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Flywheel

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Flywheel
Flywheel

Dr Jordan Dimitrov
ENGD1005
Mechanical Principles
Zubair Khan
P12227965
Dr Jordan Dimitrov
ENGD1005
Mechanical Principles
Zubair Khan
P12227965

Index

Objective…… ………………………………………………….…………...2
Introduction…………………………………………………………………..2
Theory…………………………………………….………………………..3, 4
Apparatus……………...…………………………………………………….5
Procedure……….…………………………………………………………..6
Results……………………………………………………………….…...7 - 11
Discussion……………………………………………………………….11, 12
Conclusion……………………………………………………………….…12
Reference…………………………………………………………………..12

Objective
The objective of this experiment is to determine the moment of inertia of a flywheel and axle by experiment and compare it with the theoretical value.

Introduction
A flywheel is a heavy shaft-mounted rotating disc that absorbs and stores twisting or spinning motion and then releases it as rotational kinetic energy to provide motion to a stationary or nearly stationary object. [1]
Flybrids (a variation of regular electromechanical hybrids) use a flywheel instead of a battery to store regenerative braking energy. This stored energy is used to initially propel (or assist the vehicle’s internal combustion engine) for powering and maintaining motion of the vehicle.[1]

Application of a flywheel

Flywheels can be used to store energy and used to produce very high electric power pulses for experiments, where drawing the power from the public electric network would produce unacceptable spikes. A small motor can accelerate the flywheel between the pulses [2].The phenomenon of precession has to be considered when using flywheels in moving vehicles. However in one modern application, a momentum wheel is a type of flywheel useful in satellite pointing operations, in which the flywheels are used to point the satellite's instruments in the correct directions without the use of thrusters rockets [2].Flywheels are used in punching machines and riveting machines. For internal combustion engine applications, the flywheel is a heavy wheel mounted on the crankshaft. The main function of a flywheel is to maintain a near constant angular velocity of the crankshaft [2].Theory | |

(Figure 1)

The figure above shows the setup of the flywheel. Referring to the figure 1, such an arrangement would not remain in static equilibrium because a downward force mg acts on the string. As the mass moves down, the thread unwind itself and loses its potential energy which is partly used in imparting the kinetic energy to the mass itself and partly in imparting the rotational kinetic energy to the flywheel. In addition to this, some energy would also be lost against the frictional forces. Ultimately the mass and the threads detach from the flywheel. Let the linear velocity of the mass and the angular velocity of the flywheel at the instant of the detachment of the string be respectively v and ω [3].

Then, v=rω (1)
Where r is the radius of the axle of the flywheel. As the string leaves the axle, let the mass fall through a distance. The loss of potential energy of the mass is equal to mgh where as the gain in its kinetic energy is equal to 12mv2 and the gain in rotational energy of the flywheel is equal to 12Iω2, where I is the moment of inertia of the flywheel about its axle. Hence applying the law of conservation of energy, we get the following: mgh=12mv2+12Iω2+e.n (2)
Where e represents the energy dissipated per unit revolution against the frictional forces and n is the number of revolutions for the mass to fall.

Where;

I | polar moment of inertia of fly wheel | M | attached mass | H | height mass falls | Ω | angular velocity of fly wheel | v | velocity of mass | n | number of revolutions for mass to fall | t | time for mass to fall | g | acceleration due to gravity | e | energy lost per revolution | N | total number of revolutions |

From the constant acceleration equation ( s=ut+12at²)
The acceleration of the mass can be determined by: a=2ht2
Substituting in( v=u+at ) , v=2ht (3)
Also ω=vr =v(d2) =4htd (4)
The energy lost per revolution when the mass is released and the flywheel decelerates to rest through (N-n) revolutions is 12ω²
Therefore energy lost per revolution is, e= 12ω²[ 1N-n ] (5)

Substituting equation (3), (4) and (5) into (2) gives us the following equation, mgh = N2N-nI(4htd)2+12m(2ht)2 (6)
As a result, I can be calculated from the above equation.

Apparatus

Flywheel
Flywheel

Cord
Cord

Metre Scale
Metre Scale

Masses
Masses

Stop Watch
Stop Watch

Procedure

1) Measure and record the flywheel dimensions shown in Fig 2.

2) Attach the cord to the peg on the flywheel axle. Attach a mass to the free end of the cord and note that the cord falls off the axle peg as the mass reaches the floor.

3) Life the accelerating mass by rotating the flywheel through a convenient whole number of revolutions. Note the number of revolutions n the height h of the mass above the floor.

4) Release the accelerating mass from the rest and allow it to fall freely noting the time t for the mass to reach the floor. Count the total number of revolutions of the flywheel. N from the moment of release until it stops rotating after the mass has fallen off.

5) Repeat the procedure three times and take the average of t and N.

6) Repeat steps 2 to 5 for three different masses taking three sets of readings for each mass.

7) Calculate the value of I for three set of experimental values.

Precautions * The length of the string is less than the height of the axle of the flywheel from the floor. * The loop slipped over the pin should be loose enough to be removed easily. * The first few turns of the string should overlap the others. * The stopwatch should be started just when the string is detached * The movement of the load or effort during the experiment should be even throughout the motion.

Results

Determination of I by experiment
The equation below shows how the Moment of inertia of flywheel and axle can be calculated: mgh=N2N-n I (4htd)2 +12m (2ht)2

n is 8.5 turns and other variables (d, h ,n, N, t) are measured in the experiment.

Experimental Data Table: Mass/kg | 2.5kg | 3.5kg | 4.5kg | Test 1 | N = 31revst = 14.41s | N =45.5revst =11.10s | N = 60revst =9.41s | Test 2 | N = 31revst =15.12s | N = 45.5revst =11.06s | N = 61revst =9.41s | Test 3 | N = 31revst =14.47s | N = 45.5revst =10.97s | N = 60.5revst =9.41s | Average results | Nav=31revsTav=14.67s | Nav=45.5revsTav=11.04s | Nav=60.5revsTav=9.41s | Iexp= | Iexp = 0.62Kgm2 | Iexp=0.55Kgm2 | Iexp=0.53Kgm2 |

The equation below shows how I is subjected and worked out when the masses are at 2.5kg, 3.5kg and 4.5kg: I=td4h2 2(N-n)(mgh-12m2ht2N Mass at 2.5kg:

I=14.67x0.0364x122(31-8.5)(2.5x9.81x1-12x2.52 x 114.67231

= 0.62 Kgm2

Mass at 3.5kg:
I=11.04x0.0364x122(45.5-8.5)(3.5x9.81x1-12x3.52 x 111.04245.5

= 0.55037 Kgm2

Mass at 4.5kg:
I=9.31x0.0364x122(60.5-8.5)(4.5x9.81x1-12x4.52 x 19.31260.5

= 0.5315218 Kgm2

Theoretical determination of I:
Referring to the lab sheet, density of iron is 7800 kg/m³. The dimensions of the flywheel are provided below:

D | 0.38200m | D | 0.036000m | d1 | 0.24400m | d2 | 0.091000m | L1 | 0.39000m | L2 | 0.17.600m | B | 0.049000m | B | 0.17000m | b1 | 0.01600m | b2 | 0.17500m |

1) Calculate the axle moment of inertia:
Iaxle=πρ(L1+L2)d432
Iaxle=π×7800(0.039+0.176)×0.036432=2.7653×10-4kgm2

Iaxle=πρL1+L2d⁴32

2) Calculate the flywheel moment of inertia with two annuli:
IF+A=πρBD432
IF+A=π×7800×0.049×0.382432=0.798995kgm2

3) Calculate the annuli moments of inertia:

IA=πρb32(d14-d24)
IA1=π×7800×0.016320.2844-0.00914=0.07887kgm2
IA2=π×7800×0.0175320.2844-0.0914=0.08626kgm2

4) Calculate the flywheel moment of inertia:

I=Iaxle+IF+A-IA1-IA2
2.7653×10-4+0.798995-0.0796348-0.0871005=0.6325kgm2

The percentage error between calculated theoretical value and experimental value is worked out using the formulae below:

% Error= Experimental-Theoritical Theoritical ×100 % Error =0.5673- 0.63410.6341×100 % Error =-10.31%

Discussion
Judging from the results, it can be gathered that the mass is directly proportional to the number of revolutions ad inversely proportional to the time. Therefore, meaning that if the mass is heavier, the higher the speed of the mass falling down will be and more the number of revolutions. The percentage difference between the experimental and theoretical value of moment of inertia (I) is 10.31%. These differences in the value of I could be because of too many errors such as stated below:
Equipment errors • The stop watch is not calibrated. • The masses and hangers might be aged and worn out, as results this creates inaccuracy of the masses. • The meter rule not perfectly straight.

Human errors • When reading the meter rule. • The reaction time is not accurate when using the stopwatch. • The number of revolution n that the flywheel performed cannot be accurately obtained. To concur this problem, maybe more than one student could count the number of revolution. • Student must keep hands steady and release the mass slowly and softly when releasing the mass because force may be pushed to mass.

Modelling errors * The model of the flywheel is not exactly the same as the diagram in the lab sheet. With the flywheel used in the experiment, the shape of the axle is more complicated and there is no pin to which the string is hooked. This is a big error in this experiment. * The friction is not constant throughout the experiment as it changes because of the ball bearings warming up.

Conclusion

• The moment of inertia was concluded therefore the objective of this experiment was successfully achieved.
• The moment of inertia of flywheel I is 0.6325kgm2 theoretically and the average experimental value is 0.5673kgm². • Despite of many advantages of flywheel it has some limitations for example, if it spins too fast for its materials, it will break from the centrifugal force it creates. • There were some differenced in the results which can be solved by taking precautions as stated before or could possible improvement to the apparatus.

References
[1] “what is a flywheel” by Christine and scot gable. [online] available at: http://alternativefuels.about.com/od/glossary/g/flywheel.htm Date accessed: 15/11/2013 [2] “moment of inertia of flywheel” “Amrita” [online] available at: http://amrita.vlab.co.in/?sub=1&brch=74&sim=571&cnt=1 Date accessed: 16/11/2013
[3] Practical Physics for Engineers by Ashok Kr. Mishra
Date accessed 17/11/2013

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