This pdf contains material balance of adipic acid plant. Adipic acid is used for making nylon and other products.

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

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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 1M2 = 400 kmol/h
H2O = 0.25 B

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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

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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

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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

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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

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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

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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

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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

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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

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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%