Who will solve this first ?

[Fel]

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...

I don't see how this is true in anything other than a "textbook" example. I assume this is on the idea that water is, for the most part, incompressible.

But, in a pipe, any liquid that rubs against the walls will slow down, this slow down causes further slowdowns which rob energy from the system as a whole. Furthermore, this same effect can be applied to the ends of a pipe, if the starting diameter is larger than the end diameter. We have still other effects if the end of the pipe is wider than the pipe that supplies "power." Mostly that, from the ideal, the nozzle that matters would be the smallest one, the non-ideal is dependent on the length of the second pipe.

Sure, the problem said to ignore all losses and WAS only "jimmy threw a baseball up in the air, how high did it go", but when we start talking about nozzles that all disappears, does it not?

[Bill Phil]

I don't see how this is true in anything other than a "textbook" example. I assume this is on the idea that water is, for the most part, incompressible.

But, in a pipe, any liquid that rubs against the walls will slow down, this slow down causes further slowdowns which rob energy from the system as a whole. Furthermore, this same effect can be applied to the ends of a pipe, if the starting diameter is larger than the end diameter. We have still other effects if the end of the pipe is wider than the pipe that supplies "power." Mostly that, from the ideal, the nozzle that matters would be the smallest one, the non-ideal is dependent on the length of the second pipe.

Sure, the problem said to ignore all losses and WAS only "jimmy threw a baseball up in the air, how high did it go", but when we start talking about nozzles that all disappears, does it not?

Referring to exit velocity.

[K^2]

I don't see how this is true in anything other than a "textbook" example. I assume this is on the idea that water is, for the most part, incompressible.

But, in a pipe, any liquid that rubs against the walls will slow down, this slow down causes further slowdowns which rob energy from the system as a whole.

Yes. The problem treats water as incompressible and inviscid. In truth, it's not that horrible of an approximation. Air resistance as water goes up will cause a greater reduction in maximum height than viscosity of water.

[Fel]

Yes. The problem treats water as incompressible and inviscid. In truth, it's not that horrible of an approximation. Air resistance as water goes up will cause a greater reduction in maximum height than viscosity of water.

Just a quick note, the water compressibility quib was more a memory of trying to explain that it is a fairly accurate approximation, but not a physical truth.

I do admit that this is far from my area of expertise; but I wonder just how accurate you're being here. Your assumptions want to place everything "perfect"; if the pump was a couple meters away, or designed with a different flow rate in mind (note we're only given power output on the pump)... I mean, everything seems to say that the assumptions being made are "low pressure" and "extremely short" pipes.

Okay, I am 65% certain this is correct, but

E = P * V

1e6 J/s = P * (1m^3 / 1000L) * 500L/s

P = 1e6 J/s / ((1m^3 / 1000L) * 500L/s)

P = 1e6 J/s / (0.5 m^3/s)

P = 2e6 J/m^3 = 2e6 N/m^2 = 2e6 Pa

P = 2e6 Pa * (1 atm / 101325 Pa) = 19.73 atm

19.73 atm doesn't seem like "low pressure"

*Meh, I really don't like that. Yes it is a displaced volume due to pressure... but now I'm using ideals when I know that non-ideals can change how much volume gets displaced due to pressure.

Okay, I see it. This assumes that there is no "pipe" between the output of the pump and the nozzle.

Edited April 15, 2015 by Fel

[K^2]

The question isn't if the pressure on the pump output is low, but rather if pressure differential through the pipe is low compared to that. You can use a straight pipe lamniar flow approximation after you work out the diameter of the relevant pipe to make sense with flow velocity and flow rate.

Of course, in a real pipe, flow velocity isn't going to be uniform, so it's slightly more complex than it sounds.

[Hcube]

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.

Agreed with the "making simple mistakes part ! i did not ask after what time does the car reach 36km/h but after what distance down the slope ! good try though

it is not off topic since no one has yet given the answer in meters !

(some are having fun with the fountain problem it seems )

[Rondon]

Agreed with the "making simple mistakes part ! i did not ask after what time does the car reach 36km/h but after what distance down the slope ! good try though

it is not off topic since no one has yet given the answer in meters !

(some are having fun with the fountain problem it seems )

...Well it had to be something. Well atleast it's on the right path

T=19.08

S =1/2aT² from rest

S = 1/2 (0.5124)(19.08²) = 93.2687 meters

[Fel]

The question isn't if the pressure on the pump output is low, but rather if pressure differential through the pipe is low compared to that. You can use a straight pipe lamniar flow approximation after you work out the diameter of the relevant pipe to make sense with flow velocity and flow rate.

Of course, in a real pipe, flow velocity isn't going to be uniform, so it's slightly more complex than it sounds.

I see... I think. I was quickly starting to realize that the pressure would change while maintaining the same flow rate as I increased the pipe length; but I wasn't exactly liking that result. Looking back I realize that the issue was that (differential) pressure isn't the cause of flow, (differential) pressure is the RESULT of flow. Which goes more to point out how the system is restricted by flow rather than supplied power.

*or, I could be confusing myself even more XD; thanks for this though, I learn far better by tearing down model after model until I have something more accurate than I do reading equation after equation*

[K^2]

Relationship between flow velocity and pressure gradients is always a chicken or the egg sort of question in fluid dynamics. Of course, from the perspective of solving a problem, it doesn't matter.

[Hcube]

...Well it had to be something. Well atleast it's on the right path

T=19.08

S =1/2aT² from rest

S = 1/2 (0.5124)(19.08²) = 93.2687 meters

That's about right, even though you made a crazy calculation that really wasn't needed very much, but i understand that is because you first calculated the time...

Take this rep

You could have done it much faster with just ÃŽâ€Emeca=Σ(Work of all the non conservative forces), and then knowing that initial kinetic energy is 0 and final potential energy is 0 just resolve it quickly

[Rondon]

Heck of a crazy calculation it's almost exactly how I was taught to solve this exact same scenario in secondary school, go figure:confused:. Never taught that one though, pretty neat.