Astiff, 15-cm-long tube with an inner diameter of 2.5 mm is attached to a small hole in the side of a tall beaker. the tube sticks out horizontally. the beaker is filled with 20∘c water to a level 45 cm above the hole, and it is continually topped off to it maintain that level. what is the volume flow rate through the tube?

Volume rate of flow through the pipe is 1.457 × 10⁻⁵ m³/s

Explanation:

Given:

Inner diameter of the tube, d = 2.5 mm = 2.5 × 10⁻³ m

water level above the hole, h = 45 cm = 0.45 m

now,

the velocity of the water at the level of hole, v = √(2gh)

thus,

v = √(2 × 9.8 ×.45)

or

v = 2.969 m/s

Now,

the discharge (Q) is given as:

Q = Area × velocity

area of the outlet  = πd²/4

or

area = π(2.5 × 10⁻³)²/4 = 4.90 × 10⁻⁶ m²

on substituting the values in the formula for discharge, we get

Q = 4.90 × 10⁻⁶ m² × 2.969 m/s

or

Q = 1.457 × 10⁻⁵ m³/s

hence, the volume rate of flow through the pipe is 1.457 × 10⁻⁵ m³/s

falling water hydro = water .. spectcular places these plants are

conservation of mass

mass before something happens = mass afterwards