Question

A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maximum isolator stiffness of an undamped isolator that can be used to reduce th etransmitted force to 300 N at all its operating speeds

Answer 1

maximum isolator stiffness k =1764 kN-m

Step-by-step explanation:

mean speed of rotation
`=(N_1 +N_2)/(2)`

`Nm = (500+750)/(2) = 625 rpm`

`w =(2\pi Nm)/(60)`

=65.44 rad/sec

`F_T = mw^2 e`

`F_T = mew^2`

= 0.1*(65.44)^2

F_T =428.36 N

Transmission ratio
`=(300)/(428.36) = 0.7`

also

transmission ratio
`= (1)/([(w)/(w_n)]^(2) -1)`

`0.7 =(1)/([(65.44)/(w_n)]^2 -1)`

SOLVING FOR Wn

Wn = 42 rad/sec

`Wn = \sqrt {(k)/(m)`

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

k =1764 kN-m