Radiation range

Description

Radiation is divided into that from the sun (mostly visible) and that from the earth (mostly infrared). When represented as a function of wavelength, it appears as follows:



Typical satellite measurements, such as those from the MODIS project, are taken in different channels.

The visible part is measured with three channels:

Channels Ranges [µm] Relative Weights
Blue 0.459-0.479 0.4364
Green 0.545-0.565 0.2366
Red 0.620-0.670 0.3265



The infrared part is estimated with the following channels:

Channels Ranges [µm] Relative Weights
NIR 0.841-0.876 0.5447
1.2 1.230-1.250 0.1363
1.6 1.628-1.652 0.0469
2.1 2.105-2.155 0.2536



The results from the first group are referred to as VIS, while those from the second group are referred to as NIR, although part of the observed spectrum falls within the visible range.

To understand why the separation is made around 750 nm instead of 3 microns, as is normally defined for the infrared range, one must consider the behavior of the albedo. It shows a substantial increase for wavelengths around 750 nm and above, not just from 3 microns onwards (see the albedo chart as a function of wavelength).

ID:(9921, 0)


Radiation flux and energy transport from the surface

Description

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$\kappa_c$
k_c
Coefficient Convection
J/m^3K
$\epsilon$
e
Emissivity
-
$I_d$
I_d
Energy transmitted by conduction and evaporation
W/m^2
$I_b$
I_b
Infrared Intensity emitted by the Bottom of the Atmosphere
W/m^2
$I_e$
I_e
NIR intensity emitted by the earth
W/m^2
$\sigma$
s
Stefan Boltzmann constant
J/m^2K^4s
$T_e$
T_e
Surface Temperature of the Earth
K
$T_b$
T_b
Temperature of the lower atmosphere
K
$I_{ev}$
I_ev
VIS intensity absorbed by the ground
W/m^2

Calculations


First, select the equation:   to ,  then, select the variable:   to 

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

 Variable   Given   Calculate   Target :   Equation   To be used



Equations

If the Earth is at a temperature $T_s$, it emits radiation, mainly at wavelengths $\lambda > 750$ nm, with a power given by the Stefan-Boltzmann law:

$ P = \sigma \epsilon S T_s ^4$



where $\sigma$ is the Stefan-Boltzmann constant, $\epsilon$ is the emissivity, and $S$ is the emitting surface area.

The intensity of the radiation is defined as the power per unit area, so we can express it as:

$ I =\displaystyle\frac{ P }{ S }$



where $S$ is the emitting area.

Thus, the intensity emitted from the Earth's surface $I_e$ is given by:

$ I_e = \epsilon \sigma T_e ^4 $



where $T_e$ is the temperature and $\epsilon$ is the emissivity of the surface.

(ID 4676)

Since the evaporation flow can be expressed as:

$I_E=c_aL_vC_Eu_z\displaystyle\frac{(p_{v,e}-p_{v,a})}{p_a}$



and we want to model the flow as:

$ I_d =( \kappa_l + \kappa_c ( T_e - T_b )) u $



we can determine the constant factor as:

$ \kappa_l = L_v C_E \displaystyle\frac{ p_{s,e} }{ R T_e }( RH_e - \gamma_v )$


(ID 9271)


Examples

The radiations reaching the surface of the ocean or emitted from it can be summarized in the following graph:



In summary:

- $I_{sev}$: Net solar radiation.
- $I_e$: Radiation emitted by the Earth.
- $I_H$: Exchange due to droplet input/output.
- $I_E$: Exchange due to water evaporation/condensation.
- $I_c$: Exchange due to conduction.

This graph provides a concise overview of the different forms of radiation interacting at the ocean's surface and the associated energy exchanges.

(ID 13497)

Radiation is divided into that from the sun (mostly visible) and that from the earth (mostly infrared). When represented as a function of wavelength, it appears as follows:



Typical satellite measurements, such as those from the MODIS project, are taken in different channels.

The visible part is measured with three channels:

Channels Ranges [ m] Relative Weights
Blue 0.459-0.479 0.4364
Green 0.545-0.565 0.2366
Red 0.620-0.670 0.3265



The infrared part is estimated with the following channels:

Channels Ranges [ m] Relative Weights
NIR 0.841-0.876 0.5447
1.2 1.230-1.250 0.1363
1.6 1.628-1.652 0.0469
2.1 2.105-2.155 0.2536



The results from the first group are referred to as VIS, while those from the second group are referred to as NIR, although part of the observed spectrum falls within the visible range.

To understand why the separation is made around 750 nm instead of 3 microns, as is normally defined for the infrared range, one must consider the behavior of the albedo. It shows a substantial increase for wavelengths around 750 nm and above, not just from 3 microns onwards (see the albedo chart as a function of wavelength).

(ID 9921)

If the Earth is at a temperature $T_e$, it emits radiation according to the Stefan-Boltzmann law with an intensity given by the following formula:



Where $\sigma$ is the Stefan-Boltzmann constant and $\epsilon$ is the emissivity coefficient. The Stefan-Boltzmann constant $\sigma$ has a value of approximately $5.67 \times 10^{-8} W/m^2K^4$, and the emissivity coefficient $\epsilon$ represents the efficiency with which the Earth's surface emits radiation, ranging from 0 to 1.

(ID 4676)

The energy transmitted by conduction and evaporation ($I_d$) depends on the difference between the temperature of the lower atmosphere ($T_b$) and ERROR:6516, as well as on ERROR:8094 and the constants ERROR:8093 and ERROR:6521, in the following way:

$ I_d =( \kappa_l + \kappa_c ( T_e - T_b )) u $


(ID 9270)

The Earth\'s surface receives energy from the sun $I_{ev}$ and from the lower part of the atmosphere $I_b$. All of this energy is radiated as $I_e$ and lost through convection and conduction $I_d$ with:

$ I_{ev} - I_e - I_d + I_b =0$


(ID 4692)

The radiation flux due to transport elements is given by the equation:

$ I_d =( \kappa_l + \kappa_c ( T_e - T_b )) u $



where it can be shown that the constant coefficient of evaporation takes the form:

$ \kappa_l = L_v C_E \displaystyle\frac{ p_{s,e} }{ R T_e }( RH_e - \gamma_v )$



Assuming a molar concentration of air of $c_a\sim 42.4$, a molar latent heat of $L_v\sim 40.6,kJ/mol$, a wind speed of $u_z\sim 8,m/s$, a latent heat transport constant of $C_E\sim 5.0\times 10^{-4}$, a saturated water vapor pressure of $p_{s,e}\sim 1519,Pa$, a relative humidity of $RH_e\sim 85%$, an atmospheric pressure of $p_a\sim 10^5,Pa$, and a visible coverage of $\gamma_v\sim 42%$, the constant has a magnitude of $\kappa_c\sim 5.66,J/m^3s$.

(ID 9271)

Since the transport flux is given by

$ I_H = \rho_a c_p C_H u_z ( T_e - T_b )$



and the evaporation flux is

$ I_H = \rho_a c_p C_H u_z ( T_e - T_b )$



and we aim to model the flux as

$ I_d =( \kappa_l + \kappa_c ( T_e - T_b )) u $



the temperature factor can be determined as

$ \kappa_c = \rho_a c_p C_H + C_E p_{s,e} \displaystyle\frac{ \gamma_v L_v ^2}{ R ^2 T_e ^3}$



Assuming the air density is $\rho_a\sim 1.225,kg/m^3$, the molar concentration of air is $c_a\sim 42.4$, the molar latent heat is $L_v\sim 40.6,kJ/mol$, the wind velocity is $u_z\sim 8,m/s$, the heat transport constant is $C_H\sim 1.13\times 10^{-4}$, the latent heat transport constant is $C_E\sim 5.0\times 10^{-4}$, the saturated water vapor pressure is $p_{s,e}\sim 1519,Pa$, the relative humidity is $RH_e\sim 85%$, the atmospheric pressure is $p_a\sim 10^5,Pa$, and the visible coverage is $\gamma_v\sim 42%$, the increase in flux per degree of temperature difference is on the order of $\kappa_c\sim 0.47,J/m^3s,K$.

(ID 9272)

Apart from infrared radiation, there is heat transport through convection or sensible heat flux (SHF). Both phenomena are approximately proportional to the temperature difference between the Earth $T_e$ and the lower part of the atmosphere $T_b$:

$ I_H = \rho_a c_p C_H u_z ( T_e - T_b )$



If the typical heat flux transported by convection is on the order of $I_H\sim 17 W/m^2$, and considering the density $\rho_a\sim 1.225 kg/m^3$, the specific heat capacity $c_p\sim 1006.43 J/kg K$, and the wind velocity $u_z\sim 8 m/s$, then with a temperature difference of $T_e-T_b\sim 15.2 K$, the heat transfer coefficient is on the order of $C_H\sim 1.13\times 10^{-4}$.

(ID 4678)

Since the molar concentration $c_a$ is proportional to the pressure $p_a$ according to:

$ p = c_m R_C T $



we can rewrite

$I_E=c_aL_vC_Eu_z\displaystyle\frac{(p_{v,e}-p_{v,a})}{p_a}$



as

$ I_E = L_v C_E u_z \displaystyle\frac{( p_{v,e} - p_{v,a} )}{ R T_e }$


(ID 9275)


ID:(1753, 0)