Answers to Selected Problems

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, 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.D13. a. T_drum ~ 86°C, x_C6 = 0.52, y_C6 = 0.85

b. D = 10.96 feet. Use 11.0 feet

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°C, V/F = 0.293

2.F2. x₃ = 0.22, y₃ = 0.461

2.F3. T_drum ~ 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, Q_c = -13,357 kcal/min, Q_R = 3635.3 kcal/min

3.D6. B= 1502 lb mole/hr, D = 998 lb mole/hr, Q_c = - 45,385,050 BTU/hr, Q_R = 50,861,500 BTU/hr

3.F3. B = 5211.5 kmoles/hr, D = 19,788.5 kmoles/hr, Q_c = -133,572,000 kcal/hr, Q_R = 95,030,000 kcal/hr

3.F4. B = 12.54 kg/kmole feed, D = 5.15 kg/kmole feed, Q_c = - 5562 kcal/kmole feed, Q_R = 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. 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 3^rd above partial reboiler. Need ~ 5.5 equilibrium stages + partial reboiler

4.D30. a. N ~ 5 equilibrium stages

b.

4.D32. a. x_B = z = 0.4

b. x_D ~ 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. x_B ~ 0.058

4.E2. x_side = 0.0975, Trial-and-error to find x_D ~ 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 y_i are specified. Proceed as follows:

1. Pick T_guess, 2. Find K_i, 3. Calculate Σx_i = Σ(y_i/K_i), 4. If Σx_i = 1.0, are finished. 5. If Σx_i ≠ 1.0 ± ε then K_{ref, new} = K_{ref, current}/[Σ(y_i/K_i)_calculated], 6. Determine T_new from value of K_{ref, 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. N_min = 5.97,

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 4 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 use original Gilliland curve or N = 14.1 if use Liddle’s curve

Chapter 8

8.D1. Optimum feed for recycle stream is 8^th 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, α_{water-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. n_water/n_organic = 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. 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

Chapter 9

9.C2. a.

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

Chapter 10

10.D1. E_o = 0.73

10.D2. D = 12.35 feet

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

Closed: h_Δp,valve = (0.0287) v_o² inches, for v_o < 6.83 ft/sec

Open: h_Δp,valve = (0.00478) v_o² inches, for v_o > 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 α_avg = 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; x_B ~ 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.

Chapter 11

11.D1. a. (L/D)_min = 2.44

b. N_min = 23.1

c. N_equil = 36.3 + P.R.

d. N_actual = 50 stages + P.R.

e. $1,054,000 as of Sept. 2001

11.D2. Q_c = -3.39BTU/hr, Q_R = 3.423 × 10⁷ BTU/hr, A_cond = 2850 ft², A_Reb = 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. y_out = 0.1267, x_out = 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, y_1,C4 = 7.2 × 10^-6, y_1,C3 = 2.98 × 10^-4

12.D17. L = 41.1 kmoles/hr

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

b. Stripper: G (stream B) = 723.7 moles carrier gas/day, Y_out ≈ 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. y_out = 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, y_N+1 = 6.929 E-6

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

13.D22. a. 6 2/3 equilibrium stages, y_out = 0.0264

b. x_out = 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, x_out = 0.171 (mass fraction)

13.D32. m_E = 1.313

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

Chapter 14

14.D1. a. y_AE = 0.06, y_DE = 0.05, x_AR = 0.42, x_DR = 0.48

b. E = 214 kg/hr

14.D4. a. y_A = 0.115, y_w = 0.04, x_A = 0.23, x_w = 0.73

b. S = 85.7 kg/hr

14.D6. S = 5600, E_N = 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: 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

Chapter 15

15.D1. Average H_OG = 2.1, n_OG = 12.1, height = 25.4

15.D2. a. G′_flood = 0.75 lb/ft², D = 5.8 feet

b. H_G = 0.40, H_L = 0.47feet

15.D4. Height of stripping section = 9.8 feet, height of enriching section = 14.1 feet

15.D6. a. H_OG = 1.67 feet

15.D8. height = 26.2 feet

15.D10. a. n_OG = 4.6 feet

b. n_OG = 4.6 feet

15.D12. a. h = 4.59 feet

b. h = 1.98 feet, lowest y_out = 0.00081

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

b. E_MV = 0.47

Chapter 16

16D1. a. A = 2.80 × 10⁶ cm², F_p = 0.32, F_out = 0.68 kmole/hr

16D4. a. M = 1.0689

b. α_AB = 2.297 L/(kg atm), K_solv/t_ms = 5.7501 L/(m² day atm), K_A/t_ms = 2.503 kg/(m² day)

c. k = 10,091 L/(m² day)

16.D15. x_p = 0, x_r,out = 0.125, Fʹ_out= 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. q_max = 0.10456 g anthracene/g adsorbent, K_Ac = 2.104 liter/g anthracene.

17.D7. a. From t = 0 to 161.49 minutes c_out = 0. Then c_out = c_F = 0.01 until 1200 minutes

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

17.D10. The answers are in Example 17-4.

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

b. L = 4.576 m, t_br = 389 min