# A simple logical problem regarding pressure and force

Suppose a 2 meter long pipe, with area of cross section is $1 m^2$. The pipe is vertically placed, with the bottom end closed. Inside the pipe, at the closed bottom end, there is $1 m^3$ of air, closed with a disc. This disc is movable, like a piston in a cylinder (the cylinder is the pipe).

Now, I'm pressing on the disc downwards and compressing the air to 5 bar of pressure (the volume of the air decreased).

Now I'm pouring water on to the disc, $1000$ $kg$ of water.

Now in the pipe,there is the compressed air at the very bottom, then the piston on it, and on the piston, filled with water of $1 m$ height ($1 m^3$ i.e., $1$ ton of weight).

If I release the pressure of the compressed air at the bottom very suddenly, UP TO WHAT HEIGHT will the water column will throw off(like a bullet)?

Consider the atmospheric pressure also, which is acting downwards,when it is shooting up..

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If you release the compressed air, won't the disc and the water on it fall? I'm confused by the question, in that, I may not be picturing the correct physical system. –  Alan Rominger Feb 6 '13 at 12:36
I think he means "release the force which compresses the air to 5bar". –  Christoph Feb 6 '13 at 13:35
Is the air initially in a space of 1m^3, then decreases in volume once it is compressed to 5 barr? Or is the 1m^3 after the gas is compressed? –  Mew Feb 6 '13 at 13:50
At first the air has 1m^3 volume, aftr compressing, it's volume decreased considerably. Then we release the force which compresses it,and the volume once again becomes 1m^3, having no additional pressure. –  Jayan Feb 7 '13 at 7:27

The way I see it, the air will expand, probably adiabatically ("suddenly", so that there's no time for heat transfer out of the piston), until the pressure in the air volume is 1 + 0.1 bar (i.e. atmospheric pressure + hydrostatic pressure from the water). Mind that the pressure in the air volume was already 1bar.

Also, "atmospheric pressure also, which is acting downwards" is imo incorrect - the pipe being in the atmosphere, the atmospheric pressure is acting from all sides, not only downwards.

I won't do the calculations for you, as this is homework, but you should have everything you need to apply a couple formulas from your textbook.

Some more considerations/assumptions: The piston must be assumed massless, otherwise its weight would already compress the air volume initially. The air volume after compression has been given time to cool down after the compression to 5 bar. Regarding the "sudden"ness - this needs further thought in reality - the timescale of expansion (i.e. how long will it take - you have to consider you have to accelerate 1ton of water, and the acceleration won't be constant) must be small compared to the timescale of heat transfer (conduction/convection/radiation) out of/into the volume in question. this makes it a bit more complicated to justify the assumption of adiabaticity.

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(Assumes that air occupies 1m^3 before it is compressed to 5 barr) To evaluate this problem we may use the ideal gas equation:

$PV = nRT$,

where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is a constant and $T$ is temperature. If we assume that temperature doesn't change, we may write this equation as simply:

$PV = k$, where $k$ is some constant.

Now we know the following:
Initial pressure on air = $P$ = 1 barr (from atmosphere)
Initial volume of air = $V$ = 1m^3

Therefore, we have

$PV = 1\times1 = 1 units = k$

Now once we compress the water to 5 barr, we have:
$PV = 5*V = 1$, therefore $V = 0.2 m^3$.

We then add water. The total downward pressure (excluding external force) is thus: $P_{total} = P_{water} + P_{air} = 0.1 + 1 = 1.1 barr$ Therefore,
$1.1V = 1$, therefore: $V = \frac{1}{1.1}m^3 = 0.91m^3$

Therefore the water will rise to a height of $0.91$m. Since the water occupies $1m$, there will be no spillage, which is consistent with our calculations.

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Also note the fact the water was compressed initially had no bearing on the final outcome. –  Mew Feb 6 '13 at 14:18
This is not a question of Raising of water. It is regarding the hight to which the "watetr bullet" will be shooted up. –  Jayan Feb 7 '13 at 7:37
Pretty sure you only edited in that part 3 hours ago. –  Mew Feb 7 '13 at 10:37
Ya . I didn't see the website's editors had edited it then.One more thing... You can take 1000kg weight, instead of water.Then you can neglect the air resistance.Also, assume that the piston has no considerable weight. –  Jayan Feb 8 '13 at 5:36

Assumptions:

1. Piston is massless
2. After compression, air temperature is allowed to come into equilibrium with environment
3. "Sudden release of pressure" == sudden release of piston
4. Expansion of gas below the piston is adiabatic
5. Gas below the piston is assumed to be ideal, and monatomic ($\gamma = 5/3$)
6. "Water shoots up like a bullet" - no, but the piston will move beyond the point of equilibrium, and maybe even past the end of the cylinder. Interpreting the question as "how high will the piston rise".

Now we can get to work. Initial pressure below piston (with water added) is 5.1 bar (5 + 0.1 bar for the 1000 kg of water spread over 1 $m^2$ = 100 $g/cm^2$)

Force on the water as a function of position of the piston (assuming that $x=x_0$ corresponds to the initial position of the piston):

\begin{align}\\ F &= P A\\ &= P_0 \left(\frac{V_0}{V}\right)^{5/3}A\\ &= P_0 \left(\frac{x_0}{x}\right)^{5/3}A\\ \end{align}

We can integrate this to get the work done. When the work done is equal to the increase in potential energy of the water, we have reached the top of the expansion. Note that we would expect this to happen at a point where the water does not spill out - if the air was of constant temperature, equilibrium would be reached when the piston was compressed to half a meter (since we expect the pressure at equilibrium to be 2 bar - one for the atmosphere, and one for the water). In reality the adiabatic expansion will cool the air some more, and the expansion will be less. But we're not interested in the equilibrium position, but the top of the motion - thus the need to integrate between $x_0$ and $x_t$ (the top of the expansion).

\begin{align}\\ W(x) &=\int_{x_0}^{x_t} P_0 \left(\frac{x_0}{x}\right)^{5/3}Adx\\ &=\frac{3}{2} P_0 A \left(x_0 \left(\frac{x_0}{x_t}\right)^\frac{2}{3} - x_0\right)\\ &=m g (x_t - x_0) \end{align}

Integrating numerically, I obtain $x_t = 1.13 m$ (for the bottom of the water; the top is of course 0.1 m higher). It is possible that I made a mistake, but that seems like a reasonable number. Note that if you don't make the assumption of adiabatic expansion, the water will shoot considerably higher - almost 5.4 m. To get that number, you have to recognize that the expanding air will have a pressure < 1 bar once you pass the mid point (so atmospheric pressure starts to slow down the water) and that this retarding force stops when the piston shoots out of the cylinder (at height 2m, the water will be in free flight). That might have been the intention of the question, but that would be a mistake, I think. The curve (of work done on the water as a function of height) looks like this for the non-adiabatic case ($\gamma=1$ instead of $\gamma=5/3$)

Note - I may have got my units messed up... advise that you look at this carefully using the above reasoning.

My conclusion - no bullet of water shooting out of the top unless you use "wrong physics". Disappointing really.

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