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 Reynold’s Number at Minimum Fluidization | Fluidization

The image above represents reynold’s number.

To compute for reynold’s number, four essential parameters are needed and these parameters are Fluidization at Minimum Velocity (umf), Diameter of Bed (d), Density (ρ) and Viscosity (μ).

The formula for calculating reynold’s number:

Re’mf = umf/μ

Where:

Re’mf = Reynold’s Number at Minimum Fluidization
umf = Fluidization at Minimum Velocity
d = Diameter of Bed
ρ = Density
μ = Viscosity

Let’s solve an example;
Find the reynold’s number at minimum fluidization when the fluidization at minimum is 14, the diameter of bed is 8, the density is 12 and the viscosity is 2.

This implies that;

umf = Fluidization at Minimum Velocity = 14
d = Diameter of Bed = 8
ρ = Density = 12
μ = Viscosity = 2

Re’mf = umf/μ
Re’mf = (14)(8)(12)/2
Re’mf = (1344)/2
Re’mf = 672

Therefore, the reynold’s number at minimum fluidization is 672.

Calculating the Fluidization at Minimum Velocity when the Reynold’s Number at Minimum Fluidization, the Diameter of Bed, the Density and the Viscosity is Given.

umf = Re’mf x μ /

Where;

umf = Fluidization at Minimum Velocity
Re’mf = Reynold’s Number at Minimum Fluidization
d = Diameter of Bed
ρ = Density
μ = Viscosity

Let’s solve an example;
Find the fluidization at minimum velocity when the reynold’s number at minimum fluidization is 12, the diameter of bed is 4, the density is 8 and the viscosity is 3.

This implies that;

Re’mf = Reynold’s Number at Minimum Fluidization = 12
d = Diameter of Bed = 4
ρ = Density = 8
μ = Viscosity = 3

umf = Re’mf x μ /
umf = 12 x 3 / (4)(8)
umf = 36 / 32
umf = 1.125

Therefore, the fluidization at minimum velocity is 1.125.

Continue reading How to Calculate and Solve for Reynold’s Number at Minimum Fluidization | Fluidization

How to Calculate and Solve for Fluid Velocity | Fluidization

The image above represents fluid velocity.

To compute for fluid velocity, five essential parameters are needed and these parameters are Porosity (e), Pressure (-ΔP), Distance of Bed (d), Viscosity (μ) and Length of Bed (l).

The formula for calculating fluid velocity:

uc = 0.0055(/(1 – e)²)(-ΔPd²/μl)

Where:

uc = Fluid Velocity
e = Porosity
-ΔP = Pressure
d = Distance of Bed
μ = Viscosity
l = Length of Bed

Let’s solve an example;
Find the fluid velocity when the porosity is 20, the pressure is 10, the distance of bed is 12, the viscosity is 14 and the length of bed is 11.

This implies that;

e = Porosity = 20
-ΔP = Pressure = 10
d = Distance of Bed = 12
μ = Viscosity = 14
l = Length of Bed = 11

uc = 0.0055(/(1 – e)²)(-ΔPd²/μl)
uc = 0.0055(20³/(1 – 20)²)(10(12)²/(14)(11))
uc = 0.0055(8000/(-19)²)(10(144)/154)
uc = 0.0055(8000/361)(1440/154)
uc = 0.0055(22.16)(9.35)
uc = 1.139

Therefore, the fluid viscosity is 1.139 m/s.

Continue reading How to Calculate and Solve for Fluid Velocity | Fluidization

How to Calculate and Solve for Reynold’s Number

The image above represents reynold’s number.

To compute for reynold’s number, four essential parameters are needed and these parameters are Fluid density (ρ), Discharge velocity (q), Diameter of passage way (d) and Viscosity (μ).

The formula for calculating reynold’s number:

Re = ρqd / μ

Where:

Re = Reynold’s Number
ρ = Fluid Density
q = Discharge Velocity
d = Diameter of Passage Way
μ = Viscosity

Let’s solve an example;
Find the reynold’s number when the fluid density is 12, discharge velocity is 10, diameter of passsage way is 22 and the viscosity is 16.

This implies that;

ρ = Fluid Density = 12
q = Discharge Velocity = 10
d = Diameter of Passage Way = 22
μ = Viscosity = 16

Re = ρqd / μ
Re = (12)(10)(22) / 16
Re = 2640 / 16
Re = 165

Therefore, the Reynold’s number is 165.

Calculating the Fluid Density when the Reynold’s Number, the Discharge Velocity, the Diameter of Passage Way and the Viscosity is Given.

ρ = Re x μ / qd

Where;

ρ = Fluid Density
Re = Reynold’s Number
q = Discharge Velocity
d = Diameter of Passage Way
μ = Viscosity

Let’s solve an example;
Find the fluid density when the reynold’s number is 20, the discharge velocity is 8, the diameter of passage way is 4 and the viscosity is 3.

This implies that;

Re = Reynold’s Number = 20
q = Discharge Velocity = 8
d = Diameter of Passage Way = 4
μ = Viscosity = 3

ρ = Re x μ / qd
ρ = 20 x 3 / (8)(4)
ρ = 60 / 32
ρ = 1.875

Therefore, the fluid density is 1.875.

Continue reading How to Calculate and Solve for Reynold’s Number

How to Calculate and Solve for Viscosity | Ceramics

The image above represents viscosity.

To compute for viscosity, three essential parameters are needed and these parameters are Force applied (F), Area (A) and Derivation Ratio of Velocity to Distance of Fluid Flow (dv/dy).

The formula for calculating viscosity:

η = F/A / dv/dy

Where:

η = Viscosity
F = Force Applied
A = Area
dv/dy = Derivation Ratio of Velocity to Distance of Fluid Flow

Let’s solve an example;
Find the viscosity when the force applied is 21, area is 14 and derivation ratio of velocity to distance of fluid flow is 19.

This implies that;

F = Force Applied = 21
A = Area = 14
dv/dy = Derivation Ratio of Velocity to Distance of Fluid Flow = 19

η = F/A / dv/dy
η = (21/14) / (19)
η = (1.5) / (19)
η = 0.0789

Therefore, the viscosity is 0.0789 Pa s.

Calculating Force Applied when the Viscosity, the Area and the Derivation ratio of velocity to distance of fluid flow is Given.

F = (η x dv/dy) A

Where;

F = Force Applied
η = Viscosity
A = Area
dv/dy = Derivation Ratio of Velocity to Distance of Fluid Flow

Let’s solve an example;
Find the force applied when the viscosity is 20, the area is 30 and the derivation is 8.

This implies that;

η = Viscosity = 20
A = Area = 30
dv/dy = Derivation Ratio of Velocity to Distance of Fluid Flow = 8

F = (η x dv/dy) A
F = (20 x 8) 30
F = (160) 30
F = 4800

Therefore, the force applied is 4800.

Continue reading How to Calculate and Solve for Viscosity | Ceramics

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

The image above represents pressure gradient.

To compute for pressure gradient, four essential parameters are needed and these parameters are value (α), viscosity (μ), screw rotation speed (N) and proportionality constant that depends on screw geometry (B).

The formula for calculating pressure gradient:

ΔP = αμN / B

Where;

ΔP = Pressure Gradient
α = Value
μ = Viscosity
N = Screw Rotation Speed
B = Proportionality Constant that Depends on Screw Geometry

Let’s solve an example;
Find the pressure gradient when the value is 2, viscosity is 9, screw rotation speed is 20 and proportionality constant that depends on screw geometry is 24.

This implies that;

α = Value = 2
μ = Viscosity = 9
N = Screw Rotation Speed = 20
B = Proportionality Constant that Depends on Screw Geometry = 24

ΔP = αμN / B
ΔP = (2)(9)(20) / 24
ΔP = 360 / 24
ΔP = 15

Therefore, the pressure gradient is 15.

Calculating the Value when the Pressure Gradient, Viscosity, Screw Rotation Speed and Proportionality Constant that Depends on Screw Geometry is Given.

α = ΔP x B / μN

Where;

α = Value
ΔP = Pressure Gradient
μ = Viscosity
N = Screw Rotation Speed
B = Proportionality Constant that Depends on Screw Geometry

Let’s solve an example;
Find the value when the pressure gradient is 20, viscosity is 5, screw rotation speed is 11 and the proportionality constant that depends on screw geometry is 7.

This implies that;

ΔP = Pressure Gradient = 20
μ = Viscosity = 5
N = Screw Rotation Speed = 11
B = Proportionality Constant that Depends on Screw Geometry = 7

α = ΔP x B / μN
α = 20 x 7 / (5)(11)
α = 140 / 55
α = 2.54

Therefore, the value is 2.54.

Continue reading How to Calculate and Solve for Pressure Gradient | 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 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 the Radius of Investigation in Well Testing | The Calculator Encyclopedia

The image above represents radius of investigation.

To compute for the radius of investigation, five essential parameters are needed and these parameters are permeability (k), porosity (φ), viscosity (μ), time (t) and total compressibility (CT).

The formula for calculating the radius of investigation:

rinv = 0.0325 √[Kt / φμCT]

Where:

rinv = Radius of Investigation
K = Permeability
φ = Porosity
μ = Viscosity
t = Time
CT = Total Compressibility

Let’s solve an example;
Find the radius of investigation when the permeability is 25, porosity is 18, viscosity is 12, time is 22 and total compressibility is 37.

This implies that;

K = Permeability = 25
φ = Porosity =18
μ = Viscosity = 12
t = Time = 22
CT = Total Compressibility = 37

rinv = 0.0325 √[Kt / φμCT]
rinv = 0.0325 √[25 x 22 / 18 x 12 x 37]
rinv = 0.0325 √[25 x 22 / 7992]
rinv = 0.0325 √[550 / 7992]
rinv = 0.0325 √[0.068]
rinv = 0.0325 [0.26]
rinv = 0.0085

Therefore, the radius of investigation is 0.0085 ft.

Continue reading How to Calculate and Solve for the Radius of Investigation in Well Testing | The Calculator Encyclopedia

How to Calculate and Solve for Infinite Acting Period | The Calculator Encyclopedia

The image represents the infinite acting period.

To compute the infinite acting period, six essential parameters are needed and these parameters are permeability (k), porosity (φ), well drainage area (A), viscosity (μ), Dimensionless Time to End of Infinite Acting Period ((tDA)eia) and total compressibility (CT).

The formula for calculating the infinite acting period:

teia = [φ μ CT A / 0.000263K] (tDA)eia

Where;

teia = Infinite Acting Period
K = Permeability
φ = Porosity
A = Well Drainage Area
μ = Viscosity
(tDA)eia = Dimensionless Time to End of Infinite Acting Period
CT = Total Compressibility

Let’s solve an example;
Given that the permeability is 21, porosity is 15, well drainage area is 32, viscosity is 26, dimensionless time to end of infinite acting period is 44 and total compressibility is 34.
Find the infinite acting period?

This implies that;

K = Permeability = 21
φ = Porosity = 15
A = Well Drainage Area = 32
μ = Viscosity = 26
(tDA)eia = Dimensionless Time to End of Infinite Acting Period = 44
CT = Total Compressibility = 34

teia = [φ μ CT A / 0.000263K] (tDA)eia
teia = [15 x 26 x 34 x 32 / 0.000263 x 21] 44
teia = [424320 / 0.005523] 44
teia = [76827810.972] 44
teia = 3380423682.78

Therefore, the infinite acting period is 3380423682.78 s.

Continue reading How to Calculate and Solve for Infinite Acting Period | The Calculator Encyclopedia