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Evaporation of Water Is a First Order Phase Transition

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Evaporation of Water Is a First Order Phase Transition A Simplified Overview of Relative Humidity One can readily measure temperature (T) and relative humidity (RH) using standard sensors. What is often more useful to know, however, is some measure of the absolute humidity. One can take T and RH and use a psychrometric chart to find said absolute humidity, but this is not an easy process to automate. We seek to find a procedure that will allow us to calculate absolute humidity from measured air temperature and relative humidity. Standard practice is to define the absolute humidity (denoted by W on most psychrometric 1 charts) as the ratio of the density of the water vapor (ρvapor) to the density of dry air (ρair) : W = ρvapor/ρair (1) It is also standard practice to define relative humidity as the ratio of the actual water vapor density to the maximum density at the actual temperature: RH = ρvapor/ρvapor,max (2) To get useful formulae, we need to explore a bit of physics. Our first task is to find the maximum density of water vapor at a given temperature. This will be the density when the vapor is at equilibrium with the liquid phase (we confine ourselves here to the temperature range between freezing and boiling points of water). The evaporation of water is a first order phase transition. Hence, the Clapeyron equation holds: dP()T ∆s l == (3) dT ∆v T∆v where: P is the equilibrium gas pressure (called vapor pressure) T is the temperature at equilibrium ∆s is the change in the entropy per unit mass going from liquid to gas l = T∆s is the latent heat of vaporization per unit mass ∆v = vgas – vliquid is the change in volume per unit mass going from liquid to gas The gas phase is much less dense than the liquid phase and so we can approximate ∆v by vgas. The gas phase (approximately) obeys the ideal gas law: PV = nRT = mRT / A (4) where: V is the volume of the gas n is the number of moles of gas R is the gas constant [8.314 J/K·mole] m is the mass of gas A is the mass per mole of gas [18.015 g/mole for water, 28.964 g/mole for dry air] 1 W is often given in units of grains of water per pound of dry air. There are 7,000 grains per pound. Hence, ∆≈v vgas =RT / AP We then get: dP lA dT L dT == PRT22RT where: L is the latent heat of vaporization per mole This can be directly integrated (assuming L to be essentially independent of pressure and temperature) to get: LL τ ln(PP) =+ln( 0 ) −=a− RTR0 T T where T0 is some convenient reference temperature P0 is the vapor pressure at T0 a = ln(P0) + L/RT0 and has units of log of pressure τ = L/R and has units of temperature The following table2 gives vapor pressure vs. temperature3 data: T [F] Ps [PSI] ∆s [Btu/lb·R] 32 0.08859 2.1870 35 0.09992 2.1704 40 0.12166 2.1430 45 0.14748 2.1162 50 0.17803 2.0899 60 0.25630 2.0388 70 0.36320 1.9896 80 0.50730 1.9423 90 0.69880 1.8966 A fit to these data4 yields a = 17.162 ± 0.03 [for P in PSI] and τ = 9628 R ± 14 R with a correlation coefficient of r = 0.99999. We can use τ to calculate l = 1,061.4 Btu/lb ±1.5 Btu/lb (2,469 J/g ± 4 J/g). This compares quite favorably with the average l (calculated as T∆s from the table above) of 1,062.0 Btu/lb. We now have an expression that accurately gives the pressure of saturated water vapor at equilibrium with liquid water at a given temperature: a-τ/T Psat(T) = e (5) 2 a portion of Table 12.29 Properties of Saturated Steam, Wiley Engineer’s Desk Reference, Second Edition 3 0°F = 459.67°R, 1 Btu = 1,054.8 J and 1 kg = 2.205 lb 4 ln(P) vs. 1/T using Excel’s LINEST function It is useful to introduce the concept of the dew point. If one takes a mixture of air and water vapor and cools it isobarically (at constant pressure), the gas density will increase5 and will eventually reach a point where the partial pressure of the water vapor is equal to the equilibrium pressure of water vapor (Psat). Further cooling will cause water vapor to condense into liquid (i.e., dew). This temperature is called the dew point (TD) and it is a direct measure of the absolute humidity. We can find an expression for the dew point, TD, given the actual temperature, T, and relative humidity, RH. First, we note that the relative humidity can be written as nA/V RHn==ρρ//max =nmax nAmax /V where n = the number of moles of water vapor in a given volume, V and nmax = the maximum number of moles in the volume if the air were saturated We can easily see from the ideal gas law that: nPmax = sat ()TV/RT We must be a bit more careful with n. It is important to note that the volume containing a given number of gas mixture molecules will shrink by a factor of TD/T as we cool it from T to TD (ideal gas law again). We can now see that: ⎡⎤⎛⎞TD nP==sat ()TD ⎢⎥⎜⎟V/RTD Psat (TD )V/RT ⎣⎦⎝⎠T Therefore, τ 11− ( TTD ) RH==PTsat ()D /PTsat ()e or: 11ln(RH ) =− (6) TTD τ We can finally find W, the absolute humidity: ⎛⎞A ⎛⎞PT( ) WA==ρρ//PAP=water sat D (7) vapor air water water air air ⎜⎟⎜⎟ ⎝⎠APair ⎝⎠total − Psat ()TD 5 The ideal gas law can be used to show that the product of the density times the absolute temperature is a constant for isobaric processes. Summary Here are some practical expressions for use in code: [T and TD in Fahrenheit, RH in percent, Psat in atmospheres and W in grains water per pound dry air] 1 TT( , RH) =−459.67 (8) D 1 ln(RH /100) − T + 459.67 9628 9.869−+9628/(T 459.67) Psat (,T RH )= RHe (9) 4354 WT(,RH)= (10) eR9628/(T +−459.67) 9.869 /1H− .
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