How to Calculate and Solve for Reservoir Radius and Well Spacing | The Calculator Encyclopedia

The image above represents the reservoir radius.

To compute for the reservoir radius, one essential parameter is needed and this parameter is Well Spacing (A).

The formula for calculating the reservoir radius:

re = √[43560A  / π]

Where;

re = Reservoir Radius
A = Well Spacing

Let’s solve an example;
Given that the well spacing of an reservoir radius is 42. Find the reservoir radius?

This implies that;

A = Well Spacing = 42

re = √[43560A  / π]
re = √[43560 x 42 / π]
re = √[1829520 / π]
re = √582354.3
re = 763.12

Therefore, the reservoir radius is 763.12 ft.

Continue reading How to Calculate and Solve for Reservoir Radius and Well Spacing | The Calculator Encyclopedia

How to Calculate and Solve for Radial Flow Rate in Reservoir Fluid Flow | The Calculator Encyclopedia

The image above represents the radial flow rate.

To compute for the radial flow rate, eight essential parameters are needed and these parameters are External Pressure (Pe), Flowing Bottom-Hole Pressure (Pwf), Formation Thickness (h), Oil Viscosity (μo ), Permeability (k), Oil Formation Volume Factor (Bo), Drainage Radius (reand Well Bore Radius (rw ).

The formula for calculating the radial flow rate:

Qo = 0.00708kh[Pe – Pwf] / μo Bo In[re / rw]

Where;

Qo = Radial Flow Rate
Pe = External Pressure
Pwf = Flowing Bottom-Hole Pressure
h = Formation Thickness
μo = Oil Viscosity
k = Permeability
Bo = Oil Formation Volume Factor
re = Drainage Radius
rw = Well Bore Radius

Let’s solve an example;
Find the radial flow rate when the External Pressure is 14, Flowing Bottom-Hole Pressure is 21, Formation Thickness is 7, Oil Viscosity is 35, Permeability is 50, Oil Formation Volume Factor is 13, Drainage Radius is 26 and Well Bore Radius is 15.

This implies that;

Pe = External Pressure = 14
Pwf = Flowing Bottom-Hole Pressure = 21
h = Formation Thickness = 7
μo = Oil Viscosity = 35
k = Permeability = 50
Bo = Oil Formation Volume Factor = 13
re = Drainage Radius = 26
rw = Well Bore Radius = 15

Qo = 0.00708kh[Pe – Pwf] / μo Bo In[re / rw]
Qo = 0.00708 x 50 x 7 [14 – 21] / 35 x 13 In[26 / 15]
Qo = 0.00708 x 50 x 7 [-7] / 35 x 13 In[26 / 15]
Qo = 0.00708 x 50 x 7 [-7] / 35 x 13 In[1.73]
Qo = 0.00708 x 50 x 7 [-7] /35 x 13 x 0.55
Qo = 2.478 [-7] / 35 x 13 x 0.55
Qo = -17.346 / 250.27
Qo = -0.069

Therefore, the radial flow rate is -0.069 STB/day.

Continue reading How to Calculate and Solve for Radial Flow Rate in Reservoir Fluid Flow | The Calculator Encyclopedia

How to Calculate and Solve for Linear Flow Rate in Reservoir Fluid Flow | The Calculator Encyclopedia

The image above represents the linear flow rate.

To compute for the linear flow rate, four parameters are needed and these parameters are Initial pressure (P1), Final Pressure (P2), Thickness (h) and Permeability (k).

The formula for calculating linear flow rate:

q = 0.001127kh[P1 – P2]

Where:

q = Linear Flow Rate
P1 = Initial Pressure
P2 = Final Pressure
h = Thickness
k = Permeability

Let’s solve an example;
Find the linear flow rate when the initial pressure is 12, final pressure is 22, thickness is 18 and permeability is 44.

This implies that;

P1 = Initial Pressure = 12
P2 = Final Pressure = 22
h = Thickness = 18
k = Permeability = 44

q = 0.001127kh[P1 – P2]
q = 0.001127 x 44 x 18 [12 – 22]
q = 0.001127 x 44 x 18 [-10]
q = 0.892584 [-10]
q = -8.92584

Therefore, the linear flow rate is -8.92584 bbl/day.

Continue reading How to Calculate and Solve for Linear Flow Rate in Reservoir Fluid Flow | The Calculator Encyclopedia

How to Calculate and Solve for Fluid Potential, Pressure, Datum Level and Density | The Calculator Encyclopedia

The image above represents the fluid potential.

To compute for the fluid potential, three essential parameters are needed and these parameters are pressure (P), Datum Levels (ΔZ) and Density (ρ).

The formula for calculating fluid potential:

φ = P – [ρ / 144]ΔZ

Where;
φ = Fluid Potential
P = Pressure
ΔZ = Datum Levels
ρ = Density

Let’s solve an example;
Find the fluid potential when the pressure is 24 with a datum level of 18 and the density of 30.

This implies that;
P = Pressure = 24
ΔZ = Datum Levels = 18
ρ = Density = 30

φ = P – [ρ / 144]ΔZ
φ = 24 – [30/144] 18
φ = 24 – [0.2083] 18
φ = 24 – 3.75
φ = 20.25

Therefore, the fluid potential is 20.25.

Calculating the Pressure(P) when the fluid potential, Datum levels and Density is Given.

P = φ – [ρ / 144]ΔZ

Where;
P = Pressure
φ = Fluid Potential
ΔZ = Datum Levels
ρ = Density

Lets solve an example;
Find the pressure with a fluid potential of 40 and a datum levels of 18 with density of 24.

This implies that;
φ = Fluid Potential = 40
ΔZ = Datum Levels = 18
ρ = Density = 24

P = φ – [ρ / 144]ΔZ
P = 40 – [24 / 144]18
P = 40 – [0.167]18
P = 40 – 3
P = 37

Therefore, the pressure is 37.

Continue reading How to Calculate and Solve for Fluid Potential, Pressure, Datum Level and Density | The Calculator Encyclopedia

How to Calculate and Solve for True Stress | The Calculator Encyclopedia

The image above represents the true stress.

To compute for the true stress, two essential parameters are needed and these parameters are force (F) and instantaneous area (Ai).

The formula for calculating true stress:

σT = F / Ai

Where;
T = True Stress
F = Force
Ai = Instantaneous Area

Let’s solve an example;
Find the true stress when the instantaneous area is 60 with a force of 25.

This implies that;
F = Force = 25
Ai = Instantaneous Area = 60

σT = F / Ai
σT = 25 / 60
σT = 0.416

Therefore, the true stress is 0.416 Pa.

Calculating the Force when True Stress and Instantaneous Area is Given.

F = Ai x σT

Where;
F = Force
σT = True Stress
Ai = Instantaneous Area

Let’s solve an example;
Find the force when the instantaneous area is 30 with a true stress of 15.

This implies that;
σT = True Stress = 15
Ai = Instantaneous Area = 30

F = Ai x σT
F = 30 x 15
F = 450

Therefore, the force is 450.

Continue reading How to Calculate and Solve for True Stress | The Calculator Encyclopedia

How to Calculate and Solve for Maximum Velocity to avoid Overturning of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

The image above represents the maximum velocity to avoid overturning of a vehicle moving along a level circular path.

To compute for the maximum velocity, four essential parameters are needed and these parameters are Acceleration due to Gravity (g), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the maximum velocity:

vmax = √(gra / h)

Where:
vmax = Maximum Velocity to avoid Overturning of a Vehicle moving along a Level Circular Path
g = Acceleration due to Gravity
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the maximum velocity when the Acceleration due to Gravity (g) is 10.2, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 14, Radius of Circular Path (r) is 22 and Half of the Distance between the Centre Lines of the Wheel (a) is 32.

This implies that;
g = Acceleration due to Gravity = 10.2
h = Height of Centre of Gravity of the Vehicle from Ground Level = 14
r = Radius of Circular Path = 22
a = Half of the Distance between the Centre Lines of the Wheel = 32

vmax = √(gra / h)
vmax = √((10.2)(22)(32)/14)
vmax = √((7180.79)/14)
vmax = √(512.91)
vmax = 22.647

Therefore, the maximum velocity to avoid Overturning of a Vehicle moving along a Level Circular Path is 22.647 m/s.

Continue reading How to Calculate and Solve for Maximum Velocity to avoid Overturning of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

How to Calculate and Solve for the Reaction at the Inner Wheel of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

The image represents reaction at the inner wheel of a vehicle moving along a level circular path.

To compute for the reaction, six essential parameters are needed and these parameters are Mass of the Vechicle (m), Acceleration due to Gravity (g), Velocity of the Vehicle (v), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the reaction at the inner wheel of a vehicle moving along a level circular path:

RA = mg / 2[1 – v²h / gra]

Where:
RA = Reaction at the Inner Wheel of a Vehicle moving along a Level Circular Path
m = Mass of the Vechicle
g = Acceleration due to Gravity
v = Velocity of the Vehicle
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the reaction when Mass of the Vechicle (m) is 13, Acceleration due to Gravity (g) is 9.8, Velocity of the Vehicle (v) is 11, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 5, Radius of Circular Path (r) is 7 and Half of the Distance between the Centre Lines of the Wheel (a) is 3.

This implies that;
m = Mass of the Vechicle = 13
g = Acceleration due to Gravity = 9.8
v = Velocity of the Vehicle = 11
h = Height of Centre of Gravity of the Vehicle from Ground Level = 5
r = Radius of Circular Path = 7
a = Half of the Distance between the Centre Lines of the Wheel = 3

RA = mg / 2[1 – v²h / gra]
RA = 13(9.8) / 2[1 – (11)²(5) / (9.8)(7)(3)]
RA = 127.4 / 2[1 – (121)(5) / 205.8]
RA = 63.7[1 – 605 / 205.8]
RA = 63.7[1 – 2.939]
RA = 63.7[-1.939]
RA = -123.56

Therefore, the reaction at the inner wheel of a vehicle moving along a level of circular path is -123.56 N.

Continue reading How to Calculate and Solve for the Reaction at the Inner Wheel of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

How to Calculate and Solve for Road Bank Angle, Velocity and Radius of a Body in Motion of Circular Path | The Calculator Encyclopedia

The image represents road bank angle in circular motion.

To compute for the road bank angle, three essential parameters are needed and these parameters are velocity (v), acceleration due to gravity (g) and radius (r).

The formula for calculating the road bank angle;

θ = tan-1( / gr)

Where;
θ = Road Bank Angle
v = Velocity
g = Acceleration due to Gravity
r = Radius

Let’s solve an example;
Find the road bank angle where the acceleration due to gravity is 9.8, velocity is 35 and radius is 18.

This implies that;
v = Velocity = 35
g = Acceleration due to Gravity = 9.8
r = Radius = 18

θ = tan-1( / gr)
θ = tan-1(35² / (9.8)(18))
θ = tan-1(1225 / 176.4)
θ = tan-1(6.94)
θ = 81.81°

Therefore, the road bank angle is 81.81°.

Calculating the Velocity when Road Bank Angle, Acceleration due to Gravity and Radius is Given.

v = √gr.tan θ

Where;
v = Velocity
θ = Road Bank Angle
g = Acceleration due to Gravity
r = Radius

Let’s solve an example;
Given that the road bank angle is 50, radius is 15 and acceleration due to gravity is 9.8. Find the velocity?

This implies that;
θ = Road Bank Angle = 50
g = Acceleration due to Gravity = 9.8
r = Radius = 15

v = √gr.tan θ
v = √(9.8 x 15)(tan 50)
v = √(147)(1.1917)
v = √175.1799
v = 13.235

Therefore, the velocity is 13.235.

Continue reading How to Calculate and Solve for Road Bank Angle, Velocity and Radius of a Body in Motion of Circular Path | The Calculator Encyclopedia

How to Calculate and Solve for the Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path | Nickzom Calculator

The image represents reaction at the outer wheel of a vehicle moving along a level circular path.

To compute for the reaction, six essential parameters are needed and these parameters are Mass of the Vechicle (m), Acceleration due to Gravity (g), Velocity of the Vehicle (v), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the reaction at the outer wheel of a vehicle moving along a level circular path:

RB = mg / 2[1 + v²h/gra]

Where;
RB = Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path
m = Mass of the Vechicle
g = Acceleration due to Gravity
v = Velocity of the Vehicle
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the reaction when Mass of the Vechicle (m) is 12, Acceleration due to Gravity (g) is 9.8, Velocity of the Vehicle (v) is 28, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 16, Radius of Circular Path (r) is 8 and Half of the Distance between the Centre Lines of the Wheel (a) is 4.

This implies that;
m = Mass of the Vechicle = 12
g = Acceleration due to Gravity = 9.8
v = Velocity of the Vehicle = 28
h = Height of Centre of Gravity of the Vehicle from Ground Level = 16
r = Radius of Circular Path = 8
a = Half of the Distance between the Centre Lines of the Wheel = 4

RB = mg / 2[1 + v²h/gra]
RB = 12 x 9.8 / 2[1 + 28² x 16/9.8 x 8 x 4]
RB = 117.60 / 2[1 + 784 x 16/313.6]
RB = 58.80[1 + 40]
RB = 58.80[41]
RB = 58.80[41]
RB = 2410.8

Therefore, the reaction of the outer wheel of a vehicle moving along a level circular path is 2410.8 N.

Continue reading How to Calculate and Solve for the Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path | Nickzom Calculator

How to Calculate and Solve for Centripetal Acceleration, Angular Velocity and Radius of Circular Path of a Body | The Calculator Encyclopedia

The image above represents centripetal acceleration in a motion of circular path.

To compute for the centripetal acceleration, two essential parameters are needed and these parameters are angular velocity (ω) and radius of circular path (r).

The formula for the calculating centripetal acceleration:

a = ω²r

Where:
a = Centripetal Acceleration
ω = Angular Velocity
r = Radius of Circular Path

Let’s solve an example;
Find the centripetal acceleration with an angular velocity of 33 and a radius of 21.

This implies that;
ω = Angular Velocity = 33
r = Radius of Circular Path = 21

a = ω²r
a = 33² x 21
a = 1089 x 21
a = 22869

Therefore, the centripetal acceleration is 22869 m/s².

Calculating the Angular Velocity when Centripetal Acceleration and Radius of Circular path is Given.

ω = √a / r

Where:
ω = Angular Velocity
a = Centripetal Acceleration
r = Radius of Circular Path

Let’s solve an example;
Find the angular velocity with a centripetal acceleration of 320 and a radius of 8.

This implies that;
a = Centripetal Acceleration = 320
r = Radius of Circular Path = 8

ω = √a / r
ω = √320 / 8
ω = √40
ω = 6.32

Therefore, the angular velocity is 6.32.

Continue reading How to Calculate and Solve for Centripetal Acceleration, Angular Velocity and Radius of Circular Path of a Body | The Calculator Encyclopedia