Is finding my cooling load this easy?
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If I look back at last summer and find a time that is was 92°F outside (design temp for Boulder CO) and find out how many kWh I used in that hour (condenser and blower) on cooling could I get a somewhat close on what my cooling load is for my house?
92°F for one hour outside
78°F inside
1,520 watts used in that hour for the 5 ton condenser and blower
1.52 X 3412.14 (BTU per kWh) = 5,186 BTU of cooling
So about half a ton of cooling?
That doesn’t seem right since my house is over 3,000′ sq ft. In that hour my 5 ton condenser ran only 40% of the time. The 5 ton condenser uses 3,800 watts and the blower is using 880 watts pushing 1750 CFM of air. I set the blower’s max CFM to 1750 because that is all my duct work can handle (if I increase the blower to 2000CFM that air coming out of the ducts is the same as the 1750CFM setting).
I have tried so many heating and cooling load calculators and I have received back so many wildly different numbers that I am not sure what to trust.
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Replies
You have to multiply the electricity use by the COP of the compressor.
What's also missing from that calculation is whether the AC was able to hold temperature. If the temperature was creeping up then some of the heat was being absorbed by the house. As a really rough number a 3,000 SF probably has a heat capacity in the ballpark of 50,000 BTU/degree F, so if during that hour the house crept up by half a degree that would be 25,000 BTU. So in short time periods small fluctuations in temperature can mean big differences. You need to measure over a longer duration.
What's the compressor rating in BTU? Multiply that by the 40% run time to get a ballpark.
Thanks for your reply DCcontrarian!
I have a 14 SEER 5 ton Amana Goodman ASX-140601AA condenser and a Goodman capf4860d6aa 4-5 ton A coil (both 17 years old). Is it as simple as multiplying 5 tons X 12,000 = 60,000 BTU for it's BTU rating? Since I am running it at 1750CFM perhaps it's more like 4.4 tons X 12,000 = 52,800 BTU for the rating?
I found a COP calculator online and inputted 14 seer, the result was COP 3.45. So 3.45 COP X 1,520 watts (40% usage in that hour at 92°F hour outside) = 5,244 watts.
So 5.244 X 3412.14 (BTU per kWh) = 17,892 BTU of cooling.
17,892 BTU + 25,000 BTU (0.5°F creep in temp you mentioned) = 42,892 BTU cooling load. So about 3.5 tons needed for cooling? That number seems about right since it seems like my unit short cycles.
SEER is a seasonal average. You want COP at 92F outside and 78F inside, which will be different.
If you can go back and get 1 hour, can you get multiple hours? That’ll address the points DC had.
Instead of converting SEER to COP, I think I can calculate COP since I have all the variables from last summer recorded if that helps.
BTU output = temp rise X CFM X 1.08
78°F return air - 57°F supply air = 21°F
BTU output = 21°F X 1750CFM X 1.08
BTU output = 39,690 BTU
Convert 39,690 BTU to output watts X 3.41 = 11,639 output watts
Input wattage is 4,680 for condenser and blower
That equals a COP of 2.5
If I use 2.5COP in the equation above instead of 3.5COP (14 SEER), I now get a cooling load of 37,897 BTU. So around 3 to 3.5 tons of cooling needed.
Does that sound right?
Sorry I forgot to answer your question, yes I can get more hours, how many do you need and do you need the temp at each hour since it is always moving?
Yes, one ton is 12,000 Btu/hr.
I can think of three ways of measuring the BTU output of an air conditioner based on observations:
1. Electricity usage * COP, which gives 17.9K BTU/hr.
2. Run time vs rated capacity (for a single stage). If this is a 5-ton compressor, 60K Btu/hr, and observed run time is 40%, that gives 24K BTU/hr.
3. Airflow times temperature change. The formula is 1.08 times temperature change in degrees F times air flow in CFM. Let's say your air is entering at 78F, exiting at 50F and you have 1750 CFM, that gives 22.7K BTU/hr. You then have to add in the latent heat of any condensate that was removed, that's 1K BTU per pint.
All three measures are giving roughly the same number, certainly within the precision of the measurements. I suspect you're actually getting something less than the rated 1750 CFM for air movement.
Note that method #3 is how they come up with the BTU rating used in method #2 and the COP used in method #1.
The wildcard in all of this is whether the system is able to hold a steady temperature or whether the temperature is creeping up. Temperature changes that would be imperceptible on a one-hour timeframe would have an enormous impact on the observed levels.
My initial take is that your ductwork is undersized relative to the compressor. In order to get 60K BTU at 1750 CFM your exiting air would have to be well below 30F. But if your coil temperature is below 32F the condensate forms as ice on the coil, blocking air flow and shutting the system down.
DCcontrarian,
Thank you so much for your help.
This info will help me figure out what I do next.