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- 1. 13 CHAPTER 5 INITIAL MATERIAL BALANCE
- 2. 14
- 3. 15 5.1Assumptions : 1. In KA oil, K:A=2:1 2. Purity of cyclohexanone(K) in KA oil is 99.8% 3. Purity of cyclohexanol(A) in KA oil is 98% 4. In feed, HNO3 : KA oil = 3:1 and 60% HNO3 is used. 5. Acetic acid, acetaldehyde, acetone, ethyl acetate are present as impurity in KA oil and each has 25% mole fraction in the impurity. 6. Single pass conversion in both reactors is 12%. 7. 10% excess air and 10% excess water are supplied to the NOx bleacher, HNO3 absorber respectively. 8. 90% water is evaporated in the concentrator. 9. 0.45 mol% of inert, 0.7 mol% of adipic acid in the purge stream. 10. Amount of recycle = 90% of KA oil. Basis: 100 Kmol/h of KA oil (cyclohexanone-cyclohexanol) in feed. Now, in KA oil Amount of actual cyclohexanone = 100 x (2/3) x 0.998 kmol/h = 66.53666 kmol/h Amount of actual cyclohexanol = 100 x (1/3) x 0.98 kmol/h = 32.6634 kmol/h Total impurity = 100 – (66.53666 + 32.6634) kmol/h = 0.79994 kmol/h Since, acetic acid, acetaldehyde, acetone, ethyl acetate are present as impurity in KA oil and each has 25% mole fraction in the impurity, So amount of each component = 0.79994 x 0.25 kmol/h = 0.199985 kmol/h In feed, HNO3 : KA oil = 3:1 Amount of HNO3 in feed = 3 x 100 kmol/h = 300 kmol/h But, it is 60% concentrated, so actual amt. of HNO3 = 180 kmol/h Amount of water = 120 kmol/h
- 4. 16 5.2 balance over mixer M1 M1 Fig 5.1: Balance over mixer Two reactions are taking place in the reactors : HNO3 + CH3COOC2H5 C2H5NO3 + CH3COOH----> 5.1 HNO3 + CH3COCH3 CH3NO2 + CH3COOH ---->5.2 Table 5.1.a: Composition of feed stream to the mixer(KA oil + HNO3 solution) Components Kmol/h Kg/h C5H10CO (K) 66.53666 6530.5771 C6H11OH (A) 32.6634 3271.2395 HNO3 180 11341.8 H2O 120 2161.8336 Impurity: Acetic acid Acetaldehyde Ethyl acetate Acetone 0.199985 0.199985 0.199985 0.199985 12.0095 8.8093 17.6206 11.6151 Total 400 23355.5047 ≈23355.5 Mixer KA oil = 100 kmol/h HNO3 soln = 300 kmol/h
- 5. 17 Table 5.1.b: Composition of output stream (M1) Of mixer Components Kmol/h Mole fraction Kg/h C5H10CO (K) 66.53666 0.1663 6530.5771 C6H11OH (A) 32.6634 0.0817 3271.2395 HNO3 179.6003 0.4490 11316.59789 H2O 120 0.3 2161.8336 Inert: Acetaldehyde Acetic acid CH3NO2 C2H5NO3 0.199985 0.599985 0.199985 0.199985 0.003 8.8093 35.9973 12.1991 18.1896 Total 400 1 23355.45239 ≈23355.5 Table 5.1.c : Material balance over mixer Input (kg/h) Output (kg/h) 23355.5 23355.5 5.3. calculation for recycle, purge stream: I=0.003 M2 I=0 ML Recycle (R) Fig 5.2: Recycle, Purge stream We know, amount of inert in the feed = amount of inert in the purge stream So, 400 x 0.003 = P x 0.0045 Reactor Concentrator Crystallizer Centrifuge M1= 400 kmol/h Crystal I=0 Purge(P),I=0.0045 H2O = 0.3 H2O = 0.25
- 6. 18 P = ( 400 x 0.003)/ 0.0045 = 266.64667 kmol/h Again, amount of recycle = 90% of KA oil R = 0.90 x 100 = 90 kmol/h From the fig 4.2, ML = R+P ML = (90 + 266.64667) kmol/h = 356.64667 kmol/h Let, in the recycle stream (R), x, y, z, z1 are the mole fractions of cyclohexanone (K), cyclohexanol(A), nitric acid, water respectively. Now, x + y + z + z1 + 0.007 + 0.0045 = 1 x + y + z + z1 = 0.9885 ------------------------> eqn 5.3 From fig 4.2, M2 = M1 + R = 400 + 90 = 490 kmol/h Doing H2O balance, M2 x 0.25 = (M1 x 0.3) + (R x z1) 490 x 0.25 = (400 x 0.3) + (90 x z1) z1 = (122.5 – 120)/ 90 = 0.0278 Mole fraction of water in recycle stream, z1 = 0.0278 5.4 balance over reactor1 Fig 5.3: balance over reactor 1 Reactor 1 M2 = 400 kmol/h H2O = 0.25 B
- 7. 19 Table 5.2.a: Input stream(M2) of reactor 1 Component Kmol/h C5H10CO (K) 66.53666 + 90x C6H11OH (A) 32.6634 + 90y HNO3 179.6003 + 90z H2O 120 + (90 x 0.0278) = 122.5002 Inert 1.19991 + (0.0045x90) = 1.60491 Adipic acid 0.63 Here, C5H10CO + 1.5 HNO3 (CH2)4(COOH)2 + 0.75 N2O + 0.75 H2O---->5.4 Per pass conversion is 12%. So, amount of cyclohexanone reacted = 0.12 x (66.53666 + 90x) = 7.98440 + 10.8x kmol/h Amount of HNO3 reacted = (7.98440 + 10.8x) x 1.5 = 11.9766 + 16.2x kmol/h Table 5.2.b: Output stream(B) of reactor 1 Components Kmol/h C5H10CO (K) 58.55226 + 79.2 C6H11OH (A) 32.6634 + 90y HNO3 167.62343 + 90z -16.2x H2O 128.4885 + 8.1x Adipic acid 8.6144 + 10.8x N2O 5.9883 + 8.1x Inert 1.60491 5.5 balance for reactor 2 B1 C Fig 5.4: Balance over reactor 2 Reactor 2
- 8. 20 Table 5.3.a: Input stream(B1) of reactor 2 Components Kmol/h C5H10CO (K) 58.55226 + 79.2x C6H11OH (A) 32.6634 + 90y HNO3 167.62343 + 90z -16.2x H2O 128.4885 + 8.1x Adipic acid 8.6144 + 10.8x Inert 1.60491 In reactor 2, C6H11OH + 2HNO3 (CH2)4(COOH)2 + N2O + 2 H2O---->5.5 Per pass conversion is 12%. So, amount of cyclohexanol reacted = 0.12 x (32.6634 + 90y) = 3.91961+10.8y kmol/h Amount of HNO3 reacted = (3.91961+10.8y) x 2 = 7.83922+21.6 kmol/h Table 5.3.b: Output stream(C) of reactor 2 Components Kmol/h C5H10CO (K) 58.55226 + 79.2x C6H11OH (A) 28.74379 + 79.2y HNO3 159.78421+90z-16.2x-21.6y H2O 136.32772+8.1x+21.6y Adipic acid 12.53401+10.8x+10.8y N2O 3.91961 + 10.8y Inert 1.60491
- 9. 21 B1 C1 Fig 5.5: Balance for cyclohexanol, cyclohexanone and nitric acid Doing cyclohexanone balance over envelop 1: 58.55226 + 79.2x = ML. x = (90+266.64667). x X = 0.2110 Doing cyclohexanol balance over envelop 1: 28.74379 + 79.2y = ML. y = (90+266.64667). y Y = 0.1036 From eqn 5.3, we get x + y + z + z1 = 0.9885 0.2110 + 0.1036 + z + 0.0278 = 0.9885 z = 0.6461 Since mother liq., recycle and purging stream have same composition, so, mole fractions of various components in these streams are: Reactor 2 Concentrator Crystallizer Centrifuge Crystal K,A=0 H2O Mother liq.(ML) K=x,A=y Envelop1
- 10. 22 Table 5.4: components in mother liq. Component Mole fraction Cyclohexanone (K), x Cyclohexanol (A), y Nitric acid, z Water, z1 Inert Adipic acid 0.2110 0.1036 0.6461 0.0278 0.0045 0.007 Now putting these values in table 4.2.a, table 4.2.b, table 4.3.a, table 4.3.b, we get: Table 5.5.a: Actual composition of input stream of reactor 1 Component Kmol/h Kg/h C5H10CO (K) 85.52666 8394.44168 C6H11OH (A) 41.9874 4205.03811 HNO3 237.74903 14980.56638 H2O 122.5002 2206.875403 Inert 1.60491 100.58766 Adipic acid 0.63 92.0682 Total 489.9982 29979.57742 Table 5.5.b: Actual composition of output stream of reactor 1 Component Kmol/h Kg/h C5H10CO (K) 75.26346 7387.10860 C6H11OH (A) 41.9874 4205.03811 HNO3 222.35423 14010.54003 H2O 130.1976 2345.54622 Adipic acid 10.8932 1591.93225
- 11. 23 N2O 7.6974 338.78567 Inert 1.60491 100.58766 Total 489.9982 29979.53849 Table 5.5.c: Material balance over reactor 1 Input (kg/h) Output(kg/h) 29979.5 29979.5 Table 5.6.a: Actual composition of input stream of reactor 2 Component Kmol/h Kg/h C5H10CO (K) 75.26346 7387.10860 C6H11OH (A) 41.9874 4205.03811 HNO3 222.35423 14010.54003 H2O 130.1976 2345.54622 Adipic acid 10.8932 1591.93225 Inert 1.60491 100.58766 Total 482.3008 29640.75282 Table 5.6.b: Actual composition of output stream of rector 2 Component Kmol/h Kg/h C5H10CO (K) 75.26346 7387.10860 C6H11OH (A) 36.94891 3700.43334 HNO3 212.27725 13375.00668 H2O 140.27458 2527.08584 Adipic acid 15.93169 2328.25718 N2O 5.03849 222.21252 Inert 1.60491 100.58765 Total 487.33929 29640.69181≈ 29640.7
- 12. 24 Table 5.6.c: Material balance over reactor 2 Input (kg/h) Output(kg/h) 29640.7 29640.7 5.6 balance over NOx bleacher Fig 5.6: Balance over NOx bleacher Amount of N2O coming to the bleacher = B2 + C2 = (5.03849+7.69740) kmol/h = 12.73589 kmol/h Reaction which is taking place in the bleacher is: N2O + 1.5 O2 2NO2 ---->5.6 So, 1kmol N2O required = 1.5 kmol O2 12.73589 kmol N2O required = 1.5 x 12.73589 = 19.103835 kmol O2 0.21 kmol O2 = 1 kmol air 19.103835 kmol O2 = (19.103835/0.21) = 90.97064 kmol air 10% excess air (79% N2 and 21% O2) is supplied for complete conversion. So, actual amount of air supplied = (1.1 x 90.97064) = 100.0677 kmol/h NOx Bleacher D N2O AIR
- 13. 25 Table 5.7.a: Composition of input stream of NOx bleacher Components Kmol/h Kg/h N2 79.05348 2213.49744 O2 21.01422 672.45504 N2O 12.73589 560.54473 Total 112.80359 3446.49721 ≈3446.5 Table 5.7.b: Composition of output stream(D) of NOx bleacher Components Kmol/h Kg/h N2 79.05348 2213.49744 O2 1.91039 61.13248 NO2 25.47178 1171.84198 Total 106.43565 3446.4719 ≈3446.5 5.7. balance over HNO3 absorber D1 H2O Fig 5.7 : Balance over HNO3 Absorber HNO3 absorber D D2
- 14. 26 Reaction: 3 NO2 + H2O 2HNO3 + NO ---->5.7 3 kmol NO2 required = 1 kmol H2O So, 25.47178 kmol NO2 required = (25.47178/3) = 8.49059 kmol H2O 10% excess water is supplied for complete conversion. Actual H2O supplied = (1.1 x 8.49059) = 9.339649 kmol H2O Table 5.8.a: Composition of input stream of HNO3 absorber Components Kmol/h Kg/h N2 79.05348 2213.49744 O2 1.91039 61.13248 NO2 25.47178 1171.84198 H2O 9.339649 168.25641 Total 115.72830 3614.72831 ≈3614.7 Table 5.8.b: Composition of output stream of HNO3 absorber Components Kmol/h Kg/h D1: N2 O2 NO 79.05348 1.91039 8.49059 2213.49744 61.13248 254.80261 D2: HNO3 H2O 16.98119 0.84906 1069.98478 15.29605 Total 107.28471 3614.71336 ≈3614.7 H2O (v) 5.8 balance over concentrator Concentrator D2 C1 E
- 15. 27 Fig 5.8: Balance over concentrator Table 5.9.a: Composition of input stream of concentrator Components Kmol/h Kg/h D2: HNO3 H2O 16.98119 0.84906 1069.98478 15.29605 C1: C5H10CO (K) C6H11OH (A) HNO3 H2O Adipic acid Inert 75.26346 36.94891 212.27725 140.27458 15.93169 1.60491 7387.10860 3700.43334 13375.00668 2527.08584 2328.25718 100.58765 Total 500.13105 30504.34296 90% water is evaporated. Amount of water evaporated = (0.9 x 141.12364) = 127.011276 kmol/h = 2288.1437 kg/h Water remaining in the solution = 14.112364 kmol/h = 254.23819 kg/h Table 5.9.b: Composition of output stream of concentrator Components Kmol/h Kg/h H2O (v) 127.011276 2288.1437 E: C5H10CO (K) C6H11OH (A) HNO3 H2O Adipic acid 75.26346 36.94891 212.27725 14.112364 15.93169 7387.10860 3700.43334 13375.00668 254.23819 2328.25718
- 16. 28 Inert 1.60491 100.58765 Total 500.13105 30504.34296 5.9 balance over crystalliser E Mother liq. (ML= E-W1) Crystal (W1) Fig 5.9: Balance over crystallizer Table 5.10.a: Composition of input stream of crystallizer Component Kmol/h Kg/h C5H10CO (K) 75.26346 7387.10860 C6H11OH (A) 36.94891 3700.43334 HNO3 212.27725 13375.00668 H2O 14.112364 254.23819 Adipic acid 15.93169 2328.25718 Inert 1.60491 100.58765 Total 373.119774 28216.19926 Crystallization is taking place at 10˚C. At 10˚C, solubility of adipic acid = 14 gm/ kg of water = 0.014 kg/ kg of water Doing adipic acid balance over crystallizer, 2328.25718 = W1 + (28216.19926 – W1) x (0.014/1.014) W1 = 1965.83256 So, crystals of adipic acid = 1965.83256 kg/h Crystallizer
- 17. 29 But crystals are 99.8 wt% pure. Table 5.10.b: Composition of output stream of crystallizer Component Kmol/h Kg/hr Crystal: Adipic acid HNO3 H2O 13.42480 0.0312 0.10912 1961.9009 1.96583 1.96583 Mother liq. : C5H10CO (K) C6H11OH (A) Adipic acid H2O HNO3 Inert 75.26346 36.94891 1.5752 13.89088 228.9688 1.60491 7387.10860 3700.43334 366.3569 252.27229 14443.60839 100.58765 Total 371.81028 28216.19973 Input stream of centrifuge is the output stream of crystallizer and it is given in table 4.10.b. Centrifuge sepatates the crystals from mother liq. Table 5.11.a: Composition of crystal stream from centrifuge Component Kmol/h Kg/hr Crystal: Adipic acid HNO3 H2O 13.42480 0.0312 0.10912 1961.9009 1.96583 1.96583 Total 13.56512 1965.83256
- 18. 30 Table 5.11.b: Composition of mother liq. stream from centrifuge Component Kmol/h Kg/hr C5H10CO (K) C6H11OH (A) Adipic acid H2O HNO3 Inert 75.26346 36.94891 1.5752 13.89088 228.9688 1.60491 7387.10860 3700.43334 366.3569 252.27229 14443.60839 100.58765 Total 358.24516 26250.36717 We know, purge = 266.64667 kmol/h From table 4.4, we get the mole fractions of various components So ,we can determine the composition of purge stream. Table 5.12: Composition of purge stream Component Mole fraction Kmol/h Kg/h C5H10CO (K) 0.2110 56.26245 5522.15947 C6H11OH (A) 0.1036 27.6246 2766.60369 HNO3 0.6461 172.28041 10855.9929 H2O 0.0278 7.41278 133.54331 Adipic acid 0.007 1.86653 272.77469 Inert 0.0045 1.19991 75.2043 Total 1 266.64667 19626.27836
- 19. 31 5.10 overall input – output balance Table 5.13.a: Total input Input Kg/h KA oil + HNO3 solution 23355.5047 N2 2213.49744 O2 672.45504 H2O 168.25641 Total 26409.7 Table 5.13.b: Total output Output Kg/h Crystals 1965.83256 Purge 19626.27836 N2 2213.49744 O2 61.13248 NO 254.80261 H2O 2288.1437 Total 26409.7 Conversion of adipic acid = (15.30169/100) x 100% =15.3%