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 2m^{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.

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