Posts Tagged ‘formula’

The absolute humidity formula posted in 2012 on this blog has a range of -30°C to 35°C. To expand this range I have developed a new formula to compute absolute humidity from relative humidity and temperature based on a simple but little known polynomial expression (Richards, 1971) for the saturation vapor pressure of water, valid to ±0.1% over the temperature range -50°C to 140°C.

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 Euler number 2.71828 [raised to the power of the contents of the square brackets]:

Absolute Humidity = 1013.25 × e^[13.3185t – 1.9760t^2 – 0.6445t^3 – 0.1299t^4] × rh × 18.01528
(grams/m^3)                                                   100 × 0.083145 × (273.15 + T)

where the parameter t = 1 – 373.15/(273.15 + T)

The above formula simplifies to

Absolute Humidity = 1013.25 × e^[13.3185t – 1.9760t^2 – 0.6445t^3 – 0.1299t^4] × rh × 2.1667
(grams/m^3)                                                                   273.15 + T

To cite this formula please quote: P Mander (2020), carnotcycle.wordpress.com/2020/08/01/compute-absolute-humidity-from-relative-humidity-and-temperature-50c-to-140c

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Study notes

Strategy for computing absolute humidity, defined as water vapor density in grams/m3, from temperature (T) and relative humidity (rh):

1. Water vapor is a gas whose behavior in air approximates that of an ideal gas due to its very low partial pressure.

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 polynomial expression of Richards (ref) which generates saturation vapor pressure Psat (hectopascals) as a function of temperature T (Celsius) in terms of a parameter t

Psat = 1013.25 × e^[13.3185t – 1.9760t2 – 0.6445t3 – 0.1299t4]
where t = 1 – 373.15/(273.15 + T)

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

P = 1013.25 × e^[13.3185t – 1.9760t2 – 0.6445t3 – 0.1299t4] × 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 the molecular weight of water to give the answer in grams.

Absolute humidity (grams/m3) = Psat  ×  rh  ×  mol wt
                                                          100 × R × (273.15 + T)

Saturation vapor pressure Psat is expressed in hectopascals hPa
Relative humidity rh is expressed in %
Molecular weight of water mol wt = 18.01528 g mol-1
Gas constant R = 0.083145 m3 hPa K-1 mol-1
Temperature T is expressed in degrees Celsius

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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Formula jpgs

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P Mander, July 2020

Ventus W636 Weather Station with outdoor sensor

If your weather station displays barometric pressure, temperature and relative humidity like the one pictured above, you can calculate the amount of water vapor in the air expressed either as grams of water vapor per kilogram of dry air (known as Mixing Ratio) or as grams of water vapor per kilogram of vapor-containing air (known as Specific Humidity). The two measures are very similar for cooler air; differences only become apparent for warmer air.

Formulas for calculating Mixing Ratio and Specific Humidity

In the formulas below, barometric pressure P is expressed in hectopascals (hPa), temperature T is expressed in degrees Celsius, relative humidity rh is expressed in %, and e is the Euler number 2.71828 [raised to the power of the contents of the square brackets]:


The decimal separator is shown as a full point (.) In developing these formulas, the following textbook was consulted: Atmospheric Thermodynamics by Grant W. Petty, Sundog Publishing, Madison Wisconsin. ISBN-10: 0-9729033-2-1

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P Mander, June 2020

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 dew point temperature TD

There has been a fair amount of interest in my formula 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 which computes dew point temperature TD from measured RH and T.

Formula for computing dew point temperature TD

In this formula (P Mander 2017) the measured temperature T and the computed dew point temperature TD are expressed in degrees Celsius, and the measured relative humidity RH is expressed in %

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Strategy for computing TD from RH and T

1. The dew point temperature TD is defined in the following relation where RH is expressed in %


2. To obtain values for Psat, we can use the Bolton formula[REF, eq.10] which generates saturation vapor pressure Psat (hectopascals) as a function of temperature T (Celsius)

These formulas are stated to be accurate to within 0.1% over the temperature range –30°C to +35°C

3. Substituting in the first equation yields

Taking natural logarithms

Rearranging

Separating TD terms on one side yields

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P Mander August 2017

Radon gas levels in indoor spaces are known to fluctuate considerably, so continuous monitoring is necessary to compute long-term averages. This particular radon detector, which uses continuous air sampling coupled to algorithm-based alpha spectrometry, is designed to do this job and has gained good reviews on Amazon. It is made in Norway by Corentium AS.

My unit has been in continuous operation since October 2015. Although the short-term average figure goes up and down from day to day, and to a lesser extent from week to week (the display shows alternating 1-day and 7-day figures), the long-term average figure is really quite steady.

I was thinking about this the other day, and it occurred to me that since the long-term figure varies so little in a month’s turning, I could use it to estimate the entry rate of radon gas into the enclosed space where the device is located.

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Formula for computing entry rate

1. Units in becquerels per cubic meter (Bq/m3)

If the device is showing a steady long-term average figure (n) and the enclosed space has a volume of v cubic meters, the entry rate of radon gas is computed as follows:
Entry rate = 8.78nv attomoles per month
For example, if n = 79 and v = 5
Entry rate = 8.78 × 79 × 5 = 3468 attomoles per month
(1 attomole = 10-18 moles)

2. Units in picocuries per litre (pCi/L)

If the device is showing a steady long-term average figure (n) and the enclosed space has a volume of v cubic meters, the entry rate of radon gas is computed as follows:
Entry rate = 324.74nv attomoles per month
For example, if n = 4.64 and v = 5
Entry rate = 324.74 × 4.64 × 5 = 7534 attomoles per month
(1 attomole = 10-18 moles)

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Explanatory notes
First I should clarify what I mean by entry rate. Radon is a gas that seeps into enclosed spaces through conduits, joints and cracks; it is also exhaled by diffusion through surfaces. Having infiltrated the space, some of the radon will escape, either through back-diffusion or infiltration into adjacent spaces. Without knowing the rates of ingress and escape, one can conclude that a steady long-term average figure on the detector, which implies a steady concentration of radon in the enclosed space, indicates equilibrium between the rate of radon ingress on the one hand, and the rate of radon escape and decay on the other. In other words at equilibrium

Defining entry rate as the difference between ingress rate and escape rate, we have

Given the premise that the concentration of radon gas in the enclosed space is steady, the decay rate can be taken as constant since it is determined solely by the concentration – i.e. the number of radon atoms present in a given volume. So a steady long-term average on the detector means that the entry rate, as here defined, is also constant.

Radon is a very dense gas, almost 8 times as dense as air, and this tempts many to think that radon accumulates at the bottom of an enclosed space. This is not what happens. Like any gas, radon exhibits the phenomenon of diffusion – which is the tendency of a substance to spread uniformly throughout the space available to it. What the density of a gas does affect is the rate at which the gas diffuses. But given sufficient time to reach a state of equilibrium, it can be assumed that the concentration of radon gas will be uniform throughout the enclosed space.

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Assigning a unit of time
So far I have said nothing concerning the unit of time to be applied in relation to the foregoing rate equation. Now we can address this issue, which constitutes the novelty of the computation scheme.

— Let the unit of time by which rate is measured be set equal to the half-life of the isotope (Rn-222) of which radon gas is largely composed.

Theorem
Let the entry rate of radon gas into a previously radon-free bounded space be x atoms per unit of time corresponding to the half-life of Rn-222. At the end of the first half-life period, x/2 atoms will have decayed (via α emission) while x/2 atoms remain. At the end of the second half-life period, the first x atoms will have decayed to x/4 while the second x atoms will have decayed to x/2. At the end of the third half-life period, the first x atoms will have decayed to x/8 and the second x atoms to x/4, while the third x atoms will have decayed to x/2 … and so on according to the following scheme

This process forms an absolutely convergent geometric series in which the number of radon atoms remaining in the space after n half-life periods will be

The conclusion is reached that if the entry rate of radon gas into a previously radon-free bounded space is x radon atoms in the unit of time corresponding to the half-life of Rn-222, the number of radon atoms in this space will over successive half-lives approach a steady-state value of x.

Assuming diffusion throughout the space, a steady state value of x should be realized in little more than a month since when n = 8 (equivalent to 30.6 days) the series sum is 99.6% of x.

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Computing entry rate
Given a steady long-term average figure on the detector, which implies a steady concentration of radon gas throughout the bounded space, the number of radon atoms in this space can be estimated as follows.

For decay rates measured in becquerels per cubic meter (Bq/m3)
Let the long-term average figure on the detector, measured in decays per second per cubic meter be n, and let the bounded space be v cubic meters.
In the half-life of Rn-222 (3.8235 days) the number of decays in volume v will be n × v × 330,350
This equals x/2 where x is the steady state population of radon atoms in volume v
Therefore x = n × v × 660,701 radon atoms
By the theorem, x is also the number of radon atoms entering the bounded space in a unit of time equal to the half-life of the isotope (Rn-222) of which radon gas is largely composed – 3.8235 days. The magnitude 8x is therefore the number of radon atoms entering the space in a month (8 x 3.8235 = 30.6). Dividing 8x by the Avogadro number converts the number of radon atoms into moles of radon gas:
Entry rate (moles of radon gas per month) = 8 × n × v × 660,701 / (6.022 x 10^23)

The entry rate of radon gas ≅ 8.78nv attomoles per month
(1 attomole = 10-18 moles)

For decay rates measured in picocuries per litre (pCi/L)
Let the long-term average figure on the detector, measured in picocuries per litre be n, and let the bounded space be v cubic meters.

The entry rate of radon gas ≅ 324.74nv attomoles per month
(1 attomole = 10-18 moles)

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Disclaimer
Please note that the theorem on which the above calculations are based is untested. Until the theorem has been tested and the accuracy of results obtained with it has been determined, the calculation of entry rate as herein defined can only be regarded as a theoretical prediction and should be viewed accordingly.

P Mander May 2017

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afr01

If the man who almost single-handedly invented chemical thermodynamics – the American mathematical physicist Josiah Willard Gibbs – had owned an automobile, he would have had no trouble figuring out the action of antifreeze.

“The problem reduces to consideration of a binary solution in equilibrium with solid solvent,” I can hear old Josiah saying. “Such a thermodynamic system has two degrees of freedom, so at constant pressure there must be a relation between temperature and composition.”

And indeed there is. The relation corresponds to the observed depression of the freezing point of a solvent by a solute. What’s more, its exact form confirms how antifreeze really works.

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Computing chemical potential

We have Josiah Willard Gibbs to thank for introducing the concept of chemical potential (μ) as a sort of generalized force driving the flow of chemical components between coexistent phases.

When the phases are in equilibrium at constant temperature and pressure, the chemical potential of any component has the same value in each phase

The key point to note here is that μi is the chemical potential of component i in an arbitrary state, i.e. in a mixture of components. In order to compute this potential we need to know two things: the chemical potential of the pure substance μi0 at a pressure p (such as that of the atmosphere), and the mole fraction (xi) of the component in the mixture. Assuming an ideal solution, use can then be made of the textbook formula

dce12  …(1)

With pressure and temperature fixed, this equation has a single variable (xi), from which we can draw the conclusion that the variation in chemical potential of a component in an ideal solution is determined solely by its own mole fraction.

The significance of this fact can be appreciated by considering the following diagrams

Here is water in equilibrium with ice at 273K. The chemical potentials of the solid and liquid phases are equal; there is no net driving force in either direction. Now consider the effect of adding an antifreeze agent to the liquid phase

afr04

Assuming the temperature held constant at 273K, the addition of antifreeze reduces the mole fraction of water, lowering its chemical potential in accordance with equation 1. The coexistent solid phase now has a higher potential, providing the driving force to transform ice into water. Since the temperature is held constant, this equates to the lowering of the freezing point of water in the mixture.

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Deducing a formula for freezing-point depression

To obtain a formula for the freezing point of water in a solution containing antifreeze, we start with the equilibrium relation

where the zero superscript indicates a standard potential, i.e. that the solid phase consists of pure ice whose mole fraction x is unity. Substituting the left hand side with

we obtain

which after differentiation with respect to temperature at constant pressure and subsequent integration yields the formula for the freezing point of water in a solution containing antifreeze at 1 atmosphere pressure:

The terms on the right are the molar enthalpy of fusion of water (ΔHf0), the freezing point of pure water (Tf0), the gas constant R and the mole fraction of water (xH2O) in the solution containing antifreeze.

The latter is the only variable, confirming that the freezing point of water in a solution containing antifreeze is determined solely by the mole fraction of water in the mixture – in other words the extent to which the water is diluted by the antifreeze agent.

This is how antifreeze works. There is nothing active about its action. It exerts its effect passively by being miscible and thereby reducing the mole fraction of water in the liquid mixture. There’s really nothing more to it than that.

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Using the formula

Values for constants

Enthalpy of fusion of water ΔHf0 = 6.02 kJmol-1
Freezing point of pure water Tf0 = 273.15 K
Gas constant R = 0.008314 kJmol-1K-1

Example

651 grams of the antifreeze agent ethylene glycol (molecular weight 62.07) are added to 1.5 kg of water (molecular weight 18.02). What is the freezing point of water in this solution?

Strategy

1. Calculate the mole fraction of water in the solution

Number of moles of water = 1500/18.02 = 83.2
Number of moles of ethylene glycol = 651/62.07 = 10.5
Mole fraction of water = 83.2/(83.2 + 10.5) = 0.89

2. Calculate the freezing point of water in the solution

The solution will give antifreeze protection down to 261.65K or –11.5°C

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P Mander March 2015