Initially water is coming out of a hose at a velocity of 5 m/s. You place your thumb on the opening, reducing the area the water comes through to .001 $m^2$. You then find the velocity of the water to be 20 m/s. Calculate the radius of the hose.

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.

Flow Equation

Q = v*A

where:

Q = Flow in (m³/s)

A is the surface of the cross sections of points 1 and 2 of the duct.

v is the flow velocity at points 1 and 2 of the pipe.

It can be concluded that since the Q must be kept constant throughout the entire duct, when the section (A) decreases, the speed (v) increases in the same proportion and vice versa.

Data

D₂= 0.001 m² : final hose diameter

v₁ = 5 m/s : initial speed of fluid

v₂ = 20 m/s : final speed of fluid

Area calculation

A = (π*D²)/4

A₁ = (π*D₁²)/4

A₂ = (π*D₂²)/4

Continuity equation

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁(π*D₁²)/4 = v₂(π*D₂²)/4 : We divide by (π/4) both sides of the equation

v₁ (D₁)² = v₂(D₂)²

We replace data

5 *(D₁)² = 20*(0.001)²

(D₁)² = (20/5)*(0.001)²

(D₁)² = 4*10⁻⁶ m²

[tex]D_{1} = \sqrt{4*10^{-6} } ( m)[/tex]

D₁ = 2*10⁻³ m : diameter of the hose

Radius of the hose(R)

R= D₁/2

R= (2*10⁻³ m) / 2

R= (1*10⁻³ m) = 0.001 m