# Heat Pumps and Cycles

 Question 1
In a vapour compression refrigeration cycle, the refrigerant enters the compressor in saturated vapour state at evaporator pressure, with specific enthalpy equal to 250 kJ/kg. The exit of the compressor is superheated at condenser pressure with specific enthalpy equal to 300 kJ/kg. At the condenser exit, the refrigerant is throttled to the evaporator pressure. The coefficient of performance (COP) of the cycle is 3. If the specific enthalpy of the saturated liquid at evaporator pressure is 50 kJ/kg, then the dryness fraction of the refrigerant at entry to evaporator is __
 A 0.2 B 0.25 C 0.3 D 0.35
GATE ME 2022 SET-2   Refrigeration and Air-conditioning
Question 1 Explanation:
Given data,
Saturated vapour refrigerant enters to compressor COP = 3

Specific liquid enthalpy of refrigerant corresponding to evaporator pressure $(h_f)_{P_1}=50 kJ/kg$.
Dryness fraction of refrigerant of entry to evaporator $(x_4)=?$
\begin{aligned} COP &=\frac{h_1-h_4}{h_2-h_1} \\ 3 &=\frac{250-h_4}{300-250} \\ h_4&=100kJ/kg \\ \text{But } h_4&=(h_f)_{P_1}+x_4(h_{f_g})_{P_1} \\ (h_{f_g})_{P_1}&= (h_{g})_{P_1} - (h_{f_1})_{P_1}\\ \Rightarrow h_{fg}&=250-50=200kJ/kg \\ 100&=50+x_4 \times 200 \\ \Rightarrow x_4&=0.25 \end{aligned}
 Question 2
Superheated steam at 1500 kPa, has a specific volume of 2.75 $m^3$/kmol and compressibility factor ($Z$) of 0.95. The temperature of steam is $^{\circ}C$ (round off to the nearest integer).
 A 522 B 471 C 249 D 198
GATE ME 2021 SET-1   Refrigeration and Air-conditioning
Question 2 Explanation:
\begin{aligned} P &=1500 \mathrm{kPa} \\ V &=2.75 \mathrm{~m} 3 / \mathrm{k}-\mathrm{mol} \\ Z &=0.95 \\ P V &=n \bar{R} T \\ P \bar{V} &=\bar{R} T \\ P \bar{V} &=Z \times n \bar{R} T \\ P \frac{\bar{V}}{n} &=Z \bar{R} T \\ P_{\bar{V}} &=Z \bar{R} T\\ 1500 \mathrm{KPa} \times 2.75 \mathrm{~m}^{3} / \mathrm{K}-\mathrm{mol}&=0.95 \times 8.314 \mathrm{~kJ} / \mathrm{K}-\mathrm{molK} \times T\\ T&=522.26 \mathrm{~K}\\ T&=522.26-273=249.26^{\circ} \mathrm{C} \end{aligned}

 Question 3
The thermodynamic cycle shown in figure (T-s diagram) indicates
 A reversed Carnot cycle B reversed Brayton cycle C vapor compression cycle D vapor absorption cycle
GATE ME 2015 SET-3   Refrigeration and Air-conditioning
Question 3 Explanation:
Reversed Brayton cycle is shown in the figure.
 Question 4
The COP of a Carnot heat pump operating between 6 $^{\circ}C$ and 37 $^{\circ}C$ is ___________
 A 10 B 11 C 12 D 13
GATE ME 2015 SET-2   Refrigeration and Air-conditioning
Question 4 Explanation:
Given data:
\begin{aligned} T_{2} &=6^{\circ} \mathrm{C}=(6+273) \mathrm{K} \\ &=279 \mathrm{K} \\ T_{1} &=37^{\circ} \mathrm{C}=(37+273) \mathrm{K} \\ &=310 \mathrm{K} \end{aligned}

\begin{aligned} (\mathrm{COP})_{\mathrm{HP}} &=\frac{T_{1}}{T_{1}-T_{2}}=\frac{310}{310-279} \\ &=10 \end{aligned}
 Question 5
A reversed Carnot cycle refrigerator maintains a temperature of -5$^{\circ}$C. The ambient air temperature is 35$^{\circ}$C. The heat gained by the refrigerator at a continuous rate is 2.5 kJ/s. The power (in watt) required to pump this heat out continuously is _______
 A 333.13watt B 986.65watt C 373.13watt D 984.65watt
GATE ME 2014 SET-4   Refrigeration and Air-conditioning
Question 5 Explanation:
Given data:
$\begin{array}{l} T_{2}=-5^{\circ} \mathrm{C}=(-5+273) \mathrm{K}=268 \mathrm{K} \\ T_{1}=35^{\circ} \mathrm{C}=(35+273) \mathrm{K}=308 \mathrm{K} \end{array}$

\begin{aligned} Q_{2}=2.5 \mathrm{kJ} / \mathrm{s}&=2.5 \mathrm{kW}=2500 \mathrm{W} \\ (\mathrm{COP})_{R} &=\frac{T_{2}}{T_{1}-T_{2}} \\ &=\frac{268}{308-268}=6.7 \\ \text { also } \quad(\mathrm{COP})_{R} &=\frac{Q_{2}}{W}\\ \therefore \quad 6.7&=\frac{2500}{W}\\ \text{or }\quad W&=\frac{2500}{6.7}=373.13 \text { watt } \end{aligned}

There are 5 questions to complete.