Physics Help Forum calculate the temperature due to Power loss over a res.

 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

 Sep 6th 2017, 07:53 PM #1 Junior Member   Join Date: Sep 2017 Posts: 4 calculate the temperature due to Power loss over a res. Hi All, Having the following problem I am trying to solve I am an electronics guy some of e physics eludes me in my calc. Here goes.... I am supplying 125Vac into a R which is a 20kohm resistor where I have a 12V zener in series to GND. P_loss = (125Vac - 5V)^2 / R P_loss = 0.72W So lets say we wait for 1000 secs. Total Energy (J) = 0.72kJ Now I am taking into consideration that the SPH of dry air is 0.716 KJ/Kg.K and Air Density is 1.3 Kg/m3 Therefore Temperate raise should be Tr = 0.72 / 0.716 / 1.3 = 0.77 K? Am I missing something here, or is this not he way to calculate heat loss over a resistor? Someone? THKS. Last edited by Merdidose; Sep 6th 2017 at 07:59 PM.
Sep 6th 2017, 10:04 PM   #2
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 Originally Posted by Merdidose Total Energy (J) = 0.72kJ Now I am taking into consideration that the SPH of dry air is 0.716 KJ/Kg.K and Air Density is 1.3 Kg/m3 Therefore Temperate raise should be Tr = 0.72 / 0.716 / 1.3 = 0.77 K? .
If the electronic components are putting out 0.77W , The temp raise should be 0.77 , if all the heat goes evenly into only 1m3 of air

 Originally Posted by Merdidose Am I missing something here, or is this not he way to calculate heat loss over a resistor? Someone?
You don't need to do this to "calculate the heat loss over (from) a resistor" .... ohms law is all you need , the temperature or specific heat of the air is immaterial.

Last edited by oz93666; Sep 6th 2017 at 10:13 PM.

 Sep 6th 2017, 11:48 PM #3 Junior Member   Join Date: Sep 2017 Posts: 4 Hi oz93666, 0.77W is 0.77 Deg. C per second? Doesnt sound right..
 Sep 7th 2017, 02:05 AM #4 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 708 You are an electronics guy so let's start with the electronics. First you need to review you power calculation because 1) The diode will block every other half cycle. 2) Why have you clamped the base at 5 volts not 12? When you have got the electronics of power generation in the resistor correct you should look up 'themal resistance' from an electronics point of view. The calculation of temperature rise in a component is a standard procedure using published characteristics of that component. Merdidose likes this.
Sep 7th 2017, 03:10 AM   #5
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 Originally Posted by Merdidose Hi oz93666, 0.77W is 0.77 Deg. C per second? Doesnt sound right..
No .... it isn't ..

assuming you have components putting out 0.77W ...

Now what is your question ?

 Sep 7th 2017, 04:46 AM #6 Junior Member   Join Date: Sep 2017 Posts: 4 @studiot 1) I am using a bridge so its full-wave cycle. RMS is 120V 2) Yes you are right I usually use 5V zener so its abit of force of habit. My bad. Should be (125V - 12V)^2 , but these are minor errors.. You nailed it with "thermal resistance" this is the search engine term I needed! Thank You!
 Sep 7th 2017, 06:03 AM #7 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 708 Here is a typical manufacturers pdf for high power resistors. http://www.vishaypg.com/docs/65104/tr_hmr.pdf
Sep 13th 2017, 05:27 AM   #8
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 Originally Posted by nhatminh61 You don't need to do this to "calculate the heat loss over (from) a resistor" .... ohms law is all you need , the temperature or specific heat of the air is immaterial. Now what is your question ?
I suggest you study the subject (a good place to start is the manufacturer's pdf I linked to) before making claims that might confuse others.

Yes the OP originally attempted to calculate thermal performance from first principles - all credit to him. It would work but, as you say, it does make things more complicated. Of course such calculations are required if the resistor is on a heatsink, and this is a situation of the mains being applied to a high power low ohmage resistor.

Sep 13th 2017, 05:38 AM   #9
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 Originally Posted by studiot I suggest you study the subject (a good place to start is the manufacturer's pdf I linked to) before making claims that might confuse others. Yes the OP originally attempted to calculate thermal performance from first principles - all credit to him. It would work but, as you say, it does make things more complicated. Of course such calculations are required if the resistor is on a heatsink, and this is a situation of the mains being applied to a high power low ohmage resistor.
This is ridiculous ... no need for any of it ... just operate components within their power ratings with heat sinks if required ...

OP is all too muddled .....What exactly is the question being asked ???

 Sep 13th 2017, 08:56 AM #10 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 708 Of course it's not ridiculous. If you know anything about it you would know that the 'power ratings' are only correct for specified temperature and mounting conditions. Special formulae exist for use at other operating points, particularly the ambient temperature. This is standard electronic engineering stuff.

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