Question

A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from the bottom of the tank. The water flows out of the small hole into an empty, 0.2 m-diameter container placed below the tank. The student uses a garden hose to keep the water level in the tank at the original height.

h1 = 0.50 meters, h2 = 0.03 meters. The garden hose used to keep the tank full has a diameter of 1 inch which is exactly 0.0254 meters

Required:
a. What is the velocity of the water as it leaves the hole near the bottom of the tank?
b. What is the volume flow rate of water from the garden hose into the tank?
c. After 60 seconds of operation, what is the water pressure at the bottom of the container that is being used to catch the water leaving the tank?

Answer 1

Step-by-step explanation:

a )

Depth of hole from surface of water d = .50 m - .03 m = .47 m

velocity of efflux v = √ 2gd

v = √ (2 x 9.8 x .47 )

v = 3.03 m /s

b )

Volume flow rate = π R² v where R is radius of hole at the bottom .

= 3.14 x ( .005 ) ² x 3.03 m/s

= 2.378 x 10⁻⁴ m³ /s

c )

Volume of water collected in 60 s

= 2.378 x 10⁻⁴ x 60

= 1.4268 x 10⁻² m³

If height attained in collecting container be h

π R² h = 1.4268 x 10⁻² m³ where R is radius of container

3.14 x ( .1 )² x h = 1.4268 x 10⁻²

h = .4544 m .

Pressure at the bottom of container = hρ g

where h is height of water , ρ is density of water

Pressure = .4544 x 1000 x 9.8 N /m²

= 4453.12 N /m²