1. Not finding help here? Sign up for a free 30min 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!

Pressure and Flow Rate in a Tank

  1. Sep 27, 2015 #1
    1. The problem statement, all variables and given/known data
    WjsHyUr.png
    A tank of cross-sectional area A is initially filled with fluid of density ρ and viscosity μ to height hi. The pressure above the fluid in the tank is atmospheric, Patm. At the base of the tank there are two pipes, which are both open to teh atmosphere, such that fluid flows out of them. The pipe on the left has diameter D1, length L1, and flow rate Q1. The pipe on the right has diameter D2, length L2, and flow rate Q2.
    A. Write an expression for the initial pressure that drives the flows through pipes 1 and 2.
    B. What is the ratio of the initial flow rates Q1/Q2?
    C. What is the ratio of the initial Reynold's numbers for the flow in the two pipes, Re1/Re2?
    D. Where is the shear force lowest and highest in the pipes? (a. At the wall, b. At the center line, c. Where the pipe meets the tank, d. Where the fluid flows out of the pipe to the atmosphere)
    2. Relevant equations
    0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2
    Q =(ΔPπR4)/(8μL) = Au
    Re = ρuD/μ

    3. The attempt at a solution
    A. Pinit = ρg(hi) + (Patm)

    B. Q1 =(ΔPπ(D1/2)4)/(8μL1)

    Q2 =(ΔPπ(D2/2)4)/(8μL2)

    Q1/Q2 = (D1/D2)4(L2/L1)

    C. When I got to this one, I realized that I don't know how to find the velocity of the fluid, since I had assumed it to initially be zero, so I probably did A and B wrong.
    Maybe I'm supposed to get it from Bernoulli's equation:
    0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2

    u = √((-2/ρ)(ΔP + ρgΔh))
    u = √((-2/ρ)(Patm + ρg(hi)))

    But I don't know why this would be the case, and it also isn't taking into account he difference in area or length between the pipes. Could you please send me in the right direction? Am I missing an equation?

    D. I think the shear stress would be highest where the pipes meet the tank and lowest at the center line, but I don't know why.
     
  2. jcsd
  3. Sep 27, 2015 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You forgot to write one seemingly trivial, but essential equation: the one for continuity of flow, Q = A * V.

    Since the problem specified that the flow rate at point 1 was Q1, you can calculate the velocity at this point. (BTW, a velocity of zero suggests no flow is occurring.)
     
  4. Sep 27, 2015 #3
    With regard to question C, in my judgement, they are expecting you to assume fully developed quasi-steady flow in the pipes, with the flow rate given by the Hagen-Poiseulle equation.
     
  5. Sep 27, 2015 #4
    If flow rate Q = Area * velocity, Q = Au, then A can be redone...

    A. 0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2
    Pinit = Patm + ρg(hi) + (1/2)ρ(Δuavg)2
    where uavg = (1/2)(u1 + u2)
    u1 = Q1/A1 = Q1/(π(D1/2)2)
    u2 = Q2/A2 = Q2/(π(D2/2)2)

    Would this Pinit value be correct?

    Then for part B, do I just let the ΔP1 = ρg(hi) + (1/2)ρ(Q1/A1)2
    and ΔP2 = ρg(hi) + (1/2)ρ(Q2/A2)2, keeping the rest the same? And then use the velocities from part A to calculate the Reynold's numbers?
     
  6. Sep 27, 2015 #5
    Bernoulli's equation should not be applied to this problem, unless you include the frictional pressure drop in the equation.

    Chet
     
  7. Sep 27, 2015 #6
    How do I find pressure without Bernoulli's equation? I've never seen a frictional pressure drop term in Bernoulli's equation; what would that look like?
     
  8. Sep 27, 2015 #7

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

  9. Sep 27, 2015 #8
    So the Bernoulli equation is now:
    0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2 + fD (L/D)(1/2)ρ(uavg)2
    Should the lengths and diameters be averaged too, and then plug in the velocities from my last post to calculate Pinit?
     
  10. Sep 27, 2015 #9

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    No, they shouldn't. You figure out the friction losses in each section of pipe which has a constant diameter. The friction losses don't scale in a linear fashion as diameter changes, unfortunately.

    I think incorporation of friction losses into this problem has exceeded the scope of what was intended. There's too much background knowledge which is being glossed over in the application of Darcy-Weisbach.
     
  11. Sep 27, 2015 #10
    I respectfully disagree. I think that once they mentioned viscosity in the problem statement, they expected the students to include friction losses. However, in my judgement, they probably intended for the students to neglect inertia (i.e., the kinetic energy term in Bernoulli) so that the pressure drop was determined solely by the frictional loss (i.e., Hagen-Poiseulle).

    Chet
     
  12. Sep 27, 2015 #11
    This problem came from a practice exam recommended by my professor that belongs to a course different from, but similar to, what he taught us, so I can't actually speak for the original intentions of this problem. Since I hadn't been taught about friction, I'm fine with keeping it out. I don't mind keeping it in either though, as long as I know that I'm approaching the problem correctly and using the formulas correctly.

    It sounds like if I want to keep the friction term, I should add it once for each of the pipes:
    0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2 + fD(L1/D1)(1/2)ρ(u1)2 + fD(L2/D2)(1/2)ρ(u2)2

    If friction isn't included then is the below formula correct, or is there a different issue?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Pressure and Flow Rate in a Tank
Loading...