Page 43 - Rotating Machinery Pratical Solutions to Unbalance and Misalignment

Developing the Vibration Equations

Using Equation 2.14:
f = 36.5 and F = 36.5 × 60
V = [πDF]/60,000 = [3.14159 × 3.5 × 36.5 × 60]/60000
V = .401 inches per second
Using Equation 2.24
G = .014190252 * D * [F/1000] 2
= .014190252 × 3.5 × [36.5 × 60/1000] 2
G = .2382

Using Equation (2.18)
A = 386.4 * G = 92.04 feet per second per second.
If the shaft weighed 23.6 pounds, it would have a mass of:
M = W/g = 23.6/32.2 = .7329
F = MA = .7329 * 92.04 = 67.46 pounds.

Thus the force generated by this vibration is almost three
times the weight of the shaft. Using the vibration severity table at
the end of Chapter 3, a velocity reading of .401 inches per second
rates as rough, and some action should be taken.

Example 2.3
A centrifugal pump operating at 3,450 rpm shows a filter in
vibration of .114 inches per second at operating speed. What is the
peak-to-peak displacement? If the impeller and shaft weighs 40
pounds, what is the force produced?

Using Equation (2.15)

D = 60000 × .114/(3.14159 × 3450) = .63 mils peak-to-peak
The mass of the rotating element is 40/32.2 = 1.24 pounds mass.
Using Equation (2.33)

G = 3450 × .114/3689.8484201 = .1066
The force generated is:
F = 386.4 × G = 368.4 × .1066 = 41.19 pounds force.