- Thread starter imacheezhd
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- Location
- Logan, Utah

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

- Location
- Huntsville Alabama

- Location
- Henrico County, VA

- Occupation
- Electrical Contractor

Especially without knowing the system voltage and configuration.I don't think it's possible to convert the UOM of output (BTU, tonnage) to an accurate UOM number of input (amps).

- Location
- VA

- Location
- Columbia SC

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

- Location
- Henrico County, VA

- Occupation
- Electrical Contractor

So, your constant is 5.625 kva/ton.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

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

Diesel, or unleaded plus?Especially without knowing the system voltage and configuration.

????

Hi Larry,So, your constant is 5.625 kva/ton.

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

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.

Last edited:

Bmd

Hmmm...Bad math day. 12000/8.533= 1.411 kVA/hr. Evidently 8.533 Btu represents the base or 8.533/3.412= 2.5wHi 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.

- Location
- Henrico County, VA

- Occupation
- Electrical Contractor

Do you walk to school or carry your lunch?Diesel, or unleaded plus?

Theoretically.Do you walk to school or carry your lunch?

- Location
- The Sooner Nation

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).

My take on the problem

Without the Energy Efficiency Ratio (EER), Minuteman is absolutely right here. These may help you:

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).

Code:

```
Cooling (BT/Hr)
EER = -------------------------
Electrical input (watts)
or:
Cooling (BTU/Hr)
Electrical input (watts) = ---------------------
EER
then:
Watts
Amps = -------------
Voltage
```

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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