3
$\begingroup$

A lot of times, dew point is focused primarily upon temperature and relative humidity. However, that same point of saturation is affected by the pressure, but I can't find a formula, or law even, that discusses this.

It's possible that my premise of the relationship between dew point and pressure may be inaccurate, but if so, how can we then explain that a refrigerant may be a vapor at 210°F in a 280psi container, have a dew point of 125°F, but also be a vapor at 75°F in a 70psi container.

Whether the substance is water, freon, or any liquid, there should be relationships between pressure and dew points (saturation points, condensation points, etc - they're all meaning the same thing). We can clearly see with freon there's some relationship between pressure and temperatures where the rate of condensation is greater than that of the evaporation, but I can't seem to find any laws/formulas for this.

$\endgroup$
5
  • $\begingroup$ I'm a bit confused, since in the first paragraph you talk about 'relative humidity' which by default I associate with water, while in the second paragraph you talk about a refrigerant. $\endgroup$
    – Jon Custer
    Nov 27, 2017 at 20:55
  • $\begingroup$ You're right - humidity is confined to water only, but there's lots of other substances that condense/vaporize as well. Perhaps you can substitute my use of "humidity" for "the amount of gas in the air relative to its condensation temperature." If there's a term for that, it may be beneficial to know that and I'll edit the question. The bigger question lies in the 2nd paragraph though. $\endgroup$ Nov 27, 2017 at 21:12
  • $\begingroup$ OK, so now we can clarify that second paragraph. Yes, you can have a substance that would be all vapor at 210F and still have vapor pressure at 75F (in equilibrium over liquid). $\endgroup$
    – Jon Custer
    Nov 27, 2017 at 21:21
  • $\begingroup$ Making this a comment, because I'm not certain, but I think you want the Clausius-Clapeyron relation, which allows you to find the dividing line in a phase diagram. $\endgroup$ Nov 29, 2017 at 23:00
  • 1
    $\begingroup$ @TonyDiNitto, research the Antoine equation. This equation establishes the relationship between vapor pressure and temperature for pure substances. See en.wikipedia.org/wiki/Antoine_equation $\endgroup$ Nov 30, 2017 at 1:06

4 Answers 4

4
+25
$\begingroup$

For single-component liquids, boiling point = condensation point. Both are the same temperature for a given pressure. Let's talk boiling.

Liquids boil when the vapor pressure of the liquid equals the pressure of the surrounding gas (e.g. 1 atm for open containers at sea level). As you raise the temperature of a liquid, the vapor pressure increases until it equals the pressure of the surrounding gas at which point it boils. If you reduce the pressure of the surrounding gas, then you do not need to raise the temperature of the liquid as much anymore. This is why water boils at 82 C instead of 100 C on Everest-- because the atmospheric pressure is lower:

enter image description here

If you have the vapor pressure $P_1$ of a pure substance at one temperature $T_1$ you can calculate the vapor pressure $P_2$ at a second temperature $T_2$ using the Clausius-Clapeyron equation:

$$ \ln\frac{P_2}{P_1} = \frac{-\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) $$

where
$P$ = vapor pressure
$\Delta H_{vap}$ = enthalpy of vaporization in $J/mol$
$R$ = the gas constant = $8.3145\ J/mol \cdot K$
$T$ = temperature in $K$

$\endgroup$
1
$\begingroup$

The common use of the term "dew point" is actually the "Atmospheric Dew Point" which doesn't take account the pressure of a system, as it's presumed to be at 1 atm. Many readily found formulas on dew point are formulas for atmospheric dew point, as they focus solely on temperature and the amount of gas (eg. water vapor) trapped the air.

The other important variable in what affects dew point is the pressure, which is where Pressure Dew Point (PDP) comes into play.

As a rule of thumb (without the use of complex equations), compression increases/raises the dew point temperature and expansion/decompression lowers the dew point.

This makes sense considering if you have a fixed amount of vapor in the air, and expand/decompress that vapor, the overall percentage of moisture in the air will be smaller, meaning the relative humidity percentage will decrease, and therefore lower the dew point.

At this point, the standard atmospheric dew point formulas can then be applied to find dew points at different levels of pressure.

As an actual example, the dew point for a parcel of air at 200 PSIg would have a dew point at -40°F, whereas the same parcel at 5 PSIg, significantly less compressed than the 200 PSIg parcel, would have a dew point of -77°F.

This has important applications in really understanding how things fully work like heat pumps, air dryers, food processing plants, electronics manufacturing, and more.

$\endgroup$
0
$\begingroup$

Even though I am not an expert on this, I found some information that can help. Check out this website

https://www.vaisala.com/sites/default/files/documents/Dew-point-compressed-air-Application-note-B210991EN-B-LOW-v1.pdf

and this website

https://en.wikipedia.org/wiki/File:Dewpoint.jpg

$\endgroup$
1
  • $\begingroup$ This site actually says, "changing the pressure of a gas changes the dew point temperature of the gas." Which is what I've known from observation of Freon gasses, but yet I could never find anyone ever talking about this. $\endgroup$ Dec 1, 2017 at 15:28
0
$\begingroup$

Disclaimer: I'm not a physicist or chemist, just a software developer who struggled with all those water vapor related concepts for a few weeks himself.

Dew-point vs. boiling-point

I'm not sure if this holds true for all substances, but water does not only vaporize when it reaches it's boiling point. Instead, a few molecules will always leave the surface of liquid or even frozen water. Depending on the partial water vapor pressure of the surrounding gas mixture, those might be balanced by the water molecules that condense / sublime back into liquid or solid phase. The temperature at which those two processes balance is called dew-point.

The dew-point depends on

  • the temperature (of the water in all it's states)
  • the partial water vapor pressure
  • the surface geometry of the liquid / solid water
  • presence and abundance of particles suspended in the air mixture that promote condensation on their surface
  • presence and concentration of salt or other impurities on the surface of the liquid / solid water

As user @pentane has stated in his answer, the boiling-point does depend on the total air pressure, the dew-point in contrast depends on the partial water vapor pressure instead. Unlike the boiling-point,

The dew-point does NOT depend on the total air pressure in the system!

The Pressure Dew Point (PDP) formulas that user @tony-dinitto talks about in his answer let you calculate how the dew-point changes when you change the pressure in the system while holding the percentage of water vapor in the air mixture constant.

If you're like me dealing with a system, where the sensor measures temperature, relative humidity and total air pressure "at the same time" - and you don't need to predict what would happen if you encapsulated this exact air mixture and compressed or expanded it, then Pressure Dew Point (PDP) formulas are of no use to you.

According to Wikipedia, the Arden Buck Equation is the most accurate formula to calculate the saturated water vapor pressure for a given temperature. Using this fact I came up with the following way to

calculate the dew-point:

  1. Calculate saturated water vapor pressure for current temperature using the Arden Buck Equation.
  2. Calculate the current water vapor pressure by multiplying the saturated water vapor by the relative humidity as a factor (divide by 100 if given as percent)
  3. Use the inverse of the Arden Buck Equation to calculate which temperature would have the current water vapor pressure as saturated water vapor pressure.

Since for me it was quite difficult to find the inverse of the Arden Buck Equation, I'll post what I came up with to make it easier for readers who try to follow my steps:

let (a, b, c, d) = (611.21, 18.678, 234.5, 257.14); // empirical constants of the Arden Buck Equation for temperatures > 0°C
let g = (actual_water_vapor_pressure_pa / a).ln();
-0.5 * c * (((b - g).powi(2) - (4.0 * d * g) / c).sqrt() + g - b)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.