HVAC Load

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gulkhan123

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Q:Can we assume the Killowatt converted from BTU/hr to be the electrical load of the HVAC Equipment? So that we may use it to calculate the electrical loads of a building.
It is said that the killowatt calculated from btu/hr is actually mechanical(heAting/cooling) KW and not the electrical KW.

If it is true, then we need to know the electrical load(consumption) of the HVAC Equipment from the manufacturer/catalog. Am I right?

Pls. if somebody may clarify.

gk
 
Hvac - Kw Load Calc

Hvac - Kw Load Calc

If the HVAC tons are converted to Btu/Hr (tons x 12,000) and then divided by the unit's EER (typically 8-10 for commercial equipment), one will get a KW that is usually conservative. A seasoned mentor of mine taught me this trick and it has worked the majority of the time for capacity planning,

Once the mechanical designers know there exact load and have selected equipment, one should confirm the actual load with the manufacturer's data sheets.

DPW
 
DPW said:
If the HVAC tons are converted to Btu/Hr (tons x 12,000) and then divided by the unit's EER (typically 8-10 for commercial equipment), one will get a KW that is usually conservative. A seasoned mentor of mine taught me this trick and it has worked the majority of the time for capacity planning,

Once the mechanical designers know there exact load and have selected equipment, one should confirm the actual load with the manufacturer's data sheets.

DPW
Click to expand...

Dear DPW,

Thanks. What does EER mean? My question is - can we consider the kw obtained by converting btu/Hr as the electrical load in kw?
gk
 
gulkhan123 said:
My question is - can we consider the kw obtained by converting btu/Hr as the electrical load in kw?
gk
Click to expand...

Please clarify your equation. If you're saying a 4 TON AC, which is 48000 btu/hr, divided by btu per kw (48000/3412) is 14 KW, then you'll be oversized by a factor of 2 to 3. So yes you'd be plenty conservative using this conversion, but wasting quite a bit of service capacity.

The EER is a rating from the manufacturer. It is basically the BTU/3.412/COP. The COP is Coefficient of Performance in heat pumps and is basically the factor you save by pumping heat rathing than converting it directly. It is easier to understand in heating mode -- a baseboard heater of 12K BTU uses 3 times the electricity of a 12K BTU COP=3 heat pump.

Typical COP values range from 2 to 3 depending on ambient temperature and how hot or cold you want the interior of the building compared to what they designed it to.

You could probably take the BTU rating and divide by 6800 and at least be closer.
 
Eelctrial load(KW) of HVAC

Eelctrial load(KW) of HVAC

suemarkp said:
Please clarify your equation. If you're saying a 4 TON AC, which is 48000 btu/hr, divided by btu per kw (48000/3412) is 14 KW, then you'll be oversized by a factor of 2 to 3. So yes you'd be plenty conservative using this conversion, but wasting quite a bit of service capacity.

The EER is a rating from the manufacturer. It is basically the BTU/3.412/COP. The COP is Coefficient of Performance in heat pumps and is basically the factor you save by pumping heat rathing than converting it directly. It is easier to understand in heating mode -- a baseboard heater of 12K BTU uses 3 times the electricity of a 12K BTU COP=3 heat pump.

Typical COP values range from 2 to 3 depending on ambient temperature and how hot or cold you want the interior of the building compared to what they designed it to.

You could probably take the BTU rating and divide by 6800 and at least be closer.
Click to expand...

Mark,

Do you mean the electrical load of the HVAC equipt will be= (BTU/Hr)/3412/2 KW. Is this right?

ghk
 
One horsepower per ton is the rule of thumb for comfort cooling temperatures with typical refrigerants (22, 134a or 410a).
 
gulkhan123 said:
Do you mean the electrical load of the HVAC equipt will be= (BTU/Hr)/3412/2 KW. Is this right?
Click to expand...

It would be if your COP is 2 and there is nothing unusual (like a giant fan, some fancy controls, or they use chilled water). Do you have nothing to go on -- no nameplate to look at (MCA or FLA/RLA), no manufacturers spec which lists EER or circuit ampacity?

I'd try to find data for at least one unit as a sanity check.
 
suemarkp said:
It would be if your COP is 2 and there is nothing unusual (like a giant fan, some fancy controls, or they use chilled water). Do you have nothing to go on -- no nameplate to look at (MCA or FLA/RLA), no manufacturers spec which lists EER or circuit ampacity?

I'd try to find data for at least one unit as a sanity check.
Click to expand...

Mark,

Thanks. Actually sometimes we need to estimate the electrical load of a building when we know the HVAC load in BTU/Hr,. At that time we need to estimate the electrical load of the HVAC in addition to other loads. We dont have any nameplate or say catlog etc. to determine the electrical load of the HVAC equipment. So I will use the above rule you mentioned.

I will appreciate for the data you said will try to find.

gk
 
HVAC load

HVAC load

mdshunk said:
One horsepower per ton is the rule of thumb for comfort cooling temperatures with typical refrigerants (22, 134a or 410a).
Click to expand...

Marck,

Thanks. That means the HP we get from the Tons can be onverted to KW (1 HP= .746 KW), and this KW obtained can be assumed as electrical load of the HVAC Equipment.

Pls confirm if I am right.

gk
 
Hvac Load

Hvac Load

Hi MARK

Did ou find the data you mentioned below?

Will appreciate great.

gk



suemarkp said:
It would be if your COP is 2 and there is nothing unusual (like a giant fan, some fancy controls, or they use chilled water). Do you have nothing to go on -- no nameplate to look at (MCA or FLA/RLA), no manufacturers spec which lists EER or circuit ampacity?

I'd try to find data for at least one unit as a sanity check.
Click to expand...
 
No. It was a note for you to try and find at least one respreentative unit and see if the nameplate is close to the estimating factor. There are too many types, sizes, and styles to generalize. Do you at least have a clue as to the approximate HVAC size -- are we talking 2 ton units, or 50 ton? Once you know the approximate size and the type used (chilled water, R-134a, etc), look at a few HVAC manufacturers catalogs and see what data they list.
 
A single phase A/C unit needs 1 HP compressor motor per ton.

But, you can't convert 12000 BTU's per hour into motor amperage because most motors are only about 65% efficient.

One Horse Power is 746 watts. 746 watts x 165% eff = 1230.9 watts needed to produce 1 HP of work. A 4 Ton A/C unit would need a 4 HP compressor motor.

4 x 746 x 165% = 4923.6 watts and 4923.6 / 240 volts = 20.52 amps for the compressor motor. Then we have the fan motor.

I believe most condenser fan motors are about 1/3 - 1/2 HP on a one ton. So I think a 4 ton would have about a 1 1/2 HP fan motor (I am guessing).

1.5 x 746 x 165% = 1846.35 watts and 1846.35 / 208 = 7.69 amps for the fan motor.

25.52 + 7.69 = 28.21 calculated full load amps for a 4 Ton A/C unit.

The nameplate would have something like 35 minimum amps and 50 maximum.
 
The rule of thumb I use is that over the long term an A/C unit moves three times as much heat as it consumes electricity. So an A/C consuming 10KW is moving 30KW of heat. Makes it easy to calculate running loads for data centres :)
 
suemarkp said:
Here are samples of 4 different residential 4 ton AC's that are the current top of the line models:

Goodman SSX16 25.9 MCA ( http://www.goodmanmfg.com/Portals/0/pdf/SS-SSX16.pdf )
Bryant 180A 28.7 MCA ( http://www.docs.hvacpartners.com/idc/groups/public/documents/techlit/pds180a.24.2.pdf )
York CZH 29.2 MCA ( http://www.yorkupg.com/PDFFiles/350154-YTG-A-0108.pdf )
Click to expand...

Dear Mark,
Thanks. I checked the catlog you indicated. They are helpful as having detail electrical data(min.ckt amps, max.ocp ratings etc..). Actually my query was that if we know only the BTU/Hr, what wil be the electrical load(consumption)?

I think the thumb rule given above by Debuckley can be used to find out electrical load.
What is your opinion?
gk
 
Minuteman said:
A single phase A/C unit needs 1 HP compressor motor per ton.

But, you can't convert 12000 BTU's per hour into motor amperage because most motors are only about 65% efficient.

One Horse Power is 746 watts. 746 watts x 165% eff = 1230.9 watts needed to produce 1 HP of work. A 4 Ton A/C unit would need a 4 HP compressor motor.

4 x 746 x 165% = 4923.6 watts and 4923.6 / 240 volts = 20.52 amps for the compressor motor. Then we have the fan motor.

I believe most condenser fan motors are about 1/3 - 1/2 HP on a one ton. So I think a 4 ton would have about a 1 1/2 HP fan motor (I am guessing).

1.5 x 746 x 165% = 1846.35 watts and 1846.35 / 208 = 7.69 amps for the fan motor.

25.52 + 7.69 = 28.21 calculated full load amps for a 4 Ton A/C unit.

The nameplate would have something like 35 minimum amps and 50 maximum.
Click to expand...

Dear Michael,
Thank you for your detail calculation. It is ok if we have the electrical data. My query was that if we have only the BTU/Hr, how to calculate electrical load(consumption)?
Regards
gk
 
There are no rules of thumb that will work universally. I can give you one for a 4 TON, but it won't be right for a 2 Ton or a 10 Ton. Just look at some of the examples I gave you.

For the Goodman units, MCA is:
17.4 for 2 Ton
25.9 for 4 ton
33.0 for 5 ton
===> 6.5 amps per ton (12000 BTU) kind of works, but gives too small of a number for the 2 ton unit.

For the Bryant, MCA is:
13.9 for 2 ton
28.7 for 4 ton
34.8 for 5 ton
===>7 amps per ton works well for these.

For the York units, MCA is:
15.6 for 2 ton
29.2 for 4 ton
34.8 for 5 ton
===> About 7 amps per ton works for these.

If I had data for 8, 10, and 20 ton units, I think the errors would be even worse.

The equation Minuteman gave you was related to BTU. He just started with tons (12000 BTU) and said figure 1 HP per ton. Then add some more for the fan. You're asking for a single magic equation to handle all cases. You're not going to find one. But maybe 8 amps per ton (at 240V single phase) would be OK for estimating 2 and 3 ton units and 7 amps per ton for the larger ones.

I think dbuckleys rule is too generous. I earlier said use a COP of 2. He suggests 3. Perhaps in a few more years a 3:1 ratio will be more appropriate.
 
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gulkhan123 said:
Dear Michael,
Thank you for your detail calculation. It is ok if we have the electrical data. My query was that if we have only the BTU/Hr, how to calculate electrical load(consumption)?
Regards
gk
Click to expand...
Dear GK,

I'm sorry, I thought that you knew that 12000 BTU/Hr = 1 ton. I also thought that you could interpolate from my example how to find your own answer. If you give me your BTU/HR, I would be able to help you more. Please feel free to post your units info here or send me a PM.


Oh, I agree totally with Mark, but I tend to use about 10-amps per ton for my rule-of-thumb.
 
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The answer to your question about being able to convert from tonage to electric load is NO. You need to know the type of HVAC. If it's chilled water your electric load will be much less than converting from tons to btu/hr (12,000 btu/hr per ton) to watts (3.41 btu/hr per watt). If your AC portion is chilled water you only need a fan blowing across a coil. Look for the info off the unit.
 
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