1. Instantaneous power p(t) flowing into the box
SG_140122_Power_Definitions_and_Transformers_Presentation.ppt
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.

2. Average value of periodic instantaneous power p(t)
SG_140122_Power_Def

3. Two-wire sinusoidal case
zero average
Power factor
Average power
SG_140122_Power_Def

4. 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
SG_140122_Power_Def

5. Root-mean squared value of a periodic waveform with period T
For the sinusoidal case
SG_140122_Power_Def

6. 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
SG_140122_Power_Def

7. 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
SG_140122_Power_Def

8. The average power is
SG_140122_Power_Def

9. Voltage – Current Relationships
SG_140122_Power_Def

10. (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
SG_140122_Power_Def

11. Problem 10.17
SG_140122_Power_Def

12. SG_140122_Power_Def

13. SG_140122_Power_Def

14. 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
SG_140122_Power_Def

15. 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
SG_140122_Power_Def

16. 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
SG_140122_Power_Def

17. 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
SG_140122_Power_Def

18. Resistor
also so ,
Use rms V, I
SG_140122_Power_Def

19. Inductor
also so ,
Use rms V, I
SG_140122_Power_Def

20. Capacitor
also so ,
Use rms V, I
SG_140122_Power_Def

21. 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
SG_140122_Power_Def

22. Now, demonstrate Excel spreadsheet EE411_Voltage_Current_Power.xls
to show the relationship between v(t), i(t), p(t), P, and Q
SG_140122_Power_Def

23. A Single-Phase Power Example
SG_140122_Power_Def

24. 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
SG_140122_Power_Def

25. SG_140122_Power_Def

26. SG_140122_Power_Def

27. 0.05 + j0.15 pu ohms j0.20 pu mhos PL + jQL VL = 1.020 /0 ° VR = 1.010- 1
PR + jQR IS
IcapL
IcapR
SG_140122_Power_Def

28. 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
SG_140122_Power_Def

29. RMS of common periodic waveforms, cont.
Sawtooth V T
SG_140122_Power_Def

30. 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
SG_140122_Power_Def

31. 2. Three-Phase Circuits
SG_140122_Power_Def

32. 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
SG_140122_Power_Def

33. Observe Constant Three-Phase P and Q in Excel spreadsheet
1_Single_Phase_Three_Phase_Instantaneous_Power.xls
SG_140122_Power_Def

34. SG_140122_Power_Def The p hasors are rotating counter - clockwise .
The magnitude of line to
line
voltage
phasors is 3
times the magnitude of line
neutral
phas
ors. V bn ab = an – bc b n c cn
30°
120°
Imaginary
Real ca
SG_140122_Power_Def

35. Conservation of power requires that the magnitudes of
delta
currents I ab
, I ca
, and I bc
are 3 1
times the magnitude of line currents I a b c . V an bn cn
Real
Imaginary = –
30° I
Vab
Balanced Sets Add to Zero in Both
Time and Phasor Domains
+ I
= 0
+ V n
Line currents I
Delta currents I
SG_140122_Power_Def

36. SG_140122_Power_Def

37. (i.e., same line currents I a , I b c and phase - to phase voltages V
The Two Above
Sources
are
Equivalent
in Balanced Syst
ems
(i.e., same line currents I a
, I b c
and phase - to
phase voltages V ab
, V bc ca
in both cases ) – V + I n
SG_140122_Power_Def

38. The Experiment: Opening and closing the switch has no effect because I Z a b – V ab + I c n
KCL: I
= I
+ I
But for
a balanced set,
= 0, so I
= 0
Ground (i.e., V = 0)
The Experiment: Opening and closing the switch has no effect because I is
already zero for a three -
phase
balanced set.
Since no current flows, even if there is a resistance in the grounding path, we must
conclude that
= 0 at the neutral point
(or equivalent neutral point)
of any
balanced three phase load or source in a bala
nced
system.
This allows us to draw a “one
line” diagram (typically for phase a) and solve a single
phase problem.
Solutions for phases b and c follow from the phase shifts that must exist.
SG_140122_Power_Def

39. SG_140122_Power_Def Z l ine c c I c 3Z load 3Z load b a b a Z l ine I– 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.
SG_140122_Power_Def

40. 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?
SG_140122_Power_Def

41. 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.
SG_140122_Power_Def

42. 3. Transformers
SG_140122_Power_Def

43. Transformer Core Types
SG_140122_Power_Def

44. High-Voltage Grid Transformers, 100’s of MW
SG_140122_Power_Def

45. 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
SG_140122_Power_Def

46. (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
SG_140122_Power_Def

47. (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
SG_140122_Power_Def

48. 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?
SG_140122_Power_Def

49. X / R Ratios for Three-Phase Transformers 345kV to 138kV, X/R = 10
Substation transformers (e.g., 138kV to 25kV or 12.5kV, X/R = 2, X = 12%
25kV or 12.5kV to 480V, X/R = 1, X = 5%
480V class, X/R = 0.1, X = 1.5% to 4.5% Rs
jXs
Ideal
Transformer Rm
jXm
SG_140122_Power_Def

50. 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
SG_140122_Power_Def

51. Transformer Core Saturation
No DC
Log10 Scale
Magnetizing Current for Single-Phase 25 kVA. 12.5kV/240V Transformer.
THDi = 76.1%, Mostly 3rd Harmonic.
No DC
With DC
Linear Scale
SG_140122_Power_Def

52. Apply a DC Voltage to a Transformer and Watch It Saturate
Where there is a DC current, there is a DC voltage, and vice-versa
Voltage
Current
Saturates
SG_140122_Power_Def

53. B-H Curve Constructed from V-I Measurements Shows Linear Region, Saturation, Hysteresis, and Residual Magnetism
Shape of normal hysteresis path
Severe saturation
Severe hysteresis
Residual magnetism
SG_140122_Power_Def

54. 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 54

55. Single-Phase Transformer Impedance Reflection from High-Side (H) to Low-Side (L) by the Square of the Turns Ratio Rs
jXs
Ideal
Transformer
7200:240V Rm
jXm
7200V V
Faraday’s law
Conservation of power
Ideal
Transformer
7200:240V
7200V V
SG_140122_Power_Def

56. 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
SG_140122_Power_Def

57. 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 side 1 wye ohms to side 2 wye ohms by multiplying by [N2 / N1]^2
N1:N2
Standard 345/138kV autotransformers, GY - GY, with a tertiary 12.5kV Δ winding to provide circulating 3rd harmonic current
Y - Y
SG_140122_Power_Def

58. For Delta-Delta Connection Model, Convert the Transformer to Equivalent Wye-Wye
Ideal
Transformer A N
Wye-Equivalent One-Line Model
N1:N2
Reflect side 1 delta ohms to side 2 delta ohms by multiplying by [N2 / N1]^2
Convert side 2 delta ohms to wye ohms by dividing by 3
Convert side 1 delta ohms to wye ohms by dividing by 3
Above circuit results in the proper reflection. Note that N2/Sqrt3 divided by N1/Sqrt3 is the same as N2 divided by N1
N1:N2
N1:N2
Δ - Δ
SG_140122_Power_Def

59. For Delta-Wye Connection Model, Convert the Transformer to Equivalent Wye-Wye
Standard building entrance and substation transformers. Δ high side/ GY low side
Ideal
Transformer A N
Wye-Equivalent One-Line Model
N1:N2
Reflect side 1 delta ohms to side 2 wye ohms by multiplying by [N2 / N1]^2
Convert side 1 delta ohms to wye ohms by dividing by 3
Above circuit results in the proper reflection
N1:N2
N1:N2
Δ - Y
SG_140122_Power_Def

60. For Wye-Delta Connection Model, Convert the Transformer to Equivalent Wye-Wye
Ideal
Transformer A N
Wye-Equivalent One-Line Model
N1:N2
Reflect side 1 wye ohms to side 2 delta ohms by multiplying by [N2 / N1]^2
Convert side 2 impedances from delta ohms to wye ohms by dividing by 3
Above circuit results in the proper reflection
N1:N2
N1:N2
Y - Δ
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
SG_140122_Power_Def

61. But there is no phase shift in any of the individual transformers
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
SG_140122_Power_Def