Power Requirement for Fire Hose in Urban Areas

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SUMMARY

The minimum power required for a fire hose to shoot a stream of water to a maximum height of 35 meters is determined using the equations for work and power. The diameter of the hose is 3.5 cm, leading to a radius of 0.0175 m. The volume of water needed to reach this height is calculated to be approximately 0.033674 m³, resulting in a mass of 33.67 kg. The work done against gravity is 11,550 N, and the power can be calculated by considering the time taken to achieve this height.

PREREQUISITES
  • Understanding of basic physics concepts such as force, work, and power
  • Familiarity with the equations P=W/t and W=Fd
  • Knowledge of fluid dynamics, specifically volume calculations for cylinders
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Calculate the power required for different heights using the formula P=W/t
  • Explore the effects of varying hose diameters on water flow and power requirements
  • Investigate the physics of projectile motion to understand water trajectory
  • Learn about the energy conservation principles in fluid dynamics
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Engineers, physics students, and fire safety professionals interested in the mechanics of fire hoses and water delivery systems in urban environments.

pinkybear
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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 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.


Homework Equations


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


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)...
 
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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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