APS Lecture 2

8/13/2019 APS Lecture 2

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Advanced Power Systems

Electric Power Fundamentals


2/45

Public service announcement

How many of you are EITs?

How many are registered to take the EIT?

Do yourself a favor

Enroll and take the test, our ECE students at

Rowan thus far have a 100% pass rate and are well

on their way to becoming PEs http: //www.state.nj .us/lps/ca/pels/pelsinfo.htm


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Aims of Todays Lecture

Course Training Tours (Find Potential Times)

Part One: Overview of Chapter 1 equations

Discussion of Articles reviewed

Your Carbon Footprints

A summary of ch. 2 concepts

Effective Values of V and I (rms) R, L & C in ac circuits

Power Factor

Power Triangle & Power Factor Correction

Three-Wire Single Phase Residential Wiring

Three Phase Systems

Power Supplies and Power Quality


4/45

Aims of Todays Lecture (cont)

15 minute stretch break at 6

Part Two: An intro to ch. 3 concepts

Early developments

Electric industry today (NUGS, IPPs, QFs)

Polyphase synchronous generators

Heat engines, steam cycles and efficiencies

GTs, CCs, Baseload Plants and LDCs

T&D

Regulatory impacts (PUHCA, PURPA, FERC)


5/45

Chapter 1 Equations

pdtwEnergy

vidt

dq

dt

dw

dt

dwp

v

dq

dwv

i

dt

dqi

loop

node

)(

0

0


6/45

Chapter 1 Equations (2)

2

2

21

21

22

21:

2

1:

111

LiwdtdiLvInductors

Cvwdt

dvCiCaps

A

lR

RRR

RRR

R

v

Rivip

Gvi

Riv

parallel

series


7/45

Chapter 1 Equations (3)

)(:

111

111

11

1

22

21

21

21

21

vturnsratiovN

NvrsTransforme

CCC

LLL

CCC

LLL

series

parallel

parallel

series


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What is your Carbon Footprint?

http://www.carbonfootprint.com/

http://www.safeclimate.net/calculator/

http://www.climatecrisis.net/takeaction/carboncalculator/

http://www.carboncalculator.co.uk/ http://www.bp.com/

http://www.southampton-sustainability.org/carboncalc.htm

http://carbonfund.org/site/pages/calculator/

http://www.cleanair-coolplanet.org/action/footprint.php

http://www.stopglobalwarming.org/carboncalculator.asp http://www.cambridgecarbonfootprint.org.uk/footprint_links.htm

http://www.bestfootforward.com/carbonlife.htm

http://www.mycarbonfootprint.eu/

http://www.nativeenergy.com/lifestyle_calc.html
http://www.carbonfootprint.com/http://www.safeclimate.net/calculator/http://www.climatecrisis.net/takeaction/carboncalculator/http://www.carboncalculator.co.uk/http://www.bp.com/http://www.southampton-sustainability.org/carboncalc.htmhttp://carbonfund.org/site/pages/calculator/http://www.cleanair-coolplanet.org/action/footprint.phphttp://www.stopglobalwarming.org/carboncalculator.asphttp://www.cambridgecarbonfootprint.org.uk/footprint_links.htmhttp://www.bestfootforward.com/carbonlife.htmhttp://www.mycarbonfootprint.eu/http://www.nativeenergy.com/lifestyle_calc.htmlhttp://www.nativeenergy.com/lifestyle_calc.htmlhttp://www.mycarbonfootprint.eu/http://www.bestfootforward.com/carbonlife.htmhttp://www.cambridgecarbonfootprint.org.uk/footprint_links.htmhttp://www.stopglobalwarming.org/carboncalculator.asphttp://www.cleanair-coolplanet.org/action/footprint.phphttp://www.cleanair-coolplanet.org/action/footprint.phphttp://www.cleanair-coolplanet.org/action/footprint.phphttp://carbonfund.org/site/pages/calculator/http://www.southampton-sustainability.org/carboncalc.htmhttp://www.southampton-sustainability.org/carboncalc.htmhttp://www.southampton-sustainability.org/carboncalc.htmhttp://www.bp.com/http://www.carboncalculator.co.uk/http://www.climatecrisis.net/takeaction/carboncalculator/http://www.safeclimate.net/calculator/http://www.carbonfootprint.com/


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Chapter Two

Effective Values of V and I (rms)

R, L & C in ac circuits

Power Factor

Power Triangle & Power Factor Correction

Three-Wire Single Phase Residential Wiring

Three Phase Systems Power Supplies and Power Quality


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Sinusoidal Sources

T is period of oscillation

f = 1/T (frequency)

Units of Hertz Cycles per second

= 2f

Angular frequency

Radians per second = phase angle

Offset from zero

May be degrees or rads

)sin()( tVtv m


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effective values of periodic waveforms

what is the effective voltageof a

120VAC/60Hz outlet in the walls

of your home and in this building?

What is v(t)?

What is T?

What is ?

What is Vm?

T = 1/f = 1/60Hz = 16.667msecs

= 2f = 260 = 377rads/sec

Vm= ?

)sin()( tVtv m


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To understand what Vmmight be weneed a good definition of effectivevalue.

The effective valueof a current is thesteady current (dc) that transfers thesame amount of average (real) power asthe given varying current.

The effective valueof a voltage is thesteady voltage (dc) that transfers the

same amount of average (real) power asthe given varying voltage.


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dtiTI

dtiT

RRI

RIP

RdtiT

P

T

eff

T

eff

eff

T

0

2

0

22

2

0

2

1

1

effective value: is the

square root of the mean ofthe squared values

also known as (a-k-a)

the root-mean-square

avgmavgmavgrms tVtVvV )(cos)cos()( 2222


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dtiTI

dtiT

RRI

RIP

RdtiT

P

T

eff

T

eff

eff

T

0

2

0

22

2

0

2

1

1

What is the averagevalue of the square ofthe cosine of asinusoid?

22

1)(cos2 mmavgmrms

VVtVV

By inspectio n it is ,

going f rom 0 1 andback again


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2

2

m

rmseff

mrmseff

V

VV

III

Since 120 & 240VAC areeffective valuesthis means theyare root-mean-square (rms) values

So

What are the amplitudes of thesesinusoidal waveforms if viewed on a

scope?

Vmof 120VAC = ? LM#1

Vmof 240VAC = ? LM#2


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NOTE:

In practice, electrical engineers must take care to notice (ordetermine) whether a sinusoidal voltage is being expressed interms of its effective value (rms) or its maximum (Vm) value

Generally in:

electronics and communications V is Vm

power applications and this course V is Vrms

Also in practice, we treat incoming voltage as having a zerophase angle unless otherwise specified, current angles are thenexpressed w.r.t. voltage


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R, L & C in ac circuits

Voltage across resistors is the same as the voltage

supplied by the source

Phase angle of resulting current is the same asphase angle of the supply voltage

V and I are considered in phase

And V = RI (where V and I are rmsvalues)

Power = VI (and I2R, V2/R)


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Sample problems

An electric water heater element provides 4500

watts of real power at 240VAC, what is its

resistance (and current)?

How much power will it provide if the voltage

sags to 210VAC?


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Capacitors

V= ZI= (1/jC)I (NOTE: Z = -j/C)

In a purely capacitive circuit the current leads

the voltage by exactly 90o(or voltage lags by 90o)

Average power dissipated by a capacitor iszero (stores and releases) ideal

Sample exercise: what is rmscurrent and i(t) in a 5F capacitor used to balance the power factor on amotor in the UK (assume 240V/50Hz)


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Capacitor example

Sample exercise: what is current in a 5F capacitor usedto balance the power factor on a motor in the UK (assume240V/50Hz)

V = (1/jC)I so I = VjC = (240)j(250)(5x10-6) = 0.377j A

Atti

Atti

Arms

)2

310cos(533.0)(

)90310cos(533.0)(

90377.0

I


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You try it capacitor example

Sample exercise: find the rms current and write anequation for the current as a function of time when a 500

nF capacitor is used in a 120VAC/60Hz system

Remember V = (1/jC)I so I = VjC


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Inductors

V= ZI =jLI (NOTE: Z = jL)

In a purely inductive circuit the voltage leads the current

by exactly 90o

(or current lags by 90o

) Average power dissipated by an inductor is zero

(stores and releases) ideal

if V= 1050o V, and L=2H, =100 rads/s

what does I= ?

I = V/ jL = V/ j200 = 1050o/20090o

I = 0.05-40oA


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Impedance (Z) and Reactance (X)

Z=V/I

Z= R + jX

X= L

or

X= -1/C


24/45

example problem

A home in Europe (240VAC/50Hz) that had only resistive loads(very unlikely) for heating and lighting converts to a ground-source heat pump (a motor and compressor) and compact

fluorescent lights. What happens to its current phase angle if itsimpedance (Z) prior was 2resistive and no reactance and afterhas 4of resistance and 10mH of inductance?

V = 2400o

V, Z = 20o so I = 1200

oA

V = 2400o

V, Z = 4 + j(314)0.01 = 5.0938o

so I = V/Z = 2400

o/ 5.0938

o= 47.2-38

oA


25/45

Memory aid

ELI the ICEman

For an inductor (L)

E(voltage as in EMF) leads current (I) same as Current Lags the Voltage

For a capacitor (C) I (the current) leads E(voltage)

same as Voltage Lags the Current


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R, L & C in AC circuits

SUMMARY:

Currents flowing through any of these elements

(R, L & C) will have same frequency as thevoltage supplied by the source

Phase angle shift may result in current dependingon whether there is net L or C in the circuit

Resistive elements are the only circuitcomponents that dissipate any real energy


27/45

Power factor

Typical residential and small commercial watt-hour meters measure only average power, watts,

also called real power. Utilities must generate power at the power sources

to cover all the demands for current both real andreactive.

The power factor is a measure of what part of theoverall power in the system is real


28/45

Power factor

Typically, average power, watts, also called real

poweris represented by the following equation:

Pavg= VI cos = VI x Power Factor Where is the phase angle (time shift) between V and I

Therefore, the power factor is identical to:

cos

PF = cos


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(a) The impedance triangle where Z=R+jX (b) The complex power triangle where S=P+jQ

complex power


30/45

complex impedance (Z) triangle

jL

1/jC = -j/C

Power factor phase angle


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power factor

since cosine function is even with respect topositive or negative phase angles:

i.e; cos 45o= cos -45o

we resolve difficulty by labeling power factor aseither leading or lagging:

If V - I > 0 pf is lagging

If V

- I< 0 pf is leading

NOTE: If pf is leading or lagging THEN the current (I) isleading or lagging w.r.t. voltage (V) too!


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Resistive and Reactive Loads


33/45

power factor

power factor leading or lagging:

lagging with inductive loads/circuit

leading with capacitive loads/circuit utilities generally experience lagging power factor due to

highly inductive loads of customers

motors, flourescent lights, monitors, tvs, appliances, dimmer

switches, a/c power factor correction is big part of utility planning

activities


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complex power triangle

Q =reactive power

P = real power

S = apparent power


35/45

complex power triangle

)sin(2

IVmmIVQ

2

mmIVS

Q =reactive

P = real

S = apparent

)cos(2

IVmmIVP

units: Watts

units: VARs

units: Volt-Amps

note: = V - I

j


36/45

Express S,P &Q with rms values for

V and I LM#6

)sin(2

IVmmIVQ 2

mmIVS

Q =reactive

P = real

S = apparent

)cos(2

IVmmIVP

units: Watts

units: VARs

units: Volt-Amps

j

note: = V - I


37/45

Power factor example

A 240-V induction motor draws 30 amps ofcurrent and delivers 5.4kW of power to the shaft,

show its power triangle: Real (average) power = 5,400 Watts

Apparent power = (240)(30) = 7,200 Volt-Amps

PF = Real/Apparent = 5.4kW/7.2kW = 0.75

Phase angle = cos-1(0.75) = 41.41o

Draw power triangle and compute reactive power


38/45

Power factor correction?

Why correct power factor?

1/5thof all grid losses may be due to poor power factor (>$2B/yr), the

outage of 2003 was made more severe by extremely high reactivedemand, all transformers are rated on kVA not watts, all theseeconomic, efficiency and reliability benefits can be achieved at a verylow cost


39/45

Power factor correction

Adding capacitive impedance in parallel with the load enables the current tooscillate between the inductors and capacitors rather than being drawn from

the utility system or the customer transformer.

Capacitors are rated by volt-amps-reactive VARs that they supply at thesystem voltage in which they are installed, and PF correction is astraightforward engineering design


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Power factor correction example

An industrial customers service entrance substation is rated at1MVA (1,000kVA) and is at 95% capacity. The plant nowexperiences a power factor of 80%. A new manufacturing line is

planned that will increase power demand 125kW. How manykVAR of capacitance should be added to avoid purchasingadditional substation transformer capacity?

Real power (at present) = (0.8)(0.95)(1,000kVA) = 760kW

Phase angle = cos-1(0.8) = 36.87o

Apparent power = (1,000)(0.95) = 9,500 Volt-Amps

If demand grows from 760kW to 885kW Apparent Power will grow toReal/PF = 885/(0.8) = 1106kVA > 1MVA capacity

Reactive Power = Q = VI sin = 1106(0.6) = 664kVAR


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PF correction example (cont)

An industrial customers service entrance substation is rated at 1MVA (1,000kVA) and is at 95%capacity. The plant now experiences a power factor of 80%. A new manufacturing line is plannedthat will increase power demand 125kW. How many kVAR of capacitance should be added to avoidpurchasing additional substation transformer capacity?

For substation to handle the growth, power factor must improveto at least PF = 885kW/1,000kVA = 0.885

Phase angle now will be = cos-1(0.885) = 27.75o

Reactive Power (Q) = VI sin= 1000(0.4656) = 466 kVAR

Difference in reactive power must be supplied by the capacitorbank: 664466 = 198 kVAR

Specify a >= 200 kVAR cap bank at industrial customersservice entrance switchgear


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3-phase systems

Benefits:

Generators and motors work more efficiently in

torque transfer and have higher stability (they runsmoother and have less vibration)

Transmission benefits include the reduction and/or

cancellation of neutral return currents so that less

wires and/or smaller common neutrals can be used.


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44/45

3current

240)240cos(2

120)120cos(2

0)cos(2

cmc

bmb

ama

ItIi

ItIiItIi


45/45

3neutral current

)240cos(2

)120cos(2)cos(2

tI

tItIiiii

m

m

mcban

Documents

APS Lecture 2