Lets finish this! Numbers do not lie.

reply to post by 4nsicphd

CO^2 has absorption peaks at 2.4and 4 microns, and a shoulder

Peaks at the wavelengths you mention equate to frequencies of 125,000 GHz and 75,000 GHz, respectively. That is the heat spectrum (IR). The shoulder at 13 microns equates to about 23,000 GHz, which is the lower end of the heat spectrum (far-infrared), not far above microwave radiation.

Of further consideration is the bandwidth of these peaks, each of which is les than a micron wide. Any radiation outside these narrow frequency bands is not affected.

I have come across conflicting information as to exactly what the absorption bands are. This graph shows the bands (and provides some comparison to other gases), but the shoulder shows as a wider band (maybe 6-7 microns) and the wavelengths are slightly different from those you mention. But, your figures agree with other charts.



Just some quick conversions for those more familiar with frequency than wavelength.

frequency= 300,000,000/λ when λ is in meters (1 micron = 0.000001 m).

TheRedneck

Originally posted by DjSharperimage

IT'S NOT CARBON DIOXIDE (CO2);
IT'S CARBON MONOXIDE (CO)


[edit on 2-12-2009 by DjSharperimage]

Why pray tell would you say that. Its absorption is in the wrong range and it's not a greenhouse gas. It's a product of inefficient incomplete burning. The stoichiometric formula for gasoline burning is C8H18--->9H2O+8CO2. Not a single CO molecule.
The stoichiometry for coal burning, ignoring the sulphur and water content, is 2C10H2+21O2 --->2H2O+20CO2. Nary a drop of CO
BTW, For the many whose Chemistry classes ended with High School, let me say Bravo for taking it then and Stoichiometry is just a big word for balancing. You've got so much stuff going into a reaction - you need the same amount in the product.
Anyway, CO will kill you but it's not a greenhouse gas and emissions into the3 atmosphere barely exceed 1 billion tons/year, whereas CO2 emissions are more than 9 billion tons/year per EDGAR 2000; RIVNNetherlands) 2000.)

I just though i would add this video here, as its a pretty good explaination of what is actually going on with scientific data to back it.
www.youtube.com...

A true eye-opener. no mistake about that.

S&F 4 this thread.



kind regards,

I hate to throw a monkey wrench into all those fine calculations, but no one seems to accounting for one of the biggest contributors to atmospheric heating: jet engines.

flightaware.com...

Just over the US alone, there are 4719 aircraft in flight right now as I write this.

Each of those aircraft are emitting heat into the atmosphere within the 5K-30K band for the most part. Where jet engines are concerned, you need to calculate the cubic volume of air per minute for, say, 9,000 engines, assuming 3000 aircraft are jet-powered and average 3 engines per aircraft. Each heats that volume to around roughly 450 degrees Centigrade, not sure what the cruising speed temps are on average, but I'm being very conservative I think.

Now consider that this is just a fraction of the global output, and that it is a constant, in that it never stops, hasn't since the dawn of the jet age.

Until I see someone take all that heating into account, I assume all global warming calculations to be erroneous on the low side.

reply to post by TheRedneck




Red, I don't think you should use the specific heat of air to find the contribution of CO2

Air is
Nitrogen N2 78.084 %
Oxygen O2 20.9476 %
Argon Ar 0.934 %

only 0.03% CO2, CO2 has a complete different property that allows it to trap light within the atmosphere. In fact, I'm not sure if specific heat is the right thing to use here.

Originally posted by DjSharperimage

Originally posted by downisreallyup
reply to post by DjSharperimage

 

Other than cars there are still other things that are creating more CARBON MINOXIDE (CO) (NOT CO2; THERE IS NO F'IN 2 IN IT FOREVER!!!!)
such as burning natural gas for heating, cooking, industrial companies, and also electric companies burning coal, and forest fires and etc....



[edit on 1-12-2009 by DjSharperimage]

OK, explain to me where the CO comes from when you completely burn gasoline. The stoichiometry of this oxidation-reduction reaction, ignoring additives is C8H18+12.5O2--->9H2O+8CO2. Where is any CO? Not there.Lets look at natural gas. It's primarily a simple alkane, CH4, or methane. It's CH4+2O2--->2H2O+CO2. If the oxygen comes from air, you will also have 7.52 N2s on each side of the equation and the same with any other trace elements.
Please show me a stoichiometrically correct complete oxidation-reduction reaction with oxygen and a carbon fuel which results in CO as a product and I will take it all back.
Coal fired power plants don't emit CO either, although they spew enormous amounts of CO2, and SO2, which combines with O2 and H2 to form H2SO4, which is sulphuric acid, and which causes acid rain.

Originally posted by apacheman
I hate to throw a monkey wrench into all those fine calculations, but no one seems to accounting for one of the biggest contributors to atmospheric heating: jet engines.

flightaware.com...

Just over the US alone, there are 4719 aircraft in flight right now as I write this.

Each of those aircraft are emitting heat into the atmosphere within the 5K-30K band for the most part. Where jet engines are concerned, you need to calculate the cubic volume of air per minute for, say, 9,000 engines, assuming 3000 aircraft are jet-powered and average 3 engines per aircraft. Each heats that volume to around roughly 450 degrees Centigrade, not sure what the cruising speed temps are on average, but I'm being very conservative I think.

Now consider that this is just a fraction of the global output, and that it is a constant, in that it never stops, hasn't since the dawn of the jet age.

Until I see someone take all that heating into account, I assume all global warming calculations to be erroneous on the low side.

This is relatively simply to figure out. Let's say the average jet engine is 20 feet in diameter. So the area of a engine is: pi*20=62.83 ft^2. Lets say the average jet engine is going 500 mph all the time (which is isn't, but we'll err on the high side). That means it covers 733 feet per second, so the volume of air going into a jet engine is 46,055.75 ft^3/second. Now lets just double that, again to err on the high side: 92,111.50 ft^3/second of air is heated to 400 degrees every second by every jet engine. So in a day, it's heating 7,958,433,306.52 ft^3 of air. And lets say there are an average of 9000 jet engines operating at a time, as you say. That gives us 71,625,899,758,689.33 ft^3 per day being heated over the US.

Now lets look at the total volume of air over the US. I'll just count the first 30k feet. The land area of the US is 3,717,813 miles^2, or 103,646,677,939,200 ft^2. Multiply that by 30,000 feet altitude, you get a volume of 3,109,400,338,176,000,000 ft^3. So the amount of atmosphere being heated by jet engines is, at most, 0.002% per day. Since the US has the most air traffic, this number will only go down if you look at the whole globe. Hence, the heating due to the output of jet engines is practically non existent compared to the sun.

Originally posted by erkokite

Now we can calculate how much energy it would require to raise the temperature of the troposphere by a single degree Kelvin:

1.012 J/g·°K = 1.012 kJ/kg·°K

1.012 kJ/kg·°K · 1.2 kg/m³ = 1.2144 kJ/m³·°K

1.2144 kJ/m³·°K = 1,214,400,000 kJ/km³·°K

You are assuming a constant density of 1.2 kg/m^3 for the atmosphere up to 17 km. This is entirely incorrect. At 17 km, atm density is 0.16 kg/m^3. You can find a rough average density by integrating density by height from 0 to 17 km altitude and then dividing by 17 km. So your real value you should be using is roughly:

(1.22 - 0.16) / 2 = 0.53 kg/m^3

Also the atmosphere extends out to roughly 60 km (depending on time/temp/atm conditions), with the density changing as a function of all these things (it gets a lot lower).

To do this properly, you need to use the absorptivities (throughout the optical spectrum) of CO2, and all other greenhouse gases to find heat input for each greenhouse gas.

Integrate over the absorption spectrum/irradiant solar energy like this:

Qtotal=ʃα(λ)Q(λ)dλ

where α is the absorptivity of CO2 at a given wavelength of light, and Q is the irradiant power from the sun as a function of wavelength (you essentially assumed this was a constant based on 1366 W/m^2*Earth cross sectional area).

This is probably the most important issue. You have neglected the fact that excessive CO2 and other greenhouse gases can affect the total absorptivity greatly enough to cause a temperature rise.

[edit on 1-12-2009 by erkokite]

[edit on 1-12-2009 by erkokite]

This is the best and most accurate post in the entire thread. Everyone, except you, is ignoring a fundamental formula. E=hv. The energy is directly proportional to the frequency (and wavelength) of the photon mediating the energy. h is Planck's constant (6.626X10^-34 joule seconds. The OP's energy calculations in joules/meter squared are obviously at a constant wavelength.Because of atmopheric filtering and absorption, wavelengths can change. Visable Sunlight is in the 400-7--nanometer range giving a frequency of c (3X10^8 meters/second)divided by 4-700 nanometers. Bouncing off things changes the wavelength and this energetic photon can be absorbed by CO2. CO2 has absoption peaks at 2.6 and 4 microns wavelength and a shoulder or complete blockout at 13 microns. If anywhere, this is where the OP went astray. There, and ignoring the absorption quality of the receiving substance. The reflection of visible and near-
infrared light from a surface is important for climato-
logical studies involving the energy balance . Energy is not onlycoming down from the sun, but is also coming up as reflected, radiated or convected energy.

Originally posted by DjSharperimage

Originally posted by neo5842
Thi

IT IS NOT CARBON DIOXIDE (CO2)
IT IS CARBON MONOXIDE (CO) !!!!!!
will you people please!!!!! learn your table of elements!!@!!!!!

I agree wholeheartedly with the las sentence. Neither Carbon Monoxide nor Carbon Dioxide are elements, people!!!. They are molecules!!!They are called compounds and they are either ionic or covalent. So you get a big Chemistry FAIL! Diatomic oxygen will not combine with carbon in a completed oxidation reduction combustion reaction to form CO. Carbon has 4 valence electrons. Oxygen has 6. Research oxidation reduction reactions and organic chemistry before you shout about what it is and is not.

reply to post by wx4caster

wx4caster - you get an A. Great catch on the units problem. and th disc v. sphere issue.

reply to post by die_another_day

Red, I don't think you should use the specific heat of air to find the contribution of CO2

I don't.

I use the concentration of anthropogenic carbon dioxide in the atmosphere (100 ppmv or 0.01%) to find the total amount of energy that it can account for, based on solar irradiance.

I then use that amount of heat that anthropogenic carbon dioxide can account for, under the worst possible conditions, to determine how much it will heat the troposphere and oceans. That's why I use the specific heat capacity of the air instead of the specific heat capacity of carbon dioxide. It is the entire troposphere which is heated, not just the carbon dioxide.

The specific heat capacity of air is an average value based on the average composition percentages and specific heat capacities of the components of that air.

TheRedneck

post by TheRedneck

A carbon dioxide molecule is invisible to visible light; we cannot see it. But it is not completely invisible to certain wavelengths of light. There are actually two areas of the spectrum (absorption bands) that CO2 is more translucent to than invisible. both are located in long-wavelength radiation (heat radiation).

The theory is this: sunlight can reach the earth just fine, since it is mostly shorter wavelengths and is not affected by CO2. But when the earth absorbs short wavelength light, it begins to emit longer-wavelength heat. That heat would simply escape out into space if not for the greenhouse gases in our atmosphere intercepting them. If a ray of heat strikes a greenhouse gas, it causes that gas to heat up. Then, when the gas molecule cools back down, it releases that heat again, this time in a random direction. Sometimes the emitted heat will go on into space, but other times it will head back to earth to be absorbed and re-emitted by the ground again.

Thanks, that pretty much explains everything I was wondering about.

In your equations then, how do you determine the percent chance of light that will hit a CO2 atom? It seems you are saying in your OP that if CO2 was say is 2% of all gasses in the atmosphere that it would intecept 2% of the light. Is that correct?

In my opinion, given your description of how global warming works, the way to determine how much heat will be captured would be to determine:
- The area (from a cross-section-wise perspective) of an individual CO2 atom.
- The area within that area which if hit would result in absorption of a photon... I would guess 100% but don't know the physics.
- How many CO2 atoms there are per area unit in the atmosphere.
So, by multiplying the cross-section area of CO2 by the number of atoms per unit area you can calculate what the maximum area is of CO2 per unit area that would be that is vulnerable to interaction with light rays. Using a statistical function (not sure which one) you could cancel out CO2 atoms who's area overlaps.

The next step would then be to determine the area (again as a cross-section perspective) of a photon. Since I don't really know too much about thermodynamics there is probably no point in me going on much further but the point is my idea would be to match up the area of the photon.

reply to post by 4nsicphd

although they spew enormous amounts of CO2, and SO2, which combines with O2 and H2 to form H2SO4

I just want to make a clarification to that. From the wording, it could be believed that you are saying carbon dioxide (CO2) is an ingredient in the manufacture of sulfuric acid (H2SO4), which it is not. Sulfur dioxide (SO2) is the main ingredient in atmospheric production of sulfuric acid; carbon dioxide does not enter into that equation.

I know you didn't intend for it to come across in that way, but it could to someone without chemical knowledge.

Kudos on your explanations, nonetheless.

TheRedneck

reply to post by 4nsicphd

The energy is directly proportional to the frequency (and wavelength) of the photon mediating the energy.

The average energy coming from solar irradiation is a value that is compiled by NASA (see source #3). I feel certain that NASA has thought to include the energy calculations surrounding Plank's constant.

If they had not, the resulting values would be in something like lumens per square meter instead of Joules per square meter.

TheRedneck

reply to post by nataylor


Here's a link to a nice jet engine caclulator:

www.grc.nasa.gov...

Output parameters seem to be showing a CF6 engine heats about 47 kilos worth of air from -54 Centigrade to around 1000 degrees Centigrade every minute at 3/4 throttle at 35k feet/10K meters. If we consider this average (I know, very sloppy), then we come up with 140,000 kilos of air heated from -54 to 1000 degrees centrigrade every minute of every day, not counting smaller engines and the military, all within a very narrow band of the atmosphere.

It isn't a question of merely being a small component, the real question is whether anyone's taken the cumulative amount for over fifty years worth of constant heating into account. Even a small amount can add up to a signifcant figure, especially given the exponential rise in the number of active engines over the last thirty years.

Where's Melatonin?

Licking his wounds over Hadley CRU?

Or still studying the numbers in this thread?

Conspicuous by his absence...

Anyway...
Great job Redneck - I've only just come across this, and it's going to take me quite some time to make sense of the numbers/equations, simply because I am REALLY out of practice.

But I'll have a go.

In the meantime - Kudos, S&F

But, one small point.
Using the oceans as a heat sink is all well and good, but there is a lot more to the Earth than just ocean and land mass.
Would it not be true to say that the upper mantle, mantle and crust play a part in all this?
After all if the ocean soaks up heat, that must in some way be transferred to the crust - just to start with.
Of course, the mass is much greater, but is this a factor?

For my part, I have long believed that AGW is the warmest form of bull# - my very first thread stated this.

I don't have the science to express it, but I do know when I'm being conned.

reply to post by budski

Would it not be true to say that the upper mantle, mantle and crust play a part in all this?

Yes, but to a very small amount. It is generally accepted that once one begins to move deep into the crust, the temperature rises dramatically. After all, that's where lava comes from.

I look forward to your analysis.

TheRedneck

Originally posted by nataylor

Originally posted by apacheman
I hate to throw a monkey wrench into all those fine calculations, but no one seems to accounting for one of the biggest contributors to atmospheric heating: jet engines.

flightaware.com...

Just over the US alone, there are 4719 aircraft in flight right now as I write this.

Each of those aircraft are emitting heat into the atmosphere within the 5K-30K band for the most part. Where jet engines are concerned, you need to calculate the cubic volume of air per minute for, say, 9,000 engines, assuming 3000 aircraft are jet-powered and average 3 engines per aircraft. Each heats that volume to around roughly 450 degrees Centigrade, not sure what the cruising speed temps are on average, but I'm being very conservative I think.

Now consider that this is just a fraction of the global output, and that it is a constant, in that it never stops, hasn't since the dawn of the jet age.

Until I see someone take all that heating into account, I assume all global warming calculations to be erroneous on the low side.

This is relatively simply to figure out. Let's say the average jet engine is 20 feet in diameter. So the area of a engine is: pi*20=62.83 ft^2. Lets say the average jet engine is going 500 mph all the time (which is isn't, but we'll err on the high side). That means it covers 733 feet per second, so the volume of air going into a jet engine is 46,055.75 ft^3/second. Now lets just double that, again to err on the high side: 92,111.50 ft^3/second of air is heated to 400 degrees every second by every jet engine. So in a day, it's heating 7,958,433,306.52 ft^3 of air. And lets say there are an average of 9000 jet engines operating at a time, as you say. That gives us 71,625,899,758,689.33 ft^3 per day being heated over the US.

Now lets look at the total volume of air over the US. I'll just count the first 30k feet. The land area of the US is 3,717,813 miles^2, or 103,646,677,939,200 ft^2. Multiply that by 30,000 feet altitude, you get a volume of 3,109,400,338,176,000,000 ft^3. So the amount of atmosphere being heated by jet engines is, at most, 0.002% per day. Since the US has the most air traffic, this number will only go down if you look at the whole globe. Hence, the heating due to the output of jet engines is practically non existent compared to the sun.

Yor conclusion is right on but several facts are way off. First,the initial poster uses an average of 3 engines per jet.for all aircraft flying. Way too high. The only civil four engine jets flying in the US tare the Boing 747 and the Jetstar, an ancient Lockheed business jet using 4 baby GE CJ-610-8s. Anoccasional Antonov 105might come to the US. I just did a smpling on Flightaware of all jets over Florida and the average was 2 engines/jet . There were no 4 engine jets. And the temperature figures are way off. Most jet engines made today are high-bypass which means most of the air bypasses the combustion section. The exhaust gas temperature EGT for these engines is typically 800-900 in cruise flight but only 20% of the air entering the engine inlet is heated. And the 20 foot average diameter figure for engine inlets is way off. Even the huge TF39 powering the C5 SuperGalaxy has an inlet size of 10'4". A CJ-610, which powers the 20 series Learjets is only 3'2" in diameter. The average would be about5', given the airline/GA mix.

Originally posted by TheRedneck
reply to post by die_another_day

Red, I don't think you should use the specific heat of air to find the contribution of CO2

I don't.

I use the concentration of anthropogenic carbon dioxide in the atmosphere (100 ppmv or 0.01%) to find the total amount of energy that it can account for, based on solar irradiance.

I then use that amount of heat that anthropogenic carbon dioxide can account for, under the worst possible conditions, to determine how much it will heat the troposphere and oceans. That's why I use the specific heat capacity of the air instead of the specific heat capacity of carbon dioxide. It is the entire troposphere which is heated, not just the carbon dioxide.

The specific heat capacity of air is an average value based on the average composition percentages and specific heat capacities of the components of that air.

TheRedneck

By not separating the gases, you're assuming that CO2 has the same properties as N2 and O2.