# Environment-resistor temperature

Discussion in 'Electronic Basics' started by signo, May 25, 2005.

1. ### signoGuest

Hello to everybody, first of all sorry for my English (I'm Italian).

I have a circuit, enclosed in a plastic box, where there is power
resistor 1K, 50W and Tmax=200 degrees centigrade.
On the resistor is not present any heat sink.

After few hours that the circuit is working, the temperature measured
on the case of the resistor (Tr) reaches 110 degrees, in a
environmental temperature (Te) of 20 degrees.

My boss says that if Te increases50 degrees (Te = 20+50 =70 degrees),
Tr will increase the same value (Tr = 110+50 = 160 degrees).

I would say that Tr will increase, but not so much, just a little bit.
In my opinion it will be reduced the time for the resistor of reaching
the maximum value.

Is my boss right? Am I right? Are we both wrong?

Massimo

2. ### BanGuest

Massimo,
your boss is right, what counts is the temperature difference. You shouldn't
distrust an engineer so easily, (that's what I suppose he is). )

3. ### signoGuest

well... to be honest he is not an engineer, I'm an engineer.. I am
deeply ashamed...

4. ### Larry BrasfieldGuest

I suspect that the approximation attributed to
Massimo's boss is the most useful one to be made.
However, it must be recognized as nothing more
than an approximation. The heat loss mechanisms
do not have an inherently linear relationship between
heat flow and temperature differences. For example,
convection loss partially depends on radiation which
varies with the 4th power of absolute temperature.
Chimney effect can cause more air flow and thinner
boundary layers through which the heat must flow
for convection cooling.

5. ### Larry BrasfieldGuest

Minor correction inserted below.
Should be:
total heat loss partially depends on radiation which

6. ### BanGuest

Unfortunately you cannot rely on the radiation, it will be a miniscule
amount compared to the convection, because he says the box is closed, so
radiation can only be from the outer surface, which he didn't report. I
presume it will be half way between inside and outside, if the surface is
0.1mq and e=0.8, then with 338K 0.55W is radiated and with 408K 1W is
radiated, hardly much in comparison to the total power, but it makes a good
excuse to the boss. THX to Larry
The formula to impress the boss:
surfaceArea(m^2) * e(0.5...1) * constant 5.67e-8(W/(m^4 * K^4)) *
(emitterK^4 - ambientK^4)/(emitterK - ambientK)

7. ### John PopelishGuest

About the most I can say is that Tr will not rise more than Te does,
though it is possible that Tr will rise almost as much as Te does. At
low temperatures (low not being defined) heat is transferred primarily
by by conduction and convection. Conduction is a fairly linear
process that passes energy with a temperature rise about proportional
to the energy flow. Convection is also fairly linear. But in
parallel with these two mechanisms is radiation, and it is decidedly
nonlinear. Radiation passes energy in proportion to the 4th power of
absolute temperature difference. So as the temperature of box and
resistor rise, there is some absolute temperature at which radiation
will overtake both conduction and convection as the dominate transfer
mechanism. I don't know if that temperature is any where near the
temperatures you are talking about, however.

So your is correct, in the limit for low temperatures, and you are
right above some particular temperature. An experiment is the best
way to settle this argument.

8. ### signoGuest

Yesterday I've tried to test Tr with Te variable (from 20 degrees to 35
degrees).
I can say that, with a tollerance of one degree, approximately the
variation of Te is the same of the variation of Tr.

Thanks to everybody,
Massimo