# Intensity of radiation

Do not be confused on How Is Radiation A Mode Of Heat Transfer. Intensity of radiation is the rate at which energy leaving a particular surface in a given direction per unit solid area of an emitting surface normal to the average distance in space or coordinate system.

This is associated with heat radiation of the black body concept. The energy of radiation is transmitted by waves having different wavelength in the space.

From the above definition of intensity of radiation, a solid area is a prescribe area of a space inside an enclosed spherical body by a conical surface with the vertex of the cone at the center of the sphere. This means that the normal of the enclosed sphere by a surface cone is perpendicular to the apex of the cone. The solid area subtended by a hemisphere is given as 2π with unit in steradian (sr). Solid area is given by the ratio of a given surface area of a body to the square of the radius of the body. ( A/r^{2})

The total emissive power of a hemispherical surface is given as (pie*)

Where I is the intensity of radiation.

## Lambert cosine law

This law states that the total emissive power from a radiating surface is directly proportional to the cosine of the angle of the emitted energy.

E_{θ} = E_{n}Cosθ

This law holds for diffuse surfaces and intensity of radiation of a diffuse surface is constant

Example one:

The temperature of a radiating body having an area of 0.30m^{2} is 573K. Calculate the following:

- The total rate of energy emission
- Intensity of normal radiation
- The maximum wavelength of the emissive power.

Solution

- Total rate of energy emission: E
_{b}= Α*T^{4}*α

Where α is the Steffan Boltzmann’s constant

E_{b} = 0.0567µ*0.3*573

E_{b} =974.673W

- Intensity of normal radiation: E = 3.142I

I = 974.673/3.142 = 310.248W/m^{2}.sr

- From wien’s law the maximum wavelength can calculated thus:

λ= 2898/573 =5.0576µm

Example two:

Assuming a black body of area 12.57m2, surface temperature of 400k and average distance area of 5026.548m^{2} from a receiving body of 0.126m^{2}. calculate the following:

- Total energy emitted by the black body
- The emission received per unit area just outside the atmosphere of the receiving body
- Total energy received by the received by the receiving body

Solution

Total emitted energy from the black body:

- : E
_{b}= Α*T^{4}*α = 18259W - Emission received per unit area: this is the energy received per unit area of the receiving body from the black body

E = 18250/5026.548 =3.6307W/m^{2 }

- Total energy of the receiving body from the black body: the energy released by the black body is proportional to the perpendicular projected area of the receiving body, therefore,

E = 3.630*0.126

E= 0.457W