What is a Watt???
10-12-2011, 09:00 AM
RE: What is a Watt???
I think it boils down to people tripping over the difference between power and energy.
Please excuse me, simple plots will be added soon to this post.
I have to THANK Richard T Fowler. He, in this thread has patiently explained several times the difference between power and energy.
It is a subtle difference to be honest, but it is an important difference all the same.
Once the difference between power and energy is realised then something more becomes clear.
I hope readers will back track on this thread, if they need to, to clearly understand the difference.
In short, power is a timeless figure.
This means that the Stefan Boltzman Law makes sense when applied to imaginary black bodies.
That's the rub, that is the subtle, rhetorical misunderstanding, that has escaped us so far.
Power figures in the context of K&T type budgets, and the Greenhouse Effect "theory" ONLY make sense when applied (instantaneously) to a black body.
They would not, and do not apply to grey bodies...
The earth is a grey body.
The subtle, rhetorical misunderstanding, is not so subtle or rhetorical after all,
it is actually a (imaginary black body) scientific basis that is utterly divorced from (a grey body) reality.
Ok, let us look at P/4, the starting point of K&T, and GH from an energy units, rather than a power point of view (but I will be comparing to W/m2 as well for clarity).
Rather than power ie, 1368W/m2, let us use energy units.
For example, let us say the earth's lite hemisphere receives 15 units of energy over 12 hours.
During that 12 hours the lite hemisphere looses 10 units, and the unlit hemisphere looses 5 units.
Strictly speaking it does not matter that the figures are 10 and 5,
they could be 12 and 3, or, 9 and 6, as long as they add up to 15,
so that, the unspoken assumption of IN = OUT is maintained, and not questioned.
So, we have, over 12 hours,
[in] 15 - [out] (10 + 5) = 0
OK, for 24 hours, one revolution of earth, we have.
[in] (15 + 15) - [out] ((10 + 10) + (5+5)) = 0
30 in, and 20 out on the lite side, and a further 10 out on the unlit side which equals 0
That is the way it should be done.
P/4 does it this way, using the power figure of W/m2.
First P/4 uses power rather than units of energy, so W/m2.
A power of 1368W/m2 is correctly divided by two,
(a hemisphere has twice the surface area of the same diameter disc)
giving an average received on the lit side of 684W/m2.
But then the power is divided again by 2?
The explanation is that this is the power received over 12 hours, on one hemisphere,
so for two it is simply divided by two to cover the whole planet.
Wrong, it is still one hemisphere receiving an average of 684W/m2 for all of the twenty four hours.
It may appear correct using power figures, but energy unit figures show the fault.
In other words, using units of energy P/4 takes 15 units of energy and divides it by two incorrectly, to get an answer of 7.5 units.
THEN, P/4 takes day and night, adds the losses together and divides by 2, for the whole planet.
AGAIN WRONG, day is added to night, and the 12 hour average is used, it is simply incorrect.
That is how the figures are divorced from reality, they are neither day or night, but an average of the two.
There is no day and no night in K&T, there is only an average of the two.
However, it does give P/4 the right answer, by the wrong means for the overall sum,ie,
[in](15/2) - [out] ((10+ 5) /2) = 0
more simply as,
[in] 7.5 - [out] 7.5 = 0
12 hours insolation of one hemisphere spread over the whole planet, minus the day and night losses over 12 hours, added together and then divided by two = 0.
The first question that arises is why does P/4 do the sum this way?
WHY is 7.5 as opposed to 30 so important.
7.5 energy units used in this explanation, in W/m2 terms is 240 W/m2 received at the surface, which equals a surface temperature of -18C for earth...
The answer to the question of WHY does P/4 do the sum this way is actually quite simple and obvious,
to get the earth's surface temperature low enough in the first place to REQUIRE a Greenhouse Effect.
This point, and others in regards of P/4 were explained in the first two posts of this thread,
P/4 - Why it is THE issue that destroys GH and AGW.
The question we must ask here therefore is, which is the right sum?
[in] (15 + 15) - [out] ((10 + 10) + (5 + 5)) = 0
ie, 30 in - minus 20 out day, and 10 out night = 0
or the present, P/4 "route" of,
[in] (15/2) - ((10 + 5)/2) = 0
more simply as,
[in] 7.5 - [out] 7.5 = 0
Both ways get the "right answer", if indeed 0 is the right answer (it isn't - earth has life, that is merely for starters)
but the ways they do the sum are vastly different.
One appears to reflect reality, the other appears UTTERLY divorced from reality.
In the end neither way described above to do the sum is correct, but which is the better?
Which way would, most probably, lead us to a better understanding?
Although the power route of P/4 may appear correct, the energy units way to do the sum instantly shows the faults of the P/4 "route".
The "trick" is to convert power to energy units then apply it to the P/4 "route" of how to do the sum to get the right answer..
Then the fallacy is plain to see, especially when S/B law and blackbody assumptions are taken into account as well.
The difference between the "routes" has been "covered up" by the misuse and application of S/B Law and blackbody implicit assumptions never mentioned.
Hence "they" will not do grey body.
Grey body destroys the power basis of the P/4 "route", and
that also explains why they will never use energy units to explain such budgets, they would simply never work.
As I hope the above "sums" clearly show.
The whole aim of practical politics is to keep the populace alarmed
(and hence clamorous to be led to safety)
by menacing it with an endless series of hobgoblins, all of them imaginary.
H. L. Mencken.
The hobgoblins have to be imaginary so that
"they" can offer their solutions, not THE solutions.
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RE: What is a Watt??? - Derek - 10-12-2011 09:00 AM
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