When making coil selections, it is important to know whether the coil will be designed based upon standard or actual airflow. Understanding what the difference is between standard and actual airflow — and how coil selection software uses the information supplied by its operator — will result in coils that perform as desired.

Know Your CFM

One of the key pieces of information needed to make a coil selection is the cfm (cubic feet per minute) of air passing through the coil.

Coil selection software uses this information indirectly to calculate a coil’s required capacity. Coil selection software calculates the required coil capacity based upon pounds per hour (lb/hr) of air passing over the fins of the coil — not cfm.

In order to convert cfm to lb/hr, programs must know what the density, or the number of pounds contained in one cubic foot (lb/cu ft), of the air is. If the coil is run under standard air conditions (scfm), the programs automatically recognize that the density of the air is 0.075 lb/cu ft. If the coil is run under actual conditions, programs need further information to calculate the density of the air.

The two pieces of information programs use to calculate the actual density are the entering air temperature to the coil and the altitude at which the coil is operated. The density of the air depends upon both the temperature and pressure of the air.

As stated earlier, standard air has a temperature of 70?. Air that is warmer will be less dense or lighter. Air that is cool will be more dense (heavier). Likewise, standard air pressure is defined as 29.92 in. Hg. At higher pressures the air will be denser, and at lower pressures the air will be less dense. It’s a simple matter to demonstrate how the density of the air affects the calculated required capacity for a coil.

Function of a Coil

The function of a coil is to condition air such that when delivered will satisfy the space cooling or heating load. The required capacity of a coil can be calculated using the basic thermodynamic equation:

Q = m x Delta hr

Where

q = Heat transferred to or from the air (Btuh)

m = Mass flow rate of air (lb/hr)

Delta hr = Difference between the entering and leaving air enthalpy or total heat (Btu/lb)

The mass flow rate is equal to the density of air times the face area of the coil times the velocity of the air at the coil or face velocity:

m = p x A x V

Where

p = Density (lb/cu ft) A = Face area of coil (sq ft) V = Air velocity/face velocity (ft/min) Area (A) has the units of square feet — fin height (inches) times the finned length (inches) divided by 144 sq in. per sq ft. Face velocity (V) has the units of feet per minute. The product of the two yields the volume of air passing through the coil per unit of time or volumetric flow rate. Volumetric flow rate has the units of cubic feet per minute or cfm.

Therefore, the second equation can be rewritten as m = p x cfm, which when substituted into the first equation yields q = p x cfm x Delta hr. From this, it is clear that the required coil capacity is dependent upon the density of the air entering the coil.

This equation can be further reduced for standard air. As stated earlier, the density for standard air is 0.075 lb/cu ft. Substituting this into the modified first equation yields q = 0.075 x scfm x Delta hr. This equation represents the required total capacity (total = latent + sensible) of the coil. For a sensible process, Delta hr = cp x Delta T where cp = specific heat of dry air = 0.24 Btu/lb/?F, where Delta T is the drybulb temperature difference between the entering and leaving air.

Here is the equation for standard air:

q = 0.075 x scfm x Delta hr (total heat) and q = 0.018 x scfm x Delta T (sensible heat)

A quick check of the units reveals that the equations must be multiplied by the unit conversion of 60 min/hr in order to obtain the desired units for capacity of Btuh. The final form of the equation for standard air is:

q = 4.5 x scfm x Delta hr (total heat) and q = 1.08 x scfm x Delta T (sensible heat).

Many people frequently use these equations without knowing how they were derived. Therefore, there is the potential for misuse. These equations can only be used under standard air conditions.

Another way to show how air density affects a coil’s required capacity is by comparing actual and standard cfm for the same mass flow rate (lb/hr).

Figure 1 is a graph of temperature conversion factor (FT) vs. a coil’s entering air temperature. Figure 2 is a graph of altitude conversion factor (FA) vs. altitude or feet above sea level. To convert actual cfm to standard cfm, multiply the actual cfm by FT and FA. To convert standard cfm to actual cfm, divide the standard cfm by FT and FA.

Example:

Convert 15,900 cfm of air at 95? and at 3,000-ft altitude to standard conditions.

Cfm (standard air) = cfm (actual air) x FT x FA

Cfm (standard air) = 15,900 x 0.955 x 0.896 = 13,605

What this means is that it takes 13,605 cfm of air at standard conditions to have the same pounds per hour of air passing through the coil as there are for 15,900 cfm of air at 95? and 3,000 ft altitude.

This illustrates how important it is to know whether a coil is to be designed under standard or actual conditions. It is the difference between potentially oversizing, or worse, undersizing a coil.

Purpose

Most published ratings for hvac products are at standard air conditions. This allows for the direct comparison of similar products by different manufacturers, since published ratings are all at the same design conditions.

The difference between scfm and acfm is in the density of the air or the number or the pounds per cubic foot. This is important because coil selection software calculates required coil capacities based upon pounds per hour (lb/hr) of air passing through the coil, not cfm.

So, the next time you are making a coil selection, don’t think of the air passing through the coil in terms of cfm. Start thinking in terms of pounds per hour, also known as mass flow rate.

Guariglia is an engineer with Heatcraft, a subsidiary of Lennox International.

Publication date: 09/03/2001