Tropospheric Hot Spot- Why it does not exist...

[Wuwei]

Efficiency is a big part of the real world. The most efficient way to enclose area with a defined circumference is a circle. The radius is equal in all directions. A square is not as good but still better than a rectangle.

The moon receives as much solar energy as the Earth so why isn't it as warm? The main reason is the length of a lunar day/night cycle. It is very hot in the day and very cold at night. The energy inputs and outputs are extreme, less efficient. Like how a circle has more area per circumference than a flat rectangle. Only more so, because area is a ^2 relationship and temperature is a ^4 relationship. If the lunar 'day' was half as long the average temperature would be warmer. If you halved it again, warmer still. And so on, ignoring the effects of spinning the mass.

The predicted temperature of the Earth is -18C, from the average amount of sunlight hitting it. Now hold on a minute. We were talking about efficiencies. An Earth day is shorter than a lunar day but it is still 24 hours. Shouldn't the hypothetical max (circle) be greater than the actual Earth (square) which is greater than the Moon (rectangle)?

Obviously the atmosphere has something to do with it. But how does the atmosphere raise the average temperature ABOVE the theoretical maximum derived from the solar input?

I'll talk more about this if anyone is interested. Or we can just leave it as another unanswered question.

Hey Wuwei, you like theoretical problems.

Don't you find it interesting that the GHE is large enough to more than compensate for the inefficiencies caused by unequal input and output energies?

While it is just a variation on the theme of why the surface is warmer than the Sun's input, don't you find it amazing that entropy gained by converting sunlight to IR can 'power' such a large effect?

I realize the actual energy comes from radiation not lost to space but it seems to me that entropy gained in the total area causes entropy loss by the storage mechanism in a subset area.

I could be wrong. What do you think?

I looked at the idea of entropy for a rotating absorber (the moon). The following formula works for simple problems that are closed system, a solid (no atmosphere), and at equilibrium. The change in entropy is given by

delta S = n Cp log(Tf/Ti) where,

n = moles of moon surface that is warmed,
Cp = specific heat of surface,
Ti = initial temperature,
Tf = final temperature.

The factors in n x Cp are hard to determine. They refer to the mass and specific heat of the mass being heated.. They will be replaced by M referring to the amount of material being heated.

Dark side of moon Ti = 100 K
Bright side of moon Tf = 373 K

When looking at the sun hitting the moon there are two extremes where a quasi-equilibrium exists.

1. The moon is slowly rotating, and the sun is shining on one side for an extended period.
2. The moon is rapidly rotating, and night is so short that the temperature has no time to drop.

Case 1. The delta entropy for energy hitting the light side of the moon only.

delta S = M log(373K/100K) = M x 0.572

Case 2 . This is a bit trickier because there is no temperature data for the non-existing case of both sides getting half the energy all the time, but it can be computed from what we already know.

The sun energy E results in a temperature of Tf.
From SB equation, E = K Tf ^4. Where K contains all the proportionality constants.

Let T2 be the unknown Case 2 temperature of the moon.
The whole moon will be receiving half the energy. 2 x E = K T2^4.

Using simple algebra, Tf = T2 x fourth root of 2 or, T2 = Tf x 0.841

The entropy formula for case 2 is
delta S = 2 M log(Tf x 0.841 / Ti) = 2 M log( 3.137) = 2 M x .496
Note: the leading factor of 2 is because twice as much material (both sides) are being heated.

Summary for the moon:

Case 1 delta S = M .572
Case 2 delta S = M .992

The Case 1 slow spin is almost twice as efficient in using energy from the sun.

The earth, it is more like the fast spin Case 2 since both sides of the earth are closer in temperature. Case 2 is a less efficient mechanism for gaining energy from the sun, but at least we don't freeze or boil to death.

Your intuition seems to be right, entropy shows an inefficiency to the storage mechanism.

 

[IanC]

Efficiency is a big part of the real world. The most efficient way to enclose area with a defined circumference is a circle. The radius is equal in all directions. A square is not as good but still better than a rectangle.

The moon receives as much solar energy as the Earth so why isn't it as warm? The main reason is the length of a lunar day/night cycle. It is very hot in the day and very cold at night. The energy inputs and outputs are extreme, less efficient. Like how a circle has more area per circumference than a flat rectangle. Only more so, because area is a ^2 relationship and temperature is a ^4 relationship. If the lunar 'day' was half as long the average temperature would be warmer. If you halved it again, warmer still. And so on, ignoring the effects of spinning the mass.

The predicted temperature of the Earth is -18C, from the average amount of sunlight hitting it. Now hold on a minute. We were talking about efficiencies. An Earth day is shorter than a lunar day but it is still 24 hours. Shouldn't the hypothetical max (circle) be greater than the actual Earth (square) which is greater than the Moon (rectangle)?

Obviously the atmosphere has something to do with it. But how does the atmosphere raise the average temperature ABOVE the theoretical maximum derived from the solar input?

I'll talk more about this if anyone is interested. Or we can just leave it as another unanswered question.

Hey Wuwei, you like theoretical problems.

Don't you find it interesting that the GHE is large enough to more than compensate for the inefficiencies caused by unequal input and output energies?

While it is just a variation on the theme of why the surface is warmer than the Sun's input, don't you find it amazing that entropy gained by converting sunlight to IR can 'power' such a large effect?

I realize the actual energy comes from radiation not lost to space but it seems to me that entropy gained in the total area causes entropy loss by the storage mechanism in a subset area.

I could be wrong. What do you think?

I looked at the idea of entropy for a rotating absorber (the moon). The following formula works for simple problems that are closed system, a solid (no atmosphere), and at equilibrium. The change in entropy is given by

delta S = n Cp log(Tf/Ti) where,

n = moles of moon surface that is warmed,
Cp = specific heat of surface,
Ti = initial temperature,
Tf = final temperature.

The factors in n x Cp are hard to determine. They refer to the mass and specific heat of the mass being heated.. They will be replaced by M referring to the amount of material being heated.

Dark side of moon Ti = 100 K
Bright side of moon Tf = 373 K

When looking at the sun hitting the moon there are two extremes where a quasi-equilibrium exists.

1. The moon is slowly rotating, and the sun is shining on one side for an extended period.
2. The moon is rapidly rotating, and night is so short that the temperature has no time to drop.

Case 1. The delta entropy for energy hitting the light side of the moon only.

delta S = M log(373K/100K) = M x 0.572

Case 2 . This is a bit trickier because there is no temperature data for the non-existing case of both sides getting half the energy all the time, but it can be computed from what we already know.

The sun energy E results in a temperature of Tf.
From SB equation, E = K Tf ^4. Where K contains all the proportionality constants.

Let T2 be the unknown Case 2 temperature of the moon.
The whole moon will be receiving half the energy. 2 x E = K T2^4.

Using simple algebra, Tf = T2 x fourth root of 2 or, T2 = Tf x 0.841

The entropy formula for case 2 is
delta S = 2 M log(Tf x 0.841 / Ti) = 2 M log( 3.137) = 2 M x .496
Note: the leading factor of 2 is because twice as much material (both sides) are being heated.

Summary for the moon:

Case 1 delta S = M .572
Case 2 delta S = M .992

The Case 1 slow spin is almost twice as efficient in using energy from the sun.

The earth, it is more like the fast spin Case 2 since both sides of the earth are closer in temperature. Case 2 is a less efficient mechanism for gaining energy from the sun, but at least we don't freeze or boil to death.

Your intuition seems to be right, entropy shows an inefficiency to the storage mechanism.

Thanks for your hard work. I hope you had fun doing it.

Unfortunately it doesn't seem to address the difference in average temperature between the slow and fast spinning globes. The faster spinning globe is warmer on average.

I'll read your comment through a few more times.

 

[Wuwei]

Thanks for your hard work. I hope you had fun doing it.

Unfortunately it doesn't seem to address the difference in average temperature between the slow and fast spinning globes. The faster spinning globe is warmer on average.

I'll read your comment through a few more times.

T2 is the unknown Case 2 temperature of the moon.

Using simple algebra, Tf = T2 x fourth root of 2 or, T2 = Tf x 0.841

Tf is 373K but I didn't multiply it out to get T2

T2 = 313.7K

 

[The Sage of Main Street]

This law-

And the law that energy cannot be created or destroyed.

You can only control the amount of radiation produced by an object by adding or subtracting to the amount of energy available to be converted to radiation.

Any object above absolute zero will attempt to cool down by radiating. It always radiates but it only cools if it emits more energy than it receives.

That law states no such thing...that law states that a perfect black body radiating in a vacuum devoid of any other matter radiates according to its temperature to the 4th power...an ideal, perfect black body in a perfectly empty universe...a fantasy.

Bring matter into the equation and j starts changing in accordance with the difference between the temperature of the radiator and its cooler surroundings. You keep forgetting the basics and jumping to knee jerk conclusions...think ian, I know that you are capable of it.

There is no law that says that the rate of energy emitted by an object can not change if its environment changes...quite the contrary..the application of the SB law relating to radiators that are not perfect black bodies radiating into perfectly empty space says exactly the opposite...P keeps changing as the difference between the temperature of the radiator and the temperature of its surroundings change.

This equation is obviously not for a perfect Blackbody because it has a term for emmisivity.

You should also note that it has no term for area. It is an intensive property.

If you divide the j by area then you get an actual amount of power, an extensive property. But you then have to do the calculations for every angle that is in line-of-sight of the area you defined.

Your confusion seems to stem from confusing the intensive property of j with the extensive property of P.

The object exists and radiates. The environment that surrounds the object does not change the radiation, which is always proportional to the temperature of the object.

We can turn the intensive property of radiation into the extensive property of power by defining the radiating surface area of the object. We can turn the intensive property of temperature into the extensive property of total heat content by measuring the mass and the specific heat for its constituents.

With the two extensive properties of power and heat content we can calculate the rate of cooling as time passes. Quickly at first and then progressively slower.

If we add a second object nearby, it also radiates in the same fashion as the first. Some of the radiation from the first will be intercepted by the second, and vice versa. The amount will be determined by the line-of-sight connection. The amount of radiation energy absorbed will add to the total energy content of the two objects, thus slowing the rate of their cooling. But both objects will still be cooling overall.

I will leave it there for now.

You can appeal to as much complexity as you like ian, but that law does not state that an object can not radiate less when it is in the presence of other matter..

The whole topic is moot as IR does not warm the air.

Actually I am appealing to simplicity.

j = sigmaT^4 is the most simple of the S-B equations. It makes the assumption of a Blackbody and gives the amount of radiation for any temperature. Regardless of the environment that it is imbedded in, because it doesn't have a term for the environment. The radiation is always that much. The only way you can change the radiation is to change the temperature.

The first, but still reasonably simple step up in complexity is to add the emmisivity term. Emission and absorption are still exactly equal but are less than the unity value of a blackbody. Emmisivity can be further defined by individual wavelengths. Eg, CO2 has near perfect absorption at 15 microns near perfect non-absorption at 10 microns.

The next step up is to include the environment. Any of the equations that include some variation of the term (T^4 - Tc^4). This is a huge complication, because we are now dealing with radiation both going out and coming in. We are dealing with at least two objects, and imbedding them in three dimensional space. We now have to add the area term with complex line-of-sight calculations. And emerging temperature gradients. Etc.

Only Closed Minds Theorize a Closed Universe

Fission science is a lot more important than the Greenies' greenhouse plaything. Since you know the orthodox dogma on this, wouldn't it explain a lot of things if instead the first collision after fission actually happens at the square of the speed of light (a light-year in 3 minutes). Is it possible to measure that now? Why should we believe that nothing can travel faster than light, especially if it is energized from outside this universe?

 

[SSDD]

But it places the object in a space where there is no other matter...when you bring other matter into the picture, the equation changes and J starts to change with the difference in temperature between the radiator and its surroundings.

That is incorrect. The radiation j is always there, always sigmaT^4.

Not true..unless you live in a model.

 

[SSDD]

The radiation is immediately emitted on towards cooler pastures...

You keep saying this but you never explain what you actually think is going on.

Exactly what I said is going on...radiation only moves from warm to cool...it can't spontaneously move anywhere else so if you are a theoretical photon leaving the surface of the earth, where are you constantly moving towards...

http://www.usmessageboard.com/posts/19139813/I think we all agree that the surface emits 15 micron energy. I think we all agree that this radiation is absorbed to extinction within just a few metres of the atmosphere.[/quote]

No...we are not agreed on that...if it we're absorbed to extinction, that would suggest that IR could warm the air...it can't...

Once absorbed, where does it go? SSDD has two mutually exclusive statements on this.

It is absorbed and emitted or the energy is transferred via collision...if it is emitted then it is moving towards a cooler region.

Sometimes he says it is immediately re-emitted. Sometimes he says it is transferred to nearby molecules by collision, 99.9999% of the time. So which is it?

Do you know what the term sometimes means? On rare occasions it is reemitted...in the vast majority of instances it is passed on via collision. Not sure what is so difficult to understand about that....I say what i mean to say...no interpretation required on your part.

What does 'towards cooler pastures' mean?

Geez ian, are you not able to take a cooler region from that statement...any cooler region.. Since it can't move towards a warmer region, and the CO2 molecule has no capacity to retain it at atmospheric temperatures, where else would it go?

SSDD claims that energy in all forms, but specifically radiation, can only 'spontaneously' from warm to cool.

Not my claim ian...it is the statement of the 2nd law of thermodynamics..

Therefore radiation is prohibited from being emitted toward the surface (presumably by some unknown mechanism).

Unknown being the operative word...you seem to think that science knows all...guess again. If enough IR were being radiated back towards the earth to alter the cliamte, you would not need to cool an instrument to -80F in order to measure it...and that is the only way to measure so called back radiation.

Even though radiation can be consistently measured coming from the atmosphere in the direction of the surface. Apparently we are being 'fooled by instrumentation'.

Only if you cool the instrument to a temperature lower than that of the atmosphere. Reference being fooled by instrumentation. IR can not warm the air.

 

[Toddsterpatriot]

The radiation is immediately emitted on towards cooler pastures...

You keep saying this but you never explain what you actually think is going on.

Exactly what I said is going on...radiation only moves from warm to cool...it can't spontaneously move anywhere else so if you are a theoretical photon leaving the surface of the earth, where are you constantly moving towards...

radiation only moves from warm to cool...it can't spontaneously move anywhere else

We've already determined that radiation can move from the 296 K walls of my house toward my 307 K body.

 

[SSDD]

We've already determined that radiation can move from the 296 K walls of my house toward my 307 K body.

No we haven't...we have determined that via conduction, not radiation energy absorbed by your walls can be transferred to the air. And air which is at a lower temperature than your body doesn't impart any energy to your body...your body radiates according to the temperature difference between it and its surroundings...if the air is warmer, then your body radiates less. T-Tc.

 

[Toddsterpatriot]

We've already determined that radiation can move from the 296 K walls of my house toward my 307 K body.

No we haven't...we have determined that via conduction, not radiation energy absorbed by your walls can be transferred to the air. And air which is at a lower temperature than your body doesn't impart any energy to your body...your body radiates according to the temperature difference between it and its surroundings...if the air is warmer, then your body radiates less. T-Tc.

No we haven't.

You said radiation can travel from cooler to warmer matter, if work is done.
That's why radiation is allowed to flow from the Sun's surface toward the corona. Correct?

Exactly how do you think the walls of my home reached 296 K?

 

[IanC]

We've already determined that radiation can move from the 296 K walls of my house toward my 307 K body.

No we haven't...we have determined that via conduction, not radiation energy absorbed by your walls can be transferred to the air. And air which is at a lower temperature than your body doesn't impart any energy to your body...your body radiates according to the temperature difference between it and its surroundings...if the air is warmer, then your body radiates less. T-Tc.

No we haven't.

You said radiation can travel from cooler to warmer matter, if work is done.
That's why radiation is allowed to flow from the Sun's surface toward the corona. Correct?

Exactly how do you think the walls of my home reached 296 K?

Exactly. Any object that could simply radiate freely without absorbing energy from it's surroundings would quickly approach absolute zero temperature. But that still leaves the question of how it got to its initial temperature.

Two object of unequal temperature are close to each other. They will both radiate freely except in the area defined by line-of-sight. They will both cool to absolute zero, but more slowly. The initially cooler object may not even cool at first if the net exchange is large enough to offset the radiation loss by the rest of its surface area not in line-of-sight to the warm area. But this initial stage would quickly pass, and both would then continue to cool. This is the passive 'spontaneous' part that SSDD claims is the only thing that happens in thermodynamics. That everything cools, and that only the rate of cooling can change.

In reality, there are energy sources. Especially the Sun but other sources as well in smaller systems, like the furnace in your house. That is a different thermodynamic story that does not lead to absolute zero, but instead, to equilibrium temperatures and temperature gradients.

 

[IanC]

From over at Tallbloke. Model output.

Temperature at the equator for spin rates from very fast to very slow.

Sorry Wuwei. I keep meaning to get back to this but...

Interesting that the two extreme temperatures are very close to equal for rotation rates of 24 hrs or longer.

 

[IanC]

From over at Tallbloke. Model output.

Temperature at the equator for spin rates from very fast to very slow.

Sorry Wuwei. I keep meaning to get back to this but...

Interesting that the two extreme temperatures are very close to equal for rotation rates of 24 hrs or longer.

By eyeball it looks like the average temperature would be about 280 for infinitely fast ( even radiation), just under for 0.01 (14 minutes per spin), 250 for 0.1, 220 for a 24 hour day, 190 for a lunar day, and bottoming out at about 185 for very, very long days.

The poles would be colder but so what? There is no mechanism to spread the heat out from the equator.

100K drop just from lengthening the time of the day/night cycle. A big inefficiency.

 

[Wuwei]

From over at Tallbloke. Model output.

Temperature at the equator for spin rates from very fast to very slow.

Sorry Wuwei. I keep meaning to get back to this but...

Interesting that the two extreme temperatures are very close to equal for rotation rates of 24 hrs or longer.

By eyeball it looks like the average temperature would be about 280 for infinitely fast ( even radiation), just under for 0.01 (14 minutes per spin), 250 for 0.1, 220 for a 24 hour day, 190 for a lunar day, and bottoming out at about 185 for very, very long days.

The poles would be colder but so what? There is no mechanism to spread the heat out from the equator.

100K drop just from lengthening the time of the day/night cycle. A big inefficiency.

It looks like my computation is slightly off from your graph. I got 313 K where your graph is a bit lower, at 280 K.
However the limiting values I used are different. I used a low of 100 K, and a high of 373 K.

 

[IanC]

An average temperature for the moon is a bit farcical anyways. And there is nothing to make use of the entropy either.

But it is useful as a landmark for what the Earth might look like without an atmosphere. I wonder how long the day was 500,000,000 years ago. Obviously it would have made some difference.

 

[Wuwei]

An average temperature for the moon is a bit farcical anyways. And there is nothing to make use of the entropy either.

But it is useful as a landmark for what the Earth might look like without an atmosphere. I wonder how long the day was 500,000,000 years ago. Obviously it would have made some difference.

Not totally farcical. If we want to colonize the moon we have to figure out how to speed up the rotation to maybe four revolutions per hour. That would play havoc with our circadian rhythm though.

 

[Toddsterpatriot]

An average temperature for the moon is a bit farcical anyways. And there is nothing to make use of the entropy either.

But it is useful as a landmark for what the Earth might look like without an atmosphere. I wonder how long the day was 500,000,000 years ago. Obviously it would have made some difference.

Not totally farcical. If we want to colonize the moon we have to figure out how to speed up the rotation to maybe four revolutions per hour. That would play havoc with our circadian rhythm though.

Live underground.......

 

[Billy_Bob]

Here we go again...

 

[SSDD]

They have tried every other scare word in an attempt to get people to believe in the sham...may as well try cancer...although, it is pointless as the climate sensitivity to CO2 is steadily trending towards zero...maybe when it reaches zero, where it belongs, they may get around to addressing the real cancer...pollution, and begin working on a cure.

 

[Toddsterpatriot]

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater..

Excellent! My walls were warmed because of work.
That's why they are allowed to radiate toward my warmer body.

 

[SSDD]

Excellent! My walls were warmed because of work.[/quote[

Absorbing radiation is not work. What work do you suppose your walls performed in order to emit energy towards a warmer object?


You can go on about this as much as you like..and demonstrate beyond any doubt that the topic is beyond you....but your walls are not radiating towards you...as they warm to a temperature closer to that of your skin, you lose less energy to them...the cool walls are not warming you up...