Version: 008
Date: Nov. 12, 1998
Updated: When necessary
Lines: about 720
The _Temp, Humidity & Dew Point_ ONA (Often Needed Answers)
Table-of-contents:
1) Introduction.
2) Formulae.
3) Examples.
4) Net Readings & Paper-Based Literature.
5) Committment.
6) Outlook.
7) Signature.
A plain text version of this text can also be found on:
http://www.agsci.kvl.dk/~bek/relhum.htm
1)
Introduction:
From the discussions on the newsgroup sci.geo.meteorology this is a
collection of some formulae and texts that reflect on connections
of temperature, humidity and dew point temperature (BeK):
Air will normally contain a certain amount of water vapour. The
maximum amount of water vapour, that air can contain, depends on
the temperature and, for certain temperature ranges, also on whether
the air is near to a water or ice surface. If you have a closed con-
tainer with water and air (like a beaker) then an equilibrium
will develop, where the air will contain as much vapour as it can.
The air will then be saturated with respect to water vapour.
The real world outside is not closed, so that the air normally will
contain less vapour as it could. Sources of vapour are evaporation
processes from soil-, water- and ice surfaces, transpiration from plants
and respiration from animals. The expression "evapotranspiration"
takes into consideration plants' large share of evaporation over
land areas.
Sinks of water vapour are clouds or condensation on surfaces.
"Dew" is created when a surface temperature has such a low temperature
that the air chills to the dew point and the water vapour condenses
to a liquid.
"Hoar" is the same process only occuring at such cold temperatures
that the water solidifies directly.
Physically, at the dew point temperature the vapour loses the energy
that it gained at evaporation, the latent energy, again.
The precipitable water (total column water vapor) is strongly
correlated (r > 0.9) with the surface dew point on most days.
Exceptions to the rule include days when a cold front has passed
and during other transient events. (Kerry Andersen)
[I presume that the cold air after a cold front is depleted of
water vapour, and therefore the correlation breaks down. (BeK)]
2)
Formulae:
Enough for dry physical theories; here comes the practice.
For some people skipping this and going directly to the examples
would be the most rewarding. Especially as they treat the conversion
of relative humidity and psychrometer temperatures. (BeK)
Vapor pressure (e) is the fraction of the ambient pressure that is
due to the fraction of water vapor in the air.
Saturation vapor pressure (es) is the maximum vapor pressure that the
air can support (non supersaturated) at a given temperature.
e can vary from 0 (verrry dry) to the maximum, es.
es is a function of temperature es(T), I give the formula later on.
Relative humidity (RH) is 100% times the ratio of the environmental
vapour pressure, e, to the saturation vapour pressure es(T).
RH = 100% * e/es(T)
The environmental vapour pressure is the saturation vapour pressure
at the physical dew point or
e = es(Td)
so RH becomes
RH = 100% * es(Td)/es(T)
In other words: if you have a parcel of air and cool it until the
water vapor in it condenses, then you have reached the saturation point.
At this point you will measure the same vapour pressure as in your
original air probe.
Some more elaborate expressions follow here:
e0 = reference saturation vapor pressure (=es at a certain temp,
= 6.11 hPa usually 0 deg C)
Lv = latent heat of vaporization of water (2.5 * 10^6 Joules
per kilogram)
Rv = gas constant for water vapor (461.5 Joules / Kelvin / kilogram)
T0 = reference temperature (273.15 Kelvin, Kelvin = degree C +
Td = dew point temperature (Kelvin) 273.15)
T = ambient air temperature (Kelvin)
e = e0 * exp( Lv/Rv * (1/T0 - 1/Td))
es = e0 * exp( Lv/Rv * (1/T0 - 1/T))
RH= e/es * (100%) = relative humidity !!!!
So just above is the answer to many questions in the direction of
how to calculate the relative humidity if you have the dew point
and air temperature.
There are some simple and more complicated formulas for the
saturation vapour pressure at a given temperature.
A simple first guess (assuming the latent heat of vaporization is
constant with temperature) would be:
log10(es) = 9.4041 - 2354/T
or
ln(es) = 21.564 - 5420/T
where T is in Kelvin (i.e., 273.15+T(C)). {After inverting the
logarithms, es is given in hPa.}
Another approximation (Magnus' formula) would be
log10(es) = -2937.4/T - 4.9283*log10(T) + 23.5470
In the following the input gets a little more complicated. Here we
also shall distinguish between the saturation vapour pressure over
ice or water. Both are different, as the molecular forces bind much
more in an ice crystal than in a water bobble. So the saturation
pressure esW will be larger than esI (W for water, I for ice).
1. Saturated vapor pressure (es):
es = 6.39 * exp ( A * (T -Tw) / T ) [in hPa]
A = 19.65
Tw= 273 K
This formula is physically correct and almost empirical, a better
description is given by using 6.11 and A=17.27 (Monteith&Unsworth
1990).
Or:
es = 6.1078 * 10 ** ((T * A)/(T + B))
\
temperature in C
A = 7.5 } for use in vapor pressure
B = 237.3 } with respect to WATER
* A = 9.5 } for use in vapor pressure
B = 265.5 } with respect to ICE
2. Ambient vapor pressure (e):
dew point temperature in degrees C.
/
e = 6.1078 * 10 ** ((TD * A)/(TD + B)) in hPa
So, once you have the dewpoint temperature you can establish the
actual ambient vapour pressure. But in order to do this, you need to
find the dewpoint temperature. And this is easier written than done.
3. Dewpoint temperature
How do you find out the dewpoint temperature? Let us look a little
bit at the physics.
Now, unfortunately, with the psychrometers you cannot measure the
dewpoint temperature directly. Instead you measure the "wet-bulb"
temperature, Tw, which you can infer the ambient vapour pressure
with. In order to do this, you exploit some assumptions on energy
conservation between the airflow over the thermometer and the water
evaporated.
In the literature you find s, often gamma, to indicate the so-called
"psychrometer constant". In fact (see also the warnings below), it
is by far a constant, although for most purposes it suffices to
assume that it is.
s = cp * P / Lv / epsilon
cp= air's specific heat at constant pressure
P = atmospheric pressure (this can change, can it not!)
Lv= latent heat of water (slightly temperature dependent!)
Only epsilon is really constant 0.62197 (normally taken to be 0.622).
Also the derivative of the water vapour curve with respect to
temperature, Gamma, plays a role.
The following formula is almost fully correct:
e = es(T) - (Gamma + s) * (T - Tw)
Wet Bulb
Vapour Pressure
Dew Point (P365, Smithsonian for first part)
Ew - e
------- = .000660 (1 + .00115*Tw)
P(T-Tw)
Therefore:
e = Ew - P (T-Tw) (.000660) (1 + .00115 Tw)
e = Vapor pressure in air
Ew = Saturated vapor pressure at temperature T {=es(T)}
P = Total barometric pressure (units same as Ew, e)
T = Air temperature (degrees C.)
Tw = Wet bulb temperature (degrees C.)
e is the vapor pressure in the air, which is the vapor pressure at
the dew point temperature. To solve for the dew point temperature,
let:
C = log (e/6.1078)
10
Then:
C Td + C B = A Td
B*C
Td = --- Dew point in degrees C
A-C
where A = 7.5
B = 237.3
4. Mixing ratio (W):
expresses the water vapour pressure against the pressure of
the dry air and corrected by the ratio of the gas constants for
water vapour R and dry air, R :
V L
vapor pressure
/
e grams water
W = 0.622 -------- -------------
P - e grams dry air
\ |
total pressure |
Thus 12 g/kg comes out
as .012
5. Virtual temperature (Tv) [Kelvin]:
vapor pressure
/
Tv = T /( 1 - 0.378 * e/P )
\
total pressure
The virtual temperature, Tv, takes into account that water vapour
weighs less than air. Therefore moist air is less dense than dry air
at the same temperature. So, if you could warm up the dry air to the
temperature at which it has reached the same density as the moist air
has at the actual temperature, you need to reach the virtual
temperature.
The following formula is not exact (but as e is normally much smaller
than P, the error is negligible).
Tv = T * (1 + 0.378*e/P )
The numbers 0.378 is normally described as 1-epsilon, where epsilon
is the ratio of the molar weight of water to the "molar weight" of
dry air; epsilon is 0.622.
The equivalent temperature, Te, takes into account that you could try
to condense all the water vapour in your air parcel and use the
condensation heat to warm up the air. This is a first way to
distinguish different air parcels that may have the same sensible
temperature but that have different relative humidity levels.
All the above saturation pressure temperature relationships are
relatively uncomplicated. Here one that is more mindboggling:
A saturation-pressure-curve which is valid for a total pressure of
1000 hPa. "This curve was computed by approximating the standard
steam table for pure water using the least square method by a
Bulgarian colleague. I experienced it to be quite exact, but I'd
be glad to be corrected." (Dr Haessler)
Psat = 610.710701 + 44.4293573*t + 1.41696846*t^2 +
0.0274759545*t^3 + 2.61145937E-4*t^4 + 2.85993708E-6*t^5
The pressure is in Pa, the temperature in degrees Celsius (C).
Relative Humidity then is:
Phi = Psteam/Psat = (Ptot/Psat)*x / ((R /R )+x),
L V,
where x is the absolute humidity in kilogramm water per kilogramm
of dry air,
R and R are specific gas constants for air and steam (i.e. water vapour),
L V
and their ratio has the value of 0.622.
For the handling:
1. Calculate the saturation pressure at Your dew point, giving Your
steam pressure.
2. Calculate the saturation pressure at Your temperature.
3. Divide'em (see above) to get Your relative humidity.
4. Calculate Your absolute humidity, if desired.
5. Mail me for further informations, if necessary.
6. The reverse way is possible.
For the pressure dependence of relative humidity:
If air and steam behave as ideal gases, there is no pressure
dependence.
This is so around 1000 hPa (+-100hPa, approx.).
NET reading :
http://www.ipass.net/~pjm/pmtherm.htm
(free[!] psychrometer programme that:
"does ALL the things others promise, and MUCH MORE! Calculate any process.",
so Paul Milligan - PJM@POBOX.COM)
CAUTIONS:
Good psychrometers
a) Air velocity
k (and A) don't really become (sorta) independent of the air velocity
past your wet bulb until velocities above 3 meters/ second.
Velocities greater than 1 m/s are sufficient at temperatures of 60 C
or more.
The worse your arrangement, (less adiabatic, i.e. the more extraneous
energy radiates/conducts into the water) the steeper k over velocity
becomes for lower velocities. So you can compensate poor design to
some extent by cranking up that fan.
k and A are really device-dependent. This k (and A, of course)
strictly refers to the "Assmann psychrometer" only - two radiation
shields, thermal insulation, fan downstream from the thermometers. k
should be similar for any well-made psychrometer.
b) Adiabatic wet bulb
Shield it from radiative & conductive errors, i.e. all energy to
vaporize the water must come from the air and thus be reflected in
Tw.
In wetting the wet bulb, use distilled water. Salty scale on your
"sock" can change the vapor pressure, and will really mess
measurements near zero. Use enough water to hit steady-state
conditions well before you start to dry out.
If you use a wick for continuous wetting, make it long enough so that
conductive errors are minimized, and it is cooled to the wet bulb
temperature by the time it gets near the thermometer. Make sure enough
water can reach the wet bulb, so don't overdo the "long enough" part.
Don't get anything but the wet bulb wet. Getting the radiation shield
or the thermal insulation wet will introduce errors.
Keep direct sunlight off. A great way to pump heat into your
"adiabatic" system. Don't ever paint the outside black. Many
commercial humidity meters are a pretty black finish. They will be
sensitive to indirect sunlight (and other radiative sources). Humidity
measurements are VERY sensitive to temperature!
c) Supercooled water and ice below freezing
Your measurement will become screwy below freezing, as you cannot
really distinguish between supercooled water (evaporation) and ice
(sublimation) in your wet bulb, and the vapor pressures differ. And
Lueck says supercooled water can be present as low as -12 C. It
suggests manually scraping the wet bulb to ensure that supercooled
water turns to ice.
And note that humidity measurements never are terribly accurate, 2%
error in absolute hum. are pretty good, depending on where you are in
terms of temp and water content. Anything that reads "relative
humidity=52.783 %" is guessing (if you paid less than 100k$...:-)
Thomas Prüfer
Wet bulb temperature is really defined by the psychrometer and is not
an atmospheric water vapor property (compared with Td which has a firm
definition)!!!! The above computations assume the standard
psychrometer equation, but the psychrometer constant (0.00066*P in
kPa/C) is a theoretical value that is not always matched even by very
good psychrometers. PLEASE note this psychrometer constant depends
directly on atmospheric pressure so it's value is not a "universal"
constant!
Terry Howell
NET readings:
http://www.uswcl.ars.ag.gov/exper/relhumeq.htm
http://nwselp.epcc.edu/elp/rhsc.html
http://storm.atmos.uiuc.edu/covis2/visualizer/help/general/rh.dwp.html
3)
Examples
1. EXAMPLE X M P L X M P L X M P L X M P L X M P L
>My problem is the following. I want to calculate wetbulb temperature
>(Tw) where my input is drybulb temperature (T) and relative humidity
>(rH). (Pieter Haasbroek)
Pieter:
Your problem can be solved explicitly using the methods from Jensen et
al. (1990) ASCE Manual No. 70 (see pages 176 & 177) using the
following steps and equations:
1) compute e as [es(T)*rH/100]
where es(T) = 0.611*EXP(17.27*T/(T+237.3)) in kPa
T is drybulb temp in C
e = (rH/100)* 0.611*EXP(17.27*T/(T+237.3))
where e is ambient vapor pressure in kPa
2) compute dewpoint temperature (Td)
Td = [116.9+237.3ln(e)]/[16.78-ln(e)] in C
3) compute wet bulb temperature (Tw)
Tw = [(GAMMA*T)+(DELTA*Td)]/(GAMMA+DELTA)
GAMMA = 0.00066*P where P is ambient barometric pressure in kPa
DELTA = 4098*e/(Td+237.3)^2
This method should be close, especially when Tw is close to Td (DELTA
should be evaluated at (Tw+Td)/2.
For example:
T = 25C
rH = 50%
assume elev is sea level and P = 100 kPa.
1) es(25) = 0.611*EXP(17.27*25/(25+237.3)) = 3.17 kPa
e = (50/100)* es(25) = 1.58 kPa
2) Td = [116.9+237.3*ln(1.30)]/[16.78-ln(1.30)] = 13.85 C
3) GAMMA = 0.00066*100 = 0.066 kPa/C
DELTA = 4098*(1.58)/(13.85+237.3)^2 = 0.103 kPa/C
Tw = [(0.066*25)+(0.103*13.85)]/(0.066+0.103) = 18.21 C
CHECK ANSWER:
EW(Tw) = 0.611*EXP(17.27*18.21/(18.21+237.3)) = 2.09 kPa
e = EW(Tw) - GAMMA*(T-Tw)
e = 1.58 - 0.066*(25-18.21) = 1.64 kPa
The exact answer for Tw is about 17.95C
EW(18.0) = 2.07 kPa; e = 1.60 kPa
EW(17.9) = 2.05 kPa; e = 1.58 kPa
EW(17.95) = 2.06 kPa; e = 1.59 kPa
Thus,
ERROR e = [(1.64 - 1.58)/1.58]*100 = 3.1%
ERROR Tw = [(18.2-17.95)/17.95]*100 = 1.4%
2. EXAMPLE X M P L X M P L X M P L X M P L
>Hello:-
>I am looking for the algorithm to convert wet/dry bulb temperatures to /
>from rH (and moisture content as well, for that matter).
>I know the Psychometric charts, but they are difficult to use accurately
>in software. Anyone have a pointer to appropriate equations?
>Thanks in advance! (Spehro Pefhany)
Answer
(I shall find a formula in SI units, please be patient, BeK)
pw = psf - p * A * (T - Tw)
T: dry bulb temp., Kelvin or Celsius
Tw: wet bulb temp., "
psf: Saturation pressure at temp Tw, see 1.), in Torr (mm Hg)
pw: Vapor pressure of ambient air, in Torr (mm Hg)
p: pressure of ambient air, in Torr
A: optimally (see below) 0.66 * 10e-3 * (1/C)
3.) The short way round:
We're in your backyard: p = 755 Torr, 0 C < T < 50 C.
pw = psf - k (T - Tw)
phi = pw/psf
phi: relative humidity.
k= p*A = 0.5 Torr/degree
Higher temperatures:
Tw about 60 C: k is about 0.52
Tw about 80 C: k is about 0.53
(Formula suggested by A. Sprung, 1888)
3. EXAMPLE X M P L X M P L X M P L X M P L
This question is _often_ asked:
>I have the air pressure (p), the temperature (T) and the
>relative humidity (rH) and want to calculate the specific humidity
>(i.e. the mass of water vapour to the humid air)?
First: This air pressure that you have, is actually the total pressure,
i.e. it is the sum of the pressure of the dry air (pair) PLUS the share
from the water vapour (e).
Then calculate the saturation pressure (es) from one of the formulas given
above.
Then multiply by the relative humidity (rh). This gives you the ambient
water vapour pressure, (e).
Then the specific humidity is given by the following formula:
R
L e
rho = --- -----------------
R p + e(R / R - 1)
V L V
WHERE:
R / R = 0.62197 (see the example for the mixing ratio)
L V
4. EXAMPLE X M P L X M P L X M P L X M P L
> Could somebody send or post the method, or fomula, used to calculate
> dewpoints. I have hunted the local library but am unable to find it.
Hi Tom --
Here it is:
Td = B / ln(A * 0.622 / W / p)
where:
B = 5420 K
A = 2.53 E8 kPa
W = water vapor mixing ratio
p = local pressure
5. EXAMPLE X M P L X M P L X M P L X M P L
I need some help with calculating RH. Our control system allows us to read
> dry bulb temp and enter the specific humidity (g/kg of dry air). We are
> looking for a formula to calculate a RH setpoint to use for control. As
> the dry bulb temp changes the system would calculate the new RH setpoint
> to maintain the same specific humidity.
I propose and easy solution.
We start with the formula for the mixing ratio:
e
W = 0.622 -------
P - e
and transform it with the formulas for the Saturation vapor pressure (es),
resulting in:
w0 * p
rH = ----------------
es(T)*(1 + w0)
where:
p is the total measured pressure and
w0 is the specific humidity (w) at the start of the run, which is
supposed to stay constant.
To give an example with the same starting conditions as in the example
above, see the following table:
rel.
err. w' rh' es(T) T
1.1% 0.016 60% 2.645 22
1.3% 0.016 56% 2.810 23
1.4% 0.016 53% 2.985 24
1.6% 0.016 50% 3.169 25
1.8% 0.016 47% 3.363 26
2.1% 0.016 44% 3.567 27
2.3% 0.016 42% 3.781 28
As you can see w' equals w0, but the relative humidity changes of course.
6. EXAMPLE X M P L X M P L X M P L X M P L
>Can the air temperature ever be below the dewpoint temperature ?
This question is not easy to answer.
First, completely dry temperature will not have any dewpoint whatsoever
as there simply would be no humidity to condensate.
Second, if air reaches dewpoint temperature that water condenses
and this heat should prevent further temperature decreases. However,
this energy will not be enough if other processes (radiative cooling,
a diabatic process) will cool the air ever more.
I guess that air very well can have a temperature below the dewpoint then.
NB
By now you should be able to solve your undergraduate humidity calculations
really by yourselves. But, looking at the text for the mixing ratio, given
above, most of you could have gained knowledge of this formula by
yourselves, I guess.
4)
NET readings :
http://covis2.atmos.uiuc.edu/guide/wmaps/general/rhdef.html
http://njnie.dl.stevens-tech.edu/curriculum/oceans/rel.html
http://www.mtc.com.my/fpub/lib/drying/ch11.htm
Literature hints:
For the book and paper aficionados of the readers check out this:
I found the following books and papers helpful in understanding
the physics of relative humidity measurement (BeK):
Jones, H.G.; 1992; Plants and microclimate.
A quantitative approach to environmental plant physiology.
Second edition; Cambridge University Press, Cambridge/UK; 428 pp.
Monteith, J.L., and Unsworth, M.H.; 1990; Principles of Environmental Physics;
Edward Arnold; 304 pp.
FAO; 1990; 'Annex V. FAO Penman-Monteith Formula';
Expert consultation on revision of FAO methodologies for crop water requirements;
FAO, Roma, 28-31 May 1990; 23 pp.
'If your really interested in this stuff, I (Kerry Anderson) suggest
the book "Atmospheric Thermodynamics" by Irabarne and Godson."'
But unfortunately I learned this book is out of stock
(amazon.com). Instead I could recommend:
"Fundamentals of Atmospheric Dynamics and Thermodynamics"
Paperback, Amazon.com Price: $29.00; Published by
World Scientific Pub Co. Publication date: May 1992
ISBN: 9971978873
"Most introductory texts on meteorology will have one
or two paragraphs on the matter." (K Anderson)
"My sources (other than experience) are all German books
(Thomas Prüfer):"
Lueck, Winfired: Feuchtigkeit - Grundlagen, Messen, Regeln.
Muenchen: R. Oldenbourg, 1964. Good basics.
Sonntag, D.: Hygrometrie: Ein Handbuch der Feuchtigkeitsmessung in
Luft und anderen Gasen. (6 vols.) Berlin: Akademie, 1966 - 1968
Also contains a very detailed description of nearly everything on the
market in 1966-68.
Heinze, D.: Einheitliche, methodische Beschreibung von
Gasfeuchte-Messverfahren. Dissertation an der Technischen Hochschule
Ilmenau, 1980
Comprehensive block and signal diagrams with the Laplace functions (!)
of nearly all humidity measurement methods. Nearly unobtainable,
unfortunately.
(Thomas Prüfer)
5)
Committment:
This ONA was collected and provided to you by Bernd Kuemmel
(bek@kvl.dk).
I admit to have used especially the willing help and the contributions
of the of the following people:
Pierre-Alain Dorange, Forrest M. Mims III, Kerry Anderson, Len Padilla,
Ralf Haessler, Pieter Haasbroek, Terry Howell, David F Palmer, Thomas
Prüfer,
Richard Harvey, Spehro Pefhany, and of course Ilana Stern
during the ongoing improvement of the ONA.
Yours sincerely
Bernd Kuemmel
6)
Outlook:
None-so-far
7)
Signature:
Bernd Kuemmel + bek@less.xy-gene.kvl.dk + VOX: (+45) 35 28 35 33
AgroEcology Group, RVAU/KVL, Agrovej 10, DK-2630 Taastrup
Disclaimer: They do not necessarily agree with all this.
|
