1. 1. Power and RMS Values

2. Instantaneous power p(t) flowing into the box
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.

3. Average value of periodic instantaneous power p(t)

4. Two-wire sinusoidal case
zero average
Power factor
Average power

5. Root-mean squared value of a periodic waveform with period T
Compare to the average power expression
compare
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

6. Root-mean squared value of a periodic waveform with period T
For the sinusoidal case

7. Given single-phase v(t) and i(t) waveforms for a load
Determine their magnitudes and phase angles
Determine the average power
Determine the impedance of the load
Using a series RL or RC equivalent, determine the R and L or C

8. 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

9. The average power is

10. Voltage – Current Relationships

11. (no differential equations are needed)
Thanks to Charles Steinmetz, Steady-State 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

12. Problem 10.17

15. Active and Reactive Power Form a Power TriangleQ
Projection of S on the real axis
Projection of S on the imaginary axis
Complex power S
is the power factor

16. 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
Consider a node, with voltage (to any reference), and three currents IA IB IC

17. Voltage and Current Phasors for R’s, L’s, C’s
Voltage and Current in phase
Q = 0
Resistor
Voltage leads Current by 90°
Q > 0
Inductor
Current leads Voltage by 90°
Q < 0
Capacitor

18. Projection of S on the real axis Projection of S on the imaginary axisQ
Projection of S on the real axis
Projection of S on the imaginary axis
Complex power S

19. Resistor
also so ,
Use rms V, I

20. Inductor
also so ,
Use rms V, I

21. Capacitor
also so ,
Use rms V, I

22. 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

23. Now, demonstrate Excel spreadsheet EE411_Voltage_Current_Power.xls
to show the relationship between v(t), i(t), p(t), P, and Q

24. A Single-Phase Power Example

25. 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.
j0.15
pu ohms
j0.20 pu mhos
PL + jQL
VL = 1.020 /0
VR = 1.010 / - 1
PR + jQR IS
IcapL
IcapR

28. j0.15
pu ohms
j0.20 pu mhos
PL + jQL
VL = 1.020 /0
VR = 1.010 / - 1
PR + jQR IS
IcapL
IcapR

29. RMS of some common periodic waveforms
Duty cycle controller DT T V
0 < D < 1
By inspection, this is the average value of the squared waveform

30. RMS of common periodic waveforms, cont.
Sawtooth V T

31. RMS of common periodic waveforms, cont.
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 V V -V V V V V

32. 2. Three-Phase Circuits

33. Three Important Properties of Three-Phase Balanced Systems
Because they form a balanced set, the a-b-c currents sum to zero. Thus, there is no return current through the neutral or ground, which reduces wiring losses.
A N-wire system needs (N – 1) meters. A three-phase, four-wire system needs three meters. A three-phase, three-wire system needs only two meters.
The instantaneous power is constant
Three-phase, four wire system a b c n
Reference

34. Observe Constant Three-Phase P and Q in Excel spreadsheet
1_Single_Phase_Three_Phase_Instantaneous_Power.xls

40. Z l
ine c c I c 3Z
load 3Z
load b a b a Z l
ine I a 3Z
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 Z
follow these steps : l
ine a I a 1.
Convert the entire circuit to an equivalent wye
with a 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 +
The “One -
Line”
line currents . Z
Van
load
Diagram 4.
If needed, compute l
ine - to -
neutral voltages and line currents –
for phases b and c
using
the ±120° relationships. 5.
If needed, compute l
ine - to -
line voltages
and
delta
currents n n
using
the 3
and 30
relationships.

41. Now Work a Three-Phase Motor Power Factor Correction Example
A three-phase, 460V motor draws 5kW with a power factor of 0.80 lagging. Assuming that phasor voltage Van has phase angle zero,
Find phasor currents Ia and Iab and (note – Iab is inside
the motor delta windings)
Find the three phase motor Q and S
How much capacitive kVAr (three-phase) should be connected in
parallel with the motor to improve the net power factor to 0.95?
Assuming no change in motor voltage magnitude, what will be the new phasor current Ia after the kVArs are added?

42. Now Work a Delta-Wye Conversion Example
Part c. Draw a phasor diagram that shows line currents Ia, Ib, and Ic, and load currents Iab, Ibc, and Ica.

43. 3. Transformers

44. Single-Phase Transformer
Turns ratio 7200:240
(30 : 1)
(but approx. same amount of copper in each winding) Φ Rs
jXs
Ideal
Transformer
7200:240V Rm
jXm
7200V V

45. (but approx. same amount of copper in each winding)
Short Circuit Test
Short circuit test: Short circuit the 240V-side, and raise the 7200V-side voltage to a few percent of 7200, until rated current flows. There is almost no core flux so the magnetizing terms are negligible. +
Vsc -
Isc Rs
jXs
Ideal Rm
jXm
Transformer
7200:240V
7200V V
Turns ratio 7200:240
(but approx. same amount of copper in each winding) Φ

46. (but approx. same amount of copper in each winding)
Open Circuit Test
Ioc Rs
jXs +
Voc -
Ideal Rm
jXm
Transformer
7200:240V
7200V V
Open circuit test: Open circuit the 7200V-side, and apply 240V to the 240V-side. The winding currents are small, so the series terms are negligible.
Turns ratio 7200:240
(but approx. same amount of copper in each winding) Φ

47. 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 = 0.9
%noloadloss = 0.2
%Xs = 2.2
Load loss Xs
No load loss
Magnetizing current Rs
jXs
Ideal
Transformer
7200:240V Rm
jXm
7200V V
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?

48. Distribution Feeder Loss Example
Annual energy loss = 2.40%
Largest component is transformer no-load loss (45% of the 2.40%)
Modern Distribution Transformer:
Load loss at rated load (I2R in conductors) = 0.75% of rated transformer kW.
No load loss at rated voltage (magnetizing, core steel) = 0.2% of rated transformer kW.
Magnetizing current = 0.5% of rated transformer amperes 48

49. Single-Phase Transformer Impedance Reflection by the Square of the Turns RatioRs
jXs
Ideal
Transformer
7200:240V Rm
jXm
7200V V
Ideal
Transformer
7200:240V
7200V V

50. Now Work a Single-Phase Transformer Example
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 s c
and Q s c
). Measure d P s c
= 400W, Q s c
= 200VAr .
Open circuit test (open circuit the high -
voltage side, apply rated voltage to the low -
voltage
side
, and measure P oc
and Q oc
). Measure d P oc
= 100W, Q oc
= 250VAr .
Determine the four impedance
val
ues
(in ohms) for the transformer model shown.
Turns ratio 7200:240
(30 : 1)
(but approx. same amount of copper in each winding) Φ Rs
jXs
Ideal
Transformer
7200:240V Rm
jXm
7200V V

51. A three-phase transformer can be three separate single-phase transformers, or one large transformer with three sets of windings Rs
jXs
Ideal
Transformer
N1 : N2 Rm
jXm
Wye-Equivalent One-Line Model A N
N1:N2
N1:N2
Reflect to side 2 using individual transformer turns ratio N1:N2
N1:N2
Y - Y

52. For Delta-Delta Connection Model, Convert the Transformer to Equivalent Wye-Wye
Ideal
Transformer A N
Wye-Equivalent One-Line Model
N1:N2
Convert side 1 impedances from delta to equivalent wye
Then reflect to side 2 using individual transformer turns ratio N1:N2
N1:N2
N1:N2
Δ - Δ

53. For Delta-Wye Connection Model, Convert the Transformer to Equivalent Wye-Wye
Ideal
Transformer A N
Wye-Equivalent One-Line Model
N1:N2
Convert side 1 impedances from delta to wye
Then reflect to side 2 using three-phase bank line-to-line turns ratio
N1:N2
N1:N2
Δ - Y

54. For Wye-Delta Connection Model, Convert the Transformer to Equivalent Wye-Wye
Ideal
Transformer A N
Wye-Equivalent One-Line Model
N1:N2
N1:N2
Reflect to side 2 using three-phase bank line-to-line turns ratio
So, for all configurations, the equivalent wye-wye transformer ohms can be reflected from one side to the other using the three-phase bank line-to-line turns ratio
N1:N2
Y - Δ

55. For wye-delta and delta-wye configurations, there is a phase shift in line-to-line voltages because
the individual transformer windings on one side are connected line-to-neutral, and on the other side are connected line-to-line
But there is no phase shift in any of the individual transformers
This means that line-to-line voltages on the delta side are in phase with line-to-neutral voltages on the wye side
Thus, phase shift in line-to-line voltages from one side to the other is unavoidable, but it can be managed by standard labeling to avoid problems caused by paralleling transformers

56. Saturation – relative permeability decreases rapidly after 1.7 Tesla
Relative permeability drops from about 2000 to about 1 (becomes air core)
Magnetizing inductance of the core decreases, yielding a highly peaked magnetizing current
Linear Scale
Log10 Scale

57. Residual magnetism
Hysteresis Loss is ½ the Area of the Parallelogram per AC Cycle per Cubic Meter of Core Steel