# Power of Fire hose

1. Oct 15, 2010

### pinkybear

1. The problem statement, all variables and given/known data
A fire hose for use in urban areas must be able to shoot a stream of water to a maximum height of 35 m. The water leaves the hose at ground level in a circular stream 3.5 cm in diameter.
What minimum power is required to create such a stream of water? Every cubic meter of water has a mass of 1.00*10^3 kg.

2. Relevant equations
P=W/t
W=Fd
V(of cylinder)=pi*r^2*h

3. The attempt at a solution
r=.0175m
v=pi*(.0175)^2*35
=.033674 m^3 then, mass= 33.67kg
w=(33.67)*g*35=11550 N
but then I'm stuck because I don't know what t is...
I tried v=v(0)+at, from which I got
P=m*a^2*D/v(0)
but now, I don't know how to get v(0)...

2. Oct 15, 2010

### Delphi51

It looks good down to the point where you put 35 m into the formula for volume. You do need to work with some definite amount of water, but I think a much smaller amount would be better. I would pick a small time - say one second or one tenth of a second - and figure out how much water is squirted in that time. You could just figure out how much energy it has when it reaches the maximum height. The power calc should then be easy.

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