How much force is required to pump 1900 liters of water per minute through 20 mm nozzle to spray 50m in distance?

Answer 1

V= Q/A (velocity = volumetric flow rate / area)

conversion: 1000L = 1m3; therefore 1900L/min= 1.9m3/min

V = (1.9m3/min * 1min/60s) / Pi*(.02m)2/4 = 100.798 m/s

Simplified Bernoulli Equation:

∆P/rho + 0.5*(initial velocity2-final velocity2) = 0 (rho=density)

(equation assumes no fluid height changes and no frictional losses)

∆P/(1000 kg/m3) + 0.5*(02 - (100.798 m/s)2) = 0

∆P = 5.08x106 N/m2= 5.08x106 Pascals

P= F/A (Pressure= Force/Area)

F= P*A

F= ∆P * Area = 5.08x106 * Pi*(.02m)2/4 = 1595.97N

A height of nozzle was not given; however, there is a minimum height when you specify a horizontal distance of 50m.

Minimum nozzle height above ground to reach 50 meters horizontal distance:

d=r*t (distance= rate * time)

t= d / r

t= 50m / 100.798m/s = 0.496s

(time for a droplet to leave the nozzle and travel 50m horizontally)

h= g*t2

(height of a falling object = gravity * time2 assuming spray nozzle is horizontal)

h= 9.81m/s2 * (0.496s)2 = 2.41m above grade

The minimum height will change if the nozzle is angled up or down.

Note: 9.81m/s2 is approximately the earth's gravitational acceleration at sea level at the equator.