Question

The damping factor for a shaft-turbine system is measured to be \(0.15\). When the turbine rotates at a rate equal to \(120 \%\) of the undamped natural frequency of the shaft, the system is observed to whirl with an amplitude equal to the radius of the shaft. What will be the amplitude of whirling when the rotation rate of the turbine is reduced to \(80 \%\) of the undamped natural frequency of the shaft if the radius of the shaft is 1 inch?

Short answer

Question: Determine the amplitude of whirling when the rotation rate of a turbine is reduced to 80% of its undamped natural frequency. Given a damping factor of 0.15, a whirling amplitude equal to the radius of the shaft (1 inch) at 120% of the undamped natural frequency, and a shaft radius of 1 inch. Answer: The amplitude of whirling at 80% of the undamped natural frequency is approximately 1.22 inches.

Step-by-step solution

Known Information

We know the following information from the problem statement: - Damping factor, \(\zeta = 0.15\). - At \(120\%\) of the undamped natural frequency (\(\omega_{n1} = 1.2\omega_n\)), the amplitude of whirling \(A_1\) is equal to the radius of the shaft, which is \(1\) inch.

Objective

Find the amplitude of whirling \(A_2\) when the rotation rate of the turbine is reduced to \(80 \%\) of the undamped natural frequency (\(\omega_{n2} = 0.8\omega_n\)).

Whirling Amplitude Ratio

We will use the formula for the amplitude ratio of a damped system when the rotation rate (\(\omega_r\)) is a percentage of the undamped natural frequency (\(\omega_n\)). The formula is: $$Amplitude Ratio = \frac{A_2}{A_1} = \frac{(\frac{1}{1-\frac{\omega_{n2}^2}{\omega_n^2}})}{(\frac{1}{1-\frac{\omega_{n1}^2}{\omega_n^2}})}$$

Solve for A_2

Using the given values, we can now solve for \(A_2\): Substitute \(\omega_{n1}=1.2\omega_n\) and \(\omega_{n2}=0.8\omega_n\) in the amplitude ratio equation: $$\frac{A_2}{1} = \frac{(\frac{1}{1-\frac{(0.8\omega_n)^2}{\omega_n^2}})}{(\frac{1}{1-\frac{(1.2\omega_n)^2}{\omega_n^2}})}$$ Simplify the equation and solve for \(A_2\): $$A_2 = \frac{1}{1-0.64}*\frac{1-1.44}{1} = \frac{1}{0.36} * -0.44$$ Hence, the amplitude of whirling at \(80\%\) of the undamped natural frequency is: $$A_2 = \frac{-0.44}{0.36} \approx -1.22 \, inches$$ However, the amplitude cannot be negative; the negative sign indicates the phase shift in whirling, which doesn't affect the amplitude value. So, the actual amplitude of whirling is: $$A_2 \approx 1.22 \, inches$$