Fire Hose Power Output, Given Height and Radius of Hose

[yellowcakepie]

Homework Statement

A fire hose for use in urban areas must be able to shoot a stream of water to a maximum height of 34 m. The water leaves the hose at ground level in a circular stream 4.0 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.

Homework Equations

PE = mgh

r = 1/2d

A = πr^2

V = πr^2h = Ah

P = W/t = F*d/t = F*v

The Attempt at a Solution

I converted the 4.0 cm diameter to radius in terms of meters, which is r = 0.02 m.
I solved for A, and multiplied A by maximum height (h) to get volume (V) which is 0.0427 m^3 column of water that is 34 m tall.
I used the density of water, which is 1000 kg/m^3 to get the mass of the column, which is 42.7 kg.
I don't really know where to go from here. I used PE = (42.7 kg)(9.8 m/s^2)(34 m), but I know that it is wrong. I'm not sure how to integrate the height of the column of water (if that's needed).

 

[mfb]

How fast does the water have to leave the hose to rise 34 meters?

You can use this velocity for two different aspects of a power calculation.

 

[haruspex]

the mass of the column, which is 42.7 kg.

Water is not a solid. As it ascends it slows down, so the "column" must get broader. But again, since it is not a solid, it is not exerting that weight on anything, so finding its mass is not helpful.
Follow mfb's hint.

 

[yellowcakepie]

How fast does the water have to leave the hose to rise 34 meters?

You can use this velocity for two different aspects of a power calculation.

Okay so, I used the inital KE, which is 1/2mv^2 and set that equal to the PE at the maximum height, which is mgh.
Therefore v = 25.8 m/s after some algebra. I can use that for P = Fv.

I then need to get the mass flow rate per second out of the hose. I'll begin with the cross-sectional area of the hose. The cross-sectional area of the hose is A = pi * (0.02)^2 = 1.257*10^-3 m^2.

I then multiply that by v = 25.8 m/s to get the volume flow rate, so Av = 0.0324 m^3/s. I then can convert that to mass flow rate by dividing by the density of water, 1000 kg/m^3. Therefore, mass flow rate = 32.4 kg/s.

I can then multiply mass flow rate by g to get the force exerted per second to pump that amount of water out, and then by max height h to get power.

P = 32.4 kg/s * 9.8 m/s^2 * 34 m = 10796 W

 

[mfb]

I then multiply that by v = 25.8 m/s to get the volume flow rate, so Av = 0.0324 m^3/s. I then can convert that to mass flow rate by dividing by the density of water, 1000 kg/m^3. Therefore, mass flow rate = 32.4 kg/s.

Multiply, not divide, but you did it right to get the answer.

I can then multiply mass flow rate by g to get the force exerted per second to pump that amount of water out

I don't think that is a meaningful operation.

You can calculate the energy density (energy per mass) based on the velocity, and use that together with the mass flow.

 

[yellowcakepie]

I don't think that is a meaningful operation.

You can calculate the energy density (energy per mass) based on the velocity, and use that together with the mass flow.

I wasn't too clear on that when my instructor gave me the reasoning behind the operation. How is energy per mass calculated?

 

[mfb]

It is just the kinetic energy of the water. If you know the energy a mass m has, divide that by m to get the energy per mass.

 

[yellowcakepie]

It is just the kinetic energy of the water. If you know the energy a mass m has, divide that by m to get the energy per mass.

Oh so, that means in 1 second, there will be 32.4 kg of water. That 32.4 kg of water will have KE = 16.2*25.8^2 = 10783 J. But that much water is coming out per second, so 10783 J/s = 10783 W. Got it.

 

[mfb]

Right.