B Is εσT⁴ hemispherical or full sphere?

[fizzy]

TL;DR Summary

what is the solid angle of the famous Stephan-Boltzmann equation?

Can someone advise on the εσT⁴ term of Stephan-Boltzmann law?

Is this the power radiated from a point or flat surface into a hemisphere or a sphere fully enclosing a black body?

thanks.

 

[Charles Link]

It is the power radiated per unit area from a plane surface into a hemisphere. The intensity radiated over the hemisphere is assumed to fall of as $I(\theta)=I_o \cos{\theta}$, where $\theta$ is the polar angle.

 

[fizzy]

Thanks for the reply Charles. I thought the term I asked about came from the double integral of the term you cite. That would make I the radiance (watts per square metre per steradian). I think that was part of the problem I was having with following some of the stuff I'd read on this.

So it's per unit area of the surface and you need to multiply ( or integrate ) across the surface to get total power.

How does this relate to the case of a more realistic 3D object radiating into space rather than a hemisphere?

 

[Charles Link]

A practical calculation that might interest you is to treat the spherical sun as a $T_s=6000$ K blackbody, and assume the Earth absorbs all of the light that reaches it from the sun. The Earth then radiates as a spherical blackbody at some temperature $T_e$, which is determined by equating the power absorbed by the Earth to the power radiated by the earth. For the diameter of the sun, you can use $d_s=$ 865,000 miles, and the earth-sun distance as $s_s=$ 93,000,000 miles. The problem is to compute $T_e$. This is kind of a simple calculation involving two 3D radiating objects. (To simplify the result, it is helpful to use $d_s/s_s=865,000/93,000,000 \approx .01$. It gives a nice round number result for $T_e$). $\\$ Additional item: Radiant emittance $M=\sigma T^4$, and radiance $L=\frac{\sigma T^4}{\pi}$. The effective angle of the hemisphere, because of the $\cos{\theta}$ dependence of the intensity, is $\pi$ steradians. You can compute an irradiance of $E=\frac{LA_s}{s^2}$ watts/m^2 , because the steradian, (like the radian), is dimensionless. (check the units=it looks like the steradian in the denominator of the radiance $L$ units disappears). These units can be a little tricky when you first read some of the literature.

 

[fizzy]

Thanks Charles, that is interesting but that does not seem to answer my quesiton.

I know that the integral ∫cosθ dθ dφ across the hemisphere equals π but that is still dealing with a plane emitting into a hemisphere. What I was asking is how to do a similar calculation for a sphere.

 

[Charles Link]

If you try working the problem with the sun, you compute $P_s=M_s A _s =(\sigma T_s^4)(4 \pi (d_s/2)^2)$. You can then compute the irradiance at the Earth from the sun, $E_s=P_s/(4 \pi s_s^2)$ or $E_s=L_s (\pi (d_s/2)^2)/s_s^2$, where $L_s=\sigma T_s^4/\pi$.(In the second way of computing $E_s$, you use the projected area $\pi r_s^2$).

 

[fizzy]

In E=LA_s/s² , I guess that s is the solid angle subtended by the Earth at the sun. However, the sun can hardly be regarded as point source, an implicit assumption idea of the Earth subtending a solid angle somewhere. That has to relate to a point, not a finite and large object. When the angle of the sun at the Earth is much larger, that hardly looks legit.

Measured total insolated irradiance at top of atmosphere is about 1633W/m^2 , so we can use that instead. Assuming Earth is a black body ( which it is not ) gives the following result. 4 being the ratio of emitting Earth surface to cross-section capturing TIR.

(TIR/4./SB)**0.25-273.15 = 18.14 deg C

Rather hotter than so called "average surface temperatures" , a concept which itself defies physics. Not sure any of that is particularly useful.

 

[Charles Link]

Let me write out the details: Power absorbed by Earth is $P_e=P_s (\pi r_e^2)/(4 \pi s_s^2)$. Power radiated by Earth is $P_e=(\sigma T_e^4)(4 \pi r_e^2)$. Previously $P_s=(\sigma T_s^4)(4 \pi (d_s/2)^2)$. With a little algebra, the result is $T_e=(T_s/2) (d_s/s_s)^{.5}=300$ K, if you use $d_s/s_s=.01$. $\\$ You were asking how to treat a spherical blackbody, and do calculations with it. This does precisely that. $\\$ With the sun less than a million miles in diameter, (865,000), and the Earth 93 million miles away, it's a very good approximation to treat it as a point source. Even the $d_s/s_s \approx .01$ is fairly accurate. $\\$ More refined calculations would need to assign an emissivity to the earth, which may be different for the absorbed spectrum, (with a fair amount of visible light), from the radiated spectrum, which is basically middle to long wavelength infrared. The $T_e=300$ K is a simple quick estimate. $\\$ Note: $s_s$ is the sun-earth distance.

 

[fizzy]

You were asking how to treat a spherical blackbody, and do calculations with it.

Err, no.

Can someone advise on the εσT⁴ term of Stephan-Boltzmann law?

Is this the power radiated from a point or flat surface into a hemisphere or a sphere fully enclosing a black body?

So the bottom line is that it is the radiated flux density, not power, and this applies to both.

Thanks for your help in sorting it all out.

 

[Anand Sivaram]

The temperature at the surface of the Sun is 5777K, that corresponds to a flux of 63 MW/m2. Solar flux reaching the top of our atmosphere is 1366 W/m2. That corresponds to a temperature of 393K (120C). Now considering our Earth as a sphere, the average insolation on top of our atmosphere would have to be divided by a ratio of 4 ( 4 * pi * r2 / pi * r2), that would be 341.5 W/m2. That corresponds to a temperature of 279K (6C).

That means, we take a black paper to space and keep it perpendicular to the incoming sunlight, it would get heated up to 120C. Now, we take rotating blackball in space, it would have a temperature of 6C.

Using Wien's law, the Sun has a peak wavelength of 500nm (Green). Whereas the energy radiated from our Earth at 393K has a wavelength of 7.3um (Long Infrared).

 

[fizzy]

Thanks for correcting my dyslexic 1633, I was surprised I got such a high result before adding albedo, which raises the result.
Of course the whole idea of modelling the Earth as a uniform ball with no atmosphere, clouds, ice which mean it reflects a lot of solar before it even gets absorbed, is pretty silly.

The next step is rig "effective" albedo to make sure you get the answer you expect and declare it all be "about right".

 

[DrClaude]

The temperature at the surface of the Sun is 5777K, that corresponds to a flux of 63 MW/m2. Solar flux reaching the top of our atmosphere is 1366 W/m2. That corresponds to a temperature of 393K (120C). Now considering our Earth as a sphere, the average insolation on top of our atmosphere would have to be divided by a ratio of 4 ( 4 * pi * r2 / pi * r2), that would be 341.5 W/m2. That corresponds to a temperature of 279K (6C).

That means, we take a black paper to space and keep it perpendicular to the incoming sunlight, it would get heated up to 120C. Now, we take rotating blackball in space, it would have a temperature of 6C.

You forgot a factor 2 in your calculation. That sheet of paper has two sides, so its temperature is 331 K, not 393 K.

 

[Anand Sivaram]

Thanks for pointing out @DrClaude . I could sense that you divided 1366/2 and then found the Temperature.

Since only one side is in direct contact with the sunlight, I assume the other side would radiate energy back to the space. Then the equilibrium temperature would be 331K..Is that correct?

 

[DrClaude]

Thanks for pointing out @DrClaude . I could sense that you divided 1366/2 and then found the Temperature.

Since only one side is in direct contact with the sunlight, I assume the other side would radiate energy back to the space. Then the equilibrium temperature would be 331K..Is that correct?

Yes. It is receiving 1366 W/m² and therefore emitting 1366 W/m², but since it is emitting on two sides, it emits 683 W/m² per side, which is the power you have to set in Stefan-Boltzmann's law to calculate its temperature.