How to Calculate and Solve for Shielding Factor | Polymer & Textile

The image above represents shielding factor.

To compute for shielding factor, two essential parameter are needed and these parameters are coil radius (Rs) and shielding depth (L).

The formula for calculating the shielding factor:

ε = Rs / L

Where:

ε = Shielding Factor
Rs = Coil Radius
L = Shielding Depth

Let’s solve an example;
Find the shielding factor when the coil radius is 11 and the shielding depth is 21.

This implies that;

Rs = Coil Radius = 11
L = Shielding Depth = 21

ε = Rs / L
ε = 11 / 21
ε = 0.52

Therefore, the shielding factor is 0.52.

Calculating for the Coil Radius when the Shielding Factor and the Shielding Depth is Given.

Rs = ε x L

Where;

Rs = Coil Radius
ε = Shielding Factor
L = Shielding Depth

Let’s solve an example;
Given that the shielding factor is 30 and the shielding depth is 5. Find the coil radius?

This implies that;

ε = Shielding Factor = 30
L = Shielding Depth = 5

Rs = ε x L
Rs = 30 x 5
Rs = 150

Therefore, the coil radius is 150.

Continue reading How to Calculate and Solve for Shielding Factor | Polymer & Textile

How to Calculate and Solve for Degree of Polymerization | combination | Polymer & Textile

The image above represents the degree of polymerization.

To compute for degree of polymerization, one essential parameter is needed and this parameter is parameter (V).

The formula for calculating the degree of polymerization:

Xn = 2V

Where:

Xn = Degree of Polymerization
V = Parameter

Let’s solve an example;
Find the degree of polymerization with a parameter of 28.

This implies that;

V = Parameter = 28

Xn = 2V
Xn = 2 (28)
Xn = 56

Therefore, the degree of polymerization is 56.

Continue reading How to Calculate and Solve for Degree of Polymerization | combination | Polymer & Textile

How to Calculate and Solve for Helix Angle | Polymer & Textile

The image above represents the helix angle.

To compute for helix angle, two essential parameters are needed and these parameters are screw lead (S) and barrel diameter (Db).

The formula for calculating helix angle:

θ = tan-1[S / (π x Db)]

Where;

θ = Helix Angle
S = Screw Lead
Db = Barrel Diameter

Let’s solve an example;
Find the helix angle when the screw lead is 14 and the barrel diameter is 22.

This implies that;

S = Screw Lead = 14
Db = Barrel Diameter = 22

θ = tan-1[S / (π x Db)]
θ = tan-1[14/(π x 22)]
θ = tan-1[14/69.115]
θ = tan-1[0.202]
θ = 11.45°

Therefore, the helix angle is 11.45°.

Continue reading How to Calculate and Solve for Helix Angle | Polymer & Textile

How to Calculate and Solve for Reynold’s Number | Polymer & Textile

The image above represents the Reynold’s number.

To compute for  Reynold’s number, four essential parameters are needed and these parameters are density of fluid (ρ), mean flow velocity (v), diameter of die (d) and viscosity (μ).

The formula for calculating Reynold’s number:

Re = ρvd / μ

Where:

Re = Reynold’s Number
ρ = Density of Fluid
v = Mean Flow Velocity
d = Diameter of Die
μ = Viscosity

Let’s solve an example;
Given that density of fluid is 14, mean flow velocity is 27, diameter of die is 18 and viscosity is 10. Find the Reynold’s number?

This implies that;

ρ = Density of Fluid = 14
v = Mean Flow Velocity = 27
d = Diameter of Die = 18
μ = Viscosity = 10

Re = ρvd / μ
Re = (14)(27)(18) / 10
Re = 6804 / 10
Re = 680.4

Therefore, the Reynold’s number is 680.4.

Calculating the Density of Fluid when the Reynold’s Number, Mean Flow Velocity, Diameter of Die and Viscosity is Given.

ρ = Re x μ / vd

Where:

ρ = Density of Fluid
Re = Reynold’s Number
v = Mean Flow Velocity
d = Diameter of Die
μ = Viscosity

Let’s solve an example;
Given that Reynold’s number is 26, mean flow velocity is 14, diameter of die is 8 and viscosity is 12. Find the density of fluid?

This implies that;

Re = Reynold’s Number = 26
v = Mean Flow Velocity = 14
d = Diameter of Die = 8
μ = Viscosity = 12

ρ = Re x μ / vd
ρ = 26 x 12 / (14)(8)
ρ = 312 / 112
ρ = 2.78

Therefore, the density of fluid is 2.78.

Calculating the Mean Flow Velocity when the Reynold’s Number, Density of Fluid, Diameter of Die and Viscosity is Given.

v = Re x μ / ρd

Where:

v = Mean Flow Velocity
Re = Reynold’s Number
ρ = Density of Fluid
d = Diameter of Die
μ = Viscosity

Let’s solve an example;
Given that density of fluid is 22, Reynold’s number is 32, diameter of die is 9 and viscosity is 6. Find the mean flow velocity?

This implies that;

Re = Reynold’s Number = 32
ρ = Density of Fluid = 22
d = Diameter of Die = 9
μ = Viscosity = 6

v = Re x μ / ρd
v = 32 x 6 / (22)(9)
v = 192 / 198
v = 0.96

Therefore, the mean flow velocity is 0.96.

Continue reading How to Calculate and Solve for Reynold’s Number | Polymer & Textile

How to Calculate and Solve for Moisture Regain | Polymer & Textile

The image above represents the moisture regain.

To compute for moisture regain, two essential parameters are needed and these parameters are mass of absorbed water in specimen (MW) and mass of dry specimen (MD).

The formula for calculating moisture regain:

R = MW / MD x 100

Where;

R = Moisture Regain
MW = Mass of Absorbed Water in Specimen
MD = Mass of Dry Specimen

Let’s solve an example;
Find the moisture regain when the mass of absorbed water in specimen is 13 and the mass of dry specimen is 21.

This implies that;

MW = Mass of Absorbed Water in Specimen = 13
MD = Mass of Dry Specimen = 21

R = MW / MD x 100
R = 13 / 21 x 100
R = 0.619 x 100
R = 61.90

Therefore, the moisture regain is 61.90%.

Calculating the Mass of Absorbed Water in Specimen when the Moisture Regain and the Mass of Dry Specimen is Given.

Mw = RMD / 100

Where;

MW = Mass of Absorbed Water in Specimen
R = Moisture Regain
MD = Mass of Dry Specimen

Let’s solve an example;
Find the mass of absorbed water in specimen when the moisture regain is 14 and the mass of dry specimen is 11.

This implies that;

R = Moisture Regain = 14
MD = Mass of Dry Specimen = 11

Mw = RMD / 100
Mw = 14 x 11 / 100
Mw = 154 / 100
Mw = 1.54

Therefore, the mass of absorbed water in specimen is 1.54.

Continue reading How to Calculate and Solve for Moisture Regain | Polymer & Textile

How to Calculate and Solve for Denier | Polymer & Textile

The image above represents denier.

To compute for denier, two essential parameters are needed and these parameters are mass (M) and length (L).

The formula for calculating denier:

Denier = M/L x 9000

Where;

D = Denier
M = Mass
L = Length

Let’s solve an example;
Find the denier with a mass of 24 and a length of 28.

This implies that;

M = Mass = 24
L = Length = 28

Denier = M/L x 9000
Denier = 24/28 x 9000
Denier = 0.857

Therefore, the denier is 0.857.

Calculating the Mass when the Denier and the Length is Given.

M = DL / 9000

Where;

M = Mass
D = Denier
L = Length

Let’s solve an example;
Find the mass of a denier with 30 and a length of 18.

This implies that;

D = Denier = 30
L = Length = 18

M = DL / 9000
M = 30 x 18 / 9000
M = 540 / 9000
M = 0.06

Therefore, the mass is 0.06.

Continue reading How to Calculate and Solve for Denier | Polymer & Textile

How to Calculate and Solve for Burst Pressure | Polymer & Textile

The image above represents burst pressure.

To compute for burst pressure, three essential parameters are needed and these parameters are tensile strength (T), half of outside diameter of tubing (x) and half of inside diameter of tubing (y).

The formula for calculating the burst pressure:

P = T(x² – y²) / y²(1 + /y²)

Where;

P = Burst Pressure
T = Tensile Strength
x = Half of Outside Diameter of Tubing
y = Half of Inside Diameter of Tubing

Let’s solve an example;
Find the burst pressure when the tensile strength is 12, the half of outside diameter of tubing is 11 and the half of inside diameter of tubing is 7.

This implies that;

T = Tensile Strength = 12
x = Half of Outside Diameter of Tubing = 11
y = Half of Inside Diameter of Tubing = 7

P = T(x² – y²) / y²(1 + /y²)
P = 12(11² – 7²) / 7²(1 + 11²/7²)
P = 12(121 – 49) / 49(1 + 121/49)
P = 12(72) / 49(1 + 2.469)
P = 864 / 49(3.469)
P = 864 / 170
P = 5.08

Therefore, the burst pressure is 5.08.

Continue reading How to Calculate and Solve for Burst Pressure | Polymer & Textile

How to Calculate and Solve for Flexural Stress | Polymer & Textile

The image above represents the flexural stress.

To compute for the flexural stress, three parameters are needed and these parameters are axial load at fracture point (F), width of polymer (b) and thickness of material (d).

The formula for calculating the flexural stress:

σ = F / bd

Where:

σ = Flexural Stress
F = Axial Load at Fracture Point
b = Width of Polymer
d = Thickness of Material

Let’s solve an example;
Find the flexural stress when the axial load at fracture point is 11, the width of polymer is 17 and the thickness of material is 27.

This implies that;

F = Axial Load at Fracture Point = 11
b = Width of Polymer = 17
d = Thickness of Material = 27

σ = F / bd
σ = 11 / (17)(21)
σ = 11 / 357
σ = 0.0308

Therefore, the flexural stress is 0.0308.

Calculating the Axial Load at Fracture Point when the Flexural Stress, the Width of Polymer and the Thickness of Material is Given.

F = σbd

Where:

F = Axial Load at Fracture Point
σ = Flexural Stress
b = Width of Polymer
d = Thickness of Material

Let’s solve an example;
Find the axial load at fracture point when the flexural stress is 21, the width of polymer is 10 and the thickness of material is 7.

This implies that;

σ = Flexural Stress = 21
b = Width of Polymer = 10
d = Thickness of Material = 7

F = σbd
F = 21 x 10 x 7
F = 1470

Therefore, the axial load at fracture point is 1470.

Continue reading How to Calculate and Solve for Flexural Stress | Polymer & Textile

How to Calculate and Solve for Maxwell’s Tone of Relaxation | Polymer & Textile

The image above represents maxwell’s tone of relaxation.

To compute for the maxwell’s tone of relaxation, two essential parameters are needed and these parameters are parameter (λ) and viscosity (η).

The formula for calculating the maxwell’s tone of relaxation:

G = λη

Where:

G = Maxwell’s Tone of Relaxation
λ = Parameter
η = Viscosity

Let’s solve an example;
Given that the parameter is 15 and the viscosity is 10. Find the maxwell’s tone of relaxation?

This implies that;

λ = Parameter = 15
η = Viscosity = 10

G = λη
G = (15)(10)
G = 150

Therefore, the maxwell’s tone of relaxation is 150.

Calculating the Parameter when the Maxwell’s Tone of Relaxation and the Viscosity is Given.

λ = G / η

Where;

λ = Parameter
G = Maxwell’s Tone of Relaxation
η = Viscosity

Let’s solve an example;
Given that the maxwell’s tone of relaxation is 30 and the viscosity is 15. Find the viscosity?

This implies that;

G = Maxwell’s Tone of Relaxation = 30
η = Viscosity = 15

λ = G / η
λ = 30 / 15
λ = 2

Therefore, the parameter is 2.

Continue reading How to Calculate and Solve for Maxwell’s Tone of Relaxation | Polymer & Textile

How to Calculate and Solve for Weight Average Molecular Weight | Polymer & Textile

The image above represents the weight average molecular weight.

To compute for the weight average molecular weight, two essential parameters are needed and these parameters are Weight Fraction (W) and Arithmetic Mean of the Molar Mass Distribution (M).

The formula for calculating the weight average molecular weight:

Mw = ΣWM

Where;

Mw = Weight Average Molecular Weight
W = Weight Fraction
M = Arithmetic Mean of the Molar Mass Distribution

Let’s solve an example;
Find the weight average molecular weight when the weight fraction is 9 and the arithmetic mean of the molar mass distribution is 18.

This implies that;

W = Weight Fraction = 9
M = Arithmetic Mean of the Molar Mass Distribution = 18

Mw = ΣWM

W M WM
9 18 162

ΣWM = 162

ΣWM = 162

Mw = 162

Therefore, the weight average molecular weight is 162.

Calculating the Weight Fraction when the Weight Average Molecular Weight and the Arithmetic Mean of the Molar Mass Distribution is Given.

W = Mw / M

Where;

W = Weight Fraction
Mw = Weight Average Molecular Weight
M = Arithmetic Mean of the Molar Mass Distribution

Let’s solve an example;
Find the weight fraction when the weight average molecular weight is 27 and the arithmetic mean of the molar mass distribution is 14.

This implies that;

Mw = Weight Average Molecular Weight = 27
M = Arithmetic Mean of the Molar Mass Distribution = 14

W = Mw / M
W = 27 / 14
W = 1.928

Therefore, the weight fraction is 1.928

Continue reading How to Calculate and Solve for Weight Average Molecular Weight | Polymer & Textile