Farm Tractor Questions and Answers – Piston Forces and Crank Efforts

This set of Farm Tractor Multiple Choice Questions & Answers (MCQs) focuses on “Piston Forces and Crank Efforts”.

1. What is the turning moment diagram?
a) Graphical representation of crank efforts
b) Representation of connecting rod
c) Representation of cam rotation
d) Representation of heat supply

Explanation: The turning moment of a crankshaft is graphically represented as a turning moment diagram. The diagram shows the crankshaft positions and so it is also called a crank effort diagram. In the graphical representation, the y-axis is taken as turning moment and x-axis is taken as the crank angle.

2. The working stroke (crest) loop in a turning moment diagram of the four-stroke engine is larger and positive.
a) True
b) False

Explanation: The working stroke is the combustion of gases and the work is done by gases. So it is a positive loop. During the working stroke, the expansion of gases occurs and the loop is larger compared to other loops of the strokes.

3. Find the net load on the piston of an engine with cylinder bore 10 cm and pressure 10 kg/cm2.
a) 7704.75 N
b) 7500.0 N
c) 7900 N
d) 7205.55 N

Explanation: Netload = Pressure * area
= (10 kg/cm2) * (π / 4 * 102)
= 785.398 kg
= 7704.75 N

4. Find the piston effort of an engine during acceleration having net load on the piston is 1000kg, inertia force of 0.5kN, mass of the reciprocating parts 10kg and neglecting frictional resistance.
a) 9.4 kN
b) 10kN
c) 8kN
d) 6kN

Explanation: Piston effort = FL – FI + WR – RF
= (1000 * 9.81) – 500 + (10 * 9.81) – 0
= 9408.1 N = 9.4 kN

5. What is the fluctuation of energy in an engine?
a) Variation in energy
b) Variation in torque
c) Variation in power
d) Variation in speed

Explanation: The fluctuation of energy is the variations of energy in an engine above and below the mean resisting torque. The mean resisting torque is the maximum torque of a flywheel which sets it into the rotation.

6. What is the maximum fluctuation of energy?
a) Difference between maximum and minimum energy
b) Difference between maximum and minimum power
c) Difference between maximum and minimum torque
d) Difference between maximum and minimum speed

Explanation: The maximum fluctuation of energy is the difference between the maximum and minimum energies of an engine according to the turning moment diagram. This is in accordance with the rotation of the crankshaft.

7. The coefficient of fluctuation of energy is the ratio of minimum fluctuation of energy to work done per cycle.
a) True
b) False

Explanation: The coefficient of fluctuation of energy is the ratio of maximum fluctuation of energy to work done per cycle. It has no unit. The engine with varying number of cylinders will have a different coefficient of fluctuation of energy.

8. Find the work done per cycle by a four-stroke power tiller engine having mean torque 200 N-m and angle turned by crank is 120°.
a) 5263.7 N-m
b) 5116.6 N-m
c) 5503.7 N-m
d) 5703.6 N-m

Explanation: Work done = Tmean * θ
Where Tmean – mean torque and θ – angle turned
= 200 * (120° * π / 180) * 4π
= 5263.7 N-m
4π denotes one cycle in a four-stroke engine.

9. Find the angular speed of the crankshaft if the power transmitted by it is 65 Hp and the mean torque is 2000 N-m.

Explanation: P = Tmean * ω
ω = P / Tmean
ω = (65 * 745.6) / 2000 = 24.232 rad/s

10. Find the work done per cycle of a four-stroke engine with power generation of 30 Hp and has 200 revolutions per minute.
a) 13.4 kN-m
b) 14 kN-m
c) 15 kN-m
d) 16 kN-m

Explanation: Work done per cycle = P * 60 / n
= (30 * 745.6) * 60 / (200 / 2)
= 13420.8 N-m = 13.4 kN-m
Where n – N / 2 for a four stroke engine.

11. Find the coefficient of fluctuation of the energy of having maximum fluctuation of energy 10 kN-m and work done per cycle is 12 kN-m.
a) 0.833
b) 0.950
c) 0.123
d) 0.321

Explanation: CE = ∆E/work done
= 10 / 12
= 0.833
Where ∆E – maximum fluctuation of energy, CE – coefficient of fluctuation of energy.

12. Find the piston acceleration of an engine having crank angle of 600, length of connecting rod 30 cm and N – 500 rpm.
a) 101 m/s2
b) 102 m/s2
c) 103 m/s2
d) 104 m/s2

Explanation: Ap = ω2.r (cosθ + (cos 2θ/n))
Where n = l / r and l = 2r
r = l / 2
r = 0.30 / 2 = 0.15m
n = 0.30 / 0.15
ω = 2πN / 60 = 2π * 500 / 60 52 rad/s
Ap = 522 * 0.15 (cos 60° + (cos 2(60°) / 2))
= 101.4 m/s2

13. Find the inertia force of reciprocating parts of an engine, given mass of the reciprocating parts 200 kg and acceleration of reciprocating parts is 120 m/s2.
a) 24000 kg-m/s2
b) 25000 kg-m/s2
c) 23000 kg-m/s2
d) 26000 kg-m/s2

Explanation: FI = MR * AR
= 200 * 120
= 24000 kg-m/s2

14. Find the torque on the crankshaft having piston effort 200 N and connecting rod length is 20 cm.
a) 22 N-m
b) 20 N-m
c) 23 N-m
d) 24 N-m

Explanation: T = FP * r
Where r = l / 2 = 0.20 / 2 = 0.1
T = 200 * 0.1 = 20 N-m

15. What is the crank effort?
a) T = FP * r
b) T = FP / r
c) T = FP + r
d) T = FP + FI

Explanation: The crank effort is the product of piston effort and the radius of the crank pin. It is also called as the turning moment. It is the torque on the crankshaft.

Sanfoundry Global Education & Learning Series – Farm Tractor.

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