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Fluids problem - gauge pressure

  1. Aug 9, 2007 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    Water flows steadily from an open tank. The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 0.0480 m[tex]^2[/tex]; at point 3 it is 0.0160 m[tex]^2[/tex]. The area of the tank is very large compared with the cross-sectional area of the pipe.

    Part 1- Assuming that Bernoulli's equation applies, compute the discharge rate in cubic meters per second.

    Part 2- What is the gauge pressure at point 2


    2. Relevant equations
    [tex]A_1v_1=A_2v_2[/tex]
    [tex]p_1+\rho gh_1+\frac{1}{2}\rho v_1^2=p_2+\rho gh_2+\frac{1}{2}\rho v_2^2[/tex]


    3. The attempt at a solution

    I have already calculated the discharge rate (at point 3) to be 0.200 m[tex]^3[/tex]/s which I know to be correct but I am stuck with the second part. I used Bernoulli's equation with points 2 and 3 and came up with an answer of 17.8 Pa but this appears to be wrong. I took [tex]h_2=h_3=0[/tex] to simplify Bernoulli's equation to get
    [tex]p_3-p_2=\frac{1}{2}\rho v_2^2-\frac{1}{2}\rho v_3^2[/tex]
    and used the volume flow rate equation to get the velocity at point 2 to be 0.067m/s and then used that in Bernoulli's equation to get the answer I have, am I doing something wrong here or should I even be using Bernoulli's equation to solve this part of the question. [tex]p_3-p_2[/tex] should be equal to the gauge pressure shouldn't it? Any guidance whatsoever would be greatly appreciated.
     
  2. jcsd
  3. Aug 9, 2007 #2

    Doc Al

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    Staff: Mentor

    Revisit this calculation.
     
  4. Aug 9, 2007 #3
    Oh, thank you, I accidentally put in the volume flow rate in as v instead of the velocity of the fluid, I have the correct answer now, Thanks.
     
  5. Oct 29, 2007 #4
    I still don't understand... If v_2 = 4.17 and v_3 = 12.5, then what is this p_3 that I need in order to solve for p_2?? (I also know that rho=density of water=1000 kg/(m^3))
     
  6. Oct 29, 2007 #5

    Doc Al

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    Staff: Mentor

    Note that point 3 is in an open stream.
     
  7. Oct 29, 2007 #6
    Then p_3 is atmospheric pressure .. is it 101.3 kPa? so then p_2 = -(.5(1000(4.7^2-12.5^2)))-101300) = 168380 Pa?
     
  8. Oct 29, 2007 #7
    The question asked for gauge pressure, which is the pressure at 2 minus atmospheric pressure given by p_2-p_3 in the top equation, which is 6.97*10^4. So yeah, the absolute pressure at 2 would be the gauge pressure plus atmospheric which is pretty much what you got.
     
  9. Oct 29, 2007 #8
    OHHHH i finally get it!
     
    Last edited: Oct 30, 2007
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