Question

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?

Answer 1

1). 340.41N/m^2

2). 0.743_m/s

3). 12.29_m/s

4). 0.168_m^3/s

Step-by-step 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