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An air conditioner weighs 1115N and is driven by a motor at 500 rpm. What is the required static deflection of an undamped isolator to achieve 80% isolation?
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m = 1115N

= 113.65 kg

N = 500 rpm

$w = \frac{2 \pi \times 500}{60} = 52.35$ r/sec

$\eta_{iso} = 80%$

$W_n = \sqrt{ \frac{9}{\delta st}}$

$\eta_{iso} = 1 - T_r$

0.80 = 1 – $T_r$

$T_r$ = 0.2

$T_r = \frac{\sqrt{1 + (2 \xi r)^2}}{\sqrt{ (1- r^2)^2 + (2 \xi r)^2}}$

$0.2 = \frac{1}{1 – r^2}$

$0.2(1 – r^2) = 1$

R = 2.449

$r = \frac{w}{w_n}$

$W_n = 21.37 \ r/sec$

$w_n = \sqrt{ \frac{9}{\delta_{st}}}$

$21.37 = \sqrt{ \frac{9}{\delta_{st}}}$

$\delta_{st} = 0.021 \ m$

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