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)
Ploss = ((2+sqrt(2))*P)^2*ρ*(L)^2/((Vm*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
Ploss = ((2+sqrt(2))*P)^2*ρ*(L)^2/((Vm*cos(Φ))^2*V) --> ((2+sqrt(2))*300)^2*1.7E-05*(24)^2/((230*cos(0.5235987755982))^2*60)
Evaluating ... ...
Ploss = 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 00.020 seconds)

Credits

Creator Image
Created by Urvi Rathod
Vishwakarma Government Engineering College (VGEC), Ahmedabad
Urvi Rathod has created this Calculator and 1500+ more calculators!
Verifier Image
Verified by Kethavath Srinath
Osmania University (OU), Hyderabad
Kethavath Srinath has verified this Calculator and 1200+ more calculators!

Wire Parameters Calculators

Length using Volume of Conductor Material (2 Phase 3 Wire US)
​ LaTeX ​ 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)))
Line Losses using Volume of Conductor Material (2 Phase 3 Wire US)
​ LaTeX ​ 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)
Area of X Section using Volume of Conductor Material (2 Phase 3 Wire US)
​ LaTeX ​ 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)
​ LaTeX ​ Go Constant Underground AC = Volume Of Conductor*((cos(Phase Difference))^2)/(2.914)

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

​LaTeX ​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)
Ploss = ((2+sqrt(2))*P)^2*ρ*(L)^2/((Vm*cos(Φ))^2*V)

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 vm/√2 so that r.m.s. value of voltage between them is vm/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 Ploss 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 (Vm), 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 Ploss = ((2+sqrt(2))*P)^2*ρ*(L)^2/((Vm*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 (Vm), 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.
Let Others Know
Facebook
Twitter
Reddit
LinkedIn
Email
WhatsApp
Copied!