vapor or liquid. H mix h F1 H mix h F2 Then q1 and q2 H mix h mix H mix h mix F1q1 F2 q 2 Thus q T if CMO is valid. FT 4D1.
a. Top op line: y Intersects y
b. Bottom op line: y
Plot – See diagram
V V V B 2 V V Intersects y = x = xB = 0.05 1 0.5 / 2 @y 1 x 0.683 this is convenient point to plot 32 c. See diagram for stages. Optimum feed stage is #2 above partial reboiler. 5 equilibrium stages + PR is more than sufficient.
d. Feed line goes from y = x = z = 0.55 to intersection of two operating lines. q Slope 1.0 or q 0.5 . q 1 This is a 2 phase feed which is ½ liquid & ½ vapor. 4D2. New Problem in 3rd Edition. Part a.
1.703 V V V F .37 From Table 2-1, at 84.1° C y .5089 H hF liquid at 20°C and 40 mole % ethanol. q H h The pressure in Figure 2-4 is very close to 1.0 atm, thus it can be used, but must convert to wt frac.
Basis 1kmol feed.
.4 kmole E .6 kmol Water From Figure 2-4
1.20 398 75 q 1.2 Slope 6 q 1 .2 40 mole % ethanol boils at 84.1°C (Table 2-1). Then if pick reference as saturated liquid at 40 mole % h F Cp,40%liq 20 84.1
C p vapor 120 84.1
Assume only 1st and 2nd terms in C P equations are significant. From Problem 2.D9 CPvapor .4 14.66 0.03758T .6 7.88 .0032T kcal/kmol T is C which simplifies to
C p dT is equal to CPvapor @ Tavg
84.1 120 2 102.05 . Then C Pv ,avg 398
kcal 1 kmol kmol 27.2 kg
See graph for feed lines.
a. Basis 1 mole feed. 0.4 moles EtOH × 46 = 18.4 kg EtOH 0.6 moles H2O × 18 = 10.8 kg H2O Total = 29.2 wt frac 18.4 / 29.2 0.63 wt frac EtOH Calculate all enthalpies at 0.63 wt frac. Hv = 395, HL = 65 (from Figure 2-4). hF is liquid at 200°C. Assume Cp,liq is not a function of T. Estimate, 46.1 23 h h L 60 C h 20 kcal C P ,liq .63 wt frac ~ 0.864 T 60 20 80 kg C Then h F
2.24 Hv hL 395 65 q 1 0.309 b. From Figure 2-4 at 50 wt% ethanol Hv = 446 and hL = 70. Since CMO is valid obtaining both enthalpies at 50% wt is OK. The feed is a liquid h F C P,liq TF Tref CP,liq 250 0
CP,liq CP,EtOH z EtOH CPw z w in Mole fractions Basis 100 kg solution 50 kg EtOH 46.07=1.085 kg/kgmole
50 kg W 18.016 2.775 kg moles Total 3.860 kg moles Avg M.W. 100 3.86 25.91 kg/kgmole Thus, zW = 0.719 and zE = 0.281 CP,liq 37.96 .281 18.0 .719 23.61 C P,liq in kcal kg C hF q
250 C kg C h F 446 228
0.6 is intersection.
L1 V2 Alternate Solution
For subcooled reflux, Then,
L1 D L1 D 1 D L1 1.222 0.55 V2 2.222
New Problem in 3rd Edition.
85 75 .6 100 0.4 0.1 B 0.9D Solve simultaneously. D 84.375 and B 90.625 kmol hr b) Feed 1. q1 1, vertical at y x z1 0.6 Feed 2.
60% vapor = 40% liquid q 2
Slope feed line Bottom Op. Line
0.4 2 3 through y
L V 1 x B . Through y
, Slope L V V Also intersects bot. op. line and Feed line 2. Do External Balances and Find D & B. Then V V/B B
L V B 271.875 At feed 2, L .4F L or L L 0.4F 271.875 40 231.875 V V 0.6F 181.25 60 241.25 L V 0.961 40 9.625 x 0, y 0.126 Plot Middle Op Line. 241.25
V Know that y x Also, L
xD V x D and gives through interaction Middle and Feed line 1.
231.875 75 156.875 and V
L V 156.875 241.25
c) See graph. Graph for 4D6.
4.D7*. a. Plot top op. line: slope
.9. Step off stages as shown on Figure.
2 , x y x B 0.13. Step off stages V (reboiler is an equil stage). Find y2 = 0.515. c. Total # stages = 8 + reboiler Optimum feed plate = 7 or 8 from top. Plot feed line. Goes through x = y = z = .3, and intersection of two operating lines. 9 q slope gives q = 0.692. 4 q 1
b. Plot bottom op. line:
4.D8*. The equilibrium data is plotted and shown in the figure. q 0.692 and q q 1 9 4
From the Solution to 4.D7c,
a. total reflux. Need 5 2/3 stages (from large graph) – 5.9 from small diagram shown. .9 .462 b. L V min 0.660 (see figure) .9 .236 L V min L D min 1.941 1 L V min c. In 4.D7, L D act
L Dact Multiplier Multiplier = 4/1.941 = 2.06
d. Operating lines are same as in Problem 4.D7. Start at bottom of column. Reboiler is an equilibrium contact. Then use E MV AB AC 0.75 (illustrated for the first real stage) Stage 1 is the optimum feed stage. 11 real stages plus a partial reboiler are sufficient.
New Problem in 3rd Edition. a) F1 F2 D B
Solve simultaneously, B 113.68, F1 b)
D 40, V L D 120 D V V 120 Saturated Liquid Feed L L F1 40 93.68 133.68, L V c) Top Op. Line – Normal: y Through y Bottom – Normal: y
x D , Slope 1 3, y intercept
L V 1 x B , through y
Also through intersection, F2 feed line and middle op. line.
Feed line F2 slope
0.11212 V 120 d)Opt. Feed F2 stage 1 from bottom, Opt feed F1 , Stage 2. 4 stages + PR more than sufficient. Also,
(or do around bottom) V V L V . Through intersection feed line F1 and top op. line.
4.D10*. Operating Line y
Thus, operating line is y = .8x + .192 y a. Equilibrium is x or x1 1 y
Start with y1 = .96 = x D Equilibrium:
.8x .192 .8 .9317 y2
.8x 2 .192 .8 .89476
b. Generate equilibrium data from: y
1 .76x .8 .7 .8756 .8042
Plot equilibrium curve and operating line. (See Figure). Slope = L/V = .8, y intercept (x = 0) = 0.192, y = x = x D = 0.96. Find x 6 = 0.660.
4.D11. a) Same as 4.D2 part g. q = 1.0668, slope feed line = 15.97. b) Top y L V x 1 L V x D goes through y = x = x D = 0.99
Feed line: Slope
y = (1-L/V) x D = (1-0.6969) 0.99 = 0.300
yV Lx Sy M,S Bx B But y M,S 0 (Pure steam) With CMO B L
y= 0, x = x B . Also goes through intersection of feed line and top op.line. Stages: Accuracy at top is not real high. (Expand diagram for more occupancy). As drawn opt. Feed = #6. Total = 9 is sufficient, c.
0.99 0.57 0.99 0 L V
Actual L/D is 3.12 × this value.
slope. Top op line goes throug y
0.2495 From Soln to 3.D9 or from graph. 1.169
Feed line is vertical at z = 0.6. Can also plot top and feed lines, and then find bottom from 2 points y 0, x x B & intersect top & feed . For accuracy – Use expanded portions near distillate & near bottoms. From Table 2-7 from (x = .95, y = .979) Draw straight line to (x = 1.0, y = 1.0) From (x = 0, y = 0) draw straight line to (x = 0.02, y = 0.134) or (x = 0.01, y = 0.067) Opt feed = # 9 from top. Need 13 equilibrium stages.
4.D13. New Problem in 3rd Edition. a.)
0.665 0.95 0.30 0.95
0.7808 D MIN V L 1 L V 0.5615 L L L D 1.5616 c.) 2.0 L D MIN 1.5616 , D V 1 L D 2.5616 L
0.3709 . Top operating line y Goes through y
x x B & intersection top operating line & feed line. Feed Line: Vertical (saturated liquid, q = 1). Through y x z 0.3 Plot & Step off stages. Optimal feed = 5 below PC. 6 + PC + PR more than sufficient. 0.85 0.025 d.) Slope bottom: See figure for parts c & d. L V 1.941 0.45 0.025 1 1 V B V L V 1.0625 . L V 1 0.941
Graph for problem 4.D13.
External M.B. S = B Sys = BxB . Since yS = 0 (pure water) xB = 0
Two approaches to answer. Common sense is all methanol leaks out and x MA
McCabe-Thiele diagram: This is enriching column with z horizontal feed line is at x
Bottom operating line
0 . Intersection top op. line and
0 , which is also a pinch point. Thus x M,d
4.D15. New Problem in 3rd Edition. Saturated liquid. q Top operating line y
1 3 0.6885 L V x
, feed line vertical @ z
L V 1 x B goes through y
And goes through interaction feed line and top operating line. See graph. Optimum feed is stage 2 below partial condenser. Partial condenser + Partial reboiler + 3 equilibrium stages are more than enough to obtain separation.
Graph for problem 4.D15.
Using 32°F = 0°C as reference T, h F
at feed conditions.
.4 11369 .6 13572 12691 Btu/lbmole For approx. temperature of feed stage, do bubble pt. calc. y1 1 K1x1 K1z1 Pick T = 48°C (~ 40% of way between boiling pts.) K C5 1.5, K C6 .54, K1x1 1.5 .4 .54 .6 K C5 Tnew =
Note : CP feed,liq Hv q
.99 Close enough.
46.9 is from Prob. 3-D6. 5721.8 12691 18412.8 Btu/lbmole
Note: h F is from Prob. 3.D6. 4.D17.
The vertical line at x
x D , goes through y
.9 =0.2 V 9 Plot Top. Step off 2 stages. Find x S ~ 0.81
0.81 is the withdrawal line.
Bot. Op. Line intersects Top at x
Also know it intersects feed line at x
D B S Don’t know D, B, or x B .
Feed enters as saturated vapor. Thus q
Bottoms leaves an equilibrium contact, it is saturated liquid L Do flow balances V F 100 V V 100 since S is removed as saturated liquid.
L 62.777 Fz Dx D Sx S
62.7777. L V 22.222 0.9
Plot. Op. line Step off stages. 9 is more than sufficient.
4.D18. New Problem in 3rd Edition. Feed F1 : z1
62.7777 100 15 0.81
0.6, saturated liquid, q 1, q / (q 1)
0.4, 80% vapor hence 20% liquid q q
14 q 1 .8 Part a.) Bottom operating line goes through point, y
Max L V to point intersection feed F2 line and equilibrium curve.
V B and V B 1.5 L
V 133.6 Check overall balance
180 133.6 46.4 180.0 OK
To find y D use MVC mass balance
Actual bottom op. line: y
Goes through y x x B 0.04 , Slope 5 3 2nd point y = 1, x = 0.616 (this was arbitrarily found by setting y = 1.) Plot bottom op. line Dy D F1 z1 L Top Op. line: yV F1z1 Lx Dy D . y x V V Goes through intersection feed line for F 2 and bot. op. line. Does NOT go through y x yD . Since D & F, passing streams, Point z1 , y D is on op. line.
Figure for 4D18 4.D19*.
B = 0. Then from external balance F = D + B must have D = F = 1000. Acetone balance becomes Fz Dx D or x D z 0.75 .
To predict x B need operating lines. Top: y
Bottom: L V 1.0 . Thus y = x is operating line. From Figure x B Feed line can be calculated but is not needed.
To use enthalpy composition diagram change to wt. fractions. Basis = 1 kg mole Distillate: Weight Fractions: Feed: Weight Fractions: Bottoms: Weight Fractions:
0.8 ETOH = (.8)(46.07) = 36,856 0.2 Water = (.2)(18.016) = 3.6032 Total = 40.459 EtOH = .911, Water = .089 0.32 (EtOH) = (.32)(46.07) = 14.7424 0.68 (W) = (.68)(18.016) = 12.25088 Total = 26.993 EtOH = .546, W = .454 0.04 EtOH = (.04)(46.07) = 1.8428 0.96W = (.96)(18.016) = 17.295 Total = 19.1378 EtOH = .0963, W = .9037
Condenser Energy Balance is V1H1
Dh D which can be solved for L o D .
From chart: h D 54 Kcal/kg and H1 285 Kcal/kg Need D in weight units. Convert feed to weight units. 100 kgmoles Ethanol: .32 46.07 1474.24 kg/hr hr Water: (100)(.68)(18.016) = 1225 kg/hr Total: F = 2699.328 kg/hr
2, 065,113 1489.98 54 285
Now do usual McCabe-Thiele analysis using molar units. Note L o D is the same in mass and molar units. L L L L D Top Operating Line: y x 1 x D and V V V 1 L D
.8, y int ercept x
V 6 Feed Line: Goes through y = x = z = .32 Weight fraction of feed = .546. Then, h f
Bottom Operating Line:
V V intersection top operating line and feed line.
1 x B . Goes through x
From Figure need about 8 equilibrium contacts including a reboiler. Stage 1 above reboiler is the optimum feed stage location.
4.D21. Feed 1: q1 Feed 2: q 2 Top:
0 , slope feed line = 0 0.9 , slope
x D , When x = 0, y
Since F1 is saturated vapor, V
V L 108 45.46 F1 F2 D 100 80 45.46 134.532
But B is saturated liquid. Check L L .9F2
L B 134.532 62.532 0.9 80
Draw top op line. Intersects with F2 feed line. Then draw bottom op line with slope
L V 1.3453 . Intersection bottom op & q. line gives x B Check
B Check External MB 180 F1
0.09 . 20 36 43.187 134.532
45.46 134.53 179.99 , OK
56 20.0 36.0 45.46 0.95 134.53 0.095 55.97 OK See McCabe-Thiele diagram: Optimum feed = 5, 7 equilibrium stages (6.65) more than sufficient. fraction ab/ac 0.65
4.D22*. Around top of column mass balances are: L D Solving,
xw V V V For pure entering water, x W 1.0 . With saturated liquid entering, L = C. Then from overall balance, V = D. Thus L/V = C/D = ¾ and D/V = 1.0. Operating equation becomes y 0.75x .92 .75 0.75x .17 Slope
V C and Lx Dy D
Note L/V ≠ C/D since C is subcooled. Let c = amount condensed. The energy required to heat stream W to the boiling point must come from this condensation. That is, H h c h hW C
H h L C c 1.1154C V D c D 0.1154C
In addition, C/D = ¾ or D/C = 4/3. L 1.1154C 1.1154
0.77 V D .1154C 4 3 .1154 4 3 .1154 This compares to L/V = 0.75 if entering water is a saturated liquid. Very little effect since λ is very large. L L D 3.25 0.7647 V 1 L D 4.25 Goes through
Through y x x D 0.05 and intersects top op line @ feed line Opt. Feed is #4 below partial condenser – see diagram. Need 6 equil stages + P.C. (an equil. contact) Note – Commercial columns usually operate much closer to minimum reflux ratio and have many more stages. b.)
c.) Total reflux 5 stages + PC sufficient or 4 3 4 equil contacts + PC = 5 3 4 eq. contacts ab 7.6mm where fraction 0.74 or .75 ac 10.3mm
a. 99.9% methanol is essentially pure. Pure MeOH boils 64.5°C. 1 fc Lo D L Eq. (4-66) 1 where f c CPL TBP Treflux V2 1 1 f c L o D For pure MeOH, CPL
0.07586 16.83 10 5 T , average (40 + 64.5)/2 = 52.25°C
0.084654 kJ gmole ,
0.084654 24.5 35.27
35.27 kJ mole , TBP
0.5454 or 2.59% more reflux with 24.5°C cooling!
V 1 L D 2.2 subcooling not important. a) 50% feed: q
1 F q 1 20 0.05 , Slope = L F 0.0476 20 20 q 1 1 20 1 1.05 35% feed: Saturated liquid – vertical feed line. Plot both feed lines. The one with lowest intersection point with equilibrium curve will normally control V B q
1 x B . Goes through y 1
x B with total reboiler L V 1 xB / L V
1 0.49892 0.1 0.6338 1.49892 Intersects feed line with 50% feed first. Middle operating line: Do mass balance around bottom of column. Mass balance intersects streams L & V (in column), F50% and B. yV L x F50% z50% Bx B y
V V Intersects bottom operating line & 50% feed line. @ x 0, y (F50 z 50 Bx B ) / V , Slope
External balances: 250 F1 F2 D B and F1z50 F2 z35 Dy D Bx B Find D = 103.333 and B = 146.666 moles/min Since V B 2.0034, V 293.96 and L V B 440.63 Then from q 50% : L
445.63 100 146.666
y intercept (x = 0), y
[ 100 .5 146.666 0.1 ] / 398.964 0.0885 Top Intersects feed line for 35% feed and middle op. line and goes through y x y D 0.85
actual change liquid
change at equilibrium eq. change
Figure for 4D26
5 , y x z .48 14 Bx B L Middle: V B L S , V y Bx B L x Sy s y s 0 or y x V V Does not intersect at y x x B or at y 0, x x B . Does intersect top op. line at feed line. Need another point. L V B V B 1 1.5 L L Bottom: 3, y x x B 0.08 y x 1 xB , V B V B .5 V V The steam is another feed to the column: Sat’d vapor q = 0, q/(q-1) = 0, y = x = z = ys = 0. Middle Op. Line intersects this steam at bottom op. line (see figure). F
This problem is a two-feed column with the lower feed (steam) input at a non-optimum feed stage. Otimum feed is 3rd above partial reboiler. Need 5.6 equilibrium stages plus PR.
This problem was 4.D35 in 2nd edition. L L Stripping Section: y x 1 xB, y x xB V V 4.D28.
0.02 . Feed line is vertical
V B 1 Op line y
L V 1.16 Overall Balance: 10,000 = F = D + B 6000 Fz Dy D Bx B D .695
0.865 . Intersection Op & feed lines is y D
6000 .695D .02D 200
8592.6 kgmoles/day, B 1407.4 . Need a use for impure distillate.
Figure for 4.D28
V V 4 Step off 3 stages on top op line. Find x S on middle op. line
0.76 . Point on top op line at x S
Solve simultaneously D = 50.125, B = 34.875 10 x6 = .1 In top:
L = 3D = 150.375 and V = L + D = 200.500
Since saturated liquid withdrawn, V´ = V = 200.500 and L´ = L – S = 150.375 – 15 = 135.375
Middle op. line slope Middle Op line: yV
Feed z = 0.6, 20% vapor = 80% liquid Feed line slope q q 1 0.8 0.4
11.4 45.1125 200.5
a. Subcooled Reflux:
Substituting in values, we have Step off two stages.
4 Lo 3 1 V1 Lo 3 L1 1.0 L2
L o V1 1 Lo 3 V1 1 4 54
Mixed feed to column: F S Solving for mixed feed, z M
h F (sat’d vapor), h FM
External Balances: F = D + B, Fz
Energy balance for mixed feed, Fh F Since H s
h F (sat’d vapor), and q FM
Bx B , Solving simultaneously D = 500 & B = 500
2000 (subcooled reflux), V
2000 500 1500 , V
Intersects Top Op. line at x S Plot Bot. from y x x B to intersection feed line and middle Step off stages (see figure). Need 4 8/9 equilibrium stages. b. Mass balances for mixed feed injection: V FM
2500 1500 1000 , V B 1000 500
4.D31*. Was problem 4.D36 in 2nd edition. Solution is trial & error. Need to pick L/V. Final answer shown in figure. L .63 .385 0.245 .389 V .63 0.63 L V L 0.389 .636 1 L V D 0.611 Note feed stage is not optimum.
Figure for problem 4D31 External Balance: F = B = 50, and Fz
Top operating Line through y Bottom through y
0.397 (tangent pinch)
0.1 and intersection feed and top operating lines.
L L F .25F, q 5 4, slope q q 1 5 Optimum feed is 3rd from bottom. Need 9 real stages plus partial condenser (see figure). c. From figure slope of bottom operating line L V 2.025 Feed:
Since saturated steam and CMO valid, B S L V Also have mass balances, S + F = B + D SyS Fz Bx B Dy D ys 0 Solve 3 eqs. simultaneously. S = 760 lbmoles/hr = 13,680 lb steam/hr.
Trial and Error, Feed: L
Figure for 4.D34 4-E1.
Find (L/V)min (see diagram) 0.95 0.613 L V min 0.95 0
External M.B. F = D + B, Fz or Eq. (4.3) D
0.6 0.025 0.95 0.025
V = L + D = 130.192
Top op line y 1st middle
1 L F x D normal . At x = 0, y = 0.4525
1 x B looks like usual bottom! V V Goes through y x x B , and intersection top & feed line. Slope
1.2906 130.192 At pump-around return, V V 130.192 L L P 208.030, L V V
208.030 130.192 1.5979
At pump-around removal, V V 130.192 , L L P L 168.030 Check at bottom L V B or 130.192 168.030 37.838 , OK L L Bottom Op line y x 1 x B , Same as first middle. V V Step off P.R. stage 1 & 2 above. x P is liquid from stage 2, x P 0.335 . Vertical line at x P is withdrawal line for pump-around and it is feed-line for return of pump-around. 2nd middle op line slope L V intersects x P withdrawal & feed line where bottom & 1st middle intersect.
Bx B V 32838 0.025
.335 0.0690 L L 208.030 208.030 Draw, 2nd middle – Step off stage 2 & start 3. 3 is op loc. for feed and where pump-around is returned. Need PR + 6 equil stages. (Actually 5 and a large fraction)
Bottom op. line: Since steam is saturated vapor S V and B L Thus, 1 S B L V 1.2 . Operating line goes through y 0, x
Middle op. line: V B Side L S or V L S B Side V y Bx B Side x side L x SyS Since
V V Side stream is removed as a saturated liquid so q = 1. Step off two stages (see figure) and find x side 0.0975
Find slope: V L Side B
S, B L Side Side B 1 0.4 1
Draw in the middle operating line. Step off 4 stages. Trial and error to find x D for final result).
4-E3*. This column has 4 sections. The exact shape is not known ahead of time. Plot top operating line L L D 1.86 L L 0.650, y x 1 xD V 1 L D 2.86 V V
xD .35 .8 .28, y x x D 0.8 V Step off 8 stages and find x S 0.495 yS . Feed line for this vapor is horizontal. Feed line for feed to column is vertical at z = 0.32. From figure the feed is injected below the liquid withdrawal and above vapor stream from intermediate reboiler. Can now calculate flows in each section of columns. Overall Balance: Fz Dx D F D x B
L V 1.955 (this is a check) To plot: From stage 8 draw line of slope L´/V´. From intersection of first intermediate operating line and feed line draw line of slope L″/V″. Draw line from intersection of second intermediate operating line with line y y s to y x x B 0.02 . Check if slope L V 1.955 . Optimum feed is 10th below condenser while vapor from intermediate reboiler is returned on 11 th stage. Need 12 ½ stages. Note: Small differences in stepping off stages may change column geometry.
4.E.4. New Problem in 3rd Edition. External Balance. a) F D B, Fz Dx D Bx B
L V B 148.6, L V 2.0 At feed, amount condensed = C = F/10 = 10 1 L L F C L F F 148.6 110 10 V V C 74.3 10 64.3
38.6 21.43 17.17 kgmoles hour
Envelope for top
x x D 0.9 L 0,y 1 xD V
1 L V x D . Slope
2 V B Now have somewhat redundant information. Can plot bottom. Intersection bottom and feed line should also be on middle. – Or use this pt to find middle or bottom op. line. From graph: Opt. Feed = #4. Need 6 stages + P.R.
4.E.5. New Problem in 3rd Edition. Bot. Op. Line: External M.B.: F
At feed stage: L 687.5
250 kmol day , B L
750 kmol day , V
937.50 750 1687.5 kmol day V
yD V V Goes through y x y D , and intersects feed and bot. op. lines. At side withdrawal: 687.5 L L , V S V or V 937.5 200 737.5 Op line by intermediate condenser: L D V S Dy D Sx S L L x Dy D V y Sx S or y x V V 687.5 Find from intersection L V op line @ y yS plus slope 0.932 or intersection 737.5 top op line and x x S At side stream feed point: L
Can draw, yint ercept
Since bottoms are very pure h B
Extra heat 189.4 Btu/lb S
Since y ≈ x, MW are same
Must vaporize material in column. extra heat MW 189.4 v S S MW 970.33
If super heat not included L V
2 , which is incorrect.
4-F3*. An approximate check is to compare molar latent heats of vaporization. Data is available in Perry’s and in Himmelblau. a. See Example 4-4. b. isopropanol λ = 159.35 cal/g. MW = 60.09, λ = 9.575 kcal/mole. Water λ = 9.72 kcal/mole. CMO is OK. c. CMO is not valid. AA 5.83 kcal / gmole, W 9.72 d. nC4. λ = 5.331 kcal/mole, MW = 58.12. λ = 0.0917 kcal/g nC5. λ = 6.16 kcal/mole, MW = 72.15. λ = 0.0854 kcal/g Constant mass overflow is closer than constant molar overflow. e. benzene. λ = 7.353 kcal/mole, MW = 78.11, λ = 0.0941 kcal/g toluene. λ = 8.00 kcal/mole, MW = 92.13, λ = 0.0868 kcal/g CMO is within about 6%. 4G1.
a*. Answer should be close, but not identical, to result obtained in Example 4-4.
Was 4.G4 in 2nd edition. Used Peng-Robinson. QC
44, 437,300 Btu/hr,
49,859, 400 Btu/hr.
27 (Total condenser is #1)
x D 0.9992 and x B 0.00187 . 4G3*. See answers to selected problems in back of book. 4.H.1. New Problem in 3rd Edition. Use VBA program in Appendix B of Chapter 4. xd 0.995 xb 0.011 F 250 z 0.4 L/D 3 q 0 feed stg 4 partial reboiler total condenser Ethanol-water Prob. 4H1. VLE 6th 5th 4th 3rd 2nd 1st constant -24.75 85.897 -118.03 82.079 -30.803 6.6048 0 yeqatxint 0.584177 yint 0.4 xint 0.2016667 stage x y 1 0.011 0.069033 2 0.039445 0.217358 3 0.112146 0.451872 4 0.227091 0.607193 5 0.477924 0.769849 6 0.694798 0.867005 7 0.824341 0.919132 8 0.893842 0.954863 9 0.941484 0.979257
0.974009 0.992216 0.991288 0.996557
4.H.2. New Problem in 3rd Edition. The VBA program is in the result is: xd 0.7 xb 0.0001 F 1000 L/D 6.94 q 0 partial reboiler total condenser Ethanol-water VLE 6th 5th 4th 3rd 2nd 1st -47.949 161.42 -212.43 138.68 -46.65 7.9322 yeqatxint 0.099219 yint 0.1 xint 0.0135447 stage x y Reflux rate too low Reflux rate too low
Appendix B of Chapter 4. With L/D = 6.94 z 0.1 feed stg 28 Prob. 4H3. constant 0
With L/D = 6.95 the result is given below. With feed stages below 85 the feed stage was too low. xd 0.7 xb 0.0001 F 1000 z 0.1 L/D 6.95 q 0 feed stg 85 partial reboiler total condenser Ethanol-water Prob. 4H3. VLE 6th 5th 4th 3rd 2nd 1st constant -47.949 161.42 -212.43 138.68 -46.65 7.9322 0 yeqatxint 0.100057 yint 0.1 xint 0.0136691 stage x y 1 0.0001 0.000793 2 0.000194 0.001538 3 0.000295 0.002338 4 0.000404 0.003197 5 0.000521 0.004117 6 0.000646 0.005102 7 0.000779 0.006154 8 0.000922 0.007277 9 0.001075 0.008472 10 0.001237 0.009742 11 0.00141 0.011089 12 0.001593 0.012515 13 0.001786 0.014021 14 0.001991 0.015608 15 0.002206 0.017276 16 0.002433 0.019025 17 0.00267 0.020853 18 0.002919 0.022758 19 0.003178 0.024739 20 0.003447 0.026791 21 0.003725 0.02891 22 0.004013 0.03109 23 0.004309 0.033327 24 0.004613 0.035613 25 0.004924 0.037941 26 0.00524 0.040302 27 0.005561 0.042689 28 0.005885 0.045091
29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78
0.006211 0.006538 0.006865 0.00719 0.007513 0.007831 0.008144 0.00845 0.00875 0.009041 0.009323 0.009595 0.009858 0.010109 0.01035 0.010579 0.010797 0.011004 0.0112 0.011384 0.011558 0.011721 0.011874 0.012018 0.012151 0.012276 0.012392 0.0125 0.0126 0.012693 0.012779 0.012858 0.012932 0.012999 0.013062 0.01312 0.013173 0.013222 0.013267 0.013308 0.013346 0.013381 0.013413 0.013442 0.013469 0.013494 0.013516 0.013537 0.013556 0.013573
0.0475 0.049907 0.052301 0.054674 0.057017 0.059321 0.061579 0.063782 0.065925 0.068002 0.070008 0.071938 0.07379 0.075561 0.077251 0.078857 0.08038 0.08182 0.083179 0.084459 0.085661 0.086787 0.087841 0.088826 0.089743 0.090598 0.091392 0.092129 0.092812 0.093445 0.09403 0.094571 0.09507 0.095531 0.095955 0.096346 0.096705 0.097036 0.09734 0.09762 0.097876 0.098112 0.098328 0.098526 0.098708 0.098874 0.099027 0.099167 0.099295 0.099412
79 80 81 82 83 84 85 86 87 88 89 90 91 92 93
0.013589 0.013604 0.013617 0.013629 0.013641 0.013651 0.01366 0.013665 0.013706 0.014018 0.016416 0.034534 0.155188 0.490181 0.646064
0.099519 0.099618 0.099707 0.09979 0.099865 0.099934 0.099997 0.100032 0.100305 0.102401 0.11824 0.223718 0.516574 0.652848 0.724878
4.H.3. New Problem in 3rd Edition. The Spreadsheet is: xd 0.7 xb 0.0001 F 1000 z 0.17 Multiplier 1.05 q 0.5 feed stg 17 partial reboiler total condenser Ethanol-water Problem 4.H4. VLE 6th 5th 4th 3rd 2nd 1st constant -47.949 161.42 -212.43 138.68 -46.65 7.9322 0 L/Dmin 1.687377 L/Vmin 0.62789 L/D 1.771746 L/V 0.639217 stage x y 1 0.0001 0.000793 2 0.000229 0.001812 3 0.000418 0.003309 4 0.000696 0.0055 5 0.001104 0.008697 6 0.001698 0.013331 7 0.002559 0.019994 8 0.003797 0.029452 9 0.005555 0.042644 10 0.008006 0.060585 11 0.01134 0.08415 12 0.015719 0.113686 13 0.021208 0.148522 14 0.027681 0.186647 15 0.034766 0.224911 16 0.041877 0.259918 17 0.048382 0.28916 18 0.057276 0.325158 19 0.113592 0.469947 20 0.340102 0.575494 21 0.505222 0.659654 22 0.636883 0.719124 Because the multiplier is close to 1.0, this answer is very sensitive to the data fit used. One solution to the coding is the following VBA program: Option Explicit
Sub McCabeThiele() ' Find minimum reflux ratio assuming it occurs at feed plate. Then ' L/D actual = L/D min times Multiplier. Steps off stages from the bottom up. ' Assumes that the feed stage is specified. Sheets("Sheet2").Select Range("A8", "G108").Clear Dim i, feedstage As Integer Dim D, B, xd, xb, F, z, q, LoverD, LoverV, x, y, xint, yint, yeq As Single Dim a6, a5, a4, a3, a2, a1, a0, L, V, LbaroverVbar, LoverDmin As Single Dim LoverVmin, LoverVdelta, Multiplier As Single ' Input values from spread sheet xd = Cells(1, 2).Value xb = Cells(1, 4).Value F = Cells(1, 6).Value z = Cells(1, 8).Value Multiplier = Cells(2, 2).Value q = Cells(2, 4).Value feedstage = Cells(2, 8).Value ' Fit of equilibrium data to 6th order polynomial to find y. a6 is multiplied ' by x to the 6th power. a6 = Cells(5, 1).Value a5 = Cells(5, 2).Value a4 = Cells(5, 3).Value a3 = Cells(5, 4).Value a2 = Cells(5, 5).Value a1 = Cells(5, 6).Value a0 = Cells(5, 7).Value ' Calculate intersection point of two operating lines and use this to find ' minimum L/D and L/V. Initialize LoverV = 1 LoverVdelta = 0.00001 Do LoverV = LoverV - LoverVdelta xint = ((-(q - 1) * (1 - LoverV) * xd) - z) / (((q - 1) * LoverV) - q) x = xint yint = LoverV * xint + (1 - LoverV) * xd ' Equilibrium y at value of x intersection. When yint=yeq, have minimum L/V and L/D. yeq = a6 * x ^ 6 + a5 * x ^ 5 + a4 * x ^ 4 + a3 * x ^ 3 + a2 * x ^ 2 + a1 * x + a0 Loop While yint 100 Then Cells(i + 7, 6).Value = "Too many stages" Exit Do End If Loop End Sub
SPE 3rd Ed. Solution Manual Chapter 5 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 5A15, 5C1, 5D1, 5D2, 5D9, 5D10, 5H1 to 5H5. Problems and solutions from the first edition that were not in the second edition are: 5D6, 5D8, 5D11-5D13, 5E1. 5.A6.
Ethane is less volatile than methane so it decreases toward distillate. At bottoms it is more volatile than propane and butane, so must decrease towards bottoms. Thus ethane concentrates within column. 5.A7.
1. c; 2. c; 3. a (saturated liquid feed); 4. c; 5. B
Pure LK cannot be withdrawn because LNK is present. Pure LNK can be removed at distillate if all LK is removed in side stream. However, recovery of LNK will be 100 Then Cells(i + 18, 2).Value = "Too many stages" Exit For End If ' Top operating line xA = yA / LoverV - ((1 / LoverV) - 1) * xAdist xB = yB / LoverV - ((1 / LoverV) - 1) * xBdist xC = yC / LoverV - ((1 / LoverV) - 1) * xCdist ' Test for calculations being done. Loop While yA liquid mole fraction (ethane is LK), have more ethane in vapor side stream. e. The separation of n-pentane and n-butane is much more difficult than between ethane and nbutane. Thus side stream purity is less. Also feed has lot more pentane than ethane, which makes side stream below feed less pure. 6.G.7. New Problem in 3rd edition. 1. Report the following values: Temperature of condenser = _373.28____ K. Temperature of reboiler = ___411.75___ K Qcondenser = _-829828_____cal/sec, Qreboiler = ____1012650_____cal/sec Distillate product mole fractions: M = 0.59998, E = 0.36184, NP = 0.038177, NB = 0.3087E -05 Bottoms product mole fractions: M= 0.2042E-04, E = 0.03816, NP = 0.46182, NB = 0.50000_ 2. Was the specified feed stage the optimum feed stage? Yes No X If no, the feed stage should be: a. closer to the condenser, b. closer to the reboiler. Answer a (Note: Do minimum number of simulations to answer these questions. Do not optimize.) 3. Which tray gives the largest column diameter with sieve trays when one uses the originally specified feed stage? Aspen Tray #_____18_____Column Diameter =___1.77____meters [Use the default values for number of passes (1), tray spacing (0.6096 m), minimum downcomer area (0.10), foaming factor (1), and over-design factor (1). Set the fractional approach to flooding at 0.7. Use the “Fair” design method for flooding.] 4. Which components in the original problem are the key components (label light and heavy keys) ______________LK = ethanol, HK = n-propanol_______________________
5. Change one specification in the operating conditions (keep N, feed location, feed flow, feed composition, feed pressure, feed temperature or fraction vaporized constant) to make methanol the light key and ethanol the heavy key. What operating parameter did you change, and what is its new value?_____D = 60____ Temperature of condenser = __368.66__ K, Temperature of reboiler = _404.23___ K Distillate product mole fractions: M = 0.97858, E = 0.021417, NP = 0.155 E-07, NB = 0.1 E-10_ Bottoms product mole fractions: M = 0.0091787, E = 0.27654, NP = 0.35714, NB = 0.35714__
Chapter 7 New Problems and new solutions are listed as new immediately after the solution number. These new problems are: 7.A1, 7.A4, 7.D2, 7.D10, 7.D11, 7.D14, 7.D21, 7.G1. 7.A1. New problem in 3rd edition. f. none of the above. 7.A.4. New problem in 3rd edition. a. estimate fractional recoveries nonkeys at total reflux. 7.C4.
K i x i, j 1 . Then substituting into Eq. (7-20), we have Vmin K i x i, j