I am trying to replace a water pump that we have on a confined space simulator that will pump water from an open 55 gallon tank to a pair of fire sprinklers to simulate a pressurized chemical release. I have never seen this unit in action, so I am not sure how much pressure it takes to blow the water out in 4-6 foot arches...they originally had a 1/12 hp Flotec FPOF360ac pump, but said it didnt have enough psi, and eventually they burned the pump up. I need help.
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According to this Wikipedia article,
With a magic 100% efficient motor and pump, that'd require almost 3 hp, so it's no surprise 1.5 hp didn't cut it for two sprinklers.
Discharge of 0.0063 m3/s is effectively the "speed of exhausting a volume", that is, dV/dt. Elementary volume dV can be represented as a product of cross-section area by elementary length: dV=S*dl, where the elementary length dl can be represented as a product of fluid speed by elementary time: dl=v*dt. Thus, dV/dt=S*v.
Then, power is effectively the speed of performing the action: P=dA/dt. Since the elementary action ("work") can be represented as applying the force at the elementary path: dA=F*dx, it can be presented as pressure forces acting at the elementary section area: F=p*S, so the power is P=p*S*v. We need to derive the amount of power needed from first and second: P=p*dV/dt. Substituting with p=3.4*10^5 Pa, dV/dt=6.3*10^-3 m^3/s gives P=2142 W, which is almost 2.9 hp.
The max height of the arc can then be derived: the sprinkler dispenses 6.3*10^-3 m^3 each second providing it with kinetic energy of 2142 J. Considering the dispensant to be water, this is equivalent to 6.3 kg. If the kinetic energy is fully transformed into potential energy of uplifted water, then E=m*g*h, where h is then h=E/(m*g), that is, h=2142/(6.3*9.8)~=34.7 m. Whoa, the height of a 10-floor building.