Water flows through a 2.25 cm diameter hose at 0.320 m/s. find (a) the volume flow rate and (b) the speed of the water emerging from a 0.30 cm diameter nozzle.

(a) Q = (1.27)*10⁻⁴ (m³/s)

(b) v₂ = 18 m/s

Explanation:

Flow Equation

Q = v*A  

where:

Q = Flow in (m³/s)  Formula (1)

A is the surface of the cross sections (m²)

v:   is the speed of the fluid ( m/s)

Continuity equation

The continuity equation is nothing more than a particular case of the principle of conservation of mass. It is based on the flow rate (Q) of the fluid must remain constant throughout the entire pipeline.

Q₁ = Q₂

where :

Q₁ :  Flow in the point 1 of the hose

Q₂:  Flow in the point 2 of the hose

Data

D₁= 2.25 cm =  0.0225  m

v₁ = 0.320 m/s  

D₂ = 0.30 cm = 0.003 m

Area calculation

A = (π*D²)/4

A₁ = (π*D₁²)/4  = (π*(0.0225)²)/4  = 3.976*10⁻⁴ m²

(a) Volume flow rate (Q)

Q = v₁*A₁

Q = (0.320 m/s ) (3.976*10⁻⁴ m²)

Q = (1.27)*10⁻⁴ (m³/s)

(b) Speed of the water (v₂) emerging from a 0.30 cm diameter nozzle.  

A₂ = (π*D₂²)/4 = (π*(0.003)²)/4 = 7.07*10⁻⁶ m²

Continuity equation  

Q=Q₁ = Q₂

(1.27)*10⁻⁴ = v₂A₂

(1.27*10⁻⁴) /A₂ = v₂

v₂ = (1.27)*10⁻⁴ / (7.07)*10⁻⁶

v₂ = 18 m/s

uranium fissions/breaks up giving out energy and neutrons