Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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:

enter image description here

The water tank has a height of 2 meters, length of 6 meters, and width of 4 meters. If a piston, 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.

I understand there are a lot of forces to take into consideration, such as friction, hydrostatic pressure, and more, but I'm trying to figure out if it would require a lot of energy or would it remain minimal? What is the force that will require the most energy? Hydrostatic pressure?

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!

Thanks a lot for the help!

share|improve this question
Hi @Yoan. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. In particular, please note that it doesn't have to be actual homework for the homework tag to apply. –  Qmechanic Mar 4 at 21:12
Hello Qmechanic, thank you for your message. This is not a homework, I am not a student or a professional in a scientific area, I am just trying to understand how forces work! –  Yoan Mar 5 at 16:02
add comment

3 Answers 3

up vote 0 down vote accepted

I don't think that the size of the hole has an influence on the energy required. Friction is a complicated matter here, not easy to calculate. Try to calculate the pressure imposed by the water on the piston. The force working against that pressure will require most energy.


The pressure on the piston isn't uniform along its height in this case. At any given height the pressure is $p=\rho g z$, where $z$ is the vertical coordinate. To get the force required to counteract the pressure imposed on the piston You need to integrate this formula across the piston surface:

$F=\int_Sda $ = $F=\rho g \int_0^w \int_0^h zdzdx$

Finally, after the integration:

$F_{piston}= \frac{1}{2} \rho g w h^2 $,

where $\rho$ is the density of the fluid, $g$ is the gravitational acceleration, $w$ is the tank's (and piston's) width and $h$ is the tank's (and piston's) height.

Having the force all You need to do is multiply it by tank's length to get the work (and energy) required.

share|improve this answer
add comment

Hint: What potential energy does the water have with respect to the heigh of the container valve?

share|improve this answer
If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. –  Ali Mar 4 at 12:25
It isn't meant as new question. Compute the energy needed to lift the respective layers of water above the valve. You don't even need integration, if all layers have the same surface $A= 6m\cdot 4m$. Then it's lifting the weight seen as point mass half the high => $E = m* g* h/2$ –  Lord_Gestalter Mar 4 at 13:36
I will edit your post to reflect that. –  Ali Mar 4 at 13:39
add comment

This can be treated by use of Bernoulli Equation, which says that the sum of geodetic, dynamic and static pressure is constant for every flow cross section of a frictionless fluid system. On top of that, friction can be accounted separately and represent the other side of the Eqn.. For the fluid part, friction factors depend on Reynolds Number and geometry. They are multiplied by the corresponding dynamic pressures to obtain the fluid pressure losses. For the piston the friction factor gives the friction force by multiplication with the normal force (weight force). Or perhaps more realistically by the sealing force of the piston sealing. Those friction factors depend on tribology. As pressure times area is force, all of what you are interested in can be calculated that way.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.