This is just a basic question about streamline, equation of streamline and stream function. I am used to perceiving that the stream fucntion is just another form of the equations of streamlines. Since streamlines can be obtained at an "instant time" of an unsteady flow field, then the stream function should be the same, it could be obtained at an instant time of an unsteady flow field.
However, I stuck at this point: consider an unsteady, compressible, 2D – flow field at an instant time, it has stream function $\bar {\psi}$. From textbooks, we already know:
$$\rho u = \frac{\partial \bar {\psi}}{\partial y}\space\space\space (1) $$
$$\rho v = -\frac{\partial \bar {\psi}}{\partial x}\space\space\space (2) $$
Then I do some maths:
$$(1)\Rightarrow \frac{\partial (\rho u)}{\partial x} = \frac{\partial^2 \bar {\psi}}{\partial y \partial x}$$
$$(2)\Rightarrow -\frac{\partial (\rho v)}{\partial y} = \frac{\partial^2 \bar {\psi}}{\partial x \partial y}$$
Since $\frac{\partial^2 \bar {\psi}}{\partial y \partial x} = \frac{\partial^2 \bar {\psi}}{\partial x \partial y} $, hence:
$$\frac{\partial (\rho u)}{\partial x} = -\frac{\partial (\rho v)}{\partial y} \iff \frac{\partial (\rho u)}{\partial x} + \frac{\partial (\rho v)}{\partial y}=0 \iff \nabla.(\rho\vec V)=0\space\space\space (3)$$
But continuity equation says:
$$\frac{\partial \rho}{\partial t}+ \nabla.(\rho\vec V)=0\space\space\space (4)$$
Here you can see the flow field is unsteady, then $\frac{\partial \rho}{\partial t}$ is not equal zero and the equation (4) is inconsistent with (3). Here is where I stuck… if I am wrong, so does stream function hold only for steady flow?
Best Answer
Generalizing your mathematics (which are all correct), the 2D stream function automatically satisfies continuity for any 2D case where $\partial\rho / \partial t$ is zero. In flows other than these, you must independently confirm that your results satisfy continuity.