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 [1]. 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 [2]. 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.
Advance Rotor balancing
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 userfriendly 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.
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