Power of Fire hose: a calculus approach

[Musashiaharon]

A while ago, pinkybear posted the following:

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)...
​

I was not satisfied with the sole response given, since it involved unnecessary approximation (basically, run through the calculations for a small, but finite amount of water). Here is my calculus-based approach:

It is useful to remember the following:

• Velocity: v=dy/dt
• Kinetic energy: K=1/2mv²
• Power: P=dK/dt=1/2*d/dt(mv²)

To use these, we need to calculate the velocity of the water on leaving the hose, which you can get by solving the following (remember that a is negative here):

v_f² - v₀² = 2a*Δy

We can calculate the flow rate, which is the change in volume (capital V) over time, or dV/dt:

dV/dt = dy/dt * A

Then, using the density of water, we can convert this to the change in mass over time, dm/dt.

We noted above that power is:

P=1/2*d/dt(mv²).

Since v is constant:

P = 1/2v² * dm/dt

This is enough to answer the question. But we could go further and calculate the force of the water pushing back on the hose. We'll just use a couple more equations:

• Force (Newton's original definition): F=dp/dt
• Momentum: p=mv

Again, v is constant. We substitute:

F=d/dt (mv) = v * dm/dt

Now we have the force, too!

Pressure, then, is just dividing the force by the sectional area of the nozzle.

 

[Alpharup]

A while ago, pinkybear posted the following:

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)...
​

You are asking for t.. here t is the time for this 11550 J of energy is supplied by the hose to the water. here the height h is 35 m.
In a projectile motion or a 2D motion in general, there are two types of velocities, horizontal and vertical.
Here the horizontal velocity is constant whereas vertical velocity varies with time. The height of the projectile depends on vertical velocity.
thus
h=1/2*g*t^2.
and thus you get t.

 

[Musashiaharon]

You are asking for t.. here t is the time for this 11550 J of energy is supplied by the hose to the water. here the height h is 35 m.
In a projectile motion or a 2D motion in general, there are two types of velocities, horizontal and vertical.
Here the horizontal velocity is constant whereas vertical velocity varies with time. The height of the projectile depends on vertical velocity.
thus
h=1/2*g*t^2.
and thus you get t.

Pinkybear's approach was flawed, because she assumed that the column of water is a perfect cylinder, which it isn't. If you look at a vertical water jet, you will see that the column thickens at it rises. This is because gravity makes the water move slower the higher it rises, and the faster water from below spreads out the slower water above. As a result, the column of water is no longer a cylinder, and in fact has more volume.

(If there were no acceleration, like on an orbiting space station, the column would actually keep a constant radius, neglecting the cohesive properties that make the water ball up into droplets.)

Pinkybear's value for work is also inaccurate, because her value for the mass of the water is based on the faulty cylinder idea, and because not all the water in that cylinder has reached 35m yet.

In order to work out the problem, we have to find v₀ first. Energy and power primarily depend on the initial velocity; pinkybear was right that this is the crux of the issue.

Nevertheless, if you still want to know how much time it takes a drop of water to reach 35m from the nozzle, sharan's equation is accurate. If you already have v₀, the equation is even simpler:

Δv = a*t

 

[haruspex]

Not sure that it's necessary to use any explicit calculus. With A as nozzle area and ρ as density, isn't your method equivalent to this:

KE per unit mass at nozzle exit = v²/2 = gh.
Mass rate = ρAv
Power = ρAvgh = ρAgh√(2gh)

 

[Musashiaharon]

Not sure that it's necessary to use any explicit calculus. With A as nozzle area and ρ as density, isn't your method equivalent to this:

KE per unit mass at nozzle exit = v²/2 = gh.
Mass rate = ρAv
Power = ρAvgh = ρAgh√(2gh)

You are right, but others with a lesser grasp might try to actually find a discrete value for KE, mass, and time. By dealing with dt and friends, there is a reminder that we are dealing with infinitesimally small quantities.

It is also harder to see from the outset that one must multiply "KE per unit mass" by "mass rate." If these are re-labeled dK/dm and dm/dt, it gets easier to see why we should multiply, since dm gets canceled.