# Boundary conditions for Couette flow

I'm trying to reproduce a result from a paper (T. Thatcher, Boundary Conditions for Grad's 13 moment equations, equation (32), page 6), however, I haven't been able to do so. Hopefully someone can provide some help.

The general problem concerns the Couette flow: A flow passes through two infinite parallel plates at distance $L$ move with velocity $v_{w}^{0}$ and $v_{w}^{L}$ ($w$ refers to the wall) relative to each other in their respective planes, each with temperature $\theta_{w}^{0}$ and $\theta_{w}^{L}$. The velocity $v$ of the flow only depends on the $y$ coordinate, that is the coordinate perpendicular to the direction of the flow. After solving the Navier-Stokes-Fourier equations for this problem, you can find the following equations

$$\frac{dv}{dy}=a=\text{constant}, \qquad \frac{d^{2}\theta}{dy^{2}}=-\frac{2\text{Pr}}{5}a^{2} \qquad \qquad (1)$$

which are linearized. Using as boundary conditions

\begin{array} vv(L)-v_{w}^{L} = -\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\; a, \qquad \ \ v(0)-v_{w}^{0} = \frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\; a \qquad (2) \end{array}

and

\begin{array} \theta \theta(L)-\theta_{w}^{L} = -\frac{2-\chi}{\chi}\frac{5\beta_{1}}{4 Pr}\sqrt{\frac{\pi}{2}}\text{Kn}\frac{d\theta}{dy}, \quad \ \ \theta(0)-\theta_{w}^{0} = \frac{2-\chi}{\chi}\frac{5\beta_{1}}{4 Pr}\sqrt{\frac{\pi}{2}}\text{Kn}\frac{d\theta}{dy} \qquad (3) \end{array}

the solution for the velocity should be

$$v = \frac{v_{w}^{L}}{2}+a\left(\frac{y}{L}-\frac{1}{2}\right) \qquad \text{with} \quad a = \frac{v_{w}^{L}-v_{w}^{0}}{1+\frac{2-\chi}{\chi}\sqrt{2\pi} \alpha_{1}\text{Kn}} \qquad \qquad (4)$$

Don't be concerned about $\alpha_{1}$ or $\beta_{1}$; they are correction terms. $\text{Kn}$ and $\text{Pr}$ are, as usual, the Knudsen number and the Prandtl number, respectively.

For now, I'm interested only in the solution for the velocity (equation $(4)$)

Note: First and second attempts are wrong.

First attempt:

Use equation $(2)$ as $v-v_{w}^{L}=-\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\;a L$ and $v-v_{w}^{0}=\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\;a L$. Subtract the first from the second to obtain $a = \displaystyle \frac{v_{w}^{L}-v_{w}^{0}}{\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}L}$, and from equation $(1)$ for the velocity, $v = ay+\text{constant}$, evaluating at the plate localized in $L$ we have $v_{w}^{L}=aL+\text{constant}$, which gives us the constant, and a solution but it's not the correct result.

Second attempt:

I think this can be useful:

Take equation $(2)$ and solve for $a$:

$$a=\frac{-v_{w}^{L}+v_{w}^{0}}{-\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}L}$$

Then we use the first part of equation $(1)$, so we have

$$\frac{dv}{dy}=\frac{-v_{w}^{L}+v_{w}^{0}}{-\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}L}$$

After integrating this equation, we evalute the velocity at the boundary $L$, which allows us to find the constant of integration. Substituting this constant, we obtain the following result

$$v-v_{w}^{L}=\frac{v_{w}^{L}-v_{w}^{0}}{\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}}\left(\frac{y}{L}-1\right)$$

which is

$$v-v_{w}^{L}=a\left(\frac{y}{L}-1\right)$$

Not quite the result I'm looking for, but maybe this is the right path.

Third attempt:

Taking $\displaystyle \frac{dv}{dy}=b$, it follows $v(L)=a L +b$ and $v(0)=b$. Using the boundary conditions to evalute $v(L)$ and $v(0)$ we obtain

$$v_{w}^{0}+\displaystyle\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}a = b$$

and

$$v_{w}^{L}-\displaystyle \frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}a = aL+b$$

After solving this system of equations:

$$a = \frac{v_{w}^{L}-v_{w}^{0}}{L+\text{Kn}\sqrt{2\pi}\alpha_{1}(\frac{2-\chi}{\chi})}$$ $$b = v_{w}^{0}-\frac{\left(v_{w}^{0}-v_{w}^{L}\right)\text{Kn}\sqrt{\frac{\pi}{2}}\alpha_{1}\left(\frac{2-\chi}{\chi}\right)}{L+\text{Kn}\sqrt{2\pi}\alpha_{1}\left(\frac{2-\chi}{\chi}\right)}$$

Substituting back in the original equation

$$v(y)-v_{w}^{0}=a\left(y-\text{Kn}\sqrt{\frac{\pi}{2}}\alpha_{1}\left(\frac{2-\chi}{\chi}\right)\right)$$

• Do you have a typo in equation (1)? Right now the second part of (1) involves the second derivative of theta with respect to theta...should one of those thetas be a y? – Greg P Jan 12 '11 at 22:57
• @Greg: Of course, thanks. It's fixed now. As I said, I only want to obtain the solution for the velocity. I added the temperature equation to make sense of the problem. – Robert Smith Jan 12 '11 at 23:01
• Two more typos in (4): The expression for v must be wrong since you are subtracting 1/2 (unitless) from y (unit of length). And the expression for a does not match that of the paper. More importantly, it would be better to write the boundary conditions as equations involving v(0) and v(L) instead of just a single quantity V (see answer). – Greg P Jan 12 '11 at 23:20
• @Greg: Fixed. It was more or less a typo. The solution I had in (4) is actually correct but taking the plate at 0 as the frame of reference. However, to maintain coherence, I will change it. – Robert Smith Jan 12 '11 at 23:25
• In your update, you are making another mistake. In the first step, you are applying the boundary condition incorrectly. You must distinguish between v(0) and v_{w}^0. That is, the velocity of the fluid right at the wall is not equal to the velocity at the wall. – Greg P Jan 12 '11 at 23:29

Update: There is a typo in the paper. This is clear since the solution given in the paper does not obey $dv/dy = a$ but rather $dv/dy=a/L$. I believe the correct solution has (y-L/2) rather than (y/L - 1/2).
• After following your advice, I found a solution which satisfies the equation $(1)$ but doesn't seem so neat as the proposed (even with the modification you expected). I will post an update with this. – Robert Smith Jan 13 '11 at 2:17