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Chapter 6: Problem 106

The stream function of a flow field is $ \psi=A x^{3}-B x y^{2}, \text { where } A=1 \mathrm{m}^{-1} \cdot \mathrm{s}^{-1} \text {and } B=3 \mathrm{m}^{-1} \cdot \mathrm{s}^{-1} $ and coordinates are measured in meters. Find an expression for the velocity potential.

Short Answer

Expert verified
A velocity potential does not exist for the given stream function.

Step by step solution

01

Identify the given stream function

The stream function, denoted by \( \psi \), is given as \( \psi = A x^{3} - B x y^{2} \) where \( A = 1 \mathrm{m}^{-1} \cdot \mathrm{s}^{-1} \) and \( B = 3 \mathrm{m}^{-1} \cdot \mathrm{s}^{-1} \).
02

Compute velocities in terms of stream function

In fluid dynamics, the relationship between velocities and stream function in two-dimensional Cartesian coordinates is described by \( u = \partial \psi / \partial y \) and \( v = -\partial \psi / \partial x \). Applying these expressions to the given stream function, we get \( u = -2 B x y \) and \( v = 3 A x^{2} - B y^{2} \).
03

Integrate velocities to get velocity potential

The velocity potential, denoted by \( \Phi \), is such that \( u = \partial \Phi / \partial x \) and \( v = \partial \Phi / \partial y \). So, by integrating the expressions of \( u \) and \( v \), we get \( \Phi \). Unfortunately, the velocities derived from the given stream function in Step 2 do not give a function \( \Phi \) which satisfies both \( u = \partial \Phi / \partial x \) and \( v = \partial \Phi / \partial y \) simultaneously. Hence, a velocity potential does not exist for the given stream function in this case.

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Most popular questions from this chapter

Show that any differentiable function \(f(z)\) of the complex number \(z=x+i y\) leads to a valid potential (the real part of \(f\), and a corresponding stream function (the negative of the imaginary part of \(f\) ) of an incompressible, irrotational flow. To do so, prove using the chain rule that \(f(z)\) automatically satisfies the Laplace equation. Then show that \(d f / d z=-u+i v\)

The \(\tan \mathrm{k},\) of diameter \(D,\) has a well-rounded nozzle with diameter \(d .\) At \(t=0,\) the water level is at height \(h_{0} .\) Develop an expression for dimensionless water height, \(h / h_{0},\) at any later time. For \(D / d=10,\) plot \(h / h_{0}\) as a function of time with \(h_{0}\) as a parameter for \(0.1 \leq h_{0} \leq 1 \mathrm{m} .\) For \(h_{0}=1 \mathrm{m},\) plot \(h / h_{0}\) as a function of time with \(D / d\) as a parameter for \(2 \leq D / d \leq 10\)

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 \(30 \mathrm{ft} / \mathrm{s} ?\)

A flow field is represented by the stream function $\psi=x^{5}-10 x^{3} y^{2}+5 x y^{4} .$ Find the corresponding velocity field. Show that this flow field is irrotational and obtain the potential function.

An incompressible liquid with a density of \(900 \mathrm{kg} / \mathrm{m}^{3}\) and negligible viscosity flows steadily through a horizontal pipe of constant diameter. In a porous section of length \(L=2 \mathrm{m}\) liquid is removed at a variable rate along the length so that the uniform axial velocity in the pipe is \(u(x)=U e^{-x / L},\) where \(U=20 \mathrm{m} / \mathrm{s} .\) Develop expressions for and plot the acceleration of a fluid particle along the centerline of the porous section and the pressure gradient along the centerline. Evaluate the outlet pressure if the pressure at the inlet to the porous section is \(50 \mathrm{kPa}(\text { gage })\)
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