1. Wet bulb depression, under saturated ambient air conditions:
(a) Is always positive
(b) Is always negative
(c) Is always zero
(d) May have a value depending upon the dew point temperature
(2 Mark, 1989)

Ans: c
Explanation:
At saturated condition: DBT = WBT
Wet bulb depression = DBT – WBT = 0
2. Atmospheric air from 40^{0}C and 60 percent relative humidity can be brought to 20^{0}C and 60 percent relative humidity by
(a) Cooling and dehumidification process
(b) Cooling and humidification process
(c) Adiabatic saturation process
(d) Sensible cooling process
(2 Mark, 1990)
3. If moist air is cooled by sensible heat removal, which of the following is true?
(a) Neither relative humidity nor specific humidity changes
(b) Specific humidity changes but not relative humidity
(c) Both relative humidity and specific humidity change
(d) None of these
(2 Mark, 1991)
4. Select statements from List II matching the processes in List I. Enter your answer as D, C if the correct choice for (1) is (D) and that for (2) is (C).
(2 Mark, 1999)
5. For air at a given temperature, as the relative humidity is increased isothermally,
(a) The wet bulb temperature and specific enthalpy increase
(b) The wet bulb temperature and specific enthalpy decrease
(c) The wet bulb temperature increases and specific enthalpy decreases
(d) The wet bulb temperature decreases and specific enthalpy increases
(2 Mark, 2001)
6. For air with a relative humidity of 80%
(a) The dry bulb temperature is less than the wet bulb temperature
(b) The dew point temperature is less than wet bulb temperature
(c) The dew point and wet bulb temperatures are equal
(d) The dry bulb and dew point temperatures are equal
(1 Mark, 2003)

Ans: b
Explanation:
Relative humidity is 80%. So the fluid unsaturated.
At unsaturated condition, DBT > WBT > DPT
7. Dew point temperature of air at one atmospheric pressure (1.013 bar) is 18^{0}C. The air dry bulb temperature is 30^{0}C. The saturation pressure of water at 18^{0}C and 30^{0}C are 0.02062 bar and 0.04241 bar respectively. The specific heat of air and water vapour respectively are 1.005 and 1.88 kJ/kgK and the latent heat of vaporization of water at 0^{0}C is 2500 kJ/kg. The specific humidity (kg/kg of dry air) and enthalpy (kJ/kg of dry air) of this moist air respectively, are
(a) 0.01051, 52.64
(b) 0.01291, 63.15
(c) 0.01481, 78.60
(d) 0.01532, 81.40
(2 Mark, 2004)

Ans: b
Explanation:
Saturation pressure at 18^{0}C = Vapour pressure at 30^{0}C
Specific humidity \left( \omega \right) =\frac { 0.622{ P }_{ V } }{ P{ P }_{ V } }
=> \omega =\frac { 0.622\times 0.02062 }{ 1.013250.02062 } = 0.01291 kg/kg of dry air
Enthalpy (h) = 1.005\times { t }_{ d }+\omega \left( 2500+1.88\times { t }_{ d } \right)
=> h = 1.005\times 30+0.01291\left( 2500+1.88\times 30 \right)
=> h = 63.47 kJ/kg of dry air
8. In the window air conditioner, the expansion device used is
(a) Capillary tube
(b) Thermostatic expansion valve
(c) Automatic expansion valve
(d) Float valve
(1 Mark, 2004)

Ans: a
9. During chemical dehumidification process of air
(a) Dry bulb temperature and specific humidity decrease
(b) Dry bulb temperature increases and specific humidity decreases
(c) Dry bulb temperature decreases and specific humidity increases
(d) Dry bulb temperature and specific humidity increase
(1 Mark, 2004)
10. Water at 42°C is sprayed into a stream of air at atmospheric pressure, dry bulb temperature of 40°C and a wet bulb temperature of 20°C. The air leaving the spray humidifier is not saturated. Which of the following statements is true?
(a) Air gets cooled and humidified
(b) Air gets heated and humidified
(c) Air gets heated and dehumidified
(d) Air gets cooled and dehumidified
(1 Mark, 2005)

Ans: b
Explanation:
The air gets heated because the temperature of water is more than the temperature of air.
The air get humidified because the water is added to the air and the air leaving the humidifier is not saturated.
11. Various psychrometric processes are shown in the figure below.
(a) P – i, Q – ii, R – iii, S – iv, T – v
(b) P – ii, Q – i, R – iii, S – v, T – iv
(c) P – ii, Q – i, R – iii, S – iv, T – v
(d) P – iii, Q – iv, R – v, S – i, T – ii
(2 Mark, 2005)

Ans: b
12. For a typical sample of ambient air (at 35°C, 75% relative humidity and standard atmospheric pressure), the amount of moisture in kg per kg of dry air will be approximately
(a) 0.002
(b) 0.027
(c) 0.25
(d) 0.75
(1 Mark, 2005)

Ans: b
Explanation:
At DBT = 35^{0}C and \phi = 75%
p_{s }= 0.05628 bar
Relative humidity \left( \phi \right) =\frac { { p }_{ v } }{ { p }_{ s } }
=> 0.75=\frac { { p }_{ v } }{ 0.05628 }
=> { p }_{ v } = 0.04221 bar
Specific humidity \left( \omega\right) =\frac { 0.622\times { p }_{ v } }{ p{ p }_{ v } }
=> \omega =\frac { 0.622\times 0.04221 }{ 1.013250.04221 }
=> \omega = 0.0270 kg/kg of dry air
13. A thin layer of water in field is formed after a farmer has watered it. The ambient air conditions are: temperature 20°C and relative humidity 5%. An extract of steam tables is given below.
Neglecting the heat transfer between the water and the ground, the water temperature in the field after phase equilibrium is reached equals
(a) 10.3°C
(b) 10.3°C
(c) 14.5°C
(d) 14.5°C
(2 Mark, 2006)

Ans: c
Explanation:
Given, \phi = 0.05
=> \frac { { p }_{ v } }{ { p }_{ s } } = 0.05
From the table, { p }_{ s }= 2.34 kPa
{ p }_{ v }=0.05\times 2.34 = 0.117 kPa
The temperature at which { p }_{ v } become saturated pressure can be calculated by interpolation between { p }_{ s } = 0.10 to { p }_{ s } = 0.26.
\frac { T(15) }{ 0.1170.10 } =\frac { 10(15) }{ 0.260.10 }=> T = 14.47^{0}C
14. The statements concern psychometric chart.
1. Constant relative humidity lines are uphill straight lines to the right.
2. Constant wet bulb temperature lines are downhill straight lines to the right.
3. Constant specific volume lines are downhill straight lines to the right.
4. Constant enthalpy lines are coincident with constant wet bulb temperature lines.
Which of the statements are correct?
(a) 2 and 3
(b) 1 and 2
(c) 1 and 3
(d) 2 and 4
(2 Mark, 2006)

Ans: a
15. Dew point temperature is the temperature at which condensation begins when the air is cooled at constant.
(a) Volume
(b) Entropy
(c) Pressure
(d) Enthalpy
(1 Mark, 2006)

Ans: c
16. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooling and dehumidifying coil with an enthalpy of 85 kJ/ kg of dry air and a humidity ratio of 19 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio of 8 grams/kg of dry air. If the condensate water leaves the coil with an enthalpy of 67 kJ/kg, the required cooling capacity of the coil in kW is
(a) 75.0
(b) 123.8
(c) 128.2
(d) 159.0
(2 Mark, 2007)

Ans: b
Explanation:
Mass of condensate water leaves the coil \left( { \dot { m } }_{ w } \right) ={ \dot { m } }_{ a }\left( { \omega }_{ 1 }{ \omega }_{ 2 } \right)
=> { \dot { m } }_{ w }={ 3\times \left( 0.0190.008 \right) }=0.33kg/s
By applying energy balance equation:
{ \dot { m } }_{ a }{ h }_{ 1 }={ \dot { m } }_{ a }{ h }_{ 2 }+{ \dot { m } }_{ w }{ h }_{ w }+Q=> 3 * 85 = 3 * 43 + 0.033 * 67 + Q
=> Q = 123.8 kW
So, the required coil capacity is 123.8 kW
17. A building has to be maintained at 21^{0}C (dry bulb) and 14.5^{0}C (wet bulb). The dew point temperature under these conditions is 10.17^{0}C. The outside temperature is 23^{0}C (dry bulb) and the internal and external surface heat transfer coefficients are 8 W/m^{2}K and 23 W/m^{2}K respectively. If the building wall has a thermal conductivity of 1.2 W/mK, the minimum thickness (in m) of the wall required to prevent condensation is
(a) 0.471
(b) 0.407
(c) 0.321
(d) 0.125
(2 Mark, 2007)

Ans: b
Explanation:
To prevent condensation, temperature of inner wall should be more than the dew point temperature.
The limiting condition to prevent condensation is T_{w1 }= 10.17^{0}C
Heat flux inside the wall = h_{i}(T_{i} – T_{w1}) = 86.64 W/m^{2}
Heat flux outside the wall = h_{o}(T_{ w2} – T_{o}) = 86.64 W/m^{2}
=> T_{ w2 }= 19.23^{0}C
Heat flux in the wall = \frac { k\left( { T }_{ w1 }{ T }_{ w2 } \right) }{ L } = 86.64 W/m^{2}
=> \frac { 1.2\left( 10.17+19.23 \right) }{ L } = 86.64 W/m^{2}
=> L = 0.407 m
18. Moist air at a pressure of 100 kPa is compressed to 500 kPa and then cooled to 35°C in an aftercooler. The air at the entry to the aftercooler is unsaturated and becomes just saturated at the exit of the aftercooler. The saturation pressure of water at 35°C is 5.628 kPa. The partial pressure of water vapour (in kPa) in the moist air entering the compressor is closest to
(a) 0.57
(b) 1.13
(c) 2.26
(d) 4.52
(2 Mark, 2008)

Ans: b
Explanation:
As there is no entry and exit of water vapour from the system, So the specific humidity do not change.
i.e. { \omega }_{ 1 }={ \omega }_{ 2 }
=> \frac { 0.622\times { p }_{ { v }_{ 1 } } }{ p{ p }_{ { v }_{ 1 } } } =\frac { 0.622\times { p }_{ { v }_{ 2 } } }{ p{ p }_{ { v }_{ 2 } } }
=> \frac { { p }_{ { v }_{ 1 } } }{ 100{ p }_{ { v }_{ 1 } } } =\frac { 5.628 }{ 5005.628 }
=> { p }_{ { v }_{ 1 } } = 1.13 kPa
19. Air (at atmospheric pressure) at a dry bulb temperature of 40°C and wet bulb temperature of 20°C is humidified in an air washer operating with continuous water recirculation. The wet bulb depression (i.e. the difference between the dry and wet bulb temperature) at the exit is 25% of that at the inlet. The dry bulb temperature at the exit of the air washer is closest to
(a) 10°C
(b) 20°C
(c) 25°C
(d) 30°C
(2 Mark, 2008)

Ans: c
Explanation:
In a air washer for humidification, WBT_{1} = WBT_{2}
It is given: WBD_{2} = 0.25 * WBD_{1}
=> WBD_{2 }= 0.25 * (40 – 20)
=> DBT_{1 }– WBT_{1}= 5
=> DBT_{1 }= 5 + 20 = 25°C
20. A moist air sample has dry bulb temperature of 30ºC and specific humidity of 11.5g water vapour per kg dry air. Assume molecular weight of air as 28.93. If the saturation vapour pressure of water at 30ºC is 4.24 kPa and the total pressure is 90kPa, then the relative humidity (in %) of air sample is
(a) 50.5
(b) 38.5
(c) 56.5
(d) 68.5
(2 Mark, 2010)

Ans: b
Explanatio:
Specific humidity\left( w \right) =\frac { 0.622\times { p }_{ v } }{ p{ p }_{ v } }
=> 0.0115=\frac { 0.622\times { p }_{ v } }{ 90{ p }_{ v } }
=> { p }_{ v }=1.633 kPa
Relative humidity \left( \phi \right) =\frac { { p }_{ v } }{ { p }_{ s } }
=> \phi =\frac { 1.633 }{ 4.24 } = 0.385 = 38.5%
21. If a mass of moist air in an airtight vessel is heated to a higher temperature, then
(a) Specific humidity of the air increases
(b) Specific humidity of the air decreases
(c) Relative humidity of the air increases
(d) Relative humidity of the air decreases
(1 Mark, 2011)
22. A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and 25°C respectively. Given that the saturation pressure for water at 25°C is 3.17 kPa, the relative humidity of the air in the room is
(a) 67%
(b) 55%
(c) 83%
(d) 71%
(2 Mark, 2012)

Ans: d
Explanation:
Specific humidity (\omega) =\frac { { m }_{ v } }{ { m }_{ a } }
=> 0.622\times \frac { { p }_{ v } }{ { p }_{ t }{ p }_{ v } } =\frac { { m }_{ v } }{ { m }_{ a } }
=> 0.622\times\frac { { p }_{ v } }{ 100{ p }_{ v } } =\frac { 0.5 }{ 35 }
=> { p }_{ v }=2.238 kPa
Relative humidity \left( \phi \right) =\frac { { p }_{ v } }{ { p }_{ s } } =\frac { 2.238 }{ 3.17 } = 71%
23. The pressure, dry bulb temperature and relative humidity of air in a room are 1 bar, 30°C and 70% respectively. If the saturated steam pressure at 30°C is 4.25 kPa, the specific humidity of the room air in kg water vapour/kg dry air is
(a) 0.0083
(b) 0.0101
(c) 0.0191
(d) 0.0232
(1 Mark, 2013)

Ans: c
Explanation:
Relative humidity \left( \phi \right) =\frac { { p }_{ v } }{ { p }_{ s } }
=> 0.7=\frac { { p }_{ v } }{ 4.25 }
=> { p }_{ v } = 2.975 kPa
Specific humidity \left( \omega\right) =\frac { 0.622\times { p }_{ v } }{ p{ p }_{ v } }
=> \omega =\frac { 0.622\times 2.975 }{ 1002.975 }
=> \omega = 0.0191 kg/kg of dry air
24. A sample of moist air at a total pressure of 85 kPa has a dry bulb temperature of 30°C (saturation vapour pressure of water = 4.24 kPa). If the air sample has a relative humidity of 65%, the absolute humidity (in gram) of water vapour per kg of dry air is ______.
(1 Mark, 2014[3])

Ans: 20.84
Explanation:
Relative humidity \left( \phi \right) =\frac { { p }_{ v } }{ { p }_{ s } }
=> 0.65=\frac { { p }_{ v } }{ 4.24}
=> { p }_{ v }=2.756 kPa
Absolute humidity \left( \omega \right) =\frac { { 0.622p }_{ v } }{ { p }_{ a }{ p }_{ v } } =\frac { 0.622\times 2.755 }{ 852.756}
=> \omega = 0.02084 kg/kg of dry air
=> \omega = 20.84 g/kg of dry air
25. Moist air at 35°C and 100% relative humidity is entering a psychrometric device and leaving at 25°C and 100% relative humidity. The name of the device is
(a) Humidifier
(b) Dehumidifier
(c) Sensible heater
(d) Sensible cooler
(1 Mark, 2014[4])
26. A stream of moist air (mass flow rate = 10.1 kg/s) with humidity ratio of 0.01 kg/kg dry air mixes with a second stream of superheated water vapour flowing at 0.1 kg/s. Assuming proper and uniform mixing with no condensation, the humidity ratio of the final stream (in kg/kg dry air) is ___.
(1 Mark, 2015[1])

Ans: 0.02
Explanation:
Given, Humidity ratio \left( \omega \right) =0.01
=> \frac { { m }_{ v } }{ { m }_{ a } } = 0.01
=> { m }_{ v }=0.01\times { m }_{ a }
Mass of moist air = m_{a}+m_{v}
=> 10.1 = m_{a} + 0.01 * m_{a}
=> m_{a }= 10 kg/s
and m_{v }= 10.1 – 10 = 0.1 kg/s
For second stream, m_{v1 }= 0.1 kg/s
In the mixture:
Total mass of moist air = 0.1+ 0.1 = 0.2 kg/s
Total mass of dry air = 10 + 0 = 10 kg/s
Now, { \omega }_{ mix }=\frac { 0.2 }{ 10 } =0.02 kg/kg dry air
27. Air in a room is at 35°C and 60% relative humidity (RH). The pressure in the room is 0.1 MPa. The saturation pressure of water at 35°C is 5.63 kPa. The humidity ratio of the air (in gram/kg of dry air) is _____.
(2 Mark, 2015[3])

Ans: 21.74
Explanation:
Relative humidity \left( \phi \right) =\frac { { p }_{ v } }{ { p }_{ s } }
=> 0.6=\frac { { p }_{ v } }{ 5.63 }
=> { p }_{ v } = 3.378 kPa
Humidity ratio \left( \omega\right) =\frac { 0.622\times { p }_{ v } }{ p{ p }_{ v } }
=> \omega =\frac { 0.622\times 3.378 }{ 1003.378 }
=> \omega = 0.02174 kg/kg of dry air
=> \omega = 21.74 g/kg of dry air
28. The partial pressure of water vapour in a moist air sample of relative humidity 70% is 1.6 kPa, the total pressure being 101.325 kPa. Moist air may be treated as an ideal gas mixture of water vapour and dry air. The relation between saturation temperature (T_{s} in K) and saturation pressure (P_{s} in kPa) for water is given by ln(P_{s}/P_{o}) = 14.317 – 5304/T_{s}, where P_{o} = 101.325 kpa. The dry bulb temperature of the moist air sample (in ^{0}C) is___.
(2 Mark, 2016[2])

Ans: 19.89
Explanation:
We know, \phi =\frac { { P }_{ v } }{ { P }_{ s } } at the same temperature and pressure
=> 0.7 = =\frac { 1.6 }{ { P }_{ s } }
=> P_{s }= 1.6/0.7 = 2.2857 kPa
According to the relation given in question:
ln\left( \frac { { P }_{ s } }{ { P }_{ o } } \right) =14.317\frac { 5304 }{ { T }_{ sat } }=> ln\left( \frac { 2.2857 }{ 101.325 } \right) =14.317\frac { 5304 }{ { T }_{ sat } }
=> T_{sat }= 292.89 K = 29.89°C
Now, T_{sat } = T_{db }= 29.89°C
29. In a mixture of dry air and water vapour at a total pressure of 750 mm of Hg, the partial pressure of water vapour is 20 mm of Hg. The humidity ratio of the air in grams of water vapour per kg of dry air (g_{w}/kg_{da}) is ____.
(1 Mark, 2016[3])

Ans: 17.04
Explanation:
Humidity ratio \left( \omega \right) =0.622\frac { { p }_{ v } }{ p{ p }_{ v } }
=> \omega =0.622\frac { 20 }{ 75020 }
=> \omega = 0.01704 kg_{w}/kg_{da}
=> \omega = 17.04 g_{w}/kg_{da}
30. Moist air is treated as an ideal gas mixture of water vapour and dry air (molecular weight of air = 28.84 and molecular weight of water = 18). At a location, the total pressure is 100 kPa, the temperature is 30^{0}C and the relative humidity is 55%. Given that the saturation pressure of water at 30^{0}C is 4246 Pa, the mass of water vapour per kg of dry air is ______ grams.
(2 Mark, 2017[1])

Ans: 14.87
Explanation:
Given, Relative humidity \left( \phi \right) =\frac { { p }_{ v } }{ { p }_{ s } }
=> 0.55=\frac { { p }_{ v } }{ 4.246 }
=> { p }_{ v }=2.335kPa
Specific humidity \left( \omega \right) =\frac { { 0.622p }_{ v } }{ { p }_{ a }{ p }_{ v } } =\frac { 0.622\times 2.335 }{ 1002.335}
=> \omega = 0.01487 kg/kg of dry air
=> \omega = 14.87 g/kg of dry air
31. If a mass of moist air contained in a closed metallic vessel is heated, them its
(a) Relative humidity decreases
(b) Relative humidity increases
(c) Specific humidity increases
(d) Specific humidity decreases
(1 Mark, 2017[2])
32. Ambient air is at a pressure of 100 kPa, dry bulb temperature of o 30^{0}C and 60% relative humidity. The saturation pressure of water at 30^{0}C is 4.24 kPa. The specific humidity of air (in g/kg of dry air) is ________ (correct to two decimal places).
(2 Mark, 2018[2])

Ans: 16.23
Explanation:
Given, Relative humidity \left( \phi \right) =\frac { { p }_{ v } }{ { p }_{ s } }
=> 0.6=\frac { { p }_{ v } }{ 4.24 }
=> { p }_{ v }=2.544kPa
Specific humidity \left( \omega \right) =\frac { { 0.622p }_{ v } }{ { p }_{ a }{ p }_{ v } } =\frac { 0.622\times 2.544 }{ 1002.544 }
=> \omega = 0.01623 kg/kg of dry air
=> \omega = 16.23 g/kg of dry air