# Properties of Moist Air

 Question 1
Ambient pressure, temperature, and relative humidity at a location are 101 kPa, 300 K, and 60%, respectively. The saturation pressure of water at 300 K is 3.6 kPa. The specific humidity of ambient air is _______g/kg of dry air.
 A 21.4 B 35.1 C 21.9 D 13.6
GATE ME 2021 SET-2   Refrigeration and Air-conditioning
Question 1 Explanation:
\begin{aligned} \phi &=0.6 \\ \frac{P_{v}}{P_{v s}} &=0.6 \\ P_{v s} &=3.6 \mathrm{kPa} \\ P_{v} &=0.6 \times 3.6 \mathrm{kPa} \\ P_{v} &=2.16 \mathrm{kPa} \\ \omega &=0.622\left(\frac{P_{V}}{P-P_{V}}\right) \\ &=0.622\left(\frac{2.16}{101-2.16}\right)\\ &=0.01358 \mathrm{~kg} \text { of water vapour/kg of dry air }\\ &=13.58 \mathrm{~g} / \mathrm{kg} \text { of dry air } \end{aligned}
 Question 2
The relative humidity of ambient air at 300 K is 50% with a partial pressure of water vapour equal to $p_v$. The saturation pressure of water at 300 K is $p_{sat}$. The correct relation for the air-water mixture is
 A $p_v=0.5p_{sat}$ B $p_v=p_{sat}$ C $p_v=0.622p_{sat}$ D $p_v=2p_{sat}$
GATE ME 2021 SET-1   Refrigeration and Air-conditioning
Question 2 Explanation:
\begin{aligned} \text { Relative humidity, } \phi &=\frac{p_{v}}{p_{\text {sat }}} \\ 0.5 &=\frac{p_{v}}{p_{\text {sat }}} \\ p_{v} &=0.5 p_{\text {sat }} \end{aligned}

 Question 3
Ambient air is at a pressure of 100 kPa, dry bulb temperature of $30^{\circ}C$ and and 60% relative humidity. The saturation pressure of water at $30^{\circ}C$ is 4.24 kPa. The specific humidity of air (in g/kg of dry air) is ________ (correct to two decimal places).
 A 102.25 B 8.68 C 18.68 D 16.24
GATE ME 2018 SET-2   Refrigeration and Air-conditioning
Question 3 Explanation:
\begin{aligned} P_{\mathrm{atm}} &=100 \mathrm{kPa} \\ D B T &=t=30^{\circ} \mathrm{C} \rightarrow P_{v s}=4.24 \mathrm{kPa} \\ \phi &=60 \%\\ \phi &=\frac{P_{V}}{P_{V S}} \Rightarrow 0.6=\frac{P_{V}}{4.24} \\ P_{V} &=2.544 \mathrm{kPa} \\ W &=0.622 \times \frac{P_{V}}{P_{\mathrm{atm}}-P_{V}} \\ W &=0.622 \times \frac{2.544}{100-2.544} \\ W&=16.24 \text{gram/kg of dry Air} \end{aligned}
 Question 4
Moist air is treated as an ideal gas mixture of water vapor 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 $30^{\circ}C$ and the relative humidity is 55%. Given that the saturation pressure of water at $30^{\circ}C$.is 4246 Pa, the mass of water vapor per kg of dry air is _____________ grams.
 A 13.74 B 14.92 C 15.46 D 16.98
GATE ME 2017 SET-1   Refrigeration and Air-conditioning
Question 4 Explanation:
Relative humidity:
\begin{aligned} \phi &=\frac{p_{v}}{p_{v s}} \\ 0.55 &=\frac{p_{v}}{4246}\\ \text { or } \quad p_{v}&=2335.3 \mathrm{Pa}=2.335 \mathrm{kPa} \\ \therefore \quad \omega &=\frac{(\mathrm{Mol} \cdot \mathrm{wt})_{\mathrm{H}_{2} \mathrm{O}}}{(\mathrm{Mol} \cdot \mathrm{w} \mathrm{t})_{\mathrm{air}}} \times\left(\frac{p_{v}}{p-p_{v}}\right) \\ &=\frac{18}{28.84} \times\left(\frac{2.335}{100-2.335}\right) \\ &=0.01492 \mathrm{kg} / \mathrm{kg} \text { of d.a } \\ \text { also } \quad \omega &=\frac{m_{v}}{m_{a}} \\ \therefore \quad 0.01492 &=\frac{m_{v}}{1}\\ \text{or }\quad m_{v}&=0.01492 \mathrm{kg}=14.92 \mathrm{gm} \end{aligned}
 Question 5
A stream of moist air (mass flow rate = 10.1 kg/s) with humidity ratio of 0.01 $\frac{kg}{kg \; dry \; air}$ 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 $\frac{kg}{kg \; dry \; air}$ ) is _____________
 A 0.02kg B 0.98kg C 0.45kg D 0.29kg
GATE ME 2015 SET-1   Refrigeration and Air-conditioning
Question 5 Explanation:
Mass flow rate of moist air $=10.1 \mathrm{kg} / \mathrm{s}$
$\text { Humidity ratio: } \omega=\frac{m_{v}}{m_{a}}=0.01 \mathrm{kg} / \mathrm{s}$
Mass of moist air = Mass of dry air + Mass of water vapour
\begin{aligned} 10.1 &=m_{a}+0.01 \times m_{a} \\ 10.1 m_{a} &=10.1 \end{aligned}
Mass of dry air: $m_{a}=\frac{10.1}{1.01}=10 \mathrm{kg} / \mathrm{s}$
Mass of water vapour,
\begin{aligned} m_{v_{1}} &=10.1-10=0.1 \mathrm{kg} / \mathrm{s} \\ m_{v_{2}} &=0.1 \mathrm{kg} / \mathrm{s} \\ \left(m_{\mathrm{v}}\right)_{\mathrm{total}} &=m_{v_{1}}+m_{v_{2}}=0.2 \mathrm{kg} / \mathrm{s} \end{aligned}
Humidity ratio
\begin{aligned} \omega_{\mathrm{final}}=\frac{m_{v}}{m_{\mathrm{a}}} &=\frac{0.2}{10}\\ &=0.02 \mathrm{kg} / \mathrm{kg} \text{ of dry air} \end{aligned}

There are 5 questions to complete.