Chemical Engineering Questions and Answers – Degrees of Freedom Analysis in Steady-State Process

This set of tough Chemical Engineering Questions focuses on “Degrees of Freedom Analysis in Steady-State Process”.

1. What is the work of drying process?
a) Decreases the moisture content
b) Vaporizes water
c) Increases the moisture content
d) None of the mentioned
View answer

Answer: a
Explanation: Drying decreases the moisture content of air.

2. What is the work of the humidification process?
a) Vaporization
b) Freezing
c) Condensing
d) None of the mentioned
View answer

Answer: a
Explanation: Humidification vaporizes liquid water to moist air.

3. In condensation, moist air is cooled below its which temperature?
a) Dew-point temperature
b) Dry-bulb temperature
c) Wet-bulb temperature
d) Saturation temperature
View answer

Answer: d
Explanation: In condensation, moist air is cooled below its saturation temperature.
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4. What is added to the moist air in the combustion process?
a) Water
b) Carbon dioxide
c) Oxygen
d) None of the mentioned
View answer

Answer: a
Explanation: Water is added to the moist air in the combustion process.

5. In air conditioning process, what has to be done first to the moist air?
a) Heating
b) Cooling
c) Heating & Cooling
d) None of mentioned
View answer

Answer: a
Explanation: In air conditioning process, moist air is heated and then cooled.
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6. Enthalpy of air is 15 J and enthalpy of water vapor is 20 J, if the total enthalpy of the system is 20 J, what is the humidity of the system?
a) 25%
b) 50%
c) 75%
d) 100%
View answer

Answer: a
Explanation: 20 = 15 + 20*H, => H = 0.25 = 25%.

7. Enthalpy of air is 5 J and enthalpy of water vapor is 10 J, if the total enthalpy of the system is 25 J, what is the humidity of the system?
a) 50%
b) 100%
c) 150%
d) 200%
View answer

Answer: d
Explanation: 25 = 5 + 10*H, => H = 2, => H = 200%.
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8. Enthalpy of air is 10 J and enthalpy of water vapor is 15 J, if the total enthalpy of the system is 25 J, what is the humidity of the system?
a) 20%
b) 50%
c) 100%
d) 180%
View answer

Answer: c
Explanation: 25 = 10 + 15*H, => H = 1, => H = 100%.

9. Enthalpy of air is 15 J and enthalpy of water vapor is 20 J, if the total enthalpy of the system is 25 J, what is the humidity of the system?
a) 10%
b) 50%
c) 100%
d) 200%
View answer

Answer: b
Explanation: 25 = 15 + 20*H, => H = 0.5, => H = 50%.
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10. Enthalpy of air is 20 J and enthalpy of water vapor is 6 J, if the total enthalpy of the system is 23 J, what is the humidity of the system?
a) 10%
b) 20%
c) 40%
d) 50%
View answer

Answer: d
Explanation: 23 = 20 + 6*H, => H = 0.5, => H = 50%.

11. The humidity at dry-bulb temperature of 20oC is 40% and the humidity at dry-bulb temperature of 25oC is 25%, what is the slope of adiabatic cooling line?
a) 0.03
b) -0.03
c) 0.05
d) -0.05
View answer

Answer: b
Explanation: Slope of adiabatic cooling line = (0.4 – 0.25)/(20 – 25) = -0.03.

12. The humidity at dry-bulb temperature of 23oC is 30% and the humidity at dry-bulb temperature of 28oC is 25%, what is the slope of adiabatic cooling line?
a) -0.01
b) -0.02
c) -0.03
d) -0.05
View answer

Answer: a
Explanation: Slope of adiabatic cooling line = (0.3 – 0.25)/(23 – 28) = -0.01.

13. The humidity at dry-bulb temperature of 20oC is 30% and the humidity at dry-bulb temperature of 22oC is 24%, what is the slope of adiabatic cooling line?
a) -0.01
b) -0.02
c) -0.03
d) -0.05
View answer

Answer: c
Explanation: Slope of adiabatic cooling line = (0.3 – 0.24)/(20 – 22) = -0.03.

14. The humidity at dry-bulb temperature of 20oC is 10% and the humidity at dry-bulb temperature of 23oC is 4%, what is the slope of adiabatic cooling line?
a) -0.01
b) -0.02
c) -0.03
d) -0.05
View answer

Answer: b
Explanation: Slope of adiabatic cooling line = (0.1 – 0.04)/(20 – 23) = -0.02.

15. The humidity at dry-bulb temperature of 20oC is 12% and the humidity at dry-bulb temperature of 23oC is 3%, what is the slope of adiabatic cooling line?
a) -0.01
b) -0.02
c) -0.03
d) -0.05
View answer

Answer: c
Explanation: Slope of adiabatic cooling line = (0.12 – 0.03)/(20 – 23) = -0.03.

Sanfoundry Global Education & Learning Series – Basic Chemical Engineering.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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