Download EEE 302 Lecture 9 - Keith E. Holbert

EEE 302
Electrical Networks II
Dr. Keith E. Holbert
Summer 2001
Lecture 9
1
Three-Phase Circuits
• In three-phase circuits the 3 voltages sources are
120° apart
• Polyphase generation and transmission of
electricity is more advantageous and economical
(1) three-phase instantaneous power is constant
over time
(2) single-phase line losses are 50% greater than
three-phase losses (for the same load power,
voltage, pf), i.e., PSingle=3/2×PThree
Lecture 9
2
Balanced System
• A balanced system is one in which the 3 sinusoidal
voltages have the same magnitude and frequency,
and each is 120° out-of-phase with the other two
v an (t )  VM cos t 
vbn (t )  VM cos t  120
vcn (t )  VM cos t  240  VM cos t  120
Lecture 9
3
Three-Phase Voltages
a
Van
+
–
b
Vbn
+
–
c
Vcn
+
–
n
Balanced If:
Van=Vrms 0°
Vbn=Vrms -120° Vcn=Vrms -240°
Lecture 9
4
MATLAB Exercise
Let’s create and plot balanced 3-phase voltages
EDU» t=0:0.0005:0.04;
EDU» va=120*sqrt(2)*cos(377*t);
EDU» vb=120*sqrt(2)*cos(377*t-2*pi/3);
EDU» vc=120*sqrt(2)*cos(377*t-4*pi/3);
EDU» plot(t,va,'b:',t,vb,'g--',t,vc,'r')
EDU» legend(’va',’vb',’vc')
• Don’t exit MATLAB, we shall return
Lecture 9
5
Balanced Circuit
• A balanced three-phase circuit is one in which the
loads are such that the currents produced by the
voltages are also balanced
ia (t )  I M cos t   
ib (t )  I M cos t    120
ic (t )  I M cos t    240
Lecture 9
6
MATLAB Exercise
Let’s now create and plot some corresponding
balanced 3-phase currents
EDU» figure;
EDU» ia=10*sqrt(2)*cos(377*t-pi/8);
EDU» ib=10*sqrt(2)*cos(377*t-pi/8-2*pi/3);
EDU» ic=10*sqrt(2)*cos(377*t-pi/8-4*pi/3);
EDU» plot(t,ia,'c:',t,ib,'k--',t,ic,'m')
Lecture 9
7
Three-Phase Instantaneous Power
• Recall that the instantaneous power is for one
phase is p(t) = i(t) v(t)
• The total instantaneous power for three phases is
VM I M
p(t )  pa (t )  pb (t )  pc (t )  3
cos 
2
– the instantaneous power is a constant over time!
Lecture 9
8
MATLAB Exercise
Let’s now compute the individual and total
instantaneous powers, and plot them
EDU» figure;
EDU» pa=va.*ia;
EDU» pb=vb.*ib;
EDU» pc=vc.*ic;
EDU» pt=pa+pb+pc;
EDU» plot(t,pa,'b:',t,pb,'g--',t,pc,'r',t,pt,'k-.')
Lecture 9
9
Three-Phase Connections
• Standard notation specifies the phase sequence as
abc (called positive phase sequence) where
Van=Vp 0°
Vbn=Vp -120° Vcn=Vp -240°
Vbn=Van -120° Vcn=Van -240°
– Vbn lags Van by 120°
– where Vp is the phase voltage which is the magnitude of
the phasor voltage from the neutral to any line
• The balanced voltage set means that
Van + Vbn + Vcn = 0 [Book Error]
Lecture 9
10
Wye (Y) Connected Load
Load
a
a
ZY
b
b
c
ZY
ZY
ZY
c
n
n
Load
Lecture 9
11
Delta () Connected Load
a
a
Z
b
Z
b
Z
c
Z
c
Load
Load
Lecture 9
12