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!