The equation TdS=dH-Vdp
Answer: D
Since there is no path function in the equation hence the equation holds good for any process.
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The available energy of a system ___ as its temperature or pressure decreases and approaches that of the surroundings.
Answer: B
As temperature decreases, exergy decreases.
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Which of the following is true?
Answer: D
This is because, Q for reversible=(To)*(S2-S1) and Q for irreversible<(To)*(S1-S2).
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When considering a finite energy source, its working fluid expands,
Answer: C
For a finite energy source, expansion of working fluid is reversibly and adiabatically.
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For a process from state 1 to state 2, heat transfer in an irreversible process is given by
Answer: C
To is the temperature of the surroundings and S1,S2 are the entropies at state 1 and 2 respectively ans ΔS(universe)>0.
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An insulated cylinder/piston contains R-134a at 1 MPa, 50°C, volume of 100 L. The R-134a expands, dropping the pressure in the cylinder to 100 kPa. The R-134a does 190 kJ of work against the piston during this process. Is that possible?
Answer: A
v1 = 0.02185 m^3/kg, u1 = 409.39 kJ/kg,
s1 = 1.7494 kJ/kg K, m = V1/v1 = 0.1/0.02185 = 4.577 kg
m(u2 – u1) = 1Q2 – 1W2 = 0 – 190 hence u2 = u1 − 1W2/m = 367.89 kJ/kg
T2 = -19.25°C ; s2 = 1.7689 kJ/kg K
m(s2 – s1) = ⌡⌠dQ/T + 1S2(gen) = 1S2(gen)
1S2(gen) = m(s2 – s1) = 0.0893 kJ/K
This is possible since 1S2(gen) > 0.
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For an infinitesimal reversible process at constant pressure,
Answer: D
Here m is the mass of gas flowing, Cp is its specific heat, and T is the gas temperature.
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For a process from state 1 to state 2, heat transfer in a reversible process is given by
Answer: B
To is the temperature of the surroundings and S1,S2 are the entropies at state 1 and 2 respectively and ΔS(universe)=0.
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The complete conversion of heat into shaft-work is impossible.
Answer: A
This statement can be proved by the second law of thermodynamics.
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The equation TdS=dU+pdV is obtained from which law?
Answer: C
No answer description available for this question.
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