Am i correct please?

[trigger]

Hi all,

Nearly all of the energy used on earth comes originally from the sun, and that solar radiation is intercepted at 1370 Wm^-^2 at the top of the atomsphere, and that 49%of this radiation is absorbed by the earths surface.

If a shallow dish containing ice or warte, placed on the earth surface, absorbs the same the same fraction of solar radiation as the surface itself, at what rate is energy absorbed from the sun by such a container of surface 0.039m^2 ?

 

This is how i worked it out,

 

.49*1370 = watts per square meter absorbed

Multiply that by the dish area. So is the correct answer 67130 Wm^2

 

thanks Trigger

[arkain101]

Here is some formula I came up with. Should give you a rough estimate.

 

Energy Final (watts)[Ef] = (Energy constant(w/m^2)[Ec]) - (Variable engery [Ev]) x (Area of interest [Ai])

 

So,

 

[math]Ef(w/m^2) = Ec(w/m^2) - Ev (w/m^2) * Ai (m^2)[/math]

 

Energy constant [Ec] = 1412 + 1321 / 2 = 1366avg watts / square meter

 

This energy constant is a square meter in space and is variable.

 

The energy vairable (Ev) is a value of energy lost as it travels through the atmospheric conditions and circumstance. This value I can not determine it depends on weather, angle of at which the energy enters the atmosphere like dawn or dusk. So we could make the scenario where the sun is close to perpendicular angle of attack, travelling through a cloudless sky, reaching a target at sea level. Lets Pretend that 30% of the energy is absorbed and this is our energy variable.

 

[math] Ev = (0.3) * (1366w/m^2) [/math]

 

[math] Ev = 409w/m^2 [/math]

 

So finally, if our dish is [math]0.6 m^2[/math]

 

[math]Ef (w/m^2) = Ec(w/m^2) - Ev (w/m^2) * Ai (m^2)[/math]

 

[math]Ef (w/m^2) = (1366w/m^2) - (409 w/m^2) * (0.6 m^2)[/math]

 

[math]Ef (w/m^2) = 574.2 w/(0.6m^2)[/math]

 

 

Really I have no clue if this is even close.. I just thought I would try the math ;)

[modest]

Hi all,

Nearly all of the energy used on earth comes originally from the sun, and that solar radiation is intercepted at 1370 Wm^-^2 at the top of the atomsphere, and that 49%of this radiation is absorbed by the earths surface.

If a shallow dish containing ice or warte, placed on the earth surface, absorbs the same the same fraction of solar radiation as the surface itself, at what rate is energy absorbed from the sun by such a container of surface 0.039m^2 ?

 

This is how i worked it out,

 

.49*1370 = watts per square meter absorbed

Multiply that by the dish area. So is the correct answer 67130 Wm^2

 

thanks Trigger

 

This is correct. You would multiply the area of interest by 0.49 by 1370, as you say.

 

As Arkain indicates above, and it appears you are aware, this is an average. The commonly-given figure for watts per square meter intercepting the surface of the earth is 342 watts/m^2. This is averaged for day and night at all latitudes. Multiplying 342 watts/m^2 by 2 in order to ignore parts of the earth not facing the sun and dividing by 1370 watts/m^2 gives 0.499. This appears to be where you get the figure 49%.

 

So, 49% is the average amount of energy that is available to the sunny side of the surface and not absorbed or reflected by the atmosphere. The equator at noon would obviously receive far greater energy while the poles get far less as would any experiment done in the early morning or late evening.

 

~modest

 

EDIT:

 

I'm sorry, your method described is correct, but you've apparently made a computational error. 1370 w/m^2 * .49 * 0.039 m^2 = 26.18 watts, rather than the 67130 you give. The dish would have 26.18 watts available to it.