91Ó°ÊÓ

An incompressible, inviscid fluid flows into a horizontal round tube through its porous wall. The tube is closed at the left end and the flow discharges from the tube to the atmosphere at the right end. For simplicity, consider the \(x\) component of velocity in the tube uniform across any cross section. The density of the fluid is \(\rho,\) the tube diameter and length are \(D\) and \(L,\) respectively, and the uniform inflow velocity is \(v_{0}\) The flow is steady. Obtain an algebraic expression for the \(x\) component of acceleration of a fluid particle located at position \(x,\) in terms of \(v_{0}, x,\) and \(D .\) Find an expression for the pressure gradient, \(\partial p / \partial x,\) at position \(x .\) Integrate to obtain an expression for the gage pressure at \(x=0\)

Step by step solution

01

Write the equation for acceleration

Acceleration of a fluid particle located at position \(x\) is given by \(a_{x} = v \frac{dv}{dx} \) where \(v\) is the velocity of the particle.

02

Derive an expression for velocity

Given, the fluid is going into the tube through its porous wall with a uniform inflow speed \(v_{0}\) and the flow is going out at the right end. We can suppose that the fluid enters over a length \(L\) and at any section \(x\) from the closed left end, the velocity \(v\) is constant. The conservation of mass in the fluid suggests \(v_{0}\pi(0.5D)^{2}= v\pi(0.5D - v_{0}x)^{2}\). Solving for \(v\), we obtain \(v=\frac{v_{0}D}{D - 2v_{0}x}\)

03

Substitute the velocity expression into the acceleration equation

Substitute the velocity expressions derived in Step 2 into the acceleration equation obtained in Step 1 to get the acceleration \(a_{x} = \frac{2v_{0}^{3}Dx}{(D - 2v_{0}x)^{3}}\). This is the algebraic expression for the \(x\) component of acceleration of a fluid particle located at position \(x\)

04

Write the expression for pressure gradient

The pressure gradient \(\partial p / \partial x\) can be obtained using Euler’s equation for one-dimensional steady flow as \(\partial p / \partial x= -\rho a_{x}\)

05

Substitute acceleration into the pressure gradient equation.

By substituting the expression for \(a_{x}\) obtained in Step 3 into the equation derived in Step 4, we obtain \(\partial p / \partial x= -2\rho v_{0}^{3}Dx / (D - 2v_{0}x)^{3}\)

06

Integrate to find the pressure at \(x=0\)

Finally, we can integrate the expression obtained in Step 5 from 0 to \(x\) to find the gage pressure at \(x=0\), which gives \(p = \frac{\rho v_{0}^{3}}{2v_{0}}(1 - \frac{D}{D-2v_{0}x})\). Thus, the gage pressure at \(x=0\) is given by \(p_{0}= \frac{\rho v_{0}^{3}}{2}\)

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