Cumene Production Project Plant Material Balance Calculation Example

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.

Material can refer to a balance on a system for
  •   Total mass
  •   Total moles
  •  Mass of chemical compound
  •  Moles of 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

 ASSUMED DATA:
  • 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 do not participate in the reaction, it is used as fuel.

BASIS: 300 tons /day of CUMENE PRODUCTION flow sheet
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
Stiochiometry 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

PROPYLENE BALANCE


propylene required for cumene
=104.16
propylene required for DIPB 
2X3.360
=6.7204
total propylene required
=110.887
GIVEN
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 Stiochiometry factor is 1
Moles of cumene produced = 104.16 Kmole                                          
Stiochiometry factor = 1 (from the equation)                                                                                                                                       
Moles of reactant fed = 110.88 Kmole
Yield of cumene based on propylene:
    Yield = (moles of product produced) (Stiochiometry Factor) /(Moles of reactant fed to process)
              = 104.16X1/110.88                       
                 =93.93%