Thermodynamics Questions and Answers – Availability-2

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This set of Thermodynamics online quiz focuses on “Availability-2”.

1. The compressor in a refrigerator takes R-134a in at 100 kPa, −20°C and then compresses it to 1 MPa, 40°C. With the room temperature at 20°C find the minimum compressor work.
a) -48.19 kJ/kg
b) -58.19 kJ/kg
c) -68.19 kJ/kg
d) -78.19 kJ/kg
View Answer

Answer: a
Explanation: w(c) = h1 – h2 + q(rev)
w(min) = h1 – h2 + To(s2 – s1) = 387.22 – 420.25 + 293.15 × (1.7148 – 1.7665)
= -48.19 kJ/kg.
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2. Find the specific reversible work for a steam turbine with inlet at 4 MPa, 500°C and an actual exit state of 100 kPa, x = 1.0 with a 25°C ambient temperature.
a) 550.0 kJ/kg
b) 650.0 kJ/kg
c) 750.0 kJ/kg
d) 850.0 kJ/kg
View Answer

Answer: d
Explanation: To = 25°C = 298.15 K, hi = 3445.2 kJ/kg; si = 7.090 kJ/kg K,
he = 2675.5 kJ/kg; se = 7.3593 kJ/kg K
w(rev) = (hi – Tosi) – (he – Tose) = (hi – he) + To(se – si)
= (3445.2 – 2675.5) + 298.2(7.3593 – 7.0900)
= 769.7 + 80.3 = 850.0 kJ/kg.

3. Find the specific reversible work for a compressor using R-134a with inlet state of –20°C, 100 kPa and an exit state of 50°C, 600 kPa. Use 25°C as ambient temperature.
a) -28.878 kJ/kg
b) -38.878 kJ/kg
c) -48.878 kJ/kg
d) -58.878 kJ/kg
View Answer

Answer: b
Explanation: The compressor is assumed to be adiabatic so q = 0
w(rev) = To(se – si) – (he – hi)
hi = 387.22 kJ/kg; si = 1.7665 kJ/kg K;
he = 438.59 kJ/kg; se = 1.8084 kJ/kg K
w(rev) = 298.15 (1.8084 – 1.7665) – (438.59 – 387.22)
= -38.878 kJ/kg.
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4. A steady stream of R-22 at ambient temperature of 10°C, and at 750 kPa enters a solar collector. The stream exits at 80°C, 700 kPa. Calculate the change in availability.
a) 4.237 kJ/kg
b) 5.237 kJ/kg
c) 6.237 kJ/kg
d) 7.237 kJ/kg
View Answer

Answer: c
Explanation: hi = 56.46 kJ/kg, si = 0.2173 kJ/kg K,
he = 305.91 kJ/kg, se = 1.0761 kJ/kg K
∆ψie = ψe – ψi = (he – hi) – T0(se – si)
= (305.912 – 56.463) – 283.2(1.0761 – 0.2173)
= 6.237 kJ/kg.

5. Cold water is running in a river at 2°C and the air temperature is 20°C. What is the availability of water relative to the ambient temperature?
a) 2.157 kJ/kg
b) 2.857 kJ/kg
c) 3.457 kJ/kg
d) 2.457 kJ/kg
View Answer

Answer: d
Explanation: ψ = h1 – h0 – T0(s1 – s0)
ψ = 8.392 – 83.96 – 293.15(0.03044 – 0.2966)
= 2.457 kJ/kg.
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6. Nitrogen is flowing in a pipe with a of velocity 300 m/s at 500 kPa, 300°C. What is its availability relative to an ambient at 100 kPa, 20°C?
a) 272 kJ/kg
b) 252 kJ/kg
c) 292 kJ/kg
d) 232 kJ/kg
View Answer

Answer: a
Explanation: ψ = h1 – h0 + (1/2)V^2 – T0(s1 – s0)
= Cp(T1 – T0) + (1/2)V^2 – T0[Cp ln(T1/T0) – R ln(P1/P0)] = 1.042(300-20)+(300^2)/2000 – 293.15[1.042 ln (573.15/293.15) – 0.2968ln(500/100)] = 272 kJ/kg.

7. R-12 at 30°C, 0.75 MPa enters a steady flow device and exits at 30°C, 100 kPa. Assuming the process to be isothermal and reversible, find the change in availability of the refrigerant.
a) -26.1 kJ/kg
b) -36.1 kJ/kg
c) -46.1 kJ/kg
d) -56.1 kJ/kg
View Answer

Answer: b
Explanation: hi = 64.59 kJ/kg, si = 0.2399 kJ/kg K,
and he = 210.02 kJ/kg, se = 0.8488 kJ/kg K
∆ψ = he – hi – T0(se – si) = 210.02 – 64.59 – 298.15(0.8488 – 0.2399)
= -36.1 kJ/kg.
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8. A wooden bucket(2 kg) with 10 kg hot liquid water, both at 85°C, is lowered down to 400 m into a mineshaft. What is the availability of water and bucket with respect to the surface with ambient temperature of 20°C?
a) 232.2 kJ
b) 242.2 kJ
c) 212.2 kJ
d) 252.2 kJ
View Answer

Answer: a
Explanation: φ1 – φ0 = m(wood)[u1 – u0 – T0(s1- s0)] + m(H2O)[u1- u0- T0(s1- s0)] + m(tot)g(z1- z0)
= 2[1.26(85 – 20) – 293.15× 1.26 ln{(273.15 + 85)/293.15}
+ 10[ 355.82 – 83.94 – 293(1.1342 – 0.2966)] + 12 × 9.807 × (-400) /1000
= 15.85 + 263.38 – 47.07 = 232.2 kJ.

9. Air in a piston/cylinder arrangement is at 25°C, 110 kPa with a volume of 50 L. It goes through a reversible polytropic process to final state of 500 K, 700 kPa and exchanges heat with the ambient at 25°C. Find the total work from the ambient.
a) -9.28 kJ
b) -9.38 kJ
c) -9.48 kJ
d) -9.58 kJ
View Answer

Answer: d
Explanation: ma*(u2 – u1) = 1Q2 – 1W2,(tot) ; ma*(s2 – s1) = 1Q2/T0
ma = 110 × 0.05/0.287 × 298.15 = 0.0643 kg
1Q2 = T0*ma*(s2 – s1) = 298.15 × 0.0643[7.3869 – 6.8631 – 0.287 ln (700/110)] = -0.14 kJ
1W2,(tot) = 1Q2 – ma*(u2 – u1) = -0.14 – 0.0643 × (359.844 – 213.037)
= -9.58 kJ.
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10. Find the specific reversible work for a R-134a compressor with inlet state of –20°C, 100 kPa and an exit state of 600 kPa, 50°C. Use a 25°C ambient temperature.
a) 48.878 kJ/kg
b) -38.878 kJ/kg
c) 48.878 kJ/kg
d) -38.878 kJ/kg
View Answer

Answer: b
Explanation: This is a steady state flow device
and the compressor is assumed to be adiabatic so q = 0,
w(rev) = T0(se – si) – (he – hi)
= 298.15(1.8084 – 1.7665) – (438.59 – 387.22)
= -38.878 kJ/kg.

11. A steady stream of R-22 at ambient temperature, 10°C, and at 750 kPa enters a solar collector. The stream exits at 80°C, 700 kPa. Calculate the change in availability of the R-22 between these two states.
a) 8.762 kJ/kg
b) 8.143 kJ/kg
c) 7.237 kJ/kg
d) 6.237 kJ/kg
View Answer

Answer: d
Explanation: Change in availability = (he – hi) – T0(se – si)
= (305.912 – 56.463) – 283.2(1.0761 – 0.2173)
= 6.237 kJ/kg.

Sanfoundry Global Education & Learning Series – Thermodynamics.
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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