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Why is Resistance in Fluid Flow inversely related to r^4 rather than just r^2?

  1. Oct 28, 2014 #1
    So Volume Flow Rate (Q) = (P2-P1)/R where R is the total resistance of the system.

    R is directly proportional to Length and inversely proportional to surface area, and the inherent resistance (viscosity) of the fluid. But R =nL/r^4. r^4 rather than r^2.

    So there has to be another factor other than just surface area, that is dependent on radius^2, affecting resistance right? What am I missing?

    In terms of Volume Flow Rate Q, i can understand, since Q = Av and if you change r, you change not only the surface area but also the velocity since the pipe is larger so there is less resistance. A = pi*r^2 AND v is related to radius by a squared factor.

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    Last edited: Oct 28, 2014
  2. jcsd
  3. Oct 28, 2014 #2


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    Staff: Mentor

    The second part of your post answers the question of the first part.
    The pressure difference is not related to r here, so every r-dependence of Q will appear in R as well (just taken as inverse), for exactly the same reasons.
  4. Oct 28, 2014 #3


    Staff: Mentor

    A simple units analysis shows that r^2 doesn't have the correct units and r^4 does.

    As r increases, the velocity for a given flow decreases by a factor of r^2. As r increases, the shear rate for a given velocity decreases by a factor of r. As r increases, the force for a given pressure gradient increases by a factor of r^2. Finally, as r increases the surface area (proportional to viscous drag) increases by a factor of r.
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