Cooling load per KW

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NE (9.06 miles @5.9 Degrees from Winged Horses)
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EC - retired
Small machine shop.

HVAC contractor asked what/how to calculate the cooling load for each machine should be. Supplier of the milling equipment suggested that normal operating load would be 30-50%. All have VFDs. A DuraPulse VFD for 230v 3PH 40HP is rated at 30KW. Is it to simple to just use 15KW for the cooling load?

Using that basis I come up with about 83KW of cooling load, 283K btu, or a bit shy of 24 Tons of cooling just for the machinery in the shop.
 

gar

Senior Member
Location
Ann Arbor, Michigan
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EE
130307-1403 EST

ptonsparky:

Possibly. But maybe there is a better way.

Has this shop been in operation for some time with the type of machine load that would be normal, and such that you can study the electric bills? Look at kWh used per month. Estimate the number production hours, assume no energy is used when not in production, and from this calculate average power when in production. Add a safety factor. How does this compare with the machine builders estimate?

It would be better if you could see the power profile during production for some period of time. If a smart meter is installed this may be possible. Otherwise if time is available you could monitor power use.

.

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Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
This will be a new building but the existing is close enough. Studying that would gives us the total cooling load for the existing building.

I also came across this in a CE paper for Cooling Load Calculations. (This is what electricians do, right?)
Q = 2545 * (P / Eff) * FUM * FLM

Came up to about 33 Ton for 24 hour operation at half load with .8 efficiency motors.

 

Npstewart

Senior Member
I do cooling load calculations on a regular basis. I use a program which has a list of HP & associated heat gain. Attached is a screen shot of the list in the program.
 

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Npstewart

Senior Member
How do I attach this as a downloadable image??
 

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Smart $

Esteemed Member
Location
Ohio
How do I attach this as a downloadable image??
Doesn't appear to be attaching issue. Its too low resolution of an image (size and clarity). If it is smaller and grainier than original, it may be the 'forum' automatically downsampled the image. It's one of the reasons I don't use the attach image feature of the forum. I upload to Photobucket, then insert an image link.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
130307-2042 EST

In my son's shop there are about 150 nameplate HP or more and I estimate that his average load is about 8 kW. Big disparity. Not all machines are on at the same time and some have a very low duty cycle. A wide belt sander, 20 or 25 HP, is not even used 1% of the time. The 20 HP scroll air compressor is only used when a lot of air is required, otherwise the reciprocating 5 HP compressor is run. Seldom are any of the machines run near full load.

.
 

topgone

Senior Member
Small machine shop.

HVAC contractor asked what/how to calculate the cooling load for each machine should be. Supplier of the milling equipment suggested that normal operating load would be 30-50%. All have VFDs. A DuraPulse VFD for 230v 3PH 40HP is rated at 30KW. Is it to simple to just use 15KW for the cooling load?

Using that basis I come up with about 83KW of cooling load, 283K btu, or a bit shy of 24 Tons of cooling just for the machinery in the shop.

Just estimate the amount of heat your equipment is bound to emit/lose. Bluntly, if you know your equipment efficiency, you'll know the power (watts) you lost. That loss will ultimately be coming out as heat from your equipment; watts---> BTU/hr. Better still, refer to OEM datasheets for a more accurate figure. Or use Google like this link.
 

GoldDigger

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Location
Placerville, CA, USA
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Retired PV System Designer
Just estimate the amount of heat your equipment is bound to emit/lose. Bluntly, if you know your equipment efficiency, you'll know the power (watts) you lost. That loss will ultimately be coming out as heat from your equipment; watts---> BTU/hr. Better still, refer to OEM datasheets for a more accurate figure. Or use Google like this link.

Not quite that simple. If the equipment is totally efficient, but is machining metal, running a conveyor belt on a level, or anything else that does not store the energy sent to the machine, that energy will eventually end up as heat somewhere. If it is an A/C, then the total wattage of its own compressor may end up outside.
If a motor is compressing air into a tank that will be shipped out of the building, some of the energy will not end up as heat in the building, but just compressing air heats it, so some of it will.

If you lift a bunch of gravel to the top of a chute and then dump it from the top of the conveyor onto a pile outside the building, then the energy that went into lifting the gravel will not produce heat in the building.

If you charge forklift batteries, most of the energy input will be stored in the batteries, but it will become heat when you use the forklift.

The only conservative estimate will be that, except for A/Cs, the entire electrical load will end up as heat somewhere inside.
 
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broadgage

Senior Member
Location
London, England
Yes, I would agree that with most common loads, that the entire electrical consumption will end up as heat. There are some minor and relatively rare exceptions as posted above.

One caveat is with wood working machinery, this is commonly fitted with powered dust extraction equipment.
If this is arranged so as to also extract the heated air from the drive motor, then the cooling demand will be reduced.
OTOH, if the extraction flow is substantial, then a significant volume of the conditioned air from the shop will be extracted and replaced with warmer outside air thereby adding to the cooling demand.

If compressed air is much used, then the heat output from the compressor and cooling coil is significant, it can be a paying proposition to relocate compressors to a well ventilated shed or outhouse, away from the conditioned space.
 

topgone

Senior Member
Yes, I would agree that with most common loads, that the entire electrical consumption will end up as heat. There are some minor and relatively rare exceptions as posted above.

One caveat is with wood working machinery, this is commonly fitted with powered dust extraction equipment.
If this is arranged so as to also extract the heated air from the drive motor, then the cooling demand will be reduced.
OTOH, if the extraction flow is substantial, then a significant volume of the conditioned air from the shop will be extracted and replaced with warmer outside air thereby adding to the cooling demand.

If compressed air is much used, then the heat output from the compressor and cooling coil is significant, it can be a paying proposition to relocate compressors to a well ventilated shed or outhouse, away from the conditioned space.

Agree. But when you do the heat load calculations needed for sizing of your heat pumps, you will be considering the worst-case scenario and sometimes overdo a bit. It's easier to tweak down your cooling equipment parameters than to pull out the whole thing because it's dang too small for the job, or is it the other way around?
 

GoldDigger

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Location
Placerville, CA, USA
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Retired PV System Designer
It's easier to tweak down your cooling equipment parameters than to pull out the whole thing because it's dang too small for the job, or is it the other way around?
:) Right!

If the equipment (either heat pump or straight A/C) is too oversized (50% or more?) it will not dehumidify properly and may have problems with icing. But if it is only a little oversized it can just run less frequently. A multi-stage unit can run well at 50% load, but you have spent more money that you needed to.
If it is undersized, then yes, you either add a whole new parallel system or you upsize not just the compressor and coils but the air handler and duct work. Maybe... :)
 
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