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Where does this equation come from?

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


    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


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    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]
    [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!!!!
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