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Determine the steady state amp. Of pump at harmonic excitation of maximum value of 1.5KN.
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A reciprocating pump of mass 350 kg is driven through a belt by a motor with eccentric rotor at 2500 rpm. The system is mounted on four springs, Each of stiffness 1.2MN/m damping provided is 3KNs.m. Determine the steady state amp. Of pump at harmonic excitation of maximum value of 1.5KN. Also Find the maximum response when pump is switched on and passes through its resonance condition. $\rightarrow$ r = 1.

m = 350 kg

N = 2500 rpm.

$w = \frac{2 \pi N}{60} = \frac{2 \pi \times 2500}{60} = 261.79$ r/sec.

$K_{each}$ = $1.2 \times 10^6$ N/m

N = number of springs = 4

$C = 3 \times 10^3$ N-s/m

$- \frac{X}{\frac{mo.e}{m}} = \frac{r^2}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$

Assume, springs are parallel

$Keq = 4k$

$Wn = \sqrt{ \frac{4k}{m}}$

$= \sqrt{\frac{4 \times 1.2 \times 10^6}{350}} = 117.10$ rad/sec.

$r = \frac{w}{w_n} = \frac{261.79}{117.10}$

R = 2.235.

Unbalance = mo.e = $\frac{fo}{w^2} = \frac{1.5 \times 10^3}{(261.79)^2}$

Moe = 0.022

$\xi = \frac{c}{2m w_n}$

$= \frac{3 \times 10^3}{2 \times 350 \times 117.10}$

$\xi = 0.0365$

$\frac{X}{\frac{0.022}{350}} = \frac{(2.235)^2}{\sqrt{ (1-(2.235)^2)^2 + (2 \times 0.036 \times 2.35)^2}}$

$X = 7.85 \times 10^5 \ m$

Case 2: when r = 1

$\frac{X \ max}{\frac{mo.e}{m}} = \frac{r^2}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$

$\frac{X \ max}{\frac{0.022}{0.350}} = \frac{1}{\sqrt{(1-1^2)^2 + (2 \times 0.0365)^2}}$

$X \ max = 8.6 \times 10^{-4} \ m$