The sun is known to emit ~63 Mega Watts per meter squared from its photosphere. But what is the heat flux inside this emissive photosphere?
Heat flux formula: q = k*ΔT/L
q = k * (6600-4400 Kelvin) / (500,000 meters)
What is the thermal conductivity (k) value of hydrogen at these temperatures? [1]
This is actually very difficult to find, but I managed to find something:
This y-axis needs to be divided by 10 to get units (W/m*K).
The range of pressure in the photosphere is: 0.0009 atm to 0.123 atm. I think it’s safe to say that thermal conductivity of hydrogen is definitely no more than 2.5 W/m*K in our desired range. That will be our upper limit. Thus,
q = 2.5 * 2200 / 500000 = 0.011 W/m² [2]
As you can you can see, there is no problem with 0.011 W/m² “supporting” a 63 MW/m² output.
My critics will be quick to point out that I can’t use the conduction formula because the sun only has radiative transfers in the photosphere. But that’s just their excuse for NEVER figuring out what the internal heat flow is. Any of their attempts at doing so will be embarrassing for them, and so they will avoid it at all cost. Surely there is a heat flux, and surely it can be figured out.
My critics believe in conservation of heat flow, that is: internal heat flux => emergent radiation. There must be equality of these two things, otherwise there will be rapid cooling. Well, the sun has had 4.5 billion years to reach their “steady-state” equilibrium nonsense and it’s nowhere close. Maybe despite all their chest thumping, they have no idea what they’re talking about?
What goes for the sun here goes for my geothermal theory as well.
Just as <0.011 W/m² internal heat flux can “support” a 63 MW/m² emission from the sun, so too can a ~0.9 W/m² geothermal heat flux “support” a ~334 W/m² emission from Earth’s surface.
And why is that? See here and here.
Think about it!
Enjoy 🙂 -Zoe
Note:
[1] I left out helium. I don’t care to include it, as it makes little difference. I only care about being right within an order of magnitude.
[2] I don’t include surface area of emission, because the difference in solar radius of top and bottom of photosphere is too small.
Zoe's Insights 
“Heat flux formula: q = k*ΔT/L”
This for heat transferred thru material.
If material is transparent to radiant energy the radiant energy passes thru the material.
So ocean water is fairly transparent to sunlight, though different parts of spectrum
of less transparent and others are more transparent- red light only goes a few meters
and blue light part of spectrum can go about 100 meter under surface.
Ie if put solar panel six inches under water, it would not reduce the amount sunlight energy
it gets.
And rule of thumb first couple meters of water depth absorbs most of sunlight.
And most ocean {open ocean water} is quite transparent whereas coastal waters can be
more murky. And muddy water is not very transparent.
Earth atmosphere generally more transparent than water and upper atmosphere with less
air density would more transparent than the more dense lower atmosphere.
1000 meter high by 1 meter square is about 1 ton of air, and meter depth of ocean 1 meter square
is about 1 ton, so I mean, per ton, air is more transparent than water, but per meter length, air a lot more transparent.
And per meter or per km of Sun’s low density atmosphere should be quite transparent
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