Question

Consider a stream of water falling freely at atmospheric pressure from an open faucet. At the exit of the faucet the stream has a diameter of 15 mm. The stream is laminar at all times and has a circular cross section. You also measure that the stream fills a 2.5 liter bucket in 35 seconds. The density of water is 1.0 kg/liter.

Part a: Calculate the flow rate.

Part b: Calculate the current.

Part c: Find the velocity of the stream when it exits the faucet.

Part d: Calculate the velocity of the stream 55 cm below the faucet.

Answer 1

a)

Volumetric flow rate(Q) is defined the flow of volume of fluid V through a surface per unit time t.

Q = V / t

V = 2.5 L = 2.5 x 10^-3 m³

t = 35 s

So , Q = 2.5 x 10^-3 / 35 = 7.1 x 10^-5 m³/s

b)

Mass flow rate is also known as mass current.

It is given by

m = Q x

= density of fluid = 1 Kg/L = 10³ Kg/m³

So, m = (7.1 x 10^-5 ) x 10³ = 7.1 x 10^-2 Kg/s

c)

Volumetric flow rate can also be defined by:

where v = flow velocity, A = cross-sectional area = d²/4 , d= diameter = 15 mm = 15 x 10^-3 m

So, A =  (15 x 10^-3)²/4 = 1.77 x 10^-4 m²

So, v = Q/A = 7.1 x 10^-5 / 1.77 x 10^-4 = 0.4 m/s

d)By energy conservation

The kinetic energy gained by water = the potential energy lost

v = velocity below height h

= change in height = 55 cm = 0.55 m

Hence