# Availability and Irreversibility

 Question 1
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 1 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 2
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 2 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 3
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 3 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 4
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 4 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}
 Question 5
The maximum theoretical work obtainable, when a system interacts to equilibrium with a reference environment, is called
 A Entropy B Enthalpy C Exergy D Rothalpy
GATE ME 2014 SET-1   Thermodynamics
Question 5 Explanation:
Exergy is also known as availability and it is maximum theoretical work obtainable, when a system interacts to equilibrium with dead state or reference environment.
 Question 6
The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 K and 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglect potential energy. If the pressure and temperature of the surroundings are 1 bar and 300 K, respectively, the available energy in kJ/kg of the air stream is
 A 170 B 187 C 191 D 213
GATE ME 2013   Thermodynamics
Question 6 Explanation:
Available energy
$\begin{array}{l} =\left(h_{1}-h_{0}\right)-T_{0}\left(s_{1}-s_{0}\right)+\left(\frac{V_{1}^{2}}{2}-\frac{V_{0}^{2}}{2}\right) \\ \text { where } s_{1}-s_{0}=c_{p} \ln \frac{T_{1}}{T_{0}}-R \ln \frac{p_{1}}{p_{0}}\\ =1.005 \ln \left(\frac{500}{300}\right)-0.287 \ln \left(\frac{5}{1}\right) \\ =0.05147 \mathrm{kJ} / \mathrm{kgK} \end{array}$
$\therefore$Available energy
$=1.005(500-300)-300(0.5147)+\frac{(50)^{2}}{2} \times 10^{-3}\\ =186.8 \mathrm{kJ} / \mathrm{kg} \approx 187 \mathrm{kJ} / \mathrm{kg}$
 Question 7
Nitrogen at an initial state of 10 bar, 1 $m^{3}$ and 300 K is expanded isothermally to a final volume of 2$m^{3}$. The p-v-T relation is $\left ( p+\frac{a}{v^{2}} \right )v= RT$ where a $\gt$ 0. The final pressure
 A will be slightly less than 5 bar B will be slightly more than 5 bar C will be exactly 5 bar D cannot be ascertained in the absence of the value of a
GATE ME 2005   Thermodynamics
Question 7 Explanation:
Given : $p_1=10 \text{ bar}, v_1=1m^3, T_1=300K, v_2=2m^3$
Given that Nitrogen Expanded isothermally.
So, RT = Constant
and from given relation,
\begin{aligned} \left ( p+\frac{a}{v^2} \right )v &=RT=\text{Constant} \\ p_1v_1+\frac{a}{v_1} &=p_2v_2+\frac{a}{v_2} \\ p_2v_2&=p_1v_1+\frac{a}{v_1}-\frac{a}{v_2} \\ p_2 &=p_1\left ( \frac{v_1}{v_2} \right )+a\left ( \frac{1}{v_1v_2}-\frac{1}{v_2^2} \right ) \\ &= 10\left ( \frac{1}{2} \right )+a\left ( \frac{1}{2}-\frac{1}{4} \right )=5+\frac{a}{4} \end{aligned}
Here, $a\gt 0$, so above equation shows that $p_2$ is greater than 5 and +ve.
 Question 8
A steel billet of 2000 kg mass is to be cooled from 1250 K to 450 K. The heat released during this process is to be used as a source of energy. The ambient temperature is 303 K and specific heat of steel is 0.5 kJ/kg K. the available energy of this billet is
 A 490.44 MJ B 30.95 MJ C 10.35 MJ D 0.10 MJ
GATE ME 2004   Thermodynamics
Question 8 Explanation:
\begin{aligned} Q &=m c_{p} \Delta T \\ &=2000 \times(0.5 \mathrm{kJ} / \mathrm{kgK}) \times(1250-450) \\ &=800000 \mathrm{kJ}\\ \Delta S&=m c_{p} \ln \frac{T_{1}}{T_{2}}=2000 \times 0.5 \times \ln \frac{1250}{450}\\ &=1021.165 \mathrm{kJ} / \mathrm{K} \\ \text { A.E. } &=Q-T_{0} \Delta S \\ &=490439.67 \mathrm{kJ}=490.44 \mathrm{MJ} \end{aligned}
 Question 9
Considering the relationship TdS =dU + pdV between the entropy (S), internal energy (U), pressure (p), temperature (T) and volume (V), which off the statement is correct .
 A It is applicable only for a reversible process B For an irreversible process TdS$\gt$ dU + pdV C It is valid only for an ideal gas D It is equivalent to $1^{st}$ law, for a reversible process
GATE ME 2003   Thermodynamics
Question 9 Explanation:
$T d S=d U+p d V$
This equation holds good for any process reversible or irreversible, undergone by a closed system, since it is a relation among properties which are independent of the both.
There are 9 questions to complete.