Page 42 - Rotating Machinery Pratical Solutions to Unbalance and Misalignment

Rotating Machinery: Practical Solutions

D = [G/.014190252] * [1000/F] 2 (2.25)

D = [70.47090965 * G * [1000/F] 2 (2.26)
Now using the above relations to build the relation of accel-
eration to velocity:

V = X ω (2.27)
o
And;

A = X ω 2 (2.28)
o
And;
G = A/386.4 (2.29)
Thus:
G = X ω2/386.4 (2.30)
o
G = X ω* ω/386.4 (2.31)
o

And substituting from Equation (2.27):
G = V ω/386.4 (2.32)

G = V*(2πf)/386.4 (2.30)
And:
G = 2πFV/(60 * 386.4) (2.31)
Rearranging:
G= FVπ/11,592 (2.32)
Or:

G = FV/3,689.848201 (2.33)
V = 3,689.848201 * G/F (2.34)

The use of these equations will transform the three forms of
vibration characteristics to one another.

Example 2.2
A shaft has a peak-to-peak displacement reading of 3.4 mils
@ 36.5 Hz. What is the velocity and what is the acceleration?