Fire Hose Power Output, Given Height and Radius of Hose

So your answer is correct, but the reasoning you wrote in the first post is not. In summary, the minimum power required to create a 34 m tall stream of water, with a diameter of 4.0 cm, is 10796 W. This can be calculated by multiplying the mass flow rate of 32.4 kg/s by the energy per mass of 10783 J/kg.
  • #1
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).
 
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  • #2
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.
 
  • #3
yellowcakepie said:
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.
 
  • #4
mfb said:
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
 
  • #5
yellowcakepie said:
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.
yellowcakepie said:
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.
 
  • #6
mfb said:
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?
 
  • #7
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.
 
  • #8
mfb said:
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.
 
  • #9
Right.
 

Related to Fire Hose Power Output, Given Height and Radius of Hose

What is the formula for calculating fire hose power output?

The formula for calculating fire hose power output is P = (πr2hρg)/t, where P is the power output, r is the radius of the hose, h is the height of the hose, ρ is the density of the liquid being pumped, g is the acceleration due to gravity, and t is the time it takes to pump the liquid.

How does the height of the fire hose affect the power output?

The height of the fire hose directly affects the power output. The higher the hose is raised, the greater the potential energy of the liquid, resulting in a higher power output. This is because the formula for power output takes into account the height of the hose.

Does the radius of the fire hose affect the power output?

Yes, the radius of the fire hose does affect the power output. The wider the hose, the greater the cross-sectional area for the liquid to flow through, resulting in a higher power output. This is why fire hoses are often wider than regular hoses.

What is the unit of measurement for fire hose power output?

The unit of measurement for fire hose power output is watts (W), which is the standard unit for power. However, in some cases, kilowatts (kW) may also be used.

Can the power output of a fire hose be increased by increasing the height and radius of the hose?

Yes, the power output of a fire hose can be increased by increasing the height and radius of the hose. However, there are limitations to how high and wide a hose can be before it becomes impractical to use. Therefore, the power output is often optimized by finding the ideal balance between the height and radius of the hose.

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