Hi robert
Indeed. We worked together before, more then once.
Let's do this in two installments. In this one I am posting
links to the general theory.
These pages contain informationon the density of moist air, so I
guess we do not a chart describing dry air only. It should contain
the chart you have
as a special case.
In second, we can look at specific application to the heat recovery,
But please let's use SI (metric) units for any actual calculations
Btu's give me headache :-).
So,
Today's topic, (and search term) is Equation of State,
defined e.g. in wikipedia as folows:
In physics, equations of state describes the relationship between
temperature, pressure, and volume for given substance or mixture of
substances.
http://en.wikipedia.org/wiki/Equation_of_state
In our case Dalton's law is of interest:
In physics, Dalton's law (also called Dalton's law of partial
pressure) is a law that states the total pressure exerted by a gaseous
mixture is equal to the sum of the pressures that would be exerted by
the gases if they alone were present and occupied the total volume.
So, treating the water and air as a mixture of ideal gases, we can obtain the
density w at any T and p. All we need to are molecular weights of
all the components
(O2, N2, H2O ..). Ideal gas in as aproximation, which is well
applicable near room
temperature and normal atmospheric pressure.
Since water/air is an important mixture, there is a number of sites
where the numbers are
already derived. Here is powerpoint presentation:
http://snowball.millersville.edu/~amuller/esci385/equation%20of%20state.ppt
and here a simple pages from NASA which reviews composition of air and units.
http://www.grc.nasa.gov/WWW/K-12/airplane/airprop.html
and here are the equations worked out in bothe sets of units
http://www.engineeringtoolbox.com/8_677.html
Both sites have related pages
This page describes amount of water vapor in the air, often expressed
as a vapor pressure, given the symbol e.
http://snobear.colorado.edu/Markw/SnowHydro/Phases/phases.html
For practical use, it would be useful to have a spreatsheet or program.
SEARCH TERM 'moist air' produces:
http://www.mathtools.net/Excel/Chemometrics/
page, promising:
MoistAirTab - Following our success with the SteamTab products,
ChemicaLogic is proud to introduce MoistAirTab?, a unique new
spreadsheet add-in. MoistAirTab? gives you access to a comprehensive
set of air, water, steam and ice properties - allowing you to easily
do complex simulations involving moist air.
similar tools are promised at this university site:
Welcome to TEST - a visual environment to solve thermo problems,
pursue what-if scenarios, perform numerical experiments, and continue
a life-long learning experience to challenge Sommerfield! Browse a few
slides from Slide Show (10 min) followed by a hands-on Tutorial (30
min) to put TEST to full use without much of a learning curve.
http://kahuna.sdsu.edu/testcenter/testhome/index.html
These tools would have to be tested. I have not done that (so far).
Hedgie |
Request for Answer Clarification by
3rrotec-ga
on
06 Jun 2004 14:49 PDT
Hi Hedgie! I've already declined the recovery of heat for the proposed
site. Not economically feasible to build the heat exchanger. I still
want to proceed with you. I want a mathematical understanding of heat
in air, moist and dry, and the ability to readily figure different
psychrometric events. This will entail study from me and patience from
you. Good luck working with a 57 year old machinist, engineer, and
redneck! You will find I'm full of questions and will not proceed to
step two until I understand step one. On to clarification number one.
I am working on "Humid Air and the Ideal Gas Law" (Engineering tool
box- I sure like this format!). I can't get the mass equation to come
out in SI units. I'm willing to work in SI units as long as I can
transpose them correctly. I think in English units. I can't think in
metric yet but I'm trying. Our customers also use the english system
(Ft/sec, Btu's, Lbs/cu. ft., etc).
I took the equation PV=MRT. I assumed the following:
Volume= 1 cubic ft
Temp= 492 R
Gas constant= 1716
Pressure= 29.94 inches mercury=2117.5 lbs/sq ft
Gravity constant 32.16 ft/sec*sec
Solving for M, M=.0025 Lbs since I assumed 1 cu ft, answer is .0025 lbs/cu ft
Next I took the equation W=1.3125*Barometer/Temp R
W=.080 lbs/cu ft
I suppose the mass equation is in outer space, but I'm on Earth, so I
multiplied it by the gravity constant
.0025*32.16=.080
The answers equal so I assume I'm correct. This also agrees with the
charts (I don't trust charts unless I can mathematically prove them.
That way I know I understand)
Next I tried the SI units as follows:
Volume=1 cu meter
Pressure= 29.94 in hg= 101389 newtons/sq meter
Gas constant=286.9
Temp=273 Kg
Solving for M, M=.772 Kg Since I assumed 1 cu meter answer is
.772Kg /cu.meter
converting to Lbs/cu ft
.772*.0624=.048 Lbs/cu ft
.048 does not equal .0025. Where did I screw up and what is the metric
gravity constant? As soon as we clear this up I will close this
question and post a new one. Thanks Robert
|
Clarification of Answer by
hedgie-ga
on
07 Jun 2004 12:19 PDT
hello Rob
I fully support the principle of understanding formulas before using them,
particularly when using them for some real decisions. I once had a job
in which I had to convert all my results to imperial units for use in
engineering designs, so I am confident that we can reconcile the units.
However, it is not just the units. We need to have clarity about what
system we are talking about.
When you say P * V = M * R * T we mean by M mass of what?
So let's first agree on this: We take Ma kilograms of air
Mw kilograms of water
let both equilibrate at Temperature T (measured in Kelvins K), high
enough so that all components are in gas state and let's keep them in
a vessel at pressure P (Pascals)
Question is: what will be the volume of that mixture?
We can imagine that we have it in a cylinder with a piston on the top.
As an example, consider area of piston is 2 m *m and weight on top
of it is provided by the mass of M kg.
Then the pressure under the piston is M * 9.81 /2 Pascals. OK ??
(please do ask if this is not clear)!! g=9.81 m/s .s is gravity.
That same thing can done in IU (imperial or customary units),
and checked e.g. by using this handy calculator:
http://www.allmeasures.com/conversion.asp?Submit=Converter
E.g. if mass was 20kg, , P is 98.1 Pa which, using above URL is
0.0142282020 psi (I had to check the box pressure at the bottom,
which took me to the page
http://www.allmeasures.com/conversion.asp?pressure=on&Submit=Converter
BTW this webpages can be downloaded and still work locally.
I suspect that this W=1.3125*Barometer/Temp R is the same formula as the
Equation of state P * V = M * R * T
as density w= V/M and P is the "barometer" , right? , so I am not
surprised at agrees.
BUT - equation of state of what? air, water, mix 50:50 ?? or what?
The constant R changes according to the mass ratio Ma:Mw - and Mr.
Dalton figured how. That was really your question and so I want to
finish answering that .
Answer is here:
http://www.engineeringtoolbox.com/8_680.html
namely:
Based on specific volume of moist air the density can be expressed as:
? = 1 / v = (p / Ra T) (1 + x) / (1 + x Rw / Ra) (3)
where
v = specific volume of moist air per mass unit of dry air and water vapor (m3/kg)
Ra = 286.9 - the individual gas constant air (J/kg.K)
Rw = 455 - the individual gas constant water vapor (J/kg.K)
x = specific humidity or humidity ratio (kg/kg)
p = pressure in the humid air (Pa)
When x, mass ratio, is 0 (no water) this is reduced to your formula
for da="dry air" namely
?.da = p / Ra T (4)
You cannot use it when x>0.
Now - problem with conversion: Often, people using lb to denote
pound of force and pound of mass get confused, when going to SI,
since SI is different. The NASA site above is uisng Customary Units,
but measures mass in slugs and force in pounds. That has same logic as
SI, which measures mass in kg and force in N. Once you mesure mass in
either slugs or kg, you do not need g (gravity)
(except as above, when we are using weight to create the pressure).
Gravity does not enter the problem any other way, no matter where you are.
So. I expect the conversion to SI problem to diappear if we would do that
(if we would not confuse mass and force by measuring them by same unit).
Please let me know if it does.
hedgie
|
,
0 Comments
)