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[split]Photons and MASS in calculations...Split from CO2 home experiment thread.
07-18-2011, 11:18 AM
Post: #1
[split]Photons and MASS in calculations...Split from CO2 home experiment thread.
(07-18-2011 09:26 AM)Climate Realist Wrote:  And radiation of IR is a much faster way to move energy out of a system than physical means ( such as convection, conduction, latent heat of evaporation etc).

Hmm... I disagree but I don't have the math to prove it. Photons are massless and move very fast. Yes. But at normal earth surface temperatures of say 0C to 30C they don't amount to much.

BUT! If you heat something to over 1,000C then yes indeed radiation removes heat quicker than conduction/convection.

Whoever can, please try to do the math for say 1 million square kilometres of ocean losing 1 millimetre of surface area to latent heat in 12 hours and compare heat lost in joules radiating at 10C (364W/m^2).

Hey! I might even be able to do that one myself. nod_yes

Hah! I just realised... evaporation AND radiation are a combined heat loss. Now what does that mean?

Environmentalism is based on lies and the lies reflect an agenda that regards humanity as the enemy of the Earth. - Alan Caruba
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07-18-2011, 11:30 AM
Post: #2
RE: Home experiments to test the CO2 warms hypothesis
CR - Is that strictly relevant to the experiments we are supposed to be discussing on this thread?

re photons, err they have been proven to have mass, and
because they travel as a wave, the photon must be traveling faster than the speed of light.
ie, Wave "shape" equals longer distance traveled.

" But at normal earth surface temperatures of say 0C to 30C they don't amount to much. "
Smile

Please start a new thread elsewhere within the forum if you wish to continue this line of discussion.
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07-19-2011, 12:25 AM
Post: #3
RE: Home experiments to test the CO2 warms hypothesis
It's not the speed of the photon but the number of photons that denote the amount of energy transported. For my first attempt it looks like Climate Realist is right. Blush

No need for a million square kilometres, just 1 square metre will do.

Okay, surface temperature is 10C and emissivity is 0.96 this gives a radiation level of 360W/m^2 x 60 seconds x 60 minutes x 12 hours = a grand total of 15,552,000 joules.

Over the same 12 hour period our 1 square metre of water will lose 1 millimetre of surface water by evaporation or 1 kilogram of water is converted to water vapour. The figure I have at is 2,270 kJ/kg LINK so that uses 2,270,000 joules over the 12 hour period.

So radiation wins hands down. (unless my reasoning is all wrong, not unlikely)

Environmentalism is based on lies and the lies reflect an agenda that regards humanity as the enemy of the Earth. - Alan Caruba
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07-22-2011, 11:52 AM
Post: #4
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
Richard111 am I correct in thinking that you based a lot of the reasoning of your calculation upon the further below quoted post in this thread?

http://www.globalwarmingskeptics.info/fo...-1071.html
What is a Watt???

If so, please would you expand specifically.
I am interested for 2 reasons,
a) Water is a liquid, so emits IR at specific frequencies, although this may not effect power of IR emitted I admit..
ie, http://www.globalwarmingskeptics.info/fo...efrigerant
and,
b) How much of the MASS of the water surface is emitting IR? Are you assuming all because of the below, or none?
In other words, are you treating the water surface as a solid and perfect black body?

The facts that water is liquid, and a grey body may effect your sums, and quite considerably...

(01-24-2011 07:31 PM)Richard T. Fowler Wrote:  A FEW EQUATIONS


Equation #1. Watts = "joules per second" = "power".

This is a rate of INTRODUCTION of energy into a certain defined volume of space. (Or, conversely, a rate of its REMOVAL from such a volume.)

To introduce energy into a defined volume of space from outside that defined volume, the energy has to EITHER pass through the boundary of the volume, OR be created within that volume.

For problems that involve only the transport of existing energy, the only introduction of energy is by its passing through the boundary of the volume from outside the volume.

Since the boundary consists of a finite, positive number of planes, the entire boundary has an outer surface area S[a] that can be likened to a single plane of the same area S[a] on ONE of its sides.




Equation #2.

"Watts per square meter" = ["joules per second" per square meter] = "density of power".



Equation #3.

Joules = "energy".



Equation #4.

"Joules per CUBIC meter" = "density of energy".



THOUGHTS ABOUT THESE EQUATIONS

"Density of energy" contains a reference to "Cubic" rather than "Square" because ... why??

Because a joule has a nonzero volume and thus cannot exist within/on a planar surface.

"Density of power" contains a reference to "Square" rather than "Cubic" because when we are not considering newly created energy, power (for example, a watt) has a ZERO volume and thus cannot exist OUTSIDE of a planar surface.


Question: Based on the above truths, can one introduce "energy" into a defined volume of space, from outside that defined volume, in 0 (zero) seconds of time? No, it takes time, because ... why??

Because the energy (joules, or joule, or fraction of a joule) has to be moved through space in order to introduce it into a defined volume of space.

If it could happen instantaneously, then by definition, there would be no space between the starting point and the ending point of the energy. Consequently, the starting point would by definition be part of the defined volume, a reality which we have already specified by definition NOT to be true. Therefore, the answer to the question is "No."



CONCLUSIONS TO BE DRAWN FROM THE ABOVE

Conclusion #1:

A plane (whether flat or curved) can constitute a defined volume of space, but it cannot constitute a NONZERO defined volume. The volume of a plane must, by definition, be zero. Saying it has a volume but the volume is "zero" has the same effect as saying it has "no volume"; thus, we can see that a plane can simultaneously have "a volume" and "no volume", because the word "volume" can mean two different things simultaneously. But at no time can a plane have a nonzero volume.



Conclusion #2:

Units of "Energy" (i.e. just joules, without any spatial reference) are a measurement of a specific amount of energy which exists in a NONSPECIFIC, NONZERO volume of space, such that, by "nonspecific", we mean that the expression of the amount of space containing the specified energy is capable of variation, without affecting the total amount of energy being specified.



Conclusion #3:

Units of "density of energy" are a measurement of a specific amount OR AN AVERAGE AMOUNT of energy which exists in a SPECIFIC nonzero volume of space, such that, by "specific . . . volume of space" we mean "not capable of variation in its volume".



Conclusion #4:

Based on Conclusion #3,

-- If a specific amount of energy A[e] exists within an UNFIXED, nonzero space named N, and
-- if A[e] constitutes the entire energy within N at any time, and
-- if N is redefined to include a larger volume than it did before the redefinition, and
-- if N is thereby enlarged without allowing any of the energy A[e] to exit N either before or after the redefinition, and
-- if the space that is added to N by the redefinition contained no energy prior to the instant of the redefinition, and
-- if no energy crosses into N from outside of N at any time,

THEN: the value of A[e] (the total amount of energy within N at any time) is unchanged during the redefinition, and the expression of the density of energy of N DECREASES at the time of redefinition.



Conclusion #5:

A given rate of power can EXIST for a NONZERO volume of 3-dimensional space, but only for a ZERO length of time.



Conclusion #6:

A given rate of power can PERSIST for a NONZERO length of time, but only where the power is CONSTANT throughout that entire length of time.

Such an expression of persistent power could (but does not necessarily HAVE to) constitute an AVERAGE power, i.e. an AVERAGE rate of energy INTRODUCTION per unit of time, OVER a certain TOTAL period of time.

(Such total period for which the average is expressed can be of KNOWN or UNKNOWN length, and need not be equal to 1 of the specified unit of time in the denominator of the rate.)



Conclusion #7:

Within 3-dimensional space, a quantity of energy can only exist along ALL THREE dimensions of space, AND either 1) the dimension of time AND a timeless dimension of intensity ("density of power"); OR 2) a time-containing dimension of intensity ("density of energy").



Conclusion #8:

Density of energy for a given region of space can be calculated using only the energy that exists within the defined space at the start of the time for which energy is measured; OR just the energy that is introduced/removed DURING the time for which energy is measured; OR the combination of the first two quantities.



Conclusion #9:

Based on Conclusion #7, for a given, fixed, nonzero volume of space AND a given density of power, energy varies directly with time.



Conclusion #10:

Based on Conclusion #7, for a given, fixed nonzero volume of space AND a given OVERALL power, energy varies directly with time.



Conclusion #11:

Based on Conclusion #7, for a given, fixed, nonzero volume of space, a given OVERALL power, AND a given density of ENERGY, overall energy varies directly with time. QUESTION: "RTF, how can density of energy be constant, while overall energy is variable?" Aha, the answer is ........ when your "density of energy" is an AVERAGE density of energy OVER a period of time!



RTF

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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07-23-2011, 01:44 AM (This post was last modified: 07-23-2011 01:48 AM by Richard111.)
Post: #5
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
Here is the explanation for my reasoning when I calculated the energy loss from a single cubic metre of water surface. I did not mention before that the total energy loss over the period of 12 hours equates to a temperature reduction of 1C for 44.4 cubic metres of water!!! I am still trying to verify that.

Take a volume of matter, any matter, water, air, rock you name it. The volume can have any shape. The surface areas of cubes and spheres are much the easist to calculate. Big Grin

Now, our volume of matter will have a specific heat value (LINK), (LINK) and it will have an INSTANTANEOUS surface temperature. For the purpose of this discussion let us assume that only NOW (time zero), will the volume start cooling by radiation (NOTHING ELSE!). It is a straight forward calculation to find the RADIATION FLUX from the surface of our volume in WATTS per metre squared. (LINK) So from the TOTAL surface area of our volume we will have lost a precise number of JOULES OF ENERGY in the first second. After say ten seconds it will have lost ten times that number of joules of energy.

From the specific heat value we can work out what the average temperature drop NOW, ten seconds after start, due to the LOSS of that specific number of joules from the volume. Now problems creep in. Is that calculated drop in temperature over ten seconds equall throughout the whole volume? Probably not! Energy from the interior of the volume to the surface is unlikely to equall the rate of energy loss from the surface.

If we now surround our volume with air at a specific pressure and LOWER temperature the surface of our volume will lose more energy through conduction. This energy loss is in addition to the loss via radiation. If the air contains matter that absorbs energy and reradiates back to our volume then we must adjust the for the RADIATION FLUX leaving the surface by using the formula for NET RADIATION loss (LINK)

For the present I am unable to calculate the ADDITIONAL energy loss due to conduction at the gas/solid interface but I am confident that it will be resolved using WATTS and JOULES which will have precisely the same meanings and values as used in the examples above.

As for photons... I see them as a "PULSE" of electromagnetic radiation of the same nature as the radiation from a mobile phone. That "pulse" can consist of one or more complete cycles of radiation at any EM frequency. A high frequency "pulse" will have a shorter time duration than a low frequency "pulse" thus more energy is transported over a fixed time period by high frequency radiation than low frequency radiation, i.e. more "pulses" for your bang at high frequency. Cool

Now when a molecule of CO2 in the air with an EXTERNAL kinetic energy of say 400 metres per second absorbs a photon at 15 microns IT DOES NOT CHANGE THE SPEED OF THE MOLECULE. The kinetic energy remains the same. But the INTERNAL energy of the molecule has changed! The orbital level of one or more electrons will change which change the vibration/rotation arrangement of the atoms in the molecule. This internal energy can be lost by emitting a photon or colliding with other molecules nearby.

I must stop now because the massless photon is part of e=mc^2, energy is mass, not so? Wink

Environmentalism is based on lies and the lies reflect an agenda that regards humanity as the enemy of the Earth. - Alan Caruba
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07-23-2011, 11:31 AM
Post: #6
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
Thank you Richard111.

OK, let us start with the obvious one first. Water is a liquid, not a solid, so it emits IR somewhat differently.
For instance, mothers test the water temperature of babies bath with their elbows.
Why?

The reason I quoted the post above that I did in Post 4 of this thread, was that it seems to suggest one can "divorce" mass from surface area emitting,
by never mentioning mass.

I think of it this way, take 3 objects of the same size and temperature, but different masses, ie, materials, wood, plastic, and iron for example.
Without gravity we could add a fourth object, made of water, same size, same temperature.

The three solid objects would all have the same peak frequency of IR emission.
Because that is the temperature signal.
But the three objects would emits differing amounts of IR due to the amounts of mass at the surface emitting.
So, the three solid objects would have differing W/m2 outputs. W/m2 does not necessarily equal temperature, unless
one makes a lot of assumptions first in regards of emitter, and receiver especially within a "system"...ie, they are all "equal", this ain't necessarily so...........

The fourth object of water, liquid, would have a different peak frequency of emission.
And, the amount emitted would be what? More or less than for a solid object of the same mass, and size.?
I think the answer is considerably less for water, hence mothers use their elbows to test babies bath water temperature.

As I have said before in relation to P/4, and K&T type plots, and will say again now in relation to the above quote, and the sum Richard111 uses -

" Maths can not be used to determine the physics.
Maths can only be used to describe the physics,
when applied correctly to a given situation.
"

It is certainly true that reality is a hard task master, maybe an impossibly hard task master for us at present, but
that IS the reality of the situation.
It is also no reason or justification whatsoever for mathematical and physical oversimplification of the actual physical situations that
by the simplifications, assumptions, and maths used, is utterly divorced from the actual physics.

MASS used to calculate with - Richard111, you define the mass for the latent heat part of your sum, the first 1mm, yet,
I note you do not say the mass that is emitting of the water's surface?

For the sum do you need to include mass of the surface water emitting?
In which case the actual answer to your sum will be a fraction of your present answer.
You seem to have arrived at the same "ratio" of radiation losses to latent heat losses with your sum as depicted in K&T type plots.......

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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07-24-2011, 01:28 AM
Post: #7
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
Here is a link I didn't post above: Emissivity Coefficients of some common Materials.

Water is listed down towards the bottom. What more can I say? Emissivity of water is listed along with a range of solids.
I agree that the surface of water can move and this must have some effect on the emissivity. My first thought is that if ripples appear on the surface of the water then the emissivity MUST INCREASE because there will be MORE then 1 square metre of surface exposed than for a totally flat surface.

So it would seem we need to add a "roughness factor" to the water emissivity figure. Also it would seem that as the surface waves get above a certain size the water will be emitting and absorbing back to itself which would modify that "roughness factor" depending on how turbulent the water surface is. Problem is I have not found any publications relating to this problem. If you have, please say.

As far as I am aware MASS is only relevant when we talk about the "heat content" of the substance. Take two equall area trays, put water in both but ensure one tray has more water (more mass). Being as the area of the trays are identical both masses of water will have the same radiation flux while they are both at the same temperature. There is a possibility here for a home experiment. If you can insulate the trays, and cover with cling film to keep the air away, it should be possible to detect the more rapid loss of temperature of the least mass of water through radiation alone. I am assuming humidity under the cling film will not be a big factor. Then there is the "swamp cooler" which relies on a very large surface area in relation to a small mass of water to prove efficient cooling.

So for the three identical size objects you quote above, each having a different mass and specific heat, will have a different total heat content for identical temperatures. So the peak frequency of IR emission will not be quite the same for the same temperature. They are different materials so will have different "gray body" characteristics which you will find mentioned in the Emissivity Coefficients link above.

Now for the point of my post. I have read somewhere, but cannot find the link right now, that the daily average evaporation loss from the ocean surface is 2mm of water per day. So I just picked 1mm over 12 hours and found the results interesting. 1mm loss of MASS from the surface would appear to reduce the temperature of 44 cubic metres of water by 1C. If you disagree with the results please explain. If I am using any terms incorrectly, again, please explain. I do all this for my own interest. I do not have an agenda or any desire to convert anyones thinking. Everyone is entitled to their own opinions and must live by them.

You might find this link intersting: http://www.grow.arizona.edu/Grow--GrowRe...urceId=208

Environmentalism is based on lies and the lies reflect an agenda that regards humanity as the enemy of the Earth. - Alan Caruba
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07-24-2011, 04:17 AM
Post: #8
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
Richard111, thank you again.

I will only pick up on two points at the moment. I would also confirm I do not have an agenda either,
we are both just trying to understand better.

1) Solids and liquids emit differently, so the state (solid or liquid) of the material in question, in this case, makes a big difference.

2) 2mm over 24 hours, I doubt should of been halved to 1mm in 12 hours.

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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07-25-2011, 06:30 AM (This post was last modified: 07-25-2011 06:36 AM by Richard111.)
Post: #9
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
(07-24-2011 04:17 AM)Derek Wrote:  1) Solids and liquids emit differently, so the state (solid or liquid) of the material in question, in this case, makes a big difference.

Lost me here. My understanding is IR radiation occurs from surface molecules only. The photons are only identified by their wavelength, not their source.

(07-24-2011 04:17 AM)Derek Wrote:  2) 2mm over 24 hours, I doubt should of been halved to 1mm in 12 hours.

Well, I agree this is a dodgy subject. That link I suggested you look at talks about a daily evaporation loss of 7mm in Tucson, Arizona.
So the rate of evaporation changes considerably for time and place and local conditions. The one thing that doesn't change is the "latent heat" in the water vapour from where ever it is generated. The figure I have is 2,270 kJ/kg. So if you know the volume or the mass of water that vaporised you can figure out the energy loss, or cooling if you like, for the source mass, quantity of water.

One point to bear in mind is that 'heat' is not neccessary for water to evaporate. Ice will sublimate directly to water vapour and that water vapour will be indistinguishable from water vapour from the tropics.

I cannot find any discussion about this in the IPCC "science", only that "backradiation" forcing derived from CO2 in the atmosphere was the cause of more water vapour which will drive a positive feedback system etc. etc.

Complete and utter poppycock! Look up the radiation bands for water vapour and you will see there are many bands less than three microns. these bands are driven by SUNSHINE so they shade the earth during daylight. At night time NOTHING radiates at 400C which is the middle band for CO2. Thus even CO2 has two 'shade bands', 2.7 and 4.3 microns which absorb from the sun but cannot reradiate unless air temperatures exceed 400C which simply never happens.

There is a lot more cooling going on out there than these warmists "scientists" dare admit.

Environmentalism is based on lies and the lies reflect an agenda that regards humanity as the enemy of the Earth. - Alan Caruba
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07-25-2011, 07:03 AM
Post: #10
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
(07-25-2011 06:30 AM)Richard111 Wrote:  
(07-24-2011 04:17 AM)Derek Wrote:  1) Solids and liquids emit differently, so the state (solid or liquid) of the material in question, in this case, makes a big difference.

Lost me here. My understanding is IR radiation occurs from surface molecules only. The photons are only identified by their wavelength, not their source.

A solids surface molecules are locked into place, there they vibrate, and collide with each other (I think of these as head on collisions) and then emit IR.
A liquids surface molecules are more free to move about, so not so much is emitted by (what I think of as glancing) collisions I assume.
To my understanding because of the above, liquids do not emit IR with the same strict peak frequency that is a temperature signal as solids do.
This is the point behind me using the how to test babies bath water analogy.
The fluid, water, does not emit a peak frequency according to it's temperature.

That was also shown in at least one plot Brego brought to our attention some time back in the Water is the refrigerant of the troposphere.

I would also wonder aloud, is the energy not emitted by a fluid, (because the molecules do not collide as much as in a solid),
lost by the fluid as latent heat of vapourisation.
In other words, because the molecule is not slowed by collision it escapes the surface.

I agree totally with you that, as you say,
(07-25-2011 06:30 AM)Richard111 Wrote:  The one thing that doesn't change is the "latent heat" in the water vapour from where ever it is generated.
The figure I have is 2,270 kJ/kg. So if you know the volume or the mass of water that vaporised you can figure out the energy loss, or cooling if you like, for the source mass, quantity of water.

But it was the amount you used to compare with radiation losses that I was querying.
You seemed to assume 50% during the day, and 50% during the night?

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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07-25-2011, 11:51 AM (This post was last modified: 07-25-2011 11:54 AM by Richard111.)
Post: #11
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
Derek, I wonder if we are talking at cross purposes here. Emissivity and evaporation are two entirely separate functions. Seawater radiates at better than 0.99 at around 11 microns.

[Image: seagol01.gif]

I found that pic HERE.

You can sea that seawater radiates nicely into the 10 micron window to space. It is almost pure blackbody radiation!

I picked 1mm of evaporation over 12 hours because as I explained I had read that the DAILY GLOBAL AVERAGE EVAPORATION RATE WAS 2mm. but I am well aware that evaporation rates can change hugely with changeing air humidity as well as water temperature and air temperature and down welling IR radiation. I doubt there is any computer program in the world that can calculate the total evaporation rate with all those variables changeing in a chaotic fashion.

So as far as I can see the only way to establish if energy loss from the ocean by EVAPORATION can ever exceed energy loss by RADIATION is to pick an average sea surface temperature. Calculate the radiation flux in joules from a finite area and divide by 2,270 kJ to find how many kilograms of water would have to evaporate.

Thus 15,552,000 joules divided by 2,270,000 joules gives us 6.85 kilograms of water from 1 square metre tells us 6.85 millimetres of water would have had to vaporise just to equal the energy loss from radiation.

Not that I believe anything I have said above is cast in concrete as there are far too many variables to allow such simple calculation. For example down welling IR radiation WARMS the first millimetre of water and encourages evaporation which COOLS the surface water. The amounts vary depending on the INTENSITY of the incoming radiation (think sunny day/clear night) and the temperature of the water and the temperature of the air and the speed of the wind over the surface.

Nothing is as simple as it seems.

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07-25-2011, 12:38 PM
Post: #12
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
(07-25-2011 11:51 AM)Richard111 Wrote:  Seawater radiates at better than 0.99 at around 11 microns.

You can sea that seawater radiates nicely into the 10 micron window to space. It is almost pure blackbody radiation!

So as far as I can see the only way to establish if energy loss from the ocean by EVAPORATION can ever exceed energy loss by RADIATION is to pick an average sea surface temperature.
Calculate the radiation flux in joules from a finite area and divide by 2,270 kJ to find how many kilograms of water would have to evaporate.

Thus 15,552,000 joules divided by 2,270,000 joules gives us 6.85 kilograms of water from 1 square metre tells us
6.85 millimetres of water would have had to vaporise just to equal the energy loss from radiation.

Nothing is as simple as it seems.

Please note my bold and colouring.

Richard111, you do state the mass evapourated, but
you do not state the mass radiating with an emissivity of 0.99 at around 11 microns.

So, radiative losses will be a fraction of 15,552,000 joules.

" Okay, surface temperature is 10C and emissivity is 0.96 this gives a radiation level of 360W/m^2 x 60 seconds x 60 minutes x 12 hours = a grand total of 15,552,000 joules.

Over the same 12 hour period our 1 square metre of water will lose 1 millimetre of surface water by evaporation or 1 kilogram of water is converted to water vapour.
The figure I have at is 2,270 kJ/kg LINK so that uses 2,270,000 joules over the 12 hour period.
"

In the first part of your sum, I am saying that without mass radiating at it's emissivity you have given an incomplete sum.
It appears to me to be an overestimation for a grey body AND a fluid, this part of the sum appears to me to be a maximum possible for a solid black body.
Which is not the same thing.
(Let's skip over that a black body has no volume...)

You say, 10C, and therefore emission at a power of 360W/m2 - you have assumed a black body, not a grey body which is also a fluid.
So, it appears to me the first part of the sum must include a correction for mass (at the fluids surface) emitting at that power.
I have been looking for a couple of hours now and can not find any mention of this anywhere yet....

A grey body that is a fluid differs from a solid black body in many ways, which includes that not only because it has an emissivity (that will always be lower than 1), but also because
it has a mass emitting, ie a fraction of what the same temp solid black body would emit.

For the second part of the sum the amount of evapouration for day given is surely an underestimate, and for a night an overestimation.
Let's just double the figures and call it over 24 hours.
But at least mass is included in this part of the sum.

Interestingly, you end up with a ratio between radiation losses and latent heat losses that is very similar to the one found in K&T type plots...

Is emissivity a way to get around not mentioning mass in solids, that does not apply to fluids, without a further correction not presently mentioned?

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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07-26-2011, 10:10 PM (This post was last modified: 07-26-2011 10:13 PM by Richard111.)
Post: #13
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
Quote:A grey body that is a fluid differs from a solid black body in many ways, which includes that not only because it has an emissivity (that will always be lower than 1), but also because
it has a mass emitting, ie a fraction of what the same temp solid black body would emit.

That fraction of what the same temp solid black body would emit is the EMISSIVITY of the gray body. Huh

If you believe it is something different please explain and provide some links. Also why is emission from fluids and solids different?

Environmentalism is based on lies and the lies reflect an agenda that regards humanity as the enemy of the Earth. - Alan Caruba
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07-27-2011, 11:48 AM
Post: #14
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
OK, I have already really with the babies bath water analogy, but, let us try this from a different angle.

The sum is radiation losses versus latent heat losses.
I think we need to know how much energy in total first.
I suspect it is a lot more than could be lost by radiation alone.

So, how much is input into the square meter of sea water over the 12 hours?
AND, from how many sources?
The total would be roughly?

Is the 2mm of water evapourated from the global oceans in 24 hours a net figure?

If 2mm is a net figure then how much rain falls on the oceans, and how much river water flows into the oceans in 12 or 24 hours?

How much cooling of the oceans surface does the cold of rain account for in the time period used?

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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07-27-2011, 10:10 PM
Post: #15
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
BTW - Waves, do they,
increase surface area, does amount emitted by water increase, reduce, or stay the same?

Do waves increase surface area available for evapouration?
Wind, increases losses by conduction / convection, and evapouration.

The sum needs "average" waves or ripples, and wind speed.

Sorry, rushed before shift.

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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09-25-2011, 10:52 PM
Post: #16
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
Ok, I'll try another tack to explain my "issue" with W/m2 "figures".

THE TRICK of the second 2 in P/4.
Both sides of a disc have half the surface area of a globe of the same diameter.
That is the first 2 of P/4, which all can agree with.
ie, the power of beam received by a disc on ONE SIDE, is halved, because a hemisphere (in this case a lit hemisphere) has twice the surface area of a disc.
In earth's case the globe is an illuminated yellow hemisphere, and a dark blue unlit hemisphere.
Two hemispheres, not one.

Yes, if you want you could "double" the yellow hemisphere, but that does not overwrite or replace the dark blue hemisphere.
So, there has to be a dark blue "globe" as well as a lit at half power, on average 1/4 actual solar power, illuminated yellow "globe".
If we are to look at day AND night.

But, the real trick of the second two is two fold, firstly to make divorced from reality, and therefore meaningless averages seem to apply when they do not.
And secondly to cover up the simply omitted dark blue "globe" of night.

Let us say that the lit side emits 10 units, and the unlit side emits 5 units over 12 hours. *
That would be 10 +5 = 15. But the second 2 does not do this,
it does,
the sum of 10 and 5 divided by 2 = 7.5, ie, 1/4 of the real total emitted.
A meaningless average, divorced from the physics. It is neither day or night.

The full sum of what is emitted over 24 hours would be 2 yellow hemispheres, plus two dark blue hemispheres, ie,
(10 + 10) + (5 + 5) = 30

The answer is as near as dam it zero (received - emitted), but does that make the sum 1/4 - 1/4 = 0
No.
The evidence is that the sum is
Input - [output day side + output night side] = 0
In this example that would be,
(15 + 15) - (10 + 10) - (5 + 5) = 0

Whilst we persist in a 1/4 view (7.5 - 7.5 = 0), we will never understand the physics of the full 30.

The second 2 in P/4 is misdirection from the real world physics to meaningless averages of neither day or night applied incorrectly globally.
That's all.

* = Where the lit side is illuminated by a beam supplying 15 units over 12 hours, which would be a total of 30 units over 24 hours for the whole "globe".

Please Note - I have avoided using W/m2 in this example because they confuse the issue,
please see post 21 in this thread.
What is a Watt???
W/m2 does not specify amount, it only specifies the power, anything else is just an assumption...
Which could be, but is rarely, if ever, checked....

Grey body calculations anybody????? - I thought not......
Yet, such would answer so many questions!

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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09-26-2011, 02:52 AM
Post: #17
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
I also posted the below "elsewhere" recently, to no response, other than, I had forgot to mention the mirroring of a flask, which inhibits IR emission.
I responded to that correct point, with,
It would be very interesting to compare two flasks, both with a vacuum, but one
not mirrored.
Presumably that has been done, otherwise why mirror a vacuum flask?

No one has responded with an example, if anyone knows of one I'd be very grateful.
However, back to the post.

I posted, -
The often quoted "example" of two perfect black bodies in a vacuum, radiating.

Would someone please do such an example for two grey bodies in a vacuum,
I have never seen that done.
Total amount of energy of grey body, actual amount emitted per time step, etc, etc, etc.
W/m2 seems to avoid this..........

To my mind, compared to a grey body on it's own, then two grey bodies would have to show one of the three following possible results.
As I recently posted at GWS in the sister thread to this thread,
http://www.globalwarmingskeptics.info/fo...-1057.html

Namely post 29, in this thread,
http://www.globalwarmingskeptics.info/fo...age-2.html

BUT,
a vacuum flask leads me to an interesting dilemma.
Does a vacuum flask only show radiative losses?
No, there is some sensible heat losses from the cap, and that it is probably not a pure contained vacuum.
Hence, radiative losses alone must be less than a vacuum flask rate of cooling.
But, then I get to a dilemma.....

Here is the "dilemma",
Inner flask radiates according to temperature / emissivity, OK.
Outer flask, what does it do?
Well, it radiates according to it's temperature / emissivity, yes, but what's the effect upon the inner flask?
So, could the rate of cooling be calculated for the inner flask, in a perfect vacuum, by radiative losses alone?
If so, then, this could be plotted.
If the actual rate of cooling is measured and plotted, what is the result,
is the second plot, above, below, or on the first calculated line of theoretical cooling rate?

If I have got this line of thinking correct, then where the second line is shows whether,

i) radiation is all positively absorbed, rate of cooling reduced.
I think not... - it's a relative world we live in.

ii) Radiation below average is ignored (Claes Johnson approach), ie no effect, no difference to calculated line.
iii) radiation is relatively absorbed, so the rate of cooling of the inner flask is increased.

If it is not possible to calculate the theoretical rate of cooling for the inner flask only, by radiative only means,
then,
would a change in the temperature of the outer flask help illustrate which of the three above is correct?
This one I can not quite work out in my minds eye.


It seems to me that too many too often forget that,
emitted thermal IR is merely a temperature signal of the object.
Emitted IR is a result, not necessarily THE, or only cause of the objects temperature.
A point I am trying to illustrate with this thread,
http://www.globalwarmingskeptics.info/fo...-1617.html
Home experiment to illustrate the cooling power of latent heat.

So, if anyone can tell me how to calculate the emitted IR of said mugs, or heck, any grey body....
then, such a cooling rate for IR losses could be added to the plots I will eventually make.
Intuitively, it seems obvious that (otherwise vacuum flasks would not "work"),
sensible heat loss is greater than IR,
and, more importantly,
that latent heat (heat being a process by which energy is moved about) losses are by far the most powerful.

The present "black body" and "radiation" obsessions are blinding so many, so constantly to reality.

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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02-29-2012, 07:28 AM
Post: #18
RE: [split]Photons and MASS in calculations...Split from CO2 home experiment thread.
Hi All,
I think it is fairly obvious to most that if AGW and GH are false, then,
IR in current "explanations" and it's importance must be being greatly over exaggerated.

But how?
I suggest it is our current "accepted" use of a W/m2.

How though???
Are light and IR different? If they are different, does the difference have any consequences?
If there are differences and there are consequences, how does this relate to a W/m2 in it's current and accepted use?

Light is reflected off things, that is why we can see things, by the reflected light. IR is not reflected, it is absorbed by things, because it is at a far lower energy level than light is.
Light has far more energy than IR. That is quite a difference.
If all incident light upon a thing were absorbed by a thing then the thing would reach a temperature as predicted by the Stefan Boltzman equation.
BUT, the same thing would take far longer (if ever), to reach the appropriate S/B predicted temperature if it were absorbing IR.
That is quite a difference, it means that light and IR should be calculated differently.
The ONLY way to explain this difference is in the AMOUNT (of energy) emitted at the temperature determined power.

Presently the accepted "method" is to assume a black body like amount emitted, which for a grey body is patently untrue.
A grey body in an atmosphere, which conducts and convects, and may also have latent heat losses as well, just compounds the errors.

If light has more energy than IR, and it does, that is why it is reflected, then amount has to be calculated differently for light and IR.
At present "all" is assumed to emit as "light". It is a black body like amount emitted assumption that is never named or stated.
That is an enormous error that massively over estimates the amount of IR within earth's climate system.

We see because of light, we do not see IR, therein lies the difference.
The consequence of which is that we have a totally false science based upon ignoring this difference.
It is called Greenhouse Effect "theory" and Anthropogenic Global Warming, or Man Made Climate Change.

NB - I think Robert A Ashworth's pdfs in post 10 of this thread are a must read in respect of this post.

Later edit - This post has sort of turned a light bulb on for me in regard of W/m2
Please see,
W/m2 - Why will no one openly agree with me???

Some images to hint at where I am going with this.
[Image: Slide2-1.jpg]

[Image: Slide3-1.jpg]

and today's light bulb being turned on...
[Image: Slide1-1.jpg]

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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