1 Power and RMS Values 1 Instantaneous power

1. Power and RMS Values 1

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 2 be (N – 1) voltage measurements, and (N – 1) current measurements.

Average value of periodic instantaneous power p(t) 3

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

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

Root-mean squared value of a periodic waveform with period T For the sinusoidal case 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 7

Determine voltage and current magnitudes and phase angles Using a cosine reference, Voltage cosine has peak = 100 V, phase angle = -90º Current cosine has peak = 50 A, phase angle = -135º Phasors 8

The average power is 9

Voltage – Current Relationships 10

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 11

Problem 10. 17 12

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Active and Reactive Power Form a Power Triangle Complex power S Q Projection of S on the imaginary axis P Projection of S on the real axis is the power factor 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 16

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

Complex power S Q Projection of S on the imaginary axis P Projection of S on the real axis 18

Resistor also so , Use rms V, I 19

Inductor also so , Use rms V, I 20

Capacitor also so , Use rms V, I 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 22

Now, demonstrate Excel spreadsheet EE 411_Voltage_Current_Power. xls to show the relationship between v(t), i(t), p(t), P, and Q 23

A Single-Phase Power Example 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. PL + j. QL 0. 05 + j 0. 15 pu ohms VL = 1. 020 /0° j 0. 20 pu mhos VR = 1. 010 /-10° IS Icap. L PR + j. QR Icap. R j 0. 20 pu mhos 25

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PL + j. QL 0. 05 + j 0. 15 pu ohms VL = 1. 020 /0° j 0. 20 pu mhos VR = 1. 010 /-10° IS Icap. L PR + j. QR Icap. R j 0. 20 pu mhos 28

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

RMS of common periodic waveforms, cont. Sawtooth V 0 T 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 0 0 0 -V V 0 0 0 V 0 31

2. Three-Phase Circuits 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 threephase, four-wire system needs three meters. A three-phase, three-wire system needs only two meters. a b c n Three-phase, four wire system Reference • The instantaneous power is constant 33

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

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Zline c c Ic 3 Zload a b Zline – Vab Ia Zline The “One-Line” Diagram n Ib a Ia + Van – 3 Zload + Zline a a b Zload n 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: 1. Convert the entire circuit to an equivalent wyewith a grounded neutral. 2. Draw the one-line diagram for phase a, recognizing that phase a has one third of the P and Q. 3. Solve the one-line diagram for line-to-neutral voltages and line currents. 4. If needed, compute line-to-neutral voltages and line currents for phases b and c using the ± 120° relationships. 5. If needed, compute line-to-line voltages and delta currents using the 3 and ± 30° relationships. 40

Now Work a Three-Phase Motor Power Factor Correction Example A three-phase, 460 V motor draws 5 k. W 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 k. VAr (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 k. VArs are added? 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. 42

3. Transformers 43

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

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

Open Circuit Test Ioc Rs j. Xs Rm j. Xm 7200 V Φ + Voc - Ideal Transformer 7200: 240 V Open circuit test: Open circuit the 7200 V-side, and apply 240 V to the 240 V-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) 46

1. Given the standard percentage values below for a 125 k. VA 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 240 V side, compute the new Ω values. Single Phase Transformer. Percent values are given on transformer base. Winding 1 kv = 7. 2, k. VA = 125 Winding 2 kv = 0. 24, k. VA = 125 Load loss Rs Xs No load loss Magnetizing current j. Xs Ideal Rm j. Xm Transformer 7200: 240 V %imag = 0. 5 %loadloss = 0. 9 %noloadloss = 0. 2 %Xs = 2. 2 7200 V 240 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? 47

Distribution Feeder Loss Example • Annual energy loss = 2. 40% • Largest component is transformer noload loss (45% of the 2. 40%) Modern Distribution Transformer: • Load loss at rated load (I 2 R in conductors) = 0. 75% of rated transformer k. W. • No load loss at rated voltage (magnetizing, core steel) = 0. 2% of rated transformer k. W. • Magnetizing current = 0. 5% of rated transformer amperes

Single-Phase Transformer Impedance Reflection by the Square of the Turns Ratio Rs j. Xs Rm j. Xm 7200 V Ideal Transformer 7200: 240 V 7200 V 240 V 49

Now Work a Single-Phase Transformer Example Open circuit and short circuit tests are performed on asingle-phase, 7200: 240 V, 25 k. VA, 60 Hz distribution transformer. The results are: · Short circuit test (short circuit the low-voltage side, energize the high-voltage side so that rated current flows, and measure Psc and Qsc). Measured Psc = 400 W, Qsc = 200 VAr. · Open circuit test (open circuit the high-voltage side, apply rated voltage to the low-voltage side, and measure Poc and Qoc). Measured Poc = 100 W, Qoc = 250 VAr. Determine the four impedance values (in ohms) for the transformer model shown. Φ Rs j. Xs Rm j. Xm Ideal Transformer 7200: 240 V Turns ratio 7200: 240 7200 V 240 V (30 : 1) (but approx. same amount of copper in each winding) 50

A three-phase transformer can be three separate single-phase transformers, or one large transformer with three sets of windings Wye-Equivalent One-Line Model A N 1: N 2 Rs j. Xs Rm j. Xm Ideal Transformer N 1 : N 2 N N 1: N 2 • Reflect to side 2 using individual transformer turns ratio N 1: N 2 Y-Y 51

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

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

For Wye-Delta Connection Model, Convert the Transformer to Equivalent Wye-Wye Wye-Equivalent One-Line Model N 1: N 2 A Ideal Transformer N N 1: N 2 Y-Δ Reflect to side 2 using three-phase bank line-toline turns ratio So, for all configurations, the equivalent wye-wye transformer ohms can be reflected from one side to the other using the threephase bank line-to-line turns ratio 54

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 55

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 Log 10 Scale 56

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