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### Rotor Balancing

ABSTRACT
Unbalanced mass on rotary system cause unnecessary vibrations in the system and sometimes the vibrations are too large that they can damage the system. An unbalanced rotational system typically results in vibrations on the entire system and hence affects its normal working. Hence, while imbalances are insignificant for applications where rotational velocities are low, the thus induced vibrations in fast rotating systems can be catastrophic. In this experimental research work the rotor balancing apparatus is used to balance the rotor system by adding additional masses of unknown magnitude. The mass and position of unknown masses would be calculated using theoretical balancing principles.
Introduction
In engineering designs of moving part machines the most common engineering problem is the existence of vibrations in the machine. Vibrations in any machine such as steam turbine or multi cylinder internal combustion engine are very common problem and they must be removed or at least minimized. The source of any vibration in the system is misalignment or unbalanced systems. Unbalance is the most common source of vibration in rotary machines . The unbalance in the rotary machine is not necessarily because of the addition of extra mass on the rotating part but it can come from the uneven mass distribution in any rotary part. Therefore it is necessary to balance out this uneven mass so that the resultant vibrations could be minimized . For balancing unbalance masses on the rotary system more masses can be added at precise locations (angles) and with correct magnitude. In this lab following the theory given in the lab manual we will see how to determine the position and magnitude of added masses to get balancing.
For a balanced system the sum of moments of all these mass with respect to mass 1 should be zero.  By finding moments of each balance, drawing vector diagrams and using above equations, the angular position and the horizontal position of the balances can be determined to achieve dynamic balance.

Simple Rotor Balancing
As per instructions about the experiment there are three different masses attached to disk. The masses are 1 Kg, 0.5 Kg and 0.7 Kg in weight and their respective radiuses are 120 mm 100 mm and 80 mm from the disk center axis. This system is imbalance and in order to balance this system one more mass must be added at the distance of 60 mm. The unknown mass is called Md. Now the task is to calculate the required masses and their respective angle from horizontal plane in order to make this system balance statically and dynamically.

 Mass Radius M*r 1 120 120 0.5 100 50 0.7 80 56 Md 60 60 Md

60 * Md = 143

Md = 2.4 Kg

Mass of 2.4 Kg is required to balance the system at an angle of 208 degrees counter clock wise from point A

As per instructions about the experiment there are two different masses attached to an arm of a shaft. The masses are 5 Kg and 3 Kg in weight and their respective radiuses are 20 mm and 30 mm from the shaft center axis. This system is imbalance and in order to balance this system two more masses must be added at the distance of 50 mm. The two unknown mass as are called Ma and Md. The respective distances masses from each other is resented as the mass A comes first in the line and after that mass of 5 Kg is present at the distance of 80 mm from the mass A. After that mass of 3 kg is present at the distance of 150 mm from mass A and at last the mass D is present at the distance of 250 mm from mass A. Now the task is to calculate the required masses and their respective angle from horizontal plane in order to make this system balance statically and dynamically.
 M mass kg R distance mm M*r X distance from Mass one Ma M*r*x Ma 50 50Ma 0 0 5 20 100 80 8000 3 50 150 150 22500 Md 50 50Md 250 12500Md

From the polygon for M*r*x we have
12500Md = 25000
Md = 2 Kg
Mr of point D = 50*2 = 100 at an angle of 58 degree From the polygon for M*r we have
50 Ma = 105
Ma = 2.1 Kg at an angle of 35 degree With the help of graphical representation and theoretical calculation it can be said that first mass A has to be of 2 kg and should be placed at the angle of 58 degrees from the horizontal axis. The second unknown mass D should be of 2.1 kg and should be placed at the angle of 35 degrees from the horizontal axis to make system balance statically and dynamically.

Rotor balancing experiment kit was used in this experiment performance. In this set up, the circular mass blocks were used. Four masses were used and all of them were of different masses. Additionally, the pulley extension, weight hanger and weights were also used. The weight hanger itself weighs 10 g. Weights of 10 g and 1 g would be 0.0981 N and 0.00981 N respectively. Figure 2: Measuring moment apparatus set up

Experimental Procedure:
1.      By exerting the extension pulley the weight was hanged from it
2.      Block 1 was tightened to the shaft at 0 o angle
3.      Weights were hanged from the pulley until the block reached equilibrium at 90 degrees
4.      1 g was added to the equilibrium mass to result in the free fall
5.      The same method was repeated for all other masses and the results are given in table 1
Calculations and Discussions
For block 1
Mass; m = 94 g = 0.094 kg
Weight;

W = mg = 0.094* 9.81 = 0.922 N

Moment for Weight of block 1;

M1 = W* r = 0.922* 0.040 = 0.037 Nm

For block 2

Mass; m = 89 g = 0.089 kg

Weight;

W = mg = 0.089* 9.81 = 0.873 N

Moment for Weight of block 1;

M1 = W* r = 0.873* 0.040 = 0.0349 Nm

For block 3

Mass; m = 80 g = 0.080 kg

Weight;

W = mg = 0.080* 9.81 = 0.7848 N

Moment for Weight of block 1;

M1 = W* r = 0.7848 * 0.040 = 0.0314 Nm

Block 4

Mass; m = 72 g = 0.072 kg

Weight;

W = mg = 0.072* 9.81 = 0.696 N

Moment for Weight of block 1;

M1 = W* r = 0.696* 0.040 = 0.0278 Nm

For block 1 and lock 2 initial angles and positions are given as shown in table 2 in the following

From table 1, W1r1 can be drawn on y axis in negative direction (downwards) as shown in figure 3

(a). Then W2r2 can be drawn at an angle of 110 degrees anticlockwise. The magnitude of these two vectors is calculated by taking suitable scale. The scale is simple multiplier factor of 100. That is 0.037 will become 37 mm on the scale and 0.0347 will be 34.7 mm. the figure 3 has be drawn on scale. The angles are measured using protector for each vector.

Figure 3: Vector diagram

Now the vectors from figure 3 (b) can be decomposed into component vectors along horizontal and vertical axis and the then following the theoretical principles the following equations can be written as

For horizontal components

-L2*FC2* cos10 + L3*FC3*cos 80 + L4*FC4*cos2 = 0

For vertical components

-L2*FC2*sin10 + L3*FC3* sin 80 - L4*FC4*sin2 = 0

Putting in values

-0.12*(0.0349)* cos10 + L3*(0.0314)*cos 80 + L4*(0.0283)*cos2 = 0

-0.00412 + 0.00545 L3 + 0.02828*L4 = 0

Now using values in vertical components

-0.12*(0.0349)*sin10 + L3*(0.0314)*sin 80 - L4* (0.0283)*sin2 = 0

-0.000727+ 0.0309 L3 – 0.000987* L4 = 0

Solving equation (1) and (2) simultaneously

L3 = 0.028m = 28 mm

L4 = 0.1402 m = 140.2 mm

Hence the final results are shown in table 3.

Discussion

This experiment was very helpful to understand the working of rotor balancing experiment apparatus. Initially it looked quite complicated to use the apparatus but finally it was quite user-friendly and easy to get high accuracy in results. Before doing the advanced rotor balancing tow mass, three mass systems were balanced using the same apparatus. The rotor was only run for longer period when it was fully balanced because if it is run on unbalanced condition for longer period it can cause unnecessary wear and tear in the apparatus and can reduce its precision and accuracy. On the apparatus a protractor on the pulley/rotor helps to determine the angular position of the balance blocks. The vector diagrams show that the system was exactly balanced because the vectors exactly match with each other without showing any error in the readings.