An incompressible liquid with negligible viscosity and density ρ=1.75 slug/ft ^3 flows steadily

An incompressible liquid with negligible viscosity and...

An incompressible liquid with negligible viscosity and density $\rho=1.75$ slug/ft $^{3}$ flows steadily through a horizontal pipe. The pipe cross-section area linearly varies from 15 in $^{2}$ to 2.5 in $^{2}$ over a length of 10 feet. Develop an expression for and plot the pressure gradient and pressure versus position along the pipe, if the inlet centerline velocity is $5 \mathrm{ft} / \mathrm{s}$ and inlet pressure is 35 psi. What is the exit pressure? Hint: Use relation \[ u \frac{\partial u}{\partial x}=\frac{1}{2} \frac{\partial}{\partial x}\left(u^{2}\right) \]

An incompressible liquid with negligible viscosity and density $\rho=1.75$ slug/ft $^{3}$ flows steadily through a horizontal pipe. The pipe cross-section area linearly varies from 15 in $^{2}$ to 2.5 in $^{2}$ over a length of 10 feet. Develop an expression for and plot the pressure gradient and pressure versus position along the pipe, if the inlet centerline velocity is $5 \mathrm{ft} / \mathrm{s}$ and inlet pressure is 35 psi. What is the exit pressure? Hint: Use relation
\[
u \frac{\partial u}{\partial x}=\frac{1}{2} \frac{\partial}{\partial x}\left(u^{2}\right)
\]

Key Concepts

Euler’s Equation for Inviscid Flow

Euler's equation is essentially the momentum conservation law for inviscid fluids. It relates the pressure gradient to the convective acceleration of the fluid. In one-dimensional steady flow, it provides the foundation to derive expressions for the pressure distribution along a pipe with varying cross-sectional area.

Convective Acceleration and its Mathematical Representation

In one-dimensional steady flow, the acceleration term can be expressed as u(du/dx), which, using the given hint, can be rewritten as one-half the derivative of u squared. This mathematical manipulation simplifies the integration process when deriving the pressure gradient from the momentum equation, linking the changes in velocity directly to the changes in pressure.

Continuity Equation

This principle states that the mass flow rate of an incompressible fluid remains constant along any cross?section of a pipe. When the area of the cross-section changes, the velocity of the fluid must adjust so that the product of area and velocity remains constant, ensuring conservation of mass. This is critical in problems where the pipe geometry varies along its length.

Bernoulli's Principle

In the absence of viscosity and with a steady flow, Bernoulli's equation relates pressure and velocity along a streamline. For horizontal flows where gravitational effects are negligible, an increase in the fluid’s velocity (kinetic energy gain) is accompanied by a decrease in pressure (potential energy loss), and vice versa. This concept is essential when determining how pressure varies with changing cross-sectional area.

Transcript

00:01 In this question, you have an incompressible liquid that's flowing through a horizontal pipe.

00:07 The pipe cross -sectional area changes linearly from 15 inch square to 2 .5 inch square over a length of 10 feet.

00:15 We need to find the pressure gradient.

00:19 We need to find the pressure versus and also to plot the pressure gradient and pressure versus position along the pipe.

00:26 And then and also the exit pressure.

00:31 So in this question we first write down the assumptions.

00:46 The flow is incompressible and then we assume that the flow profile remains unchanged.

01:07 So that the center line velocity represents the average velocity.

01:25 And then we write down the equations that will be using in this case, the flow rate, k equals to a times v, then the acceleration, since in this case is only horizontal direction.

01:42 So we only look at a x.

01:44 So this is u, the x, and then the momentum equation will be row u the u the x equals to minus d d d x.

01:57 Okay, right and then we have our data some given information the density is 1 .75 slug the feet cube in the pressure is 35 psi the inlet area is 15 inch square the exit area is 2 .5 inch square the length of the pipe is 10 feet the inlet velocity is 5 feet per second.

02:39 Okay, so for this 1d flow, the area along the x is ai minus ai minus ae divided by l times x varies linearly, and then the slow rate equation is constant so you are ai, equals to u times a which means that u of x is equal to u i a i a i.

03:21 I .a.

03:22 Divide by a which is equal to u i a i times one over a i minus ae ae divided by l times x.

03:44 So this is ui, a .i times l, it's divided by a.

03:52 L minus x plus aex.

04:04 Then we find our acceleration.

04:09 So ax is u udx.

04:13 So you have uiail.

04:19 Divide by ai, l minus x plus aex...