# Thread: 5 Ton AC load

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Originally Posted by Minuteman
(cut)1hp for 1ton split A/C single phase system, less for a 3 phase.
So, you are saying that a 3ph horsepower is more lb-ft/sec than a 1ph horsepower?:confused:

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

carl

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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.

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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.

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Those ratings wouldn't be unusual for a unit with electric heat, a humidifier, or a energy recycler.

Steve

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Check out this website:

http://www.unitconversion.org/power/...onversion.html

It looks like 1 ton is approx. equal to 3.5kW

1 ton = 12,000 BTU/hr
12,000 BTU/hr = 3.514kW

Therefore 1 ton = 3.514kW

and 5 tons = 17.57kW

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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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Originally Posted by suemarkp
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.
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.

8. 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.

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Originally Posted by topgone
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.
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...s652a.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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Originally Posted by suemarkp
. . . I was confusing COP and watts which requires dividing again by 3.43 to find.
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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