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}
There are 5 questions to complete.