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How to Calculate and Solve for Relationship between Thermal Resistivity and Temperature | Electrical Properties

The relationship betweeen thermal resistivity and temperature is illustrated by the image below.

To compute for the relationship between thermal resistivity and temperature, three essential parameters are needed and these parameters are Constant of the Metal (a), Constant of the Metal (ρ_o) and Temperature (T).

The formula for calculating relationship between thermal resistivity and temperature:

ρ_t = ρ_o + aT

Where:

ρ_t = Thermal Resistivity
ρ_o = Constant of the Metal
a = Constant of the Metal
T = Temperature

Let’s solve an example;
Find the relationship between thermal resistivity and temperature when the constant of the metal is 32, the constant of the metal is 8 and the temperature is 2.

This implies that;

ρ_o = Constant of the Metal = 32
a = Constant of the Metal = 8
T = Temperature = 2

ρ_t = ρ_o + aT
ρ_t = 32 + (8)(2)
ρ_t = 32 + (16)
ρ_t = 48

Therefore, the relationship between thermal resistivity and temperature is 48 Ωm.

Calculating the Constant of the Metal when the Relationship between Thermal Resistivity and Temperature, the Constant of the Metal and the Temperature is Given.

ρ_o = ρ_t – aT

Where:

ρ_o = Constant of the Metal
ρ_t = Thermal Resistivity
a = Constant of the Metal
T = Temperature

Let’s solve an example;
Find the constant of the metal when the thermal resistivity is 32, the constant of the metal is 2 and the temperature is 4.

This implies that;

ρ_t = Thermal Resistivity = 32
a = Constant of the Metal = 2
T = Temperature = 4

ρ_o = ρ_t – aT
ρ_o = 32 – 2(4)
ρ_o = 32 – 8
ρ_o = 24

Therefore, the constant of the metal is 24.

Continue reading How to Calculate and Solve for Relationship between Thermal Resistivity and Temperature | Electrical Properties

How to Calculate and Solve for Equilibrium Constant for Free Energy | Corrosion

The equilibrium constant for free energy is illustrated by the image below.

To compute for equilibrium constant for free energy, three essential parameters are needed and these parameters are Change in Free Energy (ΔG°), Gas Constant (R) and Temperature (T).

The formula for calculating equilibrium constant for free energy:

K = exp(^ΔG°/_RT)

Where:

K = Equilibrium Constant for Free Energy
ΔG° = Change in Free Energy
R = Gas Constant
T = Temperature

Let’s solve an example;
Find the equilibrium constant for free energy when the change in free energy is 14, the gas constant is 10 and the temperature is 2.

This implies that;

ΔG° = Change in Free Energy = 14
R = Gas Constant = 10
T = Temperature = 2

K = exp(^ΔG°/_RT)
K = exp(¹⁴/_(10)(2))
K = exp(¹⁴/₂₀)
K = exp(0.7)
K = 2.0137

Therefore, the equilibrium constant for free energy is 2.0137.

Continue reading How to Calculate and Solve for Equilibrium Constant for Free Energy | Corrosion

How to Calculate and Solve for Nernst Equation | Corrosion

The nernst equation is illustrated by the image below.

To compute for nernst equation, seven essential parameters are needed and these parameters are Electrochemical Cell Potential (ΔV°), Gas Constant (R), Temperature (T), Number of Electrons (n), Faraday’s Constant (F), Molar Concentration [M₁ⁿ⁺] and Molar Concentration [M₂ⁿ⁺].

The formula for caluclating nernst equation:

ΔV = ΔV° – ^RT/_nF In(^M₁n+/_M2ⁿ⁺)

Where:

ΔV = Nernst Equation | Potential
ΔV° = Electrochemical Cell Potential
R = Gas Constant
T = Temperature
n = Number of Electrons
F = Faraday’s Constant
M₁ⁿ⁺ = Molar Concentration
M₂ⁿ⁺ = Molar Concentration

Let’s solve an example;
Find the nernst equation when the electrochemical cell potential is 8, the gas constant is 2, the temperature is 5, the number of electrons is 3, the faraday’s constant is 9, the molar concentration is 10 and the molar concentration is 4.

This implies that;

ΔV° = Electrochemical Cell Potential = 8
R = Gas Constant = 2
T = Temperature = 5
n = Number of Electrons = 3
F = Faraday’s Constant = 9
M₁ⁿ⁺ = Molar Concentration = 10
M₂ⁿ⁺ = Molar Concentration = 4

ΔV = ΔV° – ^RT/_nF In(^M₁n+/_M2ⁿ⁺)
ΔV = 8 – (⁽²⁾⁽⁵⁾/₍₃₎₍₉₎) In(¹⁰/₄)
ΔV = 8 – (¹⁰/₂₇) In (2.5)
ΔV = 8 – 0.37 (0.91)
ΔV = 8 – (0.339)
ΔV = 7.66

Therefore, the nernest equation is 7.66 V.

Continue reading How to Calculate and Solve for Nernst Equation | Corrosion

How to Calculate and Solve for Entropy of Water | Enthalpy

The entropy of water is illustrated by the image below.

To compute for entropy of water, two essential parameters are needed and these parameters are Constant Pressure (C_pw) and Temprature (T).

The formula for calculating entropy of water:

S_p = C_pw ln(^T / ₂₇₃)

Where:

S_p = Entropy of Water
C_pw = Constant Pressure
T = Temprature

Let’s solve an example;
Find the entropy of water when the constant pressure is 40 and the temperature is 10.

This implies that;

C_pw = Constant Pressure = 40
T = Temprature = 10

S_p = C_pw ln(^T / ₂₇₃)
S_p = 40 x ln(¹⁰ / ₂₇₃)
S_p = 40 x ln(0.0366)
S_p = 40 x -3.306
S_p = -132.27

Therefore, the entropy of water is -132.27 J/K.

Continue reading How to Calculate and Solve for Entropy of Water | Enthalpy

How to Calculate and Solve for Infinitesimal Change in Surface Tension | Gas Absorption | Mineral Processing

The image above represents infinitesimal change in surface tension.

To compute for infinitesimal change in surface tension, five essential parameters are needed and these parameters are Gas Constant (R), Temperature (T), Surface Excess or Absorption Density (Γ), Particle Diameter (d) and Activity of Adsorbate (a_i).

The formula for calculating infinitesimal change in surface tension:

dγ = -RT.Γd In(a_i)

Where:

dγ = Infinitesimal Change in Surface Tension | Gas Absorption
R = Gas Constant
T = Temperature
Γ = Surface Excess or Absorption Density
d = Particle Diameter
a_i = Activity of Adsorbate

Let’s solve an example;
Find the infinitesimal change in surface tension when the gas constant is 11, the temperature is 21, the surface excess or absorption density is 12, the particle diameter is 14 and the activity of adsorbate is 18.

This implies that;

R = Gas Constant = 11
T = Temperature = 21
Γ = Surface Excess or Absorption Density = 12
d = Particle Diameter = 14
a_i = Activity of Adsorbate = 18

dγ = -RT.Γd In(a_i)
dγ = -(11)(21).(12)(14) In(18)
dγ = (-231).(12)(14)(2.89)
dγ = (-231).(485.58)
dγ = -112169.54

Therefore, the infinitesimal change in surface tension is -112169.54 N/m.

Continue reading How to Calculate and Solve for Infinitesimal Change in Surface Tension | Gas Absorption | Mineral Processing

How to Calculate and Solve for Urbain and Boiret Viscosity Model | Transport Phenomena

The image above represents urbain and boiret viscosity model.

To compute for urbain and boiret viscosity model, three essential parameters are needed and these parameters are Empirical Constant (A), Empirical Constant (B) and Temperature (T).

The formula for calculating urbain and boiret viscosity model:

η = AT.exp(^1000B / _T)

Where:

η = Urbain and Boiret Viscosity Model
A = Empirical Constant
B = Empirical Constant
T = Temperature

Let’s solve an example;
Find the urbain and boiret viscosity model when the empirical constant is 14, the empirical constant is 21 and the temperature is 18.

This implies that;

A = Empirical Constant = 14
B = Empirical Constant = 21
T = Temperature = 18

η = AT.exp(^1000B / _T)
η = (14)(18).exp(^1000(21) / ₁₈)
η = 252.exp(²¹⁰⁰⁰ / ₁₈)
η = 252.exp(1166.66)
η = 252.(Infinity)
η = Infinity

Therefore, the urbain and boiret viscosity model is infinity.

Continue reading How to Calculate and Solve for Urbain and Boiret Viscosity Model | Transport Phenomena

How to Calculate and Solve for Viscosity for Metallic Alloys | Transport Phenomena

The image above represents viscosity for metallic alloys.

To compute for viscosity for metallic alloys, six essential parameters are needed and these parameters are Viscosity of Metal (η), Interatomic Distance (δ), Avogadro’s Number (N_A), Atomic Weight (M), Gas Constant (R) and Temperature (T).

The formula for calculating viscosity for metallic alloys:

η* = ^ηδ²N_A / _√(MRT)

Where:

η* = Viscosity for Metallic Alloys
η = Viscosity of Metal
δ = Interatomic Distance
N_A = Avogadro’s Number
M = Atomic Weight
R = Gas Constant
T = Temperature (Kelvin)

Let’s solve an example;
Find the viscosity for metallic alloys when the viscosity of metal is 4, the interatomic distance is 12, the avogadro’s number is 6.023e+23, the atomic weight is 9, the gas constant is 2 and the temperature is 5.

This implies that;

η = Viscosity of Metal = 4
δ = Interatomic Distance = 12
N_A = Avogadro’s Number = 6.023e+23
M = Atomic Weight = 9
R = Gas Constant = 2
T = Temperature (Kelvin) = 5

η* = ^ηδ²N_A / _√(MRT)
η* = ^{(4)(12)²(6.023e+23)} / _{√((9)(2)(5))}
η* = ^{(4)(144)(6.023e+23)} / _√(90)
η* = ^3.469e+26 / _9.486
η* = 3.65e+25

Therefore, the viscosity for metallic alloys is 3.65e+25 m²/s.

Continue reading How to Calculate and Solve for Viscosity for Metallic Alloys | Transport Phenomena

How to Calculate and Solve for Chapman Enskog Theory Integral | Transport Phenomena

The image above represents chapman enskog theory integral.

To compute for chapman enskog theory integral, three essential parameters are needed and these parameters are Boltzmann’s Constant (K_B), Temperature (T) and Enskog Constant (Energy Parameter) (ε).

The formula for calculating chapman enskog theory integral:

η_n = ^K_BT / _ε

Where:

η_n = Chapman Enskog Theory Integral
K_B = Boltzmann’s Constant
T = Temperature
ε = Enskog Constant (Energy Parameter)

Let’s solve an example;
Find the chapman enskog theory integral when the boltzmann’s constant is 1.380e-23, the temperature is 20 and the enskog constant (energy parameter) is 10.

This implies that;

K_B = Boltzmann’s Constant = 1.380e-23
T = Temperature = 20
ε = Enskog Constant (Energy Parameter) = 10

η_n = ^K_BT / _ε
η_n = ^{(1.380e-23)(20)} / ₁₀
η_n = ^2.76129e-22 / ₁₀
η_n = 2.76e-23

Therefore, the chapman enskog theory integral is 2.76e-23.

Calculating the Temperature when the Chapman Enskog Theory Integral, the Boltzmann’s Constant and the Enskog Constant (Energy Parameter) is Given.

T = ^{η_n x ε} / _K_B

Where:

T = Temperature
η_n = Chapman Enskog Theory Integral
K_B = Boltzmann’s Constant
ε = Enskog Constant (Energy Parameter)

Let’s solve an example;
Find the temperature when the chapman enskog theory integral is 10, the boltzmann’s constant is 1.380e-23 and the enskog constant (energy parameter) is 8.

This implies that;

η_n = Chapman Enskog Theory Integral = 10
K_B = Boltzmann’s Constant = 1.380e-23
ε = Enskog Constant (Energy Parameter) = 8

T = ^{η_n x ε} / _K_B
T = ^{10 x 8} / _1.380e-23
T = ⁸⁰ / _1.380e-23
T = 5.79e+24

Therefore, the temperature is 5.79e+24.

Continue reading How to Calculate and Solve for Chapman Enskog Theory Integral | Transport Phenomena

How to Calculate and Solve for Viscosity | Non Polar Gases | Transport Phenomena

The image above represents viscosity.

To compute for viscosity, four essential parameters are needed and these parameters are Gram Molecular Weight (M), Temperature (Kelvin) (T), Characteristic Diameter (σ) and Collision Integral of Chapman Enskog Theory ( η_n).

The formula for calculating viscosity:

η = ^{2.67E-5√(MT)} / _σ².ηn

Where:

η = Viscosity | Non Polar Gases
M = Gram Molecular Weight
T = Temperature (Kelvin)
λ = Characteristic Diameter
η_n = Collision Integral of Chapman Enskog Theory

Let’s solve an example;
Find the viscosity when the gram molecular weight is 10, the temperature is 12, the characteristic diameter is 14 and the collision integral of chapman enskog theory is 16.

This implies that;

M = Gram Molecular Weight = 10
T = Temperature (Kelvin) = 12
λ = Characteristic Diameter = 14
η_n = Collision Integral of Chapman Enskog Theory = 16

η = ^{2.67E-5√(MT)} / _σ².ηn
η = ^{2.67E-5√((10)(12))} / _(14)².(16)
η = ^{2.67E-5√(120)} / _(196).(16)
η = ^{2.67E-5(10.954)} / ₃₁₃₆
η = ^0.000292 / ₃₁₃₆
η = 9.326e-8

Therefore, the viscosity is 9.326e-8 m²/s.

Continue reading How to Calculate and Solve for Viscosity | Non Polar Gases | Transport Phenomena

How to Calculate and Solve for Newton’s Laws of Viscosity of Gases | Transport Phenomena

The image above represents newton’s laws of viscosity of gases.

To compute for newton’s laws of viscosity of gases, four essential parameters are needed and these parameters are Number of Molecules (m), Boltzmann’s Constant (K_B), Temperature (T) and Pipe Diameter (d).

The formula for calculating newton’s laws of viscosity of gases:

η = ^2(mK_BT)0.5 / _3π^1.5.d²

Where:

η = Newton’s Law of Viscosity of Gases
m = Number of Molecules
K_B = Boltzmann’s Constant
T = Temperature
d = Pipe Diameter

Let’s solve an example;
Find the newton’s law of viscosity of gases when the number of molecules is 14, the boltzmann’s constant is 1.380E-23, the temperature is 10 and the pipe diameter is 21.

This implies that;

m = Number of Molecules = 14
K_B = Boltzmann’s Constant = 1.380E-23
T = Temperature = 10
d = Pipe Diameter = 21

η = ^2(mK_BT)0.5 / _3π^1.5.d²
η = ^{2((14)(1.380e-23)(10))0.5} / _{3π^1.5.(21)²}
η = ^{2(1.93e-21)0.5} / _{(16.70).(441)}
η = ^2(4.396e-11) / _7366.89
η = ^8.79e-11 / _7366.89
η = 1.19e-14

Therefore, the newton’s law of viscosity of gases is 1.19e-14 m²/s.

Continue reading How to Calculate and Solve for Newton’s Laws of Viscosity of Gases | Transport Phenomena