Automobile Engineering Multiple Choice Questions on “Flywheel”.
1. What is the maximum fluctuation of energy?
a) It is the sum of maximum and minimum energies
b) It is the ratio of the maximum and minimum energies
c) It is the ratio of the mean resisting torque to the work done per cycle
d) It is the difference between the maximum and minimum energies
Answer: d
Clarification: The maximum fluctuation of energy is the difference between the maximum and minimum energies. The fluctuation of energy can be found out by the turning moment diagram for one complete cycle of operation.
2. What is the coefficient of fluctuation of energy?
a) The ratio of the maximum fluctuation of energy to the minimum fluctuation of energy
b) The ratio of the maximum fluctuation of energy to the work done per cycle
c) The ratio of the minimum fluctuation of energy to the maximum fluctuation of energy
d) The ratio of the work done per cycle to the maximum fluctuation of energy
Answer: b
Clarification: The coefficient of fluctuation of energy (CE) is the ratio of the maximum fluctuation of energy to the work done per cycle. CE =(frac{Delta E}{T_{mean} theta}) where ΔE = maximum fluctuation of energy, Tmean*θ = work done per cycle, Tmean = mean torque, and θ = angle turned in radian in one cycle.
3. Which of the following is true about the flywheel?
a) Flywheel regulates speed over a period of time
b) Flywheel takes care of the quantity of fluid
c) Flywheel controls (frac{dN}{dt})
d) Flywheel regulates the speed by regulating the quantity of charge of the prime mover
Answer: c
Clarification: Flywheel controls the rate of fluctuation of the speed during a cycle. Flywheel controls the speed in one cycle. It does not control the supply of fuel to the engine.
4. Given that maximum fluctuation of energy is 2000 N-m/s and coefficient of fluctuation of speed is 0.02. What is the mean kinetic energy of flywheel?
a) 50 kN-m/s
b) 40 kN-m/s
c) 30 kN-m/s
d) 20 kN-m/s
Answer: a
Clarification: The maximum fluctuation of energy in a flywheel = ΔE, mean kinetic energy of the flywheel = E, and coefficient of fluctuation of speed = Cs. ΔE = 2*E*Cs ⇒ 2000 = 2*E*0.02 ⇒ E = 50 kN-m/s.
5. What is the coefficient of fluctuation of speed?
a) The ratio of the maximum fluctuation of speed to the mean speed
b) The ratio of the maximum fluctuation of energy to the mean speed
c) The ratio of the minimum fluctuation of energy to the mean speed
d) The ratio of the work done per cycle to the mean speed
Answer: a
Clarification: The coefficient of fluctuation of speed is the ratio of the maximum fluctuation of speed to the mean speed. Cs = (ω1 – ω2)/ω =2 (ω1 – ω2) / (ω1+ω2) or Cs = 2(N1-N2) / (N1+N2).
6. Which of the following is the expression of the maximum fluctuation of energy in a flywheel where I = mass moment of inertia of the flywheel, Cs = coefficient of fluctuation of speed, ω = mean angular speed of ω1 and ω2, and E = mean kinetic energy of the flywheel?
a) I*ω*(ω1+ω2)
b) I*ω*Cs
c) 2*E2*Cs
d) I*ω2*Cs
Answer: d
Clarification: The maximum fluctuation of energy in a flywheel = I*ω*(ω1-ω2) = I*ω2*Cs = 2*E*Cs = I*ω2*Cs. The maximum fluctuation of energy can be calculated by using the turning moment diagram.
7. In the single-cylinder, single acting, four-stroke gas engine, the total fluctuation of speed is not to exceed ±3 percent of the mean speed. What is the coefficient of fluctuation of speed?
a) 0.04
b) 0.03
c) 0.09
d) 0.06
Answer: d
Clarification: Since the total fluctuation of speed is not to exceed ±3 percent of the mean speed, therefore ω1 – ω2 = 6% ω and coefficient of fluctuation of speed = Cs = (ω1 – ω2)/ω = 0.06.
8. The steam engine has a flywheel which is having a radius of gyration of 1.1 m and mass of 2200 kg. The starting torque is 1550 Nm and assumed constant. What is the angular acceleration of the flywheel?
a) 0.54 rad/s2
b) 0.58 rad/s2
c) 0.62 rad/s2
d) 0.52 rad/s2
Answer: b
Clarification: I = mass moment of inertia of the flywheel, m = mass of the flywheel, k = radius of gyration, T = stating torque, and α = angular acceleration of the flywheel. I = m*k2 = 2200*1.12 = 2662 kg-m2. T = I*α ⇒ 1550 = 2662 * α ⇒ α = 0.58 rad/s2.
9. The engine has the flywheel of mass 7 tons and the radius of gyration is 1.7 m. The fluctuation of energy is 55 kN-m. The mean speed of the engine is 150 rpm. What is the coefficient of fluctuation of speed?
a) 0.021
b) 0.011
c) 0.031
d) 0.041
Answer: b
Clarification: The fluctuation of energy = ΔE, the mass of the flywheel = m, the radius of gyration = k, the mean speed of the flywheel = N. ΔE = (frac{π^2}{900})*m*k2*N*(N1-N2) = (frac{π^2}{900})*7000*1.72*150*(N1-N2) = 33277*(N1-N2) ⇒ 55000 = 33277*(N1-N2) ⇒ (N1-N2)= (frac{55000}{33277}) = 1.652 rpm. Coefficient of fluctuation of speed = Cs = (frac{N_1-N_2}{N}) = (frac{1.652}{150}) = 0.011.
10. The flywheel stores the potential energy.
a) True
b) False
Answer: b
Clarification: The flywheel stores the rotational energy. The amount of energy that flywheel can store is proportional to the square of the mass and its rotational velocity. The lighter flywheels are used in a sports car for easy and quick recovery of the RPM.
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