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Chapter 6: Problem 104

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

Short Answer

Expert verified
The velocity potential function from the given stream function is \(- A x^{2} y+B y^{3}+\frac{1}{3} A x^{3} + C\)

Step by step solution

01

Derive Velocities from Stream Function

First, you have to obtain the velocity components from the stream function \(\psi\). The velocities in x and y direction, denoted as \(u\) and \(v\), are '-partial derivative of \(\psi\) with respect to y' and 'partial derivative of \(\psi\) with respect to x'. The result will be \[u=-\frac{\partial \psi}{\partial y}\] and \[v=\frac{\partial \psi}{\partial x}\] By plugging the given expression of \(\psi\) into these equations, they become \[u=-2 A x y+3 B y^{2}\] and \[v=A x^{2}\]
02

Integrate velocities to obtain Potential Function

The potential function \(\phi\) is obtained by integrating the x-velocity with respect to x and adding it to the integral of the y-velocity with respect to y plus an arbitray constant \(C\). Thus the expression for \(\phi\) becomes \[\phi=\int u d x+\int v d y + C.\] Substituting for \(u\) and \(v\) gives the integral expression that needs to be solved \[\phi=\int(-2 A x y+3 B y^{2}) dx+\int(A x^{2}) dy + C\]. Solve these integrals to find the potential function.
03

Simplify the Velocity Potential Function

On integrating, \[\phi=- A x^{2} y+B y^{3}+\frac{1}{3} A x^{3} + C\]. The potential function comes to \[\phi=- A x^{2} y+B y^{3}+\frac{1}{3} A x^{3} + C\]. This is the final result.

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Most popular questions from this chapter

Consider the flow field represented by the potential function $\phi=A x^{2}+B x y-A y^{2} .$ Verify that this is an incompressible flow and determine the corresponding stream function.

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