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Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) Formula

GoVolume of Conductor Material (1-Phase 2-Wire Mid-Point Earthed) Formula

GoVolume of Conductor Material using Area and Length(1-Phase 2-Wire Mid-Point US) Formula

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Volume Of Conductor the 3-dimensional space enclosed by a conductor material. Check FAQs

V = \frac{8 \cdot (P^{2}) \cdot R \cdot A \cdot L}{P loss \cdot ({V m}^{2}) \cdot {(\cos (Φ))}^{2}}

V - Volume Of Conductor?P - Power Transmitted?R - Resistance Underground AC?A - Area of Underground AC Wire?L - Length of Underground AC Wire?P_loss - Line Losses?V_m - Maximum Voltage Underground AC?Φ - Phase Difference?

Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) Example

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Here is how the Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) equation looks like with Values.

Here is how the Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) equation looks like with Units.

Here is how the Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) equation looks like.

1043.988 = \frac{8 \cdot (300^{2}) \cdot 5 \cdot 1.28 \cdot 24}{2.67 \cdot (230^{2}) \cdot {(\cos (30))}^{2}}

1043.988m³ = \frac{8 \cdot ({300W}^{2}) \cdot 5Ω \cdot 1.28m² \cdot 24m}{2.67W \cdot ({230V}^{2}) \cdot {(\cos (30°))}^{2}}

1043.988 = \frac{8 \cdot (300^{2}) \cdot 5 \cdot 1.28 \cdot 24}{2.67 \cdot (230^{2}) \cdot {(\cos (30))}^{2}}

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Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) Solution

Follow our step by step solution on how to calculate Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed)?

FIRST Step Consider the formula

V = \frac{8 \cdot (P^{2}) \cdot R \cdot A \cdot L}{P loss \cdot ({V m}^{2}) \cdot {(\cos (Φ))}^{2}}

Next Step Substitute values of Variables

∴

V = \frac{8 \cdot ({300W}^{2}) \cdot 5Ω \cdot 1.28m² \cdot 24m}{2.67W \cdot ({230V}^{2}) \cdot {(\cos (30°))}^{2}}

Next Step Convert Units

∴

V = \frac{8 \cdot ({300W}^{2}) \cdot 5Ω \cdot 1.28m² \cdot 24m}{2.67W \cdot ({230V}^{2}) \cdot {(\cos (0.5236rad))}^{2}}

Next Step Prepare to Evaluate

∴

V = \frac{8 \cdot (300^{2}) \cdot 5 \cdot 1.28 \cdot 24}{2.67 \cdot (230^{2}) \cdot {(\cos (0.5236))}^{2}}

Next Step Evaluate

∴

V = 1043.98802064527m³

LAST Step Rounding Answer

∴

V = 1043.988m³

Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) Formula Elements

Variables

Functions

Volume Of Conductor

Volume Of Conductor the 3-dimensional space enclosed by a conductor material.

Symbol: V

Measurement: VolumeUnit: m³

Note: Value should be greater than 0.

Formulas to find Volume Of Conductor

Power Transmitted

Power Transmitted is the amount of power that is transferred from its place of generation to a location where it is applied to perform useful work.

Symbol: P

Measurement: PowerUnit: W

Note: Value can be positive or negative.

Formulas to find Power Transmitted

Resistance Underground AC

Resistance Underground AC is defined as the property of the wire or line that opposes the flow of current through it.

Symbol: R

Measurement: Electric ResistanceUnit: Ω

Note: Value can be positive or negative.

Formulas to find Resistance Underground AC

Area of Underground AC Wire

Area of Underground AC Wire is defined as the cross-sectional area of the wire of an AC supply system.

Symbol: A

Measurement: AreaUnit: m²

Note: Value can be positive or negative.

Formulas to find Area of Underground AC Wire

Length of Underground AC Wire

Length of Underground AC Wire is the total length of the wire from one end to other end.

Symbol: L

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Length of Underground AC Wire

Line Losses

Line Losses is defined as the total losses occurring in an Underground AC line when in use.

Symbol: P_loss

Measurement: PowerUnit: W

Note: Value can be positive or negative.

Formulas to find Line Losses

Maximum Voltage Underground AC

Maximum Voltage Underground AC is defined as the peak amplitude of the AC voltage supplied to the line or wire.

Symbol: V_m

Measurement: Electric PotentialUnit: V

Note: Value can be positive or negative.

Formulas to find Maximum Voltage Underground AC

Phase Difference

Phase Difference is defined as the difference between the phasor of apparent and real power (in degrees) or between voltage and current in an ac circuit.

Symbol: Φ

Measurement: AngleUnit: °

Note: Value should be greater than 0.

Formulas to find Phase Difference

cos

Cosine of an angle is the ratio of the side adjacent to the angle to the hypotenuse of the triangle.

Syntax: cos(Angle)

Other Formulas to find Volume Of Conductor

Go Volume of Conductor Material (1-Phase 2-Wire Mid-Point Earthed)

V = 8 \cdot ρ \cdot (P^{2}) \cdot \frac{L^{2}}{P loss \cdot ({V m}^{2}) \cdot ({\cos (Φ)}^{2})}

Go Volume of Conductor Material using Area and Length(1-Phase 2-Wire Mid-Point US)

V = A \cdot L \cdot 2

Other formulas in Wire Parameters category

Go Area of X-Section (1-Phase 2-Wire Mid-Point Earthed)

A = 4 \cdot ρ \cdot L \cdot \frac{P^{2}}{P loss \cdot ({(V m \cdot \cos (Φ))}^{2})}

Go Length using Line Losses (1-Phase 2-Wire Mid-Point Earthed)

L = P loss \cdot \frac{A}{2 \cdot ρ \cdot (I^{2})}

How to Evaluate Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed)?

Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) evaluator uses Volume Of Conductor = (8*(Power Transmitted^2)*Resistance Underground AC*Area of Underground AC Wire*Length of Underground AC Wire)/(Line Losses*(Maximum Voltage Underground AC^2)*(cos(Phase Difference))^2) to evaluate the Volume Of Conductor, The Volume of Conductor Material using Resistance (1-phase 2-wire Mid-point Earthed) formula is defined as the 3-dimensional space enclosed by a conductor material of a 1-phase 2-wire Mid-point Earthed underground system. Volume Of Conductor is denoted by V symbol.

How to evaluate Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) using this online evaluator? To use this online evaluator for Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed), enter Power Transmitted (P), Resistance Underground AC (R), Area of Underground AC Wire (A), Length of Underground AC Wire (L), Line Losses (P_loss), Maximum Voltage Underground AC (V_m) & Phase Difference (Φ) and hit the calculate button.

FAQs on Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed)

What is the formula to find Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed)?

The formula of Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) is expressed as Volume Of Conductor = (8*(Power Transmitted^2)*Resistance Underground AC*Area of Underground AC Wire*Length of Underground AC Wire)/(Line Losses*(Maximum Voltage Underground AC^2)*(cos(Phase Difference))^2). Here is an example- 1043.988 = (8*(300^2)*5*1.28*24)/(2.67*(230^2)*(cos(0.5235987755982))^2).

How to calculate Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed)?

With Power Transmitted (P), Resistance Underground AC (R), Area of Underground AC Wire (A), Length of Underground AC Wire (L), Line Losses (P_loss), Maximum Voltage Underground AC (V_m) & Phase Difference (Φ) we can find Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) using the formula - Volume Of Conductor = (8*(Power Transmitted^2)*Resistance Underground AC*Area of Underground AC Wire*Length of Underground AC Wire)/(Line Losses*(Maximum Voltage Underground AC^2)*(cos(Phase Difference))^2). This formula also uses Cosine (cos) function(s).

What are the other ways to Calculate Volume Of Conductor?

Here are the different ways to Calculate Volume Of Conductor-

Volume Of Conductor=8*Resistivity*(Power Transmitted^2)*(Length of Underground AC Wire^2)/(Line Losses*(Maximum Voltage Underground AC^2)*(cos(Phase Difference)^2))
Volume Of Conductor=Area of Underground AC Wire*Length of Underground AC Wire*2
Volume Of Conductor=2*Constant Underground AC/(cos(Phase Difference)^2)

Can the Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) be negative?

No, the Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed), measured in Volume cannot be negative.

Which unit is used to measure Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed)?

Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) is usually measured using the Cubic Meter[m³] for Volume. Cubic Centimeter[m³], Cubic Millimeter[m³], Liter[m³] are the few other units in which Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) can be measured.

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Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) Formula

​GoVolume of Conductor Material (1-Phase 2-Wire Mid-Point Earthed) Formula

​GoVolume of Conductor Material using Area and Length(1-Phase 2-Wire Mid-Point US) Formula

Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) Example

Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) Solution

Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed) Formula Elements

Other Formulas to find Volume Of Conductor

Other formulas in Wire Parameters category

How to Evaluate Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed)?

FAQs on Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed)

Variable Name

GoVolume of Conductor Material (1-Phase 2-Wire Mid-Point Earthed) Formula

GoVolume of Conductor Material using Area and Length(1-Phase 2-Wire Mid-Point US) Formula