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Author Topic: Heat from computer vs electric furnace  (Read 2634 times)
Juggernaut118
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December 09, 2011, 03:08:24 PM
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Is the heat, per kilowatt, coming off my computer the same as an electric furnace? I have an electric furnace and I'm wondering if I can leave what ever electric appliances on that I want and my electric bill will be the same.

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cicada
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December 09, 2011, 03:13:21 PM
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Not knowing the exact specifications of your electric heater it's difficult to give a clear answer.

The best answer is to to buy a Kill-a-watt or similar power measurement device and find out for yourself.

Your electric heater is going to be more efficient at converting electricity to heat than any computer you're running, but it won't also be generating bitcoins at the same time.

[edit] I re-read and see you said 'furnace', I thought you meant a wall-plug type electric heater, a furnace will be a little harder to measure.

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December 09, 2011, 03:46:34 PM
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Is that 'electric furnace' a heat pump type or coil type?

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December 09, 2011, 03:57:13 PM
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Not knowing the exact specifications of your electric heater it's difficult to give a clear answer.
No it isn't.  The heat coming from the computer is exactly the same per kWh as from a electric furnace.

You may argue that some of the power is wasted as sound waves, but even the sound waves will get absorbed by the walls and get converted to thermal energy.  Moving air will collide with stationary air and generate heat as the flow calms down.  Even the light from LEDs will get absorbed by the surface it hits and converted to heat.  Every single milliwatt used by the computer is converted to heat.

The calculation is different for a heat pump (inverted AC), but a heat pump is not an electric furnace.

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December 09, 2011, 03:58:53 PM
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The calculation is different for a heat pump (inverted AC), but a heat pump is not an electric furnace.

  Which is percisely why I asked and electric furnace is in quotes.

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December 09, 2011, 04:01:39 PM
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Is that 'electric furnace' a heat pump type or coil type?

It's a coil type.

I wish I had thought of this before. Now I can leave on all the lights and appliances I want.

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Juggernaut118
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December 09, 2011, 04:28:03 PM
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I called my electric company to see how their rates work. I'm paying $.039 per KWH off peak and $.081 per KWH on peak. The off peak rate only applies to the furnace when the power company is at low demand. So I still have to be careful about leaving lights and appliances on.

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December 10, 2011, 03:36:05 AM
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Not knowing the exact specifications of your electric heater it's difficult to give a clear answer.
No it isn't.  The heat coming from the computer is exactly the same per kWh as from a electric furnace.

You may argue that some of the power is wasted as sound waves, but even the sound waves will get absorbed by the walls and get converted to thermal energy.  Moving air will collide with stationary air and generate heat as the flow calms down.  Even the light from LEDs will get absorbed by the surface it hits and converted to heat.  Every single milliwatt used by the computer is converted to heat.

The calculation is different for a heat pump (inverted AC), but a heat pump is not an electric furnace.

In practical situation, it is likely that they are extremely close to each other. In the OP situation, it asks for the electric cost which is the incoming energy to the house. It is possible that the electric energy go out from other wires such as the Internet wire.
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December 10, 2011, 03:53:30 AM
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In practical situation, it is likely that they are extremely close to each other. In the OP situation, it asks for the electric cost which is the incoming energy to the house. It is possible that the electric energy go out from other wires such as the Internet wire.

And perhaps somebody is steaing your water! Watch out!

Electricity cost is calculated on wattage used as work (eg. producing heat), not current flow.  The same current flows through your light bulbs, but doesn't cost as much since you use less of it as work.

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December 10, 2011, 04:43:46 AM
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Electricity cost is calculated on wattage used as work (eg. producing heat), not current flow.  The same current flows through your light bulbs, but doesn't cost as much since you use less of it as work.

You must use different math than the rest of the world.  You should probably refrain from answering questions that involve any type of electronic circuit.

Watts = Amps * Volts... You pay per KiloWattHour, which is 1000 Watts for 1 Hour.

You can't control volts - here in the US, for example, we get roughly 110 or 220 for residential service.  Thus, the only thing you can control is current.  With volts being non-variable, the only thing you can use to control Watts is Amps (Current), thus, you effectively pay for Amps.

a lightbulb pulling RMS Current of 1 Amp is EXACTLY the same as a Computer Power Supply pulling RMS 1 Amp (assuming it has a high PF) from the electric meter's point of view.

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December 10, 2011, 06:07:08 AM
 #11

You must use different math than the rest of the world.  You should probably refrain from answering questions that involve any type of electronic circuit.

My point was to poke fun at the previous poster who's electricity is leaking onto the interwebs.  Otherwise, we're using the same math.

I fully understand that wattage is a measure of the current drawn at a particular voltage, ala Watt's law:

P (work, or watts)  = I (current) * V (voltage)

But wait! Don't forget that Ohm guy, his law says:

V (voltage) = I (current) * R (resistance)

R is the only element you can normally change.  R is the light bulb, the furnace, the computer, etc, and R determines how much current you draw and how much voltage is dropped along a circuit.  You don't change current or voltage separately.


I forgot what my point was?  Oh yeah!  We're arguing from opposite sides of a balanced equation!  That's just not smart, lets not do that.

[edit] I see, when I said "The same current flows through your light bulbs, but doesn't cost as much since you use less of it as work" I didn't mean to use 'current', but instead 'electric energy' like the last guy.  Typo, my bad.


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December 10, 2011, 09:13:58 AM
 #12

Please stop using the "word" wattage.  It is like fingernails on a blackboard to us electrical engineers.

Power not "Wattage" is measured in kilowatts
Energy is measured in kilowatt-hours
Current not "Amperage" is measured in Amps
Resistance not "Ohmage" is measured in Ohms

Now we do make one exception, electric or electrostatic potential, electric potential difference or electromotive force is measured in Volts, but that is to hard to say so we do call it Voltage.

For any given load the power factor gives the ratio of resistive to reactive load.  For residential customers the utility only charges for the resistive portion of the load, large commercial consumers are charged for both their resistive and reactive loads.

As stated above the power factor of the average computer is very close to a totally resistive load.  The power factor of a totally resistive load like an electric heater is, well, totally resistive.

So if your computer consumes 1000 Watts of power and your hairdryer consumes 1000 Watts of power and you run them both for one hour you would be charged for 1 kilowatt-hour of energy for each.

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December 12, 2011, 07:01:07 AM
 #13

Is that 'electric furnace' a heat pump type or coil type?

It's a coil type.

I wish I had thought of this before. Now I can leave on all the lights and appliances I want.
Resistance electric heating (coil type) is very expensive to run.  A heat pump uses about 1/3 the electricity for the same heat unless you are pretty far north.  There are now window AC units with true heat pumps in them for under $1000.  They may pay for themselves in one heating season. 

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December 12, 2011, 01:41:37 PM
 #14

Is that 'electric furnace' a heat pump type or coil type?

It's a coil type.

I wish I had thought of this before. Now I can leave on all the lights and appliances I want.
Resistance electric heating (coil type) is very expensive to run.  A heat pump uses about 1/3 the electricity for the same heat unless you are pretty far north.  There are now window AC units with true heat pumps in them for under $1000.  They may pay for themselves in one heating season.  

"coil type" was still confusing to me.  Made me think of heat pump compessor/evaporator coils.

To the OP:
if your heat is RESISTANCE heating (i.e. a wire has electricity flowing through it and gets hot which heats up your home) then yes your mining rig is just as efficient.  

Resistance heating = 3412 BTU of thermal energy (heat) per 1 kWh electrical power used (and billed).
Mining Rig = 3412 BTU of thermal energy (heat) per 1 kWh electrical power used (and billed).
Portable Space heater = 3412 BTU of thermal energy (heat) per 1 kWh electrical power used (and billed).

If your "electric furnace" is a heat pump they you likely are getting 2x to 3x (maybe 5x) the thermal energy (in BTUs) for the same 1kWh of electrical power.  Yes you can get more than 1 unit of heat for 1 unit of electricity.  This is possible without violating any commonly misunderstood laws because a heat pump (or AC unit) doesn't create heat (thermal energy), it moves it.

AC unit only moves heat IN ONE DIRECTION (inside house -> outside house)
Heat pump moves heat IN BOTH DIRECTIONS (depending on if you want your house warmer or cooler than ambient temp).

Still even if you have a heat pump a mining rig is superior to provide some or all of your heating needs.  Mining rig provides both heat AND cashflow.  Unless your mining rig provides 100% of your heating needs you might want to look into replacing your resistance heating w/ a heat pump. It can cut your heating costs in half (or more).
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December 12, 2011, 01:48:09 PM
 #15

I called my electric company to see how their rates work. I'm paying $.039 per KWH off peak and $.081 per KWH on peak. The off peak rate only applies to the furnace when the power company is at low demand. So I still have to be careful about leaving lights and appliances on.

I wonder if that is accurate info.  For it to only apply to the "electric furnace" then the power company would need two meters one for the electric furnaces and one for the rest of the house.  Not saying that is impossible but it is unusual.  Not sure why the power company would care.  The lower off peak rate is because they don't have enough demand off peak and are trying to stimulate demand.  A power company interested in their bottom line wouldn't care if you were smelting aluminum off peak as long as you are doing is off peak.

If that info is accurate then you only get lower rate for power used by the furnance.  Thus while a mining rig would be just as efficient it would cost more simply because it is billed at a higher rate.  If it were me I would callback and reverify that off peak rate only applies to electric furnace.

If it DOES then replacing your resistance heating "furnace" with a high efficiency heat pump could reduce your annual heating bill by 50% to 80%.  If you also have an AC the heat pump would replace that too and likely at higher efficiency.  If your AC is more than 5 years old it could cut your cooling costs by 20%-30%.

Still even your peak rate is very cheap (if that is total cost including transmission) which partially explains why your home is still using higher cost resistance heating.
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December 12, 2011, 05:07:48 PM
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Quote
I'm paying $.039 per KWH off peak and $.081 per KWH on peak

Where the heck to you live.  Next door to an old coal fired power plant with no pollution controls?  $0.04 is close to wholesale price where I live.

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December 12, 2011, 06:15:23 PM
 #17

Is that 'electric furnace' a heat pump type or coil type?

It's a coil type.

I wish I had thought of this before. Now I can leave on all the lights and appliances I want.
Resistance electric heating (coil type) is very expensive to run.  A heat pump uses about 1/3 the electricity for the same heat unless you are pretty far north.  There are now window AC units with true heat pumps in them for under $1000.  They may pay for themselves in one heating season.  

"coil type" was still confusing to me.  Made me think of heat pump compessor/evaporator coils.

To the OP:
if your heat is RESISTANCE heating (i.e. a wire has electricity flowing through it and gets hot which heats up your home) then yes your mining rig is just as efficient.  

Resistance heating = 293 BTU heat per 1 kWh electrical power used (and billed).
Mining Rig = 293 BTU heat per 1 kWh electrical power used (and billed).
Portable Space heater = 293 BTU heat per 1 kWh electrical power used (and billed).

If your "electric furnace" is a heat pump they you likely are getting 2x to 3x (maybe 5x) the thermal energy (in BTU) per unit of electricity.  Yes you can get more than 1 unit of heat for 1 unit of electricity.  This is possible without violating any commonly misunderstood laws because a heat pump (or AC unit) doesn't just create heat.  It MOVES heat (thermal energy).

AC unit only moves heat IN ONE DIRECTION (inside house -> outside house)
Heat pump moves heat IN BOTH DIRECTIONS (depending on if you want your house warmer or cooler than ambient temp).

Heat pump = AC in reverse (pumps heat into house instead of pumping it out).  Still even if you have a heat pump a mining rig is superior to provide some or all of your heating needs.  Mining rig provides both heat AND cashflow.  Unless your mining rig provides 100% of your heating needs you should look at getting a heat pump.  It can overnight cut your heating costs in half (or more).

I thought 1 kwh of resistance heating was equal to 3,412 BTU?
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December 12, 2011, 09:19:16 PM
 #18

I thought 1 kwh of resistance heating was equal to 3,412 BTU?

Reverse conversion fail on my part.  1 BTU = 0.293 Wh.  So embarrassing.  Fixed.
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December 26, 2011, 12:11:23 AM
 #19

I think everyone answered the theoretical part of this question sufficiently Smiley  The heat generated by a computer is going to be identical to a coil-based electric heater.

However.. What has not been touched on is perceived warmth.  You may be a bachelor, so this might not matter much to you... But those with wives and girlfriends will know how this works!  Assuming you have an identical power draw on both your furnace and mining rigs, I would still say you are very likely going to have to augment with the heater.

A furnace designed to create heat and move air (in theory in an efficient manner) is likely going to "feel" a lot warmer than an mining rig of the same wattage.  This is simply due to air movement and distribution throughout the house.

One thing I did was stuff my mining rigs in the furnace room, and open the cold-air return duct so they blow warm air into it.  I then set the furnace to just run the fan 24x7, even if the burner (gas here) isn't running.  End result was an extremely hot laundry room, and a slightly reduced heating bill.  I run around 6kw worth of gear, but honestly have no clue how many BTU my furnace is rated at.  I did ponder stuffing mining rigs into the cold air ducting itself (right before the burner intake) - but then realized the fire risk probably wasn't worth it!  Note to anyone contemplating this (or running wire through a heating duct) - smoke is the most hazardous part of a household fire, and putting things that burn toxic fumes into your air vent system to ensure all the fumes get quickly distributed throughout the house is an extremely poor idea.

Now, had I designed the house ground-up to be "mining rig heated" then this would be a different story!  It's very easy to neglect air distribution and simply focus on the raw math though, and modern HVAC systems are in general very efficient at taking created heat and distributing it to where it's needed.  Mining rigs not so much.

-Phil
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December 26, 2011, 12:21:55 AM
 #20

No it isn't.  The heat coming from the computer is exactly the same per kWh as from a electric furnace.

For all practical purposes, you are of course correct. However, I seem to recall from highschool some decades ago reading about irreversible deformation and iirc, not all energy was transformed in to heat. I would assume the electromigration taking place in side our chips would fall under that. I suppose its like 0.0000000001% (if that) but just curious if thats correct nonetheless?

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