Flywheel MCQ Quiz - Objective Question with Answer for Flywheel - Download Free PDF

Concept:

Kinetic energy of a rotating body is given by,

\( KE = \frac{1}{2} I \omega^2 \)

Since moment of inertia (I) is constant,

\( KE \propto N^2 \)

Given:

A flywheel absorbs 24 kJ of energy when the speed increases from 210 rpm to 214 rpm.

We need to find the kinetic energy at 220 rpm.

Calculation:

Let:

\( KE_1 = k \times (210)^2 = 44100k \)

\( KE_2 = k \times (214)^2 = 45796k \)

Given: \( KE_2 - KE_1 = 24 \) kJ

\( \Rightarrow 45796k - 44100k = 1696k = 24 \Rightarrow k = \frac{24}{1696} = 0.01415 \)

Finding KE at 220 rpm:

\( KE = k \times (220)^2 = 0.01415 \times 48400 = 685.86~\text{kJ} \)

Explanation:

Flywheel:

• A flywheel is a mechanical device specifically designed to store rotational energy and regulate the cyclic fluctuations in the speed of an engine. It is typically a heavy wheel that is mounted on the crankshaft of an engine and rotates along with it. The flywheel smoothens the variations in the speed of the crankshaft caused by the intermittent power pulses of the engine by acting as an energy reservoir.
• The flywheel operates by absorbing energy during the power strokes of the engine and releasing it during the non-power strokes. When the engine generates excess energy during the power stroke, the flywheel stores it as rotational kinetic energy. During the other strokes, such as the exhaust, intake, and compression strokes, the flywheel releases the stored energy to maintain a consistent speed. This ensures a smooth operation of the engine and reduces vibrations.

Functions of a Flywheel:

• Regulating Speed Fluctuations: The flywheel minimizes cyclic speed variations by smoothing out the energy fluctuations during different strokes of the engine.
• Energy Storage: It stores kinetic energy during the power stroke and releases it during the other strokes, maintaining a consistent engine speed.
• Starting the Engine: In some cases, the flywheel aids in starting the engine by providing the initial rotational inertia required.
• Balancing: The flywheel helps in balancing the engine and reducing vibrations, contributing to smoother operation.

Advantages:

• Improves the smoothness of engine operation by regulating speed fluctuations.
• Reduces vibrations and noise in the engine.
• Enhances the efficiency and reliability of the engine.

Applications: Flywheels are widely used in various applications, including:

• Internal combustion engines in automobiles.
• Industrial machinery and equipment.
• Energy storage systems.
• Power plants and turbines.

Explanation:

Fluctuation of Energy in a Turning Moment Diagram

• In a turning moment diagram, the term fluctuation of energy refers to the variations in energy levels above and below the mean resisting torque line. These variations occur due to the periodic nature of the torque generated by the engine during one complete cycle of operation. The fluctuation of energy is a crucial concept in the design of flywheels, which are used to smooth out these variations and ensure uniform operation.

In any reciprocating engine, the torque produced is not constant throughout the cycle. Instead, it varies depending on the position of the crank and the forces acting during different stages of the engine’s operation, such as suction, compression, expansion, and exhaust. This variation creates regions in the turning moment diagram where the energy is higher or lower than the mean resisting torque line. These regions are referred to as:

• Surplus Energy: When the torque is higher than the mean torque, the engine generates surplus energy, which is stored in the flywheel.
• Deficit Energy: When the torque is lower than the mean torque, the flywheel releases the stored energy to compensate for the deficit.

The difference between the maximum and minimum energy levels in the turning moment diagram is called the maximum fluctuation of energy. This value is critical in determining the size and mass of the flywheel required to maintain smooth engine operation.

Role of Fluctuation of Energy:

• The fluctuation of energy plays a significant role in mechanical systems that rely on reciprocating engines. Without a mechanism to manage these fluctuations, the engine's operation would be jerky, leading to inefficiencies and potential damage to the connected machinery. The flywheel is the component designed to address this issue. By absorbing surplus energy and releasing it during deficit phases, the flywheel ensures a uniform delivery of power to the output shaft.

Concept:

The coefficient of speed fluctuation is defined as:

\( C_s = \frac{N_{\text{max}} - N_{\text{min}}}{N_{\text{mean}}} \)

And \(N_{mean}~=~\frac{N_{max}~+~N_{min}}{2}\)

Then:

\( N_{\text{max}} = N_{\text{mean}} \left(1 + \frac{C_s}{2} \right) \)

\( N_{\text{min}} = N_{\text{mean}} \left(1 - \frac{C_s}{2} \right) \)

Calculation:

\( \frac{N_{\text{max}}}{N_{\text{min}}} = \frac{1 + \frac{0.05}{2}}{1 - \frac{0.05}{2}} = \frac{1.025}{0.975} \approx 1.05 \)

Concept:

Power developed by a rotating shaft is given by: \( P = T_{\text{mean}} \cdot \omega \)

Where, \( T_{\text{mean}} \) is the mean torque and \( \omega \) is the angular speed in rad/s.

Given:

Torque, \(T(\theta) = 800 + 180 \sin(3\theta) \, \text{Nm}\)

Speed, N = \(600 \, \text{rpm} , \pi = 3.14\)

Calculation:

Since the average of a sine function over a complete cycle is 0, the mean torque:

\( T_{\text{mean}} = 800 \, \text{Nm} \)

Angular velocity, \( \omega = \frac{2\pi N}{60} = \frac{2 \times 3.14 \times 600}{60} = 62.8 \, \text{rad/s} \)

Power, \( P = 800 \times 62.8 = 50240 \, \text{W} = 50.24 \, \text{kW} \)

Concept:

Moment of inertia is given by:

I = mk²

Torque (τ) is given by:

τ = I × α 

Kinetic energy (KE) of flywheel is given by:

\(KE = \frac{1}{2} × I × {\omega ^2}\)

Calculation:

Given:

Radius of gyration of flywheel, k = 1 m, Mass of the flywheel, m = 2500 kg, Starting Torque, T = 2500 N, Time of rotation, t = 10 s

Starting speed, ω₀ = 0 rad/sec

Moment of inertia is:

I = mk²

I = 2500 × (1)² = 2500 kg – m²

τ = I × α

2500 = 2500 × α

α = 1 rad/s²

ω = ω₀  + αt

ω = 0 + 1 × 10 = 10 rad/s²

Kinetic energy (KE) of flywheel is:

^{\(KE = \frac{1}{2} × I × {\omega ^2}\)}

\(K = \frac{1}{2} × 2500 × 100 = 125000 = 125~kN.m\)

Concept:

Given:

Mass moment of inertia of the flywheel = I, Mean speed of the rotation = ω, Kinetic energy = E, Fluctuation of speed = K

Coefficient of fluctuation of energy:

\({C_E} = \frac{{Maximum\;fluctuation\;of\;energy}}{{Work\;done\;per\;cycle}}\)

Coefficient of fluctuation of speed:

\(\begin{array}{l} {K} = \frac{{Maximum\;fluctuation\;of\;speed}}{{Mean\;speed}}\\ {K} = \frac{{{\omega _1} - {\omega _2}}}{\omega } = \frac{{2\left( {{\omega _1} - {\omega _2}} \right)}}{{\left( {{\omega _1} + {\omega _2}} \right)}} \end{array}\)

The maximum fluctuation of energy:

\(\begin{array}{l} {\rm{\Delta }}E = \frac{1}{2}I\omega _1^2 - \frac{1}{2}I\omega _2^2 = \frac{1}{2}I\left( {{\omega _1} + {\omega _2}} \right)\left( {{\omega _1} - {\omega _2}} \right)\\ {\rm{\Delta }}E = I\omega \left( {{\omega _1} - {\omega _2}} \right) = I{\omega ^2}\frac{{\left( {{\omega _1} - {\omega _2}} \right)}}{\omega } = I{\omega ^2}{K}\\ E = \frac{1}{2}I{\omega ^2} = \frac{1}{2}.\frac{{{\rm{\Delta }}E}}{{{K}}} = \Delta E\;=2KE {{{}}}{{{}}} \end{array}\)

Hence all of the options are correct.

Concept:

• Rotational kinetic energy: The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
• A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
• Mathematically rotational kinetic energy can be written as -

\(KE = \frac{1}{2}I{ω ^2}\)

where I = moment of inertia and ω = angular velocity, k = radius of gyration

The Moment of inertia of flywheel is given as, I = mk² 

Calculation:

Given:

m = 100 kg, ω = 10 rad/sec, k = 10 cm = 0.1 m

Therefore, \(KE = \frac{1}{2}(mk^2){ω ^2}\)

\(KE = \frac{1}{2}~\times~100~\times~0.1^2~\times~{10 ^2}=50~J\)

Concept:

The area of peak torque curve = The area of the mean torque curve.

Calculation:

Given:

T_m = 20 Nm

The area of peak torque curve = The area of the mean torque curve.

\(\frac{1}{2} \times {T_P} \times \pi = 20 \times 4\pi\)

Concept:

Coefficient of fluctuation of energy:

\({C_E} = \frac{{Maximum\;fluctuation\;of\;energy}}{{Work\;done\;per\;cycle}}\)

\({C_E} = \frac{\Delta E}{W} \Rightarrow \Delta E=C_E.W\)

Coefficient of fluctuation of speed:

\(\begin{array}{l} {C_S} = \frac{{Maximum\;fluctuation\;of\;speed}}{{Mean\;speed}}\\ {C_S} = \frac{{{ω _1} - {ω _2}}}{ω } = \frac{{2\left( {{ω _1} - {ω _2}} \right)}}{{\left( {{ω _1} + {ω _2}} \right)}} \end{array}\)

ω = mean angular speed during the cycle (rad/s)

\(ω = \frac{{{ω _1} + {ω _2}}}{2}\)

Maximum fluctuation of energy: ΔE = CE. W

\(\begin{array}{l} {\rm{\Delta }}E = \frac{1}{2}Iω _1^2 - \frac{1}{2}Iω _2^2 = \frac{1}{2}I\left( {{ω _1} + {ω _2}} \right)\left( {{ω _1} - {ω _2}} \right)\\ {\rm{\Delta }}E = Iω \left( {{ω _1} - {ω _2}} \right) = I{ω ^2}\frac{{\left( {{ω _1} - {ω _2}} \right)}}{ω } = I{ω ^2}{C_S}\\ K.E = \frac{1}{2}I{ω ^2} = \frac{1}{2}.\frac{{{\rm{\Delta }}E}}{{{C_S}}} = \frac{{{C_E}W}}{{2{C_S}}} \end{array}\)

Explanation:

Flywheel

• An engine works in a cycle and produces power only in one of the strokes, whereas it consumes power in the remaining strokes.
• It means a large fluctuation of energy takes place inside the engine within the cycle itself.
• Such a variation of energy will produce a large fluctuation in speed also.

Variation of turning moment in a four-stroke engine.

• The flywheel controls the cyclic fluctuation of speed by gaining energy during the power stroke and releasing the energy during the remaining stroke.
• A flywheel is connected to the crankshaft with help of the pin joint.
• During power stroke, speed of the engine tends to increase and since the flywheel of heavy mass is connected to the engine its speed also increases.
• The increase in speed of the flywheel is not very significant, since the weight of the flywheel is heavy and increase in kinetic energy is proportional to the square of the speed.
• The flywheel has large moment of inertia and change in its kinetic energy is equal to the power produced by the engine or power required by the engine. Thus a flywheel is basically a reservoir of energy.
• There are two types of flywheel:
  1. Disc type flywheel
  2. Rim type flywheel

The use of disc type flywheel is not preferred in an engine due to the following reasons.

• The radius of gyration is required to be high for the flywheel to absorb high energy with a small change of speed.
• For disc type flywheel, the radius of gyration is 0.707 times of the rim type flywheel.
• The stress in the disc type flywheel is not uniformly distributed. In rim type, the section of the wheel is maximum where the stress in the flywheel is maximum which is the outermost position of the flywheel.

Students often confuse in the working of governor and flywheel.

Governor – Maintains the constant mean speed of the shaft.

Flywheel – it doesn't maintain a constant speed. It simply reduces the fluctuation of speed.

Explanation:

Turning Moment Diagram:

• The turning moment diagram is also known as the crank effort diagram.
• It is the graphical representation of the turning moment or torque or crank effort (Y-axis) for the various position of the crank (X-axis).
• The area under the turning moment diagram gives the work done per cycle.
• The work done per cycle when divided by the crank angle per cycle gives mean torque Tm.

The coefficient of fluctuation of speed is given by,

\(\begin{array}{l} {C_1} = \frac{{\left( {{\omega _{max}} - {\omega _{min}}} \right)}}{{\left[ {\frac{{{\omega _{max}} + {\omega _{min}}}}{2}} \right]}}\\ \Rightarrow {C_1}{\omega _{max}} + {C_1}.{\omega _{min}} = 2{\omega _{max}} - 2{\omega _{min}}\\ \therefore \frac{{{\omega _{max}}}}{{{\omega _{min}}}} = \frac{{2 + {C_1}}}{{2 - {C_1}}} \end{array}\)

Concept:

Mass moment of inertia of the flywheel is given by:

I = mk2 

Starting torque of the engine is given by:

T = I.α

where I = moment of inertia, α = angular acceleration

Calculation:

Given:

m = 2400 kg, r = 1 m, T = 1500 N-m

Now, I = mk2 = 2400 × 12 = 2400 kg-m2

Angular acceleration of the flywheel:

\(\alpha=\frac{T}{I}=\frac{1500}{2400}=0.625\) rad/s2

A flywheel used in machines serves as a reservoir, which stores energy during the period when the supply of energy is more than the requirement and releases it during the period when the requirement of energy is more than the supply.

As the force impulse is given by the change in momentum, i.e. ΔP = F.Δt. where, P is momentum and t is time.

In the same way, the power impulse is given by the change or fluctuation in energy, i.e. ΔE = P.Δt. where, P is power and t is time.

Thus, smooth out power impulses mean the smoothness in the fluctuation of energy, which is the work of a flywheel.

Note: A flywheel stores energy when the supply of energy is less than the required but flywheel will not work, then the main supply will fail, i.e. when there is no supply of energy or zero supply of energy.