Chapter 2
2D1. a. y=0.77, x= 0.48
b. V=600 and L = 900
c. V/F = 0.25, y = 0.58
d. F = 37.5 kmole/hr
e. D = 1.705 feet. Use 2.0 feet. L ranges from 6 to 10 feet
f. V/F = 0.17
2.D2. Hint: Work backwards (start with stage 2).
a. (V/F)1 = 0.148
b. x1 = 0.51, y1 = 0.78, x2 = 0.25, y2 = 0.62
2.D8. V/F = 0.076, xmethane = 0.0077, xpropane = 0.0809
2.D9. Tdrum ~ 88.2°C, xE = 0.146, yE = 0.617, V = 326.9
2.D13. a. Tdrum ~ 86°C, xC6 = 0.52, yC6 = 0.85
b. D = 10.96 feet. Use 11.0 feet
2.D15. zethane = 0.4677, znC4 = 0.2957
2.D18. Tdrum ~ 65.6°C, V/F ~ 0.57
2.D19. Tdrum ~ 57°C, V/F = 0.293
2.F2. x3 = 0.22, y3 = 0.461
2.F3. Tdrum ~ 57.3°C, V/F = 0.513
Chapter 3
3.D2. L/D = 2.77
3.D3.
3.D5. B = 76.4 kg/min, D = 13.6 kg/min, Qc = -13,357 kcal/min, QR = 3635.3 kcal/min
3.D6. B= 1502 lb mole/hr, D = 998 lb mole/hr, Qc = - 45,385,050 BTU/hr, QR = 50,861,500 BTU/hr
3.F3. B = 5211.5 kmoles/hr, D = 19,788.5 kmoles/hr, Qc = -133,572,000 kcal/hr, QR = 95,030,000 kcal/hr
3.F4. B = 12.54 kg/kmole feed, D = 5.15 kg/kmole feed, Qc = - 5562 kcal/kmole feed, QR = 7815 kcal/kmole feed
Chapter 4
4.D3. a. slope = -2.54
b. q = 0.58
4.D4. a. slope = 0.6 and goes through y = x = z = 0.6
b. q = 0.6, slope = -1.5
c. q = -1/5, slope = 1/6
4.D5. L1/V2 = 0.55
4.D7. a. x5 = 0.515
b. y2 = 0.515. Note: it is an accident these values are the same
c. Optimum feed is seventh or eighth from the top; need 8 stages + partial reboiler. q = 0.692.
4.D8. a. Nmin ~5.67
b. (L/D)min = 1.941
c. Multiplier = 2.06
d. 11 real stages + partial reboiler
4.D10. a. y3 = 0.9078
b. x6 = 0.66
4.D16. q = 1.13
4.D19. xD = z = 0.75, xB ~ 0.01 to 0.02
4.D20. Optimum feed stage = first above partial reboiler Need ~8 equilibrium stages + partial reboiler
4.D22. y = (L/V)x + (D/V)yD - (W/V)xw = 0.75x + 0.17 intersects y = x = 0.68 (not at yD)
4.D23. L/V = 0.77
4.D27. Optimum feed is 3rd above partial reboiler. Need ~ 5.5 equilibrium stages + partial reboiler
4.D30. a. N ~ 5 equilibrium stages
b.
4.D32. a. xB = z = 0.4
b. xD ~ 0.95
4.D33. a. (L/D)min = 0.659
b. Optimum feed is third real stage from bottom; need 9 real stages + partial condenser
c. S = 760 lb moles/hr = 13,680 lb steam/hr
4.D36. L/D = 0.636
4.D37. Trial-and-error. xB ~ 0.058
4.E2. xside = 0.0975, Trial-and-error to find xD ~ 0.85
4.E3. Optimum feed is tenth below condenser, vapor from intermediate reboiler is returned at stage 11; need 12½ equilibrium stages
4.F3. a. See Example 4-4.
b. CMO is OK.
c. CMO not valid. Latent heat of acetic acid is 5.83 kcal/gmole compared to 9.72 for water.
d. n-butane λ = 5.331 kcal/gmole and n-pentane λ = 6.16. This 15% error is marginal. Constant mass overflow works better.
e. benzene λ = 7.353 kcal/gmole and toluene λ = 8.00. CMO is within ~ 6%
4G1. a.* See Example 4-4, part E.
4.G3. a. Optimum feed is eleventh below condenser; need 19.43 equilibrium contacts including the partial reboiler.
b. Optimum feed is eight from the top; need 21.9 equilibrium contacts, including the partial reboiler
Chapter 5
5.D3. a. D = 2217.8 and B = 7782.2 kg moles/day
b. Bottoms mole fractions: Methanol = 0.0006, Ethanol = 0.6011, n-Propanol = 0.2313, n-Butanol = 0.1670
5.D4. a. D = 402 and B = 598 kg mole/hr.
b. Distillate mole fractions: isopentane = 0.9851, n-hexane = 0.0149, n-C7 = 0. Bottoms mole fractions: isopentane = 0.0067, n-hexane = 0.4916, n-heptane = 0.5017
c. L = 1005, V = 1407, ,
Chapter 6
6.C1. For a dew point calculation p and yi are specified. Proceed as follows:
1. Pick Tguess, 2. Find Ki, 3. Calculate Σxi = Σ(yi/Ki), 4. If Σxi = 1.0, are finished. 5. If Σxi ≠ 1.0 ± ε then Kref, new = Kref, current/[Σ(yi/Ki)calculated], 6. Determine Tnew from value of Kref, new, and 7. Return to step 2.
6.D3. a. Pure n- octane K = 1.0 gives T = 174°C
b. Pure n-hexane K = 1.0 gives T = 110°C
6.D4. n-pentane x = 0.436, n-hexane x = 0.464
6.D5. T = -29°C
6.D6. T = 28.8°C
6.F1. T1 = 203.0°F, T2 = 212.5°F, T3 = 227.6°F, T4 = 248.7°F using data in Maxwell (1950) (see Table 2-2). The Exact answer will vary depending on the data used.
Chapter 7
7.D1. a. Nmin = 5.97,
b. (L/D)min = 1.75,
c. N = 24.6 (including reboiler) and Nfeed = 14 from top
7.D3. α = 1.287
7.D5. xB = 0.229
7.D6. Nmin = 10.8, (L/D)min = 1.75, N = 25.3 (including partial reboiler)
7.D8. a. (L/D)min = 22.83,
b. Nmin = 96.9,
c. N = 181.9. This separation would probably not be done by distillation
7.D12. a. Nmin = 10.47 and FRcumene, bot = 1.0
b. (L/D)min = 2.71
c. N = 20.24 (including partial reboiler) and optimum feed is stage 10 or 11
7.D13. (L/D)min = 0.2993 if α = 2.5 and (L/D)min = 0.3073 if α = 2.25. (L/D)min will be more sensitive to α for sharper separations
7.D15. N = 9.45 (including reboiler) and use stage 4 as the feed
7.D16. a. Nmin = 12.7
b. (L/D)min = 2.13
c. (L/D)actual = 2.4
7.D18 Part a. N = 13.2 if use original Gilliland curve or N = 14.1 if use Liddle’s curve
Chapter 8
8.D1. Optimum feed for recycle stream is 8th stage, opt. feed stage for fresh feed is ninth stage. Need 9 7/8 ~ 10 stages
8.D2. a. Butanol product = 3743.45 and water product = 1256.55 kmole/hr
b. Column 1: Optimum feed is stage 3 and need 3 stages + partial reboiler
Column 2: Need ~ 2/3 stage + partial reboiler
8.D8. Must convert wt frac to mole Frac, αwater-ether_in_ether_layer = 7.026
(L/D)min = 7.467, L/D = 11.2, Top stage is opt. feed and need ~4 3/5 equilibrium contacts
8.D10. a. T = 97.5°C
b. nwater/norganic = 10.81
8.E1. a. Column 1: Optimum feed is top stage. Need ~ 2 7/8 or 3 eq. contacts (includes P.R.)
Column 2: Optimum feed is top stage. Need 1 stage + P.R.
(L/V)2 = 2.63. There is a net flow from separator into column 2
8.F1. a. T = 95.0°C
b. nwater/norganic = 5.045
c. mole water condensed/mole nonane vaporized = 1.009
d. t = 99.9°C and nwater/norganic = 245.75
Chapter 9
9.C2. a.
9.D2. Wfinal = 35.2 and D = 64.8 moles. xD,avg = 0.86
9.D3. Wfinal = 39.7 and D = 60.3 kmoles. xD,avg = 0.85
9.D9. a. D = 3.45
b. xw,final,min = 0.21
9.D14. a. Tfinal = 99.7°C
b. Wfinal = 1.11 and D = 8.89 moles
c. nwater/norganic = 107.2
9.D16. a. (L0/D)initial = 0.47
b. (L0/D)final = 5.46
c. Wfinal = 5.79 and D = 4.21 kmoles
Chapter 10
10.D1. Eo = 0.73
10.D2. D = 12.35 feet
10.D4. At balance point: hΔp,valve = 1.34 inches of liquid
Closed: hΔp,valve = (0.0287) vo2 inches, for vo < 6.83 ft/sec
Open: hΔp,valve = (0.00478) vo2 inches, for vo > 16.73 ft/sec
10.D5. HETP = 0.31 m.
10.D6. a. For α = 2.315, HETP = 0.32 m
b. For α = 2.61, HETP = 0.37 m
c. For αavg = 2.46, HETP = 0.34 m
10.D11. a. D = 6.05 inches
b. D = 8.01 inches
c. D = 19.14 inches
10.D12. HETP = 2.5 feet; xB ~ 0.65
10.D16. D = 10 feet
10.D17. D = 15.4 feet
10.F3. Maximum diameter is 2.1 feet in enriching section. Probably use 2.5 feet since there is almost no cost penalty. Need 22 real stages plus partial reboiler. Height is approximately 36 feet.
10.F4. Maximum diameter is 2.1 feet in enriching section. Need ~ 12 feet of packing.
11.D1. a. (L/D)min = 2.44
b. Nmin = 23.1
c. Nequil = 36.3 + P.R.
d. Nactual = 50 stages + P.R.
e. $1,054,000 as of Sept. 2001
11.D2. Qc = -3.39BTU/hr, QR = 3.423 × 107 BTU/hr, Acond = 2850 ft2, AReb = 32,800 ft2, total cost = $811,000. Areas and costs are very sensitive to the values of U used.
Chapter 12
12.D5. HETP = 1.7 feet
12.D6. McCabe-Thiele gave 5 real stages, and Kremser gave 5.07 real stages. In practice, use 6 real stages
12.D10. yout = 0.1267, xout = 4.93 × 10-4
12.D11. m = 1.414. m = 1.2 is incorrect (L/mV = 1)
12.D12. Need 4 equilibrium stages
12.D15. N = 2.39, y1,C4 = 7.2 × 10-6, y1,C3 = 2.98 × 10-4
12.D17. L = 41.1 kmoles/hr
12.D23. a. Absorber: N = 8 equilibrium stages, Xout = 0.2614 mole ratio
b. Stripper: G (stream B) = 723.7 moles carrier gas/day, Yout ≈ 0.287 mole ratio
12.F1. L/G = 18.4, HETP = 1.52 feet
12.F2. T = 99.1F, T = 80.9F, T = 73.9F
Chapter 13
13.D4. yout = 0.0075, N = 33.6
13.D7. E = 639.6 kg/hr.
13.D9. N = 5.44, Recovery linoleic acid = 87.7 %
13.D11. N ~ 8.5 equilibrium stages
13.D15. a. Column 1, y1 = 0.00092693, yN+1 = 6.929 E-6
b. Column 2: R = 50.35, xN = 0.0183
13.D22. a. 6 2/3 equilibrium stages, yout = 0.0264
b. xout = 0.002
13.D24. Optimum feed is fourth stage and need 5.4 stages
13.D27. Need 8 equilibrium stages
13.D28. Recovery = 95.9%
13.D30. Solve problem in mass ratio units. Need ~ 5 1/8 equilibrium stages, xout = 0.171 (mass fraction)
13.D32. mE = 1.313
13.E1. ~ 93.5 % m-xylene recovery with 8 stages with feed on stage 4
Chapter 14
14.D1. a. yAE = 0.06, yDE = 0.05, xAR = 0.42, xDR = 0.48
b. E = 214 kg/hr
14.D4. a. yA = 0.115, yw = 0.04, xA = 0.23, xw = 0.73
b. S = 85.7 kg/hr
14.D6. S = 5600, EN = 6830, R1 = 770 kg/hr. Extract: 10.5 % acetic acid and 3.5 % water
Raffinate: 5 % acetic acid and 93 % water
14.D9. a. S = 2444 kg/hr
b. N = 2
14.D14. a. Exit raffinate 27.5 % acetic acid and 67.5 % water
b. Entering extract 13 % acetic acid and 0.0 % water
c. R1 = 655 kg/hr and Entering extract flow rate = 2155 kg/hr
14.D18. Extract: yoil = 0.238, ysolid = 0.0; Raffinate: xoil = 0.078, xsolid = 0.656
14.D19. Outlet Extract: yoil = 0.38, ysolid = 0.0; Outlet Raffinate: xoil = 0.026, xsolid = 0.66
14.D20. Stage 1 extract: yoil = 0.35, ysolid = 0.0; Stage 2 extract: yoil = 0.18, ysolid = 0.0; Stage 3 extract: yoil = 0.09, ysolid = 0.0; Stage 3 raffinate: xoil = 0.03, xsolid = 0.66
Chapter 15
15.D1. Average HOG = 2.1, nOG = 12.1, height = 25.4
15.D2. a. G′flood = 0.75 lb/ft2, D = 5.8 feet
b. HG = 0.40, HL = 0.47feet
15.D4. Height of stripping section = 9.8 feet, height of enriching section = 14.1 feet
15.D6. a. HOG = 1.67 feet
15.D8. height = 26.2 feet
15.D10. a. nOG = 4.6 feet
b. nOG = 4.6 feet
15.D12. a. h = 4.59 feet
b. h = 1.98 feet, lowest yout = 0.00081
15.D14. a. kxa = 1408.19, kya = 366.32
b. EMV = 0.47
Chapter 16
16D1. a. A = 2.80 × 106 cm2, Fp = 0.32, Fout = 0.68 kmole/hr
16D4. a. M = 1.0689
b. αAB = 2.297 L/(kg atm), Ksolv/tms = 5.7501 L/(m2 day atm), KA/tms = 2.503 kg/(m2 day)
c. k = 10,091 L/(m2 day)
16.D15. xp = 0, xr,out = 0.125, Fʹout= 80 kg/hr
16.G1. The solution is in Example 13.5-1 in Geankoplis (2003).
Chapter 17
17.D1. The answers are given in Example 17-1.
17.D2. qmax = 0.10456 g anthracene/g adsorbent, KAc = 2.104 liter/g anthracene.
17.D7. a. From t = 0 to 161.49 minutes cout = 0. Then cout = cF = 0.01 until 1200 minutes
b. For downflow (t = 0 is start of downflow), cout = cF = 0.01 from t = 0 to t= 3.496 minutes; From this time until t = 83.3 minutes cout = 0.019796 kgmole/ m3
17.D10. The answers are in Example 17-4.
17.D28. a. LMTZ,lab = 1.9774 cm
b. L = 4.576 m, tbr = 389 min