250+ TOP MCQs on Evaporators – Capacity and Answers

Heat Transfer Operations Multiple Choice Questions on “Evaporators – Capacity”.

1. What is defined by the term Evaporator Capacity?
a) Kg of fluid supplied in one day
b) Kg of water evaporated in one day
c) Kg of fluid supplied in 1hr
d) Kg of water evaporated in 1hr

Answer: d
Clarification: Evaporator Capacity is defined as the amount of water in kg evaporated in a 1hour. Evaporator economy can find by using Evaporator Capacity and Steam Consumption parameters.

2. What are the parameters to determine the evaporator economy?
a) Evaporator Capacity
b) Evaporator temperature
c) Evaporator pressure
d) Evaporator Fluid supply

Answer: a
Clarification: The parameters required to measure the evaporator economy is
i. Steam Consumption
ii. Evaporator capacity
As Economy = Capacity/Consumption.

3. What is the steam consumption of an evaporator if the evaporator capacity is 30kg/her and the steam consumption is 45kg/hr?
a) 67%
b) 70%
c) 80%
d) 65%

Answer: a
Clarification: Economy of an evaporator is defined as, E = Capacity/Consumption = (30/45) x 100 = 66.7 = 67%.

4. What is the Evaporator Consumption if we evaporate 10kg of water from a feed of 50kg at a temperature of 25℃ to a final temperature of 100℃ in an hour?
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity = 6 KJ/Kg K
a) 21kg/hr
b) 22kg/hr
c) 20kg/hr
d) 19kg/hr

Answer: c
Clarification: The total energy required is = m_TSΔT + mL = 50×6×75 + 10×2260 = 45100 KJ
Hence the amount of steam needed to heat this is = 45100/2260 = 19.9 = 20 Kg
Hence the Evaporator Consumption = 20kg/hr.

5. Find the value of Evaporator Capacity if we have a feed of 30kg at 25℃ and with evaporator consumption as 10kg/hr?
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity = 6 KJ/Kg K
a) 2kg/hr
b) 3kg/hr
c) 4kg/hr
d) 5kg/hr

Answer: c
Clarification: 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/hr.

6. What is the value of Evaporator Economy if we have a feed of 30kg at 25℃ and with evaporator consumption as 10kg/hr?
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity = 6 KJ/Kg K
a) 42%
b) 30%
c) 40%
d) 50%

Answer: c
Clarification: 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/hr = Capacity.
Hence Economy = Capacity/Consumption = (4/10) x 100 = 40%.

7. What do we mean by the term Evaporator Consumption?
a) Steam consumed in 1hr
b) Steam produced in 1hr
c) Feed supplied in 1hr
d) Feed supplied in 1day

Answer: a
Clarification: The evaporator capacity is defined as the amount in kg of steam used to heat the fluid for evaporation in 1hr.

8. Find one among the following which does not measure evaporator performance.
a) Steam Consumption
b) Amount of Vapour Produced
c) Economy
d) Amount of feed supplied

Answer: d
Clarification: The three parameters that determine the performance of the evaporator are
i. Steam Consumption
ii. Evaporator Capacity
iii. Evaporator Economy.

9. If the Evaporator Consumption is 10kg/hr then what should be the inlet feed rate if we wish to have a capacity if 4kg/hr?
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

Answer: a
Clarification: 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.

10. If evaporation takes place from a 100kg feed of 50wt% solid at 25℃ in an hour and if the steam consumption is 30Kg/hr, 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

Answer: b
Clarification: 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. The capacity of an evaporator depends upon the temperature of the feed solution.
a) True
b) False

Answer: a
Clarification: The capacity of an evaporator depends upon the temperature of the feed solution because if the feed solution is at the boiling temperature then all the heat supplied will be utilized for evaporation, thus increasing the capacity of evaporator.

12. Which one of the following method can be adopted to increase the capacity of the evaporator?
a) Increase pressure in calandria
b) Using multiple effect evaporator
c) Getting the temperature of the feed close to boiling temperature
d) Reducing the temperature of the feed

Answer: c
Clarification: The capacity of an evaporator depends upon the temperature of the feed solution because if the feed solution is at the boiling temperature then all the heat supplied will be utilized for evaporation, thus increasing the capacity of evaporator. Getting the temperature of the feed close to boiling temperature method can be adopted to increase the capacity of the evaporator.