- #1

CeilingFan

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## Homework Statement

A fire hose for use in urban areas must be able to shoot a stream of water to a maximum height of 34m . The water leaves the hose at ground level in a circular stream 3.0cm in diameter.

What minimum power is required to create such a stream of water? Every cubic meter of water has a mass of 1.00×103kg.

## Homework Equations

Ek = 0.5*m*v^2

Ep = m*g*h

Volume cylinder = A*h

P=F*v (?)

## The Attempt at a Solution

I attempted twice, and one of them was wrong (almost double the value), so I was wondering what went wrong in the second attempt.

1st attempt:

First I went to find the speed of the water needed to leave the nozzle.

I used:

Ek(leaving nozzle) = Ep(at 34m)

0.5*m*v^2 = m*g*h

and got

velocity = 25.827 ms^-1

Then I went to find the volume, then the mass of water leaving the nozzle in 1s.

Mass = A*h*1000

pi*0.015*0.015*25.827*1000

= 18.2566kg

With that, I went to find the energy per second required (hence power) to give this mass of water speed V.

Ek = 0.5*18.2566*25.827*25.827 = 6089J

So in Watts, that's the correct answer.

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**But the stupid me went and tried another way of solving it (i don't know why i'd ever do that)**

So with v = 25.827 ms^-1

I used equation P = F.v for the power needed

P = F.v = (v.dm/dt)v

= v*v*(dm/dt)

So in one second,

Energy to water -> 25.827*25.827*(18.2566) = 12177

Which is the wrong answer, and coincidentally (or not), is twice as large.

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So... I can't figure out what went wrong with the second attempt.