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Continuity Equation problem

  1. Jul 25, 2009 #1

    jdg

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    Next question: A garden hose with internal diameter of 13.5 mm lies flat on a sidewalk while water is flowing in it at a speed of 6 m/s. A person happens to step on it at the very edge of the opening of the hose and decreases its internal diameter by a factor of 9

    So D (1) = 0.0135m
    r (1) = 0.00675m
    D (2) = 0.0135/9 = 0.0015m
    r (2) = 0.00075m
    A (1) = pi*r^2 = (3.14...)(0.00675^2) = 1.4134...e-4
    A (2) = pi*r^2 = (3.14...)(0.00075^2) = 1.767...e-6

    1. What was the water flow rate in the hose prior to the person stepping on it?
    - I got this part: J(1) = A(1)V(1) = 8.59 m3/s

    2. What is the flow rate of water after the person steps on it?
    part 2 I did J = A(1)V(1) = A(2)V(2):

    So V2 = V1*(A1/A2) = 486 m/s

    Is this right?


    3. What is the speed of the water just as it exits the hose after the person steps on it?

    And for part 3 I did

    J = (A2)(V2) = 8.59e-4 m3/s
     
  2. jcsd
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