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 __
0.2 | |
0.25 | |
0.3 | |
0.35 |
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}
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).
522 | |
471 | |
249 | |
198 |
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
reversed Carnot cycle | |
reversed Brayton cycle | |
vapor compression cycle | |
vapor absorption cycle |
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 ___________
10 | |
11 | |
12 | |
13 |
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}
\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 _______
333.13watt | |
986.65watt | |
373.13watt | |
984.65watt |
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}
\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 6 |
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 _______
220.4 | |
433.3 | |
110.5 | |
880.6 |
Question 6 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 7 |
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
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
10.3 ^{\circ}C | |
-10.3 ^{\circ}C | |
-14.5 ^{\circ}C | |
14.5 ^{\circ}C |
Question 7 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}
\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 8 |
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
167 | |
100 | |
80 | |
20 |
Question 8 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}
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 9 |
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
0.14 kJ | |
0.71 kJ | |
3.5 kJ | |
7.1 kJ |
Question 9 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}
\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 10 |
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
7.5 | |
6.5 | |
4 | |
3 |
Question 10 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
\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 10 questions to complete.