Air-Conditioning Questions and Answers – Psychrometric Relations

«
»

This set of Air-Conditioning Multiple Choice Questions & Answers (MCQs) focuses on “Psychrometric Relations”.

1. What is the value of humidity ratio if pv = 0.387 bar and pa = 0.997 bar?
a) 0.482
b) 0.241
c) 0.122
d) 0.622
View Answer

Answer: b
Explanation: Humidity ratio is the mass of water vapor present in 1 kg of dry air and has a unit g / kg of dry air.
W = 0.622 pv / pa
= 0.622 x 0.387 / 0.997
= 0.241 kg / kg of dry air.
advertisement

2. What is the value of humidity ratio if mv = 1.423 and ma = 3.589?
a) 0.1
b) 0.2
c) 0.4
d) 0.5
View Answer

Answer: c
Explanation: Humidity ratio is the mass of water vapor present in 1 kg of dry air and has a unit g / kg of dry air. It is also defined as the ratio of the mass of water vapor to the mass of dry air in a given volume of the air-vapor mixture. Hence, it is given as mv / ma.
W = 1.423 / 3.589 = 0.396 ≅ 0.4.

3. What is the value of relative humidity if mv = 2.901 and ms = 9.056?
a) 0.300
b) 0.330
c) 0.310
d) 0.320
View Answer

Answer: d
Explanation: Relative humidity is the ratio of actual mass of water vapor in a given volume of moist air to the mass of water vapor in the same volume of saturated air. It can be represented as,
∅ = mv / ms = pv / ps = 2.901 / 9.056 = 0.320.
advertisement
advertisement

4. What is the value of relative humidity if pv = 0.468 bar and ps = 0.893 bar?
a) 52 %
b) 54 %
c) 56 %
d) 190 %
View Answer

Answer: a
Explanation: Relative humidity is the ratio of actual mass of water vapor in a given volume of moist air to the mass of water vapor in the same volume of saturated air. It can be represented as,
∅ = mv / ms = pv / ps = 0.468 / 0.893 = 0.524 = 52.4 %.

5. What is the value of pb if pa = 1.48 bar and pv = 1.52 bar, according to Dalton’s law of partial pressures?
a) 1 bar
b) 2 bar
c) 3 bar
d) 4 bar
View Answer

Answer: c
Explanation: The Dalton’s law states that the total pressure exerted by the mixture of air and water vapor is equal to the sum of the pressures, which each constituent would exert if it occupied the same space by itself. Mathematically, it can be represented as, pb = pa + pv = 1.48 + 1.52 = 3 bar.
advertisement

6. If DBT = 30°C, WBT = 20°C and Barometer reading is 740 mm of Hg. Find out DPT using steam tables.
a) 5°C
b) 35°C
c) 15°C
d) 25°C
View Answer

Answer: b
Explanation: For given conditions, pressure corresponding to WBT from steam table is pw = 0.02337 bar
pb = 740 mm of Hg = 740 x 133.3 / 100000 = 0.98642 bar
Pressure of water vapor = pv = pw – [(pb – pw) (td – tw) / 1544 – 1.44 tw]
= 0.02337 – [(0.98642 – 0.02337) (30 – 20) / 1544 – 1.44 x 20]
= 0.01701 bar
For 0.01701 bar water vapor pressure, dew point temperature is 15°C.

7. If DBT = 30°C, WBT = 20°C and Barometer reading is 740 mm of Hg. Find out relative humidity using steam tables.
a) 40 %
b) 50 %
c) 30 %
d) 20 %
View Answer

Answer: a
Explanation: For given conditions, pressure corresponding to WBT from steam table is pw = 0.02337 bar
pb = 740 mm of Hg = 740 x 133.3 / 100000 = 0.98642 bar
Pressure of water vapor = pv = pw – [(pb – pw) (td – tw) / 1544 – 1.44 tw]
= 0.02337 – [(0.98642 – 0.02337) (30 – 20) / 1544 – 1.44 x 20]
= 0.01701 bar
Saturation pressure for given DBT is ps = 0.04242 bar
Relative humidity = pv / ps = 0.01701 / 0.04242 = 0.40 = 40 %.

8. If DBT = 30°C, WBT = 20°C and Barometer reading is 740 mm of Hg. Find out specific humidity using steam tables.
a) 10.914 m3 / kg of dry air
b) 10.914
c) 10.914 kg / kg of dry air
d) 10.914 g / kg of dry air
View Answer

Answer: d
Explanation: For given conditions, pressure corresponding to WBT from steam table is pw = 0.02337 bar
pb = 740 mm of Hg = 740 x 133.3 / 100000 = 0.98642 bar
Pressure of water vapor = pv = pw – [(pb – pw) (td – tw) / 1544 – 1.44 tw]
= 0.02337 – [(0.98642 – 0.02337) (30 – 20) / 1544 – 1.44 x 20]
= 0.01701 bar
Specific humidity = W = 0.622 pv / pb – pv = 0.622 x 0.01701 / 0.98642 – 0.01701 = 10.914 g / kg of dry air.

9. If DBT = 30°C, WBT = 20°C and Barometer reading is 740 mm of Hg. Find out degree of saturation using steam tables.
a) 40 %
b) 39 %
c) 38 %
d) 35 %
View Answer

Answer: b
Explanation: For given conditions, pressure corresponding to WBT from steam table is pw = 0.02337 bar
pb = 740 mm of Hg = 740 x 133.3 / 100000 = 0.98642 bar
Pressure of water vapor = pv = pw – [(pb – pw) (td – tw) / 1544 – 1.44 tw]
= 0.02337 – [(0.98642 – 0.02337) (30 – 20) / 1544 – 1.44 x 20]
= 0.01701 bar
Specific humidity = W = 0.622 pv / pb – pv = 0.622 x 0.01701 / 0.98642 – 0.01701 = 10.914 g / kg of dry air
Saturation pressure for given DBT is ps = 0.04242 bar
Ws = 0.622 ps / pb – ps = 0.622 x 0.004242 / 0.98642 – 0.04242 = 27.945 g / kg of dry air
Degree of saturation, μ = W / Ws = 10.914 / 27.945 = 0.39 = 39 %.

10. If DBT = 30°C, WBT = 20°C and Barometer reading is 740 mm of Hg. Find out vapor density using steam tables.
a) 0.01216 kg / m3 of dry air
b) 0.01216 g / m3 of dry air
c) 0.01216 m3 / m3 of dry air
d) 0.01216 kg / kg of dry air
View Answer

Answer: a
Explanation: For given conditions, pressure corresponding to WBT from steam table is pw = 0.02337 bar
pb = 740 mm of Hg = 740 x 133.3 / 100000 = 0.98642 bar
Pressure of water vapor = pv = pw – [(pb – pw) (td – tw) / 1544 – 1.44 tw]
= 0.02337 – [(0.98642 – 0.02337) (30 – 20) / 1544 – 1.44 x 20]
= 0.01701 bar
Specific humidity = W = 0.622 pv / pb – pv = 0.622 x 0.01701 / 0.98642 – 0.01701 = 10.914 g / kg of dry air
ρv = W (pb – pv) / Ra Td = 0.010914 (0.98642 – 0.01701) 105 / 287 (273 + 30)
= 0.01216 kg / m3 of dry air.

11. What is the humidity ratio in terms of mass?
a) mv / ma
b) ma / mv
c) pv / pa
d) pa / pv
View Answer

Answer: a
Explanation: Humidity ratio is the mass of water vapor present in 1 kg of dry air and has a unit g / kg of dry air. It is also defined as the ratio of the mass of water vapor to the mass of dry air in a given volume of the air-vapor mixture. Hence, it is given as mv / ma.
advertisement
advertisement

12. What is the humidity ratio in terms of pressure?
a) 0.622 pv / pb
b) 0.622 pv / pa
c) 0.622 mv / mb
d) 0.622 mv / ma
View Answer

Answer: b
Explanation: Humidity ratio is the mass of water vapor present in 1 kg of dry air and has a unit g / kg of dry air. It is also defined as the ratio of the mass of water vapor to the mass of dry air in a given volume of the air-vapor mixture. Hence, it is given as mv / ma = Ra pv / Rv pa using ideal gas equation.
Ra = 0.287 and Rv = 0.461 kJ/kg K, putting these values in above equation we get,
W = 0.287 / 0.461 (pv / pa)
W = 0.622 pv / pa.

13. Relative humidity is represented as __________
a) mv / ma
b) ma / mv
c) ps / pv
d) mv / ms
View Answer

Answer: d
Explanation: Relative humidity is the ratio of actual mass of water vapor in a given volume of moist air to the mass of water vapor in the same volume of saturated air. It can be represented as,
∅ = mv / ms = pv / ps.

14. Which of the following is true, according to Dalton’s law of partial pressures?
a) pb = pa x pv
b) pb = pa – pv
c) pb = pa + pv
d) pa = pb + pv
View Answer

Answer: c
Explanation: The Dalton’s law states that the total pressure exerted by the mixture of air and water vapor is equal to the sum of the pressures, which each constituent would exert if it occupied the same space by itself. Mathematically, it can be represented as, pb = pa + pv.

15. If the relative humidity is zero, then the humidity ratio is infinity.
a) False
b) True
View Answer

Answer: a
Explanation: Relative humidity is the ratio of actual mass of water vapor in a given volume of moist air to the mass of water vapor in the same volume of saturated air. It can be represented as,
If, ∅ = mv / ms = pv / ps = 0,
Then W = 0. So, the humidity ratio is zero if the relative is zero.

Sanfoundry Global Education & Learning Series – Refrigeration & Air Conditioning.

To practice all areas of Refrigeration & Air Conditioning, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter