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CHAPTER 5 MATERIAL BALANCE A process design starts with the development of a process flow sheet or process flow diagram. For the development of such a diagram, material and energy balance calculations are necessary. These balances follow the laws of conservation of mass and energy which states that the total mass of various components involved remains constant during and unit operation or unit process. p rocess. Thus, for any unit operations or unit process, INPUT = OUTPUT + ACCUMULATION or DISAPPEARANCE
For steady state operations where accumulation is constant or nil, the above equation becomes:
INPUT = OUTPUT
During material balance some of the assumptions play a vital role which we have considered wherever it is necessary.
BASIS: 1000 kg of Molasses fermented per day.
5.1 Material balance over fermenter: Assumptions: 1. 80% 80% ccon onve verrsion sion 2. 10% 10% exc excee eess Ca( Ca(OH OH))2 3. 400 kg of of water water added added to to ferme fermente nter r 5.1.1. Material balance on molasses
C6H12O6 Molasses
+
Ca(OH)2
calcium hydroxide
(CH3CHOHCOO)2Ca calcium Lactate
22
+
2H2O water
Molasses reacted = molasses fed x conversion = 1000 x 0.80 = 800 kg reacted
180 kg of C6H12O6 produces 218 kg of Calcium lactate 1000 kg of C6H12O6 produces (?) kg of Calcium lactate
Calcium lactate produced = 218 x 1000 180 = 1211.11 kg calcium lactate. But, conversion is 80% So Calcium lactate produced = 1211.11 x 0.80 = 968.89 kg Calcium lactate
5.1.4. Material balance on Water:
180 kg of C6H12O6 produces 36 kg of water 800 kg of C6H12O6 produces (?) kg of water Water produced = 800 x36 180 = 160 kg of water. So Water at Outlet = Water fed + Water produced = 400 + 160 = 560 kg of water. Lactobacillus bacteria are also added to the fermenter.Its amount per one batch is 100 Kg and its growth depends upon the temperature and acidity of water. After the fermentation, the remaining bacteria go with the cake. 24
M o l a s (1s e0 s0 k0 g)
( 4 5 .22 2 2k g) C a(O H)2
FERMENTER
M o l a s s( 2e0s0k g) C a(O H)2 ( 1 2 .33 3K G) C a l c i u m L (a9c6t.88a9kt eg) W a t (e5r6 0k g)
W a t (e4r0 0k g)
Fig 5.1 Material balance over fermenter
Table 5.1 Material balance over fermenter Components
Material in kg per day
Material out kg per day
Molasses Ca(OH)2 Calcium Lactate
1000 452.22 0
200 123.33 968.89
Water Total
400 1852.22
560 1852.22
5.2 Material balance over filter: Assumption : 1. All the solids are removed. 2. 2 % loss of calcium lactate and water with solid cake. 5.2.1 .Material balance on Calcium lactate
Calcium lactate out = Calcium lactate in – 2 % loss of Calcium lactate = 968.89 – 0.02 x 968.89 = 949.51 kg Calcium lactate. 5.2.2. Material balance on Water
25
Water out = Water in – 2 % loss of Water = 560 – 560 x 0.02 = 548.80 kg of Water
M o l a s s es (200k g) C a(O H)2 (12 3.3 3 K G) C a l c i u m L a c t(9 a t6e8.8 9 k g) W a t e(560k r g)
CAKE
FILTER
M o l a s s e(200k s g) C a(O H)2 (12 3.3 3k g) C a l c i u m l a c t(19.3 a t e 7 k g) W a t e(11.2 r k g)
FILTRATE C a l c iu m l a c t(94 a t e9.51 k g) W a t e (54 r 8.8 k g)
Fig 5.2 Material balance over Filter Table No 5.2 Material balance over filter Components
Material in
Material out in filtrate
Material out in cake
kg per day
kg per day
kg per day
Molasses
200
0
200
Ca(OH)2
123.33
0
123.33
Calcium Lactate
968.89
949.51
19.37
Water Total
560 1852.22
548.8 1498.31
11.21 353.91
5.3. Material balance over acidulation tank Assumption: 26
1. 95 % conversion of Calcium Lactate
(CH3CHOHCOO)2Ca Calcium lactate
+ H2SO4
2CH3CHOHCOOH
Sulfuric acid
+
Lactic acid
Ca SO4 Calcium sulphate
5.3.1. Material balance on calcium lactate
Calcium lactate reacted = Calcium lactate fed x Conversion = 949.51 x 0.95 = 902.03 kg Calcium Lactate.
218.00 kg of Calcium lactate produce 180 kg of lactic acid 902.03 kg of Calcium lactate produce (?) kg of lactic acid kg of lactic acid = 902.03 x 180 218 = 744.79 kg
5.3.3. Material balance on H 2SO4
218.00 kg of Calcium lactate require 98 kg of H2SO4 949.51 kg of Calcium lactate require (?) kg of H2SO4 kg of H2SO4 = 949.51 x 98 218 27
= 426.81 kg H2SO4 fed H2SO4 reacted = 426.81 x 0.95 = 405.50 kg. H2SO4 unreacted = 426.81 – 405.50 = 21.33 kg
5.3.4. Material balance on CaSO 4
218.00 kg of Calcium lactate produce 136 kg of CaSO4 902.03 kg of Calcium lactate produce (?) kg of CaSO4 = 902.03 x 136 218 = 562.73 kg of CaSO4 produced.
5.3.5. Material balance on water
Water in = water out = 548.80 kg
H2S O4 (4 2 6.81 K G)
C a l c i u m l a c t(94 a t e9.5 1 k g) W a t e(54 r 8.8 k g)
C a l c i u m L a c t(47.4 a t e 7 K G) H2S O4 (21.3 3 K G) L a c t i c a c(74 i d 4.7 9 k g) CaSO .73 kg) 4 (5 6 2 W a t e (54 r 8.8 k g)
ACIDULATION TANK
Fig 5.3 Material balance over acidulation tank Table No 5.3 Material balance over acidulation tank
28
Material in kg per day
Material out kg per day
0
744.79
H2SO4
426.81
21.33
Calcium Lactate
949.51
47.47
Water
548.8
548.8
CaSO4 Total
0 1925.12
562.73 1925.12
Components Lactic acid
5.4. Material balance over filter Assumption : 1) 1% loss of each component with cake
1. Lactic acid out = Lactic acid in – Loss with cake = 744.79 – 0.01 x 744.79 = 737.34 kg 2. H2SO4 out = H2SO4 in - Loss with cake = 21.33 – 0.01 x 21.33 = 21.11 kg 3. Calcium Lactate out = Calcium Lactate in - Loss with cake = 47.47 – 0.01 x 47.47 = 47.00 kg
4. Water out = Water in - Loss with cake = 548.8 – 0.01 x 548.8
29
= 543.31 kg 5. CaSO4 out = CaSO4 in + 1% Lactic acid + 1% H2SO4 + 1%Calcium Lactate + 1%Water = 562.73 + 7.45 + 0.22 + 0.47 + 5.49 = 576.36 kg.
C a l c i u m L a (4 c t7a.4t7K e G) H2 S O4 (2 1.3 3K G) L a c t i c a(7 c i4d4.7 9k g) C a S 4O(5 6 2.7 3k g) W a t e(5r 4 8.8 k g)
CAKE
FILTER
C a S 4O(56 2.7 3K G) L a c t i c a(7.4 c i d4k g) H2 S O4 (0.213k g) C a l c i u m l a c(0.4 t a 7k t eg) W a t e(5.4 r 8k g)
FILTRATE L a c t i c a(7 c i3d7.3 4k g) H2 S O4 (21.1 1k g) C a l c i u m l a c(4t7a.0t0k e g) W a t e(5r 4 3.3 1k g)
Fig 5.4 Material balance over filter Table no 5.4 Material balance over filter Material in kg per day 744.79
Material out kg per day 737.34
H2SO4
21.33
21.11
Calcium Lactate
47.47
47
Water
548.8
543.31
CaSO4
562.73
576.36
Total
1925.12
1925.12
Components lactic acid
5.5. Material balance over carbon column Assumption:
30
1. 5% adsorption of water 2. 90 % adsorption of calcium lactate 3. 85% adsorption of H2SO4 1. Calcium Lactate out = Calcium Lactate in – 90% Adsorption = 47 – 0.9 x 47 = 4.7 kg Calcium Lactate Adsorbed = 47 – 4.7 = 42.3 kg 2. H2SO4 out = H2SO4 in - 85% Adsorption = 21.11 – 21.11 x 0.85 = 3.16 kg.
H2SO4 adsorbed = 21.11 – 3.16 = 17.95 kg 3. Lactic acid out = Lactic acid in = 737.34 kg
4. Water out = Water in – 5% Adsorption = 543.31 – 0.05 x 543.31 = 516.14 kg. Water Adsorbed = 543.31 – 516.14 = 27.17 kg
31
FILTRATE L a c t i c a( 7c3i .7d3 4k g) H2 S O4 (2 1.1 1k g) C a l c i u m l a(4c 7t.0a0k t eg) W a t e( 5r4 3.3 1k g)
CARBON COLUMN
C a l c i u m L a(4.7 c t ka gt) e H2 S O4 (3.1 6k g) L a c t i c a( 7c 3i .7d3 4k g) W a t (e5r1 .61 4k g)
M A T E R I AL A D S O R B E D W a t e(2r7.1 6k g) H2S O4 (1 7.9 4k g) C a l c i u m l a(4c 2t.3akt eg)
Fig 5.5 Material balance over carbon column Table No. 5.5 Material balance over carbon column
Components
Material in kg per day
Material adsorbed in kg per day
Lactic acid
737.34
0
737.34
H2SO4 Calcium Lactate
21.11
17.94
3.16
47
42.3
4.7
543.31 1348.75
27.16 87.4
516.15 1261.35
Water Total
5.6. Material balance over Evaporator Assumption: 1. 85% of evaporation of water 2. 2 % loss of Lactic acid 3. 5 % loss of H2SO4 4. 5 % loss of calcium lactate
5.6.1
Material balance of Lactic acid
32
Material out kg per day
Lactic acid out = lactic acid in - 2 % loss of Lactic acid =737.34 – 0.02 x 737.34 = 722.59 kg Lactic acid loss = 737.34 – 722.59 = 14.74 kg
5.6.2. Material balance of Calcium lactate
Calcium lactate out = Calcium lactate in - 5 % loss of Calcium lactate = 4.7 – 0.05 x 4.7 = 4.465 kg Calcium lactate loss = 4.7 – 4.465 = 0.235 kg
5.6.3. Material balance of H 2SO4
H2SO4 out = H2SO4 in - 5 % loss of H2SO4 = 3.16 – 0.05 x 3.16 = 3.002 kg H2SO4 loss = 3.16 – 3.002 = 0.158 kg 5.6.4. Material balance of Water
Water Evaporated = 0.85 X 516.14 = 438.719 kg Water out = Water in – Water evaporated = 516.14 – 0.85 x 516.14 = 77.41 kg
So, Purity of Lactic acid = Lactic acid in product Total product 33
x 100
= 722.59 x 100 807.467 = 89.48 %
W a t e r(43 8.71 kg) Lactic acid (14.74 kg) C a l c i u m l a c t a(0.235 te k g) H2S O4 (0.158kg)
C a l c i u m L a c t a(4.7 t e kg) H2 S O4 (3.16 kg) Lactic acid (737.34 kg) W a t e r(516.14 kg)
EVAPORATOR
Lactic acid (722.59 kg) C a l c i u m l a c t a(4.46 t e kg) H2S O4 (3.002 kg) W a t e r(77.41 kg)
Fig 5.6 Material balance over Evaporator Table No 5.6 Material balance over Evaporator
