# Vapor and Gas Refrigeration

 Question 1
Consider an ideal vapour compression refrigeration cycle working on R-134a refrigerant. The COP of the cycle is 10 and the refrigeration capacity is 150 kJ/kg. The heat rejected by the refrigerant in the condenser is _______kJ/kg (round off to the nearest integer).
 A 125 B 185 C 215 D 165
GATE ME 2021 SET-2   Refrigeration and Air-conditioning
Question 1 Explanation:
\begin{aligned} \mathrm{RE} &=150 \mathrm{~kJ} / \mathrm{kg} \\ \mathrm{COP} &=10 \\ \mathrm{COP} &=\frac{Q_{L}}{W_{i n}}\\ \Rightarrow\qquad W_{\text {in }} &=15 \mathrm{~kJ} / \mathrm{kg} \\ Q_{\mathrm{H}}=Q_{\mathrm{C}} &=Q_{L}+W_{\text {in }} \\ &=150+15 \\ &=165 \mathrm{~kJ} / \mathrm{kg} \end{aligned}

 Question 2
Consider an ideal vapour compression refrigeration cycle. If the throttling process is replaced by an isentropic expansion process, keeping all the other processes unchanged, which one of the following statements is true for the modified cycle?
 A Coefficient of performance is higher than that of the original cycle. B Coefficient of performance is lower than that of the original cycle. C Coefficient of performance is the same as that of the original cycle. D Refrigerating effect is lower than that of the original cycle.
GATE ME 2019 SET-1   Refrigeration and Air-conditioning
Question 2 Explanation:
Due to isentropic expansion instead of throttling
1. R.E increases
2. Work input reduces
$\therefore \mathrm{COP}=\frac{\mathrm{R} \cdot \mathrm{E}}{\mathrm{W}_{\mathrm{in}}}$

 Question 3
A standard vapor compression refrigeration cycle operating with a condensing temperature of $35^{\circ}C$ and an evaporating temperature of $-10^{\circ}C$ develops 15 kW of cooling. The p-h diagram shows the enthalpies at various states. If the isentropic efficiency of the compressor is 0.75, the magnitude of compressor power (in kW) is _________ (correct to two decimal places).
 A 8 B 10 C 12 D 6
GATE ME 2018 SET-2   Refrigeration and Air-conditioning
Question 3 Explanation:
\begin{aligned} R C &=15 \mathrm{kW} \\ R C &=\dot{m} \times\left(h_{1}-h_{4}\right) \\ 15 &=\dot{m} \times(400-250) \\ \dot{m} &=0.1 \mathrm{kg} / \mathrm{sec} \\ \omega_{\text {isentropic }} &=\left(h_{2}-h_{1}\right)=(475-400)=75 \mathrm{kJ} / \mathrm{kg}\\ \eta_{C} &=\frac{\omega_{\text {isentropic }}}{\omega_{\text {actual }}} \\ \omega_{\text {actual }} &=\frac{75}{0.75}=100 \mathrm{kJ} / \mathrm{kg} \\ \omega_{\text {actual }} &=\dot{m} \times \omega_{\text {actual }}=0.1 \times 100 \\ P_{\dot{m}} &=10 \mathrm{kW} \end{aligned}
 Question 4
In the vapour compression cycle shown in the figure, the evaporating and condensing temperatures are 260 K and 310 K, respectively. The compressor takes in liquid-vapour mixture (state 1) and isentropically compresses it to a dry saturated vapour condition (state 2). The specific heat of the liquid refrigerant is 4.8 kJ/kg-K and may be treated as constant. The enthalpy of evaporation for the refrigerant at 310 K is 1054 kJ/kg.

The difference between the enthalpies at state points 1 and 0 (in kJ/kg) is ____________
 A 1000.21kJ/kg B 1103.51kJ/kg C 1212.90kJ/kg D 1443.78kJ/kg
GATE ME 2016 SET-3   Refrigeration and Air-conditioning
Question 4 Explanation:
Given data:
\begin{aligned} T_{e} &=260 \mathrm{K} \\ T_{c} &=310 \mathrm{K} \\ c_{p l} &=4.8 \mathrm{kJ} / \mathrm{kgK} \\ h_{fg} &=1054 \mathrm{kJ} / \mathrm{kg} \text { at } T_{c}=310 \mathrm{K} \end{aligned}

Taking reference temperature T , at which entropy is $s_{T}$
\begin{aligned} \therefore \quad s_{3}-s_{T} &=c \ln \frac{T_{3}}{T} \\ s_{3} &=s_{T}+c \ln \frac{T_{3}}{T}=s_{T}+4.8 \ln \frac{310}{T} \\ \text { Similarly, } s_{0} &=s_{T}+c \ln \frac{260}{T}=s_{T}+4.8 \ln \frac{260}{T} \\ \therefore \quad s_{2}-s_{3} &=\frac{h_{f g}}{T}=\frac{1054}{310} \\ s_{2} &=s_{3}+\frac{1054}{310} \end{aligned}
Substitute the value of $s_{3}$ in above equation, we get
\begin{aligned} s_{2} &=s_{T}+4.8 \ln \frac{310}{T}+\frac{1054}{310} \\ s_{2} &=s_{1} \\ s_{1} &=s_{T}+4.8 \ln \frac{310}{T}+\frac{1054}{310} \\ h_{1}-h_{0} &=T_{e}\left[s_{1}-s_{0}\right] \\ =& 260 \times\left[s_{T}+4.8 \ln \frac{310}{T}+\frac{1054}{310}\right.\\ &\left.-s_{T}-4.8 \ln \frac{260}{T}\right] \\ &=260\left[4.8 \ln \left(\frac{310}{T} \times \frac{T}{260}\right)+\frac{1054}{310}\right] \\ &1103.51\mathrm{kJ}/\mathrm{kg} \end{aligned}
 Question 5
A refrigerator uses R-134a as its refrigerant and operates on an ideal vapour-compression refrigeration cycle between 0.14 MPa and 0.8 MPa. If the mass flow rate of the refrigerant is 0.05 kg/s, the rate of heat rejection to the environment is _________ kW.
Given data:
At P = 0.14 MPa, h =236.04 kJ/kg, s =0.9322 kJ/kg-K
At P = 0.8 MPa, h = 272.05kJ/kg (superheated vapour)
At P = 0.8 MPa, h = 93.42 kJ/kg (saturated liquid)
 A 3.43 kW B 8.93 kW C 2.45 kW D 4.56 kW
GATE ME 2016 SET-2   Refrigeration and Air-conditioning
Question 5 Explanation:
Given data:
\begin{aligned} m&=0.05 \mathrm{kg} / \mathrm{s}\\ \text{At }\quad p_{e}&=0.14 \mathrm{MPa} \\ h_{1}&=236.04 \mathrm{kJ} / \mathrm{kg} \\ s_{1}&=0.9322 \mathrm{kJ} / \mathrm{kg} \mathrm{K}\\ \text{At }\quad p_{c}&=0.8 \mathrm{MPa} \\ h_{3}&=h_{4}=h_{f}=93.42 \mathrm{kJ} / \mathrm{kg} \end{aligned}

$h_{2}=272.02 \mathrm{kJ} / \mathrm{kg}$
Rate of heat rejection to the environment,
\begin{aligned} Q_{2-3} &=m\left[h_{2}-h_{3}\right]=0.05 \times[275.02-93.42] \\ &=8.93 \mathrm{kW} \end{aligned}

There are 5 questions to complete.

### 1 thought on “Vapor and Gas Refrigeration”

1. QN-4 easy solution:
S2-S3 = (1054/310 = 3.4) as S1=S2 so, S1-S3 = 3.4
S1 = S3+3.4
But, S3 = S0 + Cpln(310/260) so S1 = S0 + 4.8ln(310/260) +3.4
S1 – S0 = 4.244
h1 – h0 = 260(S1 – S0) = 260(4.244) = 1103.44KJ/Kg