Thermodynamics Questions and Answers – Work of Compression

This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Work of Compression”.

1. A gas compression process is
a) adiabatic
b) involves heat transfer
c) both of the mentioned
d) none of the mentioned
View Answer

Answer: c
Explanation: A gas compression process can be either adiabatic or can involve heat transfer.

2. If the gas is cooled during compression, work required will be ____ the adiabatic compression work.
a) more than
b) less than
c) equal to
d) none of the mentioned
View Answer

Answer: b
Explanation: Here the work required will be less than that required for adiabatic compression.

3. Which of the following is an advantage of cooling?
a) less pipe friction losses
b) reduction in volume of gas
c) both of the mentioned
d) none of the mentioned
View Answer

Answer: c
Explanation: These are the two advantages of cooling.
advertisement
advertisement

4. We use after-coolers to cool the gas which leaves the compressor.
a) true
b) false
View Answer

Answer: a
Explanation: This is done because compression process is somewhat ineffective.

5. The work of compression is ____ the shaft work.
a) positive of
b) negative of
c) equal to
d) less than
View Answer

Answer: b
Explanation: This is true for reversible adiabatic compression.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. For ɣ>n>1 and for same pressure ratio p2/p1, the maximum work is needed for
a) isothermal compression
b) adiabatic compression
c) polytropic compression
d) all need same work
View Answer

Answer: b
Explanation: This comes when these three reversible compression processes are plotted on the p-V diagram.

7. In isothermal compression, all work done on gas is transformed into
a) heat added into system
b) heat going out of system
c) internal energy increase
d) none of the mentioned
View Answer

Answer: c
Explanation: This is the case of isothermal compression.
advertisement

8. When isothermal compression is taken as ideal process, the energy imparted
a) raises the temperature of gas
b) raises the pressure of gas
c) both of the mentioned
d) none of the mentioned
View Answer

Answer: b
Explanation: In isothermal compression considered as ideal process, no energy is imparted to the gas.

9. The adiabatic efficiency is given by
a) Ws/Wc
b) Ws/Wt
c) Wt/Wc
d) Wt/Ws
View Answer

Answer: a
Explanation: This is the efficiency of compressor working in a steady flow process.
advertisement

10. The isothermal efficiency is given by
a) Ws/Wc
b) Ws/Wt
c) Wt/Wc
d) Wt/Ws
View Answer

Answer: c
Explanation: This is the efficiency of compressor working in a steady flow process and Wt=work in reversible isothermal compression.

11. The adiabatic efficiency of real compressor can be ____
a) less than unity
b) greater than unity
c) equal to unity
d) none of the mentioned
View Answer

Answer: b
Explanation: This is due to the effects of cooling.

12. For an adiabatic machine, work of compression is greater than enthalpy rise of gas.
a) true
b) false
View Answer

Answer: b
Explanation: For an adiabatic machine, work of compression is equal to the enthalpy rise of gas.

13. Argon is kept in a 5 m3 tank at −30°C and 3 MPa. Determine the mass using compressibility factor.
a) 208.75 kg
b) 308.75 kg
c) 303.75 kg
d) 203.75 kg
View Answer

Answer: b
Explanation: Tr = 243.15/150.8 = 1.612 and Pr = 3000/4870 = 0.616 hence Z = 0.96
m = PV/ZRT = (3000 × 5)/(0.96 × 0.2081 × 243.2)
= 308.75 kg.

14. Find the error in specific volume if ideal gas model is used to represent the behaviour of superheated ammonia at 40°C and 500 kPa?
a) 1.5%
b) 3.5%
c) 4.5%
d) 2.5%
View Answer

Answer: c
Explanation: NH3, T = 40°C = 313.15 K, Tc = 405.5 K, Pc = 11.35 MPa
v = 0.2923 m3/kg
Ideal gas: v = RT/P = (0.48819 × 313)/(500) = 0.3056 m3/kg
thus error = 4.5%.

15. Find the volume of ethylene having mass of 125 kg at 7.5 MPa and 296.5 K.
a) 0.369 m3
b) 0.669 m3
c) 0.569 m3
d) 0.469 m3
View Answer

Answer: d
Explanation: For ethylene, Tc = 282.4 K and Pc = 5.04 MPa
Tr = T/Tc = 296.5 / 282.4 = 1.05 and Pr = P/Pc = 7.5 / 5.04 = 1.49
thus Z = 0.32
hence V = mZRT / P = 125 × 0.32 × 0.2964 × 296.5 / 7500 = 0.469 m3.

Sanfoundry Global Education & Learning Series – Thermodynamics.
To practice all areas of Thermodynamics, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
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.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.