Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Where does this equation come from?

  1. Aug 24, 2011 #1
    Where does this equation come from??

    Hi,

    In a text book 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 ??
     
  2. jcsd
  3. Aug 24, 2011 #2

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    Re: Where does this equation come from??

    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:
    [tex]\frac{dP}{dz} = \mu \frac{1}{r} \frac{d}{dr} \left( r \frac{d u_z}{dr} \right)[/tex]
    This is easily integrable.
    [tex]\frac{1}{\mu} \frac{dP}{dz} \int r \; dr = r \frac{d u_z}{dr}[/tex]
    [tex]\frac{1}{\mu} \frac{dP}{dz} \left( \frac{r^2}{2} + C_1 \right) = r \frac{d u_z}{dr}[/tex]
    [tex]\frac{1}{\mu} \frac{dP}{dz} \int \left( \frac{r}{2} + \frac{C_1}{r} \right) = u_z[/tex]
    [tex] u_z(r) = \frac{1}{\mu} \frac{dP}{dz} \left( \frac{r^2}{4} + C_1 \ln r + C_2 \right)[/tex]

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

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

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

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

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

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

    Note that the negative sign reflects that a negative pressure gradient will move the flow in the positive [itex]z[/itex] direction and in this derivation, [itex]z[/itex] replaces [itex]x[/itex] in your equations.
     
  4. Aug 25, 2011 #3
    Re: Where does this equation come from??

    Thanks boneh3ad!!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Where does this equation come from?
Loading...