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Water flowing through an eavestrough

  1. Apr 14, 2014 #1
    1. The problem statement, all variables and given/known data

    6. During a storm, water flows off a roof into an eaves trough and then down from a 5.0 cm diameter hole in the bottom of it. The water in the trough is 2.5 cm deep. (Torricelli’s law & Bernoulli’s equation )

    a. What is the speed (velocity) of the flow as it leaves the bottom of the trough?

    b. What is the speed of the flow just before it hits the ground, 4.0 m below the hole?

    c. What is the diameter of the flow there?

    d. What is the volumetric and mass flow rate?

    2. Relevant equations

    Equation 1
    [tex]v = \sqrt{2gh}[/tex]

    Equation 2
    [tex]P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2[/tex]

    3. The attempt at a solution

    Parts A and B seem simple enough, just use equation 1 (would h for part B be 4.025m), but I'm not sure how Bernoulli's equation comes into play. There is the continuity equation A1V1 = A2V2 that relates diameter and velocity, but the question states that this is a Torricelli’s law & Bernoulli’s equation problem. Do I maybe sub that equation into Bernoulli's equation? Part D seems like it follows pretty easily from the others so I don't really need help on that, but the others have me confused.

    Any help is appreciated, thanks!
     
  2. jcsd
  3. Apr 14, 2014 #2
    Tortecelli is just a special case of Bernoulli's, so just pick one or the other.

    And if you solve for part D), doesn't this give you the answer for Part C)?
     
  4. Apr 14, 2014 #3
    Okay, thanks. Would you mind taking a look at what I got?

    A) [tex]v = \sqrt{2gh} = \sqrt{2(9.8)(.025)} = .7\frac{m}{s}[/tex]

    B) [tex]v = \sqrt{2gh} = \sqrt{2(9.8)(4.025)} = 8.882\frac{m}{s}[/tex]

    C) [tex]A_2 = \frac{A_1 V_1}{V_2} = 38.69\times 10^{-6}[/tex]

    D) Volumetric flow rate = Velocity x Area = 343x10^-6

    Mass flow rate = density x velocity x area = .344kg/s
     
  5. Apr 14, 2014 #4
    I didn't check your numbers but:
    a) looks right
    b) small mistake/omission- maybe use Bernoulli to see where
    c) question asked for diameter
    d) only off because of mistake in b)
     
  6. Apr 14, 2014 #5
    Thanks. For b), is the mistake that I forgot to consider the velocity as it enters the hole so that the equation is

    [tex]v_2 = \sqrt{v_1^2 + g h_1}[/tex]

    where h_1 = 4m?
     
  7. Apr 14, 2014 #6
    Yeah, that's it. I think that shows why using bernoulli is in general a better approach for most of these kinds of problems.
     
  8. Apr 15, 2014 #7

    BvU

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    Science Advisor
    Homework Helper
    Gold Member

    Hello K,

    No one notices that this is exactly the same as before with h_1 = 4.025 m ?

    And: You want to check your numbers. Don't you become even a little suspicious if the are drops to 40 e-6 ? The speed ratio (= area ratio) is around 12 ! And yes, they ask for diameter!
     
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