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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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written 4.9 years ago by |
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
$r = \frac{w}{w_n}$
$W_n = 21.37 \ r/sec$
$w_n = \sqrt{ \frac{9}{\delta_{st}}}$
$21.37 = \sqrt{ \frac{9}{\delta_{st}}}$
$\delta_{st} = 0.021 \ m$