A stream function in a two-dimensional flow is A = 2xy.
Show that the flow is irrotaional(potential kind).
Let q=(u,v) be the velocity
of the two-dimensional flow.
Then w know, u = delta(A)/delta(y) and v =-[delta(A)/delta(x)] , where delta denote partial derivative.
We've been thought that the necessary and sufficient condition for a potential kind of flow is curl q= 0.
So here, I get curl q=0, which implies tgisflowis of potential kind(irrotaional)
So for this question, my professor has used an alternative appoach. Can anyone verify whether my method is an acceptable answer for this problem?
Thank you!
∇×A=0 is the way I'd do it.
So grad * A implies irrotational condition of the fluid flow??
grad * A = 0 implies irrotaional condition of the flow
Yes, your method is an acceptable approach to show that the flow is irrotational (potential kind). To verify it, we can calculate the curl of the velocity field q=(u,v) and check if it is equal to zero.
Given that u = delta(A)/delta(y) and v = -[delta(A)/delta(x)], we can calculate the curl as follows:
curl(q) = delta(v)/delta(x) - delta(u)/delta(y)
Substituting the values of u and v, we have:
curl(q) = delta(-[delta(A)/delta(x)])/delta(x) - delta(delta(A)/delta(y))/delta(y)
Expanding the derivatives, we get:
curl(q) = -delta^2(A)/delta(x)delta(x) - delta^2(A)/delta(y)delta(y)
Next, let's find the second partial derivatives of A with respect to x and y.
delta^2(A)/delta(x)delta(x) = 2y * delta^2(x)/delta(x)delta(x) = 2y * 0 = 0
delta^2(A)/delta(y)delta(y) = 2x * delta^2(y)/delta(y)delta(y) = 2x * 0 = 0
Since both second partial derivatives are equal to zero, the curl(q) is equal to zero, which implies that the flow is irrotational (potential kind).
Therefore, your method of calculating the curl of the velocity field and showing that it is equal to zero is a valid and acceptable approach to demonstrate that the flow is irrotational.