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Is there a general rule for calculating heat dissipation in electronic equipment if it's not listed in the specs?

I have a couple of projects coming I'm working on that require this. For one situation I need to provide the heat dissipated for some routers, switches, UPSs, and two-way radio repeaters I'm installing in leased rack space in a equipment room.

I also have a situation where I need to install a router and UPS in a storage cabinet in an RV type vehicle. In that case I think it's important to be reasonably accurate. As far as I can tell, this information isn't listed on spec sheets by Cisco, Motorola, or even APC, so I assume I need to calculate it myself some way.

Since the formula for power to heat is 1W = 3.41 BTU/hr my first pass at calculations was just to multiply the equipment power consumption by 3.41 to get BTU per hour. However, I guess that would only apply if the equipment were 100% efficient at being a heater. So my guess is that the correct way to calculate would be to use the power efficiency, η to calculate, where (1-η) x 3.41 x Watts = BTU/hr. For example if a 1000 Watt UPS has efficiency η=.9, then I would think its heat dissipated would be (1-.9) x 3.41 x 1000 = 341 BTU/hr. If this is correct, is there a rule of thumb for estimating the efficiency of certain types of equipment?

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    \$\begingroup\$ Your approach looks right but why convert to the quaint BTU/hr? Just calculate the dissipation in the cabinet (watts) and that's the amount of cooling power required (in watts). If you need to convert to energy then multiply by 3600 for kWh. \$\endgroup\$
    – Transistor
    Sep 2 at 20:05
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    \$\begingroup\$ @MissMulan There is no mention of any sort of system that would involve impedance matching in the question; why bring that up? \$\endgroup\$
    – Hearth
    Sep 2 at 20:45
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    \$\begingroup\$ @Miss Mulan - But you can have >50% efficiency in a power supply. 80%-90% efficiencies seem to be standard. And impedance matching doesn't apply to power systems. \$\endgroup\$
    – SteveSh
    Sep 2 at 21:28
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    \$\begingroup\$ @MissMulan Neither is anyone else. The maximum power transfer theorem is straightforward to derive when you've covered some calculus. The real trick is learning when it doesn't apply. Like here. \$\endgroup\$
    – Graham Nye
    Sep 2 at 21:48
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    \$\begingroup\$ @Transistor "why convert to the quaint BTU/hr?" Because air-conditioning equipment may be rated in BTU/hr rather than kW. (Unless it's rated in - rolls eyes - tons (i.e. of ice melted per 24 hrs).) \$\endgroup\$
    – Graham Nye
    Sep 2 at 21:57

2 Answers 2

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As far as I can tell, this information isn't listed on spec sheets

Oh but it is! The worst case is the power rating on the rating label. If a device can consume 100VA worst case, then that's how much heat it will dump, worst case. You can assume that VA=W and convert that to BTU/hr, horsepower, etc.

The only case in IT equipment where it's not true is PoE switches: they dump some of that power wherever the loads are. Other than that, all electrical devices in the racks of a data center are 100% efficient heaters.

All of the electricity you pump into a stand-alone data center gets dumped right out through the HVAC system. Thermodynamically speaking, the amount of "work" that the computers actually do by "computing" is so astronomically minuscule, that we don't need to think about it.

If you're nitpicking then yes, a few hundred watts will go out to the IP cameras on and around the building. But you're not gonna save any money by taking it into account in thermal calculations for a data center. For an office server closet - it may be important if there's lots of PoE equipment hanging off the switches. But you won't go wrong if you just go conservative and assume power into the closet = heat to be taken out of the closet.

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  • \$\begingroup\$ OK. I was looking around, and found this link below, where they were saying to estimate that 10% of the power is converted to heat. But reading closer it looks like they are talking a power distrobution system. \$\endgroup\$
    – Frank
    Sep 2 at 21:48
  • \$\begingroup\$ Here's the link I was referring to: instrumentationtools.com/… \$\endgroup\$
    – Frank
    Sep 2 at 21:56
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    \$\begingroup\$ This article is talking more about general electrical power equipment, transformers, switchgear, and so on. The important thing here is hat they are "on the way" to where the power is actually used. Those loads will very often be 100% heaters, the remaining 90% (except for spare change) but the switch-gear, transformers, etc, will steal only very little of the power. In your datacentre say you have a 1MW transformer.That might disipate 25kW of heat when under load, but its 1MW load (a load of racks) will then do the business of disipating the rest of it. \$\endgroup\$
    – Dannie
    Sep 2 at 22:53
  • \$\begingroup\$ +1 for electrical devices in the racks of a data center are 100% efficient heaters - And if you don't know all the ratings you can put a Kill-A-Watt on your whole rack. \$\endgroup\$ Sep 3 at 21:14
  • \$\begingroup\$ The question mentioned UPSes too, and they also transfer most of their power to other devices. Though if the UPS is in the same rack, it doesn't matter as much which device converts the power to heat. (But you could have a bigger UPS in another room even, or you could have an UPS in the rack feeding that PoE switch feeding some other devices around the building...) \$\endgroup\$
    – ilkkachu
    Sep 4 at 8:18
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Is there a general rule to calculate heat dissipation of electronic equipment

The rule is: analyse the output power, realize it's virtually zero and use the input power as the value that generates heat.

  • Does it emit light?
  • Does it transmit radio waves?
  • Does it emit sound?
  • Does it perform mechanical work?
  • Does it supply electrical power to something else?

If none of the above then, unfortunately, the efficiency might as well be assumed to be zero.

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    \$\begingroup\$ And even if they do, a rule of thumb that all the supplied energy ends up a heat us rarely far from the mark. \$\endgroup\$
    – Frog
    Sep 2 at 20:19
  • \$\begingroup\$ All energy is eventually converted to heat (photons), but the distinction is the power dissipated by the source versus that which goes elsewhere. \$\endgroup\$
    – PStechPaul
    Sep 2 at 22:08
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    \$\begingroup\$ @Andy aka. Ok. So in the case of the radio repeater, where power requirement is 400W,and transmit power is 100W; then 100 watts are delivered to the antenna, but the remaining 300W are converted to heat inside the equipment room. Makes sense. \$\endgroup\$
    – Frank
    Sep 2 at 22:34
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    \$\begingroup\$ @PStechPaul Heat is phonons, not photons. Photons are light or other EM radiation. \$\endgroup\$
    – Hearth
    Sep 2 at 22:50
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    \$\begingroup\$ Thermal (black body) radiation en.wikipedia.org/wiki/Black-body_radiation is infrared, which is electromagnetic radiation just like radio waves and visible light, and that radiation also involves photons. en.wikipedia.org/wiki/Thermal_radiation Phonons appear to be involved in heat conduction in crystals and other phenomena. en.wikipedia.org/wiki/Thermal_radiation \$\endgroup\$
    – PStechPaul
    Sep 2 at 23:27

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