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Basics of Fluid Mechanics, 2014a

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342 CHAPTER 10. POTENTIAL FLOW<br />

Example 10.4:<br />

Is there a potential based on the following stream function<br />

ψ =3x 5 − 2 y<br />

(10.IV.a)<br />

Solution<br />

Equation (10.74) dictates what are the requirements on the stream function. According<br />

to this equation the following must be zero<br />

∂ 2 ψ<br />

∂y 2 + ∂2 ψ<br />

∂x 2 ?<br />

=0 (10.IV.b)<br />

In this case it is<br />

0 ? =0+60x 3 (10.IV.c)<br />

Since x 3 is only zero at x =0the requirement is fulfilled and therefor this function<br />

cannot be appropriate stream function.<br />

End Solution<br />

10.3 Potential Flow Functions Inventory<br />

This section describes several simple scenarios <strong>of</strong> the flow field. These flow fields will<br />

be described and exhibits utilizition <strong>of</strong> the potential and stream functions. These flow<br />

fields can be combined by utilizing superimposing principle.<br />

Uniform Flow<br />

The trivial flow is the uniform flow in which the fluid field moves directly and<br />

uniformly from one side to another side. This flow is further simplified, that is the<br />

coordinates system aligned with to flow so the x–coordinate in the direction <strong>of</strong> the flow.<br />

In this case the velocity is given by<br />

U x = U 0<br />

U y =0<br />

(10.75)<br />

and according to definitions in this chapter<br />

Hence, it can be obtained that<br />

U x = ∂φ<br />

∂x = ∂ψ<br />

∂y = U 0 (10.76)<br />

φ = U 0 x + f y (y)<br />

ψ = U 0 x + f x (x)<br />

(10.77)

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