5 Ton AC load

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cubdh

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Hi, I am curious to know what is the approximate Power consumption of a 5 Ton AC, doesnt matter if it is single or 3 phase.

I have seen wide ranges of power consumption from different electricians.

Is is accurate to say the power from a 5 Ton AC is about 7 kW to 9 kW, with 9 kW being the starting power and it levels off to abut 6 or 7 kW?????


Is this roughly accurate???
 
dereckbc

dereckbc

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Staff member
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There is really no correct answer to your question.

The question of trying to convert a one ton A/C unit into VA or kw/hr consumed is very dependant on the make and model of the air-conditioner, as well as its EER (energy Efficiency Rate). These range from 6-14 EER. Standard average one ton A/C unit consumes 1.335 KW/hr. But you can only use that as an estimating tool.
 
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nakulak

Senior Member
I am curious as to why you are curious about the average power consumption (care to explain ?) How does that help in engineering any specific job ? (I guess what I am trying to say is that it is easy to get into trouble using a number for a project which may or may not represent reality, when it is just as easy to get a spec sheet and specify an exact piece of equipment that will work and use that value, knowing that that equipment actually exists and can be used in the installation)
 
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zbittin

New member
I could see where have a general number could be beneficial from a Utility's perspective in sizing the service conductor and/or XFMR. Very often homeowners only know the tonnage of their units and do not have a spec sheet readily availble for the power company. Thus the company would have to use judgement in sizing for the full load of the unit as well as any possible flicker from starting. I use 1.5 KVA per ton of AC for full load calculations and 1.5hp per ton for starting calculations(assuming code G). These are of course very very general and I encourage you to do more research if possible.
 
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topgone

Senior Member
cubdh said:
Hi, I am curious to know what is the approximate Power consumption of a 5 Ton AC, doesnt matter if it is single or 3 phase.

I have seen wide ranges of power consumption from different electricians.

Is is accurate to say the power from a 5 Ton AC is about 7 kW to 9 kW, with 9 kW being the starting power and it levels off to abut 6 or 7 kW?????


Is this roughly accurate???
Click to expand...

Your guesstimate is wrong; as it should be!

Here it is:
1 ton of cooling = 12,000 BTUs of heat = 3.513725 kilowatt-hours
5 tons of cooling = 5 X 3.513725 = 60,000 BTUs = 17.57 kWH
 
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suemarkp

Senior Member
Location
Kent, WA
Occupation
Retired Engineer
topgone said:
Your guesstimate is wrong; as it should be!

Here it is:
1 ton of cooling = 12,000 BTUs of heat = 3.513725 kilowatt-hours
5 tons of cooling = 5 X 3.513725 = 60,000 BTUs = 17.57 kWH
Click to expand...

You're assuming there is a 1:1 conversion between cooling tons and applied energy. If an AC worked like a resistance heater, that would be a good assumption. But they don't work that way.

The EER is a factor that tells you how to divide the BTU usage to get the actual consumed power (but look out, the 3.412 BTU conversion factor is in there too). EERs vary by temperature and the design of the equipment (newer units being more efficient than old ones, mostly because they have larger coils). The KW value above needs to be divided by a factor between 2 and 4. Power use is roughly TONS * 12000 / EER. Note that this method is not a good way to derive amps or MCA -- need the nameplate or datasheet for that.
 
Minuteman

Minuteman

Senior Member
Location
The Sooner Nation
Marc and Mark are right. A rule of thumb is the best that you are going to get, without some name plate info. 1hp for 1ton split A/C single phase system, less for a 3 phase.
 
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Cold Fusion

Senior Member
Location
way north
Minuteman said:
(cut)1hp for 1ton split A/C single phase system, less for a 3 phase.
Click to expand...
So, you are saying that a 3ph horsepower is more lb-ft/sec than a 1ph horsepower?:confused::roll:

(just a humorous poke - I translated what you said to , "3ph motors are a bit more efficient than 1ph motors."):smile:

carl
 
steve66

steve66

Senior Member
Location
Illinois
Occupation
Engineer
I have always used 1.5 KVA / ton as a rough guestimate for a typical small unit.

I think I like suemarkp's formula better (if you happen to know the EER).

Add electric heat, humidification, or energy recovery and the KVA skyrockets. As others have said, using a guestimate can lead to trouble, so use with caution.

Steve
 
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cubdh

Member
Naw i wanted to know the rough estimate. I was guessing about 6-8 KVA for a 5 ton.

The reason for this was, i have recieved load sheets from electricians who have stated that the 5 Ton ac consumes 9.9 KVA x 3 = a ridiculous 29 KVA.

I have recieved a load sheet yesterday stating that a 5 Ton AC was 9.0 kVA x 2 = 18 kVA


So when i replied back and told the guy he is " double stating " his AC loads, i wanted to make sure what i was saying was right.

It makes a huge difference considering he was alloting eight 5 ton AC's 18 kVA each. That is about 45 kVA extra just on AC's


I know about the SEER ratings and stuff but i just wanted to make sure that
1 Ton is about 1.5 KVA and not something like 4 kVA

Thanks for the responses.
 
steve66

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Those ratings wouldn't be unusual for a unit with electric heat, a humidifier, or a energy recycler.

Steve
 
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suemarkp

Senior Member
Location
Kent, WA
Occupation
Retired Engineer
Again, this doesn't work for calculating power use of an air conditioner. See post #9. An air conditioner or heat pump may seem like a perpetual motion machine, but they are not converting energy, they move it.

What I think that website is trying to say is if you have electrical equipment which uses 10KW of power, you're going to need 2.8 tons of cooling to remove that heat.
 
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topgone

Senior Member
suemarkp said:
You're assuming there is a 1:1 conversion between cooling tons and applied energy. If an AC worked like a resistance heater, that would be a good assumption. But they don't work that way.

The EER is a factor that tells you how to divide the BTU usage to get the actual consumed power (but look out, the 3.412 BTU conversion factor is in there too). EERs vary by temperature and the design of the equipment (newer units being more efficient than old ones, mostly because they have larger coils). The KW value above needs to be divided by a factor between 2 and 4. Power use is roughly TONS * 12000 / EER. Note that this method is not a good way to derive amps or MCA -- need the nameplate or datasheet for that.
Click to expand...

EER is a measurement of the thermal efficiency (not electrical efficiency, so don't mixed it up with the electrical efficiency formula), while tons is a measurement of size. One ton of air conditioning is equivalent to 12,000 BTU. A BTU is a British Thermal Unit, and it is simply a measurement of heat. One BTU is roughly equivalent to the amount of heat generated by burning one wooden kitchen match. Simply put, this means that a one-ton air conditioner can remove 12,000 BTU from a space per hour.
Also, the EER rating simply refers to how much it will cost to operate the unit when it is running, and varies with every model or manufacturer. The size (tons) of the system is the determining factor on how well the A/C will cool your room.
The Energy Efficiency Ratio (EER for air conditioners, SEER for central air conditioning systems)tells the user how much energy is consumed compared to the heat energy removed from the space. See the formula.

BTU heat removed from a space​
EER= --------------------------------
Watts input to the equipment​

Back to the OP, he wants to know the correct power needs for a 5 ton A/C unit. IMO, 18kW is correct for a 5 ton ACU.
Rules of thumbs are based on experiences; say 2kVA per ton (10kVA on your case) can accommodate your 5 tons ACU if this 5-ton size is a design bigger than the actual cooling requirement of the installation. The temp settings will correct the cycle time of the refrigerant pump and the cooling mode selector will be selectable to a lower refrigeration pump/condenser fan speed; all leading to a smaller demand factor.
Hope this clears all doubts.
 
mdshunk

mdshunk

Senior Member
Location
Right here.
I still say put the calculators down and plug in one horsepower per ton and you're going to be damn close. If you need more accurate than that, get cut sheets or look at the dataplate.
 
S

suemarkp

Senior Member
Location
Kent, WA
Occupation
Retired Engineer
topgone said:
The Energy Efficiency Ratio (EER for air conditioners, SEER for central air conditioning systems)tells the user how much energy is consumed compared to the heat energy removed from the space. See the formula.

BTU heat removed from a space​
EER= --------------------------------
Watts input to the equipment​

Back to the OP, he wants to know the correct power needs for a 5 ton A/C unit. IMO, 18kW is correct for a 5 ton ACU.
Rules of thumbs are based on experiences; say 2kVA per ton (10kVA on your case) can accommodate your 5 tons ACU if this 5-ton size is a design bigger than the actual cooling requirement of the installation. The temp settings will correct the cycle time of the refrigerant pump and the cooling mode selector will be selectable to a lower refrigeration pump/condenser fan speed; all leading to a smaller demand factor.
Hope this clears all doubts.
Click to expand...

By your own calculation, if the EER is 10 and the AC is 5 ton (60000 BTU), then the power use will be 60000 / 10 = 6000 KW. I was confusing COP and watts which requires dividing again by 3.43 to find.

For some datapoints:
http://www.goodmanmfg.com/Portals/0/pdf/SS-SSX14.pdf
From page 3 for a 14 SEER Goodman air conditioner - a 5 ton unit draws 26.4 amps for the compressor and 1.6A for the fan. So the VA for this unit at 240V is 6720.

http://www.docs.hvacpartners.com/idc/groups/public/documents/techlit/pds652a.18.3.pdf
From page 5 for a 12 SEER Bryant heat pump - a 5 ton unit draws 27.6 amps for the compressor and 1.4A for the fan. So the VA for this unit at 240V is 6960.

This is nowhere near 18 KW but real close to the 1 HP per ton which MD uses.
 
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topgone

Senior Member
suemarkp said:
. . . I was confusing COP and watts which requires dividing again by 3.43 to find.
Click to expand...

Thanks suemarkp for the correcting the mess I've made. It's time to eat back posts!
Should have divided the thermal calc results by coefficient of performance.
18 kW / 3.43 = 5.248 kW (electrical). Very close to A/C unit power consumption rates.
I just don't feel comfy with rules of thumbs.
 
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