First, let me explain that this thread is a discussion on basic heat transfer calculations. I emphasize the word BASIC. They are a means to achieve a close estimate. The formula's are not exact, they are in simple math. We should all know that a firm grasp on calculus and physics would be necessary to use the proper formulas. With that said....
We know that temperature, pressure, and humidity are relative to density, no? A psychrometric chart will show this.
As temperature rises, and/or pressure falls, air expands and takes up more space, and there is a rise in specific volume ft³/lb. As temperature falls, and/or pressure rises, the air compresses and there is a fall in specific volume.
Density is the inverse being lb/ft³. So, as temperature rises and pressure falls, density will fall. This is due to molecules having more energy bouncing off one another, or less compression; this provides more space for the molecules to move in. As we rise in altitude air loses density and gains space between molecules due to loss in pressure. So, the air is less efficient at containing heat, because there are less molecules per cubic foot. You might ask, why is there less heat at higher altitude? Because our air is heated by the Earth through convection. The Sun heats the Earth by radiation.
As humidity rises, density will fall due to hydrogen having less molecular mass than oxygen and nitrogen. Remember density is mass/Volume, or m/V or lb/ft³.
davidr
If we are calculating heat transfer, we use temperature rise in the formula. In order to identify the amount of heat transfer, we need to know the amount of air we are dealing with. This is measured in cfm. However, the same 1000cfm at a specific volume of 13.5 ft³/lb, will not be equal molecular mass to air at 17.5 ft³/lb. Thus not having equal heat content. So, in order to know the heat content we need to know the density.
Now, if there is a temperature or pressure change, then density will change.
If we were only reading heat content in the air, then one density measurement would be fine. However, we are measuring heat gain from return to supply, not just heat in the supply.
I found a handy derivation heat calc formula sheet. I've attached it to this post, and it does mention to average the two.
In a discussion on the Testo 435, Jim Bergmann explained the function of this tool. I would not quote him, because I can't find the thread, but I believe he explained this meter, equipped with two real time probes one in the return and one in the supply, would account for and compensate for the density change.
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