Pressure change from cooling liquid.

[aph]

I have an application where I need to calculate the vacuum (pressure change below atmospheric) generated from a cooling liquid in a semi-rigid container.

If the container is filled near full (e.g. 2 litres) with a liquid at e.g. 70C, sealed and left to cool to ambient (e.g. 20C) there is a pressure differential to atmospheric generated from the volume change of the cooling liquid that equalises when the seal is broken.

How do I calculate what that pressure differential (vacuum effect) is?

How much will it vary depending on the size of the container?

 

[cragar]

(p1V1)/(T1)=(p2V2)/(T2)
p=pressure v=volume T=temperature .
Combined gas law
I don't know if this is quite what you need .

 

[Bob S]

Cragar's answer is correct for liquids with a low vapor pressure. For liquids with a high vapor pressure, like boiling water, the partial pressure of the steam as the liquid cools below 100 degrees centigrade must be taken into account. The vapor pressure of water (steam) drops from 760 mm (Hg) at 100 degrees C to ~355 mm at 80 C.

Bob S

 

[aph]

Thanks very much for your help.

As I never will fill at near boiling point or above I will assume use of the combined gas law without consideration of vapor pressure changes.

So, if P1=101.3kPa, T1=345K, V1=2000ml
and T2=293K, V2=1958ml

(- V2 comes from asssumption that 1ml of water at 20C = 1g and at 72C 1ml = 0.979g so I multiplyed 2000 x 0.979)

I get, P2 = 87.9kPa

Therefore, pressure differential (vacuum effect) is 13.4kPa.

That is, when I break the seal the pressure will equalise and I will hear the air rush in.

The main purpose is to determine a negative pressure to apply to the closed container to see the effect on different container designs.

Am I on the right track or off the mark?

Thanks again.

 

[PJC01]

I can provide some insights on the pressure change that occurs when a liquid cools in a semi-rigid container. The pressure change is a result of the volume change of the cooling liquid as it transitions from a higher temperature to a lower temperature.

To calculate the pressure differential or vacuum effect, you will need to take into account the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and inversely proportional to its volume. In this case, the cooling liquid is considered a gas, and its volume is changing due to the change in temperature.

To calculate the pressure differential, you will need to know the initial volume of the liquid at 70C, the final volume of the liquid at 20C, and the atmospheric pressure. You can then use the ideal gas law equation (P1V1 = P2V2) to calculate the pressure differential.

The size of the container will also have an impact on the pressure differential. A larger container will have a larger volume change, resulting in a greater pressure differential. However, the exact amount of variation will depend on the specific dimensions of the container and the properties of the cooling liquid.

In conclusion, to calculate the pressure differential from a cooling liquid in a semi-rigid container, you will need to use the ideal gas law equation and take into account the initial and final volumes of the liquid, as well as the atmospheric pressure. The size of the container will also play a role in the pressure differential, but the exact amount of variation will depend on several factors.