Where does this equation come from?

[hermano]

Where does this equation come from??

Hi,

In a textbook I found an equation for the volumetric flow rate of air through a channel (tube). I can't find from where the equation is coming, someone an idea?

The equation which I mean is: Q = - pi*d^4/(128*viscosity)* dp/dx

I know that Q = A*v = pi*d^2/4 *v

Also from the simplified equation of motion 'v' can be calculated as:

dp/dx = viscosity * ddv/ddy (dd = second derivative)

from this equation v = ?? if I fill v in, in Q = pi*d^2/4 *v will this give Q = - pi*d^4/(128*viscosity)* dp/dx ??

 

[boneh3ad]

Well, the equation you give is for Poiseuille flow in a pipe. In other words, you need to be using the cylindrical Navier-Stokes equations. These simplify to:
$$\frac{dP}{dz} = \mu \frac{1}{r} \frac{d}{dr} \left( r \frac{d u_z}{dr} \right)$$
This is easily integrable.
$$\frac{1}{\mu} \frac{dP}{dz} \int r \; dr = r \frac{d u_z}{dr}$$
$$\frac{1}{\mu} \frac{dP}{dz} \left( \frac{r^2}{2} + C_1 \right) = r \frac{d u_z}{dr}$$
$$\frac{1}{\mu} \frac{dP}{dz} \int \left( \frac{r}{2} + \frac{C_1}{r} \right) = u_z$$
$$u_z(r) = \frac{1}{\mu} \frac{dP}{dz} \left( \frac{r^2}{4} + C_1 \ln r + C_2 \right)$$

So, we know that $C_1=0$ because at $r=0$, the velocity has to be finite, so the $\ln r$ term has to go away, leaving us with:

$$u_z(r) = \frac{1}{\mu} \frac{dP}{dz} \left( \frac{r^2}{4} + C_2 \right)$$

Now, according to the no slip condition ($u_z(R) = 0$), we have:
$$u_z(R) = 0 = \frac{1}{\mu} \frac{dP}{dz} \left( \frac{R^2}{4} + C_2 \right)$$
So,
$$C_2 = -\frac{R^2}{4}$$
And our final equation for the velocity profile is:
$$u_z(r) = \frac{1}{\mu} \frac{dP}{dz} \frac{r^2 - R^2}{4}$$

To get volumetric flow, you have to use the equation:
$$Q = \int \int V(r, \theta) dA$$

We just showed that the velocity is only a function of $r$, so the equation becomes:
$$Q = \int \int u_z(r) dA$$
Where $dA = r\; dr\; d \theta$ is the differential area of the pipe. Solving that, we get:
$$Q = \int_0^{2 \pi} \int_0^R u_z(r) r\; dr\; d \theta = \int_0^{2 \pi} \int_0^R \left( \frac{1}{\mu} \frac{dP}{dz} \frac{r^2 - R^2}{4}\right) r\; dr\; d \theta$$
This simplifies to:
$$Q = \frac{1}{\mu} \frac{dP}{dz} \int_0^{2 \pi} \int_0^R \frac{r^3 - r R^2}{4} \; dr\; d \theta$$
$$Q = \frac{2 \pi}{\mu} \frac{dP}{dz} \int_0^R \frac{r^3 - r R^2}{4} \; dr$$
$$Q = \frac{2 \pi}{\mu} \frac{dP}{dz} \left[ \frac{r^4}{16} - \frac{r^2 R^2}{8} \right]_0^R$$
$$Q = -\frac{\pi}{\mu} \frac{dP}{dz}\frac{R^4}{8}$$

Now, this is very close to what you had. Yours was in terms of diameter, however, which is $D = 2R$, so you substitute this into our flow rate equation and get your result:
$$Q = -\frac{\pi D^4}{128 \mu} \frac{dP}{dz}$$

Note that the negative sign reflects that a negative pressure gradient will move the flow in the positive $z$ direction and in this derivation, $z$ replaces $x$ in your equations.

 

[hermano]

Thanks boneh3ad!