# Power Plant Questions and Answers – Rankine Cycle

This set of Power Plant Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Rankine cycle”.

1. What is the unit of Heat rate?
a) kJ/KW
b) KW/kJ
c) kJ
d) KW

Explanation: Heat rate is the rate of input required to produce unit shaft output.

2. Rankine cycle operating on low pressure limit of p1 and high pressure limit of p2 ___________
a) has higher thermal efficiency than the Carnot cycle operating between same pressure limits
b) has lower thermal efficiency than Carnot cycle operating between same pressure limits
c) has same thermal efficiency as Carnot cycle operating between same pressure limits
d) may be more or less depending upon the magnitudes of p1 and p2

Explanation: Area under P-V curve for Rankine will be more as compared to Carnot cycle.

3. Rankine efficiency of a Steam Power Plant ___________
a) improves in Summer as compared to that in Winter
b) improves in Winter as compared to that in Summer
c) is unaffected by climatic conditions
d) none of the mentioned

Explanation: In winters, the temperature of cooling water is low, which increases Condenser’s efficiency.

4. Rankine cycle comprises of ___________
a) two isentropic processes and two constant volume processes
b) two isentropic processes and two constant pressure processes
c) two isothermal processes and two constant pressure processes
d) none of the mentioned

Explanation: Rankine cycle is a reversible cycle which have two constant pressure and two constant temperature processes.

5. In Rankine cycle, the work output from the turbine is given by ___________
a) change of internal energy between inlet and outlet
b) change of enthalpy between inlet and outlet
c) change of entropy between inlet and outlet
d) change of temperature between inlet and outlet

Explanation: Work output(turbine) = h1 – h2.
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6. Which of the following contributes to the improvement of efficiency of Rankine cycle in a Thermal Power Plant?
a) reheating of steam at intermediate stage
b) regeneration use of steam for heating Boiler feed water
c) use of high pressures
d) all of the mentioned

Explanation: The regenerative features effectively raise the nominal cycle heat input temperature, by reducing the addition of heat from the Boiler/fuel source at the relatively low feedwater temperatures that would exist without regenerative feedwater heating.

7. Match the following:

```i) Boiler                 A. reversible adiabatic expansion of steam
ii) turbine               B. constant pressure heat heat addition
iii) Condenser            C. reversible adiabatic compression
iv) pump                  D. constant pressure heat rejection```

a) i-B ii-A iii-D iv-C
b) i-A ii-C iii-D iv-A
c) i-B ii-D iii-C iv-A
d) i-A ii-D iii-B iv-C

Explanation: Working fluid in Rankine cycle undergoes 4 processes, expansion in turbine, heat addition in Boiler, heat rejection in Condenser and compression in pump.

8. What is the actual turbine inlet temperature in Rankine cycle?
a) 700C
b) 800C
c) 550C
d) 1150C

Explanation: The TIT(Turbine Inlet Temperature) is of the range 500-570C.

9. Rankine cycle efficiency of a good Steam Power Plant may be in the range of?
a) 15 to 20%
b) 35 to 45%
c) 70 to 80%
d) 90 to 95%

Explanation: Efficiency of Rankine cycle in actual working condition is found to be between 35 to 45%.

10. A simple Rankine cycle operates the Boiler at 3 MPa with an outlet temperature of 350°C and the Condenser at 50 kPa. Assuming ideal operation and processes, what is the thermal efficiency of this cycle?
a) 7.7
b) 17.7
c) 27.7
d) 37.7

Explanation: Fixing the states; h1 = 340.5 kJ/kg, h2 = h1 + v1 (P2 – P1) = 343.5 kJ/kg, h3 = 3115.3 kJ/kg, s3 = 6.7428 kJ/kg – K, x4 = 0.869, and h4 = 2343.9 kJ/kg. Thus, η = 1 – Qout / Qin = 1 – (h4 – h1) / (h3 – h2) = 27.7%.

11. A simple Rankine cycle produces 40 MW of power, 50 MW of process heat and rejects 60 MW of heat to the surroundings. What is the utilization factor of this co generation cycle neglecting the pump work?
a) 50
b) 60
c) 70
d) 80

Explanation: Application of the first law to the entire cycle gives Qin = Qp + Qreject + W = 150 MW. The utilization factor is then = (Qp + W) / Qin = 60%.

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