A piston cylinder has 2.5 kg ammonia at -20°C, 50 kPa. It is heated to 50°C at constant pressure from external hot gas at 200°C. Find the total entropy generation.
Answer: A
v1 = 2.4463 m^3/kg, h1 = 1434.6 kJ/kg, s1 = 6.3187 kJ/kg K
v2 = 3.1435 m^3/kg, h2 = 1583.5 kJ/kg, s2 = 6.8379 kJ/kg K
1Q2 = m(h2 – h1) = 2.5 (1583.5 – 1434.6) = 372.25 kJ
1S2(gen) = m(s2 – s1) – 1Q2/T(gas)
= 2.5 (6.8379 – 6.3187) – 372.25/473.15 = 0.511 kJ/K.
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Whenever heat is transferred through a finite temperature difference, there is always a decrease in the availability of energy so transferred.
Answer: A
This is because of exergy lost due to irreversible heat transfer.
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The amount of entropy generation is given by
Answer: C
Here (S2-S1) is the entropy change of the system and ∫(dQ/T) is the entropy transfer.
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Which of the following processes is used to do maximum work done on the ideal gas that is compressed to half of its initial volume?
Answer: D
No answer description available for this question.
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Which of the following is mostly neglected while doing calculations for finding maximum work?
Answer: C
The changes in KE and PE are very small, hence they are neglected.
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The lowest practicable temperature of heat rejected is the
Answer: C
Work done and hence efficiency will be maximum when heat is rejected at the temperature of surroundings.
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The equation W=∫vdp holds good for
Answer: C
The equation given here is used for steady flow process and also when the fluid undergoes reversible adiabatic expansion or compression.
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A hot metal piece is cooled rapidly to 25°C, removing 1000 kJ from the metal. Calculate the change of entropy if saturated liquid R-22 at −20°C absorbs the energy so that it becomes saturated vapor.
Answer: C
R-22 boiling at -20°C; m = 1Q2 /h(fg) = 1000/220.327 = 4.539 kg
∆S(R-22) = ms(fg) = 4.539(0.8703) = 3.950 kJ/K.
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1kg of ammonia is contained in a piston/cylinder, as saturated liquid at −20°C. Heat is added at 100°C until a final condition of 70°C, 800 kPa is reached. Assuming the process is reversible, find the entropy generation.
Answer: D
P1 = 190.08 kPa, v1 = 0.001504 m^3/kg, u1 = 88.76 kJ/kg, s1 = 0.3657 kJ/kg K
v2 = 0.199 m^3/kg, u2 = 1438.3 kJ/kg, s2 = 5.5513 kJ/kg K
1W2 =(1/2)(190.08 + 800)1(0.1990 – 0.001504) = 97.768 kJ
1Q2 = m(u2 – u1) + 1W2 = 1(1438.3 – 88.76) + 97.768 = 1447.3 kJ
1S2(gen) = m(s2 – s1) – 1Q2/T(res) = 1(5.5513 – 0.3657) – (1447.3/373.15)
= 1.307 kJ/K.
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Which of the following statement is false?
Answer: B
Entropy generation is not a thermodynamic property and depends on the path that system follows.
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