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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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1. Power and RMS Values

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1 power and rms values

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

+

Instantaneous power p(t) flowing into the box

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

Average value ofperiodic instantaneous power p(t)


Two wire sinusoidal case

Two-wire sinusoidal case

zero average

Power factor

Average power


Root mean squared value of a periodic waveform with period t

compare

Root-mean squared value of a periodic waveform with period T

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


Root mean squared value of a periodic waveform with period t1

Root-mean squared value of a periodic waveform with period T

For the sinusoidal case


1 power and rms values

  • 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


1 power and rms values

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


1 power and rms values

The average power is


1 power and rms values

Voltage – Current Relationships


1 power and rms values

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


1 power and rms values

Problem 10.17


1 power and rms values

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


1 power and rms values

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


1 power and rms values

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


1 power and rms values

Complex power S

Projection of S on the imaginary axis

Q

P

Projection of S on the real axis


1 power and rms values

Resistor

also

so

Use rms V, I

,


1 power and rms values

Inductor

also

so

Use rms V, I

,


1 power and rms values

Capacitor

also

so

,

Use rms V, I


1 power and rms values

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


1 power and rms values

Now, demonstrate Excel spreadsheet

EE411_Voltage_Current_Power.xls

to show the relationship between v(t), i(t), p(t), P, and Q


1 power and rms values

A Single-Phase Power Example


1 power and rms values

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.


1 power and rms values

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


Rms of some common periodic waveforms

V

0

0 < D < 1

DT

T

RMS of some common periodic waveforms

Duty cycle controller

By inspection, this is the average value of the squared waveform


Rms of common periodic waveforms cont

RMS of common periodic waveforms, cont.

Sawtooth

V

0

T


Rms of common periodic waveforms cont1

V

0

V

0

V

0

V

0

V

0

V

0

0

-V

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


2 three phase circuits

2. Three-Phase Circuits


Three important properties of three phase balanced systems

a

b

c

n

Three-phase, four wire system

Reference

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


1 power and rms values

Observe Constant Three-Phase P and Q in Excel spreadsheet

1_Single_Phase_Three_Phase_Instantaneous_Power.xls


1 power and rms values

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


Now work a three phase motor power factor correction example

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 supply voltage, what will be the new

  • after the kVArs are added?


Now work a delta wye conversion example

Now Work a Delta-Wye Conversion Example


3 transformers

3. Transformers


Single phase transformer

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

Single-Phase Transformer


Open circuit test

Ioc

+

Voc

-

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. Calculate

Φ

Turns ratio 7200:240

(but approx. same amount of copper in each winding)

Open Circuit Test

jXs

Rs

Ideal

Transformer

jXm

Rm

7200:240V

7200V 240V


Short circuit test

Isc

+

Vsc

-

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. Calculate

Φ

Turns ratio 7200:240

(but approx. same amount of copper in each winding)

Short Circuit Test

jXs

Rs

Ideal

Transformer

jXm

Rm

7200:240V

7200V 240V


1 power and rms values

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?


Distribution feeder loss example

Distribution Feeder LossExample

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


Single phase transformer impedance reflection by the square of the turns ratio

jXs

Rs

Ideal

Transformer

jXm

Rm

7200:240V

7200V 240V

Ideal

Transformer

7200:240V

7200V 240V

Single-Phase TransformerImpedance Reflection by the Square of the Turns Ratio


Now work a single phase transformer example

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

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

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.


1 power and rms values

Wye-Equivalent One-Line Model

A

N

jXs

Rs

Ideal

Transformer

jXm

Rm

N1 : N2

  • Values for one of the transformer windings, on side 1

  • Can reflect to side 2 using either individual transformer turns ratio N1:N2, or three-phase bank line-to-line turns ratio (which are identical ratios)

A three-phase transformer can be three separate single-phase transformers, or one large transformer with three sets of windings

N1:N2

N1:N2

N1:N2

Y - Y


1 power and rms values

Wye-Equivalent One-Line Model

A

N

Ideal

Transformer

  • Converting side 1 impedances from delta to equivalent wye

  • Can reflect to side 2 using either individual transformer turns ratio N1:N2, or three-phase

  • bank line-to-line turns ratio (which are identical ratios)

For Delta-Delta Connection Model, Convert the Transformer to Equivalent Wye-Wye

N1:N2

N1:N2

N1:N2

Δ - Δ


1 power and rms values

Wye-Equivalent One-Line Model

A

N

Ideal

Transformer

  • Converting side 1 impedances from delta to wye

  • Can then reflect to side 2 using three-phase bank line-to-line turns ratio

For Delta-Wye Connection Model, Convert the Transformer to Equivalent Wye-Wye

N1:N2

N1:N2

N1:N2

Δ - Y


1 power and rms values

Wye-Equivalent One-Line Model

A

N

Ideal

Transformer

Can then reflect to side 2 using three-phase bank line-to-line turns ratio

For Wye-Delta Connection Model, Convert the Transformer to Equivalent Wye-Wye

N1:N2

N1:N2

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

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 one side are in phase with line-to-neutral voltages on the other side

  • Thus, and phase shift in line-to-line voltages is unavoidable, but it can be managed to avoid problems


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