Calculating the Enthalpy of Air

I havn't got the formula myself but this link goes over how to calculate the enthalpy and has the formula buried in it somewhere.

http://www.engineeringtoolbox.com/enthalpy-moist-air-d_683.html

Best of luck.

Linric (www.linric.com) has some software that works very well for about $40.

Enthalpy is Sensible heat plus Latent heat.

Unless you are liquifying or freezing the air the latent heat proportion will be zero and you are left with sensible heat. That is a straight line function - calories per kilo (or what ever base units you want to use)

Since air has a varying volume pressure ratio, you need the weight; then use that to do your heat calcs.

Someone else might be able to give you the, joules, calories and watts to use.

You are forgetting to take into account the moisture content. Moisture dissolved in the air makes the equation non linear and the study of it is called psychometrics.

And here is a place you can test-em.

http://www-esl.tamu.edu/about/facilities.htm

Wow! up to 50,000 CFM. They might be able to advise?

Double Wow! if one could squeeze/dehumidify the water out from an 8 ft fan like that one, at max humidity, about 80 tonnes of water a day would result. (3% of air-weight) Well cool. sorry 4 going off thread

I did not forget the moisture and I know it is called psychrometrics.

If moisture is to be included in the formula, you need to know if there is any free water (as condensed liquid) in the space, and how much. That will muck up the formulas already given.

I doubt I will ever forget the counter-intuitive fact that humid air is less dense than dry air. It might just affect the calculations a bit. Very important for balloonists though.

Its important for all aircraft as well as balloons. You would be surprised how much effect humidity has on the take off roll of a modern jet aircraft. It not only effects the wings ability to create lift but the engines ability to produce thrust.

Some time back before security stopped it I managed to get an invite to the flight deck. The flight was a 3,000Km trip in a 727 from the tropics. Since it was the wet season and the humidity was close to 100% we needed to wait till 3:00am in the morning before the air cooled down and became dense enough to get airborne. Even then the maximum altitude we could get to was only 27,000 feet.

Slightly off thread Masu, Sorry! but back in the 1970's At Ngobit Estate Naro Moru Kenya,we made an Elephant's Mud Slide (actually a feeder for a Dam) every time the rains came all the Elephants brought their youngsters down from the Aberdare Forrest to play....Weeeeeeeeeee.....splosh! Over a MILE of twisty turny mud slide! then they would come into our garden and nosh all the Roses, leaving 'Rose-Manure' gifts behind. Our neighbour (nameless) Had a Bell Jet Ranger Helicopter. He came on a visit to watch the nocturnal fun-n-games. Ngobit estate was at about 7500ft to 11000ft elevation. Next morning in the hot humid air, could he get his Jet Ranger to take off?..... No way!

The enthalpy of air depends on the temperature and moisture content. Obviously temperature can be in Celsius, or Fahrenheit (Kelvin or Rankine). Humidity can be inferred with wetbulb temperature and dewpoint, or stated as relative and absolute. However the psychrometric chart or tables is derived from measured data.

One can use curve fitting programs to develop an equation that approximates the measured values. An equation a colleague worked up years ago is below. He also did a more complicated one that was more accurate over a wider temperature range that I have used in spreadsheets.

for DB in Deg F and RH (0 to 1.0)

A = .007468 * DB^2 - .4344 * DB + 11.1769

B = .2372 * DB + .1230

Enth = A * RH + B

It gives a result that is accurate to within .7% over the range

40 F <= DB <= 80 F

0 <= RH <= 1.0

I have the following Tdb and RH value. (77.553 F, 49.136 RH)

What does the enthalpy come out to when using the equations shown here? The result is different depending on how the terms are defined.

I have the following Tdb and RH value. (77.553 F, 49.136 RH)

What does the enthalpy come out to when using the equations shown here? The result is different depending on how the terms are defined.

do you have that curve fitted equation for different temperature ranges, or a greater temperature range centered on 60 like that one? I need to calculate a range of temperatures from 0 to 95 degrees F and this one worked beautifully for the calculations where it was applicable.

h=Cpaire·TBS+RH·(Clat+Cpvap·TBS)(KJ/Kg as) (1.8)

Cpaire= 1.005(KJ/Kg°K)Calor Específico del Aire a Pres Cte

Clat=2501.6(KJ/Kg)Calor Latente de Evaporación del Agua

Cpvap=1.86(KJ/Kg°K)Calor Específico del Vapor de Agua a Pres. Cte.

TBS = Temperatura de bulbo seco (°C)

TBH = Temperatura de bulbo húmedo (C)

Thermal properties of damp air

The heat content, usually called the enthalpy, of air rises with increasing water content. This hidden heat, called latent heat by air conditioning engineers, has to be supplied or removed in order to change the relative humidity of air, even at a constant temperature. This is relevant to conservators. The transfer of heat from an air stream to a wet surface, which releases water vapour to the air stream at the same time as it cools it, is the basis for psychrometry and many other microclimatic phenomena. Control of heat transfer can be used to control the drying and wetting of materials during conservation treatment.

The enthalpy of dry air is not known. Air at zero degrees celsius is defined to have zero enthalpy. The enthalpy, in kJ/kg, at any temperature, t, between 0 and 60C is approximately:

h = 1.007t - 0.026 below zero: h = 1.005t

The enthalpy of liquid water is also defined to be zero at zero degrees celsius. To turn liquid water to vapour at the same temperature requires a very considerable amount of heat energy: 2501 kJ/kg at 0C

At temperature t the heat content of water vapour is:

h_w = 2501 + 1.84t

Notice that water vapour, once generated, also requires more heat than dry air to raise its temperature further: 1.84 kJ/kg.C against about 1 kJ/kg.C for dry air.

The enthalpy of moist air, in kJ/kg, is therefore:

h = (1.007*t - 0.026) + g*(2501 + 1.84*t)
g is the water content in kg/kg of dry air

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Here's a short BASIC program that calculates the enthalpy using dry bulb temperature and releative humidity.

10 REM ENTHALPY CALCULATION
20 REM Assumes standard atmospheric pressure (14.7 psi), see line 110
30 REM Dry-bulb temperature in degrees F (TEMP)
40 REM Relative humidity in percentage (RH)
50 REM Enthalpy in BTU/LB OF DRY AIR (H)
60 T = TEMP + 459.67
70 N = LN( T ) : REM Natural Logrithm
80 L = -10440.4 / T - 11.29465 - 0.02702235 * T + 1.289036E-005 * T ^ 2 - 2.478068E-009 * T ^ 3 + 6.545967 * N
90 S = LN-1( L ) : REM Inverse Natural Logrithm
100 P = RH / 100 * S
110 W = 0.62198 * P / ( 14.7 - P )
120 H = 0.24 * TEMP + W * ( 1061 + 0.444 * TEMP )
---

If the atmospheric pressure is known, can it be substituted in place of 14.7 on line 110? Will i get a correct calculation if i do such a substitution? yes/no? why/why not?

Of Course, there is a formula!

But you need to have some information.

the formula is:

h = Cp*t + W*hg kJ/kg dry air

where Cp = Specific heat of dry air at constant pressure = 1.0 kJ/Kg.K

(although it actually varies from 1.006 @ 0 deg C to 1.009 @ 50 deg C, it can be taken as 1.0 kJ/kg.k for paractical purpose without appreciable error)

t = temperature of air-vapour mixture, deg C

W = humidity ratio kg of water vapour/kg of dry air

hg = enthalpy of saturated steam at tempearture of air-vapour mixture, kJ/kg

Here, W can also be calculated as,

W = {(ps*V)/(Rs*T)} / {(pa*V)/(Ra*T)}

= {ps/Rs} / {(pt-ps)/Ra}

where V = Arbitrary volume of air-vapour mixture, m3

pt = Atmospheric Pressure = pa + ps, Pascals

pa = Partial pressure of dry air, Pascals

ps = Partial pressure of water vapour, Pascals

Ra = Gas constant of dry air = 287 J/kg.K

Rs = Gas constant of water air = 461.5 J/kg.K

T = Absolute temperature of air-vapour mixture, K

Substituting the values, we get:

W = (0.622*ps)/(pt - ps)

Thus one can calculate enthalpy of moist air using steam table and above formulae.

Thanks & Regards,

Tushar Shirvalkar

Mumbai - India.

One thing that has been missing from this thread is that enthalpy is also a function of pressure. Everything stated above assumes constant presrue, probably ambient.

But if you have a gas flowing through a restriction, in an adiabatic system, the gas will lose pressure across the restriction, and by energy conservation (enthalpy constant in an adiabvatic system), the temperature has to rise.

textbook definition for gas enthalpy is:

dh = CpdT + [ v - T(dv/dT)p ]dP.

conservation of energy.

The Basic program mentioned above seemed to work for me great. I converted it to work in Excel. Enter Temp dry-bulb in cell A2, Relative Humidity % in cell B2, and this Calculation in cell C2 for Enthalpy btu/lb:

=(0.24*A2)+(0.62198*(B2/100)*(EXP((-10440.4/(A2+459.67)-11.29465-0.02702235*(A2+459.67)+0.00001289036*(A2+459.67)^2-0.000000002478068*(A2+459.67)^3+6.545967*(LN(A2+459.67)))))/(14.7-(B2/100)*(EXP((-10440.4/(A2+459.67)-11.29465-0.02702235*(A2+459.67)+0.00001289036*(A2+459.67)^2-0.000000002478068*(A2+459.67)^3+6.545967*(LN(A2+459.67))))))*(1061+0.444*A2))

Cheers.

Example:

is this proved out with empirical data from the program?

Awesome formula. I copied it into excel and it works great. Can you come up with one that can calculate enthalpy from dry bulb temp and wet bulb temp?

Thanks

Enthalpy and wet bulb temperatures are parallel on the psychometric chart. You don't need the dry bulb temp, only the wet bulb temp to determine enthalpy.

Note to Ali Jamali and jt sawyer: I used the Excel version and checked it against a dozen of my previous psyc chart readings. All were right on the money except the one where I had misread the chart. Thank you both!

Dear Mr. Shadrach

One correction pl. Please note It is PSCHROMETRIC CHART, and NOT PSYCHOMETRIC - which refers to PSYCHOLOGY related subject.

DHAYANANDHAN.S