The flow is defined by the stream function ψ(x, y) = y² - x. Draw the streamlines for ψ1(x, y) = 0, ψ2(x, y)= 2 m²/s, and ψ3(x, y) = 4 m²/s. Find the velocity of a fluid particle at y = 1 m on the streamline ψ2(x, y) = 2 m²/s and show that the flow field satisfies the continuity equation.

Question

The flow is defined by the stream function [latex] \psi(x, y)=y^2-x [/latex]. Draw the streamlines for [latex] \psi_1(x, y)=0, \psi_2(x, y)=2 m ^2 / s , ext { and } \psi_3(x, y)=4 m ^2 / s [/latex]. Find the velocity of a fluid particle at y = 1 m on the streamline [latex] \psi_2(x, y)=2 m ^2 / s [/latex] and show that the flow field satisfies the continuity equation.

Accepted Answer

Fluid Description. Since time is not involved, this is steady flow of an ideal fluid.

Stream Functions. The equations for the three streamlines are

[latex] \begin{aligned} & y^2-x=0 \ & y^2-x=2 \ & y^2-x=4 \end{aligned} [/latex]

These equations are graphed in Fig. 7–18. Each is a parabola and represents a streamline for the constant that defines it.

Velocity. The velocity components along each streamline are

[latex] \begin{aligned} u & =\frac{\partial \psi}{\partial y}=\frac{\partial}{\partial y}\left(y^2-x ight)=(2 y) m / s \ v & =-\frac{\partial \psi}{\partial x}=-\frac{\partial}{\partial x}\left(y^2-x ight)=-(-1)=1 m / s \end{aligned} [/latex]

Continuity requires Eq. 7–11 be satisfied.

[latex] \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0 [/latex] (7-11); 0 + 0 = 0

For the streamline y² - x = 2, at y = 1 m, then x = -1 m, so at this point, u = 2 m/s and [latex] v[/latex] = 1 m/s. These two components produce the resultant velocity of a fluid particle at this location, Fig. 7–18. It is

[latex] V=\sqrt{(2 m / s )^2+(1 m / s )^2}=2.24 m / s [/latex]

Notice that the directions of the velocity components also provide a means for establishing the direction of the flow, as indicated by the small arrows on this streamline, Fig. 7–18.
Although it is not part of this problem, imagine the streamlines for which [latex] \psi_1=0 ext { and } \psi_3=4 m ^2 / s [/latex] represent solid boundaries for a channel, Fig. 7–18. Then from Eq. 7–24, the volumetric flow per unit depth within this channel (or streamtube) would be

[latex] q=\int_{\psi_1}^{\psi_2} d \psi=\psi_2(x, y)-\psi_1(x, y)=C_2-C_1 [/latex] (7-24)

[latex] q=\psi_3-\psi_1=4 m ^2 / s -0=4 m ^2 / s [/latex]

Question 7.4: The flow is defined by the stream function ψ(x, y) = y² - x....

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