0
1.4kviews
A 100 kg turbine operates at 2000 rpm, what percent isolation is achieved if the turbine is mounted on four identical springs in parallel each of stiffness $3 \times 10^5 N/m$
1 Answer
0
100views

$wn = \sqrt{ \frac{4k}{m}} = \sqrt{ \frac{4 \times 3 \times 10^5}{100}} = 109.54$ rad/sec. $w = \frac{ 2 \pi N}{60} = \frac{2 \pi \times 2000}{60} = 209.44$ rad/sec $r = \frac{209.44}{109.54} = 1.91$ $Tr = \frac{1}{1-r^2} - \frac{1}{1 – (1.91)^2} = 0.377$ $\eta iso = 1 – Tr.o$

= 1 – 0.377

= 62.23%

Please log in to add an answer.