Thermodynamics Questions and Answers – Availability or Exergy Balance

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This set of Thermodynamics Questions & Answers for entrance exams focuses on “Availability or Energy Balance”.

1. Which of the following clearly defines availability or exergy?
a) it is the maximum useful work obtainable from a system as it reaches the dead state
b) it is the minimum work required to bring the closed system from the dead state to the given state
c) both of the mentioned
d) none of the mentioned
View Answer

Answer: c
Explanation: The given statements clearly explain exergy.
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2. The maximum work or exergy cannot be negative.
a) true
b) false
View Answer

Answer: a
Explanation: This is because any change in state of the closed system to the dead state can be accomplished with zero work.

3. Energy is ____ conserved and exergy is ____ conserved.
a) always, generally
b) always, not generally
c) not always, always
d) always, always
View Answer

Answer: b
Explanation: Exergy is destroyed due to irreversibilities.
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4. When the closed system is allowed to undergo a spontaneous change from a given state to a dead state, its exergy is ____ destroyed ____ producing useful work.
a) not completely, though
b) not completely, without
c) completely, though
d) completely, without
View Answer

Answer: d
Explanation: The potential to develop work which was originally present is completely wasted in such a spontaneous process.

5. The difference in exergy entering a system and that leaving out is the exergy which is destroyed.
a) true
b) false
View Answer

Answer: a
Explanation: This is because exergy is not conserved.
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6. The exergy of an isolated system can ____
a) increase
b) decrease
c) never increase
d) never decrease
View Answer

Answer: c
Explanation: It is the counterpart of the entropy principle which states that the entropy of an isolated system can never decrease.

7. Since irreversibility > 0, the only processes allowed by the second law are those for which the exergy of the isolated system
a) increases
b) decreases
c) remains constant
d) none of the mentioned
View Answer

Answer: b
Explanation: The above statement also means that the exergy of an isolated system can never increase.
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8. For a closed system, availability or exergy transfer occurs through
a) heat interactions
b) work interactions
c) both of the mentioned
d) none of the mentioned
View Answer

Answer: c
Explanation: These are the two ways in which exergy transfer can take place.

9. For an isolated system, the exergy balance gives
a) ΔA=-I
b) ΔA=I
c) ΔA=0
d) none of the mentioned
View Answer

Answer: a
Explanation: Change in availability = – irreversibility.
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10. Which law is used for exergy balance?
a) first law
b) second law
c) first law and second law
d) third law
View Answer

Answer: c
Explanation: Both first and second laws are used to balance exergy.

11. Evaluate the steady state exergy flux due to a heat transfer of 250 W through a wall with 400 K on one side and 600 K on the other side. Also find the exergy destruction in the wall. thermodynamics-questions-answers-entrance-exams-q11
a) 52 W
b) 62 W
c) 72 W
d) 82 W
View Answer

Answer: b
Explanation: Φ(Q) = [1 – (T0/T)]Q
Φ1(Q) = [1 – (T0/T1)]Q = [1-(298/600)](250) = 125.8 W
Φ2(Q) = [1 – (T0/T2)]Q = [1-(298/400)](250) = 63.8 W
Φ(destruction) = Φ1 – Φ2 = 125.8 – 63.8 = 62 W.

12. A constant pressure piston/cylinder contains 2 kg of water at 5 MPa and 100°C. Heat is added from a reservoir at 700°C to the water until it reaches 700°C. Find the total irreversibility in the process.
a) 1572 kJ
b) 1672 kJ
c) 1772 kJ
d) 1872 kJ
View Answer

Answer: a
Explanation: This process is : P = C hence 1W2 = P(V2 – V1)
1Q2 = m(u2 – u1) + 1W2 = m(h2 – h1)
= 2(3900.13 – 422.71) = 6954.8 kJ
1S2 gen = m(s2 – s1) – 1Q2/T
= 2(7.5122 – 1.303) – 6954.8273 + 700 = 5.2717 kJ/K
irreversibility = m 1i2 = T0 1S2 gen
= 298.15 K × 5.2717 kJ/K
= 1572 kJ.

13. A flow of air at 1000 kPa, 300 K is throttled to 500 kPa. What is the irreversibility?
a) 47.63 kJ/kg
b) 57.63 kJ/kg
c) 67.63 kJ/kg
d) 77.63 kJ/kg
View Answer

Answer: b
Explanation: A throttle process is a constant enthalpy process
Process: he = hi so ideal gas => Te = Ti
Entropy Eq.: se – si = sgen
sgen = – 0.287 ln (500/1000) = 0.2 kJ/kg K
i = To*s = 288.15 X 0.2 = 57.63 kJ/kg.

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