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

"Pj" <frysning@yahoo.com> wrote in message
news:2e3e0e0b-1363-424e-9751-ae2d2fc359b6@i7g2000prf.googlegroups.com...
> 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

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.

Pj wrote:
> 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?
>
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.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

On Mar 3, 12:23=A0pm, Tim Wescott <t...@seemywebsite.com> wrote:
> Pj wrote:
> > 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? =A0Say 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?
>
> You need to learn a bit of thermodynamics.
>
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.

On Mar 3, 9:57=A0am, Pj <frysn...@yahoo.com> wrote:
> 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? =A0Say 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

=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.

On Mar 3, 8:51=A0pm, gearhead <nos...@billburg.com> wrote:
> On Mar 3, 9:57=A0am, Pj <frysn...@yahoo.com> wrote:
>
> > 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? =A0Say 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
>
> From 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. =A0Look it up for water and do
> some practice calculations for a given number of joules and say a
> liter of water. =A0Then 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. =A0Others posters have
> given you the power calc. =A0Google ohm's law and look into power
> calculations, do some practice calculations.

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.

On Tue, 4 Mar 2008 15:31:17 -0800 (PST), Marra
<cresswellavenue@talktalk.net> wrote:


>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.

---
That's not much of a trick, its a refrigerator.

--
JF