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

GoLine Losses (1-Phase 2-Wire US) Formula

GoLine Losses using Area of X-Section (1-Phase 2-Wire US) Formula

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

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

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

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

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

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

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

0.003 = 8 \cdot {(300)}^{2} \cdot 1.7E-5 \cdot \frac{{(24)}^{2}}{{(230 \cdot \cos (30))}^{2} \cdot 60}

0.003W = 8 \cdot {(300W)}^{2} \cdot 1.7E-5Ω*m \cdot \frac{{(24m)}^{2}}{{(230V \cdot \cos (30°))}^{2} \cdot 60m³}

0.003 = 8 \cdot {(300)}^{2} \cdot 1.7E-5 \cdot \frac{{(24)}^{2}}{{(230 \cdot \cos (30))}^{2} \cdot 60}

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

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

FIRST Step Consider the formula

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

Next Step Substitute values of Variables

∴

P loss = 8 \cdot {(300W)}^{2} \cdot 1.7E-5Ω*m \cdot \frac{{(24m)}^{2}}{{(230V \cdot \cos (30°))}^{2} \cdot 60m³}

Next Step Convert Units

∴

P loss = 8 \cdot {(300W)}^{2} \cdot 1.7E-5Ω*m \cdot \frac{{(24m)}^{2}}{{(230V \cdot \cos (0.5236rad))}^{2} \cdot 60m³}

Next Step Prepare to Evaluate

∴

P loss = 8 \cdot {(300)}^{2} \cdot 1.7E-5 \cdot \frac{{(24)}^{2}}{{(230 \cdot \cos (0.5236))}^{2} \cdot 60}

Next Step Evaluate

∴

P loss = 0.00296166351606805W

LAST Step Rounding Answer

∴

P loss = 0.003W

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

Variables

Functions

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

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

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

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

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

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 Line Losses

Go Line Losses (1-Phase 2-Wire US)

P loss = 2 \cdot (I^{2}) \cdot R

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

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

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

Line Losses using Volume of Conductor Material (1-Phase 2-Wire US) evaluator uses Line Losses = 8*(Power Transmitted)^2*Resistivity*(Length of Underground AC Wire)^2/((Maximum Voltage Underground AC*cos(Phase Difference))^2*Volume Of Conductor) to evaluate the Line Losses, The Line Losses using Volume of Conductor Material (1-Phase 2-Wire US) formula is defined as the loss of electric energy due to the heating of line wires by the current. Line Losses is denoted by P_loss symbol.

How to evaluate Line Losses using Volume of Conductor Material (1-Phase 2-Wire US) using this online evaluator? To use this online evaluator for Line Losses using Volume of Conductor Material (1-Phase 2-Wire US), enter Power Transmitted (P), Resistivity (ρ), Length of Underground AC Wire (L), Maximum Voltage Underground AC (V_m), Phase Difference (Φ) & Volume Of Conductor (V) and hit the calculate button.

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

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

The formula of Line Losses using Volume of Conductor Material (1-Phase 2-Wire US) is expressed as Line Losses = 8*(Power Transmitted)^2*Resistivity*(Length of Underground AC Wire)^2/((Maximum Voltage Underground AC*cos(Phase Difference))^2*Volume Of Conductor). Here is an example- 0.002962 = 8*(300)^2*1.7E-05*(24)^2/((230*cos(0.5235987755982))^2*60).

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

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

What are the other ways to Calculate Line Losses?

Here are the different ways to Calculate Line Losses-

Line Losses=2*(Current Underground AC^2)*Resistance Underground AC
Line Losses=(4*Length of Underground AC Wire*Resistivity*(Power Transmitted^2))/(Area of Underground AC Wire*(Maximum Voltage Underground AC^2)*((cos(Phase Difference))^2))
Line Losses=2*Resistance Underground AC*(Current Underground AC)^2

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

Yes, the Line Losses using Volume of Conductor Material (1-Phase 2-Wire US), measured in Power can be negative.

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

Line Losses using Volume of Conductor Material (1-Phase 2-Wire US) is usually measured using the Watt[W] for Power. Kilowatt[W], Milliwatt[W], Microwatt[W] are the few other units in which Line Losses using Volume of Conductor Material (1-Phase 2-Wire US) can be measured.

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

​GoLine Losses (1-Phase 2-Wire US) Formula

​GoLine Losses using Area of X-Section (1-Phase 2-Wire US) Formula

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

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

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

Other Formulas to find Line Losses

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

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

Variable Name

GoLine Losses (1-Phase 2-Wire US) Formula

GoLine Losses using Area of X-Section (1-Phase 2-Wire US) Formula