Hi -
In researching this question, I came across a set of lectures online
from the University of Washington which address many of the issues
raised by these questions. Where applicable, I've reference the
particular note set used for the calculations. The general web-site
can be found here:
http://www.atmos.washington.edu/1998Q4/501/
Another very good website from theweatherprediction.com website
listing equations is:
http://www.theweatherprediction.com/basic/equations/
1) In the case of dry air, the normal gas law equation can be used to
calculate the density (specific volume) of air given the temperature
and pressure. As question does not relate to dry air, consideration
must be given to the effect of water vapor upon the behavior of the
air sample. Water, at molecular weight of 18, is lighter than the
rest of the air, at an average molecular weight of around 29, and thus
displaces heavier air. One method of taking this into account is
through the use of 'virtual temperatures' - this effectively
calculates a temperature at which dry air would behave as the mixture
of air in the sample performs due to the vapor. Once the 'virtual
temperature' is calculated, it can be used in the regular gas law
equation to calculate the density of the air. From the University of
Washington lecture page, begins at bottom of page 12:
http://www.atmos.washington.edu/1998Q4/501/notes1.7.pdf
Also can be found here:
http://www.theweatherprediction.com/basic/equations/
Virtual temperature is calculated by:
T 250 K
Tv = ----------------------- = --------------------------- = 251 K
e 1mb
[ 1 - --- * ( 1 - ew )] [ 1 - ----- * ( 1 - 0.621 )
p 100mb
where T = temperature (in Kelvin) Tv = virtual temperature
p = pressure e = partial pressure water vapor
ew (epsilon w) = mass fraction water/air = 18/28.97 = 0.621
And then using the gas law equation (rearranged to solve for density)
p 100mb 10000kPa 10000 kg/ms2
d = --------- = ----------------- = -------------- = ------------ =
0.139 kg/m3
R * Tv 287m2/s2K * 251K 72037 m2/s2 72037 m2/s2
where d = density (specific volume) p = pressure Tv = virtual
temperature
R = gas constant = 287 m2/s2K
2) The simplest way to calculate the entropy of bringing 5g of water
at 0 Celsius to steam at 100 Celsius is to break the problem into two
portions - the first, calculation of the change in entropy in bringing
water from 0-100 Celsius - given by texts as 23.55 J/mol*K, which for
5g of water (5g water/18 g/mole = 0.28 moles) = 6.5 J/K. The entropy
change from water to steam at 100 Celsius is given by texts as 108.98
J/mol*K which is 30.5 J/K. Adding the two together gives an entropy
change of 32 J/K for the change.
The more complicated way involves using the following equations
from:
http://wendigo.unl.edu/~gerry/CHEM471/notes/lecture_9_notes.pdf
To calculate first the change in entropy for the heating of the water
from 0-100 Celsius
dS = Cw * ln ( T2/T1 ) = 75.4 J/mol*K * ln (373/273) = 23.5 J/mol*K
where dS = change in entropy Cw = heat capacity of water (assumed
constant at 75.4 J/mol*K) T2 = temp 2 (373 Kelvin) T1 = temp 1 (273
Kelvin)
To calculate the change in entropy for the state change:
dS = Hw / T = 40667 J/mol / 373 K = 108.98 J/mol*K
where ds = change in entropy Hw = molar heat of vaporization of
water T = temperature of vaporization = 373 K
Once we have the values calculated, the computation is the same as if
we had looked up the values (first paragraph)
3) The virtual temperature can also be calculated using the mixing
ratio of water vapor to air (versus the partial pressure as in #1)
using the following equation - same reference as #1, and the saturated
mixing ratio of 7.7 g/kg:
http://www.atmos.washington.edu/1998Q4/501/notes1.7.pdf
The same equation can also be found here:
http://www.theweatherprediction.com/basic/equations/
Tv = ( 1 + qv ) * T = ( 1 + 0.0077 ) * 273 K = 275 K
where T = temperature Tv = virtual temperature qv = mixing ratio
water/air
4) The potential temperature is the temperature of a sample of air if
moved to a pressure of 1000mb via an adiabatic process
A) In the case of dry air, solution from theweatherprediction.com:
http://www.theweatherprediction.com/basic/equations/
Tp = T * ( 1000mb / p ) ^ (R/Cp) = 273K * ( 1000mb / 500 mb ) ^ 0.286
= 333 K
where T = temperature Tp = potential temperature p = pressure
R = gas constant Cp = constant process factor R/Cp ~= 0.286
B) In the case of saturated air, the equivalent potential temperature
can be calculated - this is the potential temperature of the same air
parcel if it were dry - equation from theweatherprediction.com:
http://www.theweatherprediction.com/basic/equations/
Ep = T * (1000mb/p)^(R/Cp) + 3qv = 273K *(1000mb/500 mb)^0.286+3(7.7)=
356 K
where Ep = Equivalent potential temperature T = temperature p =
pressure
R = gas constant Cp = constant process factor R/Cp ~= 0.286
qv = mixing fraction (in grams/kilogram)
5) The latent heat of evaporation is an empirically determined
formula, written in SI units, with T in degrees Celsius - from:
http://www.jgsee.kmutt.ac.th/exell/JEE661/JEE661Lecture2.html
Levap = 2501 - 2.375T kJ/kg
converting from kJ/kg to cal/g is done by dividing by 4.18 J/cal and
multiplying by 1kg/1000g. Converting from Kelvin to Celsius is
(T-273).
L evap = 597.3 - 0.566(T-273)
The latent heat of sublimation is also empirically determined
L sub = 2830 - 0.026T kJ/kg with T in Celsius
Conversion is as above to yield
L sub = 677 - 0.062(T-273) cal/g
6) The vapor pressure of water can be calculated from the pressure
and mixing ratio using a few simple relations. Also from the
University of Washington pages:
http://www.atmos.washington.edu/1998Q4/501/notes1.pdf
http://www.atmos.washington.edu/1998Q4/501/notes1.7.pdf
And from theweatherprediction.com:
http://www.theweatherprediction.com/basic/equations/
the vapor pressure is related to the total pressure and mixing ratio
by the following equation:
qv * p 0.0055 * 1026.8 mb
e = ------------ = ------------------- = 9.0 mb
(0.622 + qv) (0.621 + 0.0055)
where e = vapor pressure of water p = pressure qv = mixing ratio
==============================================================
Please let me know if you have further questions.
Search strategy:
Google: thermodynamic meteorology gas law
Google: "latent heat of vaporization" thermodynamic meteorology
Google: entropy vaporization water |
