## How to Calculate and Solve for Average Velocity | Tube-Bundle Theory | Transport Phenomena The image above represents average velocity.

To compute for average velocity, five essential parameters are needed and these parameters are Constant of Proportionality (k1), Change in Pressure (ΔP), Hydraulic Radius (Rh), Length of Pipe (L) and Viscosity (η).

The formula for calculating average velocity:

v = k1ΔPRh² /

Where:

v = Average Velocity | Tube-Bundle Theory
k1 = Constant of Proportionality
ΔP = Change in Pressure
L = Length of Pipe
η = Viscosity

Let’s solve an example;
Find the average velocity when the constant of proportionality is 10, the change in pressure is 5, the hydraulic radius is 2, the length of pipe is 4 and the viscosity is 8.

This implies that;

k1 = Constant of Proportionality = 10
ΔP = Change in Pressure = 5
Rh = Hydraulic Radius = 2
L = Length of Pipe = 4
η = Viscosity = 8

v = k1ΔPRh² /
v = (10)(5)(2)² / (4)(8)
v = (10)(5)(4) / 32
v = 200 / 32
v = 6.25

Therefore, the average velocity is 6.25 m/s.

## How to Calculate and Solve for Drag Force | Transport Phenomena The image above represents drag force.

To compute for drag force, five essential parameters are needed and these parameters are Density (ρ), Viscosity (η), Length of Tube (L), Width of Tube (W) and Velocity at Infinite Flow (v).

The formula for calculating drag force:

Fk = 0.664√(ρηLW²v³)

Where:

Fk = Drag Force
ρ = Density
η = Viscosity
L = Length of Tube
W = Width of Tube
v = Velocity at Infinite Flow

Let’s solve an example;
If density is 10, the viscosity is 12, the length of tube is 16, the width of tube is 14 and the velocity at infinite flow is 11. Find the drag force?

This implies that;

ρ = Density = 10
η = Viscosity = 12
L = Length of Tube = 16
W = Width of Tube = 14
v = Velocity at Infinite Flow = 11

Fk = 0.664√(ρηLW²v³)
Fk = 0.664√((10)(12)(16)(14)²(11)³)
Fk = 0.664√((10)(12)(16)(196)(1331))
Fk = 0.664√(500881920)
Fk = 0.664 (22380.39)
Fk = 14860.57

Therefore, the drag force is 14860.57 N.

## How to Calculate and Solve for Volume Flow Rate | Transport Phenomena The image above represents volume flow rate.

To compute for volume flow rate, five essential parameters are needed and these parameters are Width of Tube (W), Separation Distance between Atoms (δ), Change in Pressure (ΔP), Viscosity (η) and Distance between Fluid Flow (L).

The formula for calculating volume flow rate:

Q = 2Wδ³ΔP / 3ηL

Where:

Q = Volume Flow Rate
W = Width of Tube
δ = Separation Distance between Atoms
ΔP = Change in Pressure
η = Viscosity
L = Distance between Fluid Flow

Let’s solve an example;
Find the volume flow rate when the width of tube is 4, the separation distance between atoms is 8, the change in pressure is 2, the viscosity is 10 and the distance between fluid flow is 3.

This implies that;

W = Width of Tube = 4
δ = Separation Distance between Atoms = 8
ΔP = Change in Pressure = 2
η = Viscosity = 10
L = Distance between Fluid Flow = 3

Q = 2Wδ³ΔP / 3ηL
Q = 2(4)(8)³(2) / 3(10)(3)
Q = 2(4)(512)(2) / 90
Q = 8192 / 90
Q = 91.02

Therefore, the volume flow rate is 91.02 m³/s.

## How to Calculate and Solve for Velocity of Newtonian Fluid | Transport Phenomena The image above represent velocity of newtonian fluid.

To compute for velocity of newtonian fluid, five essential parameters are needed and these parameters are Change in Pressure (ΔP), Distance between Fluid Flow (L), Diameter of Pipe (y), Separation Distance of Atoms (δ) and Viscosity (η).

The formula for calculating velocity of newtonian fluid:

v = (δ² – y²)ΔP / 2η.L

Where:

v = Velocity of Newtonian Fluid
ΔP = Change in Pressure
L = Distance between Fluid Flow
y = Diameter of Pipe
δ = Separation Distance of Atoms
η = Viscosity

Let’s solve an example;
Find the velocity of newtonian fluid when the change in pressure is 22, the distance between fluid flow is 13, the diameter of pipe is 7, the separation distance of atoms is 11 and the viscosity is 5.

This implies that;

ΔP = Change in Pressure = 22
L = Distance between Fluid Flow = 13
y = Diameter of Pipe = 7
δ = Separation Distance of Atoms = 11
η = Viscosity = 5

v = (δ² – y²)ΔP / 2η.L
v = (11² – 7²)22 / 2(5).(13)
v = (121 – 49)22 / (10).(13)
v = (72)22 / 130
v = 1584 / 130
v = 12.18

Therefore, the velocity of newtonian fluid is 12.18 m/s.

Calculating the Change in Pressure when the Velocity of Newtonian Fluid, the Distance between Fluid Flow, the Diameter of Pipe, the Separation Distance of Atoms and the Viscosity is Given.

ΔP = v x 2η.L / (δ² – y²)

Where:

ΔP = Change in Pressure
v = Velocity of Newtonian Fluid
L = Distance between Fluid Flow
y = Diameter of Pipe
δ = Separation Distance of Atoms
η = Viscosity

Let’s solve an example;
Find the change in pressure when the velocity of newtonian fluid is 4, the distance between fluid flow is 2, the diameter of pipe is 3, the separation distance of atoms is 8 and the viscosity is 6.

This implies that;

v = Velocity of Newtonian Fluid = 4
L = Distance between Fluid Flow = 2
y = Diameter of Pipe = 3
δ = Separation Distance of Atoms = 8
η = Viscosity = 6

ΔP = v x 2η.L / (δ² – y²)
ΔP = 4 x 2(6).(2) / (8² – 3²)
ΔP = 4 x 24 / (64 – 9)
ΔP = 96 / 55
ΔP = 1.745

Therefore, the change in pressure is 1.745.

## How to Calculate and Solve for Kinematic Viscosity | Transport Phenomena The image above represents kinematic viscosity.

To compute for kinematic viscosity, two essential parameters are needed and these parameters are Viscosity (η) and Density (ρ).

The formula for calculating kinematic viscosity:

V = η / ρ

Where:

V = Kinematic Viscosity
η = Viscosity
ρ = Density

Let’s solve an example;
Find the kinematic viscosity when the viscosity is 44 and the density is 11.

This implies that;

η = Viscosity = 44
ρ = Density = 11

V = η / ρ
V = 44 / 11
V = 4

Therefore, the kinematic viscosity is 4 m²/s.

Calculating the Viscosity when the Kinematic Viscosity and the Density is Given.

η = V x ρ

Where:

η = Viscosity
V = Kinematic Viscosity
ρ = Density

Let’s solve an example;
Find the viscosity when the kinematic viscosity is 10 and the density is 5.

This implies that;

V = Kinematic Viscosity = 10
ρ = Density = 5

η = V x ρ
η = 10 x 5
η = 50

Therefore, the viscosity is 50.

## How to Calculate and Solve for Thermal Conductivity of Gases | Energy Transport The image above represents thermal conductivity of gases.

To compute for thermal conductivity of gases, four essential parameters are needed and these parameters are Viscosity (η), Heat Capacity at Constant Pressure (cp), Gas Constant (R) and Molar Mass of Gas (M).

The formula for calculating thermal conductivity of gases:

k = η(cp + 1.25R/M)

Where:

k = Thermal Conductivity of Gases
η = Viscosity
cp = Heat Capacity at Constant Pressure
R = Gas Constant
M = Molar Mass of Gas

Let’s solve an example;
Find the thermal conductivity of gases when the viscosity is 12, the heat capacity at constant pressure is 18, the gas constant is 22 and the molar mass of gas is 14.

This implies that;

η = Viscosity = 12
cp = Heat Capacity at Constant Pressure = 18
R = Gas Constant = 22
M = Molar Mass of Gas = 14

k = η(cp + 1.25R/M)
k = 12(18 + 1.25(22)/14)
k = 12(18 + 27.5/14)
k = 12(18 + 1.96)
k = 12(19.96)
k = 239.57

Therefore, the thermal conductivity of gases is 239.57 W/mK.

Calculating the Viscosity when the Thermal Conductivity of Gases, the Heat Capacity at Constant Pressure, the Gas Constant and the Molar Mass of Gas is Given.

η = k / (cp + 1.25R / M)

Where:

η = Viscosity
k = Thermal Conductivity of Gases
cp = Heat Capacity at Constant Pressure
R = Gas Constant
M = Molar Mass of Gas

Let’s solve an example;
Find the viscosity when the thermal conductivity of gases is 40, the heat capacity at constant pressure is 22, the gas constant is 2 and the molar mass of gas is 8.

This implies that;

k = Thermal Conductivity of Gases = 40
cp = Heat Capacity at Constant Pressure = 22
R = Gas Constant = 2
M = Molar Mass of Gas = 8

η = k / (cp + 1.25R / M)
η = 40 / (22 + 1.25(2) / 8)
η = 40 / (22 + 2.5 / 8)
η = 40 / (22 + 0.313)
η = 40 / 22.313
η = 1.79

Therefore, the viscosity is 1.79.

## 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
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.

## 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.

## 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 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.