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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter