Home Improvement Stack Exchange is a question and answer site for contractors and serious DIYers. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|improve this question

closed as off topic by BMitch, Steven, Niall C., ChrisF Jun 24 '12 at 11:17

Questions on Home Improvement Stack Exchange are expected to relate to home improvement within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

Check that any pump can cope with running dry when the tank is empty. – Walker Jun 22 '12 at 8:45

According to this Wikipedia article,

a typical Early Suppression Fast Response (ESFR) sprinkler at a pressure of 50 psi (340 kPa) will discharge approximately 100 US gallons per minute (0.0063 m3/s)

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.

share|improve this answer

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.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.