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## BTU/hr calculation

I was thinking about airflow recently and had a brain storm about calculating BTU's in an airstream. I wanted to see if it was sound thinking. Here it is.

Enthalpy tells us how many BTU's/lb that air of a particular temperature and water content contains. Knowing this, if I know the CFM's and the density of the air at a specified temperature, I ought to be able to figure out how many BTU's I am pumping around. If I then decide to change the temperature, I should be able to find out how many BTU's I will have to take out of the air or put in to change the mass of air to the new temperature. So, I made these calculations to test the speculation.

Dry Bulb Temperature = 55*F
CFM = 400
Density of air at 55*F = 0.07711 lbs/ft3
Enthalpy of dry air at 55*F = 13.2 BTUs/lb

Calculations:

400 CFM x 0.07711 lbs/ft3 = 30.844 lbs/minute of air moved
30.844 lbs/minute x 60 minutes/hour = 1850.64 lbs/hour of air moved
1850.64 lbs/hour x 13.2 BTUs/lb = 24428.45 BTUs/hour moved

New Dry Bulb Temperature = 70*F
CFM = 400
Density of air at 70*F = 0.07492 lbs/ft3
Enthalpy of dry air at 70*F = 16.8 BTUs/lb

Calculations:

400 CFM x 0.07492 lbs/ft3 = 29.968 lbs/minute of air moved
29.968 lbs/minute x 60 minutes/hour = 1798.08 lbs/hour of air moved
1798.08 lbs/hour x 16.8 BTUs/lb = 30207.74 BTUs/hour moved

Delta T = 70*F - 55*F = 15*F
Total air volume = 400 CFM * 60 minutes/hour = 24000 Cubic Feet

So if I pump 24000 cubic feet of air through my air conditioner and lower the temperature of the entire volume of air from 70*F to 55*F, the unit will have to remove 30207.74 BTU's - 24428.45 BTUs or 5779.29 BTU's from the air in the hour that it takes the unit to pump the air through it at 400 CFM. This of course is speaking of perfectly dry air.

Now I have the formula BTU/hr = 1.08 * CFM * delta T which presumably tells how many BTUs/hour it will take to change the temperature of a certain quantity of air flow a certain delta T. When I do the math with this . . .

CFM = 400
delta T = 15*F

BTU/hr = 1.08 * 400 * 15 = 6480 BTU/hr

Why the difference? 5779.29 BTU/hr versus 6480 BTU/hr? Is the 6480 actually taking into account some latent heat making it a "rough" estimate for the real world?

2. I am too wiped out to confirm your findings.
However, since you are good at figures, my ARI Text book has a formula for figuring the condenser BTUH.

Would you go over my numbers & point out errors.

http://www.udarrell.com/ac-trouble-shooting-chart.html
udarrell

3. Originally Posted by pwn01
Why the difference? 5779.29 BTU/hr versus 6480 BTU/hr? Is the 6480 actually taking into account some latent heat making it a "rough" estimate for the real world?
No. The 1.08 figure simply assumes an air density of 0.075 lb/ft3 and a specific heat of 0.24 Btu/lb-&#176;F. Nothing more.

Your calculation is flawed in one respect: at 70&#176;F dry bulb you calculate 29.968 lbs/minute of air flow. At 55&#176;F dry bulb, you calculate 30.844 lbs/minute of air flow.

So you are gaining mass flow while cooling air?

Not! Violates the law of conservation of mass. See part of the problem?

4. Let me see if I understand you guy's correctly......

Nevermind.

5. Yes, I was aware that part will skew the results somewhat, but figured it would be okay for ballpark estimations of approximate Btu/h output.

I am sure there are other aspects of it that are a bit skewed as well.
I estimated the motor heat output, that would not be precise either.

If anyone wants to be precise they can get the data & let us look at it.
http://www.udarrell.com/ac-trouble-shooting-chart.html
udarrell

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Originally Posted by Andy Schoen
Your calculation is flawed in one respect: at 70°F dry bulb you calculate 29.968 lbs/minute of air flow. At 55°F dry bulb, you calculate 30.844 lbs/minute of air flow.
Yes, I see that. I would have the same amount of air flowing out of the evaporator as into the evaporator, it would just be more dense since it would be cooler. The volumetric flow rate would remain the same. A similar thought as V1A1=V2A2.

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That solves it!!! The mass flow rate is constant through the duct. That means I must use the mass flow rate calculated with the 70 degree F air which was 1798.08 lbs/hour of air moved. If I take this and multiply it by the enthalpy of 55 degree F air which was 13.2 BTUs/lb.....

1798.08 lbs/hr x 13.2 BTUs/lb = 23734.656 BTUs/hr.

Subtract this value from the initial value of 30207.74 BTUs/hour moved.....

30207.74 BTUs/hr - 23734.656 BTUs/hr = 6473.084 BTUs/hr

which is almost exactly the same as....

1.08 x 400 * 15 = 6480 BTUs/hr

The 1.08 figure simply assumes an air density of 0.075 lb/ft3...
If I use the rounded number of 0.075 lb/ft3 as the air density, the answer comes out exactly.

400 CFM * 0.075 lb/ft3 * 60 min/hr * 16.8 BTU/lb = 30240 BTU/hr
400 CFM * 0.075 lb/ft3 * 60 min/hr * 13.2 BTU/lb = 23760 BTU/hr
__________________________________________________ _____
-------------------------------------------------> 6480 BTU/hr

Ah yes. From this we can see how the 1.08*CFM*delta T equation was derived. If we can find a simple equation relating enthalpy to dry bulb temperature (@ zero percent humidity) to substitute in place of the enthalpy terms, we could collect like terms and everything would fall into place.

Some things aren't as complicated as they look at first.

8. Originally Posted by pwn01
Some things aren't as complicated as they look at first.
That's a fact.

The 1.08 formula works fine as long as you do not have condensate. It is safe to use for sensible heating or cooling.

Condensate complicates the calculation a bit. For your consideration:

Btu/hr cooling = ma[(h1 - h2) - (W1 - W2)hw2]

where:
ma = mass of dry air, lb/hr
h = enthalpy of moist air, Btu/lb
hw = enthalpy of water, Btu/lb
W = humidity ratio
1 = state 1 (beginning state)
2 = state 2 (ending state)

Enthalpy of water between 40°F to 45°F, the temperature where condensate will form on an a/c evaporator coil, falls in the range of 8 Btu/lb to 13 Btu/lb

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hw2 is the enthalpy of the water in the air at temperature (state) 2?

10. Originally Posted by pwn01
hw2 is the enthalpy of the water in the air at temperature (state) 2?
Specifically, hw2 is the enthalpy of condensed water at state 2. It takes more cooling to condense water, and it is properly accounted for in the equation.

11. ## btu

seems like a lot of work, if you know cfm!

cfm x 4.5 x grains difference or difference between entering wet bulb and leaving wet bulb!

example 400 x 4.5 x 11

12. Originally Posted by supertek65
seems like a lot of work, if you know cfm!

cfm x 4.5 x grains difference or difference between entering wet bulb and leaving wet bulb!

example 400 x 4.5 x 11

You've just successfully mashed two equations together.

latent heat = 0.68 x cfm x grains difference

total heat = 4.5 x cfm x enthalpy difference..

(enthalpy difference can be approximated by wet bulb difference)

The student may use my previous equation to determine how the 0.68 and 4.5 constants were arrived at.

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Btu/hr cooling = ma[(h1 - h2) - (W1 - W2)hw2]
hw2*(W1-W2) is accounting for the latent heat of the condensed water (?), but why are you subtracting it from the delta h of the moist air. It seems like it ought to be added because you are wanting the total BTUs being expended in the change in enthalpy of the vapor as well as in condensing the water.

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