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Help Understanding Volume Flow Rate / Bernoulli's

  1. Oct 30, 2011 #1
    Hey guys, just have a question cuz i'm having a little trouble understanding like how exactly to use the volume flow rate with bernoulli's equation. (wasnt sure which section to put this in but is more of a concept question using hw examples)

    Like the volume flow rate i understand A1v1=A2v2
    for example i was doing this problem : https://www.physicsforums.com/showthread.php?t=80584

    Well its inversely proportional right but (im just being stupid probably cuz i'm really tired) but if A2 is 1/4 the size of A1 for example, then v2 is going to be 4 x the size of V1? like

    A1v1 = 1/4A2v2 so 4A1/v1 = A2v2 then v2 = 4(A1/A2)(v1), well like in the problem i linked, what happens to that A1/A2? why does it disappear when being plugged into bernoulli's equation?

    Also another question is the A = cross sectional area right? whats that even mean like is that the same as surface area, or regular area, or what? because in a problem like this https://www.physicsforums.com/showthread.php?t=65810 (now im not sure if the member solved it 100% correct) but the person here puts Volume Flow Rate = pir^2v1 for example, what is the volume flow rate in this case i dont understand and is A just the area of a circle on the tube , even tho its open its not used like a cylinder's area?

    thanks if anyone can help me out im like brain dead at the moment.
     
  2. jcsd
  3. Oct 30, 2011 #2
    If [itex]A_2[/itex] is 1/4 [itex]A_1[/itex] then yes, [itex]V_2[/itex] will be 4 times [itex]V_1[/itex].

    I'm not really sure what you mean when you say it "disappears?"

    Cross sectional area is the area of a cross section. Say you have a piece of pipe. Stand the pipe on it's end on top of a piece of paper, then trace the inside of the pipe onto the paper. Find the area of that circle ([itex]A=\pi r^2[/itex]) and that's the cross sectional area.

    Volumetric flow rate is a volume per unit time (e.g., [itex]\frac{m^3}{sec}[/itex]) If you do unit analysis you'll see that cross sectional area (e.g., [itex]m^2[/itex]) times fluid velocity (e.g., [itex]\frac{m}{sec}[/itex]) gives you volumetric flow rate.
     
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