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Prof Sporlan
07-06-2001, 03:42 AM
Psychrometrics is the study of moist air thermodynamics, and the conditions and processes involving moist air.

Some of the terms:

<B>Dry Bulb:</B> Actual (sensible) temperature of the air sample.
<B>Dew Point:</B> Temperature at which the air sample reaches saturation.
<B>Relative Humidity:</B> Ratio of the partial pressure of water vapor in the air sample to the pressure of saturated water at the temperature of the air sample.
<B>Pressure of Water:</B> Yes, water exerts a pressure. This explains why water will evaporate at a temperature below its boiling point. In fact, even ice exerts a pressure!
<B>Partial Pressure of Water Vapor:</B> Yes, the moisture in air also exerts a pressure. It is referred to as a partial pressure because it is part of the total pressure of the moist air sample, i.e., partial pressure of water vapor + partial pressure of the dry air pressure equals the total pressure of the moist air sample.
<B>Humidity Ratio:</B> Ratio of the mass of water vapor to the mass of dry air in the air sample.
<B>Specific Humidity:</B> Ratio of the mass of water vapor to the mass of the air sample.
<B>Absolute Humidity:</B> Ratio of the mass of water vapor to the total volume of the air sample.
<B>Degree of Saturation:</B> Ratio of the humidity ratio of the air sample to the humidity ratio of the air sample at saturation.
<B>Wet Bulb:</B> Thermodynamic wet bulb is referred to as the temperature of adiabatic saturation. It is the temperature at which water, by evaporating into air, can bring the air to saturation. Interestingly, it is a property of air which cannot be measured, only calculated. Ordinary wet bulb is the temperature read using a psychrometer, and it can only approximate thermodynamic wet bulb. For the purposes of psychrometric calculations, however, ordinary wet bulb is sufficiently close enough to thermodynamic wet bulb.

Typical problems which psychrometrics is used to solve:

1. Cooling and dehumidification loads
2. Mixing of two or more airstreams
3. Heating and humidification loads

A rule of thumb used by many in the a/c business is 400 cfm of airflow requires 1 ton of air conditioning. Shall we attempt to proved this rule? :)

Prof Sporlan
13-06-2001, 02:35 AM
But the 1 Ton load typically comprises both latent and sensible heat. We need to remove these two forms of heat at the same rate they individually enter the load space.

Absolutely! Solving a typical a/c cooling load problem involves calculating both the sensible and latent load. If we assume our return air condition is the same as the SEER 'A' test, i.e., 82°F dry bulb, 67°F wet bulb, and the evaporator cools the air down to 62°F and a 75 percent rh. Using a nifty psychrometric program the Prof found at http://walden.mvp.net/~aschoen/, the following state values are determined:

return air:
enthalpy = 31.43 Btu/lb<sub>m</sub> (dry air)
humidity ratio = 0.01067
specific volume = 13.89 ft<sup>3</sup>/lb<sub>m</sub> (dry air)
rh = 45.6 percent

supply air:
enthalpy = 24.59 Btu/lb<sub>m</sub> (dry air)
humidity ratio = 0.00890

Given we have 400 cfm of air, the mass flow of dry air becomes:
400 cfm / 13.89 ft<sup>3</sup>/lb<sub>m</sub> = 28.80 lb<sub>m</sub>/min (dry air)

Sensible load is then calculated by taking the difference in enthalpies, and multiplying by the mass flow of dry air:
28.80 * (31.43 - 24.59) = 197.0 Btu/min = 0.985 tons

But we also have a latent load, since our humidity ratio has decreased, i.e., water is condensing at the coil. Assuming the water is condensing at the final air temperature, 62°F, enthalpy to condense water is 35.11 Btu/lb<sub>m</sub> using a handy ASHRAE table. Latent load is then calculated:
28.80 * 35.11 * (0.01067 - 0.00890) = 1.790 Btu/min = 0.009 tons

Total cooling load is then: 0.985 + 0.009 = 0.994 tons

Prof Sporlan
13-06-2001, 04:28 AM
Mmmmm... the Prof has just noticed he's buggered up his solution. Perhaps he needs to relearn his psychrometrics and the PMTHERM program.... :)

To calculate sensible load, we need to establish change in enthalpy at constant humidity ratio, therefore, at 62°F dry bulb and humidity ratio = 0.01067:
enthalpy = 26.52 Btu/lb<sub>m</sub>

Sensible load then becomes:
28.8 (31.43 - 26.52) = 141.4 Btu/min = 0.707 tons

Total load is solved using the conservation of energy equation:
m<sub>a</sub> * h<sub>1</sub> = m<sub>a</sub> * h<sub>2</sub> + m<sub>w</sub> * h<sub>w</sub> + q

where:
m<sub>a</sub> = mass flow of dry air
h<sub>1</sub> = return air enthalpy
h<sub>2</sub> = supply air enthalpy
m<sub>w</sub> = mass of water removed
h<sub>w</sub> = enthalpy to condense water
q = total load

And the conservation of mass equation:
m<sub>a</sub> * W<sub>1</sub> = m<sub>a</sub> * W<sub>2</sub> + m<sub>w</sub>

where:
m<sub>a</sub> = mass flow of dry air
W<sub>1</sub> = return air humidity ratio
W<sub>2</sub> = supply air humidity ratio
m<sub>w</sub> = mass of water removed

Combining these equations, we get:
q = m<sub>a</sub> * (h<sub>1</sub> - h<sub>2</sub>) - m<sub>a</sub> * h<sub>w</sub> * (W<sub>1</sub> - W<sub>2</sub>)

Using the above equation, and the numbers established in the previous post, we get:
Total cooling load: 0.985 - 0.009 = 0.976 tons

and not: 0.985 + 0.009 = 0.994 tons

The latent load becomes: 0.976 - 0.707 = 0.269 tons

Aaargggghhhhh :(

Prof Sporlan
02-07-2001, 08:11 PM
I fine effort! The curve fit equations are different than the Prof has seen, but a quick check of a couple state calculations show them to be accurate.

Using these type of equations are fine for typical psychrometric calculations. Though rarely needed, greater precision for a wider range of conditions can be obtained using the Hyland-Wexler equation of state. As one might expect, using this equation adds a significant amount of difficulty to the programming effort.

For the budding psychrometric programmer, Psychrometrics: Theory and Practice is perhaps the best source of information on this topic, which includes an extensive overview of the Hyland-Wexler equation of state. http://www.amazon.com/exec/obidos/ASIN/1883413397/qid%3D994100997/104-7203237-1523902

Unfortunately for the Prof, this book came out after his wrote PMTHERM, and it would have saved hours of work. :rolleyes:

He'll post some alternate curve fit equations from this book for your review.

Prof Sporlan
03-07-2001, 08:30 PM
Some equations of interest to psychrometrics:

<b>saturation pressure over liquid water:</b>
ln(p<sub>ws</sub>) = -0.58002206 x 10<sup>4</sup> / T + 0.13914993 x 10<sup>1</sup> - 0.48640239 x 10<sup>-1</sup> * T + 0.41764768 x 10<sup>-4</sup> * T<sup>2</sup> - 0.14452093 x 10<sup>-7</sup> * T<sup>3</sup> + 0.65459673 x 10<sup>1</sup> * ln(T)

where:
p<sub>ws</sub> = pressure, Pa
T = temperature, °K (273.15 <= T <= 473.15)

<b>saturation pressure over ice:</b>
ln(p<sub>ws</sub>) = -0.56745359 x 10<sup>4</sup> / T - 0.63925247 x 10<sup>1</sup> - 0.9677843 x 10<sup>-2</sup> * T - 0.62215701 x 10<sup>-6</sup> * T<sup>2</sup> + 0.20747825 x 10<sup>-8</sup> * T<sup>3</sup> - 0.9484024 x 10<sup>-12</sup> * T<sup>4</sup> + 0.41635019 x 10<sup>1</sup> * ln(T)

where:
p<sub>ws</sub> = pressure, Pa
T = temperature, °K (173.15 <= T <= 273.15)

<b>barometric pressure vs altitude:</b>
p = 101.325 * (1 - 2.25577x10<sup>-5</sup> * z)<sup>5.2559</sup>

where:
p = barometric pressure, kPa
z = altitude, meters

Prof Sporlan
04-07-2001, 02:47 AM
Would I have to put in a "if" statement that when the temperature is below 0°C the vapour above ice formula is used? I'm just wondering at exactly which point it comes into play, if it does?

Marc brings up a rather interesting aspect of psychrometric calculations, and that being calculations below 32°F (0°C) dry bulb.

Many (if not most) psychrometric programs do not properly calculate relative humidity or dew point below 32°F dry bulb, which then leads to errors in calculating density and enthalpy of the air. Recall the definition of relative humidity previously posted:

Relative Humidity: the ratio of the partial pressure of water vapor in the air sample to the pressure of saturated water at the temperature of the air sample.

A perfectly logical (but incorrect) assumption is to use the pressure of saturated ice equation to calculate RH below 32°F dry bulb. Even the Prof made this error when he initially wrote the psychrometric programming for PMTHERM. Fortunately, an engineer from Kodak who used PMTHERM alerted him to this error...

RH is defined with respect to the saturation pressure of water, even below 32°F. One might wonder what sense there might be in determining saturation pressure of water below 32°F, where clearly you wouldn't have water anyway. But then one simply has to consider what we mean by "dew point" below 32°F, where RH is 100 percent.

By defining RH with respect to water, we keep the RH lines on a psych chart continuous thru 32°F dry bulb. If we used the saturation of ice to determine RH below 32°F, we would get a break in the RH lines at that point. As is turns out, when we use the saturation of ice equation, the <b>frost point</b> at "100 percent RH" is calculated, not the dew point. The <b>frost point</b> is the temperature at which visible frost forms on the surface being chilled.

Of course, frost point is irrelevant above 32°F. And interestingly, it will always be greater dew point temperature below 32°F. This makes sense when one realizes that the saturation pressure of water must be greater than that of ice. Another interesting item: dew point hygrometers below 32°F measure frost point, not dew point.

So how does one calculate the pressure of saturated water below 32°F? The simplest solution is to use the saturated water equation previously posted below its stated lower limit of 32°F. For an improved method, one can check out an article by F. A. Sando, and published in ASHRAE Transactions 96(2), pp 299-308.

Prof Sporlan
05-07-2001, 09:44 PM
But something I was trying to figure out last night, dry air specific heat at altitude. I downloaded a few other programs from around the web last night and some agree with my program calculation and others say my air enhalpy component may be too high.
One thing to keep in mind when evaluating enthalpy and entropy numbers is that you are only interested in the change of enthalpy or entropy from one state to the next. The actual value of enthalpy or entropy calculated for a given state will depend upon where one's equations "zero" these values. Standard ASHRAE I-P calculations zero enthalpy and entropy at 0°F dry air at standard atmospheric pressure. But there would be no reason why one couldn't refigure the equations to provide a zero enthalpy at say -40°F dry air. This, of course, would not affect calculations on psychrometric processes, but it would change the enthalpy numbers around.

It is very likely some of these programs you've located are using a different zero point for enthalpy. PMTHERM zeros enthalpy and entropy at 0°F dry air, and its calculations also show higher enthalpy values than your calculator. When checking an enthalpy difference (28°C DB/ 26°C WB at sea level and at 5000 ft elevation), however, values were within 2.5 percent. Not bad!


If the Prof downloads the latest version from the address posted above he can use the altitude slide bar to see if the formulae agree with reality or do I need a math curve to alter dry air specific heat at altitudes?
PMTHERM uses the Hyland-Wexler equation of state which incorporates barometric pressure. But it seems your calculator is sufficiently accurate as is for engineering calcuations... :)

Prof Sporlan
07-07-2001, 02:51 AM
Your enthalpy equation is based on the perfect gas law equation of state as applied to moist air, i.e.:

(p<sub>a</sub> + p<sub>w</sub>) * V = (n<sub>a</sub> + n<sub>w</sub>) * R * T

where: (the Prof will use SI units.... :))
p<sub>a</sub> = partial pressure of dry air, kPa
p<sub>w</sub> = partial pressure of water vapor, kPa
V = total mixture volume, m<sup>3</sup>
n<sub>a</sub> = number of moles of dry air
n<sub>w</sub> = number of moles of water vapor
R = universal gas constant, 8.31441 kJ/(kg mol-°K)
T = temperature, °K

The equation you use is very close to the one suggested in the text, Psychrometrics Theory and Practice:

H = 1.006 * T + W * (2501 + 1.805 * T)

where:
H = enthalpy, kJ/kg
T = dry bulb temperature, °C
W = humidity ratio

You might try this one out and see if it improves your calculations.

So how does one determine enthalpy from the equation of state? By definition:

H = U + P * V

where:
H = enthalpy, kJ/kg
U = internal energy, kJ/kg
P = pressure, kPa
V = volume, m<sup>3</sup>

We can deal with P and V with our perfect gas law equation of state. Internal energy is a different matter. It can be expressed in the following differential form:

dU = c<sub>v</sub> dT + [T * (<font face="Symbol">d</font>P/ <font face="Symbol">d</font>T)<sub>V</sub> - P] dV

where:
c<sub>v</sub> = isochoric specific heat, kJ/kg-°K

Mmmmm... this is where we run into a problem. The [T * (<font face="Symbol">d</font>P/ <font face="Symbol">d</font>T)<sub>V</sub> - P] dV term can be dealt with using our equation of state. The c<sub>v</sub> dT term requires additional information.

Unfortunately, the Psychrometrics Theory and Practice text does not provide a simple c<sub>v</sub> equation we can use. If we could find such an equation, we may be able to develop an enthalpy equation which is a function of both pressure and dry bulb temperature without having to resort to the Hyland-Wexler equation.

Prof Sporlan
09-07-2001, 03:55 PM
Okay, I've just finished developing a kind of imperial version of my ahucalc application. And it seems to make a good effort trying to keep up with PMtherms numbers at varying altitudes.
There is likely an easy way to develop an enthalpy equation based on dry bulb temperature, barometric pressure, and the ideal gas law without having to use Hyland-Wexler, or curve fitting calculated enthalpy data. If the Prof finds a little time, he'll look at developing a c<sub>v</sub> equation from his PMTHERM data, which would allow him to develop such an enthalpy equation.


There did seem to be a discrepancy in the cfm and duty results comaparing ahucalc's numbers to those being processed with PMtherms "Build Table Of RSHR Line" facility.
Paul Milligan wrote the process routines, and his RSHR (room specific heat ratio) table was one of the more interesting features. The Prof does recall a problem with sensible/latent load calculations in an eariler version of PMTHERM, but believes it had been corrected. Calculating the process manually would be in order here... :)

Gary
09-07-2001, 04:29 PM
Whatever happened to Paul Milligan? He and I had many fascinating discussions on alt.hvac way back when. I remember when he first introduced PMtherm.

Prof Sporlan
09-07-2001, 07:10 PM
Paul Milligan traded in his service tools for a desk job which involved writing software for controllers. Apparently, he grew tired of working on rooftop units in 100°F ambients, and he considers himself largely out of the hvac/r industry. But one shouldn't be too surprised if he were to become actively involved in the industry once again. Besides, he needs to update his portion of PMTHERM one of these days.... :)

Gary
09-07-2001, 07:51 PM
I can't say that I blame him, but it is definitely a loss for our industry. He was very sharp. :)

Prof Sporlan
01-12-2001, 07:48 PM
Relative humidity (RH) is, by definition, the ratio of the mole fraction of water vapor in a given air sample to the mole fraction of water vapor in the air sample when it is saturated.

RH can also be calculated by taking the ratio of the partial pressure of water vapor in the air sample to the pressure of saturated water, which is normally a simpler method.

One might ask what is meant by the "partial pressure of water vapor" in the air. John Dalton was the first to correctly surmise that the total pressure of a mixture of gases or vapors is simply the sum of the pressures of each gas. With air, we actually have a mixture of both air and water vapor. Therefore, we have:

Ptotal = Pdryair + Pwater

where:
Ptotal = total pressure of moist air
Pdryair = partial pressure of dry air
Pwater = partial pressure of water vapor

Assume we have a sample of air at atmospheric pressure (14.696 psia) and at 75°F dry bulb (db). If this sample had an RH of 100 percent, we would know the partial pressure of water vapor in this sample must equal the pressure of saturated water at 75°F. We can look up this value in the ASHRAE Fundamentals Handbook (Table 3, Chapter 6, 1997 issue):

Pwater = 0.430 psia

Therefore, we can determine the partial pressure of dry air:

Pdryair = 14.696 - 0.430 = 14.266 psi

If our 75°F db air were actually at 50 percent RH, we now know Pwater must be one-half the pressure of saturated water, i.e.:

Pwater = 0.5 * 0.430 = 0.215 psia

And now our partial pressure of dry air becomes:

Pdryair = 14.696 - 0.215 = 14.481 psi

The dew point temperature of the air sample is achieved thru sensible cooling until the sample reaches saturation, i.e., 100 percent RH. To determine dew point, it is helpful to know the humidity ratio of our air sample, i.e., the ratio of the mass of water to the mass of dry air in the sample. Dew point occurs when the humidity ratio, W, is equal to the humidity ratio at saturation, Ws (100 pct RH).

We can calculate W for our 75°F db 50 percent RH sample as follows:

W = Mw * Pwater / (Ma * (Ptotal - Pwater))
W = 18.01528 * 0.215 / (28.9645 * (14.696 - 0.215)) = 0.00923

where:
Mw = molecular weight of water
Ma = molecular weight of air

To locate the dew point for this sample, we need to find the temperature where Ws = W = 0.00923.

Normally finding this temperature is an iterative calculation, but we can make the observation that the partial pressure of saturated water must equal 0.215 psia at this temperature. Using the ASHRAE table, we find it corresponds to about 55.1°F. One may confirm this by looking at a psych chart and were W = 0.00923 intersects with the 100 percent RH line. So when the temperature of the air reaches 55°F, we can expect condensation to occur.

If we were to lower RH of our 75°F db air to 15 percent:

Pwater = 0.15 * 0.430 = 0.0645 psia
W = 18.01528 * 0.0645 / (28.9645 * (14.696 - 0.0645)) = 0.00274

Per the ASHRAE table, we get a dew???? point of 25.2°F

Questions for those with inquiring minds..... :)

How can we have a dew point below freezing? Isn't this actually the frost point? Are dew points and frost points below 32°F the same, or are they different? If they are different, how does one calculate each?

On a psych chart, what are the points on the 100 percent RH line below 32°F? Dew points or frost points?

Calculating heat loads of air flows for a/c applications is generally a straightforward exercise. You take the change of enthalpy of the air and multiply it by its mass flow. You then must add in the load from the amount of water being condensed multiplied by its change in enthalpy.

So how does one calculate a heat load when you are freezing moisture on an evaporator coil...... :)