B Cooling liquid in rigid container

[ABC83]

Hello, I have a sealed and rigid container, (can't change volume, can allow heat through), 1.0m x 1.0m x 1.0m in size, that is completely full with warm water, let's say at 40°C. The initial pressure in the container is 1atm and the volume of the water at 40°C perfectly matches the volume of the container. If I start to cool down the container and the water in it, to 20°C, is there any way to estimate the new pressure within the container? Thank you in advance.

 

[BvU]

Must be a sturdy container...

What have you found for $\rho(T)$ ?

If the liquid expands the pressure will rise humongously

And if it shrinks, it drops until ...

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[ABC83]

Thank you, my query is kind of theoretical, as I don't want to go in the details of the container flexibility, etc... :)
Please note that in my query the water cools down reducing its volume. My thought is that the pressure at the top of the container (so not including static head) as the water reduces its volume and creates a vacuum effect, should be... [fill the gap, as I don't want to influence the answer]

 

[BvU]

What happens if the pressure above a liquid is lowered (your 'vacuum effect') ?

 

[ABC83]

Yes, that's my question. We have a liquid that reduces in volume because of the reduction in temperature, in a container that doesn't gets smaller. So what happens and can we estimate the new pressure, presumably a lower pressure in the container that what we started with?

 

[BvU]

What will fill the 'vacuum'?

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[erobz]

You can't ignore the properties of the container. You say it shrinks with the fluid(perfectly), if that's the case pressure(at the surface) is still atmospheric in the fluid.

 

[erobz]

Uh oh @BvU is Skeptical. Maybe I'm missing something.

If no container, the fluid expands and contracts freely at atmospheric pressure?

 

[BvU]

The way I understand post #1, the container volume is constant...

 

[ABC83]

Yes, the container is sealed and rigid, so it doesn't changes shape or volume.

 

[erobz]

The way I understand post #1, the container volume is constant...

If the container is completely rigid, and is completely full with "virtually" incompressible fluid the slightest positive change in temp, and the tank would burst ( or the pressure would shoot to infinity ). If the temp decreased, it would go to absolute zero.

The reality is the "real" tank dictates the pressure by elastic deformation?

 

[BvU]

it would go to absolute zero

No!

@ABC83 :

What have you found for ρ(T) ?

 

[erobz]

No!

Are we allowing for dissolved gases to escape the fluid?

 

[ABC83]

Are we allowing for dissolved gases to escape the fluid?

It is a very good question and something I have not fully thought about. If we know the amount of gas dissolved, we could make an estimate on the resulting pressure with the gas equations, assuming that the gas volume will eventually fill the space. However... for the type of system I need to apply to answer to (heating system FYI), the goal would be to reduce the dissolved gases as much as possible before the system is in operation. We don't use vacuum degassers, but we do heat up the system to expel the trapped air, as much as feasible. Going back to our container, I would assume that all the gases have been removed from the water, as this would be the ultimate goal in a perfect system. So: assume no air dissolved in the water.

My guess is that the final pressure measured at the top of the container, will be slightly above 0 bar absolute, but I am not entirely sure...

 

[jbriggs444]

If the container is completely rigid, and is completely full with "virtually" incompressible fluid the slightest positive change in temp, and the tank would burst ( or the pressure would shoot to infinity ). If the temp decreased, it would go to absolute zero.

The reality is the "real" tank dictates the pressure by elastic deformation?

Edit: After sanity checking against a competing calculation by @TonyStewart, something smelled wrong. I've corrected a factor of 100 blunder in the computation below.

If the tank is ideally inflexible (a decent approximation for a very strong, very thick tank) then we still have to worry about the water. Real water is compressible.

If you chill the water (toward about 4 degrees Celsius) it will tend to shrink away from the container walls. However, this will relieve the atmospheric pressure to which it had been subject. It will expand as a result.

What is the bulk modulus of water? We can ask Google... 300,000 PSI. Imperial units. Bah. We can ask again and get the answer as $2 \times 10^{9} \text{ Pa}$. Even more useful for our purposes would be the same figure expressed in atmospheres. Call it 20,000 atmospheres.

This says that if we reduce the volume by a small fraction (e.g. one percent) then we will get a pressure increase of that same percentage of the bulk modulus. So one percent volume decrease is going to result in a pressure increase by 200 atmospheres. The reverse is also valid. One percent volume increase = 200 atmospheres of pressure decrease.

Or we can turn it around the other way. One atmosphere of pressure reduction is 0.5 0.005 percent of volumetric expansion. Edit: Note factor of 100 error here

How much temperature reduction do we need to get 0.5 0.005 percent of volumetric reduction? The formula for water is nasty. But we can Google up some figures.

water at 20 C: 0.000207 (1/degree C)
water at 30 C: 0.000303 (1/degree C)
water at 40 C: 0.000385 (1/degree C)

If we drop 5 degrees from 40 C, that's 0.2 percent decrease in volume. We need another 0.3 percent.
If we drop 10 degrees from 35 C, that's 0.3 percent decrease in volume. Bingo.

Edit: With the factor of 100 error corrected, the calculation is easier. It is $\frac{0.00005}{0.000385} = 0.13$ degrees Celsius. (Not 15 degrees Celsius).

If my calculations are correct, when the 40 C water has cooled to 25 39.87 degrees C, it will have reduced to near zero absolute pressure and will either be boiling or at the threshold of boiling. The vapor pressure of this water will be about 24 55 torr. The low pressure will have allowed the water to remain in contact with the vessel walls.

However, as the water cools further, it will shrink away from the walls, boiling and leaving water vapor in the head space that results. The vapor in the head space will condense away to lower and lower absolute pressures as temperature is further reduced.

At some point, freezing will ensue. The water will expand to fill in the head space. Volume would nominally increase by about 10%. The pressure will increase to somewhere in the neighborhood of 2000 atmospheres (ten percent of the bulk modulus) at which point the pressure will suffice to keep the pressurized ice confined.

According to the state table of water, it looks like 2000 atmospheres at -20 C is indeed ice.

This is all pretty much back of the envelope calculations from a fellow who is not an engineer. Errors may exist.

 

[ABC83]

If my calculations are correct, when the 40 C water has cooled to 25 C, it will have reduced to near zero absolute pressure and will either be boiling or at the threshold of boiling. The vapor pressure of this water will be about 24 torr. The low pressure will have allowed the water to remain in contact with the vessel walls.

Thank you. This is what I am interested in, but at 20°C. In particular the pressure in the container. Probably easier if we consider it at the top, so we can disregard the effect of the static pressure of the water.

 

[Bystander]

I "smell" a PPM, verboten topic.

 

[BvU]

Goodness,

I was trying to let the OP discover that water has a vapour pressure and that will be the pressure in the container at 20 $^\circ$C.

(densities question wasn't answered by OP -- would lead to 'shrinking' when cooling from 40 to 20 $^\circ$C )

There was no mention of dissolved gases thus far....

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[ABC83]

I was trying to let the OP discover that water has a vapour pressure and that will be the pressure in the container at 20°C.

Thank you. :) I thought the same, so the final result would be 0.0231 atm absolute on the low pressure zone, that is located between the water layer and the top of the container, I think. At the bottom of the 1.0m tall container, should be approx. 0.1 atm higher, because of the static pressure.

 

[jbriggs444]

Thank you. :) I thought the same, so the final result would be 0.0231 atm absolute on the low pressure zone, that is located between the water layer and the top of the container, I think. At the bottom of the 1.0m tall container, should be approx. 0.1 atm higher, because of the static pressure.

Yes. I get the same for the vapor pressure of water at 20 degrees C.

The laborious calculation with the bulk modulus and volumetric expansion coefficient is relevant only if the starting temperature turns out to be low enough so that no head space at all results from cooling.

 

[erobz]

Goodness,

I was trying to let the OP discover that water has a vapour pressure and that will be the pressure in the container at 20 $^\circ$C.

(densities question wasn't answered by OP -- would lead to 'shrinking' when cooling from 40 to 20 $^\circ$C )

There was no mention of dissolved gases thus far....

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Well, my point was that there is a minimum idealization for the solution to this problem to exist, if that's vaporization pressure, so be it. It's hard to tell what people want in here, I was thinking they are testing where a particular idealization fails to produce a meaningful result.

 

[TonyStewart]

If water density, ρ=m/V at 1 atm is:
992.25 kg/m3 @ 40'C
998.19 kg/m3 @ 20'C
But neither mass nor volume may change.
The Bulk modulus B, may be used to determine the change in pressure, but it varies with bulk stress $\sigma _b$, and is the inverse of compressibility =~ 3.3 × 10^6 (psi)
But one needs to estimate the pressure in order to determine the Bulk modulus and here the pressure is a huge on the liquid.

ρ₁ = density of water at the initial temperature, T₁
β = coefficient of thermal expansion of water at each temperature
ρ₂ = density of water at the new temperature, T₂

I ran out of gas and may finish this when time permits.

 

[JT Smith]

Thank you. :) I thought the same, so the final result would be 0.0231 atm absolute on the low pressure zone, that is located between the water layer and the top of the container, I think. At the bottom of the 1.0m tall container, should be approx. 0.1 atm higher, because of the static pressure.

In the real world, unless you have boiled the water in advance, there will also typically be dissolved gases that will add to the vapor pressure. At 20°C that contribution will be pretty small, but in general it would be part of the answer.

 

[BvU]

If water density, ρ=m/V at 1 atm is:
992.25 kg/m3 @ 40'C
998.19 kg/m3 @ 20'C

Thank you. That's what I wanted the OP to dig up. So we start with 992.25 kg/m³ and end up with 992.25 kg at $20^\circ$ degrees. If the water is incompressible that would be about 0.99405 m³.
So one expects a water vapour bubble of about 6 liter at 23 mBar .
No need to worry about the amount of water that evaporated: here one has $\rho_g = 0.0173$ meaning 0.1 gram
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[jbriggs444]

here the pressure is a huge on the liquid.

Huge pressure? For a liquid at atmospheric pressure cooling within a container of fixed volume, the pressure would likely be somewhere between 0 and 1 atm absolute.

Surprisingly large negative pressures are possible for pure degassed water. However, tap water in containers of ordinary cleanliness will cavitate/boil readily.

 

[ABC83]

Thank you everyone for the quick and useful responses. :)