Mathematical Procedure for Obtaining Velocity Profile in Laminar Flow

[Andrea Vironda]

Hi guys,
i'm trying to find the velocity profile for a laminar flow in a round pipe.
Starting from a force balance, we can obtain the first equation high in the left. I started with a procedure but i think I'm making mistakes.
Can you suggest me the mathematical procedure?

 

[stockzahn]

Hi guys,
i'm trying to find the velocity profile for a laminar flow in a round pipe.
Starting from a force balance, we can obtain the first equation high in the left. I started with a procedure but i think I'm making mistakes.
Can you suggest me the mathematical procedure?

Shouldn't the force balance look like this:

$$\underbrace{-\mu\frac{du}{dr}}_{\tau}\underbrace{2\pi r}_{P} = \frac{\partial p}{\partial x} \underbrace{\pi r^2}_{A}$$

- no gravity
- no acceleration
- no viscous normal stresses

 

[Andrea Vironda]

Hi,
i considered a ring-shaped volume element, so $$(2\pi r dr P)_{x}-(2\pi r dr P)_{x+dx}+(2\pi r dr \tau)_{r}-(2\pi r dr \tau)_{r+dr}=0$$ I divided by $2\pi r dr dx$ and took the limit $dr, dx \to 0$

 

[stockzahn]

Hi,
i considered a ring-shaped volume element, so $$(2\pi r dr P)_{x}-(2\pi r dr P)_{x+dx}+(2\pi r dr \tau)_{r}-(2\pi r dr \tau)_{r+dr}=0$$ I divided by $2\pi r dr dx$ and took the limit $dr, dx \to 0$

I don't understand your 3rd and 4th term, I would have written them as

$$(2\pi r dx \tau)_{r}-(2\pi r dx \tau)_{r+dr}=0$$

since the shear stresses act on the surface of the sleeve, not on the front and back.

 

[Chestermiller]

Hi guys,
i'm trying to find the velocity profile for a laminar flow in a round pipe.
Starting from a force balance, we can obtain the first equation high in the left. I started with a procedure but i think I'm making mistakes.
Can you suggest me the mathematical procedure?

Your 2nd integration was done incorrectly. First you solve for du/dr:
$$\frac{du}{dr}=\frac{r}{2\mu}\frac{dp}{dx}+\frac{C_1}{r}$$du/dr is zero at r = 0 so $C_1=0$. So you now have:
$$\frac{du}{dr}=\frac{r}{2\mu}\frac{dp}{dx}$$
What do you get when you integrate that, subject to the boundary condition u = 0 at r = R?

 

[Andrea Vironda]

I don't understand your 3rd and 4th term, I would have written them as

$$(2\pi r dx \tau)_{r}-(2\pi r dx \tau)_{r+dr}=0$$

since the shear stresses act on the surface of the sleeve, not on the front and back.

Yup, you're right

Your 2nd integration was done incorrectly. First you solve for du/dr:
$$\frac{du}{dr}=\frac{r}{2\mu}\frac{dp}{dx}+\frac{C_1}{r}$$du/dr is zero at r = 0 so $C_1=0$. So you now have:
$$\frac{du}{dr}=\frac{r}{2\mu}\frac{dp}{dx}$$
What do you get when you integrate that, subject to the boundary condition u = 0 at r = R?

I obtain $$\Delta u=\frac{r^2}{2}\frac{1}{2\mu}\frac{dP}{dx}+C_2$$ how can i arrange accourding to boundary conditions? is $\Delta u=u(R)-u(0)$?

 

[Chestermiller]

Yup, you're right
I obtain $$\Delta u=\frac{r^2}{2}\frac{1}{2\mu}\frac{dP}{dx}+C_2$$ how can i arrange accourding to boundary conditions? is $\Delta u=u(r)-u(0)$?

When you have a constant of integration, there is no need to write $\Delta u$. You simply can write u:
$$u=\frac{r^2}{2}\frac{1}{2\mu}\frac{dP}{dx}+C_2$$To determine the constant of integration $C_2$, you simply make use of the non-slip boundary condition at the wall u(R)=0, where R is the radius of the tube.

 

[Andrea Vironda]

When you have a constant of integration, there is no need to write $\Delta u$. You simply can write u

Ok the procedure is clear, but why i shouldn't write as $\Delta u$?

 

[Chestermiller]

Ok the procedure is clear, but why i shouldn't write as $\Delta u$?

Because you didn't use a $\Delta r$ on the other side of the equation. I guess what you did is OK if you say (as you suggested) that $\Delta u=u(r)-u(0)$, but then you could have moved u(0) over to the other side of the equation and absorbed it into $C_2$.