Answers to Selected Problems

b. V = 600.0 and L = 900.0

c. V/F = 0.25, y = 0.58

d. F = 37.5 kmol/h

e. D = 1.705 ft. Use 2.0 ft. L ranges from 6.0 to 10.0 ft.

f. V/F = 0.17

g. x = 0.756, V = 16.18 mol/h, L = 33.82 mol/h

2.D2. Hint: Work backward (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, x_methane = 0.0077, x_propane = 0.0809

2.D9. T_drum ∼ 88.2°C, x_E = 0.146, y_E = 0.617, V = 326.9

2.D15. z_ethane = 0.4677, z_nC4 = 0.2957

2.D18. T_drum ∼ 65.6°C, V/F ∼ 0.57

2.D19. T_drum ∼ 57.0°C, V/F = 0.293

2.F2. x₃ = 0.22, y₃ = 0.461

2.F3. T_drum ∼ 57.3°C, V/F = 0.513

3.D3. = 1532.4 kg/h

3.D5. a. B = 76.4 kg/min, D = 13.6 kg/min, Q_c = –13,357 kcal/min, Q_R = 3635.3 kcal/min

3.D6. B = 1502 lbmol/h, D = 998 lbmol/h, Q_c = –45,385,050 BTU/h, Q_R = 50,861,500 BTU/h

3.F1. B = 5211.5 kmol/h, D = 19,788.5 kmol/h, Q_c = –133,572,000 kcal/h, Q_R = 95,030,000 kcal/h

3.F2. B = 12.54 kg/kmol feed, D = 5.15 kg/kmol feed, Q_c = –5562 kcal/kmol feed, Q_R = 7815 kcal/kmol feed

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 eight stages + partial reboiler. q = 0.692.

4.D8. a. N_min ∼ 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. x_D = z = 0.75, x_B ∼ 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)y_D – (W/V)x_w = 0.75x + 0.17. Intersects y = x = 0.68 (not at y_D).

4.D23. L/V = 0.77

4.D27. Optimum feed is third above partial reboiler. Need ∼5.5 equilibrium stages + partial reboiler.

4.D30. a. N ∼ 5 equilibrium stages

b. /B = 20.0

4.D31. L/D = 0.636

4.D32. x_B = z = 0.4

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 = 760.0 lb mol/h = 13,680 lb steam/h

4.D34. Trial and error. x_B ∼ 0.058.

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/gmol compared to 9.72 for water.

d. n-butane λ = 5.331 kcal/gmol and n-pentane λ = 6.16. This 15.0% error is marginal. Constant mass overflow works better.

e. benzene λ = 7.353 kcal/gmol and toluene λ = 8.00. CMO is within ∼6.0%.

4.G1. 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.

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 kmol/h.

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, = 1605, = 1007 kg moles/h

5.D13. T = 28.8°C

b. (L/D)_min = 1.75

c. N = 24.6 (including reboiler) and N_feed = 14 from top

7.D3. α = 1.287

7.D5. x_B = 0.229

7.D6. N_min = 10.8, (L/D)_min = 1.75, N = 25.3 (including partial reboiler)

7.D8. a. (L/D)_min = 22.83

b. N_min = 96.9

c. N = 181.9. This separation would probably not be done by distillation.

7.D12. a. N_min = 10.47 and FR_cumene,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 3 as the feed.

7.D16. a. N_min = 12.7

b. (L/D)_min = 2.13

c. (L/D)_actual = 2.4

7.D18. Part a. N = 13.2 if using the original Gilliland curve or N = 14.1 if using Liddle’s curve.

b. Column 1: Optimum feed is stage three below condenser, and need 3 stages + partial reboiler.

Column 2: Need ∼2/3 stage + partial reboiler.

8.D8. Must convert weight fraction to mole fraction, α_{water-ether_in_ether_layer} = 7.026

(L/D)_min = 7.467, L/D = 11.2, top stage is optimum Feed, and need ∼4 3/5 equilibrium contacts.

8.D10. a. T = 97.5°C

b. n_water/n_organic = 10.81

8.F1. a. T = 95.0°C

b. n_water/n_organic = 5.045

c. mole water condensed/mole nonane vaporized = 1.009

d. T = 99.9°C and n_water/n_organic = 245.75

9.D2. W_final = 35.2 and D = 64.8 moles, x_D,avg = 0.86

9.D3. W_final = 39.7 and D = 60.3 kmoles, x_D,avg = 0.85

9.D9. a. D = 3.45

b. x_w,final,min = 0.21

9.D14. a. T_final = 99.7°C

b. W_final = 1.11 and D = 8.89 moles

c. n_water/n_organic = 107.2

9.D16. a. (L₀/D)_initial = 0.47

b. (L₀/D)_final = 5.46

c. W_final = 5.79 and D = 4.21 kmoles

9.D23. Fraction 1: W₁ = 1.09615, D₁ = 0.10384, and x_D1,avg = 0.4167

Fraction 2: W₂ = 0.9793, D₂ = 0.11685, and x_D2,avg = 0.3214

Fraction 3: W₃ = 0.8809, D₃ = 0.0984, and x_D3,avg = 0.3086

10.D2. D = 12.35 ft

10.D4. At balance point: h_Δp,valve = 1.34 in. of liquid

Closed: h_Δp,valve = (0.0287) v_o² in., for v_o < 6.83 ft/s

Open: h_Δp,valve = (0.00478) v_o² in., for v_o > 16.73 ft/s

10.D6. a. For α = 2.315, HETP = 0.32 m

b. For α = 2.61, HETP = 0.37 m

c. For α_avg = 2.46, HETP = 0.34 m

10.D9. D = 8.84 ft

10.D11. b. 5/8-in. packing, D = 6.05

c. D = 8.01 in.

10.D17. D = 15.4 ft

10.F3. Maximum diameter is 2.1 ft in the enriching section. Probably use 2.5 ft since there is almost no cost penalty. Need 22 real stages plus partial reboiler. Height is approximately 36.0 ft.

10.F4. Maximum diameter is 2.1 ft in the enriching section. Need ∼12.0 ft of packing.

b. N_min = 23.1

c. N_equil = 36.3 + P.R.

d. N_actual = 50 stages + P.R.

e. $1,354,000 as of 2013

11.D2. Q_c = –3.39 × 10⁷ BTU/h, A_cond = 2850.0 ft², A_Reb = 32,800 ft², total cost = $811,000. Areas and costs are very sensitive to the values of U used.

12.D11. m = 1.414, m = 1.2 is incorrect (L/mV = 1.0)

12.D12. Need four equilibrium stages.

12.D17. L = 41.1 kmol/h

12.D23. a. Absorber: N = 8 equilibrium stages, X_out = 0.2614 mole ratio

b. Stripper: G (stream B) = 723.7 mole carrier gas/day, Y_out ≈ 0.287 mole ratio

12.F1. L/G = 18.4, HETP = 1.52 ft

12.F2. T = 99.1°F, T = 80.9°F, T = 73.9°F

13.D7. E = 639.6 kg/h

13.D9. N = 5.44, Recovery linoleic acid = 87.7%

13.D11. N ∼ 8.5 equilibrium stages

13.D14. a. y_A = 0.115, y_w = 0.04, x_A = 0.23, x_w = 0.73

b. S = 85.7 kg/h

13.D15. a. Column 1, y₁ = 0.00092693, y_N+1 = 6.929, E-6

b. Column 2: R = 50.35, x_N = 0.0183

13.D21. a. y_AE = 0.06, y_DE = 0.05, x_AR = 0.42, x_DR = 0.48

b. E = 214 kg/h

13.D25. a. S = 2444 kg/h

b. N = 2

13.D26. S = 5600.0, E_N = 6830.0, R₁ = 770.0 kg/h. Extract: 10.5% acetic acid and 3.5% water Raffinate: 5.0% acetic acid and 93.0% water

13.D28. a. Exit raffinate 27.5% acetic acid and 67.5% water

b. Entering extract 13.0% acetic acid and 0.0% water

c. R₁ = 655 kg/h and entering extract flow rate = 2155 kg/h

13.D38. x_out = 0.0000275 and y_out = 0.0037. If your answer is x_out = 0.000102, which is incorrect, you have not been careful with your units.

13.E1. ∼93.5% m-xylene recovery with 8 stages with feed on stage 4.

b. x_out = 0.002

14.D4. Optimum feed is fourth stage and need 5.4 stages.

14.D7. Need 8 equilibrium stages.

14.D8. Recovery = 95.9%

14.D10. Solve problem in mass ratio units. Need ∼5 1/8 equilibrium stages, x_out = 0.171 (mass fraction).

14.D12. m_E = 1.313

14.D18. Extract: y_oil = 0.238, y_solid = 0.0; raffinate: x_oil = 0.078, x_solid = 0.656

14.D19. Outlet extract: y_oil = 0.38, y_solid = 0.0; outlet raffinate: x_oil = 0.026, x_solid = 0.66

14.D20. Stage 1 extract: y_oil = 0.35, y_solid = 0.0; stage 2 extract: y_oil = 0.18, y_solid = 0.0; stage 3 extract: y_oil = 0.09, y_solid = 0.0; stage 3 raffinate: x_oil = 0.03, x_solid = 0.66

15.D12. At 298.16 K, D_AB = 1.114 m²/s. Calculate δ = 0.000115282 m, v_y,avg = 0.04338 m/s, Re = 19.966, and flow is laminar. This is a long residence time with Re < 20, so there are no ripples. Sh_avg = 3.41 and k_avg = 3.295 × 10^–5 m/s, and 0.000168 kg/s carbon dioxide are absorbed.

15.D18. Answers are compared in Example 15-5. Obviously, additional intervals can be added for more accuracy.

15.D19. a. At t = 10,000 s, obtain following values of C/C₀:

15.D26. Sc_V = 0.355

16.D2. a. G′_flood = 0.75 lb/ft², D = 5.8 ft

b. H_G = 0.40, H_L = 0.47 ft

16.D4. Height of stripping section = 9.8 ft, height of enriching section = 14.1 ft

16.D6. a. H_OG = 0.504 m

16.D8. height = 26.2 ft

16.D10. a. n_OG = 4.6 ft

b. n_OG = 4.6 ft

16.D12. a. h = 4.59 ft

b. h = 1.98 ft, lowest y_out = 0.00081

16.D14. a. k_xa = 1408.19, k_ya = 366.32

b. E_MV = 0.47

16.D21. average particle diameter = 0.2524 mm, k_C = 5.628 × 10^–5 m/s, k_D = 0.001905 m/s, E_MD = 0.794

b. 842.2 kg

c. 120.4 kg

17.D3. d. Q = 135.4 kW

17.D6. a. 666.7 g methanol added

b. 78.05 g KCl crystallize

17.D14. Volume = 8.889 m³ and W_seed = 0.7758 kg/h

17.D18. 0.7989 × 10^–4 mol/L

17.H1. a. G = 7.394 × 10^–8 m/s and L_median = 0.00144 m

18.D4. a. M = 1.069

b. α′_AB = 2.29 atm^–1, K′_solv/t_ms = 0.0665 g/(m²·s·atm), K′_A/t_ms = 0.0665 g/(m²s)

c. k = 0.000117 m/s

18.D15. a. x_p = 0, x_r,out = 0.125, F′_out = 80 kg/h

18.H2. The solution is in Example 13.5-1 in Geankoplis (2003).

19.D2. q_max = 0.10456 g anthracene/g adsorbent, K_Ac = 2.104 L/g anthracene.

19.D7. a. From t = 0 to 161.49 min c_out = 0. Then c_out = c_F = 0.01 until 1200.0 min.

b. For downflow (t = 0 is start of downflow), c_out = c_F = 0.01 from t = 0 to t = 3.496 min; from this time until t = 83.3 min, c_out = 0.019796 kmol/m³.

19.D10. The answers are in Example 19-4.

19.D28. a. L_MTZ,lab = 1.9774 cm

b. L = 4.576 m, t_br = 389 min

19.D34. Switching time = 5.0248 min and D/F = 1.3115