On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
1
GEN E 123 (week 12)
Prof Catherine Gebotys, Department of Electrical and Computer Engineering, DC3514
h
I. 3-PHASE
o Why AC? Economically feasible to transmit over long distances only
if voltage is very high and its easier to raise and lower sinusoidal ac
voltages than dc systems
o (stepped up/down ac voltages with transformers –easy to do with dc
voltages have to change to ac then step up then change again.
Expensive to construct)
o also typically we generate voltages with > 1 phase. Because it gives
steadier power and components weigh less.
o 3 phase generators – produce balanced set of voltages ( same
amplitude and frequency but displaced by 120o , 3*120=360)
a
Vp0o = Van
b
+
+
Vp120o = Vcn
Vbn = Vp –120 o
n
+
c
Van + Vcn + Vbn = 0
“Wye-connected system”
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
2
Note “line voltages”
Vab = Van + Vnb
= 30.5 Vp ∠30o
Vbc = 30.5 Vp∠-90 o
Vca = 30.5 Vp ∠-210 o
VL = 30.5 Vp
These line voltages are also a balanced set
Balanced Wye-Wye system:
Van
-+
n
Zp
a
Vbn
-+
Vcn - +
Zp
b
c
Zp
Source and load are Wye connected (Wye-wye three phase four wire
system. Ignore 4th wire, n , carries no current).
KVL around loops containing n wire:
IaA = Van/Zp , IbB = Vbn/Zp = IaA ∠-120 o , IcC = IaA ∠120 o
IaA + IbB + IcC = 0 ,
IaA = Ip ∠-θ o
IbB = Ip ∠-θ o-120 o
IcC = Ip ∠-θo +120 o
Power delivered to load is
3Pp = 3 Vp /(30.5)Ip cos(θv - θi) = 30.5 VpIpcos(θv-θi)
N
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
ie.
n
- +
Zp
3
N
Since system is symmetrical, we can just choose 1 loop (“single
phase”) out of the 3 loops to do the analysis.
”Source phase voltages” ”line impedance”
”phase impedance”,Zp
Vin (VL)
i=a,b,c
The ‘n’ line is always considered to be a short circuit even if it has
impedance.
Compute line currents for Vp = 100, Zp = 3 + j3 and line impedance
of 1.
Zp = 1 + 3 + j3 = 4+j3 = 5∠ 36.9o
IaA = Vp ∠0 o /Zp = ( 100 ∠0 o )/(5 ∠36.9 o ) = 20 ∠-36.9 o