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Volume of Conductor Material using Load Current (1-Phase 2-Wire US) Formula

GoVoltage of Conductor Material (1-Phase 2-Wire US) Formula

GoVolume of Conductor Material using Area of X-Section (1-Phase 2-Wire US) Formula

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

V = 4 \cdot (I^{2}) \cdot ρ \cdot \frac{L^{2}}{P loss}

V - Volume Of Conductor?I - Current Underground AC?ρ - Resistivity?L - Length of Underground AC Wire?P_loss - Line Losses?

Volume of Conductor Material using Load Current (1-Phase 2-Wire US) Example

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Here is how the Volume of Conductor Material using Load Current (1-Phase 2-Wire US) equation looks like with Values.

Here is how the Volume of Conductor Material using Load Current (1-Phase 2-Wire US) equation looks like with Units.

Here is how the Volume of Conductor Material using Load Current (1-Phase 2-Wire US) equation looks like.

1.1882 = 4 \cdot (9^{2}) \cdot 1.7E-5 \cdot \frac{24^{2}}{2.67}

1.1882m³ = 4 \cdot ({9A}^{2}) \cdot 1.7E-5Ω*m \cdot \frac{{24m}^{2}}{2.67W}

1.1882 = 4 \cdot (9^{2}) \cdot 1.7E-5 \cdot \frac{24^{2}}{2.67}

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Volume of Conductor Material using Load Current (1-Phase 2-Wire US) Solution

Follow our step by step solution on how to calculate Volume of Conductor Material using Load Current (1-Phase 2-Wire US)?

FIRST Step Consider the formula

V = 4 \cdot (I^{2}) \cdot ρ \cdot \frac{L^{2}}{P loss}

Next Step Substitute values of Variables

∴

V = 4 \cdot ({9A}^{2}) \cdot 1.7E-5Ω*m \cdot \frac{{24m}^{2}}{2.67W}

Next Step Prepare to Evaluate

∴

V = 4 \cdot (9^{2}) \cdot 1.7E-5 \cdot \frac{24^{2}}{2.67}

Next Step Evaluate

∴

V = 1.18824269662921m³

LAST Step Rounding Answer

∴

V = 1.1882m³

Volume of Conductor Material using Load Current (1-Phase 2-Wire US) Formula Elements

Variables

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

Current Underground AC

Current Underground AC is defined as the current flowing through the overhead ac supply wire.

Symbol: I

Measurement: Electric CurrentUnit: A

Note: Value can be positive or negative.

Formulas to find Current Underground AC

Resistivity

Resistivity is the measure of how strongly a material opposes the flow of current through them.

Symbol: ρ

Measurement: Electric ResistivityUnit: Ω*m

Note: Value can be positive or negative.

Formulas to find Resistivity

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

Other Formulas to find Volume Of Conductor

Go Voltage of Conductor Material (1-Phase 2-Wire US)

V = 2 \cdot \frac{K}{{(\cos (Φ))}^{2}}

Go Volume of Conductor Material using Area of X-Section (1-Phase 2-Wire US)

V = A \cdot L \cdot 2

Go Volume of Conductor Material using Line Losses (1-Phase 2-Wire US)

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

Go Volume of Conductor Material using Constant (1-Phase 2-Wire US)

V = 2 \cdot \frac{K}{{(\cos (Φ))}^{2}}

Other formulas in Wire Parameters category

Go Area of X-Section (1-Phase 2-Wire US)

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

Go Constant (1-Phase 2-Wire US)

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

Go Area of X-Section using Volume of Conductor Material (1-Phase 2-Wire US)

A = \frac{V}{2 \cdot L}

Go Length using Area of X-Section (1-Phase 2-Wire US)

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

How to Evaluate Volume of Conductor Material using Load Current (1-Phase 2-Wire US)?

Volume of Conductor Material using Load Current (1-Phase 2-Wire US) evaluator uses Volume Of Conductor = 4*(Current Underground AC^2)*Resistivity*(Length of Underground AC Wire^2)/(Line Losses) to evaluate the Volume Of Conductor, The Volume of Conductor Material using Load Current (1-Phase 2-Wire US) formula is defined as the 3-dimensional space enclosed by a conductor material of a single-phase two-wire overhead system. Volume Of Conductor is denoted by V symbol.

How to evaluate Volume of Conductor Material using Load Current (1-Phase 2-Wire US) using this online evaluator? To use this online evaluator for Volume of Conductor Material using Load Current (1-Phase 2-Wire US), enter Current Underground AC (I), Resistivity (ρ), Length of Underground AC Wire (L) & Line Losses (P_loss) and hit the calculate button.

FAQs on Volume of Conductor Material using Load Current (1-Phase 2-Wire US)

What is the formula to find Volume of Conductor Material using Load Current (1-Phase 2-Wire US)?

The formula of Volume of Conductor Material using Load Current (1-Phase 2-Wire US) is expressed as Volume Of Conductor = 4*(Current Underground AC^2)*Resistivity*(Length of Underground AC Wire^2)/(Line Losses). Here is an example- 1.188243 = 4*(9^2)*1.7E-05*(24^2)/(2.67).

How to calculate Volume of Conductor Material using Load Current (1-Phase 2-Wire US)?

With Current Underground AC (I), Resistivity (ρ), Length of Underground AC Wire (L) & Line Losses (P_loss) we can find Volume of Conductor Material using Load Current (1-Phase 2-Wire US) using the formula - Volume Of Conductor = 4*(Current Underground AC^2)*Resistivity*(Length of Underground AC Wire^2)/(Line Losses).

What are the other ways to Calculate Volume Of Conductor?

Here are the different ways to Calculate Volume Of Conductor-

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

Can the Volume of Conductor Material using Load Current (1-Phase 2-Wire US) be negative?

No, the Volume of Conductor Material using Load Current (1-Phase 2-Wire US), measured in Volume cannot be negative.

Which unit is used to measure Volume of Conductor Material using Load Current (1-Phase 2-Wire US)?

Volume of Conductor Material using Load Current (1-Phase 2-Wire US) 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 Load Current (1-Phase 2-Wire US) can be measured.

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Volume of Conductor Material using Load Current (1-Phase 2-Wire US) Formula

​GoVoltage of Conductor Material (1-Phase 2-Wire US) Formula

​GoVolume of Conductor Material using Area of X-Section (1-Phase 2-Wire US) Formula

Volume of Conductor Material using Load Current (1-Phase 2-Wire US) Example

Volume of Conductor Material using Load Current (1-Phase 2-Wire US) Solution

Volume of Conductor Material using Load Current (1-Phase 2-Wire US) Formula Elements

Other Formulas to find Volume Of Conductor

Other formulas in Wire Parameters category

How to Evaluate Volume of Conductor Material using Load Current (1-Phase 2-Wire US)?

FAQs on Volume of Conductor Material using Load Current (1-Phase 2-Wire US)

Variable Name

GoVoltage of Conductor Material (1-Phase 2-Wire US) Formula

GoVolume of Conductor Material using Area of X-Section (1-Phase 2-Wire US) Formula