1. Power and RMS Values.

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Presentation transcript:

1. Power and RMS Values

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.

Average value of periodic instantaneous power p(t)

Two-wire sinusoidal case zero average Power factor Average power

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

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

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

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

(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

Problem 10.17

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

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

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

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

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 Single-Phase Power Example

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

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

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

RMS of common periodic waveforms, cont. Sawtooth V T

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

2. Three-Phase Circuits

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

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

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.

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?

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.

3. Transformers

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 240V

(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 240V Turns ratio 7200:240 (but approx. same amount of copper in each winding) Φ

(but approx. same amount of copper in each winding) Open Circuit Test Ioc Rs jXs + Voc - Ideal Rm jXm Transformer 7200:240V 7200V 240V 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

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 240V 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?

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

Single-Phase Transformer Impedance Reflection by the Square of the Turns Ratio Rs jXs Ideal Transformer 7200:240V Rm jXm 7200V 240V Ideal Transformer 7200:240V 7200V 240V

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 240V

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

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

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

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

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

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

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