Experiment 1

EXPERIMENT 1: MOMENT OF INERTIA FOR FLYWHEEL

ABSTRACT

This experiment is an introduction to fundamental concepts and facts of rotational dynamics. A fairly realistic analysis of the motion of a flywheel can be made, assuming only that the net frictional torque on the rotating flywheel is constant. In performing this experiment, we develop understanding of rotational dynamics, evaluation of errors in measurements that may be difficult to obtain and comparing experimental value of moment of inertia with theoretical value of moment inertia.

INTRODUCTION

For ages flywheels have used to achieve smooth operation of machines. The early models consisting of only a stone wheel attached to an axel. Nowadays flywheels are more complex constructions where energy is stored mechanically and transferred to and from the flywheel by an integrated motor/generator. The stone wheel has been replaced by a steel or composite rotor and magnetic bearings have been introduced.

A flywheel stores energy in a rotating mass. Depending on the inertia and the speed of the rotating mass, a given amount of kinetic energy is stored as the form of rotational energy. One of the uses of the flywheel is inside a vacuum containment to eliminate friction-loss from the air and suspended by bearings for a stabile operation. Kinetic energy is transferred in and out of the flywheel with an electrical machine that can function either as a motor or generator depending on the load angle (phase angle). When acting as motor, electric energy supplied to the stator winding is converted to torque and applied to the rotor, causing it to spin faster and gain kinetic energy. In generator mode kinetic energy stored in the rotor applies a torque, which is converted to electric energy. Fig. 1 shows the basic layout of a flywheel energy storage system. Apart from the flywheel, Additional power electronics is required to control the power input and power output, speed, frequency and etc.

Figure1. simple flywheel used in lab

In following part we describe theory of flywheel energy storage and method used to obtain the moment of inertia of the flywheel.

Figure2

Work output from fallen mass is given by the difference between lost in potential and kinetic energy during mass separation from flywheel.

Potential energy = mgh = mg2 rN

Where

N = number of rotation

Terminal velocity of mass, v= N r Kinetic energy, KE = ½ m ( N r)

2

Work produced on flywheel, W = mg2 rN - ½ m ( N r) 2

Flywheel starts from static condition. It will reach a total of N1 rotations before stopping, when it is allowed to rotate after load separates from flywheel. This means that all work was used to overcome the bearing friction which is assumed to be constant.

At the time the load separates from the flywheel, it will reach the maximum angular velocity N and the kinetic energy is given by KE = ½ m ( N r)

2 .

In this experiment we assumed that torsion of bearing friction is Cf . By assuming the work used to overcome the friction is equal to the output work, we will have

mg2 rN - ½ m ( N r) 2

= Cf 2 N (1) And energy equivalent after N rotation is given by,

mg2 rN - ½ m ( N r) 2

= Cf 2 N + ½ I N 2

(2)

Therefore, if N, N1 and N are measured, Cf can be determined from equation (1) and substituted into equation (2) to obtain I.

After the load has detached, its angular velocity decreases on account of friction and after some time t, the flywheel finally comes to rest. At the time of detachment of the load the angular velocity of the flywheel is N and when it comes to rest its angular velocity is zero. Hence, if the force of friction is steady the motion of the flywheel is uniformly retarded and the average angular velocity is equal to N/2. Thus,

N/2 =

So, N =

OBJECTIVE

 To compare the theoretical and experimental value of momentum of inertia for flywheel.  To investigate the variations in momentum of inertia of flywheel with different parts.

RESULTS AND CALCULATIONS

Mass of the load = 5 N Part 1

1 2 3 Average

Time t (s) 8.575 8.819 8.739 8.711

N1 52 53 50 52

Experimental value of moment of inertia;

N = = = 5.77 rad/s Cf (2N1) = mg (2r) N - 1 2 m (N r) 2 = (0.5x9.81)(2)(0.02x4) - (0.5)(0.5)(5.77x0.02)2 = 2.462 Nm Cf = 2.462_ 2 (52) = 0.00754 Nm 1 2 IN 2 = mg (2r) N - 1 2 m (N r) 2 - Cf (2N) = 2.462 - (0.00754 )(2x4) = 2.272 Nm I = 2.272x2 (5.77)2 = 0.136kgm2 Theoretical value of moment of inertia;

The equation of the moment of inertia of flywheel is given by I = 1₂ MR2

It is given that the density of steel is 7850kg/m3 and Volume of the flywheel, V =  l  R Mass of the flywheel, m =  V

So, m = 7850  (0.125)2  (0.030) = 11.56 kg Moment of inertia for flywheel, I1 = (0.5)(11.56)(0.125)

2

= 0.0903 kgm2

For component of the outer ring Moment of inertia for outer ring, I2 =

1 2 MR 2 - 1₂ m r2 Moment of inertia, I2 = (0.5)[7850  (0.125 2 _{ 0.020) ](0.125)}2 - (0.5)[7850  (0.0902  0.020) ](0.090)2 = 0.0440kgm2

For the component of the inner ring Moment of inertia, I3 = (0.5)[7850  (0.089 2 _{ 0.020) ](0.089)}2 - (0.5)[7850  (0.0202  0.020) ](0.020)2 = 0.0154 kgm2

Total moment of inertia; I = I1 + I2 + I3 I = 0.0903 + 0.0440 + 0.0154 = 0.150 kgm

Part 2 1 2 3 Average Time t (s) 7.604 7.446 7.261 7.437 N1 67 66 66 66 N = = 6.76 rad/s Cf (2N2) = mg (2r) N - 1 2 m (N r) 2 = (0.5x9.81)(2)(0.020x4) - (0.5)(0.5)(6.759x0.02)2 = 2.461 Nm Cf = 2.461 2 (66) = 0.00593 Nm 1 2 IN 2 = mg (2r) N - 1₂ m (N r) 2 - Cf (2N) = 2.461 - (0.00593)(2x4) = 2.312 Nm I = 2.312 x 2 (6.76)2 = 0.101 kgm2 Theoretical value of moment of inertia; Total moment of inertia; I = I1 + I2 I = 0.0903 + 0.0154 = 0.106 kg.m2

Part 3 1 2 3 Average Time t (s) 6.527 6.652 6.821 6.667 N1 77 80 77 78 N = = 7.54 rad/s Cf (2N3) = mg (2r) N - 1 2 m (N r) 2 = (0.5x9.81)(2)(0.020x4) - (0.5)(0.5)(7.539x0.020)2 = 2.460 Nm Cf = 2.460__ 2 (78) = 0.00502 Nm 1/2IN 2 = mg (2r) N - 1 2 m (N r) 2 - Cf (2N) = 2.460 - (0.00502)(2x4) = 2.333 Nm I = _2.333x 2_ (7.54)2 = 0.0821 kgm2 Theoretical value of moment of inertia; Total moment of inertia; I = I1

Part 4 1 2 3 Average Time t (s) 8.202 8.214 8.408 8.275 N1 62 64 60 62 N = = 5.59 rad/s Cf (2N4) = mg (2r) N - 1 2 m (N r) 2 = (0.5x9.81)(2)(0.020x4) - (0.5)(0.5)(6.074 x0.020)2 = 2.462 Nm Cf = 2.462 _ 2 (62) = 0.00632 Nm 1 2 IN 2 = mg (2r) N - 1 2 m (N r) 2 - Cf (2N) = 2.462 - (0.00632)(2x4) = 2.303 Nm I = 2.303 x 2 (5.59)2 = 0.148 kgm2 Theoretical value of moment of inertia; Total moment of inertia; I = I1+ I3 I = 0.0903 + 0.0440 = 0.134 kg.m2

DISCUSION

Part Experimental value of

moment of inertia (kg/m3) Theoretical value of moment of inertia (kg/m3) Percentage of discrepancy ( % ) 1 0.136 0.150 9.33 2 0.101 0.106 4.72 3 0.0821 0.0903 9.08 4 0.148 0.134 10.45

From the calculations, we obtained the experimental value of moment of inertia for each part of experiment as follow, o.136 kg.m2, o.101kg.m2, 0.0821kg.m2 and 0.148kg.m2 respectively which show that moment of inertia value is consistent and accurate since there is small deviation between the values calculated. Furthermore, by comparing the experimental value with theoretical value for each part of experiment, we realize that there is a small difference between them which may be caused by some errors such as;

1. Time t taken for the load till getting detached is not enough accurate because of the reaction time of human or human error.

2. Difficulty in counting the number of rotation after load detached. So it’s better to use two people to count the number of rotation in order to obtain more accurate and reliable values.

3. We made a non-ideal assumption in this experiment, that all the work done to overcome the bearing friction of flywheel is completely converted to the output work. This is because based on thermodynamics studies energy is lost in the form of heat throughout the experiment. And friction cause by air resistance that resists the motion of the falling mass is not included in the calculation. Although air resistance only causes minor discrepancy, it cannot be neglected.

4. According to the equation , angular velocity of the flywheel is assumed to be constant throughout the movement. However in the actual case, it is not true as the flywheel is starting from rest and rotates with an angular acceleration.

CONCLUSION

In conclusion, moment of inertia of the flywheel is approximately consistent with small deviations in short range for both experimental and theoretical results.

Part 1; experimental I= 0.136 kg.m2 theoretical I=0.150 kg.m2 Part 2; experimental I=0.101 kg.m2 theoretical I=0.106 kg.m2 Part 3; experimental I=0.0821 kg.m2 theoretical I=0.0903 kg.m2 Part 4; experimental I=0.148 kg.m2 theoretical I=0.134 kg.m2

Sine the percentage of discrepancy of each part from 1 to 4, 9.33%, 4.72%, 9.08% and 10.45% respectively, all are less than 15%, experimental values of moment of inertia of flywheel are considered accurate and reliable.

REFERENCES

1. D. Halliday, R. Resnick, and J. walker. Fundamentals of physics. 6

edition. Wiley,

2003.

2. Ferd Beer and Russ Johnston (2005)Vector Mechanics For Engineers: Statics, New Jersey, McGraw-Hill

3. Gabrys CW. High performance composite flywheel, US patent pub. NO : US

2001/0054856 A1; 27 Dec 2001.