Mass
|
Radius
|
M*r
|
1
|
120
|
120
|
0.5
|
100
|
50
|
0.7
|
80
|
56
|
Md
|
60
|
60 Md
|
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
|
![](file:///C:/Users/DELL/AppData/Local/Temp/msohtmlclip1/01/clip_image004.jpg)
![](file:///C:/Users/DELL/AppData/Local/Temp/msohtmlclip1/01/clip_image006.jpg)
![](file:///C:/Users/DELL/AppData/Local/Temp/msohtmlclip1/01/clip_image008.png)
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.