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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 $3 \times 10^5 N/m$

written 5.5 years ago by teamques10 ★ 68k

modified 2.8 years ago by pedsangini276 • 4.8k

applied hydraulics

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written 5.5 years ago by teamques10 ★ 68k

$wn = \sqrt{ \frac{4k}{m}} = \sqrt{ \frac{4 \times 3 \times 10^5}{100}} = 109.54$ 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%

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