# Psychrometric Process

 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).
 A 135 B 115 C 98 D 175
GATE ME 2021 SET-2   Refrigeration and Air-conditioning
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).
 A 4.96 B 2.25 C 6.25 D 8.12
GATE ME 2021 SET-1   Refrigeration and Air-conditioning
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}
 Question 3
If a mass of moist air contained in a closed metallic vessel is heated, then its
 A relative humidity decreases B relative humidity increases C specific humidity increases D specific humidity decreases
GATE ME 2017 SET-2   Refrigeration and Air-conditioning
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 __________
 A 17.04 B 23.43 C 12.54 D 19.43
GATE ME 2016 SET-3   Refrigeration and Air-conditioning
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}
 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 __________
 A 15.34 C B 13.54 C C 19.89 C D 11.56 C
GATE ME 2016 SET-2   Refrigeration and Air-conditioning
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}
 Question 6
Air in a room is at $35^{\circ}$ and 60% relative humidity (RH). The pressure in the room is 0.1 MPa. The saturation pressure of water at $35^{\circ}$ is 5.63 kPa. The humidity ratio of the air (in gram/kg of dry air) is ________
 A 21.75 B 45.23 C 15.26 D 89.56
GATE ME 2015 SET-3   Refrigeration and Air-conditioning
Question 6 Explanation:
Given data:
\begin{aligned} T_{d b} &=35^{\circ} \mathrm{C} \\ \phi &=60 \%=0.60 \\ p &=0.1 \mathrm{MPa}=100 \mathrm{kPa} \\ p_{s} &=5.63 \mathrm{kPa} \text { at } 35^{\circ} \mathrm{C} \end{aligned}
Relative humidity,
\begin{aligned} \phi &=\frac{p_{v}}{p_{s}} \\ 0.60 &=\frac{p_{v}}{5.63} \\ p_{v} &=0.60 \times 5.63=3.378 \mathrm{kPa} \end{aligned}
Humidity ratio,
$\omega=\frac{0.622 p_{v}}{p-p_{v}}=\frac{0.622 \times 3.378}{100-3.378}$
$=0.02174 \mathrm{kg} \;\text{of}\; \mathrm{w.v.}/\mathrm{kg}\;$ of dry air
$=21.74 \mathrm{gram} \; \text{of} \;\mathrm{w} . \mathrm{v} . / \mathrm{kg}\;$of dry air
 Question 7
Moist air at 35$^{\circ}$C and 100% relative humidity is entering a psychrometric device and leaving at 25$^{\circ}$C and 100% relative humidity. The name of the device is
 A Humidifier B Dehumidifier C Sensible heater D Sensible cooler
GATE ME 2014 SET-4   Refrigeration and Air-conditioning
Question 7 Explanation:
$\omega_{2} \lt \omega_{1} \quad \because \text { Dehumidifier }$ Question 8
A sample of moist air at a total pressure of 85 kPa has a dry bulb temperature of 30$^{\circ}$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 _______
 A 15.35 B 98.65 C 20.84 D 28.69
GATE ME 2014 SET-3   Refrigeration and Air-conditioning
Question 8 Explanation:
Total pressure: $p=85 \mathrm{kPa}$
Dry bulb temperature,
\begin{aligned} T_{d b} &=30^{\circ} \mathrm{C} \\ p_{s} &=4.24 \mathrm{kPa} \\ \phi &=65 \%=0.65 \end{aligned}
Relative humidity
\begin{aligned} \phi &=\frac{p_{v}}{p_{s}} \\ 0.65 &=\frac{p_{v}}{4.24} \\ \text{or}\quad p_{v} &=0.65 \times 4.24=2.756 \mathrm{kPa} \end{aligned}
Absolute humidity,
\begin{aligned} \omega &=\frac{0.622 p_{v}}{p-p_{v}} \\ &=\frac{0.622 \times 2.756}{85-2.756} \end{aligned}
=0.02084 kg of w.v/kg of dry air
=20.84 gram of w.v./kg of dry air
 Question 9
Air (at atmospheric pressure) at a dry bulb temperature of 40$^{\circ}$C and wet bulb temperature of 20$^{\circ}$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$^{\circ}$C B 20$^{\circ}$C C 25$^{\circ}$C D 30$^{\circ}$C
GATE ME 2008   Refrigeration and Air-conditioning
Question 9 Explanation:
Given data:
\begin{aligned} (\mathrm{DBT})_{i}&=40^{\circ} \mathrm{C} \\ (\mathrm{WBT})_{i}&=20^{\circ} \mathrm{C} \end{aligned}
Web bulb depression,
\begin{aligned} (\mathrm{WBD})_{i} &=(\mathrm{DBT})_{i}-(\mathrm{WBT})_{i} \\ &=40-20=20^{\circ} \mathrm{C} \end{aligned}
Wet bulb depression at outlet,
$(\mathrm{WBD})_{\mathrm{o}}=(\mathrm{DBT})_{\mathrm{o}}-(\mathrm{WBT})_{\mathrm{o}}$
By given condition, $(WBD)_{o}=25 \%(\mathrm{WBD})_{i}$
\begin{aligned} \therefore(\mathrm{DBT})_{0}-(\mathrm{WBT})_{0}&=0.25 \times 20 \\ (\mathrm{DBT})_{0} &=0.25 \times 20+(\mathrm{WBT})_{0} \\ &=5+(\mathrm{WBT})_{i} \end{aligned}
where $(\mathrm{WBT})_{o}=(\mathrm{WBT})_{i}$ for humidified in an air washer
$\therefore \quad(\mathrm{DBT})_{0}=5+20=25^{\circ} \mathrm{C}$ Question 10
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 B 123.8 C 128.2 D 159
GATE ME 2007   Refrigeration and Air-conditioning
Question 10 Explanation:
Given data:
Mass flow rate of dry air
\begin{aligned} m_{a}&=3 \mathrm{kg} / \mathrm{s} \\ h_{1}&=85 \mathrm{kJ} / \mathrm{kg} \text { of dry air } \\ \omega_{1}&=19 \mathrm{gm} / \mathrm{kg} \text { of dry air } \\ &=0.019 \mathrm{kg} / \mathrm{kg} \text { of dry air } \\ h_{2}&=43 \mathrm{kJ} / \mathrm{kg} \text { of dry air } \\ \omega_{2}&=8 \mathrm{gm} / \mathrm{kg} \text { of dry air } \\ &=0.008 \mathrm{kg} / \mathrm{kg} \text { of dry air } \end{aligned} Enthalpy of condenrate water leaves the cooling
\begin{aligned} \text { coil: } h_{f g}&=67 \mathrm{kJ} / \mathrm{kg} \\ \omega_{1}&=\frac{m_{v 1}}{m_{a}}\\ \text{or }\quad m_{v 1}&=\omega_{1} m_{a}=0.019 \times 3 \mathrm{kg} / \mathrm{s} \\ & =0.057 \mathrm{kg} / \mathrm{s}, \text{mass of water vapour at inlet} \\ \end{aligned}
Similarly,
$m_{v 2}=\omega_{2} m_{a}=0.008 \times 3$
$=0.024 \mathrm{kg} / \mathrm{s},$ mass of water vapour at outlet
Mass flow rate of condensate,
\begin{aligned} m_{w} &=m_{v 1}+m_{v 2} \\ &=0.057+0.024=0.033 \mathrm{kg} / \mathrm{s} \end{aligned}
Cooling capacity of the coil,
\begin{aligned} Q&=m\left(h_{1}-h_{2}\right)-m_{w} h_{f g}\\ \text{where }m\left(h_{1}-h_{2}\right)&= \text{sensible heat removed}\\ \therefore m_{w} h_{f g}&= \text{ Latent heat removed}\\ Q &=3(85-43)-0.033 \times 67 \\ &=123.79 \mathrm{kW} \end{aligned}
Alternatively Mass flow rate of condensate,
\begin{aligned} m_{w} &=m_{a}\left(\omega_{1}-\omega_{2}\right) \\ &=3 \times(0.019-0.008) \\ &=0.033 \mathrm{kg} / \mathrm{s} \end{aligned}
Applying energy balance equation, we get
\begin{aligned} m_{a} h_{1} &=m_{a} h_{2}+m_{w} h_{f g}+Q \\ 3 \times 85 &=3 \times 43+0.033 \times 67+Q \\ Q &=123.8 \mathrm{kW} \end{aligned}
There are 10 questions to complete.