## How to Calculate and Solve for Ratio of Linear Velocity of Gas | Fluidization The image above represents ratio of linear velocity of gas.

To compute for ratio of linear velocity of gas, three essential parameters are needed and these parameters are Porosity (e), Bubble Velocity (ub) and Sedimentation Velocity (uc).

The formula for calculating ratio of linear velocity of gas:

α = e(ub/uc)

Where:

α = Ratio of Linear Velocity of Gas
e = Porosity
ub = Bubble Velocity
uc = Sedimentation Velocity

Let’s solve an example;
Find the ratio of linear velocity of gas when the porosity is 8, the bubble velocity is 10 and the sedimentation velocity is 14.

This implies that;

e = Porosity = 8
ub = Bubble Velocity = 10
uc = Sedimentation Velocity = 14

α = e(ub/uc)
α = 8(10/14)
α = 8(0.714)
α = 5.71

Therefore, the ratio of linear velocity of gas is 5.71.

Calculating the Porosity when the Ratio of Linear Velocity of Gas, the Bubble Velocity and the Sedimentation Velocity is Given.

e = α x uc / ub

Where:

e = Porosity
α = Ratio of Linear Velocity of Gas
ub = Bubble Velocity
uc = Sedimentation Velocity

Let’s solve an example;
Find the porosity when the ratio of linear velocity of gas is 30, the bubble velocity is 5 and the sedimentation velocity is 3.

This implies that;

α = Ratio of Linear Velocity of Gas = 30
ub = Bubble Velocity = 5
uc = Sedimentation Velocity = 3

e = α x uc / ub
e = 30 x 3 / 5
e = 90 / 5
e = 18

Therefore, the porosity is 18.

## How to Calculate and Solve for Porosity | Non-Uniform Fluidization The image above represents porosity.

To compute for porosity, four essential parameters are needed and these parameters are Fraction of Fluid in Bed (f), Sedimentation Velocity (uc), Fluid Channel Velocity (uf) and Index Number (n).

The formula for calculating porosity:

e = f(uc/uf) + (uc/uf(1 – f))1/n(1 – f(uc/uf))1 – (1/n)

Where:

e = Porosity | Non-Uniform Fluidization
f = Fraction of Fluid in Bed
uc = Sedimentation Velocity
uf = Fluid Channel Velocity
n = Index Number

Let’s solve an example;
Find the porosity when the fraction of fluid in bed is 2, the sedimentation velocity is 14, the fluid channel velocity is 10 and the index number is 1.

This implies that;

f = Fraction of Fluid in Bed = 2
uc = Sedimentation Velocity = 14
uf = Fluid Channel Velocity = 10
n = Index Number = 1

e = f(uc/uf) + (uc/uf(1 – f))1/n(1 – f(uc/uf))1 – (1/n)
e = 2(14/10) + (14/10(1 – 2))1/1(1 – 2(14/10))1 – (1/1)
e = 2(1.4) + (1.4(-1))1/1(1 – 2(1.4))0
e = 2.8 + (-1.4)1/1(1 – 2.8)0
e = 2.8 + (-1.4)(-1.799)0
e = 2.8 + (-1.4)(1)
e = 2.8 + -1.4
e = 1.4

Therefore, the porosity is 1.4.

## How to Calculate and Solve for Porosity | Voidage System | Fluidization The image above represents porosity.

To compute for porosity, one essential parameter is needed and this parameter is Volumetric Fractional Concentration of Solid (c).

The formula for calculating porosity:

e = 1 – c

Where:

e = Porosity | Voidage System (c)
c = Volumetric Fractional Concentration of Solid

Let’s solve an example;
Find the porosity when the volumetric fractional concentration of solid is 15.

This implies that;

c = Volumetric Fractional Concentration of Solid = 15

e = 1 – c
e = 1 – 15
e = -14

Therefore, the porosity is -14.

## How to Calculate and Solve for Porosity | Voidage System | Fluidization The image above represents porosity.

To compute for porosity, three essential parameters are needed and these parameters are Sedimentation Velocity (uc), Corresponding Velocity (uiand Index Number (n).

The formula for calculating porosity:

e = (uc/ui)1/n

Where:

e = Porosity | Voidage System
uc = Sedimentation Velocity
ui = Corresponding Velocity
n = Index Number

Let’s solve an example;
Find the porosity when the sedimentation velocity is 22, the corresponding velocity is 10 and the index number is 14.

This implies that;

uc = Sedimentation Velocity = 22
ui = Corresponding Velocity = 10
n = Index Number = 14

e = (uc/ui)1/n
e = (22/10)(1/14)
e = (2.2)(1/14)
e = 1.057

Therefore, the porosity is 1.057.

## 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.

## How to Calculate and Solve for Pressure Drop Across Bed | Fluidization The image above represents pressure drop across bed.

To compute for pressure drop across bed, four essential parameters are needed and these parameters are Acceleration due to Gravity (g), Porosity (e), Density of Solid (ρs) and Density of Liquid (ρ).

The formula for calculating pressure drop across bed:

-ΔP = (1 – e)(ρs – ρ)g

Where:

-ΔP = Pressure Drop Across Bed
g = Acceleration due to Gravity
e = Porosity
ρs = Density of Solid
ρ = Density of Liquid

Let’s solve an example;
Find the pressure drop across bed when the acceleration due to gravity is 12, the porosity is 21, the density of solid is 18 and the density of liquid is 10.

This implies that;

g = Acceleration due to Gravity = 12
e = Porosity = 21
ρs = Density of Solid = 18
ρ = Density of Liquid = 10

-ΔP = (1 – e)(ρs – ρ)g
-ΔP = (1 – 21)(18 – 10)12
-ΔP = (-20)(8)12
-ΔP = (-20)(96)
-ΔP = -1920

Therefore, the pressure drop across bed is -1920 atm.

## How to Calculate and Solve for Porosity | Soil Mechanics and Foundation The image above represents porosity.

To compute for porosity, two essential parameters are needed and these parameters are Volume of void (Vv) and Volume of the soil (V).

The formula for calculating porosity:

n = Vv / V

Where:

n = Porosity
Vv = Volume of Void
V = Volume of the Soil

Let’s solve an example;
Find the porosity when the volume of void is 12 and the volume of the soil is 14.

This implies that;

Vv = Volume of Void = 12
V = Volume of the Soil = 14

n = Vv / V
n = 12 / 14
n = 0.85

Therefore, the porosity is 0.85.

Calculating the Volume of Void when the Porosity and the Volume of the Soil is Given.

Vv = n x V

Where;

Vv = Volume of Void
n = Porosity
V = Volume of the Soil

Let’s solve an example;
Find the volume of void when the porosity is 21 and the volume of the soil is 7.

This implies that;

n = Porosity = 21
V = Volume of the Soil = 7

Vv = n x V
Vv = 21 x 7
Vv = 147

Therefore, the volume of the void is 147.

## How to Calculate and Solve for Specific Storage | Aquifer Characteristics The image above represents specific storage.

To compute for specific storage, five essential parameters are needed and these parameters are Density of Water (ρw), Acceleration due to gravity (g), Compressibility of the Aquifer (α), Porosity (n) and Compressibility of Water (β).

The formula for calculating specific storage:

Ss = ρwg(α + nβ)

Where:

Ss = Specific Storage
ρw = Density of Water
g = Acceleration due to Gravity
α = Compressibility of the Aquifer
n = Porosity
β = Compressibility of Water

Let’s solve an example;
Find the specific storage when the density of water is 22, the acceleration due to gravity is 9.8, the comepressibility of the aquifer is 15, the porosity is 10 and the compressibility of water is 22.

This implies that;

ρw = Density of Water = 22
g = Acceleration due to Gravity = 9.8
α = Compressibility of the Aquifer = 15
n = Porosity = 10
β = Compressibility of Water = 22

Ss = ρwg(α + nβ)
Ss = (22)(9.8)(15 + 10(22))
Ss = 215.60 (15 + 220)
Ss = 215.60 (235)
Ss = 50666.0

Therefore, the specific storage is 50666.0.

## 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:

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.

## 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.