· xls file · web view · 2017-06-30... (2501+1.805*tdb-4.186*twb) x...
TRANSCRIPT
Index
Table of psichrometric functions
Group 1, functions of tdb, f and H Group 3, functions of tdb, x and H31.-
17.-
Group 2, functions of tdb, twb and H Group 4, functions of ent, x and H41.-
45.-26.-
Psichrometric functions
11.- Absolute humidity x = f(tdb, f, H)12.- Specific volume v = f(tdb, f, H) 32.- Specific volume v = f(tdb, x, H) 13.- density r = f(tdb, f , H) 33.- density r = f(tdb, x , H)14.- Dew point temperature tdp = f(tdb, f, H) 34.- Dew point temperature tdp = f(tdb, x, H)15.- Enthalpy ent = f(tdb, f, H) 35.- Enthalpy ent = f(tdb, x, H)16.- Wet bulb temperature twb = f(tdb, f, H) 36.- Wet bulb temperature twb = f(tdb, x, H)
37.- Relative humidity f = f(tdb, x, H)
21.- Absolute humidity x = f(tdb, twb ,H)22.- Specific volume v = f(tdb, twb, H) 42.- Specific volume v = f(ent, x ,H)23.- Density r = f(tdb, twb, H) 43.- Density r = f(ent, x ,H)24.- Dew point temperature tdp = f(tdb, twb,H) 44.- Dew point temperature tdp = f(ent, x ,H)25.- Enthalpy ent = f(tdb, twb, H)
46.- Wet bulb temperature twb = f(ent, x, H)27.- Relative humidity f = f(tdb, twb,H) 47.- Relative humidity f = f(ent, x ,H)
rev. cjc. 28.02.2018
Group 3, functions of tdb, x and H Group 5, functions of tdb, ent and H
45.-
Group 4, functions of ent, x and H
Psychrometric functions_Deductions.xlsrev. cjc. 28.02.2018cjcruz[at]piping-tools-net
51.- Absolute humidity x = f(tdb, ent, H)32.- Specific volume v = f(tdb, x, H) 52.- Specific volume v = f(tdb, ent, H)33.- density r = f(tdb, x , H) 53.- Density r = f(tdb, ent, H)34.- Dew point temperature tdp = f(tdb, x, H) 54.- Dew point temperature tdp = f(tdb, ent, H)35.- Enthalpy ent = f(tdb, x, H)36.- Wet bulb temperature twb = f(tdb, x, H) 56.- Wet bulb temperature twb = f(tdb, ent, H)37.- Relative humidity f = f(tdb, x, H) 57.- Relative humidity f = f(tdb, ent, H)
61.- Air & water properties42.- Specific volume v = f(ent, x ,H) Ref. 1a43.- Density r = f(ent, x ,H) Ref. 1b44.- Dew point temperature tdp = f(ent, x ,H) Ref. 1c
www.piping-tools.net46.- Wet bulb temperature twb = f(ent, x, H)47.- Relative humidity f = f(ent, x ,H)
Psychrometric Functions [1] & [2]
Dry bulb temperature tdb °C Absolute humidityRelative humidity % Specific volumeHeight above sea level m.a.s.l. Density
In the functions to calculate the wet bulb temperature (functions 16, 36, 46 and 56), some limits to the input values of the dry bulbtemperature, relative humidity and height above sea level habe been set to to obtain an acceptable maximum number of iterations. If required, the maximum values can be changed in the code (functions 16, 36, 46 and 56).The numbers in square brakets indicate the number of the function in the VBA code
GroupsGroup 1 Group 2tdb, f , H tdb, twb, H
tdb ºC tdb 96.00 °C tdb 96.0 °C % 100.0 % #VALUE! [27]
H m s.n.m. H 3.4 m H 3.4 mx kg/kg x #VALUE! [11] x #VALUE! [21]v v #VALUE! [12] v #VALUE! [22]r r #VALUE! [13] r #VALUE! [23]
tdp ºC tdp #VALUE! [14] tdp #VALUE! [24]ent kJ/kg ent #VALUE! [15] ent #VALUE! [25]twb ºC twb #VALUE! [16] twb #VALUE! ºC
Note. Some depent variables will not require the three inputs when they arecalculated individually. But because the calculation method with VBAfunctions make use of other functions, the complete set of input variablesis required.
f or fH
f f f
m3/kgkg/m3
rev. cjc. 28.02.2018
Psychrometric Functions [1] & [2]
x kg w/kg da Dew point temperature ºCv Specific enthalpy ent kJ/kg dar Wet bulb temperature twb ºC
In the functions to calculate the wet bulb temperature (functions 16, 36, 46 and 56), some limits to the input values of the dry bulb Maximum values for the input parameterstemperature, relative humidity and height above sea level habe been set to to obtain an acceptable maximum number of iterations. 96 °CIf required, the maximum values can be changed in the code (functions 16, 36, 46 and 56). 100 %
5300 m.a.s.l.
GroupsGroup 3 Group 4 Group 5tdb, x, H ent, x, H tdb, ent, H
tdb 96.00 °C tdb #VALUE! [48] tdb 96.0 °C#VALUE! [37] #VALUE! [47] #VALUE! [57]
H 3.4 m H 3.4 m H 3.4 mx #VALUE! kg/kg x #VALUE! kg/kg x #VALUE! [51]v #VALUE! [32] v #VALUE! [42] v #VALUE! [52]r #VALUE! [33] r #VALUE! [43] r #VALUE! [53]
tdp #VALUE! [34] tdp #VALUE! [44] tdp #VALUE! [54]ent #VALUE! [35] ent #VALUE! kJ/kg ent #VALUE! KJ/kgtwb #VALUE! [36] twb #VALUE! [46] twb #VALUE! [56]
Note. Some depent variables will not require the three inputs when they are
functions make use of other functions, the complete set of input variables
Return to index
tdp
m3/kg dakg /m3
tdbMax =f Max =Hmax =
f f f
Psychrometric functions_Deductions
rev. cjc. 28.02.2018
cjcruz[at]piping-tools.net
www.piping-tools.net
11.- Absolute humidity x = f(tdb, f, H)
Input data Vapor pressuretdb = 40 °C fi * Psat
f = 30 % fi = 0.30H = 200 m #VALUE!
#VALUE!Atmospheric pressure [1], Eq. (3)
1.01325 * (1 - 0.0000225577 * H) ^ 5.25588 Absolute humidityH = 200.0 m x = 0.62198 * Pvap / (Patm - Pvap)
0.9895 bar #VALUE!0.98945
Using function 1 x = #VALUE!Saturation pressure at dry bulb temperature [2]
Sicro_Saturated_vapor_pressure_t(tdb) *0.00001 Using function 11tdb = 40.0 ºC Sicro_Absolute_Humidity_tdb_f_H
#VALUE! bar tdb = 40.0f = 30.0
Relative humidity H = 200.0fi = f / 100 x = #VALUE!f = 30.0 %fi = 0.30 -
Return to index
Pvap =
Psat = Pvap =
Patm =
Patm = Pvap =Patm =
Psat =
Psat =
rev. cjc. 28.02.2018
11.- Absolute humidity x = f(tdb, f, H)
-barbar
[1], Eq. (8)0.62198 * Pvap / (Patm - Pvap)
barbarkg/kg
Sicro_Absolute_Humidity_tdb_f_H°C%mkg/kg
Microsoft Editor de ecuaciones 3.0
Input data Vapor pressuretdb = 40 °C fi * Psat
f = 30 % fi = 0.30H = 200 m #VALUE! bar
#VALUE! barAtmospheric pressure [1], Eq. (3)
1.01325 * (1 - 0.0000225577 * H) ^ 5.25588 Absolute humidity [1], Eq. (8)H = 200.0 m x = 0.62198 * Pvap / (Patm - Pvap)
0.9894532 bar #VALUE! bar0.98945 bar
Using function 1 x = #VALUE! kg/kgSaturation pressure at dry bulb temperat [2]
Sicro_Saturated_vapor_pressure_t(tdb) *0.00001 Using function 11tdb = 40.0 ºC Sicro_Absolute_Humidity_t_f_H
#VALUE! bar tdb = 40.0 °Cf = 30.0 %
Relative humidity H = 200.0 mfi = f / 100 x = #VALUE! kg/kgf = 30.0fi = 0.30 Absolute dry bulb temperature
Tdb = tdb + 273.15tdb = 40.0 ºCTdb = 313.15 K
12.- Specific volume v = f(tdb, f, H) and 13.- density
Pvap =
Psat = Pvap =
Patm =
Patm = Pvap =Patm =
Psat =
Psat =
rev. cjc. 28.02.2018
Air gas constant Density [1], Eq.(11)
8314.41 J/(kmol*K)28.9645 Kmol/kg287.06 J/(kg*K) r = (1/v) * (1 + x)
v = #VALUE! m³/kg daSpecific volume [1] Eq.(28) x = #VALUE! kg/kg
0.62198 * Pvap / (Patm - Pvap) r = #VALUE! kg/m³
v = Rair*T/Patm *(1+1.6078*x) Using function 13287.06 J/(kg*K) r = Density_twb_f_H
T = 313.15 K tdb = 40.0 °C98,945 Pa f = 30.0 %
x = #VALUE! kg/kg H = 200.0 mv = #VALUE! m³/kg da r = #VALUE! kg/m³
Using function 12v = Specific_volume_tdb_twb_H
tdb = 40.0 ºCf 30.0 %
H = 200.0 mv = #VALUE! m³/kg da
Return to index
12.- Specific volume v = f(tdb, f, H) and 13.- density r = f(tdb,f , H)
Rair = Rgen / MMair
Rgen:MMair =Rair =
Rair =
Patm =
v=Rair⋅TPatm
⋅(1+1.6078⋅x ) (28 )
Microsoft Editor de ecuaciones 3.0
ρ=1v⋅(1+x ) (11)
14.- Dew point temperature tdp = f(tdb, f, H)
Data Relative humiditytdb = 37.8 °C fi = f / 100
f = 31.5 % f = 31.5H = 3.353 m fi = 0.315
Saturation pressure of water at dry bulb temperature Vapor pressuretdb = 37.8 °C [2]
Psat =(1/100)* Exp(ca / tK + cb + cc * tK + cd * tK ^ 2 + ce * tK ^ 3 + cf * Ln(tK)) [bar] 0.0655tK = t + 273.15 °C fi = 0.315tK = 310.9277778 K 0.02065ca = -5800.22006 2065cb = -5.516256cc = -0.04864024cd = 4.17648E-05 Dew point temperaturece = -1.4452E-08 -35.97 - 1.8726 * Ln(Pvap) + 1.1689 * (Ln((Pvap) )^2cf = 6.5459673 Pvap = 2065
0.0655 bar 17.84
Using function 14Using function 1 [2] tdp = Sicro_Dew_Point_tdb_f_H
Sicro_Saturated_vapor_pressure_t tdb = 37.8t = 37.8 °C twb = 31.5
#VALUE! bar H = 3.4tdp = #VALUE!
Using equation (35) from Ashrae 1985, the resulting dew point temperature isUsing equation (39) from Ashrae 20055, the resulting dew point temperature is
(see next page)Result when using ther VBA function
Datatdb = 37.8 °C [1] Ashrae 2005
f = 31.5 %
Return to index
Pvap = Psat *fiPsat =
Pvap =Pvap =
tDP =
Psat = tDP =
Psat =
Psat =
H = 3.4 m
Saturation vapor pressureSicro_Saturated_vapor_pressure_t(t)
t = 37.8 ºC
#VALUE! Pa
#VALUE! kPa
Vapor pressure
#VALUE! kPafi = 0.315 -
#VALUE! kPa
a =#VALUE! kPa
a = #VALUE!
C14 = 6.54C15 = 14.526 C16 = 0.7389C17 = 0.09486C18 = 0.4569
a = #VALUE!
#VALUE! kPatdp = #VALUE! ºC
For 0 ºC <= tdp <= 93 ºC
pw_sat =
pw_sat =
pvap = Psat *fi
pw_sat =
pvap =
Ln(pvap)pvap =
tDP = C14+ C15 * a+ C16 * a^2 + C17 * a^3 + C18*(pvap)^0.1984
pvap =
pvap: water vapor partial pressure for the moist air sample
rev. cjc. 28.02.2018
14.- Dew point temperature tdp = f(tdb, f, H)
% -
bar -barPa
Eq. (35), [2, edition 1985], page 6.9
-35.97 - 1.8726 * Ln(Pvap) + 1.1689 * (Ln((Pvap) )^2Pa °C (approx. Equatioin)
Sicro_Dew_Point_tdb_f_H°C From graphic% tdb = 37.8m f = 31.4°C x = 0.0133
tdp = 17.8
17.84 ºC#VALUE! ºC
#VALUE! ºC
Procedure to obtain the dew point temperature using the goal seek function
tDP =tDP =
tDP =
1. With the input data tdb, f and H, obtain the absolute humidity using the function
17.8
x = Sicro_Absolute_Humidity_tdb_f_HData
37.8 °Cf = 31.5 %H = 3.4 m
x = #VALUE! kg/kg
2. With the obtained value of the absolute humidity, find the dry bulb temperatureat which the resultant relative humidity is 100%.
Using the Goal seek functionData (Initially assumed)
18.00 °C By changing valuex = #VALUE! kg/kgH = 3.4 m
f = Sicro_Relative_Humidity_tdb_x_Hf = #VALUE! % Set Cell
The required value for the relative humidity is
100.0 To value
Obtained dew point
18.004 ºC
tdb =
tdb =
tDP = tdb at f = 100%
tDP =
°C%g/kg°C
Procedure to obtain the dew point temperature using the goal seek function
, f and H, obtain the absolute humidity using the function
37.837.817.8
31.4%
2. With the obtained value of the absolute humidity, find the dry bulb temperature
Result when using the equation (35) [2]
17.836 ºC
Result when using the equation (39) [1]
#VALUE! ºC
Result when using the VBA function 14
#VALUE! ºC
tDP =
tDP =
tDP =
15.- Enthalpy ent = f(tdb, f, H)
Input data Vapor pressuretdb = 37.8 °C
f = 31.5 % fi =H = 3.4 m
Atmospheric pressure [1], Eq. (3)
1.01325 * (1 - 0.0000225577 * H) ^ 5.25588 Absolute humidityH = 3.4 m x =
1.012847288 bar
Using function 1 x = Saturation pressure at dry bulb temperature [2]
Sicro_Saturated_vapor_pressure_t(tdb) *0.00001 Enthalpytdb = 37.8 ºC ent =
#VALUE! bar tdb =x =
Relative humidity ent =fi = f / 100f = 31.5 Using the function 15fi = 0.315 x =
tdb =f =H =
ent =
Pvap =
Psat = Pvap =
Patm =
Patm = Pvap =Patm =
Psat =
Psat =
rev. cjc. 28.02.2018
15.- Enthalpy ent = f(tdb, f, H)
Vapor pressure fi * Psat
0.315#VALUE! bar#VALUE! bar
Absolute humidity [1], Eq. (8)0.62198 * Pvap / (Patm - Pvap)
#VALUE! bar1.01285 bar#VALUE! kg/kg
tdb + x * (2501 + 1.805 * tdb)
37.8 ºC#VALUE! kg/kg#VALUE! kJ/kg
Using the function 15Sicro_Enthalpy_tdb_f_H
37.8 °C31.5 %3.4 m
#VALUE! kJ/kg
Return to index
Moist air as an ideal gas [1 ] , page 6 .8
The enthalpy of a mixture of perfectgases equals the sum of the individualpartial enthalpies of the components .Therefore, the enthalpy of the moistair can be written:h=ha+ hvh=ha+ x ⋅ hg (27 )whereha : specific enthalpy of dry air kJ/kg da
hv : specific enthalpy of vapor kJ/kg da
x: moisture content kg w /kgda
hg :specific enthalpy of saturated vaporat the temperature of the mixture kJ/kg wh: specific enthalpy of mixture kJ/kg da
Microsoft Equation 3.0
Moist air as an ideal gas [1 ] , page 6 . 8
The enthalpy of a mixture of perfectgases equals the sum of the individualpartial enthalpies of the components .Therefore, the enthalpy of the moistair can be written:h=ha+ hvh=ha+ x ⋅ hg (27 )whereha : specific enthalpy of dry air kJ/kg da
hv : specific enthalpy of vapor kJ/kg da
x: moisture content kg w /kgda
hg :specific enthalpy of saturated vaporat the temperature of the mixture kJ/kg wh: specific enthalpy of mixture kJ/kg da
16.- Wet bulb temperature twb = f(tdb, f, H)
Input data Saturation pressure of water at wet bulb temperaturetdb = 37.8 °C Input
f = 31.5 % Initially assumed and solution of the wet bulb temperatureH = 3.4 m 23.89 °C
Psat =(1/100)* Exp(ca / tK + cb + cc*tK + cd*tK ^ 2 + ce * tK ^ 3 + cf*Ln(tK)) [bar]
Atmospheric pressure '[bar] tK = t + 273.15 °C
1.01325 * (1 - 0.0000225577 * H) ^ 5.25588 tK = 297.04 KH = 3.4 m ca = -5800.22006
1.012847288 bar cb = -5.516256cc = -0.048640239
Using function 1 cd = 4.17648E-05Saturation pressure at dry bulb temperature [2] ce = -1.44521E-08
Sicro_Saturated_vapor_pressure_t(tdb) *0.00001 cf = 6.5459673tdb = 37.8 ºC 0.02965 bar
#VALUE! barAbsolute humidity at saturation and at twb
Relative humidityfi = f / 100 1.013 barf = 31.5 % 0.0297 barfi = 0.315 - 0.018759 kg/kg
Vapor pressure Wet bulb temperature fi * Psat twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb*x ) (2.38* xsat_twb-4.186 x-1)
fi = 0.315 tdb = 37.8 °C#VALUE! bar 0.01876 kg/kg#VALUE! bar x = #VALUE! kg/kg
twb = #VALUE! ºCAbsolute humidity [1], Eq. (8)
x = 0.62198 * Pvap / (Patm - Pvap) Iteration. Macro uses Goal seek function
#VALUE! bar
1.01285 bar #VALUE! Kx = #VALUE! kg/kg
#VALUE!Using function 11Sicro_Absolute_Humidity_tdb_f_H Using function 16
tdb = 37.8 °C twb = Sicro_Wet_Bulb_Temperature_tdb_f_Hf = 31.5 % tdb = 37.8 °CH = 3.4 m f = 31.5 %x = #VALUE! kg/kg H = 3.4 m
twb = #VALUE! ºC
Return to index
twbold =
Patm =
Patm =
Psat = Psat_wb =
Psat =
xsat_twb = 0.62198 * Psat_twb / (Patm - Psat_twb) Patm =
Psat_wb =xsat_twb =
Pvap =
Psat = xsat_twb =Pvap =
Pvap = Dtwb = Abs(twb - twbOld)
Patm = Dtwb =
16.- Wet bulb temperature twb = f(tdb, f, H)
Saturation pressure of water at wet bulb temperature Wet bulb temperature. twb = f(tdb, x, H)
Initially assumed and solution of the wet bulb temperature
Psat =(1/100)* Exp(ca / tK + cb + cc*tK + cd*tK ^ 2 + ce * tK ^ 3 + cf*Ln(tK)) [bar]
[1], Eq. (8)
twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb*x ) (2.38* xsat_twb-4.186 x-1)
twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb * x ) / (2.381 * xsat_twb - 4.186 * x - 1 )tdb = 37.8 °C
0.019311 kg/kgx = #VALUE! kg/kg
twb = #VALUE!
Sicro_Wet_Bulb_Temperature_tdb_f_H
/ (Patm - Psat_twb)
xsat_twb =
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 .9
x=(2501−2. 381⋅twb )⋅xsat twb−(tdb−twb )
2501+1 .805⋅tdb−4 . 186⋅twbEq (33)
x⋅(2501+1. 805⋅tdb−4 .186⋅twb )=(2501−2 . 381⋅twb )⋅xsat twb−(tdb−twb )2501⋅x+1 .805⋅tdb⋅x−4 .186⋅twb⋅x=2501⋅x sat twb−2. 381⋅twb⋅xsat twb− tdb+twb2 .381⋅twb⋅x sat twb−4 . 186⋅twb⋅x−twb=2501⋅xsat twb−tdb−2501⋅x−1 . 805⋅tdb⋅xtwb⋅(2 . 381⋅xsat twb−4 . 186⋅x−1 )=2501⋅xsat twb−tdb−2501⋅x−1. 805⋅tdb⋅x
twb=2501⋅xsat twb− tdb−2501⋅x−1.805⋅tdb⋅x2 .381⋅xsat twb−4 .186⋅x−1
twb=2501⋅xsat twb−tdb−2501⋅x−1. 805⋅tdb⋅x
2. 381⋅xsat twb−4 . 186⋅x−1
rev. cjc. 28.02.2018
16.- Wet bulb temperature twb = f(tdb, f, H)
(2501*xsat_twb -tdb - 2501*x - 1.805 * tdb * x ) / (2.381 * xsat_twb - 4.186 * x - 1 )
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 .9
x=(2501−2. 381⋅twb )⋅xsat twb−(tdb−twb )
2501+1 .805⋅tdb−4 . 186⋅twbEq (33)
x⋅(2501+1. 805⋅tdb−4 .186⋅twb )=(2501−2 . 381⋅twb )⋅xsat twb−(tdb−twb )2501⋅x+1 .805⋅tdb⋅x−4 .186⋅twb⋅x=2501⋅x sat twb−2. 381⋅twb⋅xsat twb− tdb+twb2 .381⋅twb⋅x sat twb−4 . 186⋅twb⋅x−twb=2501⋅xsat twb−tdb−2501⋅x−1 . 805⋅tdb⋅xtwb⋅(2 . 381⋅xsat twb−4 . 186⋅x−1 )=2501⋅xsat twb−tdb−2501⋅x−1. 805⋅tdb⋅x
twb=2501⋅xsat twb− tdb−2501⋅x−1.805⋅tdb⋅x2 .381⋅xsat twb−4 .186⋅x−1
Microsoft Editor de ecuaciones 3.0
21.- Absolute humidity x = f(tdb,twb,H)
Input datatdb = 37.8 °C Absolute humidity at saturation and at twbtwb = 23.9 °CH = 240.0 m 0.985
0.0297Atmospheric pressure [1], Eq. (3) 0.01931
Patm = 1.01325 * (1 - 0.0000225577 * H) ^ 5.25588H = 240.0 m Absolute humidity
Patm = 0.985 bar
Saturation pressure of water at wet bulb temperatureInput [2]
twb = 23.9 °CPsat =(1/100)* Exp(ca / tK + cb + cc * tK + cd * tK ^ 2 + ce * tK ^ 3 + cf * Ln(tK)) [bar]
x = ((2501-2.381*twb)*xsat_twb - (tdb-twb)) / (2501+1.805*tdb-4.186*twb)
tK = t + 273.15 °C tdb = 37.8tK = 297.0388889 K twb = 23.9
0.01931ca = -5800.22006 x = 0.013490cb = -5.516256cc = -0.04864024 Using the function 21cd = 4.17648E-05 x = Sicro_Absolute_Humidity_tdb_twb_Hce = -1.4452E-08 tdb = 37.8cf = 6.5459673 twb = 23.9
H = 240.00.02965 bar x = #VALUE!
Return to index
xsat_twb = 0.62198 * Psat_twb / (Patm - Psat_twb) Patm =
Psat_wb =xsat_twb =
xsat_wb =
Psat_wb =
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 . 9
x=(2501−2.381⋅twb )⋅xsat twb−(tdb−twb )
2501+1 . 805⋅tdb−4 .186⋅twbEq (33)
rev. cjc. 28.02.2018
21.- Absolute humidity x = f(tdb,twb,H)
Absolute humidity at saturation and at twb [1], Eq. (8)
barbarkg/kg
((2501-2.381*twb)*xsat_twb - (tdb-twb)) / (2501+1.805*tdb-4.186*twb) [2], Eq. (33)°C°Ckg/kgkg/kg
Sicro_Absolute_Humidity_tdb_twb_H°C°Cmkg/kg
0.62198 * Psat_twb / (Patm - Psat_twb)
Microsoft Editor de ecuaciones 3.0
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 .9
x=(2501−2. 381⋅twb )⋅xsat twb−( tdb−twb )
2501+1 .805⋅tdb−4 . 186⋅twbEq (33)
x = 0.012936
37.823.9
xsat_twb = 0.01875
x = 0.012936
xsat_twb = 0.01875
Input data Absolute humidity tdb = 37.8 °C x =((2501-2.381*twb)*xsat_twb - (tdb-twb)) / (2501+1.805*tdb-4.186*twb)
twb = 23.9 °C tdb = 37.8H = 3.4 m twb = 23.9
0.01876Atmospheric pressure [1], Eq. (3) x = 0.012943
1.01325 * (1 - 0.0000225577 * H) ^ 5.25588 H = 3.4 m Using the function 21
1.013 bar x = Sicro_Absolute_Humidity_tdb_twb_Htdb = 37.8
Saturation pressure of water at wet bulb temperature twb = 23.9Input [2] H = 3.4
twb = 23.9 °C x = #VALUE!Psat =(1/100)* Exp(ca / tK + cb+cc*tK + cd*tK^ 2 + ce* tK^ 3 + cf*Ln(tK)) [bar]
Vapor pressuretK = t + 273.15 °CtK = 297.0388889 K
ca = -5800.22006cb = -5.516256cc = -0.04864024cd = 4.17648E-05ce = -1.4452E-08cf = 6.5459673 x * Patm / ( 0.62198 + x )
x = 0.0129430.0297 bar 1.013
0.02065Absolute humidity at saturation and at twb [1], Eq. (8)
Absolute dry bulb temperature1.013 bar Tdb = tdb + 273.15
0.0297 bar tdb = 37.80.01876 kg/kg Tdb = 310.93
22.- Specific volume v = f(tdb,twb,H) and 23.- Density
xsat_wb =
Patm =
Patm =
Pvap =
Psat_wb = Patm =Pvap =
xsat_twb = 0.62198 * Psat_twb / (Patm - Psat_twb) Patm =
Psat_wb =xsat_twb =
Microsoft Editor de ecuaciones 3.0
Pvap=x⋅Patm
0. 62198+x
[2], Eq. (33) Air gas constant Density
x =((2501-2.381*twb)*xsat_twb - (tdb-twb)) / (2501+1.805*tdb-4.186*twb)
°C 8314.41 J/(kmol*K)°C 28.9645 Kmol/kgkg/kg 287.06 J/(kg*K) kg/kg
Specific volume [1] Eq.(28) r =v =
Sicro_Absolute_Humidity_tdb_twb_H x =°C r =°Cm v = Rair*T/Patm *(1+1.6078*x) Using the function 23kg/kg 287.06 J/(kg*K) r =
T = 310.93 K tdb =101,285 Pa twb =
x = #VALUE! kg/kg H =[1], Eq. (8a) v = #VALUE! m³/kg da r =
Using the function 22v = Sicro_Specific_volume_tdb_twb_H
tdb = 37.8 ºCtwb = 23.9 ºCH = 3.4 m
x * Patm / ( 0.62198 + x ) v = #VALUE! m³/kg dakg/kgbarbar
ºCK
Return to index
22.- Specific volume v = f(tdb,twb,H) and 23.- Density r = f(tdb, twb, H)
Rair = Rgen / MMair
Rgen:MMair =Rair =
Rair =
Patm =
Microsoft Editor de ecuaciones 3.0
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
ρ=1v⋅(1+x ) (11 )
rev. cjc. 28.02.2018
[1], Eq.(11)
(1/v) * (1 + x)#VALUE! m³/kg da0.012943 kg/kg#VALUE! kg/m³
Using the function 23Density_twb_tdb_H
37.8 °C23.9 °C3.4 m
#VALUE! kg/m³
Microsoft Editor de ecuaciones 3.0
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
ρ= 1v⋅(1+x ) (11 )
24.- Dew point temperature tdp = f(tdb, twb,H)
Input data Absolute humidity tdb = 37.8 °C x = ((2501-2.381*twb)*xsat_twb - (tdb-twb)) / (2501+1.805*tdb-4.186*twb)twb = 23.9 °C tdb = 37.8 °CH = 3.4 m twb = 23.9 °C
0.01876 kg/kgAtmospheric pressure [1], Eq. (3) x = 0.01294 kg/kg
1.01325 * (1 - 0.0000225577 * H) ^ 5.25588 H = 3.4 m Using the function 21
1.013 bar x = Sicro_Absolute_Humidity_tdb_twb_Htdb = 37.8 °C
Saturation pressure of water at wet bulb temperature twb = 23.9 °C [2] H = 3.4 m
twb = 23.9 °C x = #VALUE! kg/kgPsat =(1/100)* Exp(ca / tK + cb + cc * tK + cd * tK ^ 2 + ce * tK ^ 3 + cf * Ln(tK)) [bar]
tK = t + 273.15 °C Water vapor partial pressure. tK = 297.0388889 K Pvap = x * Patm / ( 0.62198 + x )ca = -5800.22006 x = 0.01294 kg/lgcb = -5.516256 Patm = 1.013 barcc = -0.04864024 Pvap = 0.02065 barcd = 4.17648E-05 Pvap = 2065 Pace = -1.4452E-08cf = 6.5459673 Dew point temperature
0.0297 bar tdp = -35.97 - 1.8726 * Ln(Pvap) + 1.1689 * (Ln((PvaP) )^2Pvap = 2065 Pa
Using function 1 [2] tdp = 17.84 °CSicro_Saturated_vapor_pressure_t
twb = 23.9 °C Using function 24#VALUE! bar tdp = Sicro_Dew_Point_tdb_twb_H(tdb, twb, H)
tdb = 37.8 °CAbsolute humidity at saturation and at twb [1], Eq. (8) twb = 23.9 °C
H = 3.4 m tdp = #VALUE! °C
Patm = 1.013 bar Psat = 0.0297 bar
0.01876 kg/kg
Return to index
xsat_twb =
Patm =
Patm =
Psat_wb =
Psat_wb =
Psat_wb =
xsat_twb = 0.62198 * Psat_twb / (Patm - Psat_twb)
xsat_twb =
x=0 .62198⋅PvapPair
x=0 .62198⋅PvapPatm−Pvap
x⋅(Patm−Pvap )=0. 62198⋅Pvap
x⋅Patm−x⋅Pvap=0 . 62198⋅Pvapx⋅Patm=0 . 62198⋅Pvap+x⋅Pvapx⋅Patm=Pvap⋅(0. 62198+x )
P vap=x⋅Patm0 . 62198+x
Eq . (A )
rev. cjc. 28.02.2018
24.- Dew point temperature tdp = f(tdb, twb,H)
[2], Eq. (33), page 6.9 x = ((2501-2.381*twb)*xsat_twb - (tdb-twb)) / (2501+1.805*tdb-4.186*twb)
Sicro_Absolute_Humidity_tdb_twb_H
[1], Eq. (8a)
Eq. (35), [2], page 6.9
-35.97 - 1.8726 * Ln(Pvap) + 1.1689 * (Ln((PvaP) )^2
(approx. Equation)
Sicro_Dew_Point_tdb_twb_H(tdb, twb, H)
Using function 24From graphic Sicro_Dew_Point_tdb_twb_H
tdb = 37.8 °C tdb =f = 31.4 % twb =x = 3.4 g/kg H =
tdp = 17.8 °C tdp =
Function DewPoint(p, W)'
Microsoft Editor de ecuaciones 3.0
17.8
x⋅Patm−x⋅Pvap=0 .62198⋅Pvapx⋅Patm=0 . 62198⋅Pvap+x⋅Pvapx⋅Patm=Pvap⋅(0. 62198+x )
Pvap=x⋅Patm0 . 62198+x
Eq . (A )
' DewPoint Macro' Macro recorded 9/23/2003 by Skavanaugh'pw = (p * W) / (0.62198 + W)
alpha = Log(pw)
C14 = 100.45C15 = 33.193C16 = 2.319C17 = 0.17074C18 = 1.2063
DewPoint = C14 + C15 * alpha + C16 * alpha ^ 2 + C17 * alpha ^ 3 + C18 * (pw) ^ 0.1984
End Function
tdb = 100 °F 37.8twb = 75 °F 23.9H = 11 ftW = #NAME? lb/lb
p = 14.696 * (1 - 0.0000068753 * Elevation) ^ 5.2559p = 14.690 psi
pw = (p * W) / (0.62198 + W)p = 14.690 psiW = #NAME? lb/lb
pw = #NAME? psialpha = Ln(pw) = #NAME?
tdp = C14 + C15 * alpha + C16 * alpha ^ 2 + C17 * alpha ^ 3 + C18 * (pw) ^ 0.1984tdp = #NAME? °Ftdp = #NAME? °C
Using function 24Sicro_Dew_Point_tdb_twb_H
37.8 °C23.9 °C3.4 m
#VALUE! °C
37.823.9 37.8
0.01876
0.01292
31.4%
C14 + C15 * alpha + C16 * alpha ^ 2 + C17 * alpha ^ 3 + C18 * (pw) ^ 0.1984
14.696 * (1 - 0.0000068753 * Elevation) ^ 5.2559
C14 + C15 * alpha + C16 * alpha ^ 2 + C17 * alpha ^ 3 + C18 * (pw) ^ 0.1984
0.01876
0.01292
25.- Enthalpy ent = f(tdb, twb,H)
Data Absolute humidity at saturation and at twbtdb = 37.8 °Ctwb = 23.9 °C Patm =H = 3.4 m Psat =
Atmospheric pressure [1], Eq. (3) Absolute humidity Patm = 1.01325 * (1 - 0.0000225577 * H) ^ 5.25588
H = 3.4 mPatm = 1.013 bar
Saturation pressure of water at wet bulb temperature [2]Input
twb = 23.9 °C x =Psat =(1/100)*Exp(ca tK+cb+cc * tK + cd * tK ^ 2 + ce * tK ^ 3 + cf * Ln(tK)) [bar] tdb =
twb =tK = t + 273.15 °CtK = 297.0388889 K x =
ca = -5800.22006 Using the function 21cb = -5.516256 x =cc = -0.04864024 tdb =cd = 4.17648E-05 twb =ce = -1.4452E-08 H =cf = 6.5459673 x =
0.0297 bar Enthalpyh =
tdb =[2], Eq. (30), page 6.9 x =
h =
h = t + x * (2501 + 1.805 * t) Using the function 25h: specific enthalpy [kJ/kg da] ent = t: dry bulb temperature [°C] tdb =x: absolute humidity [kg/lg] twb =
H =h =
Return to index
xsat_twb =
xsat_twb =
xsat_wb =
Psat_wb =
h=t+ x⋅(2501+1 .805⋅t ) (30 )
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 . 9
x=(2501−2. 381⋅twb )⋅xsat twb−(tdb−twb )
2501+1 . 805⋅tdb−4 . 186⋅twbEq (33)
rev. cjc. 28.02.2018
25.- Enthalpy ent = f(tdb, twb,H)
Absolute humidity at saturation and at twb [1], Eq. (8)
1.013 bar0.0297 bar0.01876 kg/kg
Absolute humidity
((2501-2.381*twb)*xsat_twb - (tdb-twb)) / (2501+1.805*tdb-4.186*twb) [2], Eq. (33)37.8 °C23.9 °C
0.01876 kg/kg0.0129433 kg/kg
Using the function 21Sicro_Absolute_Humidity_tdb_twb_H
37.8 °C23.9 °C3.4 m
#VALUE! kg/kg
[2], Eq. (30)tdb + x * (2501 + 1.805 * tdb)
37.8 °C#VALUE! kg/kg#VALUE! kJ/kg
Using the function 25Sicro_Enthalpy_tdb_twb_H
37.778 °C23.9 °C3.4 m
#VALUE! kJ/kg
0.62198 * Psat_twb / (Patm - Psat_twb)
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 . 9
x=(2501−2.381⋅twb )⋅xsat twb−(tdb−twb )
2501+1 . 805⋅tdb−4 . 186⋅twbEq (33)
Vapor pressureInput data
tdb = 37.8 °Ctwb = 23.9 °CH = 3.4 m.a.s.l.
Relative humidity
[1], Eq. (8b)
Return to index
27.- Relative humidity
Microsoft Equation 3.0
Vapor pressure
x=0 . 62198⋅PvapPair
[1 ] , Eq .(8 )
x=0 . 62198⋅PvapPatm−Pvap
x⋅(Patm−Pvap )=0. 62198⋅Pvapx⋅Patm−x⋅Pvap=0 .62198⋅Pvapx⋅Patm=0 . 62198⋅Pvap+x⋅Pvapx⋅Patm=Pvap⋅(0. 62198+x )
Pvap=x⋅Patm0 . 62198+x
[1 ] , Eq .(8a )
Microsoft Equation 3.0
Relative humidity
x=0 . 62198⋅PvapPair
[ 1] , Eq .(8 )
P vap=x0 .62198
⋅Pair
ϕ=PvapPsat
ϕ=1Psat
⋅Pvap
ϕ=1Psat
⋅x0 .62198
⋅Pair
ϕ=x0 . 62198
⋅PairP sat
[ 1] , Eq .(8b)
ϕ= x0 .62198
⋅PairPsat
[ 1] , Eq .(8b)
Pvap=x⋅Patm
0.62198+x [1 ] , Eq .(8a )
rev. cjc. 28.02.2018
Saturation pressure of water vapor [2]tdb = 37.8 °C Pvap = x * Patm / ( 0.62198 + x )
Psat =(1/100)* Exp(ca / tK + cb + cc * tK + cd * tK ^ 2 + ce * tK ^ 3 + cf * Ln(tK)) [bar] Pvap = #VALUE! bartK = t + 273.15 °CtK = 310.9 K Air partial pressureca = -5800.22006 Patm - Pvapcb = -5.516256 #VALUE! barcc = -0.04864024cd = 4.17648E-05 Relative humidity [1], Eq. (8b)ce = -1.4452E-08 (x / 0.62198) * (Pair/Psat) * 100cf = 6.5459673 x = #VALUE!
0.0655 bar #VALUE! bar0.0655 bar
Atmospheric pressure [1], Eq. (3) #VALUE! %Patm = 1.01325 * (1 - 0.0000225577 * H) ^ 5.25588
Patm = 1.012847288 bar Using the function 27Sicro_Relative_Humidity_tdb_twb_H
Using the function 21 tdb = 37.8 °CSicro_Absolute_Humidity_tdb_twb_H twb = 23.9 °C
tdb = 37.8 ºC H = 3.4 m.a.s..l.twb = 23.9 ºC #VALUE! %H = 3.4 m.a.s.l.x = #VALUE! kg/kg
27.- Relative humidity f = f(tdb, twb,H)
Water vapor partial pressure [1], Eq. (8a)
Pair =Pair =
f =
Psat = Pair =Psat =f =
f =
32.- Specific volume and 33.- density
Using the function 11Input data Sicro_Absolute_Humidity_tdb_f_H
tdb = 37.8 °C tdb = 37.8x 0.01294 kg/kg f = #VALUE!
H = 3.4 m H = 3.4x = #VALUE!
Atmospheric pressure '[bar] [1], Eq. (3)Patm = 1.01325 * (1 - 0.0000225577 * H) ^ 5.25588 Absolute dry bulb temperature
H = 3.4 m Tdb = tdb + 273.151.01285 bar tdb = 37.8
Tdb = 310.93Saturation pressure at dry bulb temperature [2]
Sicro_Saturated_vapor_pressure_t(tdb *0.00001 Air gas constanttdb = 37.8 ºC
#VALUE! bar 8314.4128.9645
Relative humidity [1], Eq. (8b) 287.06f =
1.012847288 bar Specific volume [1] Eq.(26), page 6.9#VALUE! bar
x = 0.01294 kg/kgf = #VALUE! -f = #VALUE! %
v = Rair*T/Patm *(1+1.6078*x)Vapor pressure 287.06
fi * Psat T = 310.93fi = #VALUE! 101,285
#VALUE! bar x = #VALUE!#VALUE! bar v = #VALUE!
Absolute humidity [1], Eq. (8) Using the function (32)x = 0.62198 * Pvap / (Patm - Pvap) v = Specific_volume_tdb_x_H
#VALUE! bar tdb = 37.81.01285 bar x = 0.0129
x = #VALUE! kg/kg H = 3.4v = #VALUE!
Return to index
Patm =
Psat = Rair = Rgen / MMair
Psat = Rgen:MMair =Rair =
(Patm/Psat) * (x /(0.62198 + x ) )Patm =Psat =
Rair =Pvap =
Patm =Psat = Pvap =
Pvap =Patm =
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (26 )
rev. cjc. 28.02.2018
32.- Specific volume and 33.- density
33.- Density [1], Eq.(11)Sicro_Absolute_Humidity_tdb_f_H
°C%m kg/kg
r = (1/v) * (1 + x)Absolute dry bulb temperature v = #VALUE! m³/kg da
x = #VALUE! kg/kgºC r = #VALUE! kg/m³K
Using the function 33r = Sicro_Density_twb_x_Htdb = 37.8 °C
J/(kmol*K) x = 0.0129 kg/kgKmol/kg H = 3.4 mJ/(kg*K) r = #VALUE! kg/m³
[1] Eq.(26), page 6.9
Rair*T/Patm *(1+1.6078*x)J/(kg*K)KPakg/kgm³/kg da
Specific_volume_tdb_x_H
ºCkg/kgmm³/kg da
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (26 ) Microsoft Editor de ecuaciones 3.0
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
ρ=1v⋅(1+x ) (11 )
34.- Dew point temperature tdp = f(tdb, x, H)
Input datatdb = 37.8 ºC
x 0.01294 kg/kgH = 3.4 m
Atmospheric pressure '[bar] [1], Eq. (3)Patm = 1.01325 * (1 - 0.0000225577 * H) ^ 5.25588
H = 3.4 m1.01285 bar
Water vapor partial pressure.
x * Patm / ( 0.62198 + x ) [1], Eq. (8a) Some properties do not require the input of three variables.x = 0.01294 kg/lg The VBA functions in many cases use other functions for the
1.013 bar calculation and these functions may depend on the complete set.0.02065 bar For this reason all VBA functions will require the input of a set
2065 Pa of three variables.
Return to index
Patm =
Pvap =
Patm =Pvap =Pvap =
Pvap=x⋅Patm
0. 62198+x [1 ] , Eq .(8a )
rev. cjc. 28.02.2018
34.- Dew point temperature tdp = f(tdb, x, H)
Dew point temperature Eq. (35), [2], page 6.9tdp = -35.97 - 1.8726 * Ln(Pvap) + 1.1689 * (Ln((PvaP) )^2
2065 Pa tdp = 17.84 °C (approx. Equation)
Using the function 34tdp = Sicro_Dew_Point_x_H(tdb, x, H)tdb = 37.8 ºCx = 0.01294 °CH = 3.4 m
tdp = #VALUE! °C
Some properties do not require the input of three variables.The VBA functions in many cases use other functions for thecalculation and these functions may depend on the complete set.For this reason all VBA functions will require the input of a setof three variables.
Pvap =
35.- Enthalpy ent = ent(tdb, x, H)
Input datatdb = 37.8 °Cx = 0.01294 kg/kgH = 4000.0 m.a.s.l.
Enthalpy [2], Eq. (30)ent = tdb + x * (2501 + 1.805 * tdb)tdb = 37.8 °Cx = 0.0129433 kg/kg
ent = 71.03 kJ/kg
Using the function 35 Some properties do not require the input of three variables.Sicro_Enthalpy_tdb_x_H The VBA functions in many cases use other functions for the
tdb = 37.778 °C calculation and these functions may depend on the complete set.x = 0.01294 kg/kg For this reason all VBA functions will require the input of a setH = 4000 m.a.s.l. of three variables.
ent = #VALUE! kJ/kg
[2], Eq. (30), page 6.9
h = t + x * (2501 + 1.805 * t)h: specific enthalpy [kJ/kg da]t: dry bulb temperature [°C]x: absolute humidity [kg/lg]
h=t+ x⋅(2501+1 . 805⋅t ) (30 )
rev. cjc. 28.02.2018
35.- Enthalpy ent = ent(tdb, x, H)
Some properties do not require the input of three variables.The VBA functions in many cases use other functions for thecalculation and these functions may depend on the complete set.For this reason all VBA functions will require the input of a setof three variables.
Return to index
36.- Wet bulb temperature twb = twb(tdb, x, H)
Absolute humidity at saturation and at twb
Input datatdb = 37.8 °C 1.013
x 0.01294 kg/kg 0.0297H = 3.4 m 0.018759
Atmospheric pressure '[bar] [1], Eq. (3)Patm = 1.01325 * (1 - 0.0000225577 * H) ^ 5.25588 Wet bulb temperature
H = 3.4 m twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb*x ) (2.38* xsat_twb-4.186 x-1)
1.01285 bar tdb = 37.80.01876
Saturation pressure of water at wet bulb temperature x = 0.012943Input [2] twb = 23.89
twb = 23.89 °CPsat =(1/100)* Exp(ca / tK + cb + cc * tK + cd * tK ^ 2 + ce * tK ^ 3 + cf * Ln(tK)) [bar] Using the function 36
tK = t + 273.15 °C twb = Sicro_Wet_Bulb_temperature_tdb_x_HtK = 297.0388889 K tdb = 37.8ca = -5800.22006 x = 0.01294cb = -5.516256 H = 3.4cc = -0.04864024 twb = #VALUE!cd = 4.17648E-05ce = -1.4452E-08cf = 6.5459673
0.02965 bar
xsat_twb = 0.62198 * Psat_twb / (Patm - Psat_twb) Patm =
Psat_wb =xsat_twb =
Patm =xsat_twb =
Psat_wb =
36.- Wet bulb temperature twb = twb(tdb, x, H)
Absolute humidity at saturation and at twb
[1], Eq. (8) Wet bulb temperature. twb = f(tdb, x, H)barbarkg/kg
[2], Eq. (33)twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb*x ) (2.38* xsat_twb-4.186 x-1)
°Ckg/kgkg/kg|C
Sicro_Wet_Bulb_temperature_tdb_x_H°C%m twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb * x ) / (2.381 * xsat_twb - 4.186 * x - 1 )ºC tdb = 37.8 °C
0.019311 kg/kgx = #VALUE! kg/kg
twb = #VALUE!
Return to index
0.62198 * Psat_twb / (Patm - Psat_twb)
xsat_twb =
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 . 9
x=(2501−2. 381⋅twb )⋅xsat twb−( tdb−twb )
2501+1 . 805⋅tdb−4 . 186⋅twbEq (33)
x⋅(2501+1. 805⋅tdb−4 . 186⋅twb )=(2501−2 . 381⋅twb )⋅xsat twb−( tdb−twb )2501⋅x+1 . 805⋅tdb⋅x−4 . 186⋅twb⋅x=2501⋅x sat twb−2. 381⋅twb⋅xsat twb− tdb+twb2 .381⋅twb⋅x sat twb−4 . 186⋅twb⋅x−twb=2501⋅xsat twb−tdb−2501⋅x−1 . 805⋅tdb⋅xtwb⋅(2 . 381⋅xsat twb−4 . 186⋅x−1)=2501⋅xsat twb−tdb−2501⋅x−1.805⋅tdb⋅x
twb=2501⋅xsat twb− tdb−2501⋅x−1. 805⋅tdb⋅x2 .381⋅xsat twb−4 .186⋅x−1
twb=2501⋅xsat twb−tdb−2501⋅x−1. 805⋅tdb⋅x
2. 381⋅xsat twb−4 . 186⋅x−1
rev. cjc. 28.02.2018
36.- Wet bulb temperature twb = twb(tdb, x, H)
(2501*xsat_twb -tdb - 2501*x - 1.805 * tdb * x ) / (2.381 * xsat_twb - 4.186 * x - 1 )
Microsoft Editor de ecuaciones 3.0
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 . 9
x=(2501−2.381⋅twb )⋅xsat twb−(tdb−twb )
2501+1 . 805⋅tdb−4 . 186⋅twbEq (33)
x⋅(2501+1. 805⋅tdb−4 . 186⋅twb )=(2501−2 . 381⋅twb )⋅xsat twb−(tdb−twb )2501⋅x+1 . 805⋅tdb⋅x−4 . 186⋅twb⋅x=2501⋅x sat twb−2. 381⋅twb⋅xsat twb− tdb+twb2 .381⋅twb⋅x sat twb−4 . 186⋅twb⋅x−twb=2501⋅xsat twb−tdb−2501⋅x−1 . 805⋅tdb⋅xtwb⋅(2 . 381⋅xsat twb−4 . 186⋅x−1)=2501⋅xsat twb−tdb−2501⋅x−1. 805⋅tdb⋅x
twb=2501⋅xsat twb− tdb−2501⋅x−1. 805⋅tdb⋅x2 .381⋅xsat twb−4 .186⋅x−1
twb=2501⋅xsat twb− tdb−2501⋅x−1. 805⋅tdb⋅x
2. 381⋅xsat twb−4 . 186⋅x−1
37.- Relative humidity f = f(tdb , x, H)
Input datatdb = 37.8 °C
x 0.01294 kg/kgH = 3.4 m
Atmospheric pressure '[bar] [1], Eq. (3)Patm = 1.01325 * (1 - 0.0000225577 * H) ^ 5.25588
H = 3.4 m1.01285 bar
Saturation pressure at dry bulb temperature [2]
Sicro_Saturated_vapor_pressure_t(tdb *0.00001 tdb = 37.8 |C
#VALUE! bar
Relative humidity [1], Eq. (8b)f =
1.012847288 bar#VALUE! bar x = 0.62198 * Pvap / (Patm - Pvap)
x = 0.01294 kg/kgf = #VALUE! -
Using the function 37f = Sicro_Relative_Humidity_tdb_x_H
tdb = 37.8 °Cx 0.01294 kg/kg
H = 3.4 mf = #VALUE! %
Patm =
Psat =
Psat =
(Patm/Psat) * (x /(0.62198 + x ) )Patm =Psat =
Microsoft Editor de ecuaciones 3.0
x=0 . 62198⋅Pvap(Patm−Pvap )
[ 1 ] ,Eq . (20 ) , page 6 . 8
x⋅(Patm−Pvap )=0. 62198⋅Pvapx⋅Patm−x⋅Pvap=0 . 62198⋅Pvapx⋅Patm=0 . 62198⋅Pvap+x⋅Pvapx⋅Patm=Pvap⋅(0. 62198+x )
Pvap=x⋅Patm0 . 62198+x
φ=PvapP sat
φ=PatmP sat
x0. 62198+x
[1 ] , Eq .(8b )
rev. cjc. 28.02.2018
37.- Relative humidity f = f(tdb , x, H)
0.62198 * Pvap / (Patm - Pvap)
Return to index
x=0 . 62198⋅Pvap(Patm−Pvap )
[ 1 ] ,Eq . (20 ) , page 6 . 8
x⋅(Patm−Pvap )=0. 62198⋅Pvapx⋅Patm−x⋅Pvap=0 . 62198⋅Pvapx⋅Patm=0 . 62198⋅Pvap+x⋅Pvapx⋅Patm=Pvap⋅(0. 62198+x )
Pvap=x⋅Patm0 . 62198+x
φ=PvapP sat
φ=PatmP sat
x0. 62198+x
[1 ] , Eq .(8b )
42.- Specific volume v = v(ent, x, H) and 43.- density Rho = f(ent, x, H)
ent = 71.03 kJ/kgx 0.01294 kg/kg Relative humidity [1], Eq. (8b)
H = 3.4 m.a.s.l. f =#VALUE! bar#VALUE! bar
Dry bulb temperature [1], Eq. (32) x = 0.01294 kg/kgtdb = (h - x*2501) / (1.805*x+1) f = #VALUE! -h = 71.0 kJ/kg f = #VALUE! %x = 0.0129433 kg/kg
tdb = 37.8 kJ/kg 'Vapor presure fi * Psat
Using the function 48 fi = #VALUE!Sicro_Dry_Bulb_Temperature_ent_x_H #VALUE! bar
h = 71.0316 kJ/kg #VALUE! barx = 0.0129 kg/kgH = 3.4 m.a.s.l. 'Absolute humidity [1], Eq. (8)
tdb = #VALUE! m x = 0.62198 * Pvap / (Patm - Pvap)#VALUE! bar
Atmospheric pressure [1], Eq. (3) #VALUE! barPatm = 1.01325 * (1 - 0.0000225577 * H) ^ 5.25588 x = #VALUE! kg/kg
H = #VALUE! m#VALUE! bar Using the function 11
Sicro_Absolute_Humidity_tdb_f_HSaturation pressure at dry bulb temperature tdb = 37.8 °C
[2] f = #VALUE! %Psat = Sicro_Saturated_vapor_pressure_t(tdb *0.00001 H = 3.4 m
tdb = 37.8 |C x = #VALUE! kg/kg#VALUE! bar
(Patm/Psat) * (x /(0.62198 + x ) )
Patm =Psat =
Pvap =
Psat = Pvap =
Pvap =Patm =
Patm =
Psat =
Moist air specific enthalpyent=1 .006⋅t db+x⋅(2501+1 .86⋅tdb ) [1 ] , (32)
ent=1 .006⋅t db+2501⋅x+1 . 86⋅tdb⋅xent=2501⋅x+1.006⋅tdb+1 . 86⋅tdb⋅xent=2501⋅x+tdb⋅(1 .006+1. 86⋅x )
2501⋅x+tdb⋅(1 .006+1 .86⋅x )=enttdb⋅(1 .006+1 .86⋅x )=ent- 2501⋅x
tdb=ent- 2501⋅x1.006+1 .86⋅x
[1 ] , (32a )
ent=1 .006⋅t db+x⋅(2501+1 .86⋅tdb )x⋅(2501+1 .86⋅tdb )=ent−1 . 006⋅t db
x=ent−1 .006⋅tdb2501+1 .86⋅tdb
[1 ] , (32b )
Microsoft Equation 3.0
Moist air specific enthalpyent=1 .006⋅t db+x⋅(2501+1 .86⋅tdb ) [1 ] , (32)
ent=1 .006⋅t db+2501⋅x+1 . 86⋅tdb⋅xent=2501⋅x+1.006⋅tdb+1 . 86⋅tdb⋅xent=2501⋅x+tdb⋅(1 .006+1. 86⋅x )
2501⋅x+tdb⋅(1 .006+1 .86⋅x )=enttdb⋅(1 .006+1 .86⋅x )=ent- 2501⋅x
tdb=ent- 2501⋅x1.006+1 .86⋅x
[1 ] , (32a )
ent=1 .006⋅t db+x⋅(2501+1 .86⋅tdb )x⋅(2501+1 .86⋅tdb )=ent−1 . 006⋅t db
x=ent−1 .006⋅tdb2501+1 .86⋅tdb
[1 ] , (32b )
42.- Specific volume v = v(ent, x, H) and 43.- density Rho = f(ent, x, H)
7.- DensityAbsolute dry bulb temperature
Tdb = tdb + 273.15tdb = 37.8 ºCTdb = 310.93 K
Air gas constant r = (1/v) * (1 + x)v = #VALUE!
8314.41 J/(kmol*K) x = #VALUE!28.9645 Kmol/kg r = #VALUE!
287.06 J/(kg*K)Using the function 43
Specific volume [1] Eq.(26), page 6.9 r = Sicro_Density_ent_x_H(ent, x, H)
tdb = 71.0x = 0.0129H = 3.4r = #VALUE!
v = Rair*T/Patm *(1+1.6078*x)287.06 J/(kg*K)
T = 310.93 K#VALUE! Pa
x = #VALUE! kg/kgv = #VALUE! m³/kg da
Using the function (42)v = Specific_Volume_ent_x_H(ent, x, H)
ent = 71.0 kJ/kgx = 0.0129 kg/kgH = 3.4 mv = #VALUE! m³/kg da
Density
ent = 2501 * x + tdb * (1.006+1.86*x)x = 0.0129 kg/kg
tdb = #VALUE!ent = #VALUE! kJ/kg
Return to index
Rair = Rgen / MMair
Rgen:MMair =Rair =
Rair =
Patm =
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (26 )
Microsoft Editor de ecuaciones 3.0
ent=1 .006⋅tdb+x⋅(2501+1 .86⋅t db ) [1 ] , (32)
tdb=ent- 2501⋅x1 . 006+1 . 86⋅x
[1 ] , (32a )
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
ρ=1v⋅(1+x ) (11)
ρ=1v⋅(1+x ) (11 )
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
tdb=h−x⋅25011 . 805⋅x+1
tdb⋅(1 .805⋅x+1 )=h−x⋅2501tdb⋅(1 .805⋅x+1 )+x⋅2501=hh=tdb⋅(1.805⋅x+1 )+x⋅2501
ρ=1v⋅(1+x ) (11 )
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
Moist air specific enthalpyent=1 .006⋅t db+x⋅(2501+1 .86⋅t db ) [1 ] , (32)
ent=1 .006⋅t db+2501⋅x+1 . 86⋅tdb⋅xent=2501⋅x+1.006⋅tdb+1 . 86⋅tdb⋅xent=2501⋅x+tdb⋅(1 .006+1. 86⋅x )
2501⋅x+tdb⋅(1 .006+1 .86⋅x )=entt db⋅(1 .006+1 .86⋅x )=ent- 2501⋅x
t db=ent- 2501⋅x1. 006+1 .86⋅x
[1 ] , (32a )
ent=1 .006⋅t db+x⋅(2501+1 .86⋅t db )x⋅(2501+1 .86⋅tdb )=ent−1 . 006⋅t db
x=ent−1 .006⋅tdb2501+1 . 86⋅tdb
[1 ] , (32b )
Microsoft Editor de ecuaciones 3.0
ρ=1v⋅(1+x ) (11 )
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
tdb=h−x⋅25011 . 805⋅x+1
tdb⋅(1 .805⋅x+1 )=h−x⋅2501tdb⋅(1 .805⋅x+1 )+x⋅2501=hh=tdb⋅(1.805⋅x+1 )+x⋅2501
Moist air specific enthalpyent=1 .006⋅t db+x⋅(2501+1 .86⋅t db ) [1 ] , (32)
ent=1 .006⋅t db+2501⋅x+1 . 86⋅tdb⋅xent=2501⋅x+1.006⋅tdb+1 . 86⋅tdb⋅xent=2501⋅x+tdb⋅(1 .006+1. 86⋅x )
2501⋅x+tdb⋅(1 .006+1 .86⋅x )=entt db⋅(1 .006+1 .86⋅x )=ent- 2501⋅x
t db=ent- 2501⋅x1. 006+1 .86⋅x
[1 ] , (32a )
ent=1 .006⋅t db+x⋅(2501+1 .86⋅t db )x⋅(2501+1 .86⋅tdb )=ent−1 . 006⋅t db
x=ent−1 .006⋅tdb2501+1 . 86⋅tdb
[1 ] , (32b )x=
ent−1 .006⋅tdb2501+1 . 86⋅t db
[1 ] , (32b )
rev. cjc. 28.02.2018
42.- Specific volume v = v(ent, x, H) and 43.- density Rho = f(ent, x, H)
[1], Eq.(11)
m³/kg dakg/kgkg/m³
Sicro_Density_ent_x_H(ent, x, H)
kJ/kgkg/kgmkg/m³
Some properties do not require the input of three variables.The VBA functions in many cases use other functions for thecalculation and these functions may depend on the complete set.For this reason all VBA functions will require the input of a setof three variables.
[1], Eq.(11)
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
ρ=1v⋅(1+x ) (11 )
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
tdb=h−x⋅25011 . 805⋅x+1
tdb⋅(1 .805⋅x+1 )=h−x⋅2501tdb⋅(1 .805⋅x+1 )+x⋅2501=hh=tdb⋅(1. 805⋅x+1 )+x⋅2501
ρ=1v⋅(1+x ) (11)
v=Rair⋅TPatm
⋅(1+1.6078⋅x ) (28 )
Microsoft Editor de ecuaciones 3.0
ρ=1v⋅(1+x ) (11 )
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
tdb=h−x⋅25011 . 805⋅x+1
tdb⋅(1 .805⋅x+1 )=h−x⋅2501tdb⋅(1 .805⋅x+1 )+x⋅2501=hh=tdb⋅(1. 805⋅x+1 )+x⋅2501
44.- Dew point temperature tdp = tdp(ent, x, H)
Dataent = 71 kJ/kgx = 0.01294 kg/kg Dew point temperatureH = 3.4 m tdp =
Atmospheric pressure [1], Eq. (3) tdp =Patm = 1.01325 * (1 - 0.0000225577 * H) ^ 5.25588
H = 3.4 m Using the function 441.013 bar tdp =
ent =Water vapor partial pressure. x =
x * Patm / ( 0.62198 + x ) [1], Eq. (8a) H =x = 0.01294 kg/lg tdp =
1.013 bar0.02065 bar
2065 Pa
Some properties do not require the input of three variables.The VBA functions in many cases use other functions for thecalculation and these functions may depend on the complete set.For this reason all VBA functions will require the input of a setof three variables.
Pvap =
Patm =
Pvap =
Patm =Pvap =Pvap =
rev. cjc. 28.02.2018
44.- Dew point temperature tdp = tdp(ent, x, H)
Dew point temperature Eq. (35), [2], page 6.9 -35.97 - 1.8726 * Ln(Pvap) + 1.1689 * (Ln((PvaP) )^2
2065 Pa 17.84 °C (Approx. Equation)
Using the function 44Sicro_Dew_Point_x_H(ent,x, H)
71 kJ/kg0.01294 °C
3.4 m#VALUE! °C
Some properties do not require the input of three variables.The VBA functions in many cases use other functions for thecalculation and these functions may depend on the complete set.For this reason all VBA functions will require the input of a setof three variables.
Return to index
46.- Wet bulb temperature twb = twb(ent, x, H)
Absolute humidity at saturation and at twb
Input dataent 71.0 kJ/kg #VALUE! bar
x 0.01294 kg/kg 0.0297 barH = 3.4 m.a.s.l. #VALUE! kg/kg
Dry bulb temperature Wet bulb temperaturetdb = (h - x*2501) / (1.805*x+1) twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb*x ) (2.38* xsat_twb-4.186 x-1)
h = 71.0 kJ/kg tdb = #VALUE! °Cx = 0.0129 kg/kg #VALUE! kg/kg
tdb = 37.8 kJ/kg x = 0.012943 kg/kgtwb = #VALUE! |C
Using the function 48Sicro_Dry_Bulb_Temperature_ent_x_H Using the function 46
h = 71.0316 kJ/kg twb = Sicro_Wet_Bulb_Temperature_ent_x_Hx = 0.0129 kg/kg tdb = 71.0 °CH = 3.4 m.a.s..l. x = 0.01294 %
tdb = #VALUE! m H = 3.4 mtwb = #VALUE! ºC
Atmospheric pressure [1], Eq. (3)Patm = 1.01325 * (1 - 0.0000225577 * H) ^ 5.25588
H = #VALUE! m #VALUE! bar
Saturation pressure of water at wet bulb temperatureInput
twb = 23.89 °CPsat =(1/100)* Exp(ca / tK + cb + cc * tK + cd * tK ^ 2 + ce * tK ^ 3 + cf * Ln(tK)) [bar]
tK = t + 273.15 °CtK = 297.0388889 Kca = -5800.22006cb = -5.516256 Some properties do not require the input of three variables.cc = -0.04864024 The VBA functions in many cases use other functions for thecd = 4.17648E-05 calculation and these functions may depend on the complete set.ce = -1.4452E-08 For this reason all VBA functions will require the input of a setcf = 6.5459673 of three variables.
0.02965 bar
Return to index
xsat_twb = 0.62198 * Psat_twb / (Patm - Psat_twb) Patm =
Psat_wb =xsat_twb =
xsat_twb =
Patm =
Psat_wb =
46.- Wet bulb temperature twb = twb(ent, x, H)
Wet bulb temperature. twb = f(tdb, x, H)
twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb*x ) (2.38* xsat_twb-4.186 x-1)
Sicro_Wet_Bulb_Temperature_ent_x_H
23.89 °C
twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb * x ) / (2.381 * xsat_twb - 4.186 * x - 1 )tdb = 71.0 °C
0.019311 kg/kgx = #VALUE! kg/kg
twb = #VALUE!
Some properties do not require the input of three variables.The VBA functions in many cases use other functions for thecalculation and these functions may depend on the complete set.For this reason all VBA functions will require the input of a set
/ (Patm - Psat_twb)
xsat_twb =
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 .9
x=(2501−2. 381⋅twb )⋅xsat twb−(tdb−twb )
2501+1 .805⋅tdb−4 . 186⋅twbEq (33)
x⋅(2501+1. 805⋅tdb−4 .186⋅twb )=(2501−2 . 381⋅twb )⋅xsat twb−(tdb−twb )2501⋅x+1 .805⋅tdb⋅x−4 .186⋅twb⋅x=2501⋅x sattwb−2. 381⋅twb⋅xsat twb− tdb+twb2 .381⋅twb⋅x sat twb−4 . 186⋅twb⋅x−twb=2501⋅xsat twb−tdb−2501⋅x−1 . 805⋅tdb⋅xtwb⋅(2 . 381⋅xsat twb−4 . 186⋅x−1 )=2501⋅xsat twb−tdb−2501⋅x−1. 805⋅tdb⋅x
twb=2501⋅xsat twb− tdb−2501⋅x−1.805⋅tdb⋅x2 .381⋅xsat twb−4 .186⋅x−1
twb=2501⋅xsat twb− tdb−2501⋅x−1. 805⋅tdb⋅x
2. 381⋅xsat twb−4 . 186⋅x−1
rev. cjc. 28.02.2018
46.- Wet bulb temperature twb = twb(ent, x, H)
(2501*xsat_twb -tdb - 2501*x - 1.805 * tdb * x ) / (2.381 * xsat_twb - 4.186 * x - 1 )
Microsoft Editor de ecuaciones 3.0
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 .9
x=(2501−2. 381⋅twb )⋅xsat twb−(tdb−twb )
2501+1 .805⋅tdb−4 . 186⋅twbEq (33)
x⋅(2501+1. 805⋅tdb−4 .186⋅twb )=(2501−2 . 381⋅twb )⋅xsat twb−(tdb−twb )2501⋅x+1 .805⋅tdb⋅x−4 .186⋅twb⋅x=2501⋅x sat twb−2. 381⋅twb⋅xsat twb− tdb+twb2 .381⋅twb⋅x sat twb−4 . 186⋅twb⋅x−twb=2501⋅xsat twb−tdb−2501⋅x−1 . 805⋅tdb⋅xtwb⋅(2 . 381⋅xsat twb−4 . 186⋅x−1 )=2501⋅xsat twb−tdb−2501⋅x−1. 805⋅tdb⋅x
twb=2501⋅xsat twb− tdb−2501⋅x−1.805⋅tdb⋅x2 .381⋅xsat twb−4 .186⋅x−1
Input data Saturation pressure at dry bulb temperatureent = 71.0 kJ/kg Psat = Sicro_Saturated_vapor_pressure_t(tdb *0.00001
x 0.01294 kg/kg tdb = #VALUE!H = 3.4 m #VALUE!
Dry bulb temperature Relative humiditytdb = (ent - x*2501) / (1.805*x+1)h = 71.0 kJ/kg 1.0128x = 0.0 kg/kg #VALUE!
tdb = 37.8 kJ/kg x = 0.01294f = #VALUE!
Using the function 48 Using the function 47Sicro_Dry_Bulb_Temperature_ent_x_H f = Sicro_Relative_Humidity_ent_x_H(ent, x, H)
ent = 71.03 ent = 71.0x = 0.0129 kJ/kg x 0.01294H = 3.3528 kg/kg H = 3.4
tdb = #VALUE! m f = #VALUE!
Atmospheric pressure '[bar]Patm = 1.01325 * (1 - 0.0000225577 * H) ^ 5.25588
H = 3.4 m1.0128 bar
Some properties do not require the input of three variables.The VBA functions in many cases use other functions for thecalculation and these functions may depend on the complete set.For this reason all VBA functions will require the input of a setof three variables.
Return to index
47.- Relative humidity f = f(ent, x, H)
Psat =
f = (Patm/Psat) * (x /(0.62198 + x ) )Patm =Psat =
Patm =
rev. cjc. 28.02.2018
Saturation pressure at dry bulb temperature
Psat = Sicro_Saturated_vapor_pressure_t(tdb *0.00001
|Cbar
barbarkg/kg -
Sicro_Relative_Humidity_ent_x_H(ent, x, H)kJ/kgkg/kgm%
= f(ent, x, H)
) * (x /(0.62198 + x ) )
48.- Dry bulb temperature tdb = tdb(ent, x, H)
Datah = 71.0 kJ/kgx = 0.01294 kg/kgH = 3.4 m
Dry bulb temperature [1], Eq. (32)tdb = (h - x*2501) / (1.805*x+1)h = 71.0 kJ/kgx = 0.0129433 kg/kg
tdb = 37.8 kJ/kg
Using the function 48 [2], Eq. (30), page 6.9ent = Sicro_Dry_Bulb_Temperature_ent_x_Hh = 71.0316 kJ/kg h = t + x * (2501 + 1.805 * t)x = 0.0129 kg/kg h: specific enthalpy H = 3.3528 t: dry bulb temperature
tdb = #VALUE! m x: absolute humidity
Some properties do not require the input of three variables.The VBA functions in many cases use other functions for thecalculation and these functions may depend on the complete set.For this reason all VBA functions will require the input of a setof three variables.
Return to index
Dry bulb temperature[2 ] , Eq . (30 ) , page 6 .9
h=tdb+x⋅(2501+1. 805⋅tdb )h=tdb+x⋅2501+1 . 805⋅x⋅tdbh=1 .805⋅x⋅tdb+tdb+x⋅2501h=tdb⋅(1 . 805⋅x+1 )+ x⋅2501tdb⋅(1 . 805⋅x+1 )=h−x⋅2501
tdb=h− x⋅25011. 805⋅x+1
rev. cjc. 28.02.2018
48.- Dry bulb temperature tdb = tdb(ent, x, H)
t + x * (2501 + 1.805 * t)[kJ/kg da][°C][kg/lg]
Microsoft Editor de ecuaciones 3.0
51.- Absolute humidity x = x(tdb, ent)
Imput data (Not a function of H)tdb = 37.8 °Cent = 71.03 kJ/kgH = 0 m.a.s.l.
Absolute humidityx = (ent - tdb) / (2501 + 1.805*tdb ) [1], Eq. (32b)
ent = 71.03 kJ/kgtdb = 37.8 ºCx = 0.012943 kg/kg
Using function 51 The absolute humidity is independent of the height above sea level.x = Sicro_Absolute_Humidity_tdb_ent(tdb, ent,H) The VBA functions in many cases use other functions for the
tdb = 37.8 °C calculation and these functions may depend on the height "H".ent = 71.0 kJ/kg For this reason all VBA functions will require the input of a setH = 0.0 of three variables.x = #VALUE! kg/kg
Return to index
x=ent−1 .006⋅tdb2501+1 . 86⋅t db
[1 ] , (32b )
rev. cjc. 28.02.2018
51.- Absolute humidity x = x(tdb, ent)
The absolute humidity is independent of the height above sea level.The VBA functions in many cases use other functions for thecalculation and these functions may depend on the height "H".For this reason all VBA functions will require the input of a setof three variables.
x=ent−1 .006⋅tdb2501+1 . 86⋅t db
[1 ] , (32b )
Input data Relative humidity [1], Eq. (8b)tdb = 37.78 ºC f =ent = 71.03 kJ/kg 1.012847288 barH = 3.4 m #VALUE! bar
x = 0.01294 kg/kgAtmospheric pressure [1], Eq. (3) f = #VALUE! -Patm= 1.01325 * (1-0.0000225577 *H) ^ 5.25588 f = #VALUE! %
H = 3.4 m1.01285 bar 'Vapor presure
fi * PsatSaturation pressure at dry bulb temperature fi = #VALUE!
[2] #VALUE! bar
Psat = Sicro_Saturated_vapor_pressure_t(tdb *0.00001 #VALUE! bartdb = 37.8 |C
#VALUE! bar Absolute dry bulb temperatureTdb = tdb + 273.15
Absolute humidity [1], Eq. (32b) tdb = 37.8 ºCx = (ent - tdb) / (2501 + 1.805*tdb ) Tdb = 310.93 K
ent = 71.03 kg/kgtdb = 37.8 ºCx = 0.012943 kg/kg
Return to index
52.- Specific volume v = v(tdb, ent, H) and 53.- density
(Patm/Psat) * (x /(0.62198 + x ) )
Patm =Psat =
Patm =Pvap =
Psat = Pvap =
Psat =
Air gas constant Using the function 52 v = Sicro_Specific_Volume_tdb_ent_H
8314.41 J/(kmol*K) tdb = 37.7828.9645 Kmol/kg ent = 71.0287.06 J/(kg*K) H = 3.4
v = #VALUE!Specific volume [1] Eq.(26), page 6.9
53.- Density [1], Eq.(26), page 6.8
v = Rair*T/Patm *(1+1.6078*x)287.06 J/(kg*K)
T = 310.93 K r = (1/v) * (1 + x)101,285 Pa v = 0.900
x = 0.012943 kg/kg x = 0.012943v = 0.900 m³/kg da r = 1.126
Using the function 53Rho = Sicro_Density_tdb_ent_H(tdb, ent, H)
tdb = 37.78ent = 71.0H = 3.4r = #VALUE!
52.- Specific volume v = v(tdb, ent, H) and 53.- density r = r(tdb, ent, H)
Rair = Rgen / MMair
Rgen:MMair =Rair =
Rair =
Patm =
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (26 )
Microsoft Editor de ecuaciones 3.0
ρ=1v⋅(1+x ) (9a )
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (26 )
rev. cjc. 28.02.2018
v = Sicro_Specific_Volume_tdb_ent_H
ºCkJ/kgmm³/kg da
[1], Eq.(26), page 6.8
m³/kg dakg/kgkg/m³
Rho = Sicro_Density_tdb_ent_H(tdb, ent, H)ºC
kJ/kgm
kg/m³
ρ=1v⋅(1+x ) (9a )
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (26 )
54.- Dew point temperature tdp = tdp(tdb, ent, H)
Data Dew point temperaturetdb = 37.8 ºC tdp = -35.97 - 1.8726 * Ln(Pvap) + 1.1689 * (Ln((PvaP) )^2
ent = 71.0 kJ/kg 2066H = 3.4 m tdp = 17.84
Absolute humidity [1], Eq. (32b) Using the function 54x = (ent - tdb) / (2501 + 1.805*tdb ) tdp = Sicro_Dew_Point_tdb_ent_H(tdb,ent, H)
ent = 71.03 kg/kg tdb = 37.8tdb = 37.8 ºC ent = 71.03x = 0.012943 kg/kg H = 3.4
tdp = #VALUE!Atmospheric pressure [1], Eq. (3)Patm= 1.01325 * (1-0.0000225577 *H) ^ 5.25588
H = 0.0 m1.01325 bar
Water vapor partial pressure. [1], Eq. (8a)x * Patm / ( 0.62198 + x )
x = 0.01294 kg/lg1.013 bar
0.02066 bar2066 Pa
Return to index
Pvap =
Patm =
Pvap =
Patm =Pvap =Pvap =
rev. cjc. 28.02.2018
54.- Dew point temperature tdp = tdp(tdb, ent, H)
Eq. (35), [2], page 6.9 -35.97 - 1.8726 * Ln(Pvap) + 1.1689 * (Ln((PvaP) )^2
Pa °C (approx. Equation)
Sicro_Dew_Point_tdb_ent_H(tdb,ent, H)°CkJ/kgm°C
56.- Wet bulb temperature twb = twb(tdb, ent, H)
Input data Absolute humidity at saturation and at twbtdb = 37.8 °Cent = 71.03 kg/kg #NAME?H = 3.4 m 0.0297
#NAME?Absolute humidity [1], Eq. (32b)
x = (ent - tdb) / (2501 + 1.805*tdb ) Wet bulb temperatureent = 71.03 kg/kg twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb*x ) (2.38* xsat_twb-4.186 x-1)tdb = 37.8 ºC tdb = 37.8x = 0.012943 kg/kg #NAME?
x = 0.012943Using the function 51 twb = #NAME? Sicro_Absolute_Humidity_tdb_ent(tdb, ent)
tdb = 37.8 °C Using the function 56ent = 71.0 kg/kg twb = Sicro_Wet_Bulb_Temperature_tdb_ent_HH = 3.4 m tdb = 37.8x = #NAME? m ent 71.03163
H = 3.4Atmospheric pressure [1], Eq. (3) twb = #VALUE!Patm= 1.01325 * (1-0.0000225577 *H) ^ 5.25588
H = #NAME? m#NAME? bar
Saturation pressure of water at wet bulb temperatureInput
twb = 23.89 °CPsat =(1/100)* Exp(ca / tK + cb + cc * tK + cd * tK ^ 2 + ce * tK ^ 3 + cf * Ln(tK)) [bar]
tK = t + 273.15 °CtK = 297.0388889 Kca = -5800.22006cb = -5.516256cc = -0.04864024cd = 4.17648E-05ce = -1.4452E-08cf = 6.5459673
0.02965 bar
Return to index
xsat_twb = 0.62198 * Psat_twb / (Patm - Psat_twb) Patm =
Psat_wb =xsat_twb =
xsat_twb =
Patm =
Psat_wb =
56.- Wet bulb temperature twb = twb(tdb, ent, H)
Absolute humidity at saturation and at twb [1], Eq. (8) Wet bulb temperature. twb = f(tdb, x, H)
barbarkg/kg
twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb*x ) (2.38* xsat_twb-4.186 x-1)°Ckg/kgkg/kg|C
Sicro_Wet_Bulb_Temperature_tdb_ent_H°C%mºC
twb = (2501*xsat_twb -tdb - 2501*x - 1.805 * tdb * x ) / (2.381 * xsat_twb - 4.186 * x - 1 )
tdb = 37.8 °C0.019311 kg/kg
x = #VALUE! kg/kgtwb = #VALUE!
0.62198 * Psat_twb / (Patm - Psat_twb)
xsat_twb =
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 . 9
x=(2501−2. 381⋅twb )⋅xsat twb−( tdb−twb )
2501+1 . 805⋅tdb−4 . 186⋅twbEq (33)
x⋅(2501+1. 805⋅tdb−4 . 186⋅twb )=(2501−2 .381⋅twb )⋅xsat twb−(tdb−twb )2501⋅x+1 . 805⋅tdb⋅x−4 . 186⋅twb⋅x=2501⋅x sat twb−2.381⋅twb⋅xsat twb− tdb+twb2 .381⋅twb⋅x sat twb−4 .186⋅twb⋅x−twb=2501⋅xsat twb−tdb−2501⋅x−1 . 805⋅tdb⋅xtwb⋅(2 . 381⋅xsat twb−4 . 186⋅x−1)=2501⋅xsat twb−tdb−2501⋅x−1.805⋅tdb⋅x
twb=2501⋅xsat twb− tdb−2501⋅x−1.805⋅tdb⋅x2 .381⋅xsat twb−4 .186⋅x−1
twb=2501⋅xsat twb−tdb−2501⋅x−1. 805⋅tdb⋅x
2. 381⋅xsat twb−4 . 186⋅x−1
rev. cjc. 28.02.2018
56.- Wet bulb temperature twb = twb(tdb, ent, H)
(2501*xsat_twb -tdb - 2501*x - 1.805 * tdb * x ) / (2.381 * xsat_twb - 4.186 * x - 1 )
Microsoft Editor de ecuaciones 3.0
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 .9
x=(2501−2. 381⋅twb )⋅xsattwb−(tdb−twb )
2501+1 . 805⋅tdb−4 . 186⋅twbEq (33)
x⋅(2501+1.805⋅tdb−4 .186⋅twb )=(2501−2 . 381⋅twb )⋅xsat twb−( tdb−twb )2501⋅x+1 .805⋅tdb⋅x−4 .186⋅twb⋅x=2501⋅x sattwb−2. 381⋅twb⋅xsat twb−tdb+twb2 .381⋅twb⋅x sat twb−4 . 186⋅twb⋅x− twb=2501⋅xsat twb−tdb−2501⋅x−1 .805⋅tdb⋅xtwb⋅(2 . 381⋅xsattwb−4 . 186⋅x−1 )=2501⋅xsat twb−tdb−2501⋅x−1.805⋅tdb⋅x
twb=2501⋅xsat twb− tdb−2501⋅x−1. 805⋅tdb⋅x2 .381⋅xsattwb−4 .186⋅x−1
twb=2501⋅xsat twb−tdb−2501⋅x−1. 805⋅tdb⋅x
2. 381⋅xsat twb−4 . 186⋅x−1
57.- Relative humidity f = f(tdb, ent, H)
Input data Relative humidity [1], Eq. (8b)tdb = 37.8 °Cent = 71.0 kg/kg 1.013248445 barH = 550 m #VALUE! bar
x = 0.01294 kg/kgAbsolute humidity [1], Eq. (32b) f = #VALUE! -
x = (ent - tdb) / (2501 + 1.805*tdb )ent = 71.0 kg/kgtdb = 37.8 ºC Using the function 57x = 0.012943 kg/kg f = Sicro_Relative_Humidity_tdb_ent_H(tdb, ent, H)
tdb = 37.8 °CAtmospheric pressure [1], Eq. (3) ent = 71.0 kg/kgPatm= 1.01325 * (1-0.0000225577 *H) ^ 5.25588 H = 550 m
H = 0.0 m f = #VALUE! %1.01325 bar
Saturation pressure at dry bulb temperature[2]
Psat=Sicro_Saturated_vapor_pressure_t(tdb *0.00001
tdb = 37.8 |C#VALUE! bar
Return to index
f = (Patm/Psat) * (x /(0.62198 + x ) )Patm =Psat =
Patm =
Psat =
h=tdb+x⋅(2501+1. 805⋅tdb)h−tdb=x⋅(2501+1 . 805⋅tdb )
x=h−tdb2501+1 . 805⋅tdb
rev. cjc. 28.02.2018
57.- Relative humidity f = f(tdb, ent, H)
Sicro_Relative_Humidity_tdb_ent_H(tdb, ent, H)
Microsoft Editor de ecuaciones 3.0
Data from [4]
4.- Air and water properties at atmospheric pressure
Saturated water properties 0 ºC <= t <= 100 ºC
Input data: Temperature t = 12 ºC
Function ResultsSaturatedWaterConductivity_t k = #VALUE! W/(m*K)SaturatedWaterSpecificHeat_t Cp = #VALUE! kJ/(kg*K)SaturatedWaterPrandtl_t Pr = #VALUE! -SaturatedWaterDensity_t #VALUE!SaturatedWaterAbsoluteViscosity_t m = #VALUE! Pa*sSaturatedWaterKinematicViscosity_t n = #VALUE!SaturatedWaterThermalDiffusivity_t a = #VALUE!
r = kg/m3
m2/sm2/s
rev. cjc. 28.02.2018
4.- Air and water properties at atmospheric pressureRev. cjc. 13.07.2013
Dry air properties -73.15b ºC <= t <= 726.85 ºC
Input data: Temperature t = 34 ºC
Function ResultsAirConductivity_t k = #VALUE! W/(m*K)AirSpecificHeat_t Cp = #VALUE! kJ/(kg*K)AirPrandtl_t Pr = #VALUE! -AirDensity_t #VALUE!AirAbsoluteViscosity m = #VALUE! Pa*sAirKinematicViscosity_t n = #VALUE!AirThermalDiffusivity_t a = #VALUE!
Return to index
r = kg/m3
m2/sm2/s
[1]
Return to index
Return to index
Return to index
Atmospheric pressure as function of height above sea level H [m]
1.01325 * (1 - 0.0000225577 * H) ^ 5.25588 (3)
Return to index
patm [bar] =
Absolute humidityx = 0.62198 * Pvap / Pda (8)
r = (1/v) * (1 + x) (11)
ρ=1v⋅(1+x ) (11 )
Pvap=x⋅Patm
0. 62198+x [1 ] , Eq .(8a )
[2]
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 . 9
x=(2501−2. 381⋅twb )⋅xsat twb−(tdb−twb )
2501+1 . 805⋅tdb−4 . 186⋅twbEq (33)
Dew point temperature Eq. (35), [1, Ed. 1985], page 6.9tdp = -35.97 - 1.8726 * Ln(Pvap) + 1.1689 * (Ln((Pvap) )^2
Pvap = 2065 Pa tdp = 17.84 °C
Eq. (39)
tdp = C14 + C15 * a + C16 * a^2 + C17 * a^3 + C18 * (pvap)^0.1984Pvap = 2.065 kPa
ln(pvap) = 0.72513a = 0.72513
C14 = 6.54C15 = 14.526C16 = 0.7389C17 = 0.09486C18 = 0.4569
tdp = 18.03 ºC
Constants for (6)from [2] ca = -5800.22006cb = -5.516256cc = -0.048640239cd = 0.0000417648ce = -0.0000000144521cf = 6.5459673
v = Rair*T/Patm *(1+1.6078*x) (28)
Microsoft Editor de ecuaciones 3.0
v=Rair⋅TPatm
⋅(1+1. 6078⋅x ) (28 )
Ashrae−Fundamentals 1985 , Eq . (33 ) , page 6 . 9
x=(2501−2. 381⋅twb )⋅xsat twb−(tdb−twb )
2501+1 . 805⋅tdb−4 . 186⋅twbEq (33)
tdb=ent- 2501⋅x1 . 006+1 . 86⋅x
[1 ] , (32a )
[1] Ashrae Handbook 2005 Fundamentals
[2] Ashrae Handbook 1985 Fundamentals
[3] A quick derivation relating altitude to air pressureVersion 1.03, 12/22/20042004 Portland State Aerospace Society <http://www.psas.pdx.edu>Redistribution allowed under the terms of the GNU General Public License version 2 or later.http://psas.pdx.edu/RocketScience/PressureAltitude_Derived.pdf
[4] Fundamentals of heat and mass transferFrank P. Incropera and David P. De WittSecond editionJohn Wiley & sons1985
Psychrometric functions_Deductions.xlsrev. cjc. 30.03.2016cjcruz[at]piping-tools-net
www.piping-tools.net
Redistribution allowed under the terms of the GNU General Public License version 2 or later.
Return to index