Material Balance of 100 MTPD of Fatty Alcohol Production Process Plant from Vegetable Oils

Detailed Material Balance Calculations:
material balance block diagram of fatty alcohol production process from vegetable oils using methanol
Block diagram of Fatty Alcohol
from vegetable oils
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 desired amount of Fatty alcohol =507 X 1.01 = 512 Kmol/day

Excess Hydrogen is sent to shift the reaction equilibrium in favor 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

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 favor 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