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CHAPTER V

MASS BALANCE AND ENERGY BALANCE

5.1 MASS BALANCE AT REACTOR

5.1.1 Mass Balance Equation

A mass balance also called material balance is an application of conservative of mass to the

analysis of physical systems. By accounting for material entering and leaving a system, mass

can be identified which might have been unknown, or difficult to measure without this

technique. The exact conservation law used in the analysis of the system depends on the

context of the problem but all revolve around mass conservation, which matter cannot

disappear or be created spontaneously.

The product, formaldehyde (HCHO) is estimated to produce based on the world

demand and the world supply. Below shows the estimation of the calculation according to

world demand and supply :

World demand of HCHO at year 2011 = 42700 million kg

World supply of HCHO at year 2011 = 31500 million kg

World shortage of HCHO at year 2011 = 11200 million kg

Malaysia produce approximately 4% of world demand, and since Malaysia is

estimated to have about ten plants that produce formaldehyde, hence, this process design is

expected to produce 45 million kg of formaldehyde. The process design is assume to

working at 300 days per year which equal to 7200 hours per year.

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Mass flow rate of HCHO, F0F = 45000000

7200

= 6250 kg/kmol

Molar flow rate of HCHO, N0F = 6250

(12+2+16)

= 208.333kmol/h

According to the calculation, 208.33kmol of formaldehyde is produced per hour in

this process design. The unit operation that is selected for mass balance is reactor (R101).

Methanol and oxygen gas are feed into reactor to form formaldehyde and water. Since the

conversion of HCHO is 87.4%, the output would consist of some methanol and oxygen gas.

The chemical equation of this process is shown as below :

CH3OH +

O2 CH2O + H2O

Ni = 6 230 751 kmol/hr

Xi,M = 0.18

Xi,02 = 0.82

N0 = 1 387 695 kmol/hr

X0,F X0,02

X0,H20 X0,M

Figure 5.1 :Reactor for Mass Balance Calculation

Note : M = Methanol 02 = Oxygen =

F = Formaldehyde H20 = Water

A basis is assumed. N i,02 = 200 kmol/hour

Degree of freedom analysis : ND = NV - NE

Where, ND = number of degree of freedom

NV = number of unknown

NE = number of independent equation

Reactor

(R101 )

Conversion,X = 0.82

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From the system above, NV = 5 [ Ni,M , No,M, No,O, No.W , rk]

There are 4 components that are methanol, oxygen, formaldehyde and water which can form

5 independent equations. There is also a conversion equation. Hence, there are 5 independent

equations.

Thus, NE = 6

ND = NV - NE

= 5 5

= 0

Since the number of degree of freedom is 0, so a unique solution can be obtained.

N0,F = Ni,F + Fr208.333 = 0 + 1r

r = 208.333kmol/h

Molar flow rate of inlet stream ,

Ni,O = 200kmol/h

Ni,M = -M r

NIM

0.874 = -(-1)(208.333)

NIM

Ni,M = 238.368 kmol/h

Molar flow rate of outlet stream ,

N0,02 = N0,02 + 02r

= 200 +(-

) (208.333)

= 95.834 kmol/h

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F0,H20 = N0,H20 MR

= (208.333)(2+16)

= 3749.994 kg/h

Total mass flow rate of inlet stream ,

FiT = Fi,M + F i,02

= 7627.78 + 6400

= 14027.78 kg/h

Total mass flow rate of outlet stream ,F0T = F0,F + F0,02 + F0,M + F0,H20

= 6250 + 3066.672 + 961.12 +3749.994

= 14027.78 kg/h

We find that the total mass flow rate is exactly the same in inlet and outlet. This implies that

the system obey the law of conservation of mass.

Total molar flow rate of inlet stream,

Percentage error, % (molar flow rate) =

=

= 7.38

Percentage error, % (mass flow rate) =

=

= 0

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