Sentences with phrase «emissivity =»

* The ground is a little warmer than the atmosphere, so that factor will mean some more photons going up than down (but since the back radiation is mostly from low layers, the atmosphere emitting the back radiation will not be that much cooler than the land so the effect from temperature will not be TOO great) * The ground is close to a black body for IR (emissivity = 1 for all IR frequencies), but the atmosphere has bands where it does not emit or absorb well (emissivity ~ 0) and other bands where it does emit or absorb well (emissivity ~ 1).

If the temperature of the source radiation for which we need to know the absorptivity is different from the temperature of the emitting body then we can not assume that emissivity = absorptivity.

A blackbody has an emissivity = 1 and absorptivity = 1.

Calculated Aa, Ed / Aa, Su / Eu, Su / OLR, all with emissivity = 1, are in columns M to P.

In reality TOA DOWN emissivity = 0 because there can not be direct thermalisation of absorbed IR.

The new models assumed TOA DOWN emissivity = 1 and black body IR from the earth's surface and the lower atmosphere.

Note 2 — Emissivity = Absorptivity at a particular wavelength (and direction for «non-diffuse» surfaces).

For the up / down /» back» radiation of greenhouse theory's GHG energy loop to function as advertised earth's «surface» must radiate as an ideal black body, i.e. 16 C / 289 K, 1.0 emissivity = 396 W / m ^ 2.

the simple thought experiment demonstrates completely and absolutely that (under these restrictive conditions) emissivity = absorptivity (at a given wavelength and direction).

Emissivity is presumably relative to a black body at same temperature and through some thermodynamic argument emissivity = absorptivity.

Emissivity = proportion of emission with reference to a blackbody (it's a ratio) Emission = emissivity x what a blackbody would emit at that temperature (it's an absolute value)

The boundary conditions in the models are wrong — TOA DOWN emissivity = zero.

I.absorbed / I.incident = absorptivity; I.absorbed = I.emitted; I.incident = B.emitted (because they have the same brightness temperature, where B.emitted is what would be emitted by a blackbody, and is what would be in equilibrium with matter at that temperature), emissivity = I.emitted / B.emitted; therefore, given that absorptivity is independent of incident intensity but is fixed for that material at that temperature at LTE, and the emitted intensity is also independent of incident intensity but is fixed for that material at that temperature, emissivity (into a direction) = absorptivity (from a direction).

This is because the fundamentals of thermal radiation from an isolated slab: emissivity, absorptivity, transmissivity, are related by emissivity = absorptivity = (1 — transmissivity) where transmissivity = exp -LRB-- TAU)(neglecting directionality).

Not exact matches

UWLWIR is proportional to T ^ 4, (2) with emissivity constant, so the increase in UWLWIR, assuming that the global mean surface temperature is equal 288K, works out to delta U = (288.5 / 288) ^ 4 × 398 — 398 = 2.8 W / m ^ 2.

... For example, the formula given by JBS that T = (e + f - ef) / (e * (2 - e)-RRB--RRB- **.25 * TB can't really be correct since setting the solar input fraction f to zero does not reproduce temperature of the earth TB unless the emissivity e is equal to unity.

My understanding is that Intensity = Emissivity x Stefan - Boltzmann Constant x Surface Absolute Temperature ^ 4.

i.e. sigma Ts ^ 4 = S (1 - a) / (1 - lambda / 2) where lambda is the atmospheric emissivity, a is the albedo (0.7), S the incident solar flux (340 W / m ^ 2), sigma is the SB coefficient and Ts is the surface temperature (288K).

For those who want to check out the physics, read up the statistical thermodynamics which leads to Kirchhoff; s law of radiation and realise that «Prevost exchange energy» is needed to connect the IR density of states in the two objects in radiative equilibrium and maintain absorptivity = emissivity.

The important point is that thermodynamics considerations allow us to see that absorptivity = emissivity (both as a function of wavelength), and experimental considerations allow us to extend the results to non-equilibrium conditions.

Therefore, the emissivity of that layer of gas at that wavelength = 0.95.

So emissivity at a given wavelength = absorptivity.

The main point is that for a spherical body in radiative thermal equilibrium with the sun, where absorptivity = emissivity, then the temperature is independent of albedo and emissivity, because they cancel out of the equation.

Pekka, if absorptivity = emissivity for every wavelength separately, then they must also be equal for every possible mixture of wavelengths.

This means the real surfaceoperational emissivity is 63/396 = 0.16.

For the equal temperature surface (emissivity ~ 1)-- atmosphere (emissivity ~ 0.6 in temperate regions), the surface operational emissivity is ~ (1 — 0.6) = 0.4.

We vary the one free parameter, the emissivity of the atmosphere ea, but retain the surface of Earth as a blackbody with eE = 1.

As I understand it Kirchoff's Law states αλ = ελ, but only at thermal equilibrium and does not require that the absorbtivity and emissivity of the surroundings be the same as that of the emitting object, otherwise what you are suggesting is that everything has the same absortivity and emissivity and αλ = ελ = a constant.

A non-greenhouse earth surface would receive an average of 342 * (1 -.124 reflectivity) / 0.94 emissivity would give an average surface flux of 342 * 0.876 /.94 = 318.7 watts.

But absorptivity = emissivity at any given wavelength (Kirchhoff's Law), so a good absorber is also a good emitter.

Output is blackbody radiation, given by E (out) = a T ^ 4 where a is a constant (emissivity * Stephan - Boltzman) and T is the temperature.

However, when the temperature of source body for the radiation being absorbed is within a few Kelvin of the emitting body then to a quite accurate assumption, absorptivity = emissivity.

This would be the same as saying that absorptivity at 0.5 μm = emissivity at 10 μm.

I did all my analysis with a «total» emissivity et, but in terms of total optical thikness tau it is: tau = - ln (1 - et) or et = 1 - e ^ - tau

So now let's consider what happens when we set emissivity, ε, to a more realistic surface value = 0.967 (this is the value in the spreadsheet — in the paper it is set to ε = 0.96).

Or if fig 2 IS wrong, and the author has made a mistake, please confirm, calculate the true ratio of Ed / Aa for real world emissivities (e.g., ε = 0.96) and demonstrate that the results of the paper do not change with the corrected values propagating through all of the equations and calculations.

This equation assumes that the atmosphere has no convection, has no variation in emissivity with wavelength, has no solar absorption in the atmosphere, and where the heating rate at each level in the atmosphere = zero.

Since the temperature of an emitting object is determined by p = εσT ^ 4 it is only a matter of algebra to figure out what emissivity Nasif used to calculate the above temperature.

Set emissivity set to 1, a BB, then q = Q and p = σT ^ 4.

To set the lower bound assume the other 30 % of the earth has an emissivity of 0.0 the total emissivity of te earth would be 0.95 * 0.7 +0.0 * 0.3 = 0.665.

The full version is P / a = epsilon * sigma * (Tb ^ 4 — Ts ^ 4), where Tb is the temperature of the body and Ts is the temperature of the surroundings (assuming epsilon — the emissivity — is the same for all).

Five temperatures across the globe averaged (272 +285 +271 +301 +297) / 5 = 282.2 K so radiation to space ignoring emissivities is 5.67E - 08 • 282.2 ²² = 359.59 Wm - 2 right?