Energy needed to push a volume of water

[Lucini]

Hello! I just found this website and it looks amazing! I'm not a scientist or anything, but I love it (should've studied physics but oh well), so I think it will be fun and useful for me to join this forum.

I am trying to solve a situation, where I'd like to know how much energy would be needed to push a mass of water out of a container. Here's an image to help understand:

http://imageshack.com/a/img21/6856/1m7e.png

The container has a height of 3 meters, length of 5 meters, and width of 3 meters. If a device, much like a coffee press but watertight, pushes the water with only the hole on top of the container, as shown in the drawing, as an exit point, how much energy would be needed for that device to go all the way to the other side of the container? Assuming the hole is 1 square meter, if that can help.

If there is a formula that would enable me to figure this out, that is all I'm asking, I can try and do the math myself, but I can't figure out where to start!

I realize it's an odd situation or post, but it has been bugging me for a while!

Thanks a lot!

 

[Chestermiller]

Hi Lucini. Welcome to Physics Forums!

The answer to your question depends on how fast you push the piston. If you push it very slowly, then very little energy will be required because the water will ooze out the hole. If you push it fast, the velocity of the water out the hole will be faster, and you will have to do a little work to accelerate it out the hole. However, with a huge hole like that, the amount of work will still be very little.

Chet

 

[dauto]

What Chestermiller said above plus what he missed which was the gravitational potential energy. Just compare the center of mass of the water before (half way up) and after (all the way up).

 

[Chestermiller]

What Chestermiller said above plus what he missed which was the gravitational potential energy. Just compare the center of mass of the water before (half way up) and after (all the way up).

Yes. I considered the potential energy effect negligible because the system is horizontal, but it would contribute a little bit. Also, this potential energy effect would not depend on the speed of the piston.

Chet

 

[voko]

The water exerts pressure on the piston. The pressure varies linearly from top to bottom, but it does not vary with time. The integral of the pressure over the area of the piston gives constant force, so at least this force must be applied to the piston to squeeze the water out. The work of the force over distance is the minimal energy requirement. This is equivalent to what dauto said.

More practically, one needs to accelerate the entire mass of water to some (however small it may be) velocity. The kinetic energy of the water (and the piston) is another energy requirement.

Then, one would need to consider friction between the piston and the walls (force constant in time).

The biggest complexity here will be in taking account of the viscosity of the water. I think the Darcy–Weisbach equation might be useful here.

 

[Lucini]

Thanks a lot for those quick and helpful responses! (And thanks for the welcome message Chestermiller!)

That's a good point, I hadn't thought of the speed at which the piston is pushed. For the sake of this situation, we can say it is pushed slowly.

Voko, thanks a lot for your additions. That sounds really complex, and there are a lot of things to consider! Are these forces (i.e. water's pressure on the piston, kinetic energy of water & piston, and friction between piston and walls) important, meaning they will greatly affect the total amount of energy required, or negligible but it is important to take them into account if we want an accurate result?

As for the Darcy-Weisbach equation, I have never heard of it before, so thanks for mentioning it! I've just looked it up online, and I'm not sure as to how I should use it. Should I subtract its result, the head loss due to friction, to the total amount of energy required?

Sorry if I ask silly questions!

 

[voko]

Voko, thanks a lot for your additions. That sounds really complex, and there are a lot of things to consider! Are these forces (i.e. water's pressure on the piston, kinetic energy of water & piston, and friction between piston and walls) important, meaning they will greatly affect the total amount of energy required, or negligible but it is important to take them into account if we want an accurate result?

To be completely sure, you would need to check those things one by one. Intuitively, it seems to me that the most significant factor here is the hydrostatic pressure, everything else should be (much) smaller than it.

As for the Darcy-Weisbach equation, I have never heard of it before, so thanks for mentioning it! I've just looked it up online, and I'm not sure as to how I should use it. Should I subtract its result, the head loss due to friction, to the total amount of energy required?

Basically it is the additional pressure on the moving piston. So it gets added to the hydrostatic pressure.

 

[Lucini]

Ok great, thanks for the clarifications. So just to make sure I understand the concept of hydrostatic pressure correctly: hydrostatic pressure increases in proportion to depth, so if the water tank is large but shallow, the hydrostatic pressure will not be as important, right?

 

[voko]

Yes, as long as total volume stays constant, the work by the force against hydrostatic pressure is proportional only to the height of the container.

 

[Lucini]

Thanks again, voko.

Still in relation with this situation, I've figured out that the formula F_piston = 1/2 ρgwh² can help with determining the force acting against the piston. I am a bit confused though, because even though I know the value of each of these factors, they have a different unit.

For instance: ρ is 1000 kg/m³, g is 9.80 m/s², w is 3 m and h is 3 m. Do I keep all values as they are, or should I convert them to something else? I'm really confused when it comes to units...

Also, once I get a result, the value will be in Newton, since we're dealing with Force, right?

 

[voko]

Actually, your units are perfectly fine. 1 Newton is 1 kilogram times 1 metre divided by 1 second squared. So when you multiply all the units of your values, you should get that compound unit. If you do, you know you are on the right track. Try it.

 

[Lucini]

Aaah good, I now understand. One more thing, is F_piston in Newton meter, which means that if I need to push the piston over 5 meters, I'll have to multiply F_piston by 5 to know how much energy I used in total?

 

[voko]

A force is always in simply Newtons. Newton-metres = Joules give you work (or energy).

 

[Lucini]

Ok! So if I understand correctly, F_piston * 5 meters will give me a result in N m, since Force * Distance = Torque, which is measured in N m, right?

Sorry to ask all these questions, but I feel like I'm close to understanding it and I want to make sure I do!
Thanks voko!

 

[voko]

Yes, the torque unit is also Newton times metre, so we use Joule for work and energy, and Newton-metre for torque.

But even if they are measured in units of the same dimension, they are difference things. When you compute work, you only account for the distance in the direction of force. When you compute torque you only the distance perpendicular to force.