1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why is fluid velocity unaffected by a change of pipe roughness here?

  1. Apr 15, 2013 #1
    I had no luck in the coursework forums though I guess the question becomes simpler if I state it the way I did here.

    Assuming no complex network, or just a single pipe first feeding the network, why do the speeds in the pipe and even the rest of the network pipes remain unaffected if I change the roughness of that first pipe? The pressures do change.

    I guess I need the basic answer which I'm sure is simple so I can then expand to a very rational explanation that will make me remember it because the way I think of it now, it's a bit unclear how roughness can keep velocities unaffected (in a non-part-of-a-branch, single pipe scenario).
  2. jcsd
  3. Apr 15, 2013 #2
    OK I got an answer from somewhere, but I still can't get it to hold still in my head.

    The answer is basically that since Q = V * A then V remains the same since Q is the same (and A).

    Now, how can I make that make sense when also thinking of roughness as irrelevant?

    edit: Basically how can Q remain the same? hrm.. I'm probably trapped in a circular logic that is totally wrong but I'm not sure exactly how.
  4. Apr 15, 2013 #3
    Now I'm thinking of Darcy–Weisbach et al. complex equations and I can't figure out how while they do include coefficients for friction/roughness, they just take velocity for granted as stable (if the above is true).
  5. Apr 15, 2013 #4
    Ah, I think I got somewhere. In the simulations I was running, I guess whatever output is entered, "it must be supplied", hence Q is taken as a hard-constant no matter what hence V is satisfied to a constant based on that, provided same cross section.

    Or at least that's how far I got.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook