A/C condensing unit calculation

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Im wondering if there is a way to determine wire size and ocp if just the size of the unit is known. ie. 2 ton. or if a slide rule is available, or basic chart. Thanks in advance for any help.
 

augie47

Moderator
Staff member
Location
Tennessee
with all the new effeciency standards, I think it is difficult to pin a number on an exact conversion. I know a number of folks that use 1.5 kw/ton.
 

raider1

Senior Member
Staff member
Location
Logan, Utah
I agree with Gus.

Most of the time you use the manufactures nameplate information to size the the branch circuit and the overcurrent device. Due to the fact that different SEER ratings for the same tonnage unit could change the rated current, it would be very difficult to estimate what the actual unit will require.

Chris
 

masterinbama

Senior Member
That's one reason I'm glad that here the AC guys pull there own wire on new residential and for the most part they know what they are doing. But as in our trade we do get the hacks every now and then.
 

480sparky

Senior Member
Location
Iowegia
I don't think it's possible to convert the UOM of output (BTU, tonnage) to an accurate UOM number of input (amps).

It's like trying to convert the MPG of you car to HP.

If you don't know what the nameplate of the unit will be, submit an RFI.
 

augie47

Moderator
Staff member
Location
Tennessee
In Mike Holt's residential load calculation program (available in his web site "free stuff") 2 tons is calculated at 3750 VA. I would bow to his expertise. I would think that would be a good "round number" to use if you are doing rough calculations. From bitter experience I would certainly prefer to see the unit or a spec sheet before installing the circuit.
 

mcclary's electrical

Senior Member
Location
VA
system voltage, dual fuel, heat strips, SEER ratings all affect this rating. I always go by the MCA on the ratings. If you pull whatever the AC guys tell you,,,,,,,often end up with overkill and wasted money
 
I can offer

I can offer

You can try this it has worked for me
Ton AC to Amps
One ton of frig.=12000 BTU/hour
1 watt=8.533 BTU/hour

4 tons x 12,000=48,000 BTU/hour

48,000/8.533=5,625 watts

5,625/208 volts=27 amps
 

gndrod

Senior Member
Location
Ca and Wa
????

????

So, your constant is 5.625 kva/ton.

Have you checked the 208v current on the nameplate of a 1-ton unit?
Hi Larry,

I think 4 tons was implied. (5.625/4= 1.406 kVA) So 27A/4= 6.75A per hr.) But I agree on other factors not mentioned.
 
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gndrod

Senior Member
Location
Ca and Wa
Bmd

Bmd

Hi Larry,

I think 4 tons was implied. (5.625/4= 1.406 kVA) So 27A/4= 6.75A per hr.) But I agree on other factors not mentioned.
Hmmm...Bad math day. 12000/8.533= 1.411 kVA/hr. Evidently 8.533 Btu represents the base or 8.533/3.412= 2.5w
 

Minuteman

Senior Member
Guys, you can't accurately calculate wattage from refrigeration tonnage. A/C units have various SEER ratings. There are several types of compressors; scroll, reciprocation, and rotary. There is no standard efficiency rating on the motors. Some motors are subcooled by the refrigerant. Etc.

Best you can get is a rule of thumb. For single phase, figure 10 amps per ton or 1 hp per ton. (Both are conservative. The actual FLA will be less).
 

topgone

Senior Member
My take on the problem

My take on the problem

Guys, you can't accurately calculate wattage from refrigeration tonnage. A/C units have various SEER ratings. There are several types of compressors; scroll, reciprocation, and rotary. There is no standard efficiency rating on the motors. Some motors are subcooled by the refrigerant. Etc.

Best you can get is a rule of thumb. For single phase, figure 10 amps per ton or 1 hp per ton. (Both are conservative. The actual FLA will be less).
Without the Energy Efficiency Ratio (EER), Minuteman is absolutely right here. These may help you:
Code:
              Cooling (BT/Hr)
EER       = -------------------------
           Electrical input (watts)

or:
                               Cooling (BTU/Hr)
Electrical input (watts) = ---------------------
                                     EER

then:  
            Watts 
Amps  = -------------
           Voltage
At 2 tons with an EER of say 10, the watts input required by the A/C will be 11.53 amps (208 volts supply).
 

suemarkp

Senior Member
Location
Kent, WA
Occupation
Engineer
Amps is not = watts/voltage. You're missing power factor, and that is significant on a motor-compressor.

I have not found a single equation that works well. Dividing the BTU by 8.5 is rather close to come up with VA. But it is best to find the MCA value. I'd hate to install 10-2 thinking it would be be in the high 20's for amps only to see an MCA of 30.5. Or you install 8-2 to cover your bets and see an MCA of 24 so you could have used 12-2.
 
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