The stream function for a given two-dimensional flow field is $$\psi=5 x^{2} y-(5 / 3) y^{3}$$ Determine the corresponding velocity potential.

Short Answer

Expert verified
The velocity potential is given by \(φ=5xy^{2} - (5/3)x^{3}\)

Step by step solution

01

Compute the component velocities

The x-component of the velocity \(u\) is given by \(-\frac{∂ψ}{∂y} = -5x^{2} + 5y^{2}\) and the y-component of the velocity \(v\) is given by \(\frac{∂ψ}{∂x} = 10xy\)
02

Integrate the x-component

Now integrate \(u\) with respect to \(x\) to find a part of the velocity potential. \(∫u dx =∫(-5x^{2} + 5y^{2})dx=- (5/3)x^{3} + 5yx^{2} + G(y)\). \(G(y)\) is an arbitrary function of \(y\) that arises due to the integration.
03

Integrate the y-component

Integrate \(v\) with respect to \(y\) to find another part of the velocity potential. \(∫v dy = ∫10xy dy = 5x y^{2} + H(x)\), where \(H(x)\) is an arbitrary function of \(x\).
04

Equate the results from step 2 and step 3

The velocity potential arises from both \(x\) and \(y\) components of the velocity. Hence, equate the results from steps 2 and 3. \(-(5/3)x^{3} + 5yx^{2} + G(y) = 5xy^{2} + H(x)\). Now find an appropriate \(G(y)\) and \(H(x)\) such that these two expressions are identical for all \(x\) and \(y\). In here, we can see that \(H(x)=-(5/3)x^{3}\) and \(G(y) = 0\).
05

Combine results

The final velocity potential φ is a combination of the results of these functions: \(φ=5xy^{2} - (5/3)x^{3}\)

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