Detailed Material Balance Calculations:
Basis: 30000 Metric Tonnes per year of Fatty alcohol.
1. Material Balance for Hydrogenation step:
The equation for Hydrogenation reaction is given below
RCOOCH3 + 2H2 <---->RCH2OH + CH3OH
The molecular weight of Fatty Acid Methyl Ester = 225
The molecular weight of Hydrogen = 2
The molecular weight of Fatty alcohol =198
The molecular weight of Methanol =32
Assuming 300 working days per year the rate of production of Fatty alcohol =30000/300=100 TPD
The number of moles of Fatty alcohol produced = 100000/197= 507Kmol/day
From reaction stoichiometry 1 Kmol of FAME = 1 Kmol of Fatty alcohol
But since the yield is only 99% 1.01 Kmol of FAME = 1Kmol of Fatty alcohol
Therefore FAME needed to produce the desired amount of Fatty alcohol =507 X 1.01 = 512 Kmol/day
Excess Hydrogen is sent to shift the reaction equilibrium in favour of products in FAME: H2 ratio of 1:25
Hydrogen sent into the reactor =512 X 25 = 12800 Kmol/day
Hydrogen is consumed in the reaction and also for saturation of the unsaturated compounds in FAME due to the catalyst.
From reaction stoichiometry 1 Kmol of FAME = 2 Kmol of Hydrogen
Amount of Hydrogen consumed in the reaction =1014 Kmol/day
Amount of Hydrogen used for saturation of unsaturated compounds= 108 Kmol/day
The total amount of Hydrogen consumed =1122 Kmol/day
Amount of Hydrogen recycled= Amount of Hydrogen sent – Amount of Hydrogen Consumed.
Amount of Hydrogen recycled =12800-1122 =11678 Kmol/day
Fresh Hydrogen sent into the reactor = Hydrogen consumed during the reaction.
Fresh Hydrogen sent into the reactor =1122Kmol/day
From reaction stoichiometry 1 Kmol of FAME = 1 Kmol of Methanol
Amount of Methanol produced =507Kmol/day
MASS BALANCE FOR HYDROGENATION STEP
2. Material Balance for Transesterification step:
The equation for Transesterification reaction is given below.
2(COOCH2)-RCOOCH + 3CH3OH <----> 3RCOOCH3 +2(CH2OH)-CHOH
Amount of FAME produced =512Kmol/day
According to stoichiometry:
1Kmol FAME = 0.33Kmol of Oil.
1Kmol FAME = 1Kmol Methanol.
The yield of the reaction is 95%
Using yield
1Kmol FAME = 0.35 Kmol of Oil.
1Kmol FAME = 1.05 Kmol of Methanol.
Excess Methanol is sent to shift the equilibrium of the reaction in favour of products in the FAME: Methanol ratio of 1:1.5.
So,
1Kmol of FAME=1.5+1.05=2.55Kmol of Methanol.
1Kmol of FAME =0.35 Kmol of Oil.
1Kmol of FAME =0.33 Kmol of Glycerin.
Amount of Oil consumed = Kmol/day
Amount of Methanol sent in = Kmol/day
Amount of Methanol consumed =512Kmol/day
Amount of Methanol recycled =1305-512 =793 Kmol/day
Amount of Glycerin produced = Kmol/day
The molecular weight of Fatty Acid Methyl Ester = 225
The molecular weight of Oil = 663
The molecular weight of Glycerin =92
The molecular weight of Methanol =32
MASS BALANCE FOR TRANSESTERIFICATION STEP
 |
| Block diagram of Fatty Alcohol from vegetable oils |
1. Material Balance for Hydrogenation step:
The equation for Hydrogenation reaction is given below
RCOOCH3 + 2H2 <---->RCH2OH + CH3OH
The molecular weight of Fatty Acid Methyl Ester = 225
The molecular weight of Hydrogen = 2
The molecular weight of Fatty alcohol =198
The molecular weight of Methanol =32
Assuming 300 working days per year the rate of production of Fatty alcohol =30000/300=100 TPD
The number of moles of Fatty alcohol produced = 100000/197= 507Kmol/day
From reaction stoichiometry 1 Kmol of FAME = 1 Kmol of Fatty alcohol
But since the yield is only 99% 1.01 Kmol of FAME = 1Kmol of Fatty alcohol
Therefore FAME needed to produce the desired amount of Fatty alcohol =507 X 1.01 = 512 Kmol/day
Excess Hydrogen is sent to shift the reaction equilibrium in favour of products in FAME: H2 ratio of 1:25
Hydrogen sent into the reactor =512 X 25 = 12800 Kmol/day
Hydrogen is consumed in the reaction and also for saturation of the unsaturated compounds in FAME due to the catalyst.
From reaction stoichiometry 1 Kmol of FAME = 2 Kmol of Hydrogen
Amount of Hydrogen consumed in the reaction =1014 Kmol/day
Amount of Hydrogen used for saturation of unsaturated compounds= 108 Kmol/day
The total amount of Hydrogen consumed =1122 Kmol/day
Amount of Hydrogen recycled= Amount of Hydrogen sent – Amount of Hydrogen Consumed.
Amount of Hydrogen recycled =12800-1122 =11678 Kmol/day
Fresh Hydrogen sent into the reactor = Hydrogen consumed during the reaction.
Fresh Hydrogen sent into the reactor =1122Kmol/day
From reaction stoichiometry 1 Kmol of FAME = 1 Kmol of Methanol
Amount of Methanol produced =507Kmol/day
MASS BALANCE FOR HYDROGENATION STEP
Component
|
Kmol/day
|
Kg/day
|
Amount of FAME in feed
|
512
|
115200
|
Amount of fresh Hydrogen in feed
|
1122
|
2244
|
Amount of Hydrogen recycled
|
11678
|
23356
|
Amount of Fatty Alcohol formed
|
507
|
100000
|
Amount of Methanol formed
|
507
|
16224
|
2. Material Balance for Transesterification step:
The equation for Transesterification reaction is given below.
2(COOCH2)-RCOOCH + 3CH3OH <----> 3RCOOCH3 +2(CH2OH)-CHOH
Amount of FAME produced =512Kmol/day
According to stoichiometry:
1Kmol FAME = 0.33Kmol of Oil.
1Kmol FAME = 1Kmol Methanol.
The yield of the reaction is 95%
Using yield
1Kmol FAME = 0.35 Kmol of Oil.
1Kmol FAME = 1.05 Kmol of Methanol.
Excess Methanol is sent to shift the equilibrium of the reaction in favour of products in the FAME: Methanol ratio of 1:1.5.
So,
1Kmol of FAME=1.5+1.05=2.55Kmol of Methanol.
1Kmol of FAME =0.35 Kmol of Oil.
1Kmol of FAME =0.33 Kmol of Glycerin.
Amount of Oil consumed = Kmol/day
Amount of Methanol sent in = Kmol/day
Amount of Methanol consumed =512Kmol/day
Amount of Methanol recycled =1305-512 =793 Kmol/day
Amount of Glycerin produced = Kmol/day
The molecular weight of Fatty Acid Methyl Ester = 225
The molecular weight of Oil = 663
The molecular weight of Glycerin =92
The molecular weight of Methanol =32
MASS BALANCE FOR TRANSESTERIFICATION STEP
Component
|
Kmol/day
|
Kg/Day
|
Amount of Oil in feed
|
179
|
118677
|
Amount of fresh Methanol in feed
|
512
|
16384
|
Amount of Methanol recycled
|
793
|
25376
|
Amount of FAME formed
|
512
|
115200
|
Amount of Glycerin formed
|
179
|
16468
|
