Navier-Stokes Problem: Solving for Pressure Gradient in a 2D Rectangular Cavity

[member 428835]

Hi PF!

Assume we have a rectangular cavity (2D) filled with a liquid of dimensions $L \times H$ and that the top plate of the cavity moves with some velocity $V_0$. Also assume $L \gg H$. I'll also assume $L \gg H$ implies flow is roughly 1-dimensional, and thus a pressure gradient drives the flow near the still wall in the opposite direction rather than the $y$ component that would "bounce" off the vertical cavity walls. When computing the velocity profile, from the Navier-Stokes equation I get $$v_x \frac{\partial v_x}{\partial x} = -\frac{\partial P}{\partial x} + \mu \frac{\partial^2 v_x}{\partial y^2} + \mu \frac{\partial^2 v_x}{\partial x^2}$$ subject to $v_x(0) = v_0$ and $v_x(H) = 0$, where I assume the flow is fully developed and steady so $v_x = f(y)$. Notice now the equation reduces to $$\frac{d P}{d x} = \mu \frac{d^2 v_x}{d y^2}$$

Clearly all we now must do is solve for the pressure gradient since the boundary conditions determine the two integration constants. To find this, I imagine the net flow rate must be zero since no flow enters or leaves. Thus $$\frac{1}{HL} \int_0^L \int_0^H \, v_x dy \, dx = 0$$. Then we should be able to solve for $\frac{d P}{d x}$.

Does this look right to you? Thanks for your time!

 

[boneh3ad]

Do you mind drawing a picture? I am not sure if I follow your description of the geometry completely.

 

[Chestermiller]

Hi PF!

Assume we have a rectangular cavity (2D) filled with a liquid of dimensions $L \times H$ and that the top plate of the cavity moves with some velocity $V_0$. Also assume $L \gg H$. I'll also assume $L \gg H$ implies flow is roughly 1-dimensional, and thus a pressure gradient drives the flow near the still wall in the opposite direction rather than the $y$ component that would "bounce" off the vertical cavity walls. When computing the velocity profile, from the Navier-Stokes equation I get $$v_x \frac{\partial v_x}{\partial x} = -\frac{\partial P}{\partial x} + \mu \frac{\partial^2 v_x}{\partial y^2} + \mu \frac{\partial^2 v_x}{\partial x^2}$$ subject to $v_x(0) = v_0$ and $v_x(H) = 0$, where I assume the flow is fully developed and steady so $v_x = f(y)$. Notice now the equation reduces to $$\frac{d P}{d x} = \mu \frac{d^2 v_x}{d y^2}$$

Clearly all we now must do is solve for the pressure gradient since the boundary conditions determine the two integration constants. To find this, I imagine the net flow rate must be zero since no flow enters or leaves. Thus $$\frac{1}{HL} \int_0^L \int_0^H \, v_x dy \, dx = 0$$. Then we should be able to solve for $\frac{d P}{d x}$.

Does this look right to you? Thanks for your time!

This looks really good. This is the kind of flow that exists within the flights of screw pumps and screw extruders. See the book Processing of Thermoplastic Materials by Ernest Bernhardt

 

[member 428835]

Do you mind drawing a picture? I am not sure if I follow your description of the geometry completely.

I've attached a rough sketch. I can add detail if necessary.

This looks really good. This is the kind of flow that exists within the flights of screw pumps and screw extruders. See the book Processing of Thermoplastic Materials by Ernest Bernhardt

Thanks! I'll have to check this out. So you agree with the integral constraint?

 

[Chestermiller]

I've attached a rough sketch. I can add detail if necessary.Thanks! I'll have to check this out. So you agree with the integral constraint?

Yes.

 

[member 428835]

Yes.

Cool. So if we weren't to assume that $y$ velocity was negligible, would our equations be $$\rho\left( v_x\frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} \right) = \mu \frac{\partial^2 v_x}{\partial y^2}\\ \rho\left( v_x\frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} \right) = \mu \frac{\partial^2 v_y}{\partial y^2} + \rho g$$ where the pressure gradient is gone since it no longer drives flow. Boundary conditions would be all velocities equal zero along all walls except the top moving wall, where $v_x=v_0$. The integral constraint holds as well, but this time for both velocity components.

But then we have 5 conditions to meet (the four boundaries plus the integral constraint). This now seems over specified. Any ideas?

 

[Chestermiller]

Cool. So if we weren't to assume that $y$ velocity was negligible, would our equations be $$\rho\left( v_x\frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} \right) = \mu \frac{\partial^2 v_x}{\partial y^2}\\ \rho\left( v_x\frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} \right) = \mu \frac{\partial^2 v_y}{\partial y^2} + \rho g$$ where the pressure gradient is gone since it no longer drives flow. Boundary conditions would be all velocities equal zero along all walls except the top moving wall, where $v_x=v_0$. The integral constraint holds as well, but this time for both velocity components.

But then we have 5 conditions to meet (the four boundaries plus the integral constraint). This now seems over specified. Any ideas?

The pressure gradients are not zero. They are significant.

 

[member 428835]

The pressure gradients are not zero. They are significant.

Ok, so then this problem is solvable, since we have the unknown pressure gradient vector and the integral constraint over the velocity vector.

Pressure to me is not intuitive; how would I know to include it in this analysis? I knew it was relevant in the first (simpler) part since something had to drive flow backwards.

 

[Chestermiller]

Ok, so then this problem is solvable, since we have the unknown pressure gradient vector and the integral constraint over the velocity vector.

You don't use the integral constraint on this. The velocity components on all 4 boundaries are zero.

Pressure to me is not intuitive; how would I know to include it in this analysis?

You are asking how to solve the equations for the 2D version, including the pressure gradients?

 

[member 428835]

You don't use the integral constraint on this. The velocity components on all 4 boundaries are zero.

Why wouldn't we use the integral constraint? Also, the velocity is zero everywhere except the moving side, right?

You are asking how to solve the equations for the 2D version, including the pressure gradients?

No, I think I understand how to handle the math once it's there. The difficulty is setting up the equation, and in this instance, why the pressure gradient is even there.

 

[member 428835]

Hey Chet, did you forget about me?

 

[Chestermiller]

Ok, so then this problem is solvable, since we have the unknown pressure gradient vector and the integral constraint over the velocity vector.

Pressure to me is not intuitive; how would I know to include it in this analysis? I knew it was relevant in the first (simpler) part since something had to drive flow backwards.

The qualitative nature of the solution does not change when you just make the channel deeper.

 

[Chestermiller]

Why wouldn't we use the integral constraint?

It is automatically satisfied when you satisfy the boundary conditions on the edge faces.

Also, the velocity is zero everywhere except the moving side, right?

yes.

No, I think I understand how to handle the math once it's there. The difficulty is setting up the equation, and in this instance, why the pressure gradient is even there.

The thing to do on this problem is to first consider the case where the inertial forces are very low, so that they can be neglected. That is, at very low Reynolds number. If you can't solve it for this (linear) case, you'll never be able to solve for the case where inertia is included (non-linear case). What are your equations in the case where inertia is neglected?

 

[member 428835]

What are your equations in the case where inertia is neglected?

I believe we have two, one for each velocity component. Let's say $u$ is velocity in $x-$direction and $v$ is velocity in $y$-direction. Then we have $$\frac{\partial P}{\partial x} = \mu \left( \frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial x^2} \right)\\
\frac{\partial P}{\partial y} = \mu \left( \frac{\partial^2 v}{\partial y^2}+\frac{\partial^2 v}{\partial x^2} \right)$$
I believe both derivatives on the right hand side are relevant since we are not assuming $y \ll x$, which invalidates the scaling argument. I am neglecting gravity though.

Does this look right to you?

 

[Chestermiller]

I believe we have two, one for each velocity component. Let's say $u$ is velocity in $x-$direction and $v$ is velocity in $y$-direction. Then we have $$\frac{\partial P}{\partial x} = \mu \left( \frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial x^2} \right)\\
\frac{\partial P}{\partial y} = \mu \left( \frac{\partial^2 v}{\partial y^2}+\frac{\partial^2 v}{\partial x^2} \right)$$
I believe both derivatives on the right hand side are relevant since we are not assuming $y \ll x$, which invalidates the scaling argument. I am neglecting gravity though.

Does this look right to you?

Yes. Now, in addition to these, you have the continuity equation. Please include that.

 

[member 428835]

Yes. Now, in addition to these, you have the continuity equation. Please include that.

Of course! That's $$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$$Boundary conditions are no slip on the four walls, which implies velocity is zero along the cavity except the top wall, where $u=U_0$ which is the speed of the top shelf.

So we have three equations and three unknowns ($P,u,v$). We have 8 boundary conditions (4 for $u$ and 4 for $v$, each evaluated along the wall.

 

[Chestermiller]

Of course! That's $$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$$Boundary conditions are no slip on the four walls, which implies velocity is zero along the cavity except the top wall, where $u=U_0$ which is the speed of the top shelf.

So we have three equations and three unknowns ($P,u,v$). We have 8 boundary conditions (4 for $u$ and 4 for $v$, each evaluated along the wall.

Here's a way of proceeding further with this problem: Define a stream function $\psi$ for the flow as follows:
$$u=\frac{\partial \psi}{\partial y}$$
$$v=-\frac{\partial \psi}{\partial x}$$
These equations satisfy the continuity equation exactly. What do you get if you substitute them into your two differential equations?

What do you think the streamline pattern looks like for this flow? The streamlines are the lines of constant $\psi$.

 

[member 428835]

Here's a way of proceeding further with this problem: Define a stream function $\psi$ for the flow as follows:
$$u=\frac{\partial \psi}{\partial y}$$
$$v=-\frac{\partial \psi}{\partial x}$$
These equations satisfy the continuity equation exactly. What do you get if you substitute them into your two differential equations?

What do you think the streamline pattern looks like for this flow? The streamlines are the lines of constant $\psi$.

Oh yea, the stream function $\psi \equiv \nabla \times \vec{v}$. Plugging these in yields third derivatives and mixed partials of $\psi$.

$$\frac{\partial P}{\partial x} = \mu \left( \frac{\partial^3 \psi}{\partial y^3}+\frac{\partial^3 \psi}{\partial x^2 \partial y} \right)\\
\frac{\partial P}{\partial y} = -\mu \left( \frac{\partial^3 \psi}{\partial y^2 \partial x}+\frac{\partial^3 \psi}{\partial x^3} \right)$$

I have no idea what they look like? Maybe a circle roughly?

 

[Chestermiller]

Oh yea, the stream function $\psi \equiv \nabla \times \vec{v}$. Plugging these in yields third derivatives and mixed partials of $\psi$.

$$\frac{\partial P}{\partial x} = \mu \left( \frac{\partial^3 \psi}{\partial y^3}+\frac{\partial^3 \psi}{\partial x^2 \partial y} \right)\\
\frac{\partial P}{\partial y} = -\mu \left( \frac{\partial^3 \psi}{\partial y^2 \partial x}+\frac{\partial^3 \psi}{\partial x^3} \right)$$

Good. Now take the partial derivative of the first equation with respect to y and subtract the partial derivative of the second equation with respect to x. What do you get?

I have no idea what they look like? Maybe a circle roughly?

Very good. So you envision that there will be a recirculating flow in the cavity, and the streamlines will each be closed.

 

[member 428835]

Good. Now take the partial derivative of the first equation with respect to y and subtract the partial derivative of the second equation with respect to x. What do you get?

$$\frac{\partial^4 \psi}{\partial x^4} + \frac{\partial^4 \psi}{\partial y^4}+2\frac{\partial^4 \psi}{\partial x^2 \partial y^2}=0$$

which I believe is separable. I've worked with similar equations before, where separability exists but was unusual. The mixed partial evokes such equations. Letting $\psi = X(x)Y(y)$ implies $$\frac{d_x^4X}{X}+ \frac{d_y^4 Y}{Y}+2 \frac{X''Y''}{XY}=0$$
Hmmmmm tricks that have worked in the past fail now. Any ideas? By inspection sines and cosines would all differentiate back to original functions. Perhaps a good ansatz is to let $X=\sin(ax)$ or $X=\cos(ax)$ and $Y=\sin(cy)$ or $Y=\cos(cy)$. Then we have an algebraic expression that behaves like $a^4+c^4+2c^2c^2=0$. Quadratic formula could be used to represent one in terms of the other, but before continuing does this idea look right to you?

Another option is to let $Y$ and $X$ be linear functions. And rethinking the above, exponentials would be easier to work with than trig, but trig stood out first.

 

[Chestermiller]

$$\frac{\partial^4 \psi}{\partial x^4} + \frac{\partial^4 \psi}{\partial y^4}+2\frac{\partial^4 \psi}{\partial x^2 \partial y^2}=0$$

This is called the biharmonic equation. It can also be expressed as $$\nabla^2 (\nabla^2 \psi )=0$$

which I believe is separable. I've worked with similar equations before, where separability exists but was unusual. The mixed partial evokes such equations. Letting $\psi = X(x)Y(y)$ implies $$\frac{d_x^4X}{X}+ \frac{d_y^4 Y}{Y}+2 \frac{X''Y''}{XY}=0$$
Hmmmmm tricks that have worked in the past fail now. Any ideas? By inspection sines and cosines would all differentiate back to original functions. Perhaps a good ansatz is to let $X=\sin(ax)$ or $X=\cos(ax)$ and $Y=\sin(cy)$ or $Y=\cos(cy)$. Then we have an algebraic expression that behaves like $a^4+c^4+2c^2c^2=0$. Quadratic formula could be used to represent one in terms of the other, but before continuing does this idea look right to you?

Another option is to let $Y$ and $X$ be linear functions. And rethinking the above, exponentials would be easier to work with than trig, but trig stood out first.

I haven't thought of how I would try to approach this analytically. My inclination would be to do it numerically. I would consider writing the equation in terms of two variables:
$$\nabla^2 \psi=\omega$$
$$\nabla^2\omega = 0$$
Before trying to figure out a method of solving however, I think you should figure out the boundary conditions on psi or on psi and omega.

 

[member 428835]

Boundary conditions on $\psi$ are $\psi_y(x,L) = U_0$, $\psi_y(x,0) = 0$, $\psi_y(0,y) = 0$, $\psi_y(H,y) = 0$
$\psi_x(x,L) = 0$, $\psi_x(x,0) = 0$, $\psi_x(0,y) = 0$, $\psi_x(H,y) = 0$.

I don't know what the boundary conditions on $\omega$ would be. Just because the derivatives are zero at particular locations doesn't mean the second derivatives are, so I'm stumped. Any ideas?

 

[Chestermiller]

Boundary conditions on $\psi$ are $\psi_y(x,L) = U_0$, $\psi_y(x,0) = 0$, $\psi_y(0,y) = 0$, $\psi_y(H,y) = 0$
$\psi_x(x,L) = 0$, $\psi_x(x,0) = 0$, $\psi_x(0,y) = 0$, $\psi_x(H,y) = 0$.

I don't know what the boundary conditions on $\omega$ would be. Just because the derivatives are zero at particular locations doesn't mean the second derivatives are, so I'm stumped. Any ideas?

I think you have L and H switched.

 

[member 428835]

I think you have L and H switched.

Oh gosh, what a dumb mistake. Yes, of course, here are the actual boundary conditions:
$\psi_y(x,H) = U_0$, $\psi_y(x,0) = 0$, $\psi_y(0,y) = 0$, $\psi_y(L,y) = 0$
$\psi_x(x,H) = 0$, $\psi_x(x,0) = 0$, $\psi_x(0,y) = 0$, $\psi_x(L,y) = 0$.

 

[Chestermiller]

Oh gosh, what a dumb mistake. Yes, of course, here are the actual boundary conditions:
$\psi_y(x,H) = U_0$, $\psi_y(x,0) = 0$, $\psi_y(0,y) = 0$, $\psi_y(L,y) = 0$
$\psi_x(x,H) = 0$, $\psi_x(x,0) = 0$, $\psi_x(0,y) = 0$, $\psi_x(L,y) = 0$.

You can replace the last 4 boundary conditions with an arbitrary constant value for psi, like $\psi =0$, since the outer boundary is a streamline, and the stream function is constant on a streamline.

Doing this problem numerically, I know how I would proceed. I haven't tried figuring out how I would proceed analytically. But, certainly, since omega satisfies Laplace's equation, this is where I would start.

 

[member 428835]

You can replace the last 4 boundary conditions with an arbitrary constant value for psi, like $\psi =0$, since the outer boundary is a streamline, and the stream function is constant on a streamline.

So you're saying $\psi = 0$ along the outer boundary? How did you select 0? I know if I were doing this I wouldn't know what constant to select.

Doing this problem numerically, I know how I would proceed. I haven't tried figuring out how I would proceed analytically. But, certainly, since omega satisfies Laplace's equation, this is where I would start.

So how would we solve $\nabla^2 \omega =0$ without boundary conditions on $\omega$?

 

[Chestermiller]

So you're saying $\psi = 0$ along the outer boundary? How did you select 0? I know if I were doing this I wouldn't know what constant to select.

It doesn't matter what value is used. I just chose 0 because it's easier to work with.

So how would we solve $\nabla^2 \omega =0$ without boundary conditions on $\omega$?[/QUOTE]
I would start out by writing a general analytic solution down for omega, starting from X(x)Y(y), and getting it in terms of constants A, B, C, and D. Then I would substitute that into the Poisson equation for psi, ending up with additional constants.

 

[member 428835]

It doesn't matter what value is used. I just chose 0 because it's easier to work with.

I'm with you here, but how would I know it doesn't matter?

I would start out by writing a general analytic solution down for omega, starting from X(x)Y(y), and getting it in terms of constants A, B, C, and D. Then I would substitute that into the Poisson equation for psi, ending up with additional constants.

So I arrive at $$\omega(x,y) = Ae^{\sqrt{\lambda}y}+Be^{-\sqrt{\lambda}y}+Ce^{\sqrt{-\lambda}x}+De^{-\sqrt{-\lambda}x}:\lambda \in \mathbb{R}$$ You agree so far? (As an aside, I always assume the separation constant is real; is it ever not? This is definitely off task, but I'm curious if you've seen it?)

 

[Chestermiller]

I'm with you here, but how would I know it doesn't matter?So I arrive at $$\omega(x,y) = Ae^{\sqrt{\lambda}y}+Be^{-\sqrt{\lambda}y}+Ce^{\sqrt{-\lambda}x}+De^{-\sqrt{-\lambda}x}:\lambda \in \mathbb{R}$$ You agree so far? (As an aside, I always assume the separation constant is real; is it ever not? This is definitely off task, but I'm curious if you've seen it?)

This looks OK so far. I don't even know what the term "separation constant" means.

I've never seen this full development, but I'm sure it's been published somewhere. My inclination would have been to solve the problem numerically.

Chet

 

[member 428835]

This looks OK so far. I don't even know what the term "separation constant" means.

When separating a PDE in $XY$ as you suggested, you have $Y''/Y=-X''/X = const$ which is the separation constant. It seems like we always assume it to be a real number.

I've never seen this full development, but I'm sure it's been published somewhere. My inclination would have been to solve the problem numerically.
Chet

You're probably right, and this is where the equation gets messy anyways, as substituting $\omega$ into $\nabla^2 \psi = \omega$ yields $X''Y+Y''X = f_1(x) + f_2(y)$ which doesn't look separable. Would we let $\psi(x,y) = \phi(x,y)+g_1(x)+g_2(y) \implies \nabla^2\psi+g_1''(x)+g_2''(y)=f_1(x)+f_2(y)$ and then let $g_1''=f_1$ and $g_2''=f''_2$ which yields $\nabla^2\phi(x,y)=0$?

 

[Chestermiller]

I'm having second thoughts on your equation for $\omega$. Shouldn't it be a function of x times a function of y, not the sum of such functions?

cosh cos, cosh sin, sinh cos, sinh sin

 

[member 428835]

I'm having second thoughts on your equation for $\omega$. Shouldn't it be a function of x times a function of y, not the sum of such functions?

cosh cos, cosh sin, sinh cos, sinh sin

Shoot, you're totally right. Then $\nabla \cdot \psi = \omega$ is not separable. I can't think of a way to analytically solve. How would you do this numerically? Finite difference $\nabla^4 \psi=0$?

 

[Chestermiller]

Shoot, you're totally right. Then $\nabla \cdot \psi = \omega$ is not separable. I can't think of a way to analytically solve. How would you do this numerically? Finite difference $\nabla^4 \psi=0$?

No. Solve the equations involving both omega and psi. The only tricky part is numerically specifying the boundary conditions on omega. But I know how to do that.

 

[member 428835]

Yea, how do you do that?