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What is rms value?

B

Bob Penoyer

Jan 1, 1970
0
Bob, i think you should read this...
heat , which is a form of a energy, is measured in Jouls...

I never argued that heat isn't a form of energy, or that it isn't
measured in joules. You seem confused.
 
K

Kelvin@!!!

Jan 1, 1970
0
--
{ Kelvin@!!! }
Robert C Monsen said:
Stop it, you are making my head hurt...

The amount of power radiated from a resistor is related to its
temperature. The higher the temp, in relation to the environment, the
faster the object will radiate energy.

If you add energy by passing current through it, it will heat up until
the rate of radiation = the rate of addition, ie, it will reach a
thermal equilibrium.

If it is unable to radiate energy for some reason (maybe the
surroundings are kept at the same increasing temperature) the resistor
will continue to heat up, because energy is getting added by the
current flow.
Thus, you are both right, it's energy that heats a resistor, but in
the real world, its power that determines the final temperature,
because, given a particular resistor configuration, the resistor will
radiate at a particular rate dependent on temperature. Thus, the
temperature will satisfy a formula something like

Well, Bob argues that it's power that heats up a resistor not energy...
and one more point in your arguement...
heat is loosing by convention, conduction, and radiation...
radiation is not the only way that heat is transfered from one body to
another...
I^2*R = P(T)

where P(T) is the function of radiated power for the object in its
environment, given a temperature T.

Regards,
Bob Monsen

P(T) = I^2*R
where is the independent variable T(time) in the RHS ??? so time is not
involved in this equation...
I think you mean
P(I) = I^2*R
or
W(T) = P*T

:)
hv a nice day~~~
 
R

Robert C Monsen

Jan 1, 1970
0
Kelvin@!!! said:
--
{ Kelvin@!!! }


Well, Bob argues that it's power that heats up a resistor not energy...
and one more point in your arguement...
heat is loosing by convention, conduction, and radiation...
radiation is not the only way that heat is transfered from one body to
another...

By 'radiation' I was talking about all forms of energy loss. For this
discussion, it doesn't matter what the mechanism is. It only matters
that the rate increases as temperature wrt the surroundings increases.

P(T) = I^2*R
where is the independent variable T(time) in the RHS ??? so time is not
involved in this equation...
I think you mean
P(I) = I^2*R
or
W(T) = P*T

:)
hv a nice day~~~

the "T" in my equation above is clearly stated to be temperature. P(T)
is radiated power, which is an increasing function of temperature.

I'll try again. Heat sinks are specified in units of thermal
resistance, which is the ratio of degrees of temperature rise above
ambient to watts radiated. If you have a device that dissipates x
watts, then if you use a heat sink specified to be y degrees/watt,
your heat sink will achieve thermal equilibrium at

(x * y) degrees above ambient.

This is clearly a linear function of power. So, its power, not energy,
that determines the final temperature.

Regards
Bob Monsen
 
R

R. Steve Walz

Jan 1, 1970
0
Kelvin@!!! said:
--
{ Kelvin@!!! }


Well, Bob argues that it's power that heats up a resistor not energy...
and one more point in your arguement...
heat is loosing by convention, conduction, and radiation...
radiation is not the only way that heat is transfered from one body to
another...


P(T) = I^2*R
where is the independent variable T(time) in the RHS ??? so time is not
involved in this equation...
I think you mean
* P(I) = I^2*R
or
W(T) = P*T

:)
hv a nice day~~~
-------------------
No you guys, you have T for both time and temperature!! Silly!

Power is just Watts which is Joules per Second delivered into the
resistor by V*I = V^2/R = I^2*R = Power.

Power = Energy in Joules/Second or P(t) = E/t

The temperature contributes this way for an ideal black body:

Rate of Power dissipated as InfraRed per Area is

P/A = sigma*(T^4),

where T is absolute temperature Kelvin and sigma is Wien's
constant for blackbodies.

When a body gets hot enough that it radiates heat as fast as it
gains it it is in equilibrium at that temperature and power.

P(t) = E/t = A*(P/A) = sigma*(T^4)

or
Watts supplied as electrical Power= I^2*R
equals
(P/A)(which is Watts lost per square Area [sigma*T^4] ) times Area.

Set them equal and solve for temperature or Power.
-Steve
 
B

Bill Vajk

Jan 1, 1970
0
Attributions too complex at this stage AND I was unable
for some reason to reply to Kevin's post. That being
said....

Heat radiates from the earth out into open space with
*no* receiving body involved. Consider heat to be
photons (which it is.) If and when it hits something
it is likely to be absorbed. Otherwise it can travel
forever, just like light or radio waves or x-rays.
By 'radiation' I was talking about all forms of energy loss. For this
discussion, it doesn't matter what the mechanism is. It only matters
that the rate increases as temperature wrt the surroundings increases.

Not true. You can (and do) radiate into a surrounding medium
that's at a higher temperature. BUT the net exchange is larger
incoming than outgoing so it seems as though you're not
radiating at all.
 
R

Robert C Monsen

Jan 1, 1970
0
Bill Vajk said:
Attributions too complex at this stage AND I was unable
for some reason to reply to Kevin's post. That being
said....


Heat radiates from the earth out into open space with
*no* receiving body involved. Consider heat to be
photons (which it is.) If and when it hits something
it is likely to be absorbed. Otherwise it can travel
forever, just like light or radio waves or x-rays.
increases.

Not true. You can (and do) radiate into a surrounding medium
that's at a higher temperature. BUT the net exchange is larger
incoming than outgoing so it seems as though you're not
radiating at all.

Well, true, but not salient.

"Kelvin" (who should know better, btw, given his handle :) is arguing
that its energy that determines the heat of an object. This is true,
until one considers the steady state of the system. Thats what 'Bob'
is arguing, that it's power, which is the time rate of change of
energy, which is important in determining the final temperature T.

My attempted contribution was to show the reason why this is true by
introducing the rate of radiation, the function P(T), which is power
radiated given a particular T above ambient, and asserting that when
P(T) = IIR, the system will be in a steady state. This is the
temperature the system will end up at, given I and R.

I also brought up the notion of thermal resistance, which is the way
one predicts final T given a particular heat sink and power output of
an electronics component. If theta is thermal resistance, and p is
power, then p * theta gives a temperature, which will be the final
temperature at equilibrium.

Walz described in far more detail the physics involved by actually
giving the function P(T).
 
B

Bill Vajk

Jan 1, 1970
0
Well, true, but not salient.

snip

Not salient as regards you paerticular disagreement
however that's how wrong ideas about things get
promoted and mislead readers who aren't up to speed,
eg, learning the material as they go.

I don't give a whit about the details of your stupid
argument which is covered adequately in any high school
level general science text. All each of you has to do
is refer to any such textbook or find the definitions
using google. In the meantime you're glossing over and
misstating other concepts to the determent of or younger
readers/participants.

This is sci.electronics.basics. That's not limited to the
basics you're arguing about, but *all* the basics.
 
J

John Popelish

Jan 1, 1970
0
Bob said:
John, I am SO tempted at this point, but I'm going to let
this one pass by...it's just too much of an easy target...:)

That should have read, " No. Only by the U. S. Army."
 
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