Line Losses using Volume of Conductor Material (2 Phase 3 Wire US) Calculator

✖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.ⓘ Power Transmitted [P]			+10% -10%
✖Resistivity is the measure of how strongly a material opposes the flow of current through them.ⓘ Resistivity [ρ]			+10% -10%
✖Length of Underground AC Wire is the total length of the wire from one end to other end.ⓘ Length of Underground AC Wire [L]			+10% -10%
✖Maximum Voltage Underground AC is defined as the peak amplitude of the AC voltage supplied to the line or wire.ⓘ Maximum Voltage Underground AC [V_m]			+10% -10%
✖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.ⓘ Phase Difference [Φ]			+10% -10%
✖Volume Of Conductor the 3-dimensional space enclosed by a conductor material.ⓘ Volume Of Conductor [V]			+10% -10%

✖Line Losses is defined as the total losses occurring in an Underground AC line when in use.ⓘ Line Losses using Volume of Conductor Material (2 Phase 3 Wire US) [P_loss]

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Formula

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Line Losses using Volume of Conductor Material (2 Phase 3 Wire US)

Formula

`"P"_{"loss"} = ((2+sqrt(2))*"P")^2*"ρ"*("L")^2/(("V"_{"m"}*cos("Φ"))^2*"V")`

Example

`"0.004315W"=((2+sqrt(2))*"300W")^2*"0.000017Ω*m"*("24m")^2/(("230V"*cos("30°"))^2*"60m³")`

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Line Losses using Volume of Conductor Material (2 Phase 3 Wire US) Solution

STEP 0: Pre-Calculation Summary

Formula Used

Line Losses = ((2+sqrt(2))*Power Transmitted)^2*Resistivity*(Length of Underground AC Wire)^2/((Maximum Voltage Underground AC*cos(Phase Difference))^2*Volume Of Conductor)
P_loss = ((2+sqrt(2))*P)^2*ρ*(L)^2/((V_m*cos(Φ))^2*V)
This formula uses 2 Functions, 7 Variables

Functions Used

cos - Cosine of an angle is the ratio of the side adjacent to the angle to the hypotenuse of the triangle., cos(Angle)
sqrt - A square root function is a function that takes a non-negative number as an input and returns the square root of the given input number., sqrt(Number)

Variables Used

Line Losses - (Measured in Watt) - Line Losses is defined as the total losses occurring in an Underground AC line when in use.
Power Transmitted - (Measured in Watt) - 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.
Resistivity - (Measured in Ohm Meter) - Resistivity is the measure of how strongly a material opposes the flow of current through them.
Length of Underground AC Wire - (Measured in Meter) - Length of Underground AC Wire is the total length of the wire from one end to other end.
Maximum Voltage Underground AC - (Measured in Volt) - Maximum Voltage Underground AC is defined as the peak amplitude of the AC voltage supplied to the line or wire.
Phase Difference - (Measured in Radian) - 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.
Volume Of Conductor - (Measured in Cubic Meter) - Volume Of Conductor the 3-dimensional space enclosed by a conductor material.

STEP 1: Convert Input(s) to Base Unit

Power Transmitted: 300 Watt --> 300 Watt No Conversion Required
Resistivity: 1.7E-05 Ohm Meter --> 1.7E-05 Ohm Meter No Conversion Required
Length of Underground AC Wire: 24 Meter --> 24 Meter No Conversion Required
Maximum Voltage Underground AC: 230 Volt --> 230 Volt No Conversion Required
Phase Difference: 30 Degree --> 0.5235987755982 Radian (Check conversion here)
Volume Of Conductor: 60 Cubic Meter --> 60 Cubic Meter No Conversion Required

STEP 2: Evaluate Formula

Substituting Input Values in Formula

P_loss = ((2+sqrt(2))*P)^2*ρ*(L)^2/((V_m*cos(Φ))^2*V) --> ((2+sqrt(2))*300)^2*1.7E-05*(24)^2/((230*cos(0.5235987755982))^2*60)

Evaluating ... ...

P_loss = 0.00431545999285555

STEP 3: Convert Result to Output's Unit

0.00431545999285555 Watt --> No Conversion Required

FINAL ANSWER

0.00431545999285555 ≈ 0.004315 Watt <-- Line Losses

(Calculation completed in 00.004 seconds)

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Home » Engineering » Electrical » Power System » Underground AC Supply » 2-Φ 3-Wire System » Wire Parameters » Line Losses using Volume of Conductor Material (2 Phase 3 Wire US)

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< 17 Wire Parameters Calculators

Volume of Conductor Material using Resistance (2 Phase 3 Wire US)

Go Volume Of Conductor = ((2+sqrt(2))^2*(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)

Angle of Pf using Line Losses (2-Phase 3-Wire US)

Go Phase Difference = acos((2+(sqrt(2)*Power Transmitted/Maximum Voltage Underground AC))*(sqrt(Resistivity*Length of Underground AC Wire/Line Losses*Area of Underground AC Wire)))

Length using Volume of Conductor Material (2 Phase 3 Wire US)

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

Volume of Conductor Material (2 Phase 3 Wire US)

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

Area of X-Section using Line Losses (2-Phase 3-Wire US)

Go Area of Underground AC Wire = (2+sqrt(2))*Resistivity*Length of Underground AC Wire*(Power Transmitted)^2/(Line Losses*(Maximum Voltage Underground AC*cos(Phase Difference))^2)

Length using Line Losses (2-Phase 3-Wire US)

Go Length of Underground AC Wire = Line Losses*Area of Underground AC Wire*(Maximum Voltage Underground AC*cos(Phase Difference))^2/((2+sqrt(2))*(Power Transmitted^2)*Resistivity)

Line Losses using Volume of Conductor Material (2 Phase 3 Wire US)

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

Volume of Conductor Material using Area and Length(2 Phase 3 Wire US)

Go Volume Of Conductor = (2*Area of Underground AC Wire*Length of Underground AC Wire)+(sqrt(2)*Area of Underground AC Wire*Length of Underground AC Wire)

Volume of Conductor Material using Load Current (2 Phase 3 Wire US)

Go Volume Of Conductor = (2+sqrt(2))^2*Resistivity*(Length of Underground AC Wire^2)*(Current Underground AC^2)/Line Losses

Angle using Current in Neutral Wire (2-Phase 3-Wire US)

Go Phase Difference = acos(sqrt(2)*Power Transmitted/(Current Underground AC*Maximum Voltage Underground AC))

Length using Resistance of Natural Wire (2-Phase 3-Wire US)

Go Length of Underground AC Wire = (Resistance Underground AC*sqrt(2)*Area of Underground AC Wire)/(Resistivity)

Area using Resistance of Natural Wire (2-Phase 3-Wire US)

Go Area of Underground AC Wire = Resistivity*Length of Underground AC Wire/(sqrt(2)*Resistance Underground AC)

Angle using Current in Each Outer (2-Phase 3-Wire US)

Go Phase Difference = acos(Power Transmitted/(Current Underground AC*Maximum Voltage Underground AC))

Angle of PF using Volume of Conductor Material (2 Phase 3 Wire US)

Go Phase Difference = acos(sqrt((2.914)*Constant Underground AC/Volume Of Conductor))

Area of X Section using Volume of Conductor Material (2 Phase 3 Wire US)

Go Area of Underground AC Wire = Volume Of Conductor/((2+sqrt(2))*Length of Underground AC Wire)

Constant using Volume of Conductor Material (2 Phase 3 Wire US)

Go Constant Underground AC = Volume Of Conductor*((cos(Phase Difference))^2)/(2.914)

Volume of Conductor Material using Constant(2 Phase 3 Wire US)

Go Volume Of Conductor = 2.194*Constant Underground AC/(cos(Phase Difference)^2)

Line Losses using Volume of Conductor Material (2 Phase 3 Wire US) Formula

What is the value of maximum voltage and volume of conductor material in 2-phase 3-wire system?

The volume of conductor material required in this system is 2.914/cos²θ times that of 2-wire d.c.system with the one conductor earthed. The maximum voltage between conductors is v_m/√2 so that r.m.s. value of voltage between them is v_m/2.

How to Calculate Line Losses using Volume of Conductor Material (2 Phase 3 Wire US)?

Line Losses using Volume of Conductor Material (2 Phase 3 Wire US) calculator uses Line Losses = ((2+sqrt(2))*Power Transmitted)^2*Resistivity*(Length of Underground AC Wire)^2/((Maximum Voltage Underground AC*cos(Phase Difference))^2*Volume Of Conductor) to calculate the Line Losses, The Line Losses using Volume of Conductor Material (2 phase 3 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 calculate Line Losses using Volume of Conductor Material (2 Phase 3 Wire US) using this online calculator? To use this online calculator for Line Losses using Volume of Conductor Material (2 Phase 3 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. Here is how the Line Losses using Volume of Conductor Material (2 Phase 3 Wire US) calculation can be explained with given input values -> 0.004315 = ((2+sqrt(2))*300)^2*1.7E-05*(24)^2/((230*cos(0.5235987755982))^2*60).

FAQ

What is Line Losses using Volume of Conductor Material (2 Phase 3 Wire US)?

The Line Losses using Volume of Conductor Material (2 phase 3 wire US) formula is defined as the loss of electric energy due to the heating of line wires by the current and is represented as P_loss = ((2+sqrt(2))*P)^2*ρ*(L)^2/((V_m*cos(Φ))^2*V) or Line Losses = ((2+sqrt(2))*Power Transmitted)^2*Resistivity*(Length of Underground AC Wire)^2/((Maximum Voltage Underground AC*cos(Phase Difference))^2*Volume Of Conductor). 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, Resistivity is the measure of how strongly a material opposes the flow of current through them, Length of Underground AC Wire is the total length of the wire from one end to other end, Maximum Voltage Underground AC is defined as the peak amplitude of the AC voltage supplied to the line or wire, 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 & Volume Of Conductor the 3-dimensional space enclosed by a conductor material.

How to calculate Line Losses using Volume of Conductor Material (2 Phase 3 Wire US)?

The Line Losses using Volume of Conductor Material (2 phase 3 wire US) formula is defined as the loss of electric energy due to the heating of line wires by the current is calculated using Line Losses = ((2+sqrt(2))*Power Transmitted)^2*Resistivity*(Length of Underground AC Wire)^2/((Maximum Voltage Underground AC*cos(Phase Difference))^2*Volume Of Conductor). To calculate Line Losses using Volume of Conductor Material (2 Phase 3 Wire US), you need Power Transmitted (P), Resistivity (ρ), Length of Underground AC Wire (L), Maximum Voltage Underground AC (V_m), Phase Difference (Φ) & Volume Of Conductor (V). With our tool, you need to enter the respective value for Power Transmitted, Resistivity, Length of Underground AC Wire, Maximum Voltage Underground AC, Phase Difference & Volume Of Conductor and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.

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