How to Calculate and Solve for Net Heat Flux for Radiation | Radiation Heat Transfer

The image above represents net heat flux for radiation.

To compute for net heat flux for radiation, four essential parameters are needed and these parameters are Emissivity (ε), Area (A), Emitting Power of Black Body (eb) and Radiosity (J).

The formula for calculating net heat flux for radiation:

qnet = εAeb / 1 – ε – J

Where:

qnet = Net Heat Flux for Radiation
ε = Emissivity
A = Area
eb = Emitting Power of Black Body
J = Radiosity

Let’s solve an example;
Find the net heat flux for radiation when the emissivity is 22, the area is 2, the emitting power of black body is 10 and the radiosity is 14.

This implies that;

ε = Emissivity = 22
A = Area = 2
eb = Emitting Power of Black Body = 10
J = Radiosity = 14

qnet = εAeb / 1 – ε – J
qnet = (22)(2)(10) / 1 – 22 – 14
qnet = 440 / -21 – 14
qnet = -20.9523 – 14
qnet = -34.95

Therefore, the net heat flux for radiation is -34.95 W.

Continue reading How to Calculate and Solve for Net Heat Flux for Radiation | Radiation Heat Transfer

How to Calculate and Solve for Radiosity | Radiation Heat Transfer

The image above represents radiosity.

To compute for radiosity, three essential parameters are needed and these parameters are Total Emissive Power (e), Reflectivity (ρ) and Total Irradiation (G).

The formula for calculating radiosity:

J = e + ρG

Where:

J = Radiation Heat Transfer
e = Total Emissive Power
ρ = Reflectivity
G = Total Irradiation

Let’s solve an example;
Find the radiation heat transfer when the total emissive power is 12, the reflectivity is 10 and the total irradiation is 20.

This implies that;

e = Total Emissive Power = 12
ρ = Reflectivity = 10
G = Total Irradiation = 20

J = e + ρG
J = 12 + (10)(20)
J = 12 + 200
J = 212

Therefore, the radiosity is 212 W/m².

Calculating the Total Emissive Power when the Radiosity, the Reflectivity and the Total Irradiation is Given.

e = J – ρG

Where:

e = Total Emissive Power
J = Radiation Heat Transfer
ρ = Reflectivity
G = Total Irradiation

Let’s solve an example;
Find the total emissive power when the radiation heat transfer is 32, the reflectivity is 10 and the total irradiation is 2.

This implies that;

J = Radiation Heat Transfer = 32
ρ = Reflectivity = 10
G = Total Irradiation = 2

e = J – ρG
e = 32 – (10)(2)
e = 32 – 20
e = 12

Therefore, the total emissive power is 12.

Continue reading How to Calculate and Solve for Radiosity | Radiation Heat Transfer