Balanced poly phase circuits

Two and four phase systems  A two phase system is an electrical system in which the voltages of the phases are 90 degree out of time phase.  The emf’s Eac and Ebd are 90 degree out of time phase.  A two phase system is the equivalent of two separate single-phase systems that are separated 90 degree in time phase.  If the connection is made between the two windings at n and n`, the system would be called a four-phase system.  The voltages Eda, Eab, Ebc and Ecd are called the line voltages, while voltages E0a, E0b, E0c and E0d are called the phase voltages or voltages to neutral.  Eda=Ed0+E0a. Thus, the line voltages= times the phase voltages in the four phase star.

Two and four phase systems  The line voltages are 45 degree or 135 degree out of time phase from the phase voltages.  For four-phase mesh, the aa` line current is Iaa`=Ida+Iba.  Thus, the line currents= times the phase currents degree out of time phase in the four phase mesh

Three phase four wire systems  The line voltages are 45 degree or 135 degree out of time phase from the phase voltages.  For four-phase mesh, the aa` line current is Iaa`=Ida+Iba.  Thus, the line currents= times the phase currents degree out of time phase in the four phase mesh

Three phase four wire system N is the neutral wire. Lighting loads are placed from line to neutral. Motors and other three phase power loads are connected between the three lines.

Three phase four wire system

The current in any line is the same as the current in the corresponding phase. Current in the neutral wire is obtained through the application of KCL. These currents are equal in magnitude and displaced from one another in time phase by 120 degree. Thus the current in the neutral wire is 0 since

Three phase three wire system

Balanced Wye loads Given the line voltages as 220 volts balanced three phase and R and X of each phase 6 ohms resistance and 8 ohms inductive reactance. Find the line current, power per phase and total power. Draw the vector diagram

Balanced Delta loads Given the line voltages as 220 volts balanced three phase and R and X of each phase 6 ohms resistance and 8 ohms inductive reactance. Find the line current, power per phase and total power. Draw the vector diagram self

Power calculation in balanced systems Wye connection Dealta connection

Volt-Amperes Wye connection Dealta connection Reactive Volt-Amperes Wye connection Dealta connection

Single phase and balanced three phase power For any phase a

Three wattmeter method

Two wattmeter method

Copper required to transmit power under fixed conditions

Harmonics in balanced three phase wye loads harmonics Phase A Phase B Phase C

Harmonics in balanced three phase wye loads  The fundamental and all harmonics that obtained by adding a multiple of 6 to the fundamental will have the same sequence. These are first, seventh, thirteenth, nineteenth, twenty-fifth and so on.  The fifth, elevenths, seventeenths, twenty-thirds have the same sequences but opposite to that of fundamentals.  The third, ninth and all multiples of the third will be found to be in phase.  Eba is 30 degree ahead of ena. For the third harmonic, eba3=0. fifth, eba5 lags ena5 by 30 degree.

Harmonics in balanced three phase wye loads

Unbalanced system Unbalanced system

 An unbalanced system is due to unbalanced voltage sources or unbalanced load. In a unbalanced system the neutral current is NOT zero. Unbalanced three phase Y connected load. Line currents DO NOT add up to zero. I n = -(I a + I b + I c ) ≠ 0

Exam-1

Exam-3

The Wye-Wye system with neutral connection

Methods of checking voltage phase sequence The voltage across a lamp The voltage across “c” lamp

Example:5

Two wattmeter method W 1 reads W 2 reads

Three Phase Power Measurement  Three-wattmeter method for measuring three-phase power

Power factor in unbalanced three phase system 283vars 283 watts Phase a -433vars 250 watts Phase c 200 watts Phase b

Example:11