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.
21.4 | |
35.1 | |
21.9 | |
13.6 |
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
p_v=0.5p_{sat} | |
p_v=p_{sat} | |
p_v=0.622p_{sat} | |
p_v=2p_{sat} |
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).
102.25 | |
8.68 | |
18.68 | |
16.24 |
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.
13.74 | |
14.92 | |
15.46 | |
16.98 |
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}
\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 _____________
0.02kg | |
0.98kg | |
0.45kg | |
0.29kg |
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}
\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}
Question 6 |
The pressure, dry bulb temperature and relative humidity of air in a room are 1 bar, 30^{\circ}C and 70%, respectively. If the saturated steam pressure at 30^{\circ}C is 4.25 kPa, the specific humidity of the room air in kg water vapour/kg dry air is
0.0083 | |
0.0101 | |
0.0191 | |
0.0232 |
Question 6 Explanation:
\begin{aligned} p_{\mathrm{atn}} &=1 \mathrm{bar}=100 \mathrm{kPa} \\ \mathrm{DBT} &=30^{\circ} \mathrm{C} \\ \phi &=70 \%=0.7 \\ p_{\mathrm{vs}} &=4.25 \mathrm{kPa} \end{aligned}
Specific humidity: \omega=?
\begin{aligned} \phi &=\frac{p_{v}}{p_{v s}} \\ 0.7 &=\frac{p_{v}}{4.25} \\ p_{v} &=2.975 \mathrm{kPa} \end{aligned}
Specific humidity,
\begin{aligned} \omega&=0.622 \times \frac{p_{v}}{p-p_{v}}\\ &=0.622 \times \frac{2.975}{100-2.975} \\ &=00.0191 \end{aligned}
Specific humidity: \omega=?
\begin{aligned} \phi &=\frac{p_{v}}{p_{v s}} \\ 0.7 &=\frac{p_{v}}{4.25} \\ p_{v} &=2.975 \mathrm{kPa} \end{aligned}
Specific humidity,
\begin{aligned} \omega&=0.622 \times \frac{p_{v}}{p-p_{v}}\\ &=0.622 \times \frac{2.975}{100-2.975} \\ &=00.0191 \end{aligned}
Question 7 |
A room contains 35kg of dry air & 0.5 kg water vapor. The total pressure and temperature of air in the room are 100kPa and 25^{\circ}C
respectively. Given that the saturation pressure for water at 25^{\circ}C is 3.17kPa, the relative humidity of the air in the room is
67% | |
55% | |
83% | |
71% |
Question 7 Explanation:
\begin{aligned} m_{a} &=35 \mathrm{kg} ; \quad m_{v}=0.5 \mathrm{kg} \\ \omega &=\frac{m_{v}}{m_{a}}=\frac{0.5}{35}=0.01428 \\ a \mathrm{so} & \omega=0.622 \frac{p_{v}}{p_{a}} \\ p_{t} &=100 \mathrm{kPa} \\ \therefore \quad \omega &=0.622 \frac{p_{v}}{p_{t}-p_{v}} \end{aligned}
Now,
\begin{aligned} 0.01428 &=0.622 \frac{p_{v}}{100-p_{v}} \\ p_{v} &=2.238 \mathrm{kPa} \end{aligned}
Relative humidity
\begin{aligned} \phi &=\frac{p_{v}}{p_{v s}}=\frac{2.238}{3.17} \times 100 \\ &=70.6 \% \approx 71 \% \end{aligned}
Now,
\begin{aligned} 0.01428 &=0.622 \frac{p_{v}}{100-p_{v}} \\ p_{v} &=2.238 \mathrm{kPa} \end{aligned}
Relative humidity
\begin{aligned} \phi &=\frac{p_{v}}{p_{v s}}=\frac{2.238}{3.17} \times 100 \\ &=70.6 \% \approx 71 \% \end{aligned}
Question 8 |
If a mass of moist air in an airtight vessel is heated to a higher temperature, then
Specific humidity of the air increases | |
Specific humidity of the air decreases | |
Relative humidity of the air increases | |
Relative humidity of the air decreases |
Question 8 Explanation:
The moist air in an airtight vessel is heated to a higher temperature. This process is treated as sensible heating. In sensible heating process, the relative humidity of the air decreases with increase in dry bulb temperature at constnat specific humidity
This process is shown on psychrometic chart in Fig.
This process is shown on psychrometic chart in Fig.
Question 9 |
A moist air sample has dry bulb temperature of 30^{\circ}C 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 30^{\circ}C is 4.24kPa and the total pressure is 90kPa, then the relative humidity (in %) of air sample is
50.5 | |
38.5 | |
56.5 | |
68.5 |
Question 9 Explanation:
Relative humidity,
\phi=\frac{p_{v}}{p_{v s}}
Specific humidity
\begin{aligned} \omega &=0.622\left(\frac{p_{v}}{p-p_{v}}\right) \\ 11.5 \times 10^{-3} &=0.622\left(\frac{p_{v}}{90-p_{v}}\right) \\ p_{v} &=1.63378 \mathrm{KPa} \end{aligned}
Relative humidity
\begin{aligned} \phi &=\frac{p_{v}}{p_{v s}}=\frac{1.63378}{4.24} \\ &=0.385=38.5 \% \end{aligned}
\phi=\frac{p_{v}}{p_{v s}}
Specific humidity
\begin{aligned} \omega &=0.622\left(\frac{p_{v}}{p-p_{v}}\right) \\ 11.5 \times 10^{-3} &=0.622\left(\frac{p_{v}}{90-p_{v}}\right) \\ p_{v} &=1.63378 \mathrm{KPa} \end{aligned}
Relative humidity
\begin{aligned} \phi &=\frac{p_{v}}{p_{v s}}=\frac{1.63378}{4.24} \\ &=0.385=38.5 \% \end{aligned}
Question 10 |
Moist air at a pressure of 100 kPa is compressed to 500 kPa and then cooled to 35^{\circ}C in an after
cooler. The air at the entry to the after cooler is unsaturated and becomes just saturated at the exit
of the after cooler. The saturation pressure of water at 35^{\circ}C is 5.628 kPa. The partial pressure of
water vapour (in kPa) in the moist air entering the compressor is closest to
0.57 | |
1.13 | |
2.26 | |
4.52 |
Question 10 Explanation:
Given : p_1=100 kPa, \;p_2=500kPa,\; p_{v1}=?
p_2=5.628 kPa\text{ (Saturated pressure at }35^{\circ}C)
We know that,
Specific humidity W=0.622\left ( \frac{p_v}{p-p_v} \right )
For case II:
W=0.622\left ( \frac{5.628}{500-5.628} \right )=7.08 \times 10^{-3}\; kg/kg \text{ of dry air}
For saturated air specific humidity remains same. So, for case (I) :
W=0.622\left ( \frac{p_{v1}}{p_1-p_{v1}} \right )
On substituting the values, we get
\begin{aligned} 7.08 \times 10^{-3}&=0.622\left ( \frac{p_{v1}}{100-p_{v1}} \right ) \\ 11.38 \times 10^{-3}&(100-p_{v1}) =p_{v1} \\ 1.138&= 1.01138p_{v1}\\ p_{v1}&=1.125\;kPa\simeq 1.13 \;kPa \end{aligned}
p_2=5.628 kPa\text{ (Saturated pressure at }35^{\circ}C)
We know that,
Specific humidity W=0.622\left ( \frac{p_v}{p-p_v} \right )
For case II:
W=0.622\left ( \frac{5.628}{500-5.628} \right )=7.08 \times 10^{-3}\; kg/kg \text{ of dry air}
For saturated air specific humidity remains same. So, for case (I) :
W=0.622\left ( \frac{p_{v1}}{p_1-p_{v1}} \right )
On substituting the values, we get
\begin{aligned} 7.08 \times 10^{-3}&=0.622\left ( \frac{p_{v1}}{100-p_{v1}} \right ) \\ 11.38 \times 10^{-3}&(100-p_{v1}) =p_{v1} \\ 1.138&= 1.01138p_{v1}\\ p_{v1}&=1.125\;kPa\simeq 1.13 \;kPa \end{aligned}
There are 10 questions to complete.