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Chapter 6: Problem 11

An incompressible liquid with a density of \(1250 \mathrm{kg} / \mathrm{m}^{3}\) and negligible viscosity flows steadily through a horizontal pipe of constant diameter. In a porous section of length \(L=\) \(5 \mathrm{m},\) liquid is removed at a constant rate per unit length so that the uniform axial velocity in the pipe is \(u(x)=U(1-x / L)\) where \(U=15 \mathrm{m} / \mathrm{s}\). Develop expressions for and plot the pressure gradient along the centerline. Evaluate the outlet pressure if the pressure at the inlet to the porous section is \(100 \mathrm{kPa}(\text { gage })\)

Short Answer

Expert verified
The pressure gradient along the centerline is described by the equation \(p = -\rho U^2 \frac{x^2}{2L} + 100kPa\), and the pressure at the outlet is \(56.875 kPa\)

Step by step solution

01

Identify Fluid Dynamics Concepts

This problem involves the concept of fluid dynamics in a system in which an incompressible fluid is flowing through a porous pipe. It's crucial to remember that fluid continuity, which is given by the equation \(A_1v_1 = A_2v_2\), must be preserved if the system is to remain steady. The Bernoulli equation, \(p1 + .5蟻v1^2 + 蟻gA_1 = p2 + .5蟻v2^2 + 蟻gA_2\), is also valuable 鈥� a fundamental principle stating that the total mechanical energy of the flowing fluid, system density, and gravitational acceleration remains constant.
02

Derive the Equation for the Pressure Gradient

We can begin by calculating the pressure gradient from the Bernoulli principle with respect to \(x\), which leads to: \[-\frac{dp}{dx} = \rho\frac{du}{dx}U(1-\frac{x}{L})\]Solving for the pressure yields:\[p = -\rho U^2 \frac{x^2}{2L} + C\]where \(C\) is the integration constant. Given that at \(x=0\), \(p=p_i=100kPa=100000Pa\) we can find the value of \(C\). Hence \(C=100000Pa\). Thus our equation for pressure becomes \(p = -\rho U^2 \frac{x^2}{2L} + 100000Pa\).
03

Evaluate the Outlet Pressure

Lastly, the outlet pressure is evaluated when \(x = L = 5m\). Substituting these values into the pressure equation:\[p = -1250 kg/m^3 * (15 m/s)^2 * 5^2/(2*2*5 m) + 100000Pa = 56875 Pa = 56.875 kPa\]
04

Visualize the Pressure Gradient

A graph with x-values ranging from \(0-5m\) could be plotted using the pressure equation. It gives an idea of how the pressure changes across the length of the porous part of the pipe. It should be a downward parabola beginning at \(100kPa\) and terminating at around \(56.875kPa\) at \(x = 5m\).

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Most popular questions from this chapter

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