Material Balance for 300 tons/day Cumene Plant
A material balance is an exact accounting of all materials that enters, leaves accumulates or depleted in an industrial process or a particular process equipment in a given time interval of operation. The material balance, when done properly, gives an estimate of the material requirement and products formed during a process. All material balance calculations are based on the principle of conservation of mass, which states that matter can neither be created nor be destroyed, but they may undergo physical or chemical changes.
The material can refer to a balance on a system for
- Total mass
- Total moles
- Mass of chemical compound
- Moles of the chemical compound
- Moles of an atomic species
- Volume
The law of conservation of mass states that the total mass of various components involved remains constant during a unit process. Thus any unit process,
Rate of mass input = Rate of mass output + Rate of mass accumulation
For steady state operation where accumulation is constant,
Rate of mass input = Rate of mass output
- Conversion is 99% of propylene.
- Product molar selectivity cumene to p-DIPB is 31:1.
- Propylene feed is 95% pure and rest of propane.
- Benzene is about >99.9% purity.
- Molar feed ratio of benzene to propylene 2:1.
- Top product of benzene column is 98.1% benzene and 1.9% cumene.
- Top product of cumene separation column is 99.9 mole % cumene.
- Bottom product of cumene separation column is 100mole % p-DIPB.
Let be propane is tie substance which does not participate in the reaction, it is used as fuel.
300 tons/24 per hr of cumene
12.5 tons/ hr of cumene
12500 kg/hr of cumene
12500kg/ hr /120.19 = 104 Kmole/hr of cumene
Stoichiometry equation:
Primary reaction:
CH2-CH2-CH2(propylene ) + C2H2-C2H2-C2H2 (benzene) ® C6H5-C3H7(cumene
Side reaction:
CH2-CH2-CH2(propylene ) + C9H12 (cumene) ® C3H7-C6H4-C3H7(DIPB)
For primary reaction
1 Kmole of benzene = 1kmole of propylene = 1kmole of cumene
For side reaction
1 Kmole of benzene =2kmole of propylene = 1kmole of cumene
Propylene used for the cumene and DIPB:
So for cumene, propylene required is 104 kmole/hr
Selectivity is 31: 1 of cumene to DIPB, i.e., catalyst converts propylene and benzene to the ratio of 31 moles of cumene and 1-mole DIPB.
Conversion of propylene is 99%
Let X Kmole of DIPB formed
Then, cumene formed is 31X
31 X = 104
X = 3.35
DIPB formed is = 3.35 kmole/hr
GIVEN
PROPANE PRESENT:
PROPYLENE BALANCE
| ||||
propylene required for cumene
|
=104.16
| |||
propylene required for DIPB
|
2X3.360
|
=6.7204
| ||
total propylene required
|
=110.887
| |||
propyleneconversion
|
99%
| |
so,propylene inFeed
|
110.8871/.99=
|
112.0071
|
Unreacted =
|
112.0-110.8=
|
1.11997
|
PROPANE PRESENT:
As 110.88 Kmole of propylene
Total feed is =112.0071 / 0.95 = 117.9022 Kmole
Propane in feed =117.9022 – 112.0071 = 5.89 Kmole
BENZENE BALANCE
| |||
benzene required for cumene
|
104.1667
| ||
benzene required for DIPB
|
3.360215
| ||
total benzene required
|
107.5269
| ||
GIVEN
| |||
benzene to propylene feed ratio =
|
2:1
| ||
so, benzene to be feed
|
110.8X2=
|
221.774
|
Unreacted = 221.7-107.5
|
114.247
|
Feed vessel:
Fresh feed = 107.52 Kmole
Recycle feed = 114.24 Kmole
Mixed feed = 107.52+114.24 = 221.77 Kmole
Balance around reactor:
Feed = mixed feed of benzene+ propylene feed
Separator
Bottom = 223.987 Kmole
Top = propylene + propane
Feed = top + bottom (feed would be reactor outlet 231 Kmole)
Top = 231- 223.987 = 7.01 Kmole
Benzene distillation column
Feed =223.987 kmole
Give: Top product contains 98.1 mole % benzene; assume all benzene is collected in top only then;
Benzene balance:
223.987 X 0.51 = D X 0.981
Top product = 116.4 Kmole
Cumene present in top product = 116.4 X 0.019 = 2.21 kmole
Bottom = feed – top
= 223.987- 116.4 = 107.52 Kmole
Cumene distillation column
Feed =107.52 Kmole
Given: top product contain 99.9 mole% cumene and 100 mole % p-DIPB in bottom.
Cumene balance:
107.52 X 0.968 = D X 0.999
Top product = 104.16 kmole
Bottom product = feed – top
= 107.52-104.16 = 3.36 kmole
Overall plant material balance:
INPUT
|
OUTPUT
| ||||
molwt
|
kmol/h
|
kg/h
|
kmole/h
|
kg/h
| |
BENZENE
|
78
|
221.7742
|
17298.3876
|
114.2473
|
8911.2894
|
PROPYLENE
|
42
|
112.0071
|
4704.2982
|
1.11997
|
47.03874
|
PROPANE
|
44
|
5.8951
|
259.3844
|
5.8951
|
259.3844
|
CUMENE
|
120
|
0
|
0
|
104.166667
|
12500.00004
|
DIPB
|
162
|
0
|
0
|
3.360215
|
544.35483
|
TOTAL
|
339.6764
|
22262.0702
|
228.789252
|
22262.06741
| |
Yield:
As 1 mole of cumene is produced from 1 mole of propylene the Stoichiometry factor is 1
Moles of cumene produced = 104.16 Kmole
Stoichiometry factor = 1 (from the equation)
Moles of reactant fed = 110.88 Kmole
The yield of cumene based on propylene:
Yield = (moles of product produced) (Stoichiometry Factor) /(Moles of reactant fed to process)
= 104.16X1/110.88
=93.93%
= 104.16X1/110.88