Rob,
I will give here some additional formules, for dew point etc
specific humidity
definition of s here
( using previously defined Ma Mw (mass of air, mass of water))
http://www.taftan.com/thermodynamics/SPHUMID.HTM
says s= Ma/ (Ma +Mw)
ergo
1/s = 1+ 1/x
ergo
s= x/( 1+x)
x being (as before) relative humidity Ma/ Mw
or something like that..
I hope you are not expecting me to do actual calculations.
If you look at my past answers, to other askers as well,
you will see that as a rule, I am good at explaining physics,
but not good in doing numerical calculations, particularly
when feet and inches and BTUs are involved.
If I must do that, I rather write a computer program.
That I suggest you have someone do for you, particularly
if you need do theese calculations on a repetitive basis.
It would be a fairly simple program and once tested,
it's use would greatly reduce number of errors time spent etc.
It should be custom written, so inputs and printout is formated the
way you want.
here are some additional formylas for dry/wet bulb T
partial pressure
We aleady covered that: You have equation of state
for dry air, for water vapour
each with its own gas constants Rw, Ra,
you use each separately, with Ma and Mw,
and as a check, you add the pressures and get total pressure of moist air
In more detail.
http://members.aol.com/profchm/dalton.html
-------------------------------------------------
The simplified formula WBGT= Wet Bulb Globe Temperatures
WBGT = 0.567 Td + 0.393 e + 3.94
where:
Td = Dry bulb temperature (°C)
e = Water vapour pressure (hPa) [humidity
http://www.bom.gov.au/products/IDS65004.shtml
-----------------------------
Dewpoint calculated from: Dry Bulb Temperature
Relative Humidity
B = (ln(RH / 100) + ((17.27 * T) / (237.3 + T))) / 17.27
D = (237.3 * B) / (1 - B)
-------------------------------
Dewpoint calculated from: Dry Bulb Temperature
Wet Bulb Temperature
Barometric Pressure
Es = 6.108 * (e^((17.27 * T) / (237.3 + T)))
Ew = 6.108 * (e^((17.27 * W) / (237.3 + W)))
E = Ew - (0.00066 * (1 + 0.00115 * W) * (T - W) * P)
RH = 100 * (E / Es)
B = (ln(E / 6.108)) / 17.27
D = (237.3 * B) / (1 - B)
reference and details:
http://ag.arizona.edu/azmet/dewpoint.html
book reference
http://www.madsci.org/posts/archives/mar99/920561789.Es.r.html
evaporation and drying
-----------------------
During a wet-bulb process, air and water vapor coexist.
Some of the vapor is condensed, or water vapor is evaporated in air, saturating it
governing equation:
Md Cpd + Mv Cpv) dT = - Lv dMv [1]
details
http://www.enter.net/~jbartlo/articles/070297.htm
http://www.faqs.org/faqs/meteorology/temp-dewpoint/
search terms, psychometrics,
temperature wet bulb dry bulb formula
----------------------------------
Note on sludge drying
It looks like trend is to use hot air to completely dry sludge and
produce powder or pellets for use as fertiliser.
Here, Fort Worth, TX (just like Madrid)
reports, they stopped using open air drying beds
http://www.fortworthgov.org/water/Wastewater/bioland.htm
http://www.texasep.org/html/wst/wst_2mtx_sldg.html
http://nett21.gec.jp/JSIM_DATA/WASTE/WASTE_5/html/Doc_489.html
I wonder in what cases the heat exchanger is needs to be used and when
the hot air can be used directly - It looks like lot of companies are
dewatering completely, using variety of technologies:filter press,
heat, wastewater evaporation, ion exchange, reverse osmosis, or
dialysis..
http://www.finishing.com/66/32.html
But that is a different and more complex topic.
hedgie |
Request for Answer Clarification by
3rrotec-ga
on
10 Jun 2004 07:58 PDT
Thanks Hecgie! I'm real busy in shop the rest of this week. I'll
"digest" this weekend and get back. Thanks again! Robert
|
Request for Answer Clarification by
3rrotec-ga
on
14 Jun 2004 11:47 PDT
Still digesting-making headway! Will be in touch. Robert
|
Clarification of Answer by
hedgie-ga
on
14 Jun 2004 11:47 PDT
Hi,
Are there more questions related to this last set of formulas?
There is no rush, but I need to respond to any RFC in reasonable time.
Your note is formally a RFC - so this is just
a note saying: I will be watching this page for any futher RFCs.
Hedgie
|
Clarification of Answer by
hedgie-ga
on
16 Jun 2004 22:56 PDT
Rob,
I will be away from the computer for couple days.
I will check here for any RFCs when I come back.
hedgie
|
Request for Answer Clarification by
3rrotec-ga
on
22 Jun 2004 09:44 PDT
Hi Hedgie! I think I have a hold on the situation (formulas) now. I
have been using the humidity formulas at the site:
<www.gorhamschaffler.com/humidity_formulas.htm>
I have one more clarification to clear up (a simple I think) and I
will close this question and go to next. Is the following statement
true:
THE VAPOR PRESSURE OF THE AIR WILL EQUAL THE VAPOR PRESSURE OF THE
VAPOR IN THE AIR AT DEW POINT TEMPERATURE. Thanks Robert P.s. I write
my on programs using basic, that is why I must understand the
formulas. Plus the fact I don't use formulas or charts unless I can
prove to my satisfaction they are in fact correct! Been stung before
taking "someone's word" for it!
|
Clarification of Answer by
hedgie-ga
on
23 Jun 2004 21:22 PDT
Hi
It is good to know that you are programming in Basic.
I can use pseudocode etc. It may be worth mentioning
(or may be obvious?) that many 'on line calculators' such as
used on the following two pages
http://www.bom.gov.au/lam/humiditycalc.shtml
http://physics.holsoft.nl/physics/ocmain.htm
have formulas done in Javascript, which can be inspected
e.g. by saving the page and loading it into a text editor
(instead of browser).
For example, second page (below), used this javascript function
to Get Density of Water (at given Temperature)
function getDensWa() {
var T = parseFloat(document.frmDenswa.temp.value);
var T2 = (T-3.98)*(T-3.98);
var T4 = T2*T2;
var tmp = 1000-0.0067*T2+5.2e-7*T4;
document.frmDenswa.dnswa.value = " "+Math.floor(tmp*10+0.5)/10;
}
http://physics.holsoft.nl/physics/ocmain.htm
http://www.gorhamschaffler.com/humidity_formulas.htm
The algorithm for T.d (dew point) calculation is given here:
http://www.campbellsci.co.uk/support/technts/technt16.pdf
http://www.paroscientific.com/dewpoint.htm
It has discussion of errors, range of validity and
sample results, which can be used for program verification.
It is also important to consider the assumption that water/air are
in equilibrium, often made implicitely, in these calculations.
This is pointed out in the following text:
"............
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 there 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 water and ice surfaces and 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.
Physically at the dew point temperature the vapour loses the energy
that it gained at evaporation, the latent energy, again.
............."
Which also has a collection of practical aproximations, suitable
for numerical calculations, such as:
"..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).
The _Temp, Humidity & Dew Point_ ONA (Often Needed Answers)
http://www.faqs.org/faqs/meteorology/temp-dewpoint/
A plain text version of this text can also be found on:
http://mmf.ruc.dk/~bek/relhum.htm
Physicist's viewpoint: I do not like the common expression "air can
HOLD that much water at the given T.." for this reason:
In the mix of (ideal) gases, molecules will quickly reach the SAME
temperature (since they keep colliding) but are otherwise independent
of each other. Air does not 'hold water' any more then pure nitrogen
(or other gas would).
At high temperature two water molecules collide at high speed and fly
appart again. At lower temperature they may stick together and so
start a droplet. Presence of precipitation nuclei (see cloud seeding)
or cool surfaces is more relevant then presence of other has
molecules. This is well described here:
Characteristics of Air-Vapour Mixtures
In the normal range of atmospheric temperatures and pressures, water
can exist in three different states: gas, liquid and solid. The
maximum amount of water that can exist in the gaseous state (vapour)
is limited by the temperature. Thus, if any air-vapour mixture is
cooled, a temperature will be reached at which it will be saturated,
and if cooling is continued below this point water will condense. If
the temperature at which the air becomes saturated, i.e. the
dew-point, is above the freezing point, the vapour will condense to a
liquid, if it is below freezing, it will condense as ice in the form
of hoar frost.
http://irc.nrc-cnrc.gc.ca/cbd/cbd057e.html
(Of course other molecules (air) are very important in NON-EQULIBRUIM
situation (e.g. draft) when they carry heat away from the system).
Practical consequence of this theoretical fact (Dalton law) is that
we can give P.s(T), a saturation pressure of water, which would be
valid for any mixture (at the given pressure) -
it is property of water molecules, not property of mixture.
Such curve is given e.g. here:
http://www.thermexcel.com/english/tables/eau_atm.htm
Another consequence is this: PRESSURE of (moist) AIR is SUM OF THE (
THE VAPOR PRESSURE OF THE VAPOR IN THE AIR) and ( PRESSURE OF THE DRY
AIR) .
(Dalton law - again).
The pressure of dry air is zero only at absolute zero. That (T=0) can
be considered DEW POINT TEMPERATURE at P=0, which however does not
happen under the usual situations on the Earth surface. (So, the
answer is NO).
http://www.cimms.ou.edu/~cortinas/1014/l11_2.html
SEARCH TERMS: Saturated Vapor pressure, Dew point, humidity
saturated vapour pressure,program, calculation
hedgie
|
Request for Answer Clarification by
3rrotec-ga
on
24 Jun 2004 12:24 PDT
Thanks Hedgie! I'll digest. Might be a week or so for next question.
I'll close this one now, and thanks alot!! Robert
|
Clarification of Answer by
hedgie-ga
on
24 Jun 2004 14:41 PDT
Thank you for your comment and rating.
Hedgie
|
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