gulkhan123
Member
Q: Dividing BTUH by 3413 gives to KW. Is this mechanical or electrical KW? I mean is this the electrical KW the machine will consume?
Hvac electrical load
- Thread starter gulkhan123
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KW or Btu/h?
KW or Btu/h?
This is cooling capacity of the HVAC equipment not electrical load. This is S.I unit for Cooling capacity just as FPS unit which is BTulh.
The same unit is used in calculating electrical load KW i.e. we usually not use BTu/h for electrical loads as we don't have to remove heat from any space but we have to supply power to the equipment. Although electrical loads cause heating effect too but this effect is not significant here. So we use KW unit for Electrical load.
KW or Btu/h?
This is cooling capacity of the HVAC equipment not electrical load. This is S.I unit for Cooling capacity just as FPS unit which is BTulh.
The same unit is used in calculating electrical load KW i.e. we usually not use BTu/h for electrical loads as we don't have to remove heat from any space but we have to supply power to the equipment. Although electrical loads cause heating effect too but this effect is not significant here. So we use KW unit for Electrical load.
gulkhan123
Member
Thanks. Thats fine. Actually what we are looking for is the electrical power consumption required for the given BTUH of a HVAC unit where we dont have any other data sheet available. There must be some relationship between the two.
Hameedulla-Ekhlas
Senior Member
- Location
- AFG
please check its data sheet and you will find the required kW
gulkhan123
Member
Dear Hameed
The problem is that the data sheet is not available.then what will u do?
The problem is that the data sheet is not available.then what will u do?
shamsdebout
Senior Member
- Location
- Macon,GA
I found this excerpt from the IEEE Gray Book, I hope it helps.
With a known heat loss, the electrical load in kW can be obtained by dividing the estimated heat loss (Btu/hour) by
3413 since there are 3413 Btu in 1. kWh of electricity. Usually, it is necessary to use a demand factor of 100% for
electrical heating loads.
Hameedulla-Ekhlas
Senior Member
- Location
- AFG
I had faced same problem like you but I found a solution for this. It is night here 2moro in the morning I will go Office I have the solution I will inform you.
skeshesh
Senior Member
- Location
- Los Angeles, Ca
Shamsdebout's method is what I've also used in the past. Having said that it is not exactly accurate, especially with packaged heating products that are popular nowadays because other factors than a simple conversion between electrical rating to mechanical output effect the BTU/hr rating. But I guess if you have no other way of approaching this you got no choice right?
By the way, can you provide a bit more detail about what kinda unit this is? is it just a small standalone or does it serve a building, etc.
By the way, can you provide a bit more detail about what kinda unit this is? is it just a small standalone or does it serve a building, etc.
gulkhan123
Member
A electric nickel chrome heater element produces 3.4 BTU/Hour
The very efficient mechanical refrigeration compressor produces 8.533 BTU/Hour+/-
I had a 4 ton unit 10 SEER that pulled 28 amps. That's 28x240 = 6720 kva /4ton = 1.7 kva per ton
One ton of frig.=12000 BTU/hour
1 watt=8.533 BTU/hour compressor
4 tons x 12,000=48,000 BTU/hour
48,000/8.533=5,625 watts 1.4 kw per ton 1.75 kva per ton
5,625 w/240 volts=23 amps more efficient machine
If you include the SEER, then you can do it another way. SEER is defined as the heat transferred per unit of energy, that is in BTU/watthour.
P = AC load/SEER
For your 4 ton unit and SEER = 10,
P = (48,000 BTU/hour)/(10 BTU/watthour) = 4,800 watts,
then throwing in PF,
I = P/(VxPF) = 4800W/(240V x 0.8) = 25A 1.2 kw per ton or 1.5 kva per ton
This is comparable to the other result.
The very efficient mechanical refrigeration compressor produces 8.533 BTU/Hour+/-
I had a 4 ton unit 10 SEER that pulled 28 amps. That's 28x240 = 6720 kva /4ton = 1.7 kva per ton
One ton of frig.=12000 BTU/hour
1 watt=8.533 BTU/hour compressor
4 tons x 12,000=48,000 BTU/hour
48,000/8.533=5,625 watts 1.4 kw per ton 1.75 kva per ton
5,625 w/240 volts=23 amps more efficient machine
If you include the SEER, then you can do it another way. SEER is defined as the heat transferred per unit of energy, that is in BTU/watthour.
P = AC load/SEER
For your 4 ton unit and SEER = 10,
P = (48,000 BTU/hour)/(10 BTU/watthour) = 4,800 watts,
then throwing in PF,
I = P/(VxPF) = 4800W/(240V x 0.8) = 25A 1.2 kw per ton or 1.5 kva per ton
This is comparable to the other result.
gulkhan123
Member
Thanks Bob. Excellent calculation. But I just want to know what do we conclude from this? I mean what is the relation between a HVAC unit(BTUH) and the electrical power it consumes? Or, in other words, if we know the BTUH of an HVAC unit, what would be the input elect. power consumption?thanks
Sorry did not see this post. Consumption is hard to define because you would have to know the insulation factor, area of the room(s), temp differential. But as a ball park figure use 33.33% duty cycle in a 24 hour period or 8 hours. So 4.8 Kw x 8 Hours = 38.4 Kwh per day usage. To figure the load you would need the PF so if it was say .8 then 4.8 KW / .8 = 6 KVA.
Actually it is correct that one ton of air-conditioning is equal to 12,000 btu/hr. However the other question trying to convert a one ton A/C unit into 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 about 1.4 KW/hr.
Check http://www.engineeringtoolbox.com/cop-eer-d_409.h
gulkhan123
Member
Thanks Bob. Excellent calculation. So what do we conclide from this? What we are looking for is that if we know BTUH of an HVAC unit, what will be its power consumption, if we dont know SEER or we dont have any data sheet. In other words, converting BTUH value to KW(BTUH/3413) gives us mechanical KW and not electrical KW. If it is right, what would be electrical input KW?
Thanks.
gulkhan123
Member
Thanks Bob. Excellent calculation. So what do we conclude from this? I mean if we know BTUH of a HVAC unit but we dont know SEER, then how to find the electrical power consumption. In other words, converting BTUH to KW(BTUH/3413) will provide only mechanical KW but not electrical KW. Is that right? So what will be the electrical KW? Any relation?
thanks
thanks
kwired
Electron manager
- Location
- NE Nebraska
if you are trying to do a service or feeder load calculation I would assume a low SEER to make sure I allow enough capacity. Not sure if there are any new units being made anymore with a SEER below 12 or 13 pretty sure there are none below 10. If not a new unit sorry better figure for higher VA.
gulkhan123
Member
Thanks KWIRED. Agree to use lower SEER to get higher VA. Now, is EER is the same SEER?And one said EER will be from 6-14 . While other say SEER is never below 10. What is correct?
kwired
Electron manager
- Location
- NE Nebraska
according to this link since 2006 all new units needed to be 13 SEER or higher. I believe SEER is for seasonal energy efficiency rating and will not work directly for an electric load calculation. I know that most if not all of the higher SEER units will use a multistage compressor which means if the required demand is low they will use less energy by switching a valve to regulate refrigerant flow at a different rate, this will then put less load on the compressor. So for a maximum load you will need to know what the unit uses at the max demand stage which is probably EER. A good HVAC guy is probably a lot of help here, the guys that just do installs are usually not very sharp on sizing of units and other calculations.
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