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Chapter 4: Problem 200

A pipe branches symmetrically into two legs of length \(L,\) and the whole system rotates with angular speed \(\omega\) around its axis of symmetry. Each branch is inclined at angle \(\alpha\) to the axis of rotation. Liquid enters the pipe steadily, with zero angular momentum, at volume flow rate \(Q\). The pipe diameter, \(D,\) is much smaller than \(L\) Obtain an expression for the external torque required to turn the pipe. What additional torque would be required to impart angular acceleration \(\dot{\omega} ?\)

Short Answer

Expert verified
The expression for external torque required to turn the pipe is obtained by considering the change in angular momentum of the liquid as it moves in the pipe. Additional torque required to impart angular acceleration involves the moment of inertia of the fluid.

Step by step solution

01

Compute the Fluid Velocity

First, we need to determine the velocity of the fluid. We know that the fluid enters the pipe with a volume flow rate \(Q\). We can convert this into a velocity(\(v\)) by dividing it over the area of the pipe, i.e., \(v = \frac{Q}{A}\). The area(\(A\)) of the pipe can be calculated by \(\pi (\frac{D}{2})^2\). Therefore, \(v = \frac{Q}{\pi (\frac{D}{2})^2}\).
02

Calculate the Angular Momentum Change After Fluid Enters the Pipe

The fluid enters the pipe with zero angular momentum, but gains angular momentum as it exits the pipe at the end of the branches. The final angular momentum of the fluid is \(mvr\sin{\alpha}\), where \(m\) is the mass flow rate(\(Q/\text{Fluid Density}\)), \(v\) is the velocity of the fluid, \(r\) is the radius of curvature (equal to \(L\sin{\alpha}\)), and \(\alpha\) is the angle. The pipe branches symmetrically, so the total angular momentum change, \(\Delta L\), can be written as \(2 \cdot mvr\sin{\alpha}\).
03

Compute the External Torque

The external torque needed to keep the system rotating at angular speed \(\omega\) is equal to the rate of change of angular momentum, which we can write as \(\tau = \frac{dL}{dt}\). By replacing \(\Delta L\) from step 2 we get, \(\tau = 2 \cdot \frac{Q}{\text{Fluid Density}} \cdot v \cdot L \sin^2 \alpha\). Substituting the velocity \(v = \frac{Q}{\pi (\frac{D}{2})^2}\) and simplifying further will provide the final expression for the torque.
04

Compute the Additional Torque

The additional torque to impart angular acceleration \(\dot{\omega}\) would be equal to the moment of inertia of the fluid in the pipe times \(\dot{\omega}\). This step requires the consideration of the total mass of the liquid in the rotating pipe and the moment of inertia.

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Most popular questions from this chapter

Consider the steady adiabatic flow of air through a long straight pipe with \(0.05 \mathrm{m}^{2}\) cross-sectional area. At the inlet, the air is at \(200 \mathrm{kPa}\) (gage), \(60^{\circ} \mathrm{C}\), and has a velocity of \(150 \mathrm{m} / \mathrm{s}\). At the exit, the air is at \(80 \mathrm{kPa}\) and has a velocity of \(300 \mathrm{m} / \mathrm{s}\). Calculate the axial force of the air on the pipe. (Be sure to make the direction clear.)

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