In this article weâ€™re going to look at the Sensible Heat Transfer Equation for Air. This is useful when trying to determine either of the three variables, Btuâ€™s, CFM or Delta-T. When you know two of these values you can determine the remaining missing value.

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## Sensible Heat Transfer Equation for Air

**q = m x C _{p }x âˆ†T**

q = CFM x 0.075 lb/ft^{3} x 60 min/hour x 0.24 btu/lbÂ°F x âˆ†T

**q = CFM x 1.08 x âˆ†T**

CFM is ft^{3}/minute

m is the overall mass flow rate of air

C_{p} is the specific heat of air 0.24 btu/lbÂ°F

Weâ€™ll begin by calculating for the BTUâ€™s when given the CFM and temperatures of the Entering and Leaving Air. Remember weâ€™re dealing with sensible heat only. That means there is no latent heat which involves a change of state, like moisture in the air condensing into water or condensate. All weâ€™re doing is changing the temperature of air, while not removing any moisture.

## Example:

2,000 CFM of outdoor air (Ventilation Air) at 55Â°F and 70% relative humidity (RH) is heated to 90Â°F

Step #1 â€“ Determine the âˆ†T (90Â°F – 55Â°F = 35Â°F)

Step #2 â€“ Enter all values into equation.

q = CFM x 1.08 x âˆ†T

q = 2,000 x 1.08 x 35 = 75,600 Btu/hour

Looking at this heating hot water coil we see that there is 2,000 Cubic Feet per minute of air flowing through this coil with a Leaving temperature of 90Â°F and the Entering Air at 55Â°F.

With this information we can solve for how many btuâ€™s are being supplied to the air.

Step one is to subtract the entering air temperature of 55Â°F from the leaving air temperature of 90Â°F to arrive at the Temperature difference or Delta-T. 90Â°F – 55Â°F, gives us a 35Â°F Delta-T.

Step two is to put all the known values into our formula and make the calculation.

We have our formula of q = CFM x 1.08 x âˆ†T

Now we enter our values, we get

q = 2,000 x 1.08 x 35 = 75,600 Btu/hour

Now we quickly explain where the value of 1.08 in the calculation is derived from. First, we have the weight of Air at 0.075 pounds per Ft3, then we have the conversion of minutes into hours, and lastly the specific heat of Air at 0.24 btu/lbÂ°F. This is what that looks like

0.075 lb/ft^{3} x 60 min/hour x 0.24 btu/lbÂ°F = 1.08

With all these units, we can see which units of value remain by crossing out those that are eliminated in the formula, such as

q = 2,000 ~~Ft3~~/~~minute~~ x 0.075 ~~lb~~/~~ft~~ x 60 ^{3}~~min~~/**hour** x 0.24 **btu**/~~lbÂ°F~~ x 35~~Â°F~~ = 75,600 Btu/hour