This set of Thermodynamics question bank focuses on “Irreversibility and Gouy-Stondola Theorem”.
1. The actual work done by a system is always ____ than the reversible work, and the difference between the two is called ____ of the process.
a) more, irreversibility
b) less, irreversibility
c) more, reversibility
d) less, reversible
View Answer
Explanation: Irreversibility=Maximum work – Actual work.
2. Irreversibility(I) is also called
a) degradation
b) dissipation
c) both of the mentioned
d) none of the mentioned
View Answer
Explanation: These are other names of irreversibility.
3. For a non-flow process between equilibrium states, when the system exchanges heat only with the environment
a) I=0
b) I>0
c) I<0
d) I>=0
View Answer
Explanation: I>0 for all processes bit for a reversible process, I=0.
4. For irreversibility, same expression applies to both flow and non-flow processes.
a) true
b) false
View Answer
Explanation: I=To*(sum of change in entropy of the system and surroundings).
5. The quantity [To*(ΔSsystem + ΔSsurroundings)] represents an increase in
a) available energy
b) unavailable energy
c) exergy
d) all of the mentioned
View Answer
Explanation: The quantity [To*(ΔSsystem + ΔSsurroundings)] represents an increase in unavailable energy.
6. Which of the following is true?
a) rate of loss of exergy does not depend on the rate of entropy generation
b) rate of loss of exergy is inversely proportional to the rate of entropy generation
c) rate of loss of exergy is directly proportional to the rate of entropy generation
d) none of the mentioned
View Answer
Explanation: This comes from the Gouy-Stondola theorem.
7. A thermodynamically efficient process would involve ____ exergy loss with ____ rate of entropy generation.
a) minimum, minimum
b) maximum, maximum
c) minimum, maximum
d) maximum, minimum
View Answer
Explanation: This is because rate of loss of exergy is directly proportional to the rate of entropy generation.
8. Heat transfer through a finite temperature difference is equivalent to the destruction of its exergy.
a) true
b) false
View Answer
Explanation: When heat transfers through a final temperature difference, all of its exergy is lost.
9. The decrease in availability or lost work is proportional to
a) pressure drop
b) mass flow rate
c) both of the mentioned
d) none of the mentioned
View Answer
Explanation: Lost work = (mass flow rate)*R*To*Δp/p1.
10. Entropy generation number can be given by
a) (rate of entropy generation)*(mass flow rate/specific heat)
b) (rate of entropy generation)/(mass flow rate/specific heat)
c) (rate of entropy generation)*(mass flow rate*specific heat)
d) (rate of entropy generation)/(mass flow rate*specific heat)
View Answer
Explanation: Entropy generation number is a dimensionless quantity and given by above formula.
11. If two streams with equal temperature are mixing, then the entropy generation number becomes
a) 0
b) 1
c) -1
d) infinity
View Answer
Explanation: Putting T1=T2 in the relation for entropy generation number, we get the value as zero.
11. A flow of air at 1000 kPa, 300 K is throttled to 500 kPa. What is the irreversibility?
a) 39.6 kJ/kg
b) 49.6 kJ/kg
c) 59.6 kJ/kg
d) 69.6 kJ/kg
View Answer
Explanation: A throttle process is constant enthalpy if we neglect kinetic energies.
Process: he = hi, so ideal gas => Te = Ti
se – si = s(gen), s(gen) = 0 – R*ln(Pe/Pi)
s(gen) = – 0.287 ln (500 / 1000) = 0.2 kJ/kg K
i = (T0)*s(gen) = 298*0.2 = 59.6 kJ/kg.
12. A heat exchanger increases the availability of 3 kg/s water by 1650 kJ/kg by using 10 kg/s air which comes in at 1400 K and leaves with 600 kJ/kg less availability. What is the irreversibility?
a) 1020 kW
b) 1030 kW
c) 1040 kW
d) 1050 kW
View Answer
Explanation: The irreversibility is the destruction of exergy (availability) so
I = Φ(destruction) = Φ(in) – Φ(out) = 10 × 600 – 3 × 1650 = 1050 kW.
13. A 2-kg piece of iron is heated from temperature 25°C to 400°C by a heat source which is at 600°C. What is the irreversibility in the process?
a) 96.4 kJ
b) 86.4 kJ
c) 76.4 kJ
d) 66.4 kJ
View Answer
Explanation: 1Q2 = m(h2 – h1) = mC(T2 – T1) = 2 × 0.42 × (400 – 25) = 315 kJ
S(gen) = m(s2 – s1) – 1Q2/T@res = mC ln (T2/T1) – 1Q2/T@res
= 2 × 0.42 × ln (673.15/298.15) – (315/873.15) = 0.3233 kJ/K
I = To (S gen ) = 298.15 × 0.3233 = 96.4 kJ.
14. A rock bed(at 70°C) consists of 6000 kg granite. A house with mass of 12000 kg wood and 1000 kg iron is at 15°C. They are brought to a uniform final temperature. Find the irreversibility of the process, assuming an ambient temperature of 15°C.
a) 17191 kJ
b) 18191 kJ
c) 19191 kJ
d) 20191 kJ
View Answer
Explanation: (mC)(rock)(T2 – 70) + [mC(wood) + mC(Fe)](T2 – 15) = 0
hence T2 = 29.0°C = 302.2 K
S2 – S1 = ∑mi(s2 – s1)i = 0 + Sgen
Sgen = ∑mi(s2 – s1)i = 5340 ln(302.2/343.15) + 15580 ln(302.2/288.15) = 63.13 kJ/K
I = (T0)Sgen = 288.15 × 63.13 = 18191 kJ.
15. 7. A compressor is used to bring saturated water vapour initially at 1 MPa up to 17.5 MPa, where the actual exit temperature is 650°C. Find the irreversibility.
a) 40.48 kJ/kg
b) 41.48 kJ/kg
c) 43.48 kJ/kg
d) 44.48 kJ/kg
View Answer
Explanation: hi = 2778.1 kJ/kg, si = 6.5864 kJ/kg K
Actual compressor: h(e,ac) = 3693.9 kJ/kg, s(e,ac) = 6.7356 kJ/kg K
-w(c,ac) = h(e,ac) – hi = 915.8 kJ/kg
i = T0[s(e,ac) – si] = 298.15 (6.7356 – 6.5864) = 44.48 kJ/kg.
Sanfoundry Global Education & Learning Series – Thermodynamics.
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