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Bernolli's Equation: vb^2-va^2 = ?

  1. Nov 3, 2012 #1
    Bernolli's Equation: vb^2-va^2 = ?????

    1. The problem statement, all variables and given/known data:

    Use Bernoulli’s principle to obtain a relation between the speeds of water at points A and B, vA and vB, in terms of the acceleration of gravity g and the height of the water column h. Your expression should be of the form: vb^2-va^2 =
    Let’s do a numerical check that your equation is correct. If the height of the water level is h = 27 cm, what is the value of vb^2-va^2, in SI units?

    2. Relevant equations



    3. The attempt at a solution
    I really have no idea where to start, to me, the question doesn't seem really obvious. I just need this last problem solved to be permitted in to lab on Monday morning. Please help me in any way, thank you so much!
    This is what I have as bernolli's equation in my notes, but how do I use any of this to solve for vb^2-va^2????
    P1+1/2pv1^2+pgy = P2+1/2pv2^2+pgy
    I know P=pressure
    p=density
    y= height
    Please please please help!!!
     
  2. jcsd
  3. Nov 3, 2012 #2
    Re: Bernolli's Equation: vb^2-va^2 = ?????

    You don't know the algebraic manipulations required to move all terms with a "v" in it to one side, and all the terms without "v" in it to the other? Also, some context (read: a picture) of the problem would be appreciated.
     
  4. Nov 3, 2012 #3
    Re: Bernolli's Equation: vb^2-va^2 = ?????

    Assume P1 = P2, and let va = v1 and vb=v2. Incidentally, the y's on the two sides of the equation should have subscripts.
     
  5. Nov 3, 2012 #4
    Re: Bernolli's Equation: vb^2-va^2 = ?????

    Thank you Chestermiller, I feel really stupid that I can't do simple algebra.
    I think based on P1=P2 you could just take that out of the equation.
    To move pgy you would subtract from both sides so you'd be left with:
    1/2pva^2 = 1/2pvb^2
    How would I get va^2 to the right side of the equation. Wouldn't I have to move it by dividing both side by it? So I wouldn't be left with vb^2 - va^2 but instead vb^2/va^2.
    What am I doing wrong?
     
  6. Nov 4, 2012 #5
    Re: Bernolli's Equation: vb^2-va^2 = ?????

    Your starting equation should be

    P1+1/2pv1^2+pgy1 = P2+1/2pv2^2+pgy2

    As I said in my previous response, there should be subscripts on the two y's,

    and y2 - y2 should be called h.
     
  7. Nov 4, 2012 #6
    Re: Bernolli's Equation: vb^2-va^2 = ?????

    Okay, so should P1 and P2 be taken out of the equation if they equal eachother?
    If so, you'd be left with 1/2pv1^2+pgy1 = 1/2pv2^2+pgy2
    So, to get the v2^2-v1^2 = ???? as the question asks; how do I get everything else to one side? The farthest I get to is:
    pgy1-pgy2 = 1/2pv2^2 - 1/2pv1^2
    to simplify wouldn't it be:
    y1-y2 = 1/2pv2^2 - 1/2pv1^2
    Which would lead to:
    (y1-y2)/(1/2p * -1/2p) = v2^2-v1^2
    Is that correct?
     
  8. Nov 4, 2012 #7
    Re: Bernolli's Equation: vb^2-va^2 = ?????

    No. Your algebra is wrong.
     
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