Understanding Velocity from Stream Function

In fluid dynamics, the stream function $\psi$ is a useful tool for describing incompressible, two-dimensional flow. It is defined such that the velocity components $u$ (in the x-direction) and $v$ (in the y-direction) can be obtained by differentiating the stream function. The relationships are given by:

• $u = \frac{\partial \psi}{\partial y}$
• $v = -\frac{\partial \psi}{\partial x}$

The magnitude of the velocity vector at any point $(x, y)$ is then calculated using the Pythagorean theorem for vectors: $|\vec{V}| = \sqrt{u^2 + v^2}$.

Calculating Velocity Components from $\psi = xy$

The given stream function is $\psi = xy$. We need to find the velocity components $u$ and $v$ from this function.

First, let's find the velocity component $u$ by differentiating $\psi$ with respect to $y$, treating $x$ as a constant:

$u = \frac{\partial \psi}{\partial y} = \frac{\partial (xy)}{\partial y}$

Applying the differentiation rule, we get:

$u = x$

Next, let's find the velocity component $v$ by differentiating $\psi$ with respect to $x$, treating $y$ as a constant, and then taking the negative:

$v = -\frac{\partial \psi}{\partial x} = -\frac{\partial (xy)}{\partial x}$

Applying the differentiation rule and the negative sign, we get:

$v = -y$

So, the velocity vector components at any point $(x, y)$ are $u=x$ and $v=-y$.

Evaluating Velocity at the Point (2, -2)

We are asked to find the magnitude of the velocity vector at the specific point (2, -2). We substitute $x=2$ and $y=-2$ into our expressions for $u$ and $v$.

• At $(2, -2)$, $u = x = 2$.
• At $(2, -2)$, $v = -y = -(-2) = 2$.

Thus, the velocity vector components at the point (2, -2) are $u=2$ and $v=2$.

Calculating the Magnitude of the Velocity Vector

Now that we have the velocity components $u=2$ and $v=2$ at the point (2, -2), we can calculate the magnitude of the velocity vector $|\vec{V}|$.

The formula for the magnitude is:

$|\vec{V}| = \sqrt{u^2 + v^2}$

Substitute the values of $u$ and $v$ at the point (2, -2):

$|\vec{V}| = \sqrt{(2)^2 + (2)^2}$

$|\vec{V}| = \sqrt{4 + 4}$

$|\vec{V}| = \sqrt{8}$

The magnitude of the velocity vector at the point (2, -2) is $\sqrt{8}$.

Final Answer for Velocity Magnitude

The magnitude of the velocity vector at the point (2, -2) for the given stream function $\psi = xy$ is $\sqrt{8}$.

Revision Table: Stream Function and Velocity

Additional Information on Stream Functions

The stream function $\psi$ is constant along a streamline. This property is very useful for visualizing flow patterns, as lines of constant $\psi$ represent the paths that fluid particles follow.

• The difference in stream function values between two streamlines represents the volume flow rate per unit width between those streamlines in 2D flow.
• For a flow to be irrotational (meaning the fluid elements do not rotate), the condition $\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = 0$ must be satisfied. If we substitute the expressions for $u$ and $v$ from the stream function, we get $-\frac{\partial^2 \psi}{\partial x^2} - \frac{\partial^2 \psi}{\partial y^2} = 0$, which simplifies to $\nabla^2 \psi = 0$. This is Laplace's equation, indicating that the stream function for an irrotational, incompressible flow satisfies Laplace's equation.
• In our case, for $\psi = xy$, $\frac{\partial u}{\partial y} = \frac{\partial (x)}{\partial y} = 0$ and $\frac{\partial v}{\partial x} = \frac{\partial (-y)}{\partial x} = 0$. Since $0 - 0 = 0$, the flow described by $\psi = xy$ is irrotational.