Who will solve this first ?

[Hcube]

A small physics thing i thought I'd share here : our physics teacher (3rd year of highschool, france, we students are about 18) gave us this :

a fountain is powered by a 10^6 W pump. The flow in the fountain is 500L of water per second. What height does the water reach ? ( Neglect all air resistance, jupiter's gravity influence and such and the fountain is vertical)

It's not very difficult but kinda fun (somehow). It took me a few minutes, i'm sure most guys here will get it in seconds !

[Elthy]

Im not 100% sure (did this just in my head) but it should a bit more than 200m, right?

[K^2]

That's what I got.

[Hcube]

yup thats it

well if you get lazy and use 10m.s^-2 for g instead of 9,81 to do it more easily in your head you get exactly 200m (obviously, duh)

this exercise is supposed to be about the foutain of geneve wich IRL is about 140m high because of energy "losses"

[cpast]

Yep. For someone who's confused on how to do this fast: 10^6 W is 10^6 J/s. That's 10^6 J on 500kg of water, which means you have g*h*500=10^6. As g=10, h=10^6/5000=200. Because we're physicists, water is 1 kg/L and g=10.

[Hcube]

or, if you're really into useless , additional, superfluous calculations, you can get the velocity of the "exhaust" water at the base of the fountain using the kinetic energy, and then F=ma, resolve the time equation, have some fun with primitives and get the same result

[Bill Phil]

So I did get it right?!! Sorry. I'm a bit rusty with this kind of stuff.

[Tex_NL]

Without the width of the nozzle this question is impossible to answer.

Through a narrow nozzle 500l/s will move very fast and the fountain can go very high. Through a wide nozzle it will flow much slower and won't go nearly as high.

[cpast]

Without the width of the nozzle this question is impossible to answer.

Through a narrow nozzle 500l/s will move very fast and the fountain can go very high. Through a wide nozzle it will flow much slower and won't go nearly as high.

No, this question is very possible (in fact, kinda easy) to answer. You have the power output of the fountain; that means you know how much energy the water gets. This tells you how high it goes. You don't need to bother worrying about the nozzle or how fast the water goes when it leaves.

[Linear]

Without the width of the nozzle this question is impossible to answer.

Through a narrow nozzle 500l/s will move very fast and the fountain can go very high. Through a wide nozzle it will flow much slower and won't go nearly as high.

You know the energy and mass of the water...

[arkie87]

Yep. For someone who's confused on how to do this fast: 10^6 W is 10^6 J/s. That's 10^6 J on 500kg of water, which means you have g*h*500=10^6. As g=10, h=10^6/5000=200. Because we're physicists, water is 1 kg/L and g=10.

Power = Flowrate * pressure_drop

pressure drop is rho*g*h

[Kerbart]

Without the width of the nozzle this question is impossible to answer.

Through a narrow nozzle 500l/s will move very fast and the fountain can go very high. Through a wide nozzle it will flow much slower and won't go nearly as high.

That could actually be the next question. Making all the usual assumptions, what is the diameter of the nozzle?

[Kerbin Dallas Multipass]

If the nozzle is a 1x1x1 m cube and I pump 500l/s through it I get a velocity of 0.5 m/s. I somehow don't see that getting up to 200m.

[Kerbart]

If the nozzle is a 1x1x1 m cube and I pump 500l/s through it I get a velocity of 0.5 m/s. I somehow don't see that getting up to 200m.

No, the volume flow is .5m³/_s

Since the height is 200m (assuming g=10m/s like the rest of the thread), velocity has to be 63 ^m/_s.

For .5m³/_s that means the area of the muzzle is 0.0079m² and that puts the diameter at 10cm if I'm right.

[andrewas]

If the nozzle is a 1x1x1 m cube and I pump 500l/s through it I get a velocity of 0.5 m/s. I somehow don't see that getting up to 200m.

Attach a 1MW pump to that nozzle, and you'll get much more than 500l/s of flow.

Edited April 13, 2015 by andrewas

[Kerbin Dallas Multipass]

No, the volume flow is .5m³/_s

Since the height is 200m (assuming g=10m/s like the rest of the thread), velocity has to be 63 ^m/_s.

For .5m³/_s that means the area of the muzzle is 0.0079m² and that puts the diameter at 10cm if I'm right.

Passing .5m³ through a 1m² opening in 1 second gives me a flow speed of .5m/s.

The 200m height is not specified in the question, neither is the nozzle size.

- - - Updated - - -

Attach a 100KW pump to that nozzle, and you'll get much more than 500l/s of flow.

Yea but why is the 500l/s specified in the question?

[Kerbart]

Yea but why is the 500l/s specified in the question?

Right smack in the middle:

a fountain is powered by a 10^6 W pump. The flow in the fountain is 500L of water per second. What height does the water reach ? ( Neglect all air resistance, jupiter's gravity influence and such and the fountain is vertical)

[Kerbin Dallas Multipass]

I asked why, not where

[Bill Phil]

I asked why, not where

So you know how much and how fast.

[andrewas]

Yea but why is the 500l/s specified in the question?

Because the question is designed to evaluate whether a student can determine how high an object will go if you dump a given quantity of energy into it.

[Kerbart]

I asked why, not where

It was a high school physics question. Not a philosophy exam.

[Kerbin Dallas Multipass]

It was a high school physics question. Not a philosophy exam.

Thanks for pointing out where in the text i can find it as a reply to my question about why it is mentioned at all. I still don't understand why, seems I suck at physics.

[andrewas]

The question isn't about fountains. Its about kinetic and potential energy.

The actual question is - How high can you throw a 500Kg object with 1MJ?

But if its stated that simply, the student can apply the correct formula without understanding it. So the question is dressed up, and talks about a 500L/s flow rate and a power supply of 1MW.

Edited April 13, 2015 by andrewas

[K^2]

It has practical applications, though. If you understand all of that and a few ideal gas formulas, you can easily do math on water rockets, keeping in mind that power is equal to pressure differential times flow rate.

[YNM]

Actually, instead of height you can get the velocity of the water as it leaves the tube (or as it leaves the pump). And as current is velocity times cross-section area, you can get the width of the pipe ! (which is a stunning... 12 m diameter ! what ?)