Posts Tagged ‘vapor pressure’

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The Clapeyron equation

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This equation is exact and applies to equilibrium processes between any two phases, i.e melting, boiling and sublimation. It gives the slope of the coexistence curve at (saturated) vapor pressure P and temperature T.

The original derivation by Clapeyron is a geometrical method in which an expression for the PV work done in a Carnot cycle is equated with Carnot’s principle δW = δQ (δT)/T. This classic proof is given in Appendix I.

Derivation

If two phases A and B of the same pure substance are in equilibrium with each other, and GA and GB are the respective molar Gibbs free energies, then the condition of equilibrium is GA = GB (if this were not the case, substance in the phase with higher molar free energy would pass into the phase with lower molar free energy, which is a non-equilibrium condition).

At equilibrium dG = 0, which is also the condition for the thermodynamically reversible change dGA = dGB. Since dG = VdP – SdT, we can equate the differential expressions for the two phases:

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Since we are dealing with a thermodynamically reversible process at constant pressure and temperature, ΔS =(Qp)/T which we can write ΔS = ΔH/T where ΔH is the molar enthalpy of phase transition. Hence

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Applications of the Clapeyron equation

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1. Calculate the enthalpy of phase transition ΔH
If both dP/dT and ΔV at temperature T are known, ΔH can be calculated. Scroll down to Worked example 1

2. Estimate the effect of pressure change on the boiling point temperature
If both ΔV and ΔH are known at temperature T, dP/dT can be calculated to estimate the effect of pressure change on the boiling point temperature. Scroll down to Worked example 2

3. Estimate the effect of pressure change on the melting point temperature
If both ΔV and ΔH are known at temperature T, dP/dT can be calculated to estimate the effect of pressure change on the melting point temperature. Scroll down to Worked example 3

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Worked Example 1

The vapor pressure of water changes by 27.17 mm Hg from 372.15K to 373.15K, and the corresponding volume change (water>vapor) is 3.0114 x 10-2 m3mol-1. Estimate the enthalpy of vaporization ΔHvap of water at 373.15K.

Strategy

Use the Clapeyron equation to solve for ΔHvap

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The SI units are:
P (vapor) = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
T = K
V = m3mol-1
H = Jmol-1

Calculation

1. At 373.15 K, the vapor pressure of water changes by 27.17 mm Hg per kelvin. This may be taken as the value of dP/dT at 373.15K.

2. Convert dP/dT from mm Hg per kelvin to Nm-2K-1

dP/dT = 27.17 x (105/760) = 3575 Nm-2K-1

3. Compute ΔH = (dP/dT) . TΔV

ΔH = 3575 x 373.15 x 3.0114 x 10-2 = 40172 Jmol-1

The enthalpy of vaporization of water ΔHvap at 373.15 is estimated to be 40.17 kJmol-1.

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Worked Example 2

The enthalpy of vaporization ΔHvap of water at 373.15K is 40,657 Jmol-1 and the corresponding volume change (water>vapor) is 3.0114 x 10-2 m3mol-1. Estimate the boiling point temperature at 770 mm Hg.

Strategy

Use the inverted Clapeyron equation to solve for dT/dP

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The SI units are:
T = K
P = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
ΔV = m3mol-1
ΔH = Jmol-1

Note that P is the vapor pressure. Since T is the boiling point temperature, the vapor pressure and the externally applied pressure will be the same.

Calculation

1. Compute dT/dP

dT/dP = (373.15 x 3.0114 x 10-2)/40657 = 2.76 x 10-4 K(Nm-2)-1

2. Compute the pressure increase in Nm-2

Water boils at 373.15K at a pressure of 760 mm Hg. The pressure increase under consideration is therefore 770-760 = 10 mm Hg.

10 mm Hg = 10 x (105/760) Nm-2 = 1316 Nm-2

3. Put the two facts together

dT/dP tells us that the boiling point is raised by 2.76 x 10-4 K for 1 Nm-2 increase in vapor pressure.

So a 1316 Nm-2 increase in vapor pressure will raise the boiling point temperature by 1316 x 2.76 x 10-4 = 0.36K

The boiling point of water at 770 mm Hg is estimated at 373.15 + 0.36 = 373.51K

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Worked Example 3

At 273.15K the molar enthalpy of fusion of water ΔHfusion is 6 x 103 Jmol-1 and the corresponding volume change (ice>water) is –1.6 x 10-6 m3mol-1. Estimate the melting point of ice at a pressure of 150 atmospheres.

Strategy

Use the inverted Clapeyron equation to solve for dT/dP

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The SI units are:
T = K
P = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
ΔV = m3mol-1
ΔH = Jmol-1

Note that P is the vapor pressure. Since T is the melting point temperature, the vapor pressure and the externally applied pressure will be the same.

Calculation

1. Compute dT/dP

dT/dP = (273.15 x –1.6 x 10-6)/(6 x 103) = –7.28 x 10-8 K(Nm-2)-1

2. Compute the pressure increase in Nm-2

Ice melts at 273.15K at a pressure of 1 atmosphere. The pressure increase under consideration is therefore 150-1 = 149 atmospheres.

149 atmospheres = 1.49×107 Nm-2

3. Put the two facts together

dT/dP tells us that the melting point is lowered by –7.28 x 10-8 K for 1 Nm-2 increase in vapor pressure.

So a 1.49×107 Nm-2 increase in vapor pressure will lower the melting point by

(1.49 x 107) x (–7.28 x 10-8) = –1.08K

The melting point of ice at a pressure of 150 atmospheres is estimated at 273.15 – 1.08 = 272.07K

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The Clausius-Clapeyron equation

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The Clausius-Clapeyron equation is not exact, since approximations are used in its derivation. Its use is confined to processes involving vapor phase equilibrium, but nevertheless the equation is very useful.

Derivation

Beginning with the Clapeyron equation

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use is made of the fact that for sublimation and vaporization processes, there is generally a large difference in molar volume between the two phases. In this case, ΔV can be approximated as the molar volume of the vapor Vvap.

It is then assumed that the behavior of the vapor approximates that of an ideal gas, so that the molar volume Vvap can in turn be replaced with RT/P where P is the vapor pressure, giving

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This is the Clausius-Clapeyron equation, which approximates the slope of the liquid-vapor coexistence curve at vapor pressure P and boiling point temperature T. Using the mathematical identity

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the Clausius-Clapeyron equation can also be written in the form

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If we further assume that ΔHvap is independent of temperature, integration of the above equation can be performed

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This is the integrated form of the Clausius-Clapeyron equation. If the vapor pressure P1 is known at boiling point temperature T1, this equation can be used to estimate the boiling point temperature T2 at another pressure P2. Alternatively, if the boiling point temperatures are known at vapor pressures P1 and P2, the enthalpy of vaporization ΔHvap can be estimated.

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Applications of the Clausius-Clapeyron equation

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or

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and its integrated form

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1. Estimate the enthalpy of vaporization ΔHvap
If both dP/dT and the vapor pressure P are known at boiling point temperature T, the enthalpy of vaporization ΔHvap can be estimated. The value of R needs to be known. Scroll down to Worked example 4

2. Estimate the effect of pressure change on the boiling point temperature
If the vapor pressure P1 is known at boiling point temperature T1, the boiling point temperature T2 at another pressure P2 can be estimated. The values of R and ΔHvap need to be known. Scroll down to Worked example 5

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Worked Example 4

dP/dT is 27.17 mm Hg per kelvin for water at 373.15K. Estimate the enthalpy of vaporization ΔHvap of water at this temperature. The vapor pressure of water at 373.15K is 105 Nm-2, and the gas constant R is 8.3145 JK-1mol-1.

Strategy

Use the Clausius-Clapeyron equation to solve for ΔHvap

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The SI units are:
T = K
P (vapor) = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
R = JK-1mol-1
ΔH = Jmol-1

Calculation

1. Convert dP/dT from mm Hg per kelvin to Nm-2K-1

dP/dT = 27.17 x (105/760) = 3575 Nm-2K-1

2. Compute ΔHvap = 1/P x dP/dT x RT2

ΔHvap = 10-5 x 3575 x 8.3145 x (373.15)2 = 41388 Jmol-1

The enthalpy of vaporization of water ΔHvap at 373.15K is estimated to be 41.39 kJmol-1 (which is about 2% higher than the experimental value of 40.66 kJmol-1).

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Worked Example 5

Water boils at 373.15K at a pressure of 1 atmosphere. At what temperature will water boil in a pressure cooker operating at a pressure of 2 atmospheres? The enthalpy of vaporization ΔHvap of water at 373.15K is 40657 Jmol-1 and the gas constant R is 8.3145 JK-1mol-1.

Strategy

Use the integrated Clausius-Clapeyron equation to solve for T2

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The SI units are:
T = K
R = JK-1mol-1
ΔH = Jmol-1

The units of P1 and P2 are immaterial, so long as they are the same.

Note that P1 and P2 are vapor pressures. Since T1 and T2 refer to boiling point temperatures, the vapor pressures P1 and P2 will be the same as the externally applied pressures.

Calculation

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Water in a pressure cooker operating at a pressure of two atmospheres boils at 394K, or 121°C. This explains why vegetables cook so fast in these devices.

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Appendix 1

The Clapeyron equation – Clapeyron’s proof

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Émile Clapeyron was the first to use the PV diagram as an analytical tool

Émile Clapeyron’s 1834 paper Mémoire sur la Puissance Motrice de la Chaleur (Memoire on the Motive Power of Heat) contains the first analytical representation of the Carnot cycle, depicted above as the quadrilateral ABCD. By assigning infinitely small values to the variations of volume V and vapor pressure P during the four successive operations of the cycle, Clapeyron renders the quadrilateral a parallelogram.

He then demonstrates that the parallelogram ABCD, representing the work done in a complete cycle, is equal in area to BCEF since both stand on the same base BC and lie between the same parallels T and T–δT.

The base-altitude product BF.PQ therefore equates to the work done in a complete cycle.

Now BF=δP is the increase in vapor pressure per δT rise of temperature at constant volume, while PQ=δV equates to δQ/Lv where δQ is the heat absorbed and Lv is the latent heat of expansion per unit increase in volume

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Since by Carnot’s principle the work δW done in a complete cycle is δQ (δT)/T

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This is the famous Clapeyron equation.

In modern notation Lv is ΔH/ΔV and since vapor pressure is independent of volume the subscript can be omitted and the equation written

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Our pressure cooker was a Prestige Skyline, very popular in the 1950s. Photo credit wisekitchen.wordpress.com

Our pressure cooker was a Prestige Skyline from the 1950s. The regulator is sticking up at the back.

A familiar object in the kitchen of my youth was our pressure cooker. It cooked the vegetables in a fraction of the time, saving significant amounts of energy and therefore cost.

Our pressure cooker was a stainless steel device equipped with a regulator that maintained an internal pressure of 2 atmospheres. I was thinking back to it the other day, and started wondering about the temperature of the superheated water inside the container. This can be calculated using the Clausius-Clapeyron equation

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which approximates the slope of the liquid-vapor coexistence curve at vapor pressure P and boiling point temperature T. Using the mathematical identity

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the Clausius-Clapeyron equation can also be written in the form

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If we further assume that ΔHvap is independent of temperature, integration of the above equation can be performed

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If the vapor pressure P1 is known at boiling point temperature T1, this equation can be used to estimate the boiling point temperature T2 at another pressure P2.

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Worked Example

Water boils at 373.15K at a pressure of 1 atmosphere. At what temperature will water boil in a pressure cooker operating at a pressure of 2 atmospheres? The enthalpy of vaporization ΔHvap of water at 373.15K is 40657 Jmol-1 and the gas constant R is 8.3145 JK-1mol-1.

Strategy

Use the integrated Clausius-Clapeyron equation to solve for T2

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The SI units are:
T = K
R = JK-1mol-1
ΔH = Jmol-1

The units of P1 and P2 are immaterial, so long as they are the same.

Note that P1 and P2 are vapor pressures. Since T1 and T2 refer to boiling point temperatures, the vapor pressures P1 and P2 will be the same as the externally applied pressures.

Calculation

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Water in a pressure cooker operating at a pressure of two atmospheres boils at 394K, or 121°C. This explains why vegetables cook so fast in these devices.

 

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I have a digital weather station with a wireless outdoor sensor. In the photo, the top right quadrant of the display shows temperature and relative humidity for outdoors (6.2°C/94%) and indoors (21.6°C/55%).

I find this indoor-outdoor thing fascinating for some reason and revel in looking at the numbers. But when I do, I always end up asking myself if the air outside has more or less water vapor in it than the air inside. Simple question, which is more than can be said for the answer. Using the ideal gas law, the calculation of absolute humidity from temperature and relative humidity requires an added algorithm that generates saturated vapor pressure as a function of temperature, which complicates things a bit.

Formula for calculating absolute humidity

In the formula below, temperature (T) is expressed in degrees Celsius, relative humidity (rh) is expressed in %, and e is the base of natural logarithms 2.71828 [raised to the power of the contents of the square brackets]:

Absolute Humidity (grams/m3) = 6.112 x e^[(17.67 x T)/(T+243.5)] x rh x 18.02
                                                                            (273.15+T) x 100 x 0.08314

which simplifies to

Absolute Humidity (grams/m3) = 6.112 x e^[(17.67 x T)/(T+243.5)] x rh x 2.1674
                                                                                        (273.15+T)

This formula is accurate to within 0.1% over the temperature range –30°C to +35°C

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Additional notes for students

Strategy for computing absolute humidity, defined as density in g/m^3 of water vapor, from temperature (T) and relative humidity (rh):

1. Water vapor is a gas whose behavior approximates that of an ideal gas at normally encountered atmospheric temperatures.

2. We can apply the ideal gas equation PV = nRT. The gas constant R and the variables T and V are known in this case (T is measured, V = 1 m3), but we need to calculate P before we can solve for n.

3. To obtain a value for P, we can use the following variant[REF, eq.10] of the Magnus-Tetens formula which generates saturated vapor pressure Psat (hectopascals) as a function of temperature T (Celsius):

Psat = 6.112 x e^[(17.67 x T)/(T+243.5)]

4. Psat is the pressure when the relative humidity is 100%. To compute the pressure P for any value of relative humidity expressed in %, we multiply the expression for Psat by the factor (rh/100):

P = 6.112 x e^[(17.67 x T)/(T+243.5)] x (rh/100)

5. We now know P, V, R, T and can solve for n, which is the amount of water vapor in moles. This value is then multiplied by 18.02 – the molecular weight of water ­– to give the answer in grams.

6. Summary:
The formula for absolute humidity is derived from the ideal gas equation. It gives a statement of n solely in terms of the variables temperature (T)  and relative humidity (rh). Pressure is computed as a function of both these variables; the volume is specified (1 m3) and the gas constant R is known.

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UPDATES

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Formula for computing dewpoint temperature TD from RH and T

Relative humidity (RH) and temperature (T) data from an RH&T sensor like the DHT22 can be used to compute not only absolute humidity AH but also dewpoint temperature TD

July 2017: There has been a lot of interest in my formula (P Mander 2012) which computes AH from measured RH and T, since it adds value to the output of RH&T sensors. To further extend this value, I have developed another formula (P Mander 2017) which computes dewpoint temperature TD from measured RH and T. In this formula the measured temperature T and the computed dewpoint temperature TD are expressed in degrees Celsius, and the measured relative humidity RH is expressed in %

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If an object whose temperature is at or below TD is present in the local space, the thermodynamic conditions are satisfied for water vapor to condense (or freeze if TD is below 0°C) on the surface of the object.

Further details, including the derivation of the formula and copy-and-paste spreadsheet formulas for computing TD will soon be available.

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Formula cited in two recent academic research papers

July 2017

Czech Republic: Brno University of Technology, Faculty of Mechanical Engineering
Thesis: The effect of climate conditions on wheel-rail contact adhesion
http://dl.uk.fme.vutbr.cz/zobraz_soubor.php?id=3392

Sweden: Linköping University, Institute for Economic and Industrial Development
Case study: Effect of seasonal ventilation on energy efficiency and indoor air quality
http://www.navic.se/images/Exjobb/rstidsanpassad_ventilation.pdf

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Formula computes real time AH with DHT22 sensor on single board computer

June 2017: Single board computers provide low-cost solutions to automation and testing. On element14.com a BeagleBone Black Wireless equipped with a DHT22 RH&T sensor has been used to monitor outdoor and indoor temperature and humidity using my formula to enable AH computations to be processed in real time.

https://www.element14.com/community/roadTestReviews/2398/l/BeagleBoard.org-BBB-Wireless-BBBWL-SC-562

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Formula features in Russian Arduino project on YouTube

April 2017: My formula makes its first live appearance on YouTube. The presentation concerns a humidity/temperature monitoring and management system installed in a cellar affected by mould problems. If you don’t speak Russian don’t worry, the images of the installation give you the gist of what this project is about.

See the YouTube video here:
https://www.youtube.com/watch?v=SO1yugxahpk

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Formula recommended for use in monitoring comfort levels for exotic pets

March 2017: A post has appeared on Reddit concerning an Arduino Uno with T&RH sensor and LCD screen, which the poster is using to improve temperature and humidity monitoring of a pet’s habitat – in this particular case a Bearded Dragon (not the one illustrated).

The post has attracted much interested discussion and comment, including a recommendation from one participant to use AH rather than RH, citing my conversion formula. The rationale for the change is so neatly expressed that I would like to quote it:

“May I recommend absolute humidity instead of relative? Relative humidity only tells you how “full” the air is of moisture, and it’s entirely dependent on temperature; the same amount of moisture will read lower relative humidity at higher temperatures, and vice versa. Whereas absolute humidity is measured in grams of water per cubic meter of air. You can implement this simple conversion formula in your code: (URL for this blogpost)
0-2 is extremely dry, 6-12 is your average indoors, and 30 is like an Amazon rainforest.”

See the Reddit post here:
https://www.reddit.com/r/arduino/comments/5ysmo5/i_noticed_my_bearded_dragons_habitat_could_use_a/

See the Arduino project here:
https://create.arduino.cc/projecthub/ThothLoki/portable-arduino-temp-humidity-sensor-with-lcd-a750f4

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Igor uses my formula to keep his cellar dry

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October 2016: I am impressed by this basement humidity control system developed by Igor and reported on Amperka.ru forum.

Inside the short pipe is a fan equipped with a 3D-printed circumferential seal. The fan replaces basement air with outdoor air, and is activated when absolute humidity in the cellar is 0.5 g/m^3 higher than in the street, subject to the condition that the temperature of the outdoor air is lower. This ensures that water in the cellar walls is drawn into the vapor phase and pumped out; the reverse process cannot occur. на русском здесь.

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Formula powers AH measurements from high-precision RH&T sensor

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The SHT75 RH&T sensor from SENSIRION

April 2016: Prof. Antonietta Frani has made a miniature device for measuring absolute humidity, using my formula to power an Arduino Uno microcontroller board equipped with an SHT75 RH&T sensor which connects to a computer via a USB cable. Systems Integrator Roberto Valgolio has developed an interface to transfer the data to Excel spreadsheets with their associated graphical display functions.

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Formula powers online RH←→AH calculator

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March 2016: German website rechneronline.de is using my formula to power an online RH/AH conversion calculator.

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Formula cited in academic research paper

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January 2016: A research article in Landscape Ecology (October 2015) exploring microclimatic patterns in urban environments across the United States has used my formula to compute absolute humidity from temperature and relative humidity data.

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Formula finds use in humidity control unit

August 2015: Open source software/hardware project Arduino is using my absolute humidity formula in a microcontroller designed to control humidity in basements:

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“The whole idea is to measure the temperature and relative humidity in the basement and on the street, on the basis of temperature and relative humidity to calculate the absolute humidity and make a decision on the inclusion of the exhaust fan in the basement. The theory for the calculation is set forth here – carnotcycle.wordpress.com/2012/08/04/how-to-convert-relative-humidity-to-absolute-humidity.” на русском здесь.

More photos on this link (text in Russian):http://arduino.ru/forum/proekty/kontrol-vlazhnosti-podvala-arduino-pro-mini

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AH computation procedure applied in calibration of NASA weather satellite

June 2015: My general procedure for computing AH from RH and T has been applied in the absolute calibration of NASA’s Cyclone Global Navigation Satellite System (CYGNSS), specifically in relation to the RH data provided by Climate Forecast System Reanalysis (CFSR). The only change to my formula is that Psat is calculated using the August-Roche-Magnus expression rather than the Bolton expression.

The CYGNSS system, comprising a network of eight satellites, is designed to improve hurricane intensity forecasts and was launched on 15 December 2016.

Reference: http://ddchen.net/publications (Technical report “An Antenna Temperature Model for CYGNSS” June 2015)

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Formula cited in draft paper on air quality monitoring

May 2015: Metal oxide (MO) sensors are used for the measurement of air pollutants including nitrogen dioxide, carbon monoxide and ozone. A draft paper concerning the Air Quality Egg (AQE) which cites my formula in relation to MO sensors can be seen on this link:

MONITORING AIR QUALITY IN THE GRAND VALLEY: ASSESSING THE USEFULNESS OF THE AIR QUALITY EGG

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Formula used by US Department of Energy in Radiological Risk Assessment

June 2014: In its report on disused uranium mines, Legacy Management at DoE used my formula for computing absolute humidity as one of the meteorological parameters involved in modeling radiological risk assessment.

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