This set of Thermal Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Otto Cycle (or) Constant Volume Cycle”.

1. Which cycle is idealized cycle for the spark ignition internal combustion engines?

a) Otto cycle

b) Diesel cycle

c) Dual cycle

d) Bryton cycle

View Answer

Explanation: An internal combustion engine that uses spark ignition for combustion. An air standard otto cycle is the idealized cycle for the spark ignition internal combustion engines.

2. What is the efficiency of the otto cycle?

a) 1 – \(\frac{1}{r^{γ-1}} \)

b) 1 – r^{γ-1}

c) r^{γ-1}

d) 1 – \(\frac{1}{r^{γ+1}} \)

View Answer

Explanation: Efficiency of otto cycle = 1-(v1/v2)

^{γ-1}.compression ratio, r=(v

_{1}/v

_{2}).

Therefore efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \).

3. An air standard otto cycle consist of ________

a) One constant pressure process and three adiabatic process

b) One constant pressure process, one constant volume process and two adiabatic process

c) Two constant volume process and two adiabatic process

d) Two constant pressure process and two adiabatic process

View Answer

Explanation: The sequence of processes in otto cycle is isentropic compression, isochoric heat addition, isentropic expansion and isochoric heat rejection. Therefore an air standard otto cycle consist of two constant volume and two adiabatic process.

4. Otto cycle is air standard cycle of _____________

a) S.I. engine

b) C.I. engine

c) Both SI and CI engine

d) Neither SI nor CI engine

View Answer

Explanation: An internal combustion engine that uses spark ignition for combustion is called SI engine. An air standard otto cycle is idealized for the spark ignition internal combustion engine.

5. In otto cycle heat rejection occurs at __________

a) Reversible constant volume process

b) Reversible constant pressure process

c) Irreversible constant volume process

d) Irreversible constant pressure process

View Answer

Explanation: Otto cycle consist of two isochoric process and two adiabatic process. Heat rejection process is isochoric process.

6. Which of the following is not a use of gasoline blend like gasoline mixed with tetraethyl lead in internal combustion engine?

a) Increases the octane rating fuel

b) Allows engine to operate at high compression ratio

c) Avoid auto ignition of fuel

d) Allows engine to operate at low compression ratio

View Answer

Explanation: Gasoline blend in IC engine increases the octane rating fuel and allows engine to operate at high compression ratio. Gasoline blend is use to avoid auto ignition in internal combustion engine.

7. How the efficiency of SI engine is affected by change in specific heat ratio of working fluid?

a) The efficiency of SI engine increases with increase in specific heat ratio of working fluid

b) The efficiency of SI engine decreases with increase in specific heat ratio of working fluid

c) The efficiency of SI engine increases with decrease in specific heat ratio of working fluid

d) The efficiency of SI engine does not affected by change in specific heat ratio of working fluid

View Answer

Explanation: Efficiency = 1 – \(\frac{1}{r^{γ-1}} \)

As specific heat ratio of working fluid γ increases (\(\frac{1}{r^{γ-1}} \)) decreases hence efficiency increases.

8. The compression ratio (r_{k}) of otto cycle is equal to __________

a) r_{k} = volume of the cylinder at the beginning of the compression/volume of the cylinder at the end of the compression

b) r_{k} = volume of the cylinder at the end of the compression/ volume of the cylinder at the beginning of the compression

c) r_{k} = clearance volume/ volume of the cylinder at the beginning of the compression

d) r_{k} = volume of the cylinder at the end of the compression/clearance volume

View Answer

Explanation: r

_{k}= \(\frac{swept \, volume+clearance \, volume}{clearance \, volume} \)

r

_{k}= volume of the cylinder at the beginning of the compression/volume of the cylinder at the end of the compression.

9. What is the value of working fluid in otto cycle?

a) 1.0

b) 1.2

c) 1.4

d) 1.8

View Answer

Explanation: In air standard otto cycle working fluid is air. The specific heat ratio of air is equal to 1.4.

10. For same compression ratio and heat supplied

a) Otto cycle is more efficient

b) Diesel cycle is more efficient

c) Dual cycle is more efficient

d) Both diesel and otto cycle are equally efficient

View Answer

Explanation: Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

Efficiency of diesel cycle = 1 – \(\frac{ρ^γ-1}{(ρ-1)r^{γ-1}}\).

11. What is the efficiency of an ideal air standard otto cycle if the compression ratio is 8.5?

a) 67.5%

b) 57.5%

c) 48.5%

d) 47.5%

View Answer

Explanation: Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

Efficiency of otto cycle = 1 – \(\frac{1}{8.5^{1.4-1}} \) = 57.5%.

12. What is the efficiency of an ideal air standard otto cycle if the clearance volume is 10% of the swept volume?

a) 39.8%

b) 39.2%

c) 61.7%

d) 61.1%

View Answer

Explanation: r = \(\frac{swept \, volume+clearance \, volume}{clearance \, volume} \)

r = 1.1/0.1=11

Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

Efficiency of otto cycle = 1 – \(\frac{1}{11^{1.4-1}} \)=0.6167=61.7%.

13. In an ideal otto cycle works between minimum and maximum temperature of 300K and 1800K. What is the compression ratio?

a) 4

b) 6

c) 8

d) 5

View Answer

Explanation: Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

r = T

_{max}/T

_{min}= 6.

14. What is the efficiency of an ideal air standard otto cycle if the clearance volume is 20% of the swept volume?

a) 52.1%

b) 50.1%

c) 55.1%

d) 51.1%

View Answer

Explanation: r = \(\frac{swept \, volume+clearance \, volume}{clearance \, volume} \)

r = 1.2/0.20=6

Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

Efficiency of otto cycle = 1 – \(\frac{1}{6^{1.4-1}} \)=0.511=51.1%.

15. What is the efficiency of an ideal air standard otto cycle if the compression ratio is 7?

a) 51.5%

b) 54%

c) 52.1%

d) 50%

View Answer

Explanation: Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

Efficiency of otto cycle = 1 – \(\frac{1}{7^{1.4-1}} \)=54%.

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