Heat Pumps and Cycles

 Question 1
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 1 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 2
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 2 Explanation:
Reversed Brayton cycle is shown in the figure.
 Question 3
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 3 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 4
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 4 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}
 Question 5
A heat pump with refrigerant R22 is used for space heating between temperature limits of -20$^{\circ}$C and 25$^{\circ}$C. The heat required is 200 MJ/h. Assume specific heat of vapour at the time of discharge as 0.98 kJ/kg.K. Other relevant properties are given below. The enthalpy (in kJ/kg) of the refrigerant at isentropic compressor discharge is _______
 A 220.4 B 433.3 C 110.5 D 880.6
GATE ME 2014 SET-2   Refrigeration and Air-conditioning
Question 5 Explanation:

\begin{aligned} s_{2}-s_{2^{\prime}}&=c_{p} \log _{e} \frac{T_{2}}{T_{2^{\prime}}} \\ s_{1}&=s_{2} \quad(\text { for isentropic process }) \\ s_{2} &=1.7841 \mathrm{kJ} / \mathrm{kgK} \\ s_{2} &=1.7183 \mathrm{kJ} / \mathrm{kgK} \\ c_{p} &=0.98 \mathrm{kJ} / \mathrm{kg} \mathrm{K} \\ \therefore \log _{e} \frac{T_{2}}{T_{2^{\prime}}} &=\frac{s_{2}-s_{2^{\prime}}}{c_{p}} \\ &=\frac{1.7841-1.7183}{0.98}=0.0671\\ \text { or } \quad T_{2} &=1.0694 \times T_{2^{\prime}}=\sqrt{.0694 \times 298} \\ &=318.695 \mathrm{K} \\ h_{2}-h_{2^{\prime}} &=c_{p}\left(T_{2}-T_{2}\right) \\ h_{2} &=413.02+0.98(318.695-298) \\ h_{2} &=433.3 \mathrm{kJ} / \mathrm{kg} \end{aligned}
 Question 6
A thin layer of water in field is formed after a farmer has watered it. The ambient air conditions are: temperature 20 $^{\circ}$C and relative humidity 5%.
An extract of steam tables is given below.

Neglecting the heat transfer between the water and the ground, the water temperature in the field after phase equilibrium is reached equals
 A 10.3 $^{\circ}$C B -10.3 $^{\circ}$C C -14.5 $^{\circ}$C D 14.5 $^{\circ}$C
GATE ME 2006   Refrigeration and Air-conditioning
Question 6 Explanation:
Given data:
\begin{aligned} \phi &=5 \%=0.05 \\ T_{d b} &=20^{\circ} \mathrm{C} \end{aligned}
From table,
\begin{aligned} \text { At } 20^{\circ} \mathrm{C}, p_{s}&=2.34 \mathrm{kPa} \\ \phi&=\frac{p_{v}}{p_{s}} \\ 0.05&=\frac{p_{v}}{2.34}\\ \text{Or }\quad p_{v}&=0.05 \times 2.34=0.117 \mathrm{kPa} \end{aligned}
The dew point temperature is the saturation temperature corresponding to $p_{v}$from given table
by interpolation
\begin{aligned} T_{d p} &=-15+\frac{(-10+15)}{0.26-0.10} \times(0.117-0.10) \\ &=-15+0.531=-14.469^{\circ} \mathrm{C} \approx-14.5^{\circ} \mathrm{C} \end{aligned}
 Question 7
A vapor absorption refrigeration system is a heat pump with three thermal reservoirs as shown in the figure. A refrigeration effect of 100W is required at 250 K when the heat source available is at 400 K. Heat rejection occurs at 300 K. the minimum value of heat required (in W) is
 A 167 B 100 C 80 D 20
GATE ME 2005   Refrigeration and Air-conditioning
Question 7 Explanation:
Given data:
$T_{1}=300 \mathrm{K} ; T_{2}=250 \mathrm{K} ; \quad T_{3}=400 \mathrm{K}$
Refrigerating effect,
$Q_{2}=100 \mathrm{W}$

For heat engine,
\begin{aligned} \mathrm{COP} &=\left(\frac{T_{2}}{T_{1}-T_{2}}\right)\left(\frac{T_{3}-T_{1}}{T_{3}}\right) \\ &=\left(\frac{250}{300-250}\right)\left(\frac{400-300}{400}\right) \\ &=1.25 \\ \text { also } \quad \mathrm{COP} &=\frac{Q_{2}}{Q_{3}} \\ \therefore \quad 1.25 &=\frac{100}{Q_{3}} \\ \text { or } \quad Q_{3} &=\frac{100}{1.25}=80 \mathrm{W} \end{aligned}
 Question 8
A heat engine having an efficiency of 70% is used to drive a refrigerator having a co-efficient of performance of 5. The energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine is
 A 0.14 kJ B 0.71 kJ C 3.5 kJ D 7.1 kJ
GATE ME 2004   Refrigeration and Air-conditioning
Question 8 Explanation:
Given data:
\begin{aligned} \eta &=70 \%=0.70 \\ (\mathrm{COP})_{R} &=5 \\ Q_{3} &=? \\ Q_{1} &=1 \mathrm{kJ} \end{aligned}

For heat engine,
\begin{aligned} \eta &=1-\frac{Q_{2}}{Q_{1}} \\ 0.70 &=1-\frac{Q_{2}}{1}\\ \text{or } \quad Q_{2}&=0.3 \mathrm{kJ} \\ W &=Q_{1}-Q_{2}=1-0.3 \\ &=0.7 \mathrm{kJ}\\ \text{For refrigerator,}\\ (\mathrm{COP})_{\mathrm{R}} &=\frac{Q_{3}}{W} \\ 5 &=\frac{Q_{3}}{0.7} \\ \text{or}\quad Q_{3} &=0.7 \times 5=3.5 \mathrm{kJ} \end{aligned}
 Question 9
An industrial heat pump operates between the temperatures of $27^{\circ}C$ and $-13^{\circ}C$ . The rates of heat addition and heat rejection are 750 W and 1000 W, respectively. The COP for the heat pump is
 A 7.5 B 6.5 C 4 D 3
GATE ME 2003   Refrigeration and Air-conditioning
Question 9 Explanation:
Given data:
\begin{aligned} T_{1} &=27^{\circ} \mathrm{C} \\ &=(27+273) \mathrm{K}=300 \mathrm{K} \\ T_{2} &=-13^{\circ} \mathrm{C} \\ &=(-13+273) \mathrm{K}=260 \mathrm{K} \\ Q_{2} &=750 \mathrm{W} \\ Q_{1} &=1000 \mathrm{W} \end{aligned}

$(\mathrm{COP})_{\mathrm{HP}}=\frac{Q_{1}}{Q_{1}-Q_{2}}$ for actual heat pump
$=\frac{1000}{1000-750}=\frac{1000}{250}=4$
There are 9 questions to complete.