# Calculating saturation and dew point given pressure and temperature

1. Dec 23, 2016

### InquisitiveOne

Hi guys and gals. Hobbyist, not a student. I've had great input here in the past. Thank you in advance for any direction and help figuring this out.

-Seth-

1. The problem statement, all variables and given/known data

I'd like to learn how to calculate relative humidity, saturation, and dew point under differing pressures and temperatures. All the information I've been able to find on-line seems to ignore one or the other and assumes a constant condition for the ignored value. Maybe there is a good reason for this?

Anyway, my inquisition is turbo charged engine related. I'd like to learn to calculate:
For given atmospheric conditions of, let's say, Sea Level, 59ºF, 90%RH

If this air were compressed to 29.4psia (14.7psig) and after compression the air was 221ºF, what would it's resulting relative humidity be, whats the saturation point, and what's the dew point?

If the compressed air (above), 29.4psia & 221ºF, is cooled to 84ºF, what would it's resulting relative humidity be, whats the saturation point, and what's the dew point?

What if the pressure were lowered slightly too, to say... 28psia and 84ºF?

2. Relevant equations

This is my biggest problem. I'm looking high and low for an equation and can't seem to find one that does what I want.

I want to input pressure and temperature and have the solution indicate how much water air can hold under those conditions.

Any direction at all is greatly appreciated.

3. The attempt at a solution

I have no idea where to start. I'm almost wondering if pressure doesn't have anything to do with it. But it must. Mustn't it?

If this inquiry is better suited in a different section, by all means move it please.

Any links to something that might explain this or just an equation I can try out is greatly appreciated. Or, if you can kind of explain in layman's terms the relationship and effects temperature and pressure have on how much water air can hold.

2. Dec 23, 2016

### Staff: Mentor

It's been a long time since I worked with them, but have you looked at "Steam Tables" to see if that helps? You could do a Google Images search on that term, and follow some of the images (of the tables of numbers and graphs, not the metal tables that are used to steam vegetables...) to their associated web pages...

Last edited: Dec 23, 2016
3. Dec 31, 2016

### InquisitiveOne

Thanks berkeman. Gave 'em a look. Nothing clicked. Perhaps it will become more clear to me as I learn.

So, from what I've read, and this part is starting to make sense... The statement "how much water air can hold" is more of an over simplified point of view than a statement of fact. It seems that statement assumes that the air and water in the air are the same temperature, which for all intense purposes wouldn't matter much because it would be mostly true. However, what's of real importance is what temperature the water vapor is. In that, the air doesn't "hold" water so much as it's displaced by water (open circuit) based on the partial pressure of the water, based on the water temperature. Or, in constant volume, will raise the pressure of the air to the point that their partial pressures are in equilibrium. Still wrapping my mind around the partial pressure thing.

Been trying to figure out and transpose the Arden Buck Equation for predicting Vapor Pressure ( https://en.wikipedia.org/wiki/Arden_Buck_equation ) into a spreadsheet and am missing something. Could use a little help. Apparently, I'm not as math equation savvy as I once was.

$$P_s(T) = 0.61121~ exp \left( \left( 18.678 - \frac{T}{234.5} \right) \left( \frac{T}{257.14+T} \right) \right)$$

where:

• Ps(T) is the saturation vapor pressure in kPa
• exp is the natural (i.e. base e exponential function
• T is the air temperature in degrees Celsius

Pretty sure I've got everything in the brackets right, but I'm a little confused with the "exp". I've tried it as meaning ^ or to the power of, and I've tried it as being the constant 2.71828 in a few configurations... Not getting the calculation to spit out anything even close to 101.325 given using 100 for T. How should this look in a spreadsheet?

4. Dec 31, 2016

### Cutter Ketch

The exponential function means the constant "e" which is 2.71828 etc raised to th power of the argument (what's in the parentheses). So exp(5) means e5. The spreadsheet will have an exponential function so you should be able to enter "exp(5)" and it will work fine.

It seems the saturation pressure is only part of what you want to know. Find the saturation pressure at the intake (1 atm and ambient temperature). Multiply that by the relative humidity. That is the partial pressure of water vapor in the intake air. (Don't worry about "partial". It just means you are only considering one part of the mixture. It's just pressure). You can find the partial pressure of water vapor in the compressed volume using the ideal gas law PV = nRT. (Ideal, but probably close enough). You can also find the saturation pressure under compression. If the actual pressure from step 1 is greater than the saturation pressure water will condense out (become a fog of droplets) and the partial pressure will be the saturation pressure. If the partial pressure is less than the saturation pressure, the ratio is your new relative humidity. (For whatever that is worth)

5. Jan 1, 2017

### InquisitiveOne

BINGO! Thank you.

So in the spreadsheet it looks like =0.61121*2.71828^((18.678-(T/234.5))*(T/(257.14+T))) Where T is the cell in the spreadsheet that I enter the value for temperature.
As stated in the description of the equation, it's close but not 100% accurate. Good enough for what I'm doing. One step closer to figuring out what I want.

I'll let the rest of what you said sink in a little, fiddle with this a bit, and get back to you guys on progress. I really appreciate the help guys.