Availability and Irreversibility


Question 1
The Clausius inequality holds good for
A
any process
B
any cycle
C
only reversible process
D
only reversible cycle
GATE ME 2022 SET-1   Thermodynamics
Question 1 Explanation: 
The Clausius inequality holds good for any cycle.
\oint \frac{dQ}{T}=0\Rightarrow Reversible cycle
\oint \frac{dQ}{T} \lt 0\Rightarrow Irreversible cycle
\oint \frac{dQ}{T} \gt 0\Rightarrow Impossible cycle
Question 2
Keeping all other parameters identical, the Compression Ratio (CR) of an air standard diesel cycle is increased from 15 to 21. Take ratio of specific heats = 1.3 and cut-off ratio of the cycle r_c = 2.

The difference between the new and the old efficiency values, in percentage,

(\eta _{new}|_{ CR = 21})-(\eta _{old}|_{CR = 15})= _______ %. (round off to one decimal place)
A
4.8
B
2.4
C
6.2
D
2.8
GATE ME 2020 SET-2   Thermodynamics
Question 2 Explanation: 
\begin{aligned} \eta_{d, r=21} &=1-\left(\frac{1}{r}\right)^{\gamma-1} \times \frac{\left(\rho^{\gamma}-1\right)}{\gamma(\rho-1)}=54.87 \% \\ \eta_{d, r=15} &=1-\left(\frac{1}{r}\right)^{\gamma-1} \times \frac{\left(\rho^{\gamma}-1\right)}{\gamma(\rho-1)}=50.08 \% \\ \eta_{d, r=21}-\eta_{d, r=15} &=4.8 \% \end{aligned}


Question 3
For an ideal gas with constant properties undergoing a quasi-static process, which one of the following represents the change of entropy (\Delta s) from state 1 to 2?
A
\Delta s=C_{p}ln\left ( \frac{T_{2}}{T_{1}} \right )-R ln\left ( \frac{p_{2}}{p_{1}} \right )
B
\Delta s=C_{v}ln\left ( \frac{T_{2}}{T_{1}} \right )-C_{p} ln\left ( \frac{v_{2}}{v_{1}} \right )
C
\Delta s=C_{p}ln\left ( \frac{T_{2}}{T_{1}} \right )-C_{v} ln\left ( \frac{p_{2}}{p_{1}} \right )
D
\Delta s=C_{v}ln\left ( \frac{T_{2}}{T_{1}} \right )+R ln\left ( \frac{v_{1}}{v_{2}} \right )
GATE ME 2018 SET-2   Thermodynamics
Question 3 Explanation: 
\begin{array}{l}\mathrm{Tds}=\mathrm{dH}-\mathrm{VdP}\\ \mathrm{dS}=\frac{\mathrm{dH}}{\mathrm T}-\frac{\mathrm V}{\mathrm T}\mathrm{dP}\\ \mathrm{For}\;\mathrm{an}\;\mathrm{ideal}\;\mathrm{gas}\\ \mathrm{PV}=\mathrm{RT}\\ \frac{\mathrm V}{\mathrm T}=\frac{\mathrm R}{\mathrm P}\\ \mathrm{dS}={\mathrm C}_{\mathrm P} \frac{\mathrm{dT}}{\mathrm T}-\frac{\mathrm R}{\mathrm p}\mathrm{dP}\\ {\mathrm S}_2-{\mathrm S}_1={\mathrm C}_\mathrm P\ln\frac{{\mathrm T}_2}{{\mathrm T}_1}-\mathrm{Rln}\frac{{\mathrm P}_2}{{\mathrm P}_1}\\\\\end{array}
Question 4
Which one of the following statements is correct for a superheated vapour?
A
Its pressure is less than the saturation pressure at a given temperature.
B
Its temperature is less than the saturation temperature at a given pressure.
C
Its volume is less than the volume of the saturated vapour at a given temperature.
D
Its enthalpy is less than the enthalpy of the saturated vapour at a given pressure.
GATE ME 2018 SET-1   Thermodynamics
Question 4 Explanation: 


P_{\text {sat }} @ T_{1} \rightarrow saturation pressure at T_{1} temperature P_{1} \rightarrow pressure of superheated vapour at state 1
P_{1} \lt P_{\text {sat }} @_{T_{1}}
Question 5
One side of a wall is maintained at 400 K and the other at 300 K. The rate of heat transfer through the wall is 1000 W and the surrounding temperature is 25^{\circ}C. Assuming no generation of heat within the wall, the irreversibility (in W) due to heat transfer through the wall is ________
A
288.32W
B
125.36W
C
248.23W
D
485.6W
GATE ME 2015 SET-3   Thermodynamics
Question 5 Explanation: 


Given data:
\begin{aligned} T_{1}=400 \mathrm{K} ; & T_{2}=300 \mathrm{K} ; Q=1000 \mathrm{W} \\ T_{0} &=25^{\circ} \mathrm{C}=(25+273) \mathrm{K} \\ &=298 \mathrm{K} \\ \Delta S_{\mathrm{sys}} &=-\frac{Q}{T_{1}}+\frac{Q}{T_{2}} \\ &=-\frac{1000}{400}+\frac{1000}{300} \\ &=-2.5+3.333=0.833 \mathrm{W} / \mathrm{K}\\ & Irreversibility,\\ I &=T_{0} \Delta S_{\text {uni }} \\ &=T_{0}\left(\Delta S_{\text {sys }}+\Delta S_{\text {surr }}\right) \\ &=298(0.833+0) \qquad \because \Delta S_{\text {surf }}=0 \\ &=248.23 \mathrm{W} \end{aligned}


There are 5 questions to complete.

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