# Cooling Loads from Electrical Equipment

| Posted in General Questions on

Hello –

I am doing a Manual J on a commercial building I operate – a small food processing facility in Rhode Island, climate zone 5. The building envelope is pretty straightforward and is outputting a pretty low cooling and heating load, which is great. However, I know I need to take into account the heat produced by our equipment, especially for getting cooling load correct.

We, for better or worse (I think for the better), utilize off-the-shelf air conditioning units controlled by a Coolbot to cool our cold spaces (walk-in coolers). Two of the air conditioners are fully inside the building envelope, dumping their hot air into an adjacent room that we we need to keep at 95 degrees for a part of our food processing. (We use an exhaust fan on a thermostat to regulate that hot room temperature).

Since the A/Cs are inside the building envelope, they are just moving inside air around between 2 rooms. But the mechanical operation of the A/C’s obviously causes waste heat to be produced (as opposed to the heat moved out of the walk-in cooler and into the hot room).

What is a good way to estimate the various variables (particularly the Efficiency) in the internal load calc, which I may or may not have right:

Heat Load = Power Rating x (1 – Efficiency) x Duty Cycle

Power rating for each A/C is 3300 Watts.
Duty cycle, roughly observed, is about 25% run time.
I really don’t know how to estimate the efficiency of an A/C’s components…

Andrew

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### Replies

1. | | #1

It sounds like almost all the heat from the coolers is kept in the building. I would simply measure the watt hours used and assume it all becomes heat inside the conditioned space that the new system will need to remove. Every kWh =3412 BTUs

Walta

1. | | #2

Right, that makes a lot of sense put that way.

Thanks Walta

1. | | #4

I am with Walta in this that the electric energy is what the A/C has to take care of - IF you have a tight house with no air exchange through the building walls.
Mind this is for Manual J only, if you have to do Manual D etc., then it is not that simple anymore.

edit: if you have latent loads to deal with then it is different again..

1. | | #5

Yeah, this makes sense to me now. All the work is being done inside the building envelope. I have a decent estimate for electricity usage for the units so getting from kw/hr to btu/hr is straightforward.

I was getting hung up on the efficiency component, confusing the equipment's cooling efficiency (~2.5 COP) for the general work efficiency needed for that heating load calc. Walta's comment cleared that up.

Latent load analysis and deciding if ductwork is needed or affordable is next...

Thanks!

2. | | #3

So you've got window units in the wall between the cooler and the hot room? Theoretically the heat generated by those units either stays inside the hot room or is exhausted outdoors.

The loads for the main space are the coolers and the hot room. That could be calculated but if the spaces are accessed frequently that would introduce a variable you'd have to guesstimate.

Where does the make up air for the hot room exhaust come from? If it's the main space that could be a significant load.

1. | | #6

Yes, window units are in a through-wall set up. The disadvantages of this are obvious, but the advantages - being able to repair my cooling system by myself at 11pm on a Friday night by swapping in a new A/C unit, without losing any temp sensitive product - have made the tradeoffs acceptable so far.

The make up air alternates based on the season. During the cooling season, the hot room make up air comes from an open window in an adjacent room, with the exhaust venting outside the building. In the heating season, the hot air exhausts to the storage loft above, and makeup air comes from inside the envelope.

Other internal heat loads are from the people and opening and closing of doors, lighting and small plug loads, etc. But I think A/C run time data should catch that real-world energy load.

What I am aiming to do with this project is provide some cooling to an adjacent un-conditioned (and not temp sensitive, except worker comfort) space, and to manage internal dew point too.

3. Expert Member
| | #7

For the A/Cs moving heat from the cold room to the hot room, you have two energy sources contributing to the heat in the hot room: the energy MOVED by the A/C (which is acting as a heat pump here), and the electric power consumed by the A/C to move that heat. Assuming the A/C is in good working order, I would take the rated capacity of the A/C in BTUs and add that to the BTU equivalent of the electricity used by the A/C while it's running, then use the sum of those two numbers as the amount of heat put into the hot room. That will get you very close.

You do not need to estimate the efficiency of anything here. ALL energy that goes into that system within the building envelope STAYS within the building envelope. Every watt of electricity is converted into additional heat that goes into that hot room. ALL energy that goes into ANY electric device INSIDE the cold room contributes heat to the cold room, again the total of the electricity going in becomes heat. This means if you have a 100w light bulb, that actually uses 100w, it puts 100w of heat into that room, or about 341 BTU. It doesn't matter what kind of light you have, you just need to know how much electricity it uses. ALL of the electricity that goes into the room heats up the room.

This is a question that comes up often for me at work designing datacenter facilities, which we air condition heavily all the time. ALL electricity that goes into the room is counted as heat, since that's where it all ends up. We use only the sensible cooling capacity of the air conditioners to deal with the heat load of the equipment for that same reason. The ONLY energy that goes into the room that does NOT come out as heat is the teensy weensy little bit of energy that was shot out of the room as pulses of light on the fiber optic cables leaving the datacenter. That amount of energy might be a few watts total for a facility using 100s of kilowatts, so the absolute energy that leaves as pules of light is inconsequential to the facility's overall cooling load.

For some more example, I had a customer once argue with me that some of that input energy went into spinning hard drive platters and fan blades. But think about that for a moment: if you spin something with a motor, and then turn off the motor, what happens? It slows down. The reason it slows down is because the energy stored in the spinning platter is lost to the bearings in the form of heat, since the bearings aren't perfect, and they cause some drag. That drag heats up the bearings, and the source of the energy is the spinning object. When the system is running at constant speed, the motor is putting in enough energy to overcome the losses in the bearings, so that input energy is all ultimately being lost as heat! In a closed space, this is almost always what happens: ALL energy that goes in heats up the space. Efficiencies of the stuff inside the space don't matter, because both their useful output and their energy losses all contribute to heating the interior of that closed space.

Bill

1. | | #8

Thanks, Bill!