How to Calculate and Solve for Stefan Boltzmann’s Constant | Radiation Heat Transfer

The image above represents stefan boltzmann’s constant.

To compute for stefan boltzmann’s constant, three essential parameters are needed and these parameters are Boltzmann’s Constant (KB), Planck’s Constant (h) and Speed of Light (c).

The formula for calculating stefan boltzmann’s constant:

σ = 2π5KB4 / 15c²h5

Where:

σ = Stefan Boltzmann’s Constant
KB = Boltzmann’s Constant
h = Planck’s Constant
c = Speed of Light

Let’s solve an example;
Find the stefan boltzmann’s constant when the boltzmann’s constant is 42, the planck’s constant is 34 and the speed of light is 26.

This implies that;

KB = Boltzmann’s Constant = 42
h = Planck’s Constant = 34
c = Speed of Light = 26

σ = 2π5KB4 / 15c²h5
σ = 2π5(42)4 / 15(26)²(34)5
σ = 2(306.019)(3111696) / 15(676)(45435424)
σ = 1904480458.135 / 460715199360
σ = 0.00413

Therefore, the stefan boltzmann’s constant is 0.00413.

Continue reading How to Calculate and Solve for Stefan Boltzmann’s Constant | Radiation Heat Transfer

How to Calculate and Solve for Monochromatic Emissive Power | Radiation Heat Transfer

The image above represents monochromatic emissive power.

To compute for monochromatic emissive power, five essential parameters are needed and these parameters are Planck’s Constant (h), Velocity of Light (c), Wavelength (λ), Boltzmann’s Constant (KB) and Temperature (T).

The formula for calculating monochromatic emissive power:

ebx = 2πhc²λ-5 / exp(ch/KBλT) – 1

Where:

ebx = Monochromatic Emissive Power | Planck’s Equation
h = Planck’s Constant
c = Velocity of Light
λ = Wavelength
KB = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Find the monochromatic emissive power when the planck’s constant is 6.626E-24, the velocity of light is 3E8, the wavelength is 22, the boltzmann’s constant is 1.380E-23 and the temperature is 10.

This implies that;

h = Planck’s Constant = 6.626E-24
c = Velocity of Light = 3E8
λ = Wavelength = 22
KB = Boltzmann’s Constant = 1.380E-23
T = Temperature = 10

ebx = 2πhc²λ-5 / exp(ch/KBλT) – 1
ebx = 2π(6.62607004e-34)(300000000)²(22)-5 / exp((300000000)(6.62607004e-34)/(1.38064852e-23)(22)(10)) – 1
ebx = 2π(6.62607004e-34)(90000000000000000)(1.94037e-7) / exp(1.987821012e-25/3.037426744e-21) – 1
ebx = 7.270512005456302e-23 / exp(0.00006544424539379114) – 1
ebx = 7.270512005456302e-23 / 1.0000654463869152 – 1
ebx = 7.270512005456302e-23 / 0.00006544638691519111
ebx = 1.11e-18

Therefore, the monochromatic emissive power is 1.11e-18.

Continue reading How to Calculate and Solve for Monochromatic Emissive Power | Radiation Heat Transfer

How to Calculate and Solve for Stokes-Einstein Equation of Diffusivity | Mass Transfer

The image above represents strokes-einstein equation of diffusivity.

To compute for strokes-einstein equation of diffusivity, four essential parameters are needed and these parameters are Boltzmann’s Constant (KB), Temperature (T), Radius of Sphere (R) and Viscosity (η).

The formula for calculating strokes-einstein equation of diffusivity:

D = KBT / 6πRη

Where:

D = Diffusivity
KB = Boltzmann’s Constant
T = Temperature
R = Radius of Sphere
η = Viscosity

Let’s solve an example;
Find the diffusivity when the boltzmann’s constant is 1.3806E-23, the temperature is 22, the radius of sphere is 12 and the viscosity is 10.

This implies that;

KB = Boltzmann’s Constant = 1.3806E-23
T = Temperature = 22
R = Radius of Sphere = 12
η = Viscosity = 10

D = KBT / 6πRη
D = (1.3806e-23)(22) / 6π(12)(10)
D = 3.037e-22 / 2261.94
D = 1.342

Therefore, the diffusivity is 1.342e-25 cm²/s.

Continue reading How to Calculate and Solve for Stokes-Einstein Equation of Diffusivity | Mass Transfer

How to Calculate and Solve for Relationship between Electrical Conductivity and Diffusivity | Mass Transfer

The image above represents relationship between electrical conductivity and diffusivity.

To compute for relationship between electrical conductivity and diffusivity, five essential parameters are needed and these parameters are Electrical conductivity (σ ), Coordination Number (Z), Electron Charge (e), Boltzmann’s Constant (KB) and Temperature (T).

The formula for calculating relationship between electrical conductivity and diffusivity:

σ/D = n(Ze)² / KBT

Where:

σ/D = Relationship between Electrical Conductivity and Diffusivity
σ = Electrical Conductivity
D = Diffusivity
Z = Coordination Number
e = Electron Charge
KB = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Find the relationship between electrical conductivity and diffusivity when the electrical conductivity is 2, the coordination number is 3, the electron charge is 4, the boltzmann’s constant is 1.3806e-23 and the temperature is 7.

This implies that;

σ = Electrical Conductivity = 2
Z = Coordination Number = 3
e = Electron Charge = 4
KB = Boltzmann’s Constant = 1.3806e-23
T = Temperature = 7

σ/D = n(Ze)² / KBT
σ/D = 2(3(4))² / 1.3806e-23(7)
σ/D = 2(12)² / 9.66e-23
σ/D = 2(144) / 9.66e-23
σ/D = 288 / 9.66e-23
σ/D = 2.97

Therefore, the relationship between electrical conductivity and diffusivity is 2.97e+24.

Continue reading How to Calculate and Solve for Relationship between Electrical Conductivity and Diffusivity | Mass Transfer

How to Calculate and Solve for Shear Modulus of Rubber | Fracture Mechanics

The image above represents shear modulus of rubber.

To compute for shear modulus of rubber, three essential parameters are needed and these parameters are Density of Network Cross Links (N), Boltzmann’s Constant (KBand Temperature (T).

The formula for calculating shear modulus of rubber:

G = NKBT

Where:

G = Shear Modulus of Rubber
N = Density of Network Cross Links
KB = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Find the shear modulus of rubber when the density of network cross links is 21, the boltzmann’s constant is 1.38064852E-23 and the temperature is 10.

This implies that;

N = Density of Network Cross Links = 21
KB = Boltzmann’s Constant = 1.38064852E-23
T = Temperature = 10

G = NKBT
G = (21)(1.38064852e-23)(10)
G = 2.89

Therefore, the shear modulus of rubber is 2.89e-23 Pa.

Calculating the Density of Network Cross Links when the Shear Modulus of Rubber, Boltzmann’s Constant and the Temperature is Given.

N = G / KBT

Where;

N = Density of Network Cross Links
G = Shear Modulus of Rubber
KB = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Find the density of network cross links when the shear modulus of rubber is 14, the boltzmann’s constant is 1.380e-23 and temperature is 4.

This implies that;

G = Shear Modulus of Rubber = 14
KB = Boltzmann’s Constant = 1.380e-23
T = Temperature = 4

N = G / KBT
N = 14 / 1.380e-23 x 4
N = 14 / 5.52e-23
N = 2.54e-23

Therefore, the network cross links is 2.54e-23.

Continue reading How to Calculate and Solve for Shear Modulus of Rubber | Fracture Mechanics

How to Calculate and Solve for Schottky Defect | Ceramics

The image above represents schottky defect.

To compute for schottky defect, four essential parameters are needed and these parameters are N, Activation energy (Qs), Boltzmann’s Constant (K) and Temperature (T).

The formula for calculating schottky defect:

Ns = N exp (-Qs/2KT)

Where:

Qs = Activation Energy
K = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Find the schottky defect when the activation energy is 44, N is 22, boltzmann’s constant is 1.38064852E-23 and the temperature is 30.

This implies that;

N = 22
Qs = Activation Energy = 44
K = Boltzmann’s Constant = 1.38054852E-23
T = Temperature = 30

Ns = N exp (-Qs/2KT)
Ns = (22)exp(-(44)/2(1.38064852e-23)(30))
Ns = (22)exp((-44)/(8.283891119e-22))
Ns = (22)exp(-5.3115135583771414e+22)
Ns = (22)(0)
Ns = 0

Therefore, the schottky defect is 0.

Calculating the N when the Schottky Defect, the Activation Energy, the Boltzmann’s Constant and the Temperature is Given.

N = Ns / e (-Qs / 2KT)

Where;

Ns = Schottky Defect
Qs = Activation Energy
K = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Find the N when the schottky defect is 40, the activation energy is 24, the boltzmann’s constant is 1.38064852E-23 and the temperature is 10.

This implies that;

Ns = Schottky Defect = 40
Qs = Activation Energy = 24
K = Boltzmann’s Constant = 1.38064852E-23
T = Temperature = 10

N = Ns / e (-Qs / 2KT)
N = 40 / e (-24 / 2 x 1.38064852E-23 x 10)
N = 40 / e (-24 / 2.76129704E+23)
N = 40 / e (8.691567e-23)
N = 40 / 8.691567e+23
N = 4.602e-23

Therefore, the is 4.602e-23.

Continue reading How to Calculate and Solve for Schottky Defect | Ceramics

How to Calculate and Solve for Frenkel Defect | Ceramics

The image above represents frenkel defect.

To compute for frenkel defect, four essential parameters are needed and these parameters are N, activation energy (Qfr), Boltzmann’s Constant (K) and temperature (T).

The formula for calculating the frenkel defect:

Nfr = N exp (-Qfr / 2KT)

Where:

Nfr = Frenkel Defect
Qfr = Activation Energy
K = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Find the frenkel defect when the activation energy is 34, N is 22, Temperature is 12 and the boltzmann’s constant is 1.38064852e-23.

This implies that;

Qfr = Activation Energy
K = Boltzmann’s Constant
T = Temperature

Nfr = N exp (-Qfr / 2KT)
Nfr = (22)exp(-(34) / 2(1.38064852e-23)(14))
Nfr = (22)exp((-34) / (3.865815856e-22))
Nfr = (22)exp(-8.795038684325786e+22)
Nfr = (22)(0)
Nfr = 0

Therefore, the frenkel defect is 0.

Calculating the Activation Energy when the Frenkel Defect, the Boltzmann’s Constant and the Temperature is Given.

Qfr = – (In (Nfr / N) x 2KT)

Where;

Qfr = Activation Energy
Nfr = Frenkel Defect
K = Boltzmann’s Constant
T = Temperature

Let’s solve an example;
Given that the frenkel defect is 20, the boltzmann’s constant is 5, the temperature is 2 and N is 10. Find the activation energy?

This implies that;

Nfr = Frenkel Defect = 20
K = Boltzmann’s Constant = 5
T = Temperature = 2
N = 10

Qfr = – (In (Nfr / N) x 2KT)
Qfr = – (In (20 / 10) x 2 x 5 x 2)
Qfr = – (In 2 x 20)
Qfr = – (In 40)
Qfr = – 3.688

Therefore, the activation energy is – 3.688.

Continue reading How to Calculate and Solve for Frenkel Defect | Ceramics