NTNUJAVA Virtual Physics Laboratory
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Easy Java Simulations (2001- ) => Electronic => Topic started by: Fu-Kwun Hwang on March 06, 2011, 11:26:05 am

Title: Three-phase electric power and load
Post by: Fu-Kwun Hwang on March 06, 2011, 11:26:05 am
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.

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

Title: Re: Three-phase electric power and load
Post by: Fu-Kwun Hwang on March 13, 2011, 12:12:02 pm
$V_{RS}=V_{12}=V_{RN}-V_{SN}$,   for root mean square value $|V_{RS}|=\sqrt{3}|V_{RN}|$

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

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

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

$I_{RS}$ $=V_{RS}/R,I_{ST}$ $=V_{ST}/R,I_{TR}$ $=V_{TR}/R$
$I_{1}=I_{RS}-I_{TR}$ root mean square $|I_{1}|=\sqrt{3} |I_{RS}|$
$I_2=I_{ST}-I_{RS}$ $, I_3=I_{TR}-I_{ST}$

Title: Re: Three-phase electric power and load
Post by: ahmedelshfie on March 14, 2011, 08:47:30 am
Very nice applet, great work prof  ;)