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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