Angle of Pf using Line Losses (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%
✖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%
✖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%
✖Line Losses is defined as the total losses occurring in an Underground AC line when in use.ⓘ Line Losses [P_loss]			+10% -10%
✖Area of Underground AC Wire is defined as the cross-sectional area of the wire of an AC supply system.ⓘ Area of Underground AC Wire [A]			+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.ⓘ Angle of Pf using Line Losses (2-Phase 3-Wire US) [Φ]

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Formula

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

Formula

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

Example

`"86.91777°"=acos((2+(sqrt(2)*"300W"/"230V"))*(sqrt("0.000017Ω*m"*"24m"/"2.67W"*"1.28m²")))`

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

STEP 0: Pre-Calculation Summary

Formula Used

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)))
Φ = acos((2+(sqrt(2)*P/V_m))*(sqrt(ρ*L/P_loss*A)))
This formula uses 3 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)
acos - The inverse cosine function, is the inverse function of the cosine function. It is the function that takes a ratio as an input and returns the angle whose cosine is equal to that ratio., acos(Number)
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

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.
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.
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.
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.
Line Losses - (Measured in Watt) - Line Losses is defined as the total losses occurring in an Underground AC line when in use.
Area of Underground AC Wire - (Measured in Square Meter) - Area of Underground AC Wire is defined as the cross-sectional area of the wire of an AC supply system.

STEP 1: Convert Input(s) to Base Unit

Power Transmitted: 300 Watt --> 300 Watt No Conversion Required
Maximum Voltage Underground AC: 230 Volt --> 230 Volt 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
Line Losses: 2.67 Watt --> 2.67 Watt No Conversion Required
Area of Underground AC Wire: 1.28 Square Meter --> 1.28 Square Meter No Conversion Required

STEP 2: Evaluate Formula

Substituting Input Values in Formula

Φ = acos((2+(sqrt(2)*P/V_m))*(sqrt(ρ*L/P_loss*A))) --> acos((2+(sqrt(2)*300/230))*(sqrt(1.7E-05*24/2.67*1.28)))

Evaluating ... ...

Φ = 1.51700118373287

STEP 3: Convert Result to Output's Unit

1.51700118373287 Radian -->86.9177653442595 Degree (Check conversion here)

FINAL ANSWER

86.9177653442595 ≈ 86.91777 Degree <-- Phase Difference

(Calculation completed in 00.020 seconds)

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Home » Engineering » Electrical » Power System » Underground AC Supply » 2-Φ 3-Wire System » Wire Parameters » Angle of Pf using Line Losses (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)

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

How are power factor and power angle related?

Power angles are generally caused due to voltage drop due to impedance in the transmission line. The power factor is caused due to phase angle between reactive and active power.

How to Calculate Angle of Pf using Line Losses (2-Phase 3-Wire US)?

Angle of Pf using Line Losses (2-Phase 3-Wire US) calculator uses 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))) to calculate the Phase Difference, The Angle of Pf using Line Losses (2-Phase 3-Wire US) formula is defined as the phase angle between reactive and active power. Phase Difference is denoted by Φ symbol.

How to calculate Angle of Pf using Line Losses (2-Phase 3-Wire US) using this online calculator? To use this online calculator for Angle of Pf using Line Losses (2-Phase 3-Wire US), enter Power Transmitted (P), Maximum Voltage Underground AC (V_m), Resistivity (ρ), Length of Underground AC Wire (L), Line Losses (P_loss) & Area of Underground AC Wire (A) and hit the calculate button. Here is how the Angle of Pf using Line Losses (2-Phase 3-Wire US) calculation can be explained with given input values -> 4980.021 = acos((2+(sqrt(2)*300/230))*(sqrt(1.7E-05*24/2.67*1.28))).

FAQ

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

The Angle of Pf using Line Losses (2-Phase 3-Wire US) formula is defined as the phase angle between reactive and active power and is represented as Φ = acos((2+(sqrt(2)*P/V_m))*(sqrt(ρ*L/P_loss*A))) or 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))). 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, Maximum Voltage Underground AC is defined as the peak amplitude of the AC voltage supplied to the line or wire, 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, Line Losses is defined as the total losses occurring in an Underground AC line when in use & Area of Underground AC Wire is defined as the cross-sectional area of the wire of an AC supply system.

How to calculate Angle of Pf using Line Losses (2-Phase 3-Wire US)?

The Angle of Pf using Line Losses (2-Phase 3-Wire US) formula is defined as the phase angle between reactive and active power is calculated using 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))). To calculate Angle of Pf using Line Losses (2-Phase 3-Wire US), you need Power Transmitted (P), Maximum Voltage Underground AC (V_m), Resistivity (ρ), Length of Underground AC Wire (L), Line Losses (P_loss) & Area of Underground AC Wire (A). With our tool, you need to enter the respective value for Power Transmitted, Maximum Voltage Underground AC, Resistivity, Length of Underground AC Wire, Line Losses & Area of Underground AC Wire and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.

How many ways are there to calculate Phase Difference?

In this formula, Phase Difference uses Power Transmitted, Maximum Voltage Underground AC, Resistivity, Length of Underground AC Wire, Line Losses & Area of Underground AC Wire. We can use 3 other way(s) to calculate the same, which is/are as follows -

Phase Difference = acos(sqrt((2.914)*Constant Underground AC/Volume Of Conductor))
Phase Difference = acos(Power Transmitted/(Current Underground AC*Maximum Voltage Underground AC))
Phase Difference = acos(sqrt(2)*Power Transmitted/(Current Underground AC*Maximum Voltage Underground AC))

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