How to Calculate and Solve for Effective Heat of Fusion | Solidification of Metals

The image above represents effective heat of fusion.

To compute for effective heat of fusion, three essential parameters are needed and these parameters are Latent Heat of Fusion (Hf), Heat Capacity at Constant Pressure (cp) and Change in Temperature (ΔT).

The formula for calculating effective heat of fusion:

H = Hf + cpΔT

Where:

H = Effective Heat of Fusion
Hf = Latent Heat of Fusion
cp = Heat Capacity at Constant Pressure
ΔT = Change in Temperature

Let’s solve an example;
Find the effective heat of fusion when the latent heat of fusion is 12, the heat capacity at constant pressure is 14 and the change in temperature is 10.

This implies that;

Hf = Latent Heat of Fusion = 12
cp = Heat Capacity at Constant Pressure = 14
ΔT = Change in Temperature = 10

H = Hf + cpΔT
H = 12 + 14(10)
H = 12 + 140
H = 152

Therefore, the effective heat of fusion is 152 J/Kg.

Calculating the Latent Heat of Fusion when the Effective Heat of Fusion, the Heat Capacity at Constant Pressure and the Change in Temperature is Given.

Hf = H – cpΔT

Where:

Hf = Latent Heat of Fusion
H = Effective Heat of Fusion
cp = Heat Capacity at Constant Pressure
ΔT = Change in Temperature

Let’s solve an example;
Find the latent heat of fusion when the effective heat of fusion is 42, the heat capacity at constant pressure is 10 and the change in temperature is 2.

This implies that;

H = Effective Heat of Fusion = 42
cp = Heat Capacity at Constant Pressure = 10
ΔT = Change in Temperature = 2

Hf = H – cpΔT
Hf = 42 – (10)(2)
Hf = 42 – 20
Hf = 22

Therefore, the latent heat of fusion is 22.

How to Calculate and Solve for Heat Diffusivity | Solidification of Metals

The image above represents heat diffusivity.

To compute for heat diffusivity, three essential parameters are needed and these parameters are Thermal Conductivity (K), Density (ρ) and Heat Capacity at Constant Pressure (cp).

α = Kρcp

Where:

α = Heat Diffusivity
K = Thermal Conductivity
ρ = Density
cp = Heat Capacity at Constant Pressure

Let’s solve an example;
Find the heat diffusivity when the thermal conductivity is 10, the density is 12 and the heat capacity at constant pressure is 18.

This implies that;

K = Thermal Conductivity = 10
ρ = Density = 12
cp = Heat Capacity at Constant Pressure = 18

α = Kρcp
α = (10)(12)(18)
α = 2160

Therefore, the heat diffusivity is 2160 m²/s.

Calculating the Thermal Conductivity when the Heat Diffusivity, the Density and the Heat Capacity at Constant Pressure is Given.

K = α / ρcp

Where:

K = Thermal Conductivity
α = Heat Diffusivity
ρ = Density
cp = Heat Capacity at Constant Pressure

Let’s solve an example;
Find the thermal conductivity when the heat diffusivity is 15, the density is 10 and the heat capacity at constant pressure is 14.

This implies that;

α = Heat diffusivity = 15
ρ = Density = 10
cp = Heat Capacity at Constant Pressure = 14

K = α / ρcp
K = 15 / 10 (14)
K = 15 / 140
K = 0.107

Therefore, the thermal conductivity is 0.107.

Calculating the Density when the Heat Diffusivity, the Thermal Conductivity and the Heat Capacity at Constant Pressure is Given.

ρ = α / K x cp

Where:

ρ = Density
α = Heat Diffusivity
K = Thermal Conductivity
cp = Heat Capacity at Constant Pressure

Let’s solve an example;
Find the density when the heat diffusivity is 18, the thermal conductivity is 12 and the heat capacity at constant pressure is 4.

This implies that;

α = Heat Diffusivity = 18
K = Thermal Conductivity = 12
cp = Heat Capacity at Constant Pressure = 4

ρ = α / K x cp
ρ = 18 / 12 x 4
ρ = 18 / 48
ρ = 0.375

Therefore, the density is 0.375.

Calculating the Heat Capacity at Constant Pressure when the Heat Diffusivity, the Thermal Conductivity and the Density is Given.

cp = α / K x ρ

Where:

cp = Heat Capacity at Constant Pressure
α = Heat Diffusivity
K = Thermal Conductivity
ρ = Density

Let’s solve an example;
Find the heat capacity at constant pressure when the heat diffusivity is 10, the thermal conductivity is 20 and the density is 6.

This implies that;

α = Heat Diffusivity = 10
K = Thermal Conductivity = 20
ρ = Density = 6

cp = α / K x ρ
cp = 10 / 20 x 6
cp = 10 / 120
cp = 0.083

Therefore, the heat capacity at constant pressure is 0.083.

Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the heat diffusivity.

To get the answer and workings of the heat diffusivity using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app.

You can get this app via any of these means:

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

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 Solidification of Metal under Materials and Metallurgical

Now, Click on Heat Diffusivity under Solidification of Metal

The screenshot below displays the page or activity to enter your values, to get the answer for the heat diffusivity according to the respective parameter which is the Thermal Conductivity (K), Density (ρ) and Heat Capacity at Constant Pressure (cp).

Now, enter the values appropriately and accordingly for the parameters as required by the Thermal Conductivity (K) is 10, Density (ρ) is 12 and Heat Capacity at Constant Pressure (cp) is 18.

Finally, Click on Calculate

As you can see from the screenshot above, Nickzom Calculator– The Calculator Encyclopedia solves for the heat diffusivity and presents the formula, workings and steps too.

How to Calculate and Solve for Heat Flux | Solidification of Metals

The image above represents heat flux.

To compute for heat flux, three essential parameters are needed and these parameters are Thermal Conductivity (K), Density (ρ) and Heat Capacity at Constant Pressure (cp).

The formula for calculating heat flux:

q = K / ρcp

Where:

q = Heat Flux
K = Thermal Conductivity
ρ = Density
cp = Heat Capacity at Constant Pressure

Let’s solve an example;
Find the heat flux when the thermal conductivity is 21, the density is 12 and the heat capacity at constant pressure is 24.

This implies that;’

K = Thermal Conductivity = 21
ρ = Density = 12
cp = Heat Capacity at Constant Pressure = 24

q = K / ρcp
q = 21 / (12)(24)
q = 21 / 288
q = 0.0729

Therefore, the heat flux is 0.0729 W.

Calculating the Thermal Conductivity when the Heat Flux, the Density and the Heat Capacity at Constant Pressure is Given.

K = q (ρ cp)

Where:

K = Thermal Conductivity
q = Heat Flux
ρ = Density
cp = Heat Capacity at Constant Pressure

Let’s solve an example;
Find the thermal conductivity when the heat flux is 15, the density is 10 and the heat capacity at constant pressure is 14.

This implies that;

q = Heat Flux = 15
ρ = Density = 10
cp = Heat Capacity at Constant Pressure = 14

K = q (ρ cp)
K = 15 (10 x 14)
K = 15 (140)
K = 2100

Therefore, the thermal conductivity is 2100.