Flat Plate in Space is Warmer than a Flat Plate with a Greenhouse Connected
Firstly, in order to understand this comparison, I will explain the science involved so you can see how thinking a greenhouse works because of trapped radiation is just plain stupid. I will briefly go over some of the basic information needed to understand this issue.
Steady State Temperature Equation
EQ1: 0 = ά E A(a) – έ A (e) σ T4
This is the equation which initially explains steady state temperatures of objects in space. This shows that the energy in, must balance with the energy out, the principle of energy conservation. This is the equation you need to know. The zero tells us that both sides of the equation will balance. The first part of the equation is the absorptivity side of the equation. This tells us the amount of solar radiation which is incident upon an object is absorbed by that object.
First Part of the Equation the Absorption Side
ά E A(a). This part of the equation lets us know how much energy is being absorbed by the object. It can be modified to take account of incidence angles, but I am assuming perpendicular conditions.
ά = Solar Absorptivity of the object, in this case it is going to be a flat plate, which is calculated by 1. (Black body object) Absorptivity is expressed as a ratio of absorption. Where 1 is full absorption and 0 is none. If a surface reflected 10{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} of incoming then an absorption ratio would be 0.90 The wavelength range for determining Solar Absorptivity is usually 0-2,500 nanometres at Earth orbit.
E = Solar constant emission value of 1367 W/m2
Solar Constant
The output of the sun when expressed in energy terms is between 64,169,058 W/m2 and 73,490,000 W/m2 depending upon what the surface temperature is, some sources say 5,800K others say 6,000K. This gives a radiant heat output 73.4 Million W/m2. When the strongly reductive effect of the Inverse Square Law is applied to radiation, we find we arrive at a Solar Constant of 1,367 W/m2 at the edge of the Earth’s Atmosphere. It varies a little bit depending upon the position of the earth during a year, but this is generally an agreed average value.
A (a) = Area of absorption, which is the area of the flat plate. We are going to say that the flat plate is directly perpendicular with the Sun’s rays and that the plate is just 1 square meters. This is to make the numbers easy to follow.
So on the energy in side we see that:-
ά E A(a) = 1 (black body plate) x 1367 w/m2 x 1m2.
So the amount of energy absorbed by the flat plate is quite simply 1367 w/m2. A black body object achieves full absorption of energy available 1367 watts. Simples.
The law of energy conversation means Energy In must Equal Energy out. So we know that the amount of energy out must be 1367 watts, but we don’t know what temperature the flat plate is to achieve that so we do some backwards engineering to figure that out.
Second Part of Equation the Emission Side
The second part of the equation έ A (e) σ T4 is the emissivity side of the equation. This tells us the amount of radiation which is being emitted by an object at a certain temperature.
If amount of energy being absorbed of an object equals the amount of energy being emitted by an object at steady state conditions then using these two total figures will enable us to figure out what the temperature of the object is.
έ = Infra-red emissivity of the surface. The emissivity of the plate is considered to be 1, a perfect black body. Again like absorptivity emissivity is expressed as a ratio where 1 is full emission across the spectra and 0 is none. The wavelength range for determining infra-red emissivity’s is normally 2,500-60,000 nanometres.
A (e) = Area of emission, which is the area of a flat plate at 1 m2 = 2 square meters. (A plate has two sides)
σ = Stefan Boltzmann Constant = 5.67 x 10-8 W / (m2K4)
T = Temperature of Emissive body at steady state conditions.
We are aiming to find T so the equation needs to be re-arranged to enable us to do this.
EQ 1 : 0 = ά E A(a) – έ A (e) σ T4
EQ 2 : έ A (e) σ T4 = ά E A(a)
EQ 3 : T = (A (a) / A(e) * άE / έ σ) 1/4
EQ 4 : T =( (1m2)/ 2m2)( (1.01367 W/ m2) / (1.0 * 5.67 x 10-8 W / (m2K)) ) ¼
EQ 5 : T = ( 1/2 * ( 1367 / 5.67 x 10-8 K )¼
EQ 6 : T = ( 0.50 * 24,198,347,443 ) ¼
EQ 7 : T = 12,054,673,721¼
EQ 8 : T = 331K or 58 C°
So what we have is the temperature of a two sided flat plate is 331K.
What does this equation tell us?
If the absorptivity and emissivity remain the same, i.e. 1, the only factors which will affect emissive temperature are the areas. So, if we increase the area of absorption, the temperature can increase, and if we increase the area of emission, the temperature will decrease. For a physically connected object, it can be no other way.
Some brief examples are as below.
If we had one sided plate, we find it would be much hotter. I.e. The Moon’s surface is miles deep, so there’s no radiation on other side. Only emits on one side, so it’s warmer. T= 394 K.
Figure 2.2 A Well Insulated One Sided Plate
What we see is that if we increase the emissive area, the temperature will reduce and if we reduce the emissive area the temperature will rise, all other things being equal. This approach is an approximate one and works in a vacuum and assumes Isothermal conditions will be achieved and ignores thermal conductivity of the material.
So what happens if we stick a greenhouse onto a flat plate in space, simple, the temperature lowers, because the “energy in” side remains the same, but the “energy out” side has more surface area, so therefore an equivalent amount of energy is now spread across more material, and so this material vibrates less because the energy is more spread out, resulting in lower temperatures.
Let’s say the greenhouse had 1 metre tall sides. (4 sides = 4m2) and a roof with (2 sides) total area 3m2. So that’s a total area of 7 m2 we then need to add the side of emission, to get a total area of 8m2. If we say that the emissivity of glass is 0.90.
We will end up with a temperature of approximately 240K. This value is less than the cube, despite the glass having a lower emissivity than a black body.
This is because energy transfers up the plane of the glass, from being in direct contact with the surface and the energy is transferred from the warm plate to the cooler glass and results in lower temperatures than just the plate by itself because it has nothing to transfer energy to and so can only radiate to space to lose its heat.
If we turned the greenhouse by 180 degrees the temperature would be lower still, because glass is slightly reflective so this means not all the energy available for absorption would be absorbed so less energy would be present on the in-side of the equation. If we assumed 10{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} reflectivity then energy in would only be 1230 watts. Resulting in even lower temperatures as less energy is available.
Conclusion
A flat plate in space is warmer T = 331K than a plate with a greenhouse on it T = 240K.
For information 273K = 0 Degrees Celsius.
So, when someone says Greenhouse effect causes warming.
You know now, the correct response would be “WHAT ARE YOU TALKING ABOUT?”
In my illumination three I will rip apart a commonly used experiment, to hoodwink non-believers into becoming believers. This experiment is a trick and the professors using it to explain climate change, really should know better. Their “Religio Science” explanations are quite frankly childish. It makes me cringe watching these “so called professionals” and hearing their nonsense. It’s so obvious they have no idea what’s really going on.
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