How to Calculate and Solve for Heat Extraction by Radiation | Solidification of Metals

The image above represents heat extraction by radiation.

To compute for heat extraction by radiation, four essential parameters are needed and these parameters are Emissivity (ε), Stefan-Boltzmann Constant (σ), Surface Temperature of Metal (Tsurf) and Ambient Temperature (Tamb).

The formula for calculating heat extraction by radiation:

qx = εσ(Tsurf4 – Tamb4)

Where:

qx = Heat Extraction by Radiation
ε = Emissivity
σ = Stefan-Boltzmann Constant
Tsurf = Surface Temperature of Metal
Tamb = Ambient Temperature

Let’s solve an example;
Find the heat extraction by radiation where the emissivity is 20, the stefan-boltzmann constant is 5.67E-8, the surface temperature of metal is 14 and the ambient temperature is 12.

This implies that;

ε = Emissivity = 20
σ = Stefan-Boltzmann Constant = 5.67E-8
Tsurf = Surface Temperature of Metal = 14
Tamb = Ambient Temperature = 12

qx = εσ(Tsurf4 – Tamb4)
qx = 20(5.67e-8)(144 – 124)
qx = 20(5.67e-8)(38416 – 20736)
qx = 20(5.67e-8)(17680)
qx = 0.02004912

Therefore, the heat extraction by radiation is 0.02004912 W.

Continue reading How to Calculate and Solve for Heat Extraction by Radiation | Solidification of Metals

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 Energy Transfer by Radiation | Radiation Heat Transfer

The image above represents energy transfer by radiation.

To compute for energy transfer by radiation, three essential parameters are needed and these parameters are Emissivity (ε1), Absorbivity (α1) and Emissive Power of Black Body (eb).

The formula for calculating energy transfer by radiation:

qnet = eb1 – α1)

Where:

qnet = Energy Transfer by Radiation
ε1 = Emissivity
α1 = Absorbivity
eb = Emissive Power of Black Body

Let’s solve an example;
Find the energy transfer by radiation when the emissivity is 30, the absorbivity is 10 and the emissive power of black body is 4.

This implies that;

ε1 = Emissivity = 30
α1 = Absorbivity = 10
eb = Emissive Power of Black Body = 4

qnet = eb1 – α1)
qnet = 4(30 – 10)
qnet = 4(20)
qnet = 80

Therefore, the energy transfer by radiation is 80 J.

Calculating the Emissivity when the Energy Transfer by Radiation, the Aborbivity and the Emissive Power of Black Body is Given.

ε1 = (qnet / eb) + α1

Where:

ε1 = Emissivity
qnet = Energy Transfer by Radiation
α1 = Absorbivity
eb = Emissive Power of Black Body

Let’s solve an example;
Find the emissivity when the energy transfer by radiation is 28, the absorbivity is 3 and the emissive power of black body is 7.

This implies that;

qnet = Energy Transfer by Radiation = 28
α1 = Absorbivity = 3
eb = Emissive Power of Black Body = 7

ε1 = (qnet / eb) + α1
ε1 = (28 / 7) + 3
ε1 = 4 + 3
ε1 = 7

Therefore, the emissivity is 7

Calculating the Aborbivity when the Energy Transfer by Radiation, the Emissivity and Emissive Power of Black Body is Given.

α1 = ε1 – (qnet / eb)

Where:

α1 = Absorbivity
qnet = Energy Transfer by Radiation
ε1 = Emissivity
eb = Emissive Power of Black Body

Let’s solve an example;
Find the absorbivity when the energy transfer by radiation is 8, the emissivity is 18 and the emissive power of black body is 4.

This implies that;

qnet = Energy Transfer by Radiation = 8
ε1 = Emissivity = 18
eb = Emissive Power of Black Body = 4

α1 = ε1 – (qnet / eb)
α1 = 18 – (8 / 4)
α1 = 18 – 2
α1 = 16

Therefore, the absorbivity is 16.

Calculating the Emissive Power of Black Body when the Energy Transfer by Radiation, the Emissivity and Absorbivity is Given.

eb = qnet / ε1 – α1

Where:

eb = Emissive Power of Black Body
qnet = Energy Transfer by Radiation
ε1 = Emissivity
α1 = Absorbivity

Let’s solve an example;
Find the emissive power of black body when the energy transfer by radiation is 50, the emissivity is 15 and the absorbivity is 10.

This implies that;

qnet = Energy Transfer by Radiation = 50
ε1 = Emissivity = 15
α1 = Absorbivity = 10

eb = qnet / ε1 – α1
eb = 50 / 15 – 10
eb = 50 / 5
eb = 10

Therefore, the emissive power of black body is 10.

Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the energy transfer by radiation.

To get the answer and workings of the energy transfer by radiation using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app.

You can get this app via any of these means:

Web – https://www.nickzom.org/calculator-plus

To get access to the professional version via web, you need to register and subscribe for NGN 2,000 per annum to have utter access to all functionalities.
You can also try the demo version via https://www.nickzom.org/calculator

Android (Paid) – https://play.google.com/store/apps/details?id=org.nickzom.nickzomcalculator
Android (Free) – https://play.google.com/store/apps/details?id=com.nickzom.nickzomcalculator

Apple (Paid) – https://itunes.apple.com/us/app/nickzom-calculator/id1331162702?mt=8
Once, you have obtained the calculator encyclopedia app, proceed to the Calculator Map, then click on Materials and Metallurgical under Engineering.

Now, Click on Radiation Heat Transfer under Materials and Metallurgical

Now, Click on Energy Transfer by Radiation under Radiation Heat Transfer

The screenshot below displays the page or activity to enter your values, to get the answer for the energy transfer by radiation according to the respective parameter which is the Emissivity (ε1), Absorbivity (α1) and Emissive Power of Black Body (eb).

Now, enter the values appropriately and accordingly for the parameters as required by the Emissivity (ε1) is 30, Absorbivity (α1is 10 and Emissive Power of Black Body (eb) is 4.

Finally, Click on Calculate

As you can see from the screenshot above, Nickzom Calculator– The Calculator Encyclopedia solves for the energy transfer by radiation and presents the formula, workings and steps too.

How to Calculate and Solve for %Heat Loss from Openings in Furnace | Fuel and Furnaces

The image above represents %heat loss from openings in furnace.

To compute for %heat loss from openings in furnace, five essential parameters are needed and these parameters are Black Body Radiation (BR), Emissivity (E), Radiation Factor (RF), Quantity of Oil (Q) and Gross Calorific Value of Oil (GCVoil).

The formula for calculating %heat loss from openings in furnace:

%QL = 100(BR x E x RF/Q x GCVoil)

Where:

%QL = %Heat Loss from Openings in Furnace
BR = Black Body Radiation
E = Emissivity
RF = Radiation Factor
Q = Quantity of Oil
GCVoil = Gross Calorific Value of Oil

Let’s solve an example;
Find the %heat loss from openings in furnace when the black body radiation is 15, the emissivity is 6, the radiation factor is 3, the quantity of oil is 24 and the gross calorific value of oil is 9.

This implies that;

BR = Black Body Radiation = 15
E = Emissivity = 6
RF = Radiation Factor = 3
Q = Quantity of Oil = 24
GCVoil = Gross Calorific Value of Oil = 9

%QL = 100(BR x E x RF/Q x GCVoil)
%QL = 100(15 x 6 x 3/24 x 9)
%QL = 100(270/216)
%QL = 100(1.25)
%QL = 125

Therefore, the %heat loss from openings in furnace is 125%.

Continue reading How to Calculate and Solve for %Heat Loss from Openings in Furnace | Fuel and Furnaces