Question 17.16: A 60 kg compressor rotor is mounted on a shaft of stiffness ...

A 60 kg compressor rotor is mounted on a shaft of stiffness 15 MN/m. Determine the critical speed of the rotor assuming the bearings to be rigid. If the rotor has an eccentricity of 2 mm and its operating speed is 6500 rpm, determine the unbalance response. The damping factor in the system can be taken as 0.06. If the compressor is started from rest, what will be the maximum whirl amplitude of the rotor before it reaches its full operational speed?

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\text { Given: } m=60 kg , k=15 MN / m , e=2 mm , N=6500 rpm , \zeta=0.06 .

\omega_{n}=\left(\frac{k}{m}\right)^{\frac{1}{2}} .

=\left(\frac{15 \times 10^{6}}{60}\right)^{\frac{1}{2}}=500 rad / s .

\omega=\frac{2 \pi N}{60} .

=\frac{2 \pi \times 6500}{60}=680.678 rad / s .

\beta=\frac{\omega}{\omega_{n}}=\frac{680.678}{500}=1.36 .

Whirl amplitude,        r=\frac{e \beta^{2}}{\left[\left(1-\beta^{2}\right)^{2}+(2 \zeta \beta)^{2}\right]^{1 / 2}} .

=\frac{2 \times(1.36)^{2}}{\left.\left[\{1.36)^{2}\right\}^{2}+(2 \times 0.06 \times 1.36)^{2}\right]^{1 / 2}}=4.276 mm .

\text { Maximum whirl amplitude, } r_{\max }=\frac{e}{2 \zeta}=\frac{2}{2 \times 0.06}=16.67 mm
17.34

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