Who will solve this first ?

[BlueCosmology]

Thank you for explaining that. It makes sense. And it answers why both, the flow rate and the power are described in the task.

But is it realistic? As I illustrated above, pumping 500l/s through a 1m² pipe will not get us anywhere near a 200m fountain, independent of the power at hand. I think that confused me.

The flow rate is not independent of the power. You won't get 500l/s through a 1m² pipe with that power.

Edited April 14, 2015 by BlueCosmology

[cpast]

Thank you for explaining that. It makes sense. And it answers why both, the flow rate and the power are described in the task.

But is it realistic? As I illustrated above, pumping 500l/s through a 1m² pipe will not get us anywhere near a 200m fountain, independent of the power at hand. I think that confused me.

If I were to just write down "Since we're pumping water, it depends on the nozzle"... would that be wrong? (I see it's not the desired answer, but isn't that true?)

Yes, that would be wrong. There is no "independent of the power at hand;" if you pump 500 L/s through a 1 m^2 pipe (and fill the pipe), you are using significantly less than 1 MW of power. The nozzle is *irrelevant*. If you take a huge nozzle, you will have the water stream not filling the nozzle, because you're putting much less water through than the pipe supports.

[Kerbin Dallas Multipass]

The flow rate is not independent of the power. You won't get 500l/s through a 1m² pipe with that power.

why not?

[BlueCosmology]

why not?

Work out what the speed of the water would have to be to produce a flow rate through a nozzle.

[Kerbin Dallas Multipass]

Work out what the speed of the water would have to be to produce a flow rate through a nozzle.

.5 m/s for a 1m² nozzle.

That's where my confusion started:)

[cpast]

.5 m/s for a 1m² nozzle.

That's where my confusion started:)

Now work out how much energy is in 1 L of water flowing at .5 m/s. Now work out how much power it takes to supply that much energy to 500 L of water every second.

You seem to be missing the part where the pump power is the energy output per second, which is equal to the energy imparted to each 500 L water.

[Kerbin Dallas Multipass]

Now work out how much energy is in 1 L of water flowing at .5 m/s. Now work out how much power it takes to supply that much energy to 500 L of water every second.

You seem to be missing the part where the pump power is the energy output per second, which is equal to the energy imparted to each 500 L water.

I imagine it is significantly less than 1 MW?

[cpast]

I imagine it is significantly less than 1 MW?

Correct. 500 L water at .5 m/s have 62.5 J of energy. This is several orders of magnitude less than 1 MW.

[Kerbin Dallas Multipass]

Correct. 500 L water at .5 m/s have 62.5 J of energy. This is several orders of magnitude less than 1 MW.

What if I make the nozzle smaller? So I can only get 250l/s through it but the pump still applies 1 MW?

[Bill Phil]

What if I make the nozzle smaller? So I can only get 250l/s through it but the pump still applies 1 MW?

The flow rate determines how big the nozzle is. If it's the flow rate at the nozzle...

[Kerbin Dallas Multipass]

The flow rate determines how big the nozzle is. If it's the flow rate at the nozzle...

Yea fair enough. So if you change the variables in this task you magically get different nozzle sizes?

Is that where my common sense went astray?

[Bill Phil]

Yea fair enough. So if you change the variables in this task you magically get different nozzle sizes?

Is that where my common sense went astray?

You're given a volume of water and thus only certain dimensions can account for that much water.

[cpast]

Yea fair enough. So if you change the variables in this task you magically get different nozzle sizes?

Is that where my common sense went astray?

Your common sense went awry when you assumed you could put a huge pump on a big nozzle and get a tiny flow rate.

[Duxwing]

Not very high:

Watts = Joules / Second, Watts = 10^6

Therefore, 10^6 W = 10^6 J/s

Flow = 500L/s, 1L Water = 1kg Water

Therefore, flow = 500kg/s

Therefore, energy per kilogram per second = 10^6 J / 500 = 200 J

Kinetic energy = ½ * mass * velocity^2, Kinetic energy = 200 J, mass = 500kg

Therefore, 200J = ½ * 500kg * v^2

Therefore, v = (200J * 2 / 500kg)^1/2 = 0.4m/s

h = h₀ + v₀ÃŽâ€t + ½ a(ÃŽâ€t)², h₀ = 0, v₀= 0.4m/s, ÃŽâ€t=0.04s, a = 10m/s/s

Therefore, h = 0m + 0.4m/s * 0.04s + ½ 10m/s/s(0.04s)² = 0.024m

Moving a half-ton of water every second requires lots of energy.

-Duxwing

[BlueCosmology]

Not very high:

Watts = Joules / Second, Watts = 10^6

Therefore, 10^6 W = 10^6 J/s

Flow = 500L/s, 1L Water = 1kg Water

Therefore, flow = 500kg/s

Therefore, energy per kilogram per second = 10^6 J / 500 = 200 J

Kinetic energy = ½ * mass * velocity^2, Kinetic energy = 200 J, mass = 500kg

Therefore, 200J = ½ * 500kg * v^2

Therefore, v = (200J * 2 / 500kg)^1/2 = 0.4m/s

h = h₀ + v₀ÃŽâ€t + ½ a(ÃŽâ€t)², h₀ = 0, v₀= 0.4m/s, ÃŽâ€t=0.04s, a = 10m/s/s

Therefore, h = 0m + 0.4m/s * 0.04s + ½ 10m/s/s(0.04s)² = 0.024m

Moving a half-ton of water every second requires lots of energy.

-Duxwing

Why have you randomly decided to work out the power supplied per kg?

Then after that you've decided that the power is no longer power but just energy.

And then ontop of it you've then decided that the energy supplied per kg is now the energy supplied per 500kg.

[Bill Phil]

Not very high:

Watts = Joules / Second, Watts = 10^6

Therefore, 10^6 W = 10^6 J/s

Flow = 500L/s, 1L Water = 1kg Water

Therefore, flow = 500kg/s

Therefore, energy per kilogram per second = 10^6 J / 500 = 200 J

Kinetic energy = ½ * mass * velocity^2, Kinetic energy = 200 J, mass = 500kg

Therefore, 200J = ½ * 500kg * v^2

Therefore, v = (200J * 2 / 500kg)^1/2 = 0.4m/s

h = h₀ + v₀ÃŽâ€t + ½ a(ÃŽâ€t)², h₀ = 0, v₀= 0.4m/s, ÃŽâ€t=0.04s, a = 10m/s/s

Therefore, h = 0m + 0.4m/s * 0.04s + ½ 10m/s/s(0.04s)² = 0.024m

Moving a half-ton of water every second requires lots of energy.

-Duxwing

Are you using the energy ratio as the KE?

Don't know about that....

I think you forgot to remove the 500 kg from the mass. (Technically you divided both sides by the mass, and got an energy per mass. If it's per single mass, you use 1 as your mass...)

[K^2]

Dux, you divided by mass twice. Once you got energy per kg, then your kinetic energy formula just has 1kg in it. Or use the full 1MJ against these 500kg. Either way, the answer will be the same.

A thermal particle is one that's energy is within a standard deviation of the energy of a particle with mean energy.

By "at least" I mean, at least. The answer is very sensitive to available energy yes.

Fair enough. But it's still just a matter of doing the math. I mean, if you're at the level where you know what a differential cross-section is, this is still a boring problem. And if not, then no amount of thinking is going to get it solved.

Edited April 14, 2015 by K^2

[cpast]

Not very high:

Watts = Joules / Second, Watts = 10^6

Therefore, 10^6 W = 10^6 J/s

Flow = 500L/s, 1L Water = 1kg Water

Therefore, flow = 500kg/s

Therefore, energy per kilogram per second = 10^6 J / 500 = 200 J

Kinetic energy = ½ * mass * velocity^2, Kinetic energy = 200 J, mass = 500kg

Therefore, 200J = ½ * 500kg * v^2

Therefore, v = (200J * 2 / 500kg)^1/2 = 0.4m/s

h = h₀ + v₀ÃŽâ€t + ½ a(ÃŽâ€t)², h₀ = 0, v₀= 0.4m/s, ÃŽâ€t=0.04s, a = 10m/s/s

Therefore, h = 0m + 0.4m/s * 0.04s + ½ 10m/s/s(0.04s)² = 0.024m

Moving a half-ton of water every second requires lots of energy.

-Duxwing

You made several math errors. 10^6 / 500 = 2000, not 200. So it's 2000 J/kg. You then factored in the mass *again* when you calculated velocity, which is wrong -- it's either 10^6 J = (.5)*(500kg)*(v^2) OR 2000 J/kg = (.5)*(v^2); you don't divide energy by mass to get specific energy and then divide specific energy by mass again to get v^2/2. So the actual velocity is 20*sqrt(10) or about 63 m/s, not .4 m/s. Then t=6.3s, not .04s, and a= -10m/s^2 (not 10 m/s^2); so h=0+63*6.3+(1/2)(-10)(6.3)^2=400-5*40=200.

[Bill Phil]

Huh.

Would Potential Energy be a good estimate?

PE=m*g*h

Solve for h:

h= PE/(M*g)

Would that work out pretty close? Or was that just a fluke distinct to this problem?

[BlueCosmology]

Huh.

Would Potential Energy be a good estimate?

PE=m*g*h

Solve for h:

h= PE/(M*g)

Would that work out pretty close? Or was that just a fluke distinct to this problem?

Yeah that's the way to do it.

[cpast]

Huh.

Would Potential Energy be a good estimate?

PE=m*g*h

Solve for h:

h= PE/(M*g)

Would that work out pretty close? Or was that just a fluke distinct to this problem?

Potential energy is in fact the easiest way to solve this problem. If you compute the speed of the water leaving the nozzle, you're doing things the hard way. You have the energy imparted at the bottom as kinetic energy; this is all potential energy at the top, so you calculate how high the energy gets you. It's not just close, it's exact (down to problems of reality differing from spherical cow-land).

[BlueCosmology]

Dux, you divided by mass twice. Once you got energy per kg, then your kinetic energy formula just has 1kg in it. Or use the full 1MJ against these 500kg. Either way, the answer will be the same.

Fair enough. But it's still just a matter of doing the math. I mean, if you're at the level where you know what a differential cross-section is, this is still a boring problem. And if not, then no amount of thinking is going to get it solved.

There's really no reason whatsoever to consider differential cross-sections. It's either scattered or it isn't, the direction of scattering is irrelevant.

[Bill Phil]

Potential energy is in fact the easiest way to solve this problem. If you compute the speed of the water leaving the nozzle, you're doing things the hard way. You have the energy imparted at the bottom as kinetic energy; this is all potential energy at the top, so you calculate how high the energy gets you. It's not just close, it's exact (down to problems of reality differing from spherical cow-land).

It would have to lose the same amount of energy on the way up as it gains going down. I just wanted some confirmation...

Thanks.

[K^2]

There's really no reason whatsoever to consider differential cross-sections. It's either scattered or it isn't, the direction of scattering is irrelevant.

The probability of absorption is trivially related to the angle it would have been scattered to from a point charge. Then you just integrate that over differential cross section to get total absorption probability. That's the easy way to solve it. How the hell were you planning to go about it?

Unless you want proper QCD treatment, of course. In which case, you have to solve the relevant 4-point DSE. (Lepton vertex can be taken out without loss of precision.)

[Rondon]

well no rep for you my friend

the calculation is math actually, and it's another interesting thing (although i must agree that right there the calculation is easy and boring)

Eh I did it out, so i may as well answer.

About 19.08 seconds though I have to say I can't get any good result out of Blue Cosmologys method.

I'm terrible for making simple mistakes so I'll go step by step.

So find T where T = 10m/s / A

Where A is the acceleration horizontal to the plane,

The acceleration of the car under Gravity: a = Gsin(6â°)

The Force horizontal to the plane is Fh = Ma - Fr

Fh = (1600 kg)(9.8 m/s²)(0.1045) - 800N = 836.56N

A = Fh / m = 0.5241 m/s²

T =10m/s / 0.5241 m/s² = 19.08 s

An easy enough calculation, though I suppose off topic at this rate.