# I Second derivative of friction force question

1. Apr 29, 2017

### fahraynk

I'm studying boundary layers. I am confused by what I am reading in this book.

The book says the friction force (F) per unit volume = $$\frac{dF}{dy}=\mu\frac{d^2U}{dy^2}$$
They say $$\frac{dU}{dy}=\frac{U_\infty}{\delta}$$

This makes sense to me, delta is the thickness in the y direction, and the velocity is 0 at the airfoil (or whatever object) and it equals the freestream velocity after the boundary thickness delta. so du/dy being proportional to the free stream velocity divided by the thickness makes sense.
But then they say this : $$\frac{d^2U}{dy^2} =\mu\frac{U_\infty}{\delta^2}$$

Can someone explain why the second derivative is proportional to delta^2 ?
Is it that, delta represents the thickness y, and so the second derivative of 1/y = -1/y^2, and we just ignore the negative?

#### Attached Files:

File size:
12.6 KB
Views:
24
2. Apr 29, 2017

Well for one, friction force per unit volume is not $dF/dy$ as you have written it. It is $\partial \tau/\partial y$. The difference is important, because $F$ is a force, and $\tau$ is a stress. The units don't work out if you use $F$.

So then the friction force per unit volume is
$$\dfrac{\partial \tau}{\partial y} = \mu\dfrac{\partial^2 u}{\partial y^2}.$$
So, if you've already come to terms with the fact that
$$\dfrac{\partial u}{\partial y} \sim \dfrac{U_{\infty}}{\delta},$$
Then the same logic leads to
$$\dfrac{\partial \tau}{\partial y} = \mu\dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial y}\right) \sim \mu\dfrac{\partial}{\partial y}\left(\dfrac{U_{\infty}}{\delta}\right) \sim \mu\dfrac{U_{\infty}}{\delta^2}.$$

3. May 3, 2017

### fahraynk

Ah okay, I think I figured it out.
I see how $$\frac{\partial u}{\partial y} = \frac{U\infty}{\delta}$$ because... it U=0 at the boundary and U_infty after a distance delta, so it should be proportional.

So I think :
$$\frac{\partial^2 u}{\partial y^2} = \frac{\frac{\partial u}{\partial y}_\infty}{\delta} = \frac{U\infty}{\delta^2}$$ Or basically du/dt/delta = u_infty/delta^2
Is this the correct interpretation?

4. May 3, 2017

You got it.