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Chapter 6: Problem 6

The \(x\) component of velocity in an incompressible flow field is given by \(u=A x,\) where \(A=2 \mathrm{s}^{-1}\) and the coordinates are measured in meters. The pressure at point \((x, y)=(0,0)\) is \(p_{0}=190 \mathrm{kPa}\) (gage). The density is \(\rho=1.50 \mathrm{kg} / \mathrm{m}^{3}\) and the \(z\) axis is vertical. Evaluate the simplest possible \(y\) component of velocity. Calculate the fluid acceleration and determine the pressure gradient at point \((x, y)=(2,1) .\) Find the pressure distribution along the positive \(x\) axis.

Short Answer

Expert verified
The simplest \(y\) component of velocity is \(v = -2y + C\). The fluid acceleration is \(\frac{du}{dt} = 4x - 4y\). The pressure gradient at point (2,1) is 6 Pa/m. The pressure distribution along the positive x axis is given by the equation \(p2 = p0 - \rho g u^2 + \rho g u2^2\).

Step by step solution

01

Finding the simplest y-component of velocity

In incompressible flow, the divergence of the velocity field is zero. Hence, the \(y\) component of the velocity, \(v\), can be found by solving the continuity equation \(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0\). With \(u = A x = 2x\) and \(\frac{\partial u}{\partial x} = 2\), it is implied \(v\) must be a function of \(y\) only and direct integration of \(\frac{\partial v}{\partial y} = -\frac{\partial u}{\partial x}\) gives \(v = -2y + C\), where \(C\) is the constant of integration.
02

Fluid Acceleration

The acceleration in incompressible flow is determined by the formula \(\frac{du}{dt} = \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}\). Given that the flow is steady, we have \(\frac{\partial u}{\partial t} = 0\). Also \(\frac{\partial u}{\partial x} = A = 2\). Substituting \(u = 2x\) and \(v = -2y\) gives fluid acceleration as \(\frac{du}{dt} = 4x - 4y\).
03

Pressure Gradient

The pressure gradient at any point can be found by applying the Euler's equation for inviscid flow in \(x\)-direction \(\rho \frac{du}{dt} = - \frac{dp}{dx}\), substituting the variables gives \(-\frac{dp}{dx} = 6\). Hence the pressure gradient is 6 Pa/m.
04

Pressure Distribution Along x-axis

Applying Bernoulli's equation along a streamline from \(x = 0\) to \(x = 2\) gives \(\frac{p0}{\rho g} + \frac{u^2}{2g} = \frac{p2}{\rho g} + \frac{u2^2}{2g} \), where \(p0\) and \(p2\) are the pressures at \(x = 0\) and \(x = 2\) respectively and \(u2 = 2x = 4m/s\). Solving for \(p2 = p0 - \rho g u^2 + \rho g u2^2\), gives the pressure distribution.

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Most popular questions from this chapter

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 \(100 \mathrm{kPa}(\text { gage })\)

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 \(x=0\)

Water flows steadily up the vertical 1-in.-diameter pipe and out the nozzle, which is 0.5 in. in diameter, discharging to atmospheric pressure. The stream velocity at the nozzle exit must be \(30 \mathrm{ft} / \mathrm{s}\). Calculate the minimum gage pressure required at section (1). If the device were inverted, what would be the required minimum pressure at section (1) to maintain the nozzle exit velocity at \(30 \mathrm{ft} / \mathrm{s} ?\)

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) \\]

\(\mathrm{A}\) horizontal flow of water is described by the velocity field \(\vec{V}=(-A x+B t) \hat{i}+(A y+B t) \hat{j},\) where \(A=1 \mathrm{s}^{-1}\) and \(B=2 \mathrm{m} / \mathrm{s}^{2}, x\) and \(y\) are in meters, and \(t\) is in seconds. Find expressions for the local acceleration, the convective acceleration, and the total acceleration. Evaluate these at point (1,2) at \(t=5\) seconds. Evaluate \(\nabla p\) at the same point and time.
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