Material Bal 02
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Transcript of Material Bal 02
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. 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% conversion
2. 10% excees Ca(OH)2
3. 400 kg of water added to fermenter
5.1.1. Material balance on molasses
C6H12O6 + Ca(OH)2 (CH3CHOHCOO)2Ca + 2H2O Molasses calcium hydroxide calcium Lactate water
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Molasses reacted = molasses fed x conversion = 1000 x 0.80 = 800 kg reacted
Molasses unreacted = molasses fed – molasses reacted
= 1000 – 800
= 200 kg unreacted
5.1.2. Material balance on Ca(OH)2
From the above chemical reaction
180 kg of C6H12O6 require 74 kg of Ca(OH)2
1000 kg of C6H12O6 require (?) kg of Ca(OH)2
= 1000 x 74 180 = 411.11 kg Ca(OH)2 require
But we are using 10% excess Ca(OH)2
Ca(OH)2 fed = (1 + % excess ) x Theoretically required Ca(OH)2
100
= 1.1 x 411.11
= 452.22 kg Ca(OH)2 fed
Ca(OH)2 reacted = Ca(OH)2 required x conversion
= 411.11 x 0 .80
= 328.88 kg reacted
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Ca(OH)2 unreacted = Ca(OH)2 fed - Ca(OH)2 reacted
=452.22 - 328.88
=123.33 kg unreacted
5.1.3. Material balance on calcium Lactate:
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.
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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.
Fig 5.1 Material balance over fermenter
Table 5.1 Material balance over fermenter
Components Material in Material out kg per day kg per day
Molasses 1000 200Ca(OH)2 452.22 123.33
Calcium Lactate 0 968.89Water 400 560Total 1852.22 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.
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5.2.2. Material balance on Water
Water out = Water in – 2 % loss of Water
= 560 – 560 x 0.02
= 548.80 kg of Water
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 560 548.8 11.21Total 1852.22 1498.31 353.91
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5.3. Material balance over acidulation tank
Assumption:
1. 95 % conversion of Calcium Lactate
(CH3CHOHCOO)2Ca + H2SO4 2CH3CHOHCOOH + Ca SO4
Calcium lactate Sulfuric acid Lactic acid 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.
Calcium Lactate unreacted = Calcium lactate fed - Calcium lactate reacted
= 949.51 – 902.03
= 47.47 kg of Calcium lactate.
5.3.2. Material balance on lactic acid
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 H2SO4
218.00 kg of Calcium lactate require 98 kg of H2SO4
949.51 kg of Calcium lactate require (?) kg of H2SO4
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kg of H2SO4 = 949.51 x 98 218
= 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 CaSO4
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
Fig 5.3 Material balance over acidulation tank
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Table No 5.3 Material balance over acidulation tank
ComponentsMaterial in kg per
dayMaterial out kg per
day
Lactic acid 0 744.79
H2SO4 426.81 21.33
Calcium Lactate 949.51 47.47
Water 548.8 548.8
CaSO4 0 562.73Total 1925.12 1925.12
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
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4. Water out = Water in - Loss with cake
= 548.8 – 0.01 x 548.8
= 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.
Fig 5.4 Material balance over filter
Table no 5.4 Material balance over filter
ComponentsMaterial in kg
per dayMaterial out kg per
daylactic acid 744.79 737.34
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
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5.5. Material balance over carbon column
Assumption:
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
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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
Material out kg per day
Lactic acid 737.34 0 737.34
H2SO4 21.11 17.94 3.16Calcium Lactate 47 42.3 4.7Water 543.31 27.16 516.15Total 1348.75 87.4 1261.35
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
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5.6.1 Material balance of Lactic acid
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 H2SO4
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
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So, Purity of Lactic acid = Lactic acid in product x 100 Total product
= 722.59 x 100 807.467
= 89.48 %
Fig 5.6 Material balance over Evaporator
Table No 5.6 Material balance over Evaporator
ComponentsMaterial in kg
per dayMaterial out in liquid stream
Material out in vapors stream
kg per day kg per day
Lactic acid 737.34 722.59 14.74
H2SO4 3.16 3.002 0.158
Calcium Lactate 4.7 4.465 0.235
Water 516.14 77.41 438.719
Total 1261.34 807.467 453.92
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