Heat generated from a resistive load

Does anyone happen to know of a chart, or a method to calculate the approximate heat generated from a near pure resistive load across a

110V AC line? Say I have a 5K 10 watt low inductance wire wound resistor and I have this in a tightly closed insulated container; how much heat could I expect to be generated in K, C or F degrees?

PJ

Reply to
Pj
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The resistor will dissipate 2.42 watts. As far as the temperature rise in your container, that will depend upon the volume of the container and how well your insulation works.

Reply to
Lord Garth

You need to learn a bit of thermodynamics.

The _generated_ heat depends on the power and the time; the power in turn depends on the voltage and resistance: P = E^2/R (about 2-1/2 watts, in your case, except that you should design for a range of line voltages from 100 to 125, or even 90 to 135 if you want to be conservative). The power rating of the resistor doesn't determine the power consumed, it only determines whether the resistor will survive the experience, and then only if the resistor is getting sufficient environmental cooling.

The ultimate _temperature_ inside your box depends on the amount of insulation and the temperature outside the box. Generally you find the thermal resistance of your assembly (i.e. so many degrees C rise/watt), multiply it by your power, and add that to the ambient temperature of the box.

The time-dependent temperature characteristics of the box are more complex, but they more or less depend on the thermal mass that you're heating, and the power.

Your thermal resistance will vary greatly. In theory, if your thermal resistance were infinite your temperature inside the box would continue to grow for as long as you applied power to the resistor. In practice you'll find that it's hard to get high electrical conductivity and low thermal conductivity -- most good electrical conductors are good thermal conductors, and visa versa.

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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

Thanks.

I'm seeing the problem a lot clearer. I was thinking that the heat generated from the interaction of the power and load would eventually balance out. Sort of like the heat generated by a chemical reaction. In my case the balance point is detemined more or less by enviromental factors. Using a temp. sensor and turning the power on and off will be far simpler. Thanks again.

Reply to
Pj

=46rom a theoretical standpoint, let's assume your box has perfect insulation (perhaps that's what you mean by "tightly closed"). You can calculate the temperature rise of the medium (presumably a gas or liquid) contained in the caloric box for a given amount of energy imparted if you know the mass and the specific heat of that medium. You can find this info easy on the web. Look it up for water and do some practice calculations for a given number of joules and say a liter of water. Then you would know how much temperature rise you get in an insulated liter container of water for a certain number of joules (energy).

to find the number of joules, multiply watts by the amount of time (in seconds) that you have the heater turned on. Others posters have given you the power calc. Google ohm's law and look into power calculations, do some practice calculations.

Reply to
gearhead

The resistor is permanently generating heat while power is applied to it.

The trick is to remove more heat from the box than the components generate.

I do quite a bit of work with high power amps and you are into the realms of big heatsinks and fans there.

Clearly any holes you put in the enclosure need to meet safety requirements if you are selling on the gear.

Reply to
Marra

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That\'s not much of a trick, its a refrigerator.
Reply to
John Fields

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