Bioprocess Engineering Questions and Answers – Material Balances with Recycle, By – Pass and Purge Systems

This set of Bioprocess Engineering Interview Questions and Answers for freshers focuses on “Material Balances with Recycle, By – Pass and Purge Systems”.

1. Which of the following represents the figure corresponding to the stream?

a) Recycle system
b) By-pass system
c) Purge system
d) Recover stream

Explanation: A by-pass is one where a portion of the inlet to a process unit is split from the feed and instead of entering the process is combined with the outlet from that process. This practice is far less common than recycle, but may be used if your ultimate goal is a material with properties” in-between” the untreated reactant and the process outlet product.

2. Consider the following labeled flowchart for a simple chemical process based on reaction A -> B and predict the overall and single-pass conversion of the process?
a) 100%, 70%
b) 100%, 50%
c) 100%, 75%
d) 100%, 55%

Explanation: Overall conversion: Based on the streams that enter and leave the overall process.
Overall conversion = (moles of reactants in fresh feed-moles in the output of the overall process)/(moles of reactant in fresh feed)

Therefore, the overall conversion of A is from equation:
Overall conversion = ((75 mol A/min)in –(0 mol/min)out)/((75 mol A/min)out) × 100% = 100%
Single-pass conversion: Based on streams that enter and leave the reactor.
Overall conversion = (moles of the reactants fed into the reactor-moles that exiting the reactor)/(moles of reactant fed into reactor)
Therefore, the single-pass conversion from equation is:
((100 mol A/min)in-(25 mol A/min)out)/((100 mol A/min)in) × 100% = 75%.

3. What is the combined feed ratio from the following?
a) Quantity of mixed feed stream to the quantity of fresh feed stream
b) Quantity of recycle stream to the quantity of fresh feed stream
c) Quantity of fresh feed stream to the quantity of recycle stream
d) Quantity of fresh feed stream to the quantity of mixed feed stream

Explanation: Combined feed ratio is the ratio of the quantity of mixed feed stream to the quantity of fresh feed stream.
Combined feed ratio = M/F.
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4. A single effect evaporator is fed with 10000 kg / h of weak liquor containing 15 % caustic by weight and is concentrated to get thick liquor containing 40 % by weight caustic. Calculate:
(i) kg / h of water evaporated and
(ii) kg / h of thick liquor
a) 4560 kg/h, 3720 kg/h
b) 3460 kg/h, 7680 kg/h
c) 4350 kg/h, 6732 kg/h
d) 6250 kg/h, 3750 kg/h

Explanation: 10000 kg / h of weak liquor.
Let x be the kg / h thick liquor obtained and y be the kg / h water evaporated.

Overall Material balance:
Total mass Input = Total mass output
kg / h weak liquor = kg / h water evaporated + kg / h thick liquor
10000 = x + y
Material Balance of NaOH:
NaOH in the liquid stream = NaOH in output stream
NaOH in the weak liquor = NaOH in thick liquor
0.15 × 10000 = 0.40x
x = 3750 kg/hr
Hence, y= 6250 kg/hr
Water evaporated = 6250 kg/h
Thick liquor obtained = 3750 kg/h

5. A solution of potassium dichromate in water contains 15% w/w Kr2Cr2O7. Calculate the amount of Kr2Cr2O7 that can be produced from 1500 kg of solution if 700 kg of water is evaporated and remaining solution is cooled to 293K.
Data: Solubility of Kr2Cr2O7 at 293 K is 115 kg per 1000 kg of water.
a) 158.875 kg
b) 156.678 kg
c) 145.478 kg
d) 148.875 kg

Explanation: 1500 kg of solution.
Kr2Cr2O7 content of solution = 0 15×1500 = 225 kg
Water in 15 % solution = 1500 – 225 = 1275 kg
M.B of water:
Water in final solution = 1275 – 700 = 575 kg
Solubility of Kr2Cr2O7 at 293 K = 115 kg / 1000 kg water
Amount of Kr2Cr2O7 at 293 K = 115/1000 × 575 = 66.125 kg
M.B of Kr2Cr2O7:
[Kr2Cr2O7 in feed solution] = [Kr2Cr2O7 in solution at 293k] + [Kr2Cr2O7 produced as crystals] Amount of Kr2Cr2O7 crystals produced = 225 – 66.125 = 158.875 kg.

6. Which of the following is not the application of recycle system?
a) Increased reactant conversion
b) Decreased reactant conversion
c) Continuous catalyst regeneration
d) Circulation of the working fluid

Explanation: The most common application of recycle for systems involving chemical reaction is the recycle of reactants, an application that is used to increase the overall conversion in a reactor and not the decreased reactant conversion.

7. The fresh feed to an ammonia synthesis reactor contains nitrogen, hydrogen and 2.0 mole per cent inerts. The molar ratio of H2:N2 is 3:1. The product stream consists of pure ammonia. Since conversion in the reactor is only 15%, a recycle stream is used and in order to avoid build-up of inerts, a purge stream is withdrawn. The rate of purge stream is adjusted to keep inert concentration in the recycle stream at 8 mole per cent. For a fresh feed rate of 100 moles/hr. Note that recycle stream contains only nitrogen, hydrogen and inerts. The N2:H2 ratio of 1:3 is maintained in every process stream. Calculate the moles of ammonia produced.

a) 38.90 moles/hr
b) 28.90 moles/hr
c) 37.50 moles/hr
d) 27.50 moles/hr

Explanation: The concentration of inerts in the purge stream is 8 mole percent. The amount of inerts must be 2 moles/hr in order to prevent accumulation of inerts. Therefore the flow rate of purge stream is 25 moles/hr.

The flow rate of nitrogen and hydrogen in the purge stream is 23 moles/hr (5.75 nitrogen, 17.25 hydrogen). Therefore moles of ammonia produced:
(24.50 – 5.75) x 2 = 37.50 moles/hr of ammonia.

8. The fresh feed to an ammonia synthesis reactor contains nitrogen, hydrogen and 2.0 mole per cent inerts. The molar ratio of H2:N2 is 3:1. The product stream consists of pure ammonia. Since conversion in the reactor is only 15%, a recycle stream is used and in order to avoid build-up of inerts, a purge stream is withdrawn. The rate of purge stream is adjusted to keep inert concentration in the recycle stream at 8 mole per cent. For a fresh feed rate of 100 moles/hr. Note that recycle stream contains only nitrogen, hydrogen and inerts. The N2:H2 ratio of 1:3 is maintained in every process stream, and calculate the moles of nitrogen entering the reactor and in the recycle stream?

a) 125 moles/hr, 100.50 moles
b) 135 moles/hr, 50 moles
c) 125 moles/hr, 50 moles
d) 185 moles/hr, 100.50 moles

Explanation: Moles of nitrogen entering the reactor = (37.50/2)/0.15 = 125 moles/hr
Therefore there are 125 – 24.50 = 100.50 moles of nitrogen in the recycle stream.

9. The fresh feed to an ammonia synthesis reactor contains nitrogen, hydrogen and 2.0 mole per cent inerts. The molar ratio of H2:N2 is 3:1. The product stream consists of pure ammonia. Since conversion in the reactor is only 15%, a recycle stream is used and in order to avoid build-up of inerts, a purge stream is withdrawn. The rate of purge stream is adjusted to keep inert concentration in the recycle stream at 8 mole per cent. For a fresh feed rate of 100 moles/hr. Note that recycle stream contains only nitrogen, hydrogen and inerts. The N2:H2 ratio of 1:3 is maintained in every process stream, and calculate the number of moles, moles of inerts and moles of hydrogen in the recycle stream?

a) 437 moles/hr, 35 moles/hr, 301.5 moles/hr
b) 237 moles/hr, 30 moles/hr, 200 moles/hr
c) 567 moles/hr, 35 moles/hr, 205 moles/hr
d) 347 moles/hr, 30 moles/hr, 500 moles/hr

Explanation: Total number of moles in the recycle stream = 100.50×4/0.92 = 437 moles/hr
Moles of inerts in the recycle stream = 437 – 100.50×4=35 moles/hr
Moles of hydrogen in the recycle stream = 100.50×3 = 301.50 moles/hr.

10. What do you mean by the splitting point?
a) The two streams split with different composition
b) The two streams split with equal composition
c) Assuming it’s not a reactor and there’s only 2 streams
d) Assuming it’s not a reactor and there’s only 1 stream

Explanation: Splitting Point: 6 variables – 2 mass balances – 1 knowing compositions are the same – 1 splitting ratio = 2 DOF (Degree of freedom). The splitting point is special because when a stream is split, it generally is split into two streams with equal composition.

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