This set of Heat Transfer Operations Multiple Choice Questions & Answers (MCQs) focuses on “Evaporators – Single Effect Operations”.
1. Why do we use Single effect evaporators when we already have multiple effect one?
a) To have liberty to use fouling liquids
b) To create low pressure
c) To have large temperature
d) To have higher efficiency with cheap material
View Answer
Explanation: Single effect evaporators are better than multiple effect because expensive materials of construction can be used as is the case with corrosive feeds, and when the vapour is too contaminated to be reused.
2. Which one of the following do we prefer to use for single effect over multiple effect evaporator?
a) Low pressure operations
b) High pressure operations
c) Corrosive materials use
d) High temperature operations
View Answer
Explanation: Single effect evaporators are better than multiple effect because expensive materials of construction can be used as is the case with corrosive feeds, and when the vapour is too contaminated to be reused. So best way to use for single effect over multiple effect evaporator is Corrosive materials.
3. Multiple effect evaporators can be constructed with cheap materials than single effect evaporators.
a) True
b) False
View Answer
Explanation: Single effect evaporators are better than multiple effect because expensive materials of construction can be used as is the case with corrosive feeds, and when the vapour is too contaminated to be reused. Multiple effect evaporators are constructed with cheap materials.
4. Which one of the following is the most common mode of operation of single effect evaporators?
a) Batch
b) Semi-Batch
c) Continuous batch
d) Continuous
View Answer
Explanation: The operation of single effect evaporators is usually done in continuous batches, wherein a particular amount of feed is fed continuously after which the setup is opened to be cleaned and then again the batch is repeated. Hence it is continuous batch that is more favourable to its operation.
5. What is amount of heat that needs to be supplied in order to evaporate 10kg of water from a feed of 50kg at a temperature of 25℃ to a final temperature of 100℃?
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity = 6 KJ/Kg K
a) 52600 KJ
b) 56200 KJ
c) 2272.6 MJ
d) 45100 KJ
View Answer
Explanation: The total energy required is = mTSΔT + mL = 50×6×75 + 10×2260 = 45100 KJ.
6. What is mass of Steam that needs to be supplied in order to heat the tubes and evaporate 10kg of water from a feed of 50kg at a temperature of 25℃ to a final temperature of 100℃?
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity = 6 KJ/Kg K
a) 21kg
b) 22kg
c) 20kg
d) 19kg
View Answer
Explanation: The total energy required is = mTSΔT + mL = 50×6×75 + 10×2260 = 45100 KJ
Hence the amount of steam needed to heat this is = 45100/2260 = 19.9 = 20 Kg.
7. How much evaporation can take place from a feed of 30kg at 25°C in an hour if there is constant supply of 10kg steam in that hour?
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity = 6 KJ/Kg K
a) 2kg
b) 3kg
c) 4kg
d) 5kg
View Answer
Explanation: Total energy = 2260×10 = 22600 KJ, hence if we have 30 kg feed at 25℃ the energy used to bring it to 100℃ is = 30×6×75 = 13500 KJ. Hence the remaining energy goes to evaporation = 9100/2260 = 4.02 = 4Kg.
8. If we have a steam flow rate of 10kg/hr then what should be the inlet feed rate if we wish to evaporate 4kg of water from the dilute solution in an hour?
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity = 6 KJ/Kg K
a) 30kg
b) 32kg
c) 31kg
d) 33kg
View Answer
Explanation: Total energy = 2260×10 = 22600 KJ, hence if we have X kg feed at 25℃ the energy used to bring it to 100℃ is = X×6×75 = 450X KJ. Hence the remaining energy goes to evaporation = (22600 – 450X) / 2260 = 4 or X = 30kg/hr.
9. Increasing the pressure of the setup will result in larger evaporation rate.
a) True
b) False
View Answer
Explanation: Lowering the pressure will lower their vapour pressure, hence the solution would boil at a temperature less than its normal and hence result in larger evaporation.
10. If evaporation takes place from a feed of 50wt% solid at 25℃ in an hour if there is constant supply of 30kg steam per 100kg feed in that hour, then what is the final concentration?
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity = 6 KJ/Kg K
a) 57
b) 56
c) 55
d) 54
View Answer
Explanation: Initially wt% = 50, let us assume we have 100kg feed, solid wt = 50, and water = 50, total energy = 2260×30 = 67800 KJ, hence if we have 100 kg feed at 25℃ the energy used to bring it to 100℃ is = 100×6×75 = 45000 KJ. Hence the remaining energy goes to evaporation = 22800/2260 = 10Kg steam. Hence the weight of solid = 50 and total weight = 90, wt% = 56%.
11. Which one of the following would not benefit the evaporator if employed?
a) Pressure reduced
b) Pressure increased
c) Surface area increased
d) Steam supply increased
View Answer
Explanation: Lowering the pressure will lower their vapour pressure; hence the solution would boil at a temperature less than its normal and hence result in larger evaporation.
12. If we have a feed flow rate of 30kg/hr at 25℃ then what should be the heating steam rate if we wish to evaporate 4kg of water from the dilute solution in an hour?
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity = 6 KJ/Kg K
a) 10kg
b) 12kg
c) 11kg
d) 13kg
View Answer
Explanation: Let us assume the steam inlet rate is X kg/hr, the total energy = 2260×X = 2260X KJ, and we have 30 kg at 25℃ the energy used to bring it to 100℃ is = 30×6×75 = 13500 KJ. Hence the remaining energy goes to evaporation = (2260X – 13500) / 2260 = 4 or X = 10kg/hr.
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