Question

A fire hydrant delivers water of density Ï at a volume rate q. the water travels vertically upward through the hydrant and then makes 90° turn to emerge horizontally at a speed v. the pipe has uniform cross-sectional area throughout. the force exerted by water on the corner of the pipe is

Answer 1

Answer:

F=ρL *$\frac{v^{2}A}{2V}$

Step-by-step explanation:

Density of water = ρ

speed = v

Area is uniform = A

F =?

We know the formula for force applied by liquid is

F=ρ * L * a          ................(i)

where ρ is density

L is length

a is acceleration

but for a liquid we know

v ² = 2 a h         .................(ii)

Also

$h= \frac{V}{A}$

where V is volume and A is area

putting this in equation (ii)

v ² = 2 a *$\frac{V}{A}$

a = $\frac{v^{2}A}{2V}$

putting this value in equation (i)

F=ρL *$\frac{v^{2}A}{2V}$

This will be the force applied at the corner