 # Material Balance for Process Involving Recycle

Recycle process is process in which a flow stream is returned to an earlier stage in the processing stage. The process is frequently used in the production facility. If the conversion of valuable reactant is considerably less than 100%, the unreacted material is usually separated and recycled. An example of a recycle process in which there is no reaction is the return of the reflux to the top of distillation column.

What do you think about the existence of recycle in material balance calculation?

In my opinion, recycle makes material balance calculation is more difficult. In this post, we will learn material balance for process involving recycle. I hope it will give you a little insight.

Without recycle process, the material balances are simple. The material balance can be carried out in serial, in which the calculated flows out of one unit become the feeds to the next. But if a recycle stream is present, then at the point when the recycle is returned, the flow will not be known as it will depend on downstream flows not yet calculated. Without knowing the recycle flow, the sequence of calculations cannot be continued to the point where recycle flow can be determined.

There are two possible approaches to solve the problems.

1. Use cut and try method. This is a trial-and-error method. The recycles streams flows can be estimated and the calculations continued to the point where the recycle is calculated. The estimated flows are then compared with the calculated and a better estimate made. The procedure is continued until the difference between the estimated and the calculated flows is within acceptable limits.
2. Use formal, algebraic method. The presence of recycle implies that some of the mass balance equations will have to be solved simultaneously. The equations are set up with the recycle flows as unknowns and solved using standard methods for the solution of simultaneous equations.

When there is only one or two recycle loops, the calculation can often be simplified by careful selection of basis of calculation and the system boundaries.

Let see example below.

The block diagram shows the main steps in the balanced process for the production of vinyl chloride from ethylene. Each block represents a reactor and several other processing units. The main reactions are:

Block A – Chlorination:

C2H4 + Cl2 → C2H4Cl2, yield on ethylene 98%

Block B – Oxyhydrochlorination:

C2H4 + 2HCl + ½ O2 → C2H4Cl2 + H2O, yields on ethylene 95%; yield on HCl 90%

Block C – Pyrolysis:

C2H4Cl2 → C2H3Cl + HCl, yields on DCE 99%; yield on HCl 99.5%

The HCl from the pyrolysis step is recycled to the oxyhydrochlorination step. The flow of ethylene to the chlorination and oxyhydrochlorination reactors is adjusted so that the production of HCl is in balance with the requirement. The conversion in the pyrolysis reactor is limited to 55%, and the unreacted dichloroethane (DCE) separated and recycled.

Using the yield figures given, and neglecting any other losses, calculate the flow of ethylene to each reactor and the flow of DCE to the pyrolysis reactor, for a production rate of 12,500 kg/h vinyl chloride (VC).

## Solution

Let X is mol of ethylene entering Block A.

Y is mol of ethylene entering Block B.

Z is mol of recycled HCl. ### At Block A, yield on ethylene is 98%

Based on previous post, yield = mol of product × stoichiometric factor / mol of reagent fed

Therefore,

Yield on ethylene = mol of product DCE × stoichiometric factor / mol of ethylene fed

98% = mol of product DCE × 1 / X

(Stoichiometric factor is 1, 1 mol ethylene is required to produce 1 mol of DCE)

Mol of product DCE entering Block A = 0.98 X  (Equation 1)

### At Block B, yield on ethylene is 95%

Yield on ethylene = mol of product DCE × stoichiometric factor / mol of ethylene fed

95% = mol of product DCE × 1 / Y

(Stoichiometric factor is 1, 1 mol ethylene is required to produce 1 mol of DCE)

Mol of product DCE entering Block B = 0.95 (Equation 2)

Therefore, total of DCE produced = DCE produced from Blok A + DCE produced from Block B = 0.98 X + 0.95 Y  (Equation 3)

### At Block C, yield on HCl is 99.5%

Yield on HCl = mol of product HCl × stoichiometric factor / mol of DCE fed

99.5% = mol of product HCl × 1 / (0.98 X + 0.95 Y)

(Stoichiometric factor is 1, 1 mol DCE is required to produce 1 mol of HCl)

Mol of product HCl = Z = 0.995 (0.98 X + 0.95 Y)    (Equation 4)

### At Block B, yield on HCl is 90%

Yield on HCl = mol of product DCE × stoichiometric factor / mol of HCl fed

90% = mol of product DCE × 2 / Z

(Stoichiometric factor is 2, 2 mol HCl is required to produce 1 mol of DCE)

Mol of product DCE = 0.9 Z / 2  (Equation 5)

Therefore, DCE produced in Block B can be expressed as:

0.95 = 0.9 Z / 2  (Equation 2 and Equation 5)

Z = 2 (0.95 Y) / 0.9   (Equation 6)

Based on Equation 3 and Equation 6:

0.995 (0.98 X + 0.95 Y) = 2 (0.95 Y) / 0.9

Y = 0.803 X    (Equation 7)

### At Block C, yield on DCE is 99%

Yield on DCE = mol of product VC × stoichiometric factor / mol of DCE fed

*mol of DCE fed see Equation 3

99% = mol of product VC × 1 / (0.98 X + 0.95 Y)

(Stoichiometric factor is 1, 1 mol DCE is required to produce 1 mol of VC)

Mol of product VC = 0.99 (0.98 X + 0.95 Y)

Mol of VC = 12,500 kg/h / 98.96 = 200 kmol/h

Therefore,

0.99 (0.98 X + 0.95 Y) = 200

Substitute Y with Equation 7 to get:

X = 113.29 kmol/h

⸫ Therefore, ethylene entering Block A – Chlorination unit is 113.29 kmol/h.

Substitute X to Equation 7 to get:

Y = 90.97 kmol/h

⸫ Therefore, ethylene entering Block B – Oxyhydrochlorination unit is 90.97 kmol/h.

Substitute Y to Equation 6 to get:

Z = 192 kmol/h

⸫ Therefore, HCl recycle is 192 kmol/h.

Overall yield on ethylene = mol of product (VC) / mol of ethylene fed = 200 / (113.29 + 90.97) = 97.91%

I hope you find this post useful.