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Consider fully developed laminar flow in a circular tube. Evaluate the kinetic energy coefficient for this flow.

Short Answer

Expert verified
The kinetic energy coefficient for fully developed laminar flow in a circular tube is 2.

Step by step solution

01

Identify the flow profile

For fully developed laminar flow in a circular tube, the velocity profile is parabolic, also known as Hagen-Poiseuille flow. The profile is defined as \(u(r) = \frac{1}{4\mu}\left(P_{1} - P_{2}\right)\left(R^{2}-r^{2}\right)\). Where \(u(r)\) is the fluid velocity at a distance \(r\) from the center of the tube, \(\mu\) is the dynamic viscosity of the fluid, \(P_{1} - P_{2}\) represents the pressure drop across the length of the tube, \(R\) is the tube radius and \(r\) is the radial distance from the tube center.
02

Derive the kinetic energy coefficient

The kinetic energy coefficient \(\alpha\) can be obtained by the ratio between the actual kinetic energy and the kinetic energy if the flow were plug flow. Start by calculating the mass flow rate, \(\dot{m} = \rho \int_{0}^{R} u(r) 2\pi r dr\), where \(r\) represents the distance from the radius and \(\rho\) is the fluid density. Subsequently, calculate the kinetic energy as \(KE = \frac{1}{2}\rho \int_{0}^{R} u(r)^{2} 2\pi r dr\), using the velocity profile formulated in step 1. By combining these equations, we can obtain \(\alpha = \frac{2 \times KE}{\dot{m} \times \bar{u}}\), where \(\bar{u}\) is the average velocity derivative of the mass flow rate. The kinetic energy coefficient \(\alpha\) for laminar flow in a circular tube can then be evaluated as \(\alpha = 2\)
03

Interpreting the result

For laminar flow, the kinetic energy coefficient \(\alpha\) is equal to 2, meaning that kinetic energy for fully developed laminar flow in a circular tube has double the kinetic energy of the averaged velocity flow. This is due to the parabolic velocity profile, where the velocity (and thus kinetic energy) is higher towards the center of the tube.

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