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

Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation $=\frac{N_1 +N_2}{2}$

$Nm = \frac{500+750}{2} = 625 rpm$

$w =\frac{2\pi Nm}{60}$

  =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 $=\frac{300}{428.36} = 0.7$

also

transmission ratio $= \frac{1}{[\frac{w}{w_n}]^{2} -1}$

$0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}$

SOLVING FOR Wn

Wn = 42 rad/sec

$Wn = \sqrt {\frac{k}{m}$

k = m*W^2_n

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

k =1764 kN-m