Warning about cheap TVs currently on sale

In article ,
DM wrote:

We are talkiing about the rate at which the energy supplied increases
the temperature of the room.
....
It is not a question as to whether all the energy eventaully ends up as
heat- but how long it takes to do so

That's not what your previous messages appear to say. For example,
you said:

"You believe light output all contributes to heat, but again this is
incorrect"

"Not all light or sound energy will heat up your room"

"some of this ends up as heat, but not all"

But let's move on and consider the rate.

Since virtually all the energy is converted to heat in a fraction of a
second, the only question is of how quickly that heat is released into
the room.

Once steady-state is reached, the heat will be released into the room
at the same rate that it's consumed by the TV. So the issue is just
how much latency is there in the heating of the room by the TV. This
will certainly be a few minutes, mostly while the TV itself heats up,
but the same is true if you turn on your central heating - that water
and those metal radiators have a substantial heat capacity.

If you want to be instantly heated, probably the quickest way is to
use a hairdryer.

-- Richard
--
Please remember to mention me / in tapes you leave behind.

In article ,
DM wrote:

If it is absobed then eth ernergy transfer is 1nS saym if it is
reflected and absorbed at a later time then the energy transfer is 2ns
say. Now in case 1 the power trasfer is doubele case 2.

No. The rate of energy transfer is the same, it just starts 1nS later.

-- Richard
--
Please remember to mention me / in tapes you leave behind.

Richard Tobin wrote:
In article ,
DM wrote:

If it is absobed then eth ernergy transfer is 1nS saym if it is
reflected and absorbed at a later time then the energy transfer is 2ns
say. Now in case 1 the power trasfer is doubele case 2.

No. The rate of energy transfer is the same, it just starts 1nS later.

-- Richard

Yes the energy transfer is eventully the same, but we need to consider
the rate of transfer. That is what will determine how whot it gets
within a period of time.

In case 1 you supply energy in 1nS, so you add energy , and you also
loose some energy during that period.

In case 2 you add the same amount of energy to but you do it in 2nS,
however during that time you have lost twice as much energy as in case 1.

In case 1 the temperature will increase more than in case 2.

The energy transfer from you energy source to heat is less efficeint.

Richard Tobin wrote:
In article ,
DM wrote:

We are talkiing about the rate at which the energy supplied increases
the temperature of the room.
...
It is not a question as to whether all the energy eventaully ends up as
heat- but how long it takes to do so

That's not what your previous messages appear to say. For example,
you said:

"You believe light output all contributes to heat, but again this is
incorrect"

"Not all light or sound energy will heat up your room"

If you stand outside a window, you see and hear the TV- the energy has
left the building.


"some of this ends up as heat, but not all"

But let's move on and consider the rate.

Since virtually all the energy is converted to heat in a fraction of a
second, the only question is of how quickly that heat is released into
the room.

If all the energy is virtually instantaneously converted to heat how do
you get it to do a useful function?.


Once steady-state is reached, the heat will be released into the room
at the same rate that it's consumed by the TV. So the issue is just
how much latency is there in the heating of the room by the TV. This
will certainly be a few minutes, mostly while the TV itself heats up,
but the same is true if you turn on your central heating - that water
and those metal radiators have a substantial heat capacity.

Radiators are a good example.
How quickly you get that stored energy out as useful heat depends on how
you design the radiator. The most important aspect being what the
surface area is. AS you reduce surface area you get less transfer from
the radiator to the air. But the actual water in the radiator will be
hotter than in the case of a more efficeintly designed radiator with a
larger surface area. Effectively more of the energy is stored and less
going towards heating up the room.


The will both consume the same amount of power, but one of them will
heat up your room to a higher temerpature.

DM wrote:

Every eletronic element within a TV has mass, all mass when heated up
stores that heat as potential energy relative to teh surrounding
environment. The heat is stored, and not dissipated efficeintly.

No.

All the components in the TV will have a heat capacity - that is
dictated by the specific heat capacity of the material in question,
and the amount you have of it.

So for a simple carbon resistor that will be around 700J /kg/K if you
ignore its leads for the moment.

As a current flows through the resister it will get hotter, so as you
say its potential energy will increase and it will be storing a small
amount of energy as heat. However the rate of heat loss to its
surroundings will also increase since this is dictated by the
temperature differential to its surroundings. If the current remains
constant, the the power being developed in the resister will also
remain constant and the temperature will continue to rise. After a
time one of two things could happen. The resistor could over heat and
fail open circuit, or more likely, equilibrium will be reached. At
this point the rate of heat loss to the surroundings will exactly
match the power input to it. When you turn the set off, the resister
will then lose its stored heat to the surroundings as well.

Hence ultimately all the energy dissipated by the resistor is lost as
heat.

Ultimately yes - but the issue if power and energy conversion efficeiny

So, if you agree that all the energy that you put in will ultimately
come out as heat, you would also have to agree that if you stick 1kJ
into a TV, the amount of energy released to the room will also be 1kJ.

Now efficiency in the case of a heater is defined in terms of the ratio
of energy put in to it that is usefully dissipated where you want it. A
gas boiler will not be 100% efficient since some of the heat generated
will pass out of the flue and not into your central heating. An electric
heater however is pretty much 100% efficient (at point of use) - even
the voltage drops on its connecting leads will typically result in heat
being deposited where you want it.

You belive light output all contributes to heat, but again this is
incorrect- everything would appear black if this was true. -
evidently that is mot so.

How do you come to that conclusion?

When light strikes an object, some of it is reflected, and some will
be absorbed (and this will happen no matter how reflective the
surface, since no surface it perfectly reflective). The proportion of
the light which is absorbed will result in a rise in temperature of
the object absorbing it. Of the reflected light, it will also in turn
hit an object where some will be absorbed and some reflected. This
will continue to happen until all the light has been absorbed. Now in
your closed box, the light will bounce around for few nano seconds -
but ultimately it will all be absorbed - how many reflections are
sustainable before absorption of all the photons will depend on the
reflectivity or the surfaces and the amount of light in the first place.

So unless the light keeps on being generated, it gets dark very
quickly. Assuming the light stays on, then steady state is also
reached very quickly, with all the light being generated, being
reabsorbed within a few reflections, but the reflections prevent
everything appearing black.

Again a simepl matter of time.

I think you be confusing yourself by thinking in terms of rate of
transfer of energy (i.e. power), when it is the total amount of energy
that matters.

If it is absobed then eth ernergy transfer is 1nS saym if it is
reflected and absorbed at a later time then the energy transfer is 2ns

So in both cases, 100% of the energy is absorbed.

say. Now in case 1 the power trasfer is doubele case 2. So it is a

Say your turn on a light for 1 second in a sealed box. The box is such
that it takes 10nS for the first photons to all be absorbed. That means
if you plot a graph of the energy absorbed over time, you will see a
ramp up starting at time = 0, that will reach maximum at t = 10ns. It
will stay at a steady state until you switch the light off, in which
case you will see a ramp down that finishes at 1 sec + 10ns.

The energy in = energy out, the rate of flow of energy in = rate of flow
of energy out once equilibrium is reached. The "storage" of the system
(i.e. the flight time of the light prior to absorption) will introduce a
delay, but not ultimately have any effect on the total energy radiated
and absorbed.

more efficeint case to heaqt up the matter. We are talking about the
effciency of differnet sources for heat- we can not wait indefiently

You seem to be attempting to shift ground a little here. The discussion
was whether the TV consuming power at 100W would also be radiating
energy to the room at 100W. I take it you now accept that it would once
it is warmed up?

So in the case of the TV, you may stick 100W into it for 10 hours. 1kWh
of energy all told. It may take 15 mins to reach full operating
temperature. During which time its apparent dissipation will appear to
ramp up from nil to 100W. However after warm up, it will continue
dissipating at 100W until you turn it off. At which point it will
continue to dissipate at a decaying rate loosing any stored heat energy.

Any time delay or lag in response does not change the efficiency,
although it lowers the responsiveness. A TV is probably comparable to a
normal radiator in terms of response time.



--
Cheers,

John.

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DM wrote:
Richard Tobin wrote:
In article ,
DM wrote:

If it is absobed then eth ernergy transfer is 1nS saym if it is
reflected and absorbed at a later time then the energy transfer is
2ns say. Now in case 1 the power trasfer is doubele case 2.

No. The rate of energy transfer is the same, it just starts 1nS later.

-- Richard

Yes the energy transfer is eventully the same, but we need to consider
the rate of transfer. That is what will determine how whot it gets
within a period of time.

In case 1 you supply energy in 1nS, so you add energy , and you also
loose some energy during that period.

In case 2 you add the same amount of energy to but you do it in 2nS,
however during that time you have lost twice as much energy as in case 1.

In case 1 the temperature will increase more than in case 2.

The energy transfer from you energy source to heat is less efficeint.

No, just suffering a phase lag. Something every first order[1] system
will demonstrate

Efficiency is the ratio of energy in vs energy out.

[1] i.e. one energy storage component.

--
Cheers,

John.

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| Internode Ltd - http://www.internode.co.uk |
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| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

DM wrote:

Radiators are a good example.

How quickly you get that stored energy out as useful heat depends on how

Storage is a misdirection here - remember the radiator is supposed to be
transferring energy from the water to the room.

you design the radiator. The most important aspect being what the
surface area is. AS you reduce surface area you get less transfer from
the radiator to the air. But the actual water in the radiator will be

So the power output of the radiator falls.

hotter than in the case of a more efficeintly designed radiator with a

The water entering the rad will be the same temperature as with any
other rad. That of it leave will be higher, indicating the rad has
dissipated less heat during the transit of the water.

larger surface area. Effectively more of the energy is stored and less
going towards heating up the room.

If you like.

The will both consume the same amount of power, but one of them will
heat up your room to a higher temerpature.

No, this would violate the law of conservation of energy.

They most certainly *do not* consume the same power.

The rad with lower output will consume less and emit less. If you room
was lossless, then either could heat it to the same temperature (i.e.
that of the water). If the room has losses, then both will heat it to a
steady state, but the rad with the higher power output will be able to
get closest to the water temperature.

--
Cheers,

John.

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| Internode Ltd - http://www.internode.co.uk |
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\================================================= ================/

"Richard Tobin" wrote in message
...
In article ,
If you want to be instantly heated, probably the quickest way is to
use a hairdryer.

No, climb into the microwave.

Bill

In article ,
DM wrote:

Since virtually all the energy is converted to heat in a fraction of a
second, the only question is of how quickly that heat is released into
the room.

If all the energy is virtually instantaneously converted to heat how do
you get it to do a useful function?.

I didn't say "virtually instantaneously", I said "in a fraction of a
second".

It does all its useful functions within that fraction of a second.
The light leaves the TV screen, reaches your eyes a few nanoseconds
later (the speed of light is one foot per nanosecond), and is
absorbed. The sound leaves the speaker, reaches your ears a few
milliseconds later (the speed of sound is about one foot per
millisecond), and is absorbed.

Very tiny amounts will last longer: some light might go out of the
window and if it's a clear night reach Alpha Centauri four years
later. Some of the sound might persist for thousands of feet. But
these are completely insignificant.

Once again, if you believe that an appreciable amount of energy
remains in non-heat forms for more than a fraction of a second, tell
us what that form is.

(Actually, I can think of another form, but it's again just a latency
issue, and is again negligible: some energy is stored as an electric
field in capacitors in the TV. They won't affect the heat output
after the first few seconds at most.)

-- Richard
--
Please remember to mention me / in tapes you leave behind.

In article ,
Bill Wright wrote:

In article ,
If you want to be instantly heated, probably the quickest way is to
use a hairdryer.

No, climb into the microwave.

I was going to say that, but I thought there might be small children
reading.

-- Richard
--
Please remember to mention me / in tapes you leave behind.