Hello carolkardon!!
You will find this problem solved at this page:
http://www.engr.colostate.edu/~ramirez/ce_old/classes/ce520_ramirez/Homework_1_Sln.html
The differences with the results are product of the rounds up and the
little differences with the constants values:
a)Virtual Temperature:
Tv = T / (1 - ((1 - epsilon)/epsilon)*w) ,
w: mixing ratio [kg/kg]
w = Mv/Md , where Mv is the mass of vapor and Md is the mass of
dry air.
w = 5 [g/kg] = 0.005 [kg/kg]
epsilon: if mv is the molecular weight of water vapor (18), and md is
the molecular weight of dry air (28.94), the ratio mv/md = 0.622 is a
constant, often denoted by the greek letter epsilon.
(1 - epsilon) / epsilon = 0.608
Tv = 280 / (1 - 0.608*0.005) = 280 / (1 - 0,00304) = 280 / 0,99696 =
= 280.8538 K
For references about how to obtain the Tv formula, and its
justification see the following essay: "On Using the Virtual
Temperature Correction" by Chuck Doswell:
http://www.cimms.ou.edu/~doswell/virtual/virtual.html
-----------------------------------------------------
b)Absolute Humidity:
Absolute Humidity = mass of water vapor/ volume of air, this is
equivalent to the density of water vapor in the sample.
Absolute humidity is a measure of the amount of moisture in the air;
weight of water vapor per unit volume or air.
First we need e = partial vapor pressure [Pascals]
If p is the air pressure in Pa,
e = p.w/epsilon = 0.005*90000 / 0.622 = 723.473 Pa
We can justify this using the Ideal gas law, applied to vapor and dry
air:
V = ((Mv/mv) * R * T) / e = ((Md/md) * R * T) / p
where Mv is the mass of water vapor, Md is the mass of dry air, mv is
the molecular weight of water vapor, md is the molecular weight of dry
air, R is the Gas Constant and T is temperature.
Then follows:
e = p*(md/mv)*(Mv/Md) = p*(1/epsilon)*w;
Now we can use again the Ideal gas Law applied to the vapor
rho[v] = e / (R[v].T) = 723.473 / (461.5 * 280) = 0.0056 [kg/m3],
where rho[v] is the density of the vapor of water in the sample.
c)Relative Humidity
The amount of water vapor in the air at any given time is usually less
than that required to saturate the air. The relative humidity is the
percent of saturation humidity, generally calculated in relation to
saturated vapor density. Relative humidity is a function of the
temperature at a given pressure.
RH = 100 *(actual vapor density)/(saturation vapor density)=
100*rho[v]/rho[es]
Using again the Ideal Gas Law applied to the vapor of water we have
that
e = rho[v]*Rv*T and es = rho[es]*Rv*T , where es is the
saturation vapor pressure;
then at the same temperature is e/es = rho[v]7rho[es] ; then:
RH = 100*e/es
Now we need to find es; we can use the Clausius-Clapeyron equation:
This equation gives the relationship between saturation vapor pressure
and the temperature in Kelvins. This equation is used also to
calculate relative humidity and other moisture variables."
"IMPORTANT EQUATIONS AND INTERPRETATION " by METEOROLOGIST JEFF HABY:
http://www.theweatherprediction.com/basic/equations/
ln(es/611.2) = (L/Rv )*(1/273.15 - 1/T)
e0 = 611.2 Pa = reference saturation vapor pressure at T0 = 273.15 K
(0ºC)
ln(x)= natural logarithm
es = Saturation vapor pressure in Pa
L = Latent heat of vaporization = 2.453 × 10^6 J/kg
Rv = Gas constant for moist air = 461 J/kg
T = Temperature in Kelvins
Then
es = 611.2*exp((L/Rv)*(1/273.15-1/T)) = 984.364 Pa
Rh = 100 * e/es = 73.5%
References for Relative humidity:
"CALCULATING RELATIVE HUMIDITY" by METEOROLOGIST JEFF HABY:
http://www.theweatherprediction.com/habyhints/186/
"Saturation Vapor Pressure" article at Planetary Atmospheres Node
website:
http://atmos.nmsu.edu/education_and_outreach/encyclopedia/sat_vapor_pressure.htm
"Moisture Conversions" article at Yankee Environmental Systems
website:
http://www.yesinc.com/education/moist-convert.html
-------------------------------------------
d) Dew Point
Dew point (Td) temperature is defined as the temperature to which the
air would have to cool (at constant pressure and constant water vapour
content) in order to reach saturation. A state of saturation exists
when the air is holding the maximum amount of water vapour possible at
the existing temperature and pressure.
Condensation of water vapour begins when the temperature of air is
lowered to its dew point and beyond.
This quantity is used on the tephigram, and for forecasting formation
of dew, fog and clouds: when air cools to its dew point temperature,
condensation occurs.
Remember that the Clausius-Clapeyron equation gives the relationship
between saturation vapor pressure and the temperature in Kelvins:
ln(es/611.2) = (L/Rv )*(1/273.15 - 1/T)
If we use the partial vapor pressure (e) (and keep it constant)
instead of the saturation vapor pressure (es), the temperature wich
satisfied the equation will be Td !!, because Td (dew point) is
defined as the temperature to which the air would have to cool (at
constant pressure and constant water vapour content) in order to reach
saturation.
In order to clarify, if you input the temperature of the air at the
Clausius-Clapeyron equation, the pressure that you will obtain from it
is es. if you input the partial vapor pressure you will obtain Td.
In this case the equation is:
ln(e/611.2) = (L/Rv )*(1/273.15 - 1/Td) then:
Td = 1 / (1/273.15 - (Rv/L )*ln(e/611.2)) = 275.535 K
---------------------------------------------------
e) Wet-bulb temperature:
Wet Bulb Temperature
The wet bulb temperature of moist air at pressure P, temperature T and
mixing ratio w is the temperature which the air assumes when water is
introduced gradually by infinitesimal amounts at the current
temperature and evaporated into the air by an adiabatic process at
constant pressure until saturation is reached.
"Wet-bulb temperature is measured using a standard mercury-in-glass
thermometer, with the thermometer bulb wrapped in muslin, which is
kept wet. The evaporation of water from the thermometer has a cooling
effect, so the temperature indicated by the wet bulb thermometer is
less than the temperature indicated by a dry-bulb (normal, unmodified)
thermometer. The rate of evaporation from the wet-bulb thermometer
depends on the humidity of the air - evaporation is slower when the
air is already full of water vapour. For this reason, the difference
in the temperatures indicated by the two thermometers gives a measure
of atmospheric humidity."
"Commonwealth of Australia 2002, Bureau of Meteorology - Wet-bulb
temperature":
http://www.bom.gov.au/climate/glossary/wetbulb.shtml
"Wet Bulb Temperature: The temperature an air parcel would have if
cooled adiabatically to saturation at constant pressure by evaporation
of water into it, all latent heat being supplied by the parcel."
The above definition is taken from the Glossary of Meteorology.
Huschke, R. E., Ed., 1959: Glossary of Meteorology. Amer. Meteor.
Soc., 638 pp.
There aren´t a direct way to obtain the wet-bulb temperature, it is
found using an iterative technique.
Please refer to the following page in order to see the method:
"Wet Bulb Temperature"
http://meted.ucar.edu/awips/validate/wetblb.htm
Using the method of that page I found the following values:
es = 9,84364 mb
ed = e = 7.23473 mb
(we have calculated these before)
s = 0.58430235
First guess for Tw = 277.74365 K
de = 0.05043402
Verification = Tw/de = 5,507.07 < 10,000
der = -1.1653603
Second guess for Tw = 277.786928 K
ew2 = 8.5 mb
de2 = -3.441*10e(-5)
Verification2 = -8,071,956 > 10,000
Then the second guess is a good aproximation
Tw = 277.786928 K
An easy way is found here:
"SHORTCUT TO CALCULATING WET-BULB" by METEOROLOGIST JEFF HABY:
http://www.theweatherprediction.com/habyhints/170/
Also you can see
"Dew-Point Temperature and Wet-Bulb Temperature: Two Measures of
Humidity "
http://squall.sfsu.edu/courses/metr302/F96/handouts/dwpt_wtblb.html
In the folowing link you will find a brief description of an iterative
technique and how it could be aplicated to the calculation of wet-bulb
temperature (see example 9.1)
"Other iterative procedures" from NCAR/Advanced Studies website:
http://www.asp.ucar.edu/colloquium/1992/notes/part1/node81.html
--------------------------------------------------
f) Potential temperature:
"Potential temperature is the temperature a parcel of air will have if
raised or lowered to the 1000-millibar level. Potential temperature is
the same for a parcel of air, as it rises or sinks, assuming adiabatic
conditions."
"IMPORTANT EQUATIONS AND INTERPRETATION " by METEOROLOGIST JEFF HABY:
http://www.theweatherprediction.com/basic/equations/
Tp = T.(1000/P)^Rd/Cp = T(1000/P)^(0.286)
T = temperature in Kelvins
P = pressure in millibars
Rd = gas constant for dry air = 287.05
Cp = constant pressure process or = specific heat of dry air at
constant pressure = 1004
Tp = 280 * (1000/900)^(0.286) = 288.566 K
For reference:
"Potential Temperature"
http://san.hufs.ac.kr/~gwlee/session3/potential.html
-----------------------------------------
g) Equivalent temperature:
The isobaric equation for equivalent temperature is:
Te = T * (1 + (L.w)/(cp.T))
T = temperature (K)
w= mixing ratio (unit less)
L = 2.453x10^6 J/Kg = latent heat of vaporization
Cp = 1004 = specific heat of dry air at constant pressure
Te = 280 * (1 + (12265)/(281120) = 280 * (1.04363) = 292.216 K
-----------------------------------------------
h) Equivalent potential temperature:
"Equivalent potential temperature is the potential temperature
corresponding to the adiabatic equivalent temperature. Adiabatic
equivalent temperature is the temperature a parcel of air would have
if it undergoes the following physically unrealizable process:
dry-adiabatic expansion until saturated; pseudo-adiabatic expansion
until all the moisture has precipitated out; and finally,
dry-adiabatic compression to the initial pressure."
The above definition is taken from the Glossary of Meteorology.
Huschke, R. E., Ed., 1959: Glossary of Meteorology. Amer. Meteor.
Soc., 638 pp
EPT = Te * (1000/P)^(Rd/Cp) = 292.216 * (1000/900)^(287.05/1004) =
301.152 K
For references:
"Equivalent Potential Temperature"
http://meted.ucar.edu/awips/validate/thetae.htm
-------------------------------------------------
It was a hard job, espacially with the wet-bulb temperature. But funny
in some way.
I found one more useful text that will help you understanding this
problems:
"The _Temp, Humidity & Dew Point_ ONA (Often Needed Answers)"
http://www.agsci.kvl.dk/~bek/relhum.htm
At "JEE 661 ATMOSPHERIC BOUNDARY LAYER SCIENCE" you can find two
texts, related to this subject:
"Lecture 1"
http://www.jgsee.kmutt.ac.th/exell/JEE661/JEE661Lecture1.html
"Lecture 2"
http://www.jgsee.kmutt.ac.th/exell/JEE661/JEE661Lecture2.html
---------------------------------------------------------------
I hope this helps, if you have any doubt please feel free to request a
clarification, may be to correct a litle mistake in calculations if
appears or to solve problems with dead links, etc.
Thank you for asking to Google Answers
Best Regards
livioflores-ga |
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