AC TONS TO CURRENT

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kwired

kwired

Electron manager
Location
NE Nebraska
florida-sparkey said:
Thanks kwired!
The 1K per ton is not even close to what some others or the calculator link were saying (they were at 30A up to 60A). That was my confusion and I was trying to understand how it could be so different. Ultimately, I used 4.4KVA and that seemed close to the spec sheets I read online.
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I think 1.5 kVA per ton may be more realistic value, but 4.4 is pretty high. You sure the 30 to 60 amp you mentioned isn't somehow related to overcurrent protection device setting? 5 ton single phase unit - typically needs a 60 amp circuit breaker, but actual rated current draw of the compressor is more like 26-32 amps max.
 
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florida-sparkey

Senior Member
Location
Pinellas Park, Florida
Occupation
Master Electrician
kwired said:
I think 1.5 kVA per ton may be more realistic value, but 4.4 is pretty high. You sure the 30 to 60 amp you mentioned isn't somehow related to overcurrent protection device setting? 5 ton single phase unit - typically needs a 60 amp circuit breaker, but actual rated current draw of the compressor is more like 26-32 amps max.
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kwired,
Yes, I agree with with you on the current draw vs MCA. I think when others are speaking they are talking about MCA. That was a point I brought up to the gentleman but did not get a response. The calculator however is a loss on me.
 
gadfly56

gadfly56

Senior Member
Location
New Jersey
Occupation
Professional Engineer, Fire & Life Safety
florida-sparkey said:
kwired,
Yes, I agree with with you on the current draw vs MCA. I think when others are speaking they are talking about MCA. That was a point I brought up to the gentleman but did not get a response. The calculator however is a loss on me.
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The calculator was simply converting the 12,000 BTU/hr to raw kW. It doesn't take into account that with a little thermodynamic jujitsu you can move heat from point A to point B without expending as much energy in work as heat you are moving. The ratio is call the "coefficient of performance" or COP. Heat pumps or air conditioning units in cooling can operate at COP's of 5-7 or higher. Their effectiveness drops as the relative humidity or outside air temperature increases.
 
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florida-sparkey

Senior Member
Location
Pinellas Park, Florida
Occupation
Master Electrician
Ingenieur said:
mca ~ 30
ocp ~40-45
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Thank you,
Typically, MCA is the largest motor at 125% plus the fan.
RLA and FLA are the actual loads and are listed on the spec sheets you so nicely posted for me. Add one to the other and that's the load. In the case of a load calc you add the 25% to the largest motor for the total calc and in this case that motor is a 3HP pool pump.
 
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florida-sparkey

Senior Member
Location
Pinellas Park, Florida
Occupation
Master Electrician
gadfly56 said:
The calculator was simply converting the 12,000 BTU/hr to raw kW. It doesn't take into account that with a little thermodynamic jujitsu you can move heat from point A to point B without expending as much energy in work as heat you are moving. The ratio is call the "coefficient of performance" or COP. Heat pumps or air conditioning units in cooling can operate at COP's of 5-7 or higher. Their effectiveness drops as the relative humidity or outside air temperature increases.
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Now that I can understand.
I kinda thought it was an energy in / out thing and that it did not relate to electric usage. If it matters, here in FL I think AC must be 16 SEER or higher in new installs but that means nothing to me. So, for my purpose moving forward the calculator is basically useless because the numbers can wildly vary.
 
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florida-sparkey

Senior Member
Location
Pinellas Park, Florida
Occupation
Master Electrician
Ingenieur said:
mca ~ 30
ocp ~40-45
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Please see below

tw1156 said:
The following link on Page 2 shows 1.5tons - 5 tons with their MCAs, MOCP's etc, for typical residential condensing units.https://www.master.ca/documents/regroupements/1York_YHJD_571209_ytg_c_1012.pdf
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Thank you,
Looks like 4 ton (part number ending in 48 for 48000 BTU)is RLA 19.4A compressor plus FLA 1.3A fan or total of 21A

If you do the math 19.4 @1.25= 24.25A plus 1.3 for the fan so 25.55A MCA
 
I

Ingenieur

Senior Member
Location
Earth
florida-sparkey said:
Please see below



Thank you,
Looks like 4 ton (part number ending in 48 for 48000 BTU)is RLA 19.4A compressor plus FLA 1.3A fan or total of 21A

If you do the math 19.4 @1.25= 24.25A plus 1.3 for the fan so 25.55A MCA
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I assumed based on tonnage you needed to cost estimate a ckt and ocpd size
 
tw1156

tw1156

Senior Member
Location
Texas
Besoeker, I see what you're getting at and why not just avoid asking for tons in the first place. Here's a typical Utility load form sheet that we have to fill out and provide the local POCO. See Sheet (2) for the typical equipment they're interested in for typical commercial development. http://www.oncor.com/EN/Documents/A...mer Requirements - Commercial Industrial.pdf

The Utility is interested in Tons / SEER rating info, but ours also is interested in the kW output.
 
T

topgone

Senior Member
gadfly56 said:
The calculator was simply converting the 12,000 BTU/hr to raw kW. It doesn't take into account that with a little thermodynamic jujitsu you can move heat from point A to point B without expending as much energy in work as heat you are moving. The ratio is call the "coefficient of performance" or COP. Heat pumps or air conditioning units in cooling can operate at COP's of 5-7 or higher. Their effectiveness drops as the relative humidity or outside air temperature increases.
Click to expand...

COP is used for heaters while EER is used for cooling unit computations. The minimum EER IIRC is 11 for 2015. (EER = COP x 3.41).
In the OPs case with EER =11, KW = (48,000)/11 = 4363.63 kW or Amps = (4363.63/(240) =18.2 amps! Very near!
 
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Ingenieur

Senior Member
Location
Earth
from the link
[FONT=&quot]EER = Output cooling in BTUh / input electrical energy in Wh
I guess we assume eer/cop/seer
Wh = 48000/11 = 4360
at 240 ~ 18.2a[/FONT]
 
Besoeker

Besoeker

Senior Member
Location
UK
tw1156 said:
Besoeker, I see what you're getting at and why not just avoid asking for tons in the first place. Here's a typical Utility load form sheet that we have to fill out and provide the local POCO. See Sheet (2) for the typical equipment they're interested in for typical commercial development. http://www.oncor.com/EN/Documents/A...mer Requirements - Commercial Industrial.pdf

The Utility is interested in Tons / SEER rating info, but ours also is interested in the kW output.
Click to expand...
Much appreciated, thank you. But it still leaves the question of why they need the rating in tons. The POCO need to size the supply in Volts and Amps.
 
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