There are many examples of teaching using metaphor to help someone get a grasp of how something works without being EXACTLY correct.

Some examples are how we often use water flow to explain electrical flow or refrigerant circuit dynamics. It’s enough like the way it works to get our heads wrapped around it but there are many differences and the metaphors eventually break down.

This is definitely the case with air and nitrogen “absorbing” water

I’ve done podcasts and videos about how air can “hold” less moisture when it is cooler and more when it is hotter. You have likely heard old school techs talk about triple evacuation and sweeping with nitrogen to “absorb” the moisture from the system.

News Flash, Air and Nitrogen DO NOT absorb or hold moisture… They ignore one another at parties and they certainly don’t shake hands.

Water vapor in the air behaves much like all the other gasses contained in the air with the notable exception that water exists in both vapor and liquid states at atmospheric pressure and temperature.

When the temperature of water vapor is higher, a higher percentage of the air by volume can CONTAIN water vapor, but the air itself isn’t what is holding it. It does interact with it as the molecules move and bounce around and the percentage of water vapor in the air does impact the mass/weight of the air by volume (water vapor weighs less than dry air) so there are certainly impacts to the makeup of the air based on moisture content.

The percentage of the air around us that is moisture can vary from almost zero In cold arctic & Antarctic climates to nearly 4% in hot, tropical climates.

When teaching it we speak as though the air is a sponge and the hotter the air the bigger the sponge. This certainly helps us remember but it isn’t really how it works. In reality water in the air is all about the saturation temperature and pressure of the water and the air has little to do with it.

By Greg Benson

This is the same sort of thinking when a tech is having a hard time pulling a vacuum and they add dry nitrogen to the system to “absorb” the moisture. First off, you will want to sweep the nitrogen through the system, not just pressurize. Secondly, the nitrogen has no special properties that allow it to “grab” moisture. It can entrain the water vapor using Bernoulli’s principle, it will warm up the system a bit, it will certainly add in a bit of turbulence which can help move the oil around and potentially release some trapped moisture… but nothing more than that.

Don’t get me wrong, there is nothing wrong with sweeping with dry nitrogen, even better to use a heat gun and warm the compressor crankcase, receivers and accumulator and coils during a deep vacuum on a large system to help speed up the vaporization of moisture.

It doesn’t change the fact that air and nitrogen don’t “hold” moisture.

— Bryan

 

Here is part 2 from Michael Housh from Housh Home Energy in Ohio. Thanks Michael!


This is part two in a series on a deeper look into the Sensible Heat Rate Equations.  The first article can be found here if you missed it.  In this article, I will dive deeper into the density of “standard air” and water based on a change in temperature. “Standard air” is air with 0% RH and at sea-level (14.7 psia).

 

As mentioned in the previous article density can be defined as its mass per unit of volume (or weight per unit of volume).  Also mentioned before is that the Sensible Heat Rate equations stem from a lower level equation that is about moving a mass of a fluid.

 

Let’s look at the equation to find the density of air first.  The following equation can be used to determine the density of air for a given pressure and temperature.  For all of our equations and graphs, we will use the absolute pressure of 14.7 (the pressure at sea-level).

 

Formula:

We must first convert our temperature to an absolute temperature (Rankine).

Where:
T = Temperature (°F)

Next, we can solve for the density for the given temperature and atmospheric pressure.

Where:
D = density of air (lb/ft3)

Pa = absolute pressure of air (psia) = 14.7

Tr = temperature (°R) converted in the previous step


As you can see, we are dealing with pretty minute changes in the density of air based on temperature change when the humidity is 0%.

The above formula can also be used to solve for the density of air at different altitudes, by substituting Pa with the absolute pressure at a given altitude, but that will be left for you to figure out.

 

Next, let’s look at the density of water equation.  This is a little more complex of an equation. It is also only valid for water temperatures between 50°F and 250°F.

 

Where:

D = density of water (lb/ft3)

T = temperature (°F)

As you can see the density of water has a little more of a drastic change based on temperature than that of air.

 

First off, when we relate these to the Sensible Heat Rate equations (that are about moving pounds of a fluid), we can see that the mass can be affected by a significant amount (at least in the case of water).  Secondly, 1 cubic foot of water weighs significantly more than 1 cubic foot of air.

 

Let’s envision 2  – 1’ x 1’ x 1’ cubes, one is full of air the other is full of water.  Both of these are at 50°F. Let’s envision we have a fan or pump moving 1 of these cubes each minute. Next, let’s look at how we would compare the Btu’s carried/rejected from both of these cubes with 20° Delta T.

 

We are going to use the lower level equation found in the first article to make this a 

little easier to understand.

Q = M * C * T

Where:

Q = BTU’s absorbed or rejected from the fluid (BTU/hr)

M = mass / density of the fluid (lb/ft3)

C = specific heat (BTU/lb) : Air = .24 & water = 1.0

T = temperature change dry bulb (°F)

 

When we calculate the density of a fluid that has a temperature change to it, we want to use the density for the average temperature.  So for our experiment since we are starting at 50° and we end at 70°, we will use the Density for water and for air at 60°.

Air (Density @ 60° = .076,

     Specific Heat = .24)

Water (Density @ 60° = 62.37,

           Specific Heat = 1.0)

Q = .076 * .24 * 20 * 60 (min) Q = 62.37 * 1.0 * 20 * 60 (min)
Q = 21.89 BTU/hQ = 74,844 BTU/h

 

At these parameters, we’d have to move 3,419 more cubic feet of air to achieve the same as 1 cubic foot of water.

 

So let’s see how we can use this information to build a little bit of a better equation than the default “standard air” and “standard water” Sensible Heat Rate Equations.

Since the density of air is pretty tricky and is affected by both altitude and humidity I would recommend using the online calculator at HVACR School.  With water, I would use the above equation to solve for density or graph above.  Remember for both of these we want to use the average temperature to solve for the density.

 

Now, let’s rewrite the Sensible Heat Rate equations into a slightly more accurate way to calculate the BTU/h of both air and water.

Let’s run through a little comparison of these equations from the “standard air” and “standard water” equations.  I’ll start with the air-side first. Let’s imagine we have a 70° return air temperature and a 130° supply air temperature (a 60° Delta T).  The relative humidity is 40% and I used the elevation of my home which is approximately 800 ft. above sea level. By using the online calculator I figured the density of the average temperature of the air at these parameters is 0.06857.  Now let’s assume we have 800 CFM moving across our appliance, so now we can solve for the BTU transfer of our appliance into the air.

 

Standard AirImproved Air
 Q=1.08 x 800 x 60 = 51,840 btu/hr Q = .06857 x 14.4 x 800 x 60 = 47,395 btu/hr

That is about a 10% difference by correcting for the density of the air.

 

Next, for the water-side, let’s assume that we have an incoming water temperature of 170° and outlet water temperature of 190° across a boiler (a 20° Delta T).  Our pump is moving 5 gallons per minute. The density of water at the average temperature of 180° is 60.59814, so now we can solve for the BTU transfer from the boiler into the water.

 

Standard WaterImproved Water
 Q = 500 x 5 x 20 = 50,000 btu/hr Q = (60.59814 / 7.48) x 60 x 5 x 20 = 48,605 btu/hr

This is about a 5% difference by correcting for the density of water.

 

For some, the difference in accuracy may not be worth it in the field, but it depends on what you are trying to solve.  Stay tuned for more details on better Sensible Heat Rate Equations.

— Michael

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