Question 1 |
A rigid tank of volume 50\; m^3 contains a pure substance as a saturated liquid vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are : T_{sat}=143.61^{\circ}C,\; v_f=0.001084\; m^3/kg, v_g=0.46242\; m^3/kg. The total mass of liquid vapour mixture in the tank is _______kg (round off to the nearest integer).
135 | |
115 | |
98 | |
175 |
Question 1 Explanation:
\begin{aligned} V_{\text {total }} &=v_{l}+v_{v} \\ &=v_{f} \times m_{l}+v_{g} \times m_{v} \\ 50 &=0.001084 \times 0.2 \mathrm{~m}+0.46242 \times 0.8 \mathrm{~m} \\ m &=135.08 \mathrm{~kg}=135 \mathrm{~kg} \end{aligned}
Question 2 |
An air-conditioning system provides a continuous flow of air to a room using an intake duct and an exit duct, as shown in the figure. To maintain the quality of the indoor air, the intake duct supplies a mixture of fresh air with a cold air stream. The two streams are mixed in an insulated mixing chamber located upstream of the intake duct. Cold air enters the mixing chamber at 5^{\circ} C, 105 kPa with a volume flow rate of 1.25 m^3/s during steady state operation. Fresh air enters the mixing chamber at 34^{\circ}C and 105 kPa. The mass flow rate of the fresh air is 1.6 times of the cold air stream. Air leaves the room through the exit duct at 24^{\circ}C.
Assuming the air behaves as an ideal gas with c_p= 1.005 \;\; kJ/kg.K and R= 0.287 kJ/kg.K, the rate of heat gain by the air from the room is ________kW(round off to two decimal places).
Assuming the air behaves as an ideal gas with c_p= 1.005 \;\; kJ/kg.K and R= 0.287 kJ/kg.K, the rate of heat gain by the air from the room is ________kW(round off to two decimal places).
4.96 | |
2.25 | |
6.25 | |
8.12 |
Question 2 Explanation:
1: Cold air
2: Hot air
\begin{aligned} P_{1} V_{1} &=\dot{m}_{1} R T_{1} \\ 105 \times 1.25 &=\dot{m}_{1} \times 0.287 \times 278 \\ \dot{m}_{1} &=1.645 \mathrm{~kg} / \mathrm{sec} \\ \dot{m}_{2} &=1.6 \times 1.645=2.632 \mathrm{~kg} / \mathrm{sec} \\ \dot{m}_{3} &=4.277 \mathrm{~kg} / \mathrm{sec} \end{aligned}
After mixing: 0
\begin{aligned} \dot{m}_{1} t_{1}+\dot{m}_{2} t_{2} &=\dot{m}_{3} t_{3} \\ 1.645 \times 5+2.632 \times 34 &=4.277 t_{3} \\ t_{3} &=22.85^{\circ} \mathrm{C}\\ \text{Heat gain }:&=h_{4}-h_{3}\\ &=\dot{m}_{3} c_{p}\left(t_{4}-t_{3}\right) \\ &=4.277 \times 1.005(24-22.85) \\ &=4.963 \mathrm{~kW} \end{aligned}
2: Hot air
\begin{aligned} P_{1} V_{1} &=\dot{m}_{1} R T_{1} \\ 105 \times 1.25 &=\dot{m}_{1} \times 0.287 \times 278 \\ \dot{m}_{1} &=1.645 \mathrm{~kg} / \mathrm{sec} \\ \dot{m}_{2} &=1.6 \times 1.645=2.632 \mathrm{~kg} / \mathrm{sec} \\ \dot{m}_{3} &=4.277 \mathrm{~kg} / \mathrm{sec} \end{aligned}
After mixing: 0
\begin{aligned} \dot{m}_{1} t_{1}+\dot{m}_{2} t_{2} &=\dot{m}_{3} t_{3} \\ 1.645 \times 5+2.632 \times 34 &=4.277 t_{3} \\ t_{3} &=22.85^{\circ} \mathrm{C}\\ \text{Heat gain }:&=h_{4}-h_{3}\\ &=\dot{m}_{3} c_{p}\left(t_{4}-t_{3}\right) \\ &=4.277 \times 1.005(24-22.85) \\ &=4.963 \mathrm{~kW} \end{aligned}
Question 3 |
If a mass of moist air contained in a closed metallic vessel is heated, then its
relative humidity decreases | |
relative humidity increases | |
specific humidity increases | |
specific humidity decreases |
Question 3 Explanation:
Question 4 |
In a mixture of dry air and water vapor at a total pressure of 750 mm of Hg, the partial pressure of water vapor is 20 mm of Hg. The humidity ratio of the air in grams of water vapor per kg of dry air (g_{w}/kg_{da}) is __________
17.04 | |
23.43 | |
12.54 | |
19.43 |
Question 4 Explanation:
Given data:
Total pressure,
p=750 \mathrm{mm} \text { of } \mathrm{Hg}
Partial pressure,
p_{v}=20 \mathrm{mm} \text { of } \mathrm{Hg}
We know that humidity ratio,
\begin{aligned} \omega &=\frac{0.622 p_{v}}{p-p_{v}} \mathrm{kg} w.v./ \mathrm{kg} d.a. \\ &=\frac{0.622 \times 20}{750-20} \\ &=0.01704 \;\mathrm{kg} \mathrm{w} . \mathrm{v} . \mathrm{kg} \text { of } \mathrm{d.a} \\ &=17.04 \;\mathrm{g} \text { of } \mathrm{w} . \mathrm{v} . / \mathrm{kg} \text { of } \mathrm{d.a} \end{aligned}
Total pressure,
p=750 \mathrm{mm} \text { of } \mathrm{Hg}
Partial pressure,
p_{v}=20 \mathrm{mm} \text { of } \mathrm{Hg}
We know that humidity ratio,
\begin{aligned} \omega &=\frac{0.622 p_{v}}{p-p_{v}} \mathrm{kg} w.v./ \mathrm{kg} d.a. \\ &=\frac{0.622 \times 20}{750-20} \\ &=0.01704 \;\mathrm{kg} \mathrm{w} . \mathrm{v} . \mathrm{kg} \text { of } \mathrm{d.a} \\ &=17.04 \;\mathrm{g} \text { of } \mathrm{w} . \mathrm{v} . / \mathrm{kg} \text { of } \mathrm{d.a} \end{aligned}
Question 5 |
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 ^{\circ} C) is __________
15.34 C | |
13.54 C | |
19.89 C | |
11.56 C |
Question 5 Explanation:
Given data:
Relative humidity,
\phi=70 \%=0.70
Partial pressure,
p_{v}=1.6 \mathrm{kPa}
Total pressure,
p_{0}=101.325 \mathrm{kPa}
We know that,
\begin{aligned} \phi&=\frac{p_{v}}{p_{s}}\\ 0.70 &=\frac{1.6}{p_{s}} \\ \text{or}\quad p_{s} &=\frac{1.6}{0.70}=2.2857 \mathrm{kPa} \end{aligned}
Temperature corresponding to saturation pressure
is dry bulb temperature,
\begin{aligned} \ln \left\{\frac{p_{s}}{p_{o}}\right\} &=14.317-\frac{5304}{T_{s}} \\ \ln \left\{\frac{2.2857}{101.325}\right\} &=14.317-\frac{5304}{T_{s}} \\ -3.79166 &=14.317-\frac{5304}{T_{s}} \\ \frac{5304}{T_{s}} &=14.317+3.79166 \\ \frac{5304}{T_{s}} &=18.10866 \\ \text{or}\quad T_{s} &= 292.89 \mathrm{K}\\ &=(292.89-273)^{\circ} \mathrm{C} \\ &=19.89^{\circ} \mathrm{C} \end{aligned}
Relative humidity,
\phi=70 \%=0.70
Partial pressure,
p_{v}=1.6 \mathrm{kPa}
Total pressure,
p_{0}=101.325 \mathrm{kPa}
We know that,
\begin{aligned} \phi&=\frac{p_{v}}{p_{s}}\\ 0.70 &=\frac{1.6}{p_{s}} \\ \text{or}\quad p_{s} &=\frac{1.6}{0.70}=2.2857 \mathrm{kPa} \end{aligned}
Temperature corresponding to saturation pressure
is dry bulb temperature,
\begin{aligned} \ln \left\{\frac{p_{s}}{p_{o}}\right\} &=14.317-\frac{5304}{T_{s}} \\ \ln \left\{\frac{2.2857}{101.325}\right\} &=14.317-\frac{5304}{T_{s}} \\ -3.79166 &=14.317-\frac{5304}{T_{s}} \\ \frac{5304}{T_{s}} &=14.317+3.79166 \\ \frac{5304}{T_{s}} &=18.10866 \\ \text{or}\quad T_{s} &= 292.89 \mathrm{K}\\ &=(292.89-273)^{\circ} \mathrm{C} \\ &=19.89^{\circ} \mathrm{C} \end{aligned}
There are 5 questions to complete.