Angle using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed) Solution

STEP 0: Pre-Calculation Summary
Formula Used
Phase Difference = acos(sqrt(4*(Power Transmitted^2)*Resistivity*Length of Underground AC Wire/(Area of Underground AC Wire*Line Losses*(Maximum Voltage Underground AC^2))))
Φ = acos(sqrt(4*(P^2)*ρ*L/(A*Ploss*(Vm^2))))
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.
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.
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.
Line Losses - (Measured in Watt) - Line Losses is defined as the total losses occurring in an Underground AC line when in use.
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.
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
Area of Underground AC Wire: 1.28 Square Meter --> 1.28 Square Meter No Conversion Required
Line Losses: 2.67 Watt --> 2.67 Watt No Conversion Required
Maximum Voltage Underground AC: 230 Volt --> 230 Volt No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
Φ = acos(sqrt(4*(P^2)*ρ*L/(A*Ploss*(Vm^2)))) --> acos(sqrt(4*(300^2)*1.7E-05*24/(1.28*2.67*(230^2))))
Evaluating ... ...
Φ = 1.54228931446658
STEP 3: Convert Result to Output's Unit
1.54228931446658 Radian -->88.3666685070769 Degree (Check conversion here)
FINAL ANSWER
88.3666685070769 88.36667 Degree <-- Phase Difference
(Calculation completed in 00.004 seconds)

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14 Wire Parameters Calculators

Volume of Conductor Material using Resistance (1-Phase 2-Wire Mid-Point Earthed)
Go 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)
Angle using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed)
Go Phase Difference = acos(sqrt(4*(Power Transmitted^2)*Resistivity*Length of Underground AC Wire/(Area of Underground AC Wire*Line Losses*(Maximum Voltage Underground AC^2))))
Line Losses using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed)
Go Line Losses = 4*Resistivity*Length of Underground AC Wire*(Power Transmitted^2)/(Area of Underground AC Wire*(Maximum Voltage Underground AC^2)*(cos(Phase Difference)^2))
Length using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed)
Go Length of Underground AC Wire = Area of Underground AC Wire*Line Losses*((Maximum Voltage Underground AC*cos(Phase Difference))^2)/(4*(Power Transmitted^2)*Resistivity)
Area of X-Section (1-Phase 2-Wire Mid-Point Earthed)
Go Area of Underground AC Wire = 4*Resistivity*Length of Underground AC Wire*(Power Transmitted^2)/(Line Losses*((Maximum Voltage Underground AC*cos(Phase Difference))^2))
Volume of Conductor Material (1-Phase 2-Wire Mid-Point Earthed)
Go 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))
Angle using Load Current (1-Phase 2-Wire Mid-Point Earthed)
Go Phase Difference = acos(sqrt(2)*Power Transmitted/(Current Underground AC*Maximum Voltage Underground AC))
Length using Line Losses (1-Phase 2-Wire Mid-Point Earthed)
Go Length of Underground AC Wire = Line Losses*Area of Underground AC Wire/(2*Resistivity*(Current Underground AC^2))
Area using Line Losses (1-Phase 2-Wire Mid-Point Earthed)
Go Area of Underground AC Wire = 2*Resistivity*Length of Underground AC Wire/(Line Losses*(Current Underground AC^2))
Volume of Conductor Material using Load Current (1-Phase 2-Wire Mid-Point Earthed)
Go Volume Of Conductor = 16*Resistivity*(Length of Underground AC Wire^2)*(Current Underground AC^2)/Line Losses
Volume of Conductor Material using Constant(1-Phase 2-Wire Mid-Point Earthed)
Go Volume Of Conductor = 2*Constant Underground AC/(cos(Phase Difference)^2)
Length using Volume of Conductor Material (1-Phase 2-Wire Mid-Point Earthed)
Go Length of Underground AC Wire = Volume Of Conductor/(2*Area of Underground AC Wire)
Area using Volume of Conductor Material (1-Phase 2-Wire Mid-Point Earthed)
Go Area of Underground AC Wire = Volume Of Conductor/(2*Length of Underground AC Wire)
Volume of Conductor Material using Area and Length(1-Phase 2-Wire Mid-Point US)
Go Volume Of Conductor = Area of Underground AC Wire*Length of Underground AC Wire*2

Angle using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed) Formula

Phase Difference = acos(sqrt(4*(Power Transmitted^2)*Resistivity*Length of Underground AC Wire/(Area of Underground AC Wire*Line Losses*(Maximum Voltage Underground AC^2))))
Φ = acos(sqrt(4*(P^2)*ρ*L/(A*Ploss*(Vm^2))))

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

Angle using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed) calculator uses Phase Difference = acos(sqrt(4*(Power Transmitted^2)*Resistivity*Length of Underground AC Wire/(Area of Underground AC Wire*Line Losses*(Maximum Voltage Underground AC^2)))) to calculate the Phase Difference, The Angle using Area Of X-Section (1-phase 2-wire Mid-point Earthed) formula is defined as the phase angle between reactive and active power. Phase Difference is denoted by Φ symbol.

How to calculate Angle using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed) using this online calculator? To use this online calculator for Angle using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed), enter Power Transmitted (P), Resistivity (ρ), Length of Underground AC Wire (L), Area of Underground AC Wire (A), Line Losses (Ploss) & Maximum Voltage Underground AC (Vm) and hit the calculate button. Here is how the Angle using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed) calculation can be explained with given input values -> 5063.037 = acos(sqrt(4*(300^2)*1.7E-05*24/(1.28*2.67*(230^2)))).

FAQ

What is Angle using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed)?
The Angle using Area Of X-Section (1-phase 2-wire Mid-point Earthed) formula is defined as the phase angle between reactive and active power and is represented as Φ = acos(sqrt(4*(P^2)*ρ*L/(A*Ploss*(Vm^2)))) or Phase Difference = acos(sqrt(4*(Power Transmitted^2)*Resistivity*Length of Underground AC Wire/(Area of Underground AC Wire*Line Losses*(Maximum Voltage Underground AC^2)))). 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, Area of Underground AC Wire is defined as the cross-sectional area of the wire of an AC supply system, Line Losses is defined as the total losses occurring in an Underground AC line when in use & Maximum Voltage Underground AC is defined as the peak amplitude of the AC voltage supplied to the line or wire.
How to calculate Angle using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed)?
The Angle using Area Of X-Section (1-phase 2-wire Mid-point Earthed) formula is defined as the phase angle between reactive and active power is calculated using Phase Difference = acos(sqrt(4*(Power Transmitted^2)*Resistivity*Length of Underground AC Wire/(Area of Underground AC Wire*Line Losses*(Maximum Voltage Underground AC^2)))). To calculate Angle using Area of X-Section (1-Phase 2-Wire Mid-Point Earthed), you need Power Transmitted (P), Resistivity (ρ), Length of Underground AC Wire (L), Area of Underground AC Wire (A), Line Losses (Ploss) & Maximum Voltage Underground AC (Vm). With our tool, you need to enter the respective value for Power Transmitted, Resistivity, Length of Underground AC Wire, Area of Underground AC Wire, Line Losses & Maximum Voltage Underground AC 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, Resistivity, Length of Underground AC Wire, Area of Underground AC Wire, Line Losses & Maximum Voltage Underground AC. We can use 1 other way(s) to calculate the same, which is/are as follows -
  • Phase Difference = acos(sqrt(2)*Power Transmitted/(Current Underground AC*Maximum Voltage Underground AC))
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