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If I have a 1m3 sealed rigid water-filled container with one outlet through which a piston acts, how much force does the piston need to exert to raise the water pressure 100kpa? Does the area through which the piston acts make a difference? Also how far would the piston move during this? Thanks very much to everyone in advance

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closed as off-topic by Steeven, stafusa, John Rennie, Emilio Pisanty, JMac Nov 2 '17 at 11:07

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  • $\begingroup$ F/a. I still don't understand how to apply this to the problem $\endgroup$ – Oliver Walters Nov 1 '17 at 20:25
  • $\begingroup$ Look apologies if this is a simple question to you but it's not simple to me. $\endgroup$ – Oliver Walters Nov 1 '17 at 20:28
  • $\begingroup$ I can see it doesn't make a difference the are - but then I don't understand the first main question hence it puts me in doubt about that. It's the virtually incompressible liquid part that confuses me. $\endgroup$ – Oliver Walters Nov 1 '17 at 20:31
  • $\begingroup$ *area I meant to say $\endgroup$ – Oliver Walters Nov 1 '17 at 20:32
  • $\begingroup$ I suppose I meant does it matter the volume of water, rather than the acting area? Cheers $\endgroup$ – Oliver Walters Nov 1 '17 at 20:49
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You are given the pressure, and remember that Pascal=Newton/meter^2=pressure. The water exerts the same pressure everywhere, including on the face of the piston. Multiply the area (which you need to be given) of the piston by the pressure to get the piston force. Use the Bulk Modulus of water (=2.2 GPa) to calculate the fractional volume change of the 1 m^3 of water. This volume change is the volume the piston moves down its bore.

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  • $\begingroup$ Ok cheers. That answers my question. I'm still confused by it - for example that the same force gives the same pressure throughout the liquid whatever volume of liquid you are compressing. Also that, by using a piston of say 1 molecule width, a force of just a few grams but on that tiny piston could be enough, going on what you're saying, to make the pressure of a million cubic km of water, contained in a thick planet sized steel sphere, to rupture it??? There must be more to it that I am not understanding. Anyway thanks $\endgroup$ – Oliver Walters Nov 2 '17 at 0:05
  • $\begingroup$ Force is a vector; if you add the forces on all the walls enclosing a pressurized bit of water, you'd get... zero. The question requires you to choose ONE wall. Or something equivalent to one wall. $\endgroup$ – Whit3rd Nov 2 '17 at 6:11
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Gary Godfrey answered the calculation portion of your question, so I will answer you additional questions with regards to the pressurization of the liquid.


So then if I apply the same force but over half the area, the pressure throughout the water is doubled?

Think about it this way:

At all times the liquid is exerting some outwards pressure on the boundaries of its container, and, if we do this in empty space (away from massive bodies for simplicities sake), the pressure is the same in all directions—over every square meter of the surface.

Let’s say you cut a whole in one side and stick a piston in. Given that the liquid is exerting some outwards pressure, let’s say $P=10\text{kPa}$, the piston must then also exert the same pressure $P$ to stay level with the wall. From this you can see that the force needed to support the piston is proportional to its area:

$$F=P\cdot A$$

Now let’s examine what will happen if we half the area of the piston while maintaining its force.

Given that we know $P\propto\frac{1}{A}$, if $A$ is halved, $P$ will be doubled, resulting in $P_{\text{piston}}=20\text{kPa}$. This does not, however, mean that the water pressure also raises to $20\text{kPa}$.

The result of such an event will depend on the volume of the container compared to the area of the piston’s surface. If the surface of the piston had a large enough area, relative to the container, it could compress the liquid sufficiently such that the pressure of the liquid would raise to balance—and stop the motion of—the piston; however, if the surface area of the piston were relatively small, the piston could continue extending through the entire container without the compression of the liquid being great enough to raise the pressure of the liquid to the balancing point.


TL:DR

Changing the area of the piston does not change the liquid’s hydrostatic pressure—at least not directly—it only changes the pressure being exerted by the piston onto the liquid.

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  • $\begingroup$ Right. I think that shows me where I'm going wrong, though it will take a while to get my head around it. Cheers! $\endgroup$ – Oliver Walters Nov 2 '17 at 9:49

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