Estimate the temperature change due to vaporizing liquid

In summary, the pressure in the chamber will be less than the vapor pressure of the liquid, so the liquid will boil off and the pressure will drop.
  • #1
men5j2s
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TL;DR Summary
Spraying a compressed gas canister (like a deodorant spray) into a larger chamber causes cooling. How can I estimate the temperature drop is the system is closed?
I'm releasing pressurized propellant into another chamber which is at atmospheric pressure, I want to estimate the temperature to which that chamber will drop to and the rate at which the liquid (collecting in the new chamber) will boil.

The second chamber is open to atmosphere via a small orifice above the bottom of the chamber, therefore the liquid will collect in this 'sump' and boil off as the vapor escapes through the orifice.

Any ideas?
 
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  • #2
men5j2s said:
Summary: Spraying a compressed gas canister (like a deodorant spray) into a larger chamber causes cooling. How can I estimate the temperature drop is the system is closed?

I'm releasing pressurized propellant into another chamber which is at atmospheric pressure, I want to estimate the temperature to which that chamber will drop to and the rate at which the liquid (collecting in the new chamber) will boil.

The second chamber is open to atmosphere via a small orifice above the bottom of the chamber, therefore the liquid will collect in this 'sump' and boil off as the vapor escapes through the orifice.

Any ideas?
I'm unclear: are you spraying a gas or a liquid? You seem to have said both. A diagram might help.

You'll need to set up an equation describing the heat transfer: masses and temperature changes (and maybe heat of vaporization), then solve it.
 
  • #3
Hi Russ...

I am getting a bit overwhelmed by the theory in this field...

I have a liquefied gas (high pressure - 6bar)
Can is upside down (stem down), so gravity ensures that when I spray, I spray liquid.
I am spraying into a second container with a volume a little larger than that of the initial container
The liquid will vaporize instantly, then liquefy again as the pressure increases in the second container and keep vaporizing as the pressure decreases due to the hole in the second container that is open to atmospheric.

Please see the diagram to help explain.

I've calc's mass flow rate between chamber 1 and 2, but don't know how to get pressure, viscocity, temperature in chamber 2, to calc mass flow rate from chamber 2 to atmosphere.
 

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  • #4
Are you wanting to evaluate the transient conditions or the ultimate stabilized flow condition?
Either way the analyses are not going to be straight forward and, so much so, for the transient case that I have no suggestions for determining that initial flow case.

This is a two sequential orifice problem and the result depends upon the the ratio of the two orifice areas and the pressure differential across each of those orifices and matching the the mass flow across those two orifices.
For the stabilized flow condition the mass flow into the chamber will equal the mass flow out of the chamber and by establishing the required pressure in the chamber to balance those mass flows you will determine the pressure and therefore thermodynamic state of the fluid or vapor in that chamber.
Because you initially do not know either the stabilized pressure and therefore liquid vs vapor state in the second chamber and the mass flow rate into the chamber is liquid it is only dependent upon the pressure differential across the inlet nozzle; but, the discharge chamber discharge nozzle mass flow rate is dependent upon either the liquid vs vapor condition and the chamber pressure. As a result, this analysis will require a trial and error method for its solution. (With respect to your "known first orifice flow"; that, in fact cannot be known until you know the stabilized pressure in the chamber because that flow rate is a function of the differential pressure P1 - P2 across that orifice.)
I would suggest that if you have two already chosen orifice sizes, you do an analysis to determine if the selected balanced orifice mass flows of the two orifices results in a sufficient P2 pressure in the chamber that insures a sustained liquid chamber discharge flow; that way, the chamber temperature effect will be basically eliminated in your flow balance determination because, unlike the vapor flow, it will have little effect for on the discharge nozzle mass flow rate and you will have a more straight froward P1 vs P2 vs P3 analysis with P2 being the only unknown pressure. If that is the case, then you can immediately determine the thermodynamic state in the chamber.
Assuming you are starting with your supply bottle pressure equal to the vapor pressure then fortunately the inlet pressure should remain relatively constant until the liquid in that bottle is depleted. However, depending upon the height of the bottle and the weight of the liquid will be a liquid head (ρgh) pressure that must be added to the bottle pressure and will decline as the bottle level falls for a fully accurate analysis.
If you determine the balanced flow P2 is below the vapor pressure of your liquid then the analysis becomes more difficult because a vapor discharge orifice flow will need to be used for that orifice and the vapor temperature in the chamber must be included in that orifice's flow rate and mass flow determination.
 
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  • #5
Hi JBA,

Thank you for the detailed response..

I'm hoping to model a 'steady state' scenario.

I am assuming uniformity between the two orifices.

I have calculated mass flow rate across the first orifice, but to do that I had P1, T1 and the viscosity of the liquid in chamber 1, I have no clue how to get P2, T2 or viscosity 2 (presumably gas). I used 1bar for the pressure as it is initially at atmospheric.

I think (if I read you correctly) that my assumption of 1bar in chamber 2 is incorrect as I am assessing a steady state - If I have boiling liquid at the bottom of my chamber (2) with vaporized liquid escaping from the nozzle, does that mean that I must be almost at the vapor pressure in chamber 2 too?

I'm trying to put this into a step by step solution, but can't see the full story. so far I've used Poiseuilles law to give me flow rate (but I've probably used the wrong P2). I think I need to use it again to get flow rate at the second orifice, but am missing some inputs.
 
  • #6
As I stated above, for your type of an situation, you require a trial and error loop style of analysis.
Because your inlet nozzle mass flow will vary with any change in the chamber pressure and your outlet nozzle mass flow will vary with any change of of the chamber's fluid/vapor state and pressure.

Start this process for your selected nozzle sizes by assuming some arbitrary chamber pressure; and, since you have chosen the chamber pressure you can also determine the fluid/vapor state in the chamber; and, therefore, at the discharge nozzle inlet.

With this information, you can now select the correct equation for calculating the discharge nozzle flow i.e. either liquid or vapor, and determine both the inlet nozzle liquid mass flow and the exhaust nozzle mass flow for your selected chamber pressure and chamber fluid/vapor state.

You now compare the resulting two nozzles' mass flows to see if they are equal; if not, then select a new higher or lower chamber pressure and repeat the above analysis; and, continue in this trial and error manner until your newly selected chamber pressure finally results in an equal mass flow for both nozzles.

This is an extended and time consuming process but is the only way to manually solve the problem when you start with three unknowns (i.e inlet nozzle mass flow, fluid state and exhaust nozzle mass flow, all dependent upon only one common element, the chamber pressure.
 
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1. What is the process of vaporizing liquid?

Vaporization is the process of converting a liquid into a gas or vapor. This occurs when the temperature of the liquid reaches its boiling point, causing the molecules to gain enough energy to break free from the liquid's surface and enter the gas phase.

2. How does vaporizing liquid affect temperature?

When a liquid vaporizes, it absorbs heat energy from its surroundings in order to overcome the intermolecular forces that hold the molecules together. This results in a decrease in temperature of the remaining liquid and its surroundings.

3. Can the temperature change due to vaporizing liquid be accurately estimated?

Yes, the temperature change can be estimated using the equation Q = mL, where Q is the heat energy absorbed, m is the mass of the liquid, and L is the latent heat of vaporization. This equation takes into account the specific properties of the liquid and the amount of heat energy required for vaporization.

4. Does the temperature change due to vaporizing liquid depend on the type of liquid?

Yes, the temperature change will vary depending on the specific properties of the liquid, such as its boiling point and latent heat of vaporization. Different liquids will require different amounts of heat energy to vaporize, resulting in varying temperature changes.

5. Is the temperature change due to vaporizing liquid reversible?

Yes, the temperature change due to vaporizing liquid is reversible. When the vaporized gas or vapor condenses back into a liquid, it releases the same amount of heat energy that was absorbed during vaporization. This is known as the heat of condensation and results in a temperature increase in the surroundings.

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