Electrical Equipment Heat Load Calculation

The answer to your questions is basically the difference between Input and Output in most cases.

That difference is in the friction of moving parts, the load being moved (water, air, goods), back EMF etc...

In an IT environment Most of the input comes back as waste heat...up to 80%.
From some perspective, there is a valid argument that IT equipment does no "Work", if you treat it as a machine.

In a factory or mechanical environment, the simplistic way to work it out would be: -

((Output/Input)/1)*Input=Waste Heat

Keep in mind that all figures should be in kW.

Regards,
Sapper

Of course there's a variation in how much electric power gets converted into a heat load. People do different things with electric power. You can even get over 100% heat loading if most of your electric usage is refrigeration of one kind or another.

Heat transferred from electrical equipment can be calculated as

H_eq = P_eq K₁ K₂ (2)

where

H_eq = heat transferred from electrical equipment (W)

P_eq = electrical power consumption (W)

K₁ = load coefficient

K₂ = running time coefficient

When machines run heat may be transferred to the room from the motor and/or the machine.

H_m = heat transferred from the machine to the room (W)

P_m = electrical motor power consumption (W)

h_m = motor efficiency

If the motor and the machine is in the room - the heat transferred can be calculated as

H_m = P_m / h_m (3c)

In this situation the total power is transferred as heat to the room.

http://www.engineeringtoolbox.com/indoor-temperature-humidity-d_114.html

Motor control centres, power distribution boards, electrical control panels, PLC panels are generally installed in air conditioned rooms in tropical/ subtropical countries. What is the basis of calculating heat loads for these equipment? This heat load needs to be considered by HVAC engineer for designing the air conditioning system.

Thanks.

The answer is, almost one hundred per cent.

Even a fan adds to the heat load of a confined space. There is resistive heating in the motor and its electrical supply wires, the heat produced by bearing friction, heat produced by flexing the belt if it is belt-driven, and the heat produced by the work of moving the air. If an electrically powered device produces sound or light, moves anything, heats anything, cools anything, all of the electrical energy that produces these desirable outcomes will eventually be converted into heat.

The long and the short of it is, that all of the electric power used by anything at all is eventually converted to heat, and will add to the heat load in a confined space. There is an exception in the case of a device that produces emissions that are not absorbed and re-radiated as heat. A radio transmitter or an X-ray machine or something else that produces radiation to which part of the enclosure is completely transparent would be an example. In the event, that part of the electrical energy would escape and not add to the heat load, but in an enclosure like a server room the part that escaped as radiation would be negligible.

Could add that for a pump, the fluid power (Q x ΔP) is not released into the enclosure (OP is probably thinking about temperature rise in an enclosure).

Also the extra power needed because the pump efficiency is < 100% causes temperature rise of the liquid and is not (usually) released into the enclosure.

More precise answer depends on the system details.

IEEE publishes tables that show the heat load characteristics of many kinds of electrical equipment. Obtain and consult those tables, and you'll be off to a good start on your heat load calcs.

Ooops not so I'm afraid. The answer is almost always more than 100%.

Furore! Uproar! Not possible! etc etc

The reason is that all equipment is rated according to output power. All equipment is inefficient and this inefficiency must be added to get input power. Eventually, all the input power is manifested as heat, and usually (but not always) entirely in the air conditioned space. For example;

Motors (hard ones first)

For example, a 3 kW motor has 3kW output. Roughly a 3ph motor will be ~85% efficient, and single phase ~75%. This translates into:

1ph 3kW /0.75 = 4kW (waste = 1kW)

3ph 3kW /0.85 = 3.5kW (waste = 0.5kW)

(EASY) If all the work the motor does is in the air conditioned space (e.g. a fan) then all this power manifests as heat in the space (i.e. 4kW 1ph, 3.5kW 3ph).

(EASY) If the motor is elsewhere (e.g. outdoors) then all the heat manifests in that place.

(HARDER) If it is located in an a/c space, but does work on a medium that goes elsewhere (e.g. pumps water somewhere), (a) all the waste will appear in the a/c space (b) all the mechanical loss in the pump will appear as heat in the a/c space (c) all the hydraulic loss will appear as heat in the a/c space. The remainder is transferred to the medium. How much depends on the pump curve, which is available from the manufacturer. e.g.

Motor power = 3kW

Eff = 85 % = 500 W waste

Mech loss (depends on manufacturer) = 10% = 3000 x 0.1 = 300W (NB % of output)

Hyd loss (from pump curve) = 50% = 3000 x 0.5 = 1500W (NB % of output)

Power given to the water = 1500 W. Heat in a/c space = 2300W.

Lighting

All the power goes to heat + inefficiency. Typically I would use 5-10%. So a 4 x 14W fitting would be 4 x 14 x 1.1 = 62W. Note only a very small percentage is radiated as light, but this radiation is absorbed by surfaces and creates heat. (OK, some goes out the window and creates heat outside ).

Comms

All the power goes to heat + inefficiency. Typically I would use 20%. So a 300W network switch would be 300 x 1.2 = 360W. A very small percentage is conducted as signals along comms cables, and this creates heat somewhere else. But this is mW per Cat 5/6 cable. So all the power goes into cooking the silicon chips.

However, the latest switches may be power over Ethernet (POE). This allows the Cat 5/6 cable to provide up to 15W or for POE+ 25W, at the device end, and this creates heat somewhere else (e.g. on the office floor). This is useful because say a VOIP phone or CCTV camera does not need a mains dongle-thing. If the switch is UPS backed, even better, as your device will keep working on mains failure.

So lets take a 96-port Cisco beast with POE+. 96 x 25 = 2400 W which leaves the comms room and goes to the floor. This guy is powered by twin 6000 W power supplies operating in A+B plane mode i.e. each takes half the equipment load, so if one fails the other takes over. But that means they are not operating at max load and so their efficiencies drop. I have measured this to be 70%. So;

Input power = 6000 / 0.7 = 8500 W

Less power going outside the comms room = 2400 W

Power cooking the chips and lost as heat in room = 6100 W.

All true however you're making an assumption of what the OP means by the ambiguous phrase electrical equipment. If you consider the difficult to tally aggregate sum of each and every piece of electric load that gets plugged into the power grid and do a statistical analysis of what will be the maximum load combination that will likely plugged into the facility, then your analysis is correct albeit ludicrously difficult to implement. If instead one considers instead the electrical equipment service provided by the utility as the absolute maximum electric power possibly provided via the wires, then one starts with a very manageable, easy to calculate number. I suspect the easy to find number is what the OP is considering.

I don't agree with any part of this post.

It infers you can suspect what someone is asking, but not make an assumption.

The phrase "electrical equipment" is not ambiguous in any way. It refers to equipment which uses electric.

It may be difficult to tally aggregate sums etc but it is not ludicrous. It is a legal obligation. The US NEC, UK BS7671, Aust/NZ AS/NZS3000 et alia all mandate this process as law, and contain the necessary statistical information.

What is ludicrous is assuming that the electrical equipment service provided by the utility uses the maximum of its wired capacity. This may be easy but completely erroneous.

So if you don't agree with any part of my post, then you don't agree that you originally made truthful statements. Ignoring that paradox, I see that I have the NEC NFPA 70 code book on my shelf behind me in my work office. Will you please site where in the code book it says that all load equipment (light bulbs, motors, computers, stereo equipment, refrigerators, toasters, televisions) must be counted by law. I hope to get a new TV soon and want to not break the law.

Yes we all know (well most of us do) that the rating of a motor is the shaft output power. After all that's what you need to know to assess whether it will drive your machine. But that doesn't mean all the input electrical energy finishes up as heat in the enclosure.