Power System Engineering - Rowan Universityusers.rowan.edu/~krchnavek/Rowan_University/Power_files/Intro_and... · Power System Engineering Robert R. Krchnavek Rowan University Glassboro,

Power System Engineering

Robert R. KrchnavekRowan University

Glassboro, New Jersey

IntroductionRobert R. Krchnavek

Rowan UniversityGlassboro, New Jersey

Objectives

• Gain an understanding of how the electric grid originated – brief history of power.

• Understand some of the challenges facing the U.S. with respect to power generation, transmission, distribution, and reliability.

Electricity Generation, Transmission,

Distribution

Source: http://www.eia.gov/energyexplained/index.cfm?page=electricity_delivery

A Brief History• Numerous references on the subject including

video documentaries.

• We will consider the following: http://inventors.about.com/od/timelines/a/electricity_timeline.htm

• Earliest systems were DC, not AC.

• Common problem: different electrical applications required different voltages.

• Edison vs Westinghouse – “War of Currents” http://en.wikipedia.org/wiki/War_of_Currents

DC Systems

• Because different devices required different voltages, several different generating facilities were required.

• Power loss along the transmission lines was significant so generating facilities needed to be close by.

• What is the problem with DC?

• Is there anything better with DC vs AC?

AC Systems

• The ability for an AC system to easily change voltages (transformers) made AC the system of choice.

• Why does a higher voltage produce less power loss?

• Is this equally important today?

What has happened since the early days?

• Early 1900s, over 4000 isolated electric utilities.

• Post WWII, utilities started to interconnect.

• Eventually, there were 3 large interconnected systems.

• The 3 interconnected systems have over 3200 distribution utilities, 10,000 generating facilities, and tens of thousands of transmission and distribution lines.

• The North American Electric Reliability Corporation (NERC) sets standards, mandatory procedures, guarantees supply, and maintains security under the oversight of U.S. DOE and Federal Energy Regulatory Commission.

Source: http://www.eia.gov/energy_in_brief/power_grid.cfm

The “National” Power Grid

Major Challenges

• New transmission lines – NIMBY !

• Economic recovery for transmission lines crossing one area to serve another.

• Interconnecting remote but large renewable generation facilities.

• Addressing uncertainty in the Federal regulatory process for cost of upgrading the grid.

Regulation and Infrastructure

• The terminology and abbreviations related to the grid are confusing.

• This must be understood from a business and policy perspective.

• We will concentrate on the technical aspects.

Source: Energy Information Adminsitration. Map shows NERC (North American Electric Reliability Corporation) regions.

Source: http://www.ferc.gov/industries/electric/indus-act/rto.asp

Some interesting information ....

Energy Use

Energy Consumption/Source

Power Flow in North America

Source: http://www.eia.gov/todayinenergy/detail.cfm?id=4270

Sources of Electric Energy

Renewables

Electricity in Homes

Cost of Electricity

Future Demand Growth

Source: http://www.eia.gov/forecasts/archive/aeo11/MT_electric.cfm

A 1% demand growth from 2009-2035 results

in a 31% increase in demand for electricity.

Fuel for Electricity Generation Projections

Source: http://www.eia.gov/forecasts/archive/aeo11/MT_electric.cfm

Residential Use and Commercial Growth Projections

Source: http://www.eia.gov/forecasts/archive/aeo11/MT_electric.cfm

Summary• Electrical energy use continues to grow in the U.S. and

an annual rate of ≈1%. (International growth is larger.)

• Even at this modest growth rate, the demand will increase by ≈30% by 2035.

• The national grid will require significant upgrades to meet this additional demand.

• New generating facilities, higher transmission voltages, distributed generation, smart grid, and increased security will all be required in the future.

FundamentalsRobert R. Krchnavek

Rowan UniversityGlassboro, New Jersey

Objectives

• Relearn/understand the use of phasors to analyze AC voltages and currents.

• Understand real, reactive, and complex powers and the significance of power factor.

• Develop a working knowledge of 3-phase systems including their fundamental advantages over single-phase systems.

Phasorsv(t) = Vm cos(!t+ �)

i(t) = Im cos(!t+ �)

For sinusoids, the root-mean-square value,sometimes called the effective value, is given by

Vrms = V =Vmp2

Irms = I =Imp2

Phasorsv(t) = <

hVme|(!t+�)

i= v(t) = Vm cos(!t+ �)

The phasor is a complex number that represents the magnitude and phase of the sinusoid.

To get back to the time domain, multiply by and take the real part:e|!t

v(t) = <⇥Ve|!t

⇤

V = Vme|� = Vm �

Instantaneous Power Resistor

p(t) = VmIm cos

2(!t+ �)

p(t) =1

2

VmIm {1 + cos [2(!t+ �)]}

p(t) = V I {1 + cos [2(!t+ �)]}

And the average power is simply

Note:

p(t) = v(t) i(t) = Vm cos(!t+ �) Im cos(!t+ �)

pavr = V I

IR =VR

R

Instantaneous Power Inductor

p(t) = v(t) i(t) = Vm cos(!t+ �) Im cos(!t+ � � 90

�)

p(t) =1

2

VmIm cos [2(!t+ �)� 90

�]

p(t) = V I sin [2(!t+ �)]

And the average power is simplypavr = 0

Note: IL =VL

|!L

Instantaneous Power Capacitor

p(t) = v(t) i(t) = Vm cos(!t+ �) Im cos(!t+ � + 90

�)

p(t) =1

2

VmIm cos [2(!t+ �) + 90

�]

p(t) = �V I sin [2(!t+ �)]

And the average power is simply

pavr = 0

Note: IC =VC1

|!C

Instantaneous Power General RLC Load

v(t)

i(t)

RLCcircuit

+–

v(t) = Vm cos(!t+ �)

i(t) = Im cos(!t+ �)

i(t) =v(t)

Z Z

p(t) = v(t)i(t) = Vm cos(!t+ �)Im cos(!t+ �)

p(t) = V IR {1 + cos [2(!t+ �)]}+ V IX sin [2(!t+ �)]

after some algebra . . . .

pR(t) pX(t)

where IR = I cos(� � �) and IX = I sin(� � �)

Real Poweris power dissipated in the resistance in the circuit.pR(t)

The average power, also called the real power or active power,

is given by

pavr = P = V IR = V I cos(� � �)

Power Factorcos(� � �)The term is the power factor.

� � � is the power factor angle.

Recall: v(t) = Vm cos(!t+ �)

i(t) = Im cos(!t+ �)

� � � is the angle between the voltage and current.

Inductive loads: voltage ( ) leads current ( ). Lagging power factor.

Capacitive loads: current ( ) leads voltage ( ). Leading power factor.

� �

� �

Note: By convention, the power factor is positive.

Reactive Poweris power absorbed in the reactive part of the circuit.pX(t)

Q = V IX = V I sin(� � �)

The average power is 0.

The magnitude of the reactive power is given by

Units: var which stands for volt-amperes-reactive.

Q is the reactive power.

Complex PowerInstead of working in the time domain with instantaneous power, we now switch to phasors and define Complex Power, S, for sinusoidal, steady-state circuits.

P = V I cos(� � �)

Q = V I sin(� � �)

S = P + |Q

S = V � I �� = V I � � � = VI⇤

S = V Ie|(���) = V e|�Ie�|�

S = V I cos(� � �) + |V I sin(� � �) =

V I [cos(� � �) + | sin(� � �)]

Complex Power

|S| = |VI⇤| = V I

Apparent power is given by

and has units of voltamperes or VA.

P = V I cos(� � �) W

Q = V I sin(� � �) var

S = V I VA

(� � �)

S =p

P 2 +Q2

(� � �) = tan�1(Q/P )

Q = P tan(� � �)

power factor = p.f. = cos(� � �) =P

S=

PpP 2

+Q2

Balanced 3-Phase Circuits

• Assuming ideal sources.

• Assuming ideal lines.

• Assuming loads are all equal.

Balanced 3-Phase Circuits Line-to-Neutral Voltages

• Ean, Ebn, and Ecn are line-to-neutral voltages.

• 120˚ out of phase.

• abc or positive sequence is when Ean leads Ebn by 120˚ and Ebn leads Ecn by 120˚.

• Could have a negative or acb sequence.

� �

�

Ean

Ebn

Ecnn

a

b

c

Ean = 480 0

Ebn = 480 �120

Ecn = 480 120

Ean

Ecn

Ebn

120˚

Phasor Diagram

Balanced 3-Phase Circuits Line-to-Line Voltages

� �

�

Ean

Ebn

Ecnn

a

b

c

Ean � Eab � Ebn = 0

Eab = Ean � Ebn

Eab = 480 0� 480 �120 =p3 480 30

Similarly,

Ebc =p3 480 �90

Eca =p3 480 150

The line-to-line voltages in a balanced, 3Φ, Y-connected positive sequence are times the line-to-neutral voltages and 30˚ ahead.p

3

Balanced 3-Phase Circuits Phasor Representation

Ean

EabEca

Ebc

Ecn

Ebn

� �

�

Ean EabEbc

Eca

Ebn

Ecnn

a

b

c

Balanced 3-Phase Circuits Balanced Line Currents

Since we are neglecting line impedances, EnN = 0.

� �

�

EanZY

ZY

ZYEbn

Ecnn N

a

b

c C

A

B

Ic

Ia

In

Ib

Ia =Ean

ZY

Ib =Ebn

ZY

Ic =Ecn

ZY

In a balanced 3Φ circuit, the line currents are equal and the neutral current is 0.

Balanced 3-Phase Circuits Balanced ∆ Loads

� �

�

EanZ∆Z∆

Z∆Ebn

Ecnn

a

b

c

C A

B

Ic

ICA

IBC IAB

Ia

Ib

IAB =Eab

Z�Ia + ICA � IAB = 0

Ia = IAB � ICA

with similar expressions for the other currents ....

Balanced 3-Phase Circuits Balanced ∆ Loads

For a balanced ∆-load, supplied by a balanced, positive-sequence Y-source, the line currents into the load are times the ∆-load currents and lag by 30˚.

p3

Ia =p3IAB �30 � �

�

EanZ∆Z∆

Z∆Ebn

Ecnn

a

b

c

C A

B

Ic

ICA

IBC IAB

Ia

Ib

Δ-Y Conversion Balanced Loads

ZYZY

ZY

N

a

b

c C

A

B

Ic

Ia

Ib

Z∆Z∆

Z∆

a

b

c

C A

B

Ic

ICA

IBC IAB

Ia

Ib

For the loads to look equivalent from the source perspective, for equal applied line voltages, the line currents must be equal.

IA =EAN

ZY=

EAB �30p3ZY

IA =p3IAB �30 =

p3Eab �30

Z�

3-Phase Circuits Power

van(t) = Vm,LN cos(!t+ �) ia(t) = Im,L cos(!t+ �)

pa(t) = Vm,LNIm,L cos(!t+ �) cos(!t+ �)

pa(t) = VLNIL cos(� � �) + VLNIL cos(2v!t+ � + �)

Similarly, for the B and C phases:

pb(t) = VLNIL cos(� � �) + VLNIL cos(2v!t+ � + � � 240

�)

pc(t) = VLNIL cos(� � �) + VLNIC cos(2v!t+ � + � + 240

�)

3-Phase Circuits Power

The total power delivered by the source, as a function of time, is

ptotal

(t) = pa

(t) + pb

(t) + pc

(t) =

3VLNIL cos(� � �)+ VLNIL cos(2v!t+ � + �)+

VLNIL cos(2v!t+ � + � � 240

�)+

VLNIC cos(2v!t+ � + � + 240

�)

ptotal

(t) = 3VLN

IL

cos(� � �)

Constant power delivered as a function of time !

3-Phase Circuits Power

ptotal

(t) = 3VLN

IL

cos(� � �)

VLL =p3VLNRecall

ptotal

(t) =p3V

LL

IL

cos(� � �)

Note: Line-to-Line voltages are more common than Line-to-Neutral voltages.

Note: Refer to a textbook for a complex power analysis for generators and balanced Y and loads.�

3-Phase vs Single Phase Advantages of 3

• 1/2 the number of conductors in a 3-phase system for the same delivered power.

• 1/2 of the line loss in the 3-phase system.

• 1/2 of the line voltage drop in the 3-phase system.

• Constant power as a f(t). Requires constant mechanical input power and shaft torque for the generator.

�

Summary• Phasors are the preferred method of

mathematically describing (linear) AC circuits.

• Real, reactive, and complex powers are useful concepts for describing an AC power system.

• The power factor is a measure of the phase relationship between voltage and current and therefore real and reactive power.

• Balanced 3-phase systems are common in AC power systems. There are numerous advantages over single-phase systems.