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Bernoulli's Principle of a pipe

  1. Apr 28, 2005 #1
    I have two problems that I can not figure out. Thanks for your help.

    Problem #1: A horizontal pipe 10.0 cm in diameter has a smooth reduction to a pipe 5.00 cm in diameter. If the pressure of the water in the larger pipe is 8.00 x 10^4 Pa and the pressure in the smaller pipe is 6.00 x 10^4 Pa, at what rate does water flow through the pipe?

    Problem #2: A large storage tank, open at the top and filled with water, develops a small hole in its side at a point 16.0 m below the water level. If the rate of flow from the leak is equal to 2.50 x 10^-3 m^3/min, determine (a) the speed at which the water leaves the hole and (b) the diameter of the hole.
  2. jcsd
  3. Apr 28, 2005 #2


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    #1: You'll have 2 equations with 2 unknowns. The first will be Bernoulli equation. The second will involve the fact that the volumetric flow rate through the pipe is constant.

    #2: Use the surface of the water in the tank as a point. Open at the top is your hint on this one (V=0, p=0). Pt a) Use Bernoulli to calculate the velocity. Pt b) Use the given flow rate to calculate the area and then the diameter.
  4. Apr 28, 2005 #3
    Two unknowns? There is V1 and V2 in the Bernoulli's equation. Then there is rate. Or am I doing something wrong?

    In the second problem when you say p=0, that's the density, correct? If so, then what is P1 and P2?
    Last edited: Apr 28, 2005
  5. Apr 28, 2005 #4
    Ok you know that
    [tex]P_1+\frac{1}{2} \rho_1 V_1^2+\rho_1 g h_1 = P_2+\frac{1}{2} \rho_2 V_2^2+\rho_2 g h_2[/tex]


    [tex]Q_1 = Q_2 = V_1 A_1=V_2 A_2[/tex]

    Because of steady flow

    So you have the P1 and P2 you know the density and the height I assume would be the same. Therefore from Bernoulli's Equation you have the two unkowns V1 and V2 however, you can solve the second equatoin for either V1 or V2 and because you know the diameter you know the the area
    [tex] A= \frac{\pi}{4} D^2[/tex]

    For Part two the p fred was referring to is pressure I like to use big p (P)
    P=0 becaues P=Patm or 0 gauge. Therefore if you are calling the top of the tank point 1 then P1 and V1 are both 0. Now call your hole point 2. This is a free jet so again P2=Patm=0 gauge. Therefore the only engery coming giving the fluid its velocity is the change in height. Depending upon where you select your coordinate system will determine what your h1 and h2 terms are. After you do that you can easily solve Bernoulli's equation and then use the relation Q=VA to find the diameter of the hole.
  6. Apr 29, 2005 #5


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    and I didn't even that last post supersix. I just went ahead and solved it not knowing you already clearly laid it out for him, haha.

    One thing to point out. Since you do have two unknowns in the first equation (v1 and v2 both being unknown), simply solve the equation for either of those. Now you have an equation of say v2 in terms of v1. Now you can plug that into your Q formula to solve for v1.
  7. Apr 29, 2005 #6


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    Sorry for being a bit on the cryptic side. I was hoping to give you general direction and then you'd get the idea on the approach for solving the problems. I didn't want to make it more confusing for you.
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