1. Power and RMS Values. + −. Circuit in a box, two wires. + −. Circuit in a box, three wires. + −. Instantaneous power p(t) flowing into the box. Any wire can be the voltage reference.
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+
−
Circuit in a box, two wires
+
−
Circuit in a box, three wires
+
−
Any wire can be the voltage reference
Works for any circuit, as long as all N wires are accounted for. There must be (N – 1) voltage measurements, and (N – 1) current measurements.
zero average
Power factor
Average power
compare
Compare to the average power expression
The average value of the squared voltage
Apply v(t) to a resistor
rms is based on a power concept, describing the equivalent voltage that will produce a given average power to a resistor
For the sinusoidal case
Determine voltage and current magnitudes and phase angles
Using a cosine reference,
Voltage cosine has peak = 100V, phase angle = 90º
Current cosine has peak = 50A, phase angle = 135º
Phasors
The average power is
Voltage – Current Relationships
Thanks to Charles Steinmetz, SteadyState AC Problems are Greatly Simplified with Phasor Analysis
(no differential equations are needed)
Time Domain
Frequency Domain
Resistor
voltage leads current
Inductor
current leads voltage
Capacitor
Problem 10.17
Complex power S
Projection of S on the imaginary axis
Q
P
Projection of S on the real axis
is the power factor
Active and Reactive Power Form a Power Triangle
Consider a node, with voltage (to any reference), and three currents
IA
IB
IC
Question: Why is there conservation of P and Q in a circuit?
Answer: Because of KCL, power cannot simply vanish but must be accounted for
Voltage and Currentin phase
Q = 0
Voltage leads Current by 90°
Q > 0
Current leads Voltage by 90°
Q < 0
Voltage and Current Phasors for R’s, L’s, C’s
Resistor
Inductor
Capacitor
Complex power S
Projection of S on the imaginary axis
Q
P
Projection of S on the real axis
Resistor
also
so
Use rms V, I
,
Inductor
also
so
Use rms V, I
,
Capacitor
also
so
,
Use rms V, I
Active and Reactive Power for R’s, L’s, C’s
(a positive value is consumed, a negative value is produced)
Active Power P
Reactive Power Q
Resistor
Inductor
Capacitor
source of reactive power
Now, demonstrate Excel spreadsheet
EE411_Voltage_Current_Power.xls
to show the relationship between v(t), i(t), p(t), P, and Q
A SinglePhase Power Example
0.05 + j0.15
pu ohms
PL + jQL
PR + jQR
/0
°
VR = 1.010
/

1
0
°
VL = 1.020
IS
IcapL
IcapR
j0.20 pu mhos
j0.20 pu mhos
A Transmission Line Example
Calculate the P and Q flows (in per unit) for the loadflow situation shown below,
and also check conservation of P and Q.
0.05 + j0.15
pu ohms
PL + jQL
PR + jQR
/0
°
VR = 1.010
/

1
0
°
VL = 1.020
IS
IcapL
IcapR
j0.20 pu mhos
j0.20 pu mhos
V
0
0 < D < 1
DT
T
Duty cycle controller
By inspection, this is the average value of the squared waveform
Sawtooth
V
0
T
V
0
V
0
V
0
V
0
V
0
V
0
0
V
Using the power concept, it is easy to reason that the following waveforms would all produce the same average power to a resistor, and thus their rms values are identical and equal to the previous example
a
b
c
n
Threephase, four wire system
Reference
Observe Constant ThreePhase P and Q in Excel spreadsheet
1_Single_Phase_Three_Phase_Instantaneous_Power.xls
3
Z
l
ine
c
c
I
c
3Z
3Z
load
load
a
a
b
b
Z
l
ine
I
3Z
a
load
–
V
+
ab
Z
l
ine
I
b
Balanced three

phase systems, no matter if they are delta
connected, wye connected, or a mix, are easy to solve if you
follow these steps
:
Z
l
ine
1.
Convert the entire circuit to an equivalent wye
with a
a
a
I
a
ground
ed
neutral
.
2.
Draw the one

line diagram for phase a
, recognizing that
phase a has one third of the P and Q
.
3.
Solve th
e one

line diagram
for
line

to

neutral voltages and
+
line currents
.
The “One

Line”
Z
load
Van
4.
If needed, compute l
ine

to

neutral voltages and line currents
Diagram
–
for phases b and c
using
the ±120° relationships.
5.
If needed, compute l
ine

to

line voltages
and
delta
currents
n
using
the
and
±
30
°
relationships.
n
Φ
jXs
Rs
Ideal
Transformer
jXm
Rm
7200:240V
Turns ratio 7200:240
(30 : 1)
(but approx. same amount of copper in each winding)
7200V 240V
Ioc
+
Voc

Open circuit test: Open circuit the 7200Vside, and apply 240V to the 240Vside. The winding currents are small, so the series terms are negligible. Calculate
Φ
Turns ratio 7200:240
(but approx. same amount of copper in each winding)
jXs
Rs
Ideal
Transformer
jXm
Rm
7200:240V
7200V 240V
Isc
+
Vsc

Short circuit test: Short circuit the 240Vside, and raise the 7200Vside voltage to a few percent of 7200, until rated current flows. There is almost no core flux so the magnetizing terms are negligible. Calculate
Φ
Turns ratio 7200:240
(but approx. same amount of copper in each winding)
jXs
Rs
Ideal
Transformer
jXm
Rm
7200:240V
7200V 240V
jXs
Rs
Ideal
Transformer
jXm
Rm
7200:240V
7200V 240V
1. Given the standard percentage values below for a 125kVA transformer, determine the R’s and X’s in the diagram, in Ω.
2. If the R’s and X’s are moved to the 240V side, compute the new Ω values.
Single Phase Transformer. Percent values are given on transformer base.
Winding 1
kv = 7.2, kVA = 125
Winding 2
kv = 0.24, kVA = 125
%imag = 0.5
%loadloss = 0.75
%noloadloss = 0.2
%Xs = 2.2
Load loss
Xs
No load loss
Magnetizing current
3. If standard open circuit and short circuit tests are performed on this transformer, what will be the P’s and Q’s (Watts and VArs) measured in those tests?
Annual energy loss = 2.40%
Largest component is transformer noload loss (45% of the 2.40%)
jXs
Rs
Ideal
Transformer
jXm
Rm
7200:240V
7200V 240V
Ideal
Transformer
7200:240V
7200V 240V
Φ
jXs
Rs
Ideal
Transformer
jXm
Rm
7200:240V
Turns ratio 7200:240
(30 : 1)
(but approx. same amount of copper in each winding)
7200V 240V
Open circuit and short circuit tests are performed on a
single

phase,
7200:240V, 25kVA, 60Hz
distribution transformer. The results are:
·
Short circuit test (short circuit the low

voltage side, energize the high

voltage side
so that
rated current flows
, an
d measure P
and Q
). Measure
d
P
= 400W, Q
= 200VAr
.
s
c
s
c
s
c
s
c
·
Open circuit test (open circuit the high

voltage side, apply rated voltage to the low

voltage
side
, and measure P
and Q
). Measure
d
P
= 100W, Q
= 250VAr
.
oc
oc
oc
oc
Determine the four impedance
val
ues
(in ohms) for the transformer model shown.
WyeEquivalent OneLine Model
A
N
jXs
Rs
Ideal
Transformer
jXm
Rm
N1 : N2
A threephase transformer can be three separate singlephase transformers, or one large transformer with three sets of windings
N1:N2
N1:N2
N1:N2
Y  Y
WyeEquivalent OneLine Model
A
N
Ideal
Transformer
For DeltaDelta Connection Model, Convert the Transformer to Equivalent WyeWye
N1:N2
N1:N2
N1:N2
Δ  Δ
WyeEquivalent OneLine Model
A
N
Ideal
Transformer
For DeltaWye Connection Model, Convert the Transformer to Equivalent WyeWye
N1:N2
N1:N2
N1:N2
Δ  Y
WyeEquivalent OneLine Model
A
N
Ideal
Transformer
Can then reflect to side 2 using threephase bank linetoline turns ratio
For WyeDelta Connection Model, Convert the Transformer to Equivalent WyeWye
N1:N2
N1:N2
So, for all configurations, the equivalent wyewye transformer ohms can be reflected from one side to the other using the threephase bank linetoline turns ratio
N1:N2
Y  Δ