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water steam pressure at ambient temp

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pardal

Automotive
Oct 17, 2001
444
Talking about psichrometry
Given T in Kelvin or absolute temp
For Pws [presure water steam]
Is this formula valid ?
for Water air from 0 to 200 C degree
ln Pws= a/t + b+c*t+d*t^2+e*t^3+f*lnT

for the followings values

a -5800.2206
b -5.5162256
c -0.048640239
d 4.18E-05
e -1.45E-08
f 6.5459673

I used it for 100 C dgree and give
p mbar 1014.217979 mbar , that is a normal atmospheric pressure

My concern is if it is valid for the air ambient temperature

I wan to calc the Relative humidity by reading the DryBulb and WETbulb temperature.

I have 2 data loggers and I modify one of them to read wet bulb temperature, so I can charge the data from it to a spread sheet and get the HR % .

maybe it is not a conventional way of doing , but is it what I have and will have for ever. (lack of money to buy a HR datalogger)

Hope you can give me any advice.

PS I have the spread sheet done.





Pardal
 
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I get 10 times lower values with your formula but i think this is a question of units. If I compare your values to my values calculated with the formula from International Critical values I get a 0,9% deviation at zero C and 0,2% deviation at 100 C.However I do not claim that my formula is better-both are good enough for technical purposes.I keep some other polynomial expression to calculate the Pws, but at the moment I cannot remember where they are. By the way if you are interested in a small VBa code to calculate air-steam properties I can send it to you. Contact me at dixie.m@email.si.
M777182
 
It may give you only one part of the equation i.e saturation vapor pressure. Now you have to find out the actual vapor pressure based on DBT and WBT.

Check this link,


Regards,


Eng-Tips.com : Solving your problems before you get them.
 
One approximation taken from Holmans book on heat transfer is:

Tdb = dry bulb temp = 40C
Twb = wet bulb temp = 25C

From steam tables
sat press at Tdb = Pgd = 73.75mbar
sat press at Twb = Pgw = 31.66mbar

Atmos press = Pa

Partial press of vapour = Pv
= Pgw - ((Pa-Pgw)(Tdb-Twb)/(1537.778-Twb)) = 21.927mbar

Relative humidity = Pv*100/Pgd = 29.73%

Specific humidity = 0.622*Pv/(Pa-Pv) = 0.01376 kgH2O/kg dry air

athomas236
 
Dears all people, thanks for your tip.



Pardal
 
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