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31 May, 21:46

A cylindrical water-storage tank has a height of 7.7 m and a radius of 6.0 m. The storage tank is full of water but is vented to the atmosphere. The bottom of the tank is 26 m off the ground. A 13 cm diameter pipe runs vertically down from the tank and goes 1.0 m underground before turning horizontal. The water flow in the horizontal pipe is 84 L/s.

1 What is the water pressure (gauge) in the horizontal pipe underground?

2 How fast does the water level in the tank drop?

3 bUTch Jones drills a 6.5 mm diameter hole near the bottom of the tank. How fast does the water shoot out?

4 What volume flow of water is lost out the hole?

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Answers (1)
  1. 1 June, 00:16
    0
    1). 340.41N/m^2

    2). 0.743_m/s

    3). 12.29_m/s

    4). 0.168_m^3/s

    Explanation:

    Applying Bernoulli's equation

    P₁ + 0.5ρv₁² + ρgh₁ = P₂ + 0.5ρv₂² + ρgh₂

    the velocity in the 13 cm diameter pipe is given by

    flow rate = 84 L/s

    1). Height=7.7+26+1=34.7_m

    Density = 1_Kg/m^3

    Acceleration due to gravity = 9.81_m/s^2

    Pressure = density*height*gravity

    = 340.41N/m^2

    2). Level of the tank drops by

    Taking the tank as a pipe

    v = pi * r^2 * h = flow rate through the pipe

    = 84_L/s

    r = 6.0_m, h = 7.7_m

    Level drop by 84 / (6*7.7) _m/s

    = 0.743_m/s

    3). Speed of water shoot out v = Sqr (2*g*h) = Sqr (2*9.81*7.7) = 12.29_m/s

    4). Volume flow rate = v * a = 12.29*0.014 = 0.168_m^3/s
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