2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You already answered your own question. $\endgroup$ – Chet Miller Jul 15 '17 at 2:29
  • $\begingroup$ By virtue of equations 1 and 2 stream function is defined for any 2D flow of constant density, steady or not. $\partial\rho/\partial t$ can be zero in an unsteady flow. However in an unsteady flow streamline patterns change from one instant to another. $\endgroup$ – Deep Jul 15 '17 at 4:00
  • $\begingroup$ @Chester so you mean stream functions hold only for steady flow. But I always thought it holds for unsteady flows at an insant time. What's wrong with this? $\endgroup$ – Dat Jul 15 '17 at 4:46
  • $\begingroup$ @Deep no, equations (1) and (2) hold for incompressible or compressible flow. And I am not sure whether they hold for steady or unsteady flow. But I think they hold for an unsteady flow at an instant time $\endgroup$ – Dat Jul 15 '17 at 4:50
  • $\begingroup$ Equation 1 and 2 define stream function $\psi$ irrespective of the kind of 2D flow under consideration. But if $\psi$ so defined is also required to satisfy continuity equation, then it is necessary that $\partial\rho/\partial t=0$. $\endgroup$ – Deep Jul 15 '17 at 5:49
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Could we have an unsteady, compressible flow which has $\partial\rho / \partial t = 0$ ? $\endgroup$ – Dat Jul 17 '17 at 2:56
  • $\begingroup$ @Dat: Yes, I suppose so. See this answer for a good point about terminology. I've edited my answer. $\endgroup$ – Peter Schilling Jul 17 '17 at 11:26
  • $\begingroup$ the comment was too long so I cannot post it here so I answered my question $\endgroup$ – Dat Jul 17 '17 at 16:36
0
$\begingroup$

@Peter: I have clicked your link and read it. I think we should forget temporarily about true meaning of these words: steady or unsteady, imcompressible or compressible...let's the equations speak words. Does your answer above mean stream function only hold for flow which has $\partial\rho / \partial t = 0$, despite it is steady or unsteady or whatever ? If the answer is yes, so is this true: consider a 2D flow has:

V = u$\vec i$ + v$\vec j$ ; u = u(x, y, t) ; v = v(x, y, t);

$\rho = (x+y)t^2$.

Suppose we already have u and v so that the equation (4) is true (at any point, any instant time of the flow, satisfying continuity equation). Now, consider this flow at instant time t = 1 . At this instant time, $\partial\rho / \partial t = 2(x+y)t = 2(x+y) \ne 0$. So at this instant time, we can not obtain stream function for this flow. Isn't it true?

$\endgroup$
  • $\begingroup$ Like you say, let the math speak for itself. I encourage you to define some different flows and see if their stream functions satisfy continuity. $\endgroup$ – Peter Schilling Jul 17 '17 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.