NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/
January 24, 2020, 04:35:25 am

"If I have a thousand ideas and only one turns out to be good, I am satisfied." ..."Alfred Nobel(1833-1896, Swedish inventor, chemist, philanthropist)"

 Pages: [1]   Go Down
 Author Topic: Three-phase electric power and load  (Read 20548 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3085

 « Embed this message on: March 06, 2011, 11:26:05 am » posted from:Taipei,T\'ai-pei,Taiwan

Three-phase electric power is a common method of alternating-current electric power generation, transmission, and distribution. A three-phase system is generally more economical than others because it uses less conductor material to transmit electric power than equivalent single-phase or two-phase systems at the same voltage.
In a three-phase system, three circuit conductors carry three alternating currents (of the same frequency) which reach their instantaneous peak values at different times. Taking one conductor as the reference, the other two currents are delayed in time by one-third and two-thirds of one cycle of the electric current. This delay between phases has the effect of giving constant power transfer over each cycle of the current and also makes it possible to produce a rotating magnetic field in an electric motor.
* $V_a(t)=V_p \sin(\omega\,t-\tfrac{2}{3}\pi)=V_p (-\tfrac{1}{2}\sin \omega t-\tfrac{\sqrt{3}}{2}\cos\omega t)$
* $V_b(t)=V_p \sin(\omega\,t)$
* $V_c(t)=V_p \sin(\omega\,t+\tfrac{2}{3}\pi)=V_p (-\tfrac{1}{2}\sin \omega t+\tfrac{\sqrt{3}}{2}\cos\omega t)$

And you will find $V_a(t)+V_b(t)+V_c(t)=0$.

Three-phase systems may have a neutral wire. A neutral wire allows the three-phase system to use a higher voltage while still supporting lower-voltage single-phase appliances.
The phase currents tend to cancel out one another, summing to zero in the case of a linear balanced load. This makes it possible to eliminate or reduce the size of the neutral conductor; all the phase conductors carry the same current and so can be the same size, for a balanced load.

Power transfer into a linear balanced load is constant, which helps to reduce generator and motor vibrations.

Three-phase systems can produce a magnetic field that rotates in a specified direction, which simplifies the design of electric motors.

The following simulation let you play with such 3 phase electric power system, power source and load either in Y or $\Delta$ connection mode.
Use checkbox to select AC Y mode (AC $\Delta$ mode) or R Y mode (R $\Delta$ mode)
For AC Y mode:
Phase voltage referer to VRN=Va,VSN=Vb,VTN=Vc,
Line voltage referer to VRS=Va-Vb,VST=Vb-Vc,VTR=Vc-Va.

Y-connected sources and loads always have line voltages greater than phase voltages, and line currents equal to phase currents. If the Y-connected source or load is balanced, the line voltage will be equal to the phase voltage times the square root of 3.

$V_a-V_b=V_p (-\tfrac{3}{2}\sin \omega t-\tfrac{\sqrt{3}}{2}\cos\omega t)=\sqrt{3}\,V_p (-\tfrac{\sqrt{3}}{2}\sin \omega t-\tfrac{1}{2}\cos\omega t)=\sqrt{3} V_p \sin(\omega\,t-\tfrac{5}{6}\pi)$

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)
Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
• Please feel free to post your ideas about how to use the simulation for better teaching and learning.
• Post questions to be asked to help students to think, to explore.
• Upload worksheets as attached files to share with more users.
Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
 electronic_3phasepowerNload.gif (13.65 KB, 856x367 - viewed 851 times.)  3-phase_flow.gif (293.66 KB, 412x263 - viewed 2269 times.)  423px-CenterTappedTransformer.svg.png (44.96 KB, 423x436 - viewed 967 times.)  3phase-rmf-320x240-180fc.gif (216.42 KB, 320x240 - viewed 2704 times.)  ejs_electronic_3phasepowerNloadYY.jpg (34.67 KB, 756x318 - viewed 3478 times.)  ejs_electronic_3phasepowerNloadYD.jpg (35.96 KB, 756x318 - viewed 3284 times.)  ejs_electronic_3phasepowerNloadDY.jpg (35.5 KB, 756x318 - viewed 3214 times.)  ejs_electronic_3phasepowerNloadDD.jpg (36.96 KB, 756x318 - viewed 3223 times.) Logged
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3085

 « Embed this message Reply #1 on: March 13, 2011, 12:12:02 pm » posted from:Taipei,T\'ai-pei,Taiwan

$V_{RS}=V_{12}=V_{RN}-V_{SN}$, for root mean square value $|V_{RS}|=\sqrt{3}|V_{RN}|$
$I_{10}=V_{RN}/R$

$V_{RS}=V_{12}=V_{RN}-V_{SN}$, for root mean square value $|V_{RS}|=\sqrt{3}|V_{RN}|$

$I_{12}$ $=V_{RS}/R=(V_{RN}-V_{SN})/R$,$I_{23}$ $=V_{ST}=(V_{SN}-V_{TN})/R$, $I_{31}=$ $V_{TS}=(V_{TN}-V_{SN})/R$
$I_R=I_{12}-I_{31}=$ $(V_{RN}$ $-V_{SN})/R-(V_{TN}-V_{SN})/R$ $=(2V_{RN}$ $-V_{SN}-V_{TN})/R=3V_{RN}/R$
$V_{RN}+V_{SN}+V_{TN}=0$, $I_S=3V_{SN}/R, I_T=3V_{TN}/R$

$V_{12}=V_{RS}$
$I_{RS}$ $=V_{RS}/(3R),I_{ST}$ $=V_{ST}/(3R),I_{TR}$ $=V_{TR}/(3R)$
$I_1=I_{10}=I_{RS}-I_{TR}$ root mean square $|I_{10}|=\sqrt{3} |I_{RS}|$
$I_{12}=I_{10}-I_{20}$ root mean square $|I_{12}|=\sqrt{3} |I_{10}|=3 |I_{RS}|=|V_{RS}|/R$

$V_{12}=V_{RS}$
$I_{RS}$ $=V_{RS}/R,I_{ST}$ $=V_{ST}/R,I_{TR}$ $=V_{TR}/R$
$I_{1}=I_{RS}-I_{TR}$ root mean square $|I_{1}|=\sqrt{3} |I_{RS}|$
$I_2=I_{ST}-I_{RS}$ $, I_3=I_{TR}-I_{ST}$
 Logged
ahmedelshfie
Ahmed
Hero Member

Offline

Posts: 954

 « Embed this message Reply #2 on: March 14, 2011, 08:47:30 am » posted from:Uberaba,Minas Gerais,Brazil

Very nice applet, great work prof
 Logged
 Pages: [1]   Go Up
"If I have a thousand ideas and only one turns out to be good, I am satisfied." ..."Alfred Nobel(1833-1896, Swedish inventor, chemist, philanthropist)"