# Second Law, Carnot Cycle and Entropy

 Question 1
An adiabatic vortex tube, shown in the figure given below is supplied with 5 kg/s of air (inlet 1) at 500 kPa and 300 K. Two separate streams of air are leaving the device from outlets 2 and 3. Hot air leaves the device at a rate of 3 kg/s from outlet 2 at 100 kPa and 340 K, and 2 kg/s of cold air stream is leaving the device from outlet 3 at 100 kPa and 240 K. Assume constant specific heat of air is 1005 J/kg.K and gas constant is 287 J/kg.K. There is no work transfer across the boundary of this device. The rate of entropy generation is ________kW/K (round off to one decimal place).
 A 1.2 B 4.3 C 2.2 D 3.8
GATE ME 2021 SET-2   Thermodynamics
Question 1 Explanation: \begin{aligned} \left(\frac{d S}{d t}\right)_{C . V} &=\dot{S}_{i}+\dot{S}_{g e n}-\dot{S}_{\theta} \\ \dot{S}_{\text {gen }} &=\dot{S}_{e}-\dot{S}_{i} \\ &=\dot{m}_{2} s_{2}+\dot{m}_{3} s_{3}-\dot{m}_{1} s_{1} \\ &=3\left(s_{2}-s_{1}\right)+2\left(s_{3}-s_{1}\right) \\ &=3 \times 0.587+2(0.237) \\ &=2.235 \mathrm{~kW} / \mathrm{K} \simeq 2.2 \mathrm{~kW} / \mathrm{K} \end{aligned}
 Question 2
The fundamental thermodynamic relation for a rubber band is given by $dU=TdS+\tau dL$, where $T$ is the absolute temperature, $S$ is the entropy, $\tau$ is the tension in the rubber band, and $L$ is the length of the rubber band. Which one of the following relations is CORRECT:
 A $\tau =\left ( \frac{\partial U}{\partial S}\right )_L$ B $\left ( \frac{\partial T}{\partial L}\right )_S =\left ( \frac{\partial \tau}{\partial S}\right )_L$ C $\left ( \frac{\partial T}{\partial S}\right )_L =\left ( \frac{\partial \tau}{\partial L}\right )_S$ D $T=\left ( \frac{\partial U}{\partial S}\right )_ \tau$
GATE ME 2021 SET-1   Thermodynamics
Question 2 Explanation:
\begin{aligned} d U&=T d s+\tau d L \\ \left(\frac{\partial T}{\partial L}\right)_{S}&=\left(\frac{\partial \tau}{\partial S}\right)_{L}\\ \text { Comparing with } M=\left(\frac{\partial z}{\partial x}\right)_{y}& \\ T&=\left(\frac{\partial U}{\partial S}\right)_{L}\\ \text { Comparing with } N=\left(\frac{\partial z}{\partial y}\right)_{x} \\ \tau&=\left(\frac{\partial U}{\partial L}\right)_{S} \\ M&=\left(\frac{\partial z}{\partial x}\right)_{y} \\ N&=\left(\frac{\partial z}{\partial y}\right)_{x} \\ d z&=M d x+N d y \end{aligned}
If z is exact different
$\left(\frac{\partial M}{\partial y}\right)_{x}=\left(\frac{\partial N}{\partial x}\right)_{y}$
 Question 3
In which of the following pairs of cycles, both cycles have at least one isothermal process?
 A Diesel cycle and Otto cycle B Carnot cycle and Stirling cycle C Brayton cycle and Rankine cycle D Bell-Coleman cycle and Vapour compression refrigeration cycle
GATE ME 2021 SET-1   Thermodynamics
Question 3 Explanation:
1.Brayton $\begin{array}{ll} 1 \text { to } 2: & S=C \\ 2 \text { to } 3: & P=C \\ 3 \text { to } 4: & S=C \\ 4 \text { to } 1: & P=C \end{array}$

2. VCRS 1 to 2 : Isentropic compression
2 to 3: constant pressure (P=C)
3 to 4: Isenthalpic expansion $\left(h_{3}=h_{4}\right)$
4 to 1: Constant pressure (P=C)

3. Carnot 1 to 2 : lsentropic compression
2 to 3 : Isothermal heat addition
3 to 4 : lsentropic expansion
4 to 1 : Isothermal heat rejection

4. Bell Coleman 1 to 2 : lsentropic compression
2 to 3 : Constant pressure
3 to 4 : lsentropic expansion
4 to 1 : Constant pressure

5. Rankine 1 to 2 : lsentropic compression
2 to 3 : Constant pressure
3 to 4 : lsentropic expansion
4 to 1 : Constant pressure

6. Diesel 1 to 2 : lsentropic compression
2 to 3 : Constant pressure (P = C)
3 to 4 : lsentropic expansion
4 to 1 : Constant volume

7. Otto 1 to 2 : lsentropic compression
2 to 3 : Constant pressure (V = C)
3 to 4 : lsentropic expansion
4 to 1 : Constant volume (V = C)
 Question 4
One kg of air in a closed system undergoes an irreversible process from an initial state of $p_1 = 1$ bar (absolute) and $T_1 = 27^{\circ}C$, to a final state of $p_2 = 3$ bar (absolute) and $T_2= 127^{\circ}C$. If the gas constant of air is 287 J/kg.K and the ratio of the specific heats $\gamma = 1.4$, then the change in the specific entropy (in J/kg.K) of the air in the process is
 A -26.3 B 28.4 C 172 D indeterminate, as the process is irreversible
GATE ME 2020 SET-2   Thermodynamics
Question 4 Explanation:
$\begin{array}{c} P_{1}=1 \text { bar, } P_{2}=1 \text { bar, } T_{1}=300 \mathrm{K}, T_{2}=400 \mathrm{K} \\ c_{p}=\frac{\gamma R}{\gamma-1}=1005 \mathrm{J} / \mathrm{kgK} \\ \Delta S=c_{p} \ln \frac{T_{2}}{T_{1}}-R \ln \frac{P_{2}}{P_{1}}=1005 \ln \frac{400}{300}-287 \mathrm{ln} \frac{3}{1}=-26.32 \mathrm{J} / \mathrm{kgK} \end{array}$
 Question 5
If a reversed Carnot cycle operates between the temperature limits of $27^{\circ}C$ and $-3^{\circ}C$, then the ratio of the COP of a refrigerator to that of a heat pump (COP of refrigerator/ COP of heat pump) based on the cycle is __________ (round off to 2 decimal places).
 A 0.2 B 0.8 C 0.9 D 1.5
GATE ME 2020 SET-2   Thermodynamics
Question 5 Explanation:
\begin{aligned} T_{H}&=27^{\circ} \mathrm{C}=300 \mathrm{K} \\ T_{L}&=-3^{\circ} \mathrm{C}=270 \mathrm{K} \\ \frac{\mathrm{COP}_{\mathrm{ref}}}{\mathrm{COP}_{\mathrm{HP}}}&=\frac{\frac{T_{L}}{T_{H}-T_{L}}}{\frac{T_{H}}{T_{H}-T_{L}}}=\frac{T_{L}}{T_{H}}=\frac{270}{300}=0.9 \end{aligned} Question 6
Air of mass 1 kg, initially at 300 K and 10 bar, is allowed to expand isothermally till it reaches a pressure of 1 bar. Assuming air as an ideal gas with gas constant of 0.287 kJ/kg.K, the change in entropy of air (in kJ/kg.K, round off to two decimal places) is____
 A 0.22 B 0.82 C 0.98 D 0.66
GATE ME 2019 SET-1   Thermodynamics
Question 6 Explanation:
Given data:
\begin{aligned} \mathrm{m}&=1 \mathrm{kg}, \mathrm{T}_{1}=300 \mathrm{K}, \mathrm{P}_{1}=10 \mathrm{bar} \\ \mathrm{T}_{2}&=300 \mathrm{K}, \mathrm{P}_{2}=1 \mathrm{bar} \\ \mathrm{dS}&=\frac{\mathrm{d} \mathrm{Q}}{\mathrm{T}}=\frac{\mathrm{mRT} \ell \mathrm{n} \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}}{\mathrm{T}}=\mathrm{mR} \ell \mathrm{n} \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}} \\ &=1 \times 0.287 \times \ell n 10=0.66 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K} \end{aligned}
 Question 7
A tank of volume 0.05 $m^{3}$ contains a mixture of saturated water and saturated steam at $200^{\circ}C$ . The mass of the liquid present is 8 kg. The entropy (in kJ/kg K) of the mixture is ___________ (correct to two decimal places).

Property data for saturated steam and water are:
At $200^{\circ}C$ , $p_{sat}$ = 1.5538 MPa
$v_{f}$ = 0.001157 $m^{3}$ /kg, $v_{g}$ = 0.12736 $m^{3}$ /kg
$s_{fg}$ =4.1014 kJ/kg K, $s_{f}$= 2.3309 kJ/kg K
 A 1.53 B 2.12 C 2.48 D 3.59
GATE ME 2018 SET-1   Thermodynamics
Question 7 Explanation:
Total volume of tank (V) = 0.05 m3
Means of liquid ($m_{2}$) = 8 kg \begin{aligned} x &=\frac{m_{v}}{m_{v}+m_{L}}(\text { dryness fraction }) \\ V &=V_{L}+v_{v} \\ \text{where, } &V_{L}-\text{ liquid in tank} \\&V_{v} - \text{ vapour in tank}\\ V_{V} &=m_{v} v_{v} \\ v_{v} &=v_{g} @ 200^{\circ} \mathrm{C} \\ &=0.12736 \mathrm{m}^{3} / \mathrm{kg} \\ V_{V} &=m_{L} v_{L} \\ v_{L} &=v_{f} @ 200^{\circ} \mathrm{C} \\ &=0.001157 \mathrm{m}^{3} / \mathrm{kg} \\ V_{L} &=8 \times 0.001157=9.256 \times 10^{-3} \mathrm{m}^{3} \\ 0.05 &=9.256 \times 10^{-3}+\mathrm{V}_{V} \\ V_{V} &=0.0474 \mathrm{m}^{3}\\ \text{So,}\qquad 0.04074 &=m_{v} \times 0.12736 \\ m_{v} &=0.3198 \mathrm{kg} \\ x &=\frac{0.3198}{0.3198+8} \\ \Rightarrow \quad x &=0.0384 \end{aligned}
Specific entropy of mixture(s)
\begin{aligned} s&=s_{f}+x s_{f g} \\ s&=2.3309+0.0384 \times 4.1014 \\ s&=2.4884 \mathrm{kJ} / \mathrm{kg}-\mathrm{K} \end{aligned}
 Question 8
An ideal gas undergoes a process from state 1 $(T_{1} = 300K, p_{1} = 100kPa)$ to state 2 $(T_{2} = 600K, p_{2} = 500kPa)$. The specific heats of the ideal gas are : $C_{p} = 1kJ/kg-K$ and $C_{v} = 0.7kJ/kg-K$ . The change in specific entropy of the ideal gas from state 1 to state 2 (in kJ/kg-K) is ________ (correct to two decimal places).
 A 0.15 B 0.21 C 0.45 D 0.84
GATE ME 2018 SET-1   Thermodynamics
Question 8 Explanation:
Ideal gas
\begin{array}{l} \text { State-1: } T_{1}=300 \mathrm{K}, P_{1}=100 \mathrm{kPa} \\ \text { State-2: } T_{2}=600 \mathrm{K}, P_{2}=500 \mathrm{kPa} \\ \begin{aligned} c_{p}=1 \mathrm{kJ} / \mathrm{kg}-\mathrm{K}, c_{p}-c_{v}&=R, c_{v}=0.7 \mathrm{kJ} / \mathrm{kg}-\mathrm{K} \\ \Rightarrow \qquad c_{p}-c_{v}&= 1-0.7=R \\ R &=0.3 \mathrm{kJ} / \mathrm{kg}-\mathrm{K} \end{aligned} \end{array}
Change in specific entropy
\begin{aligned} s_{2}-s_{1} &=c_{p} \ln \frac{T_{2}}{T_{1}}-R \ln \frac{P_{2}}{P_{1}} \\ &=1 \times \ln \frac{600}{300}-0.3 \ln \frac{500}{100}=0.21 \mathrm{kJ} / \mathrm{kg}-\mathrm{K} \end{aligned}
 Question 9
One kg of an ideal gas (gas constant R=287 J/kg.K) undergoes an irreversible process from state-1 (1 bar,300K) to state -2 (2 bar, 300 K). The change in specific entropy ($s_{2}-s_{1}$) of the gas (in J/$kg\cdot K$) in the process is ______
 A $-198.93\, J/kg\! \cdot \! K$ B $-175.90\, J/kg\! \cdot \! K$ C $-210.00\, J/kg\! \cdot \! K$ D $-140\, J/kg\! \cdot \! K$
GATE ME 2017 SET-2   Thermodynamics
Question 9 Explanation:
Change in specific entropy of ideal gas
\begin{aligned} S_{f}-s_{i} &=c_{p} \ln \frac{T_{f}}{T_{i}}-R \ln \frac{P_{f}}{P_{i}} \\ &=1005 \ln \frac{300}{300}-287 \ln \frac{2}{1} \\ &=0-287 \ln 2=-198.93 \mathrm{J} / \mathrm{kg}^{-k} \end{aligned}
 Question 10
One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at constant volume,$c_{\textrm{v}}$ =1000 J / kg .K) at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.
 A 287.682 B 274.127 C 295.427 D 300.315
GATE ME 2017 SET-1   Thermodynamics
Question 10 Explanation:
$m=1 \mathrm{kg} ; \quad R=400 \mathrm{J} / \mathrm{kg}-\mathrm{k}$ \begin{aligned} c_{v}&=1000 \mathrm{J} / \mathrm{kg}-\mathrm{K} \\ T_{i}&=300 \mathrm{K} ; P_{i}=100 \mathrm{kPa} \\ W&=100 \mathrm{kJ} \end{aligned}
Applying first law:
\begin{aligned} Q&=\Delta U+W\\ \Rightarrow \quad 0&=m c_{v}\left(T_{f}-T_{i}\right)+(-100) \times 10^{3} J \\ \Rightarrow \quad 0&=1 \times 1000 \times\left(T_{f}-300\right)-100 \times 10^{3} \\ \Rightarrow \quad T_{f}&=400 \mathrm{K} \end{aligned}
Entropy change for ideal gas
\begin{aligned} \Delta S &=m\left[C_{v} \ln \frac{T_{f}}{T_{i}}+R \ln \frac{V_{f}}{V_{i}}\right] \quad \because V_{i}=V_{f} \\ &=1 \times\left[1000 \ln \frac{400}{300}+0\right] \\ &=287.682 \mathrm{J} / \mathrm{K} \end{aligned}
There are 10 questions to complete.

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