Advanced Power Systems

1
Advanced Power Systems

• ECE 0909.402-02, 0909.504-02
• Lecture 2 Electric Power Fundamentals
• 29 January 2007
• Dr. Peter Mark Jansson PP PE
• Associate Professor Electrical and Computer
  Engineering

2
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

3
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

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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
• TD
• Regulatory impacts (PUHCA, PURPA, FERC)

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Chapter 1 Equations
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Chapter 1 Equations (2)
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Chapter 1 Equations (3)
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Articles Reviewed

• What did you find?
• What did you learn?
• What do you think?

9
What is your Carbon Footprint?

• http//www.carbonfootprint.com/
• http//www.safeclimate.net/calculator/
• http//www.climatecrisis.net/takeaction/carboncalc
  ulator/
• http//www.carboncalculator.co.uk/
• http//www.bp.com/
• http//www.southampton-sustainability.org/carbonca
  lc.htm
• http//carbonfund.org/site/pages/calculator/
• http//www.cleanair-coolplanet.org/action/footprin
  t.php
• http//www.stopglobalwarming.org/carboncalculator.
  asp
• http//www.cambridgecarbonfootprint.org.uk/footpri
  nt_links.htm
• http//www.bestfootforward.com/carbonlife.htm
• http//www.mycarbonfootprint.eu/
• http//www.nativeenergy.com/lifestyle_calc.html

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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
• ? 2?f
• Angular frequency
• Radians per second
• ? phase angle
• Offset from zero
• May be degrees or rads

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effective values of periodic waveforms

• what is the effective voltage of 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
Vm ?
? 2?f 2?60 377rads/sec
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effective values of periodic waveforms
To understand what Vm might be we need a good
definition of effective value. The effective
value of a current is the steady current (dc)
that transfers the same amount of average (real)
power as the given varying current. The effective
value of a voltage is the steady voltage (dc)
that transfers the same amount of average (real)
power as the given varying voltage.
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effective values of periodic waveforms
effective value is the square root of the mean
of the squared values also known as (a-k-a) the
root-mean-square
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effective values of periodic waveforms
What is the average value of the square of the
cosine of a sinusoid?
By inspection it is ½, going from 0? 1 and back
again
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effective values of periodic waveforms
Since 120 240VAC are effective values this
means they are root-mean-square (rms) values
So What are the amplitudes of these sinusoidal
waveforms if viewed on a scope?
Vm of 120VAC ? LM1 Vm of 240VAC ? LM2
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NOTE

• In practice, electrical engineers must take care
  to notice (or determine) whether a sinusoidal
  voltage is being expressed in terms 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 zero phase angle unless otherwise
  specified, current angles are then expressed
  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 as
  phase angle of the supply voltage
• V and I are considered in phase
• And V RI (where V and I are rms values)
• 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?

Write your answers as LM3
Write your answer as LM4
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Capacitors

• V ZI (1/j?C)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 is zero
  (stores and releases) ideal
• Sample exercise what is rms current and i(t) in
  a 5 ?F capacitor used to balance the power factor
  on a motor in the UK (assume 240V/50Hz)

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Capacitor example

• Sample exercise what is current in a 5?F
  capacitor used to balance the power factor on a
  motor in the UK (assume 240V/50Hz)
• V (1/j?C)I so I Vj?C
• (240)j(2?50)(5x10-6) 0.377j A

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You try it capacitor example

• Sample exercise find the rms current and write
  an equation for the current as a function of time
  when a 500 nF capacitor is used in a 120VAC/60Hz
  system
• Remember V (1/j?C)I so I Vj?C

Write your answer as LM5
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Inductors

• V ZI j?LI (NOTE Z j?L)
• 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 10?50o V, and L2H, ?100 rads/s
• what does I ?
• I V/ j?L V/ j200 10?50o/200?90o
• I 0.05?-40o A

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Impedance (Z) and Reactance (X)
Z V/I
Z R jX
X ?L or X -1/?C
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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 its impedance (Z) prior was 2? resistive and
  no reactance and after has 4? of resistance and
  10mH of inductance?
• V 240?0o V, Z 2?0o ? so I 120?0o A
• V 240?0o V, Z 4 j(314)0.01 5.09?38o ?
• so I V/Z 240?0o/ 5.09?38o 47.2?-38o A

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Memory aid

• ELI the ICE man
• 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 the
  voltage supplied by the source
• Phase angle shift may result in current depending
  on whether there is net L or C in the circuit
• Resistive elements are the only circuit
  components that dissipate any real energy

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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 and reactive.
• The power factor is a measure of what part of the
  overall power in the system is real

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Power factor

• Typically, average power, watts, also called real
  power is 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
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complex impedance (Z) triangle
j?L
Power factor phase angle
1/j?C -j/?C
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power factor

• since cosine function is even with respect to
  positive or negative phase angles
• i.e cos 45o cos -45o
• we resolve difficulty by labeling power factor
  as either leading or lagging
• If ?V - ?I gt 0 ? pf is lagging
• If ?V - ?I lt 0 ? pf is leading
• NOTE If pf is leading or lagging THEN the
  current (I) is leading or lagging w.r.t. voltage
  (V) too!

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Resistive and Reactive Loads
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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
? ?S? apparent power
Q reactive power
P real power
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complex power triangle
note ? ?V - ?I
?S? apparent
j
units Volt-Amps
Q reactive
units VARs
P real
units Watts
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Express S,P Q with rms values for V and I LM6
note ? ?V - ?I
?S? apparent
j
units Volt-Amps
Q reactive
units VARs
P real
units Watts
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Power factor example

• A 240-V induction motor draws 30 amps of current
  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 as
  LM7 on blank area next to todays work

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Power factor correction?
Why correct power factor? 1/5th of all grid
losses may be due to poor power factor (gt2B/yr),
the outage of 2003 was made more severe by
extremely high reactive demand, all transformers
are rated on kVA not watts, all these economic,
efficiency and reliability benefits can be
achieved at a very low cost
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Power factor correction
Adding capacitive impedance in parallel with the
load enables the current to oscillate 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 the
system voltage in which they are installed, and
PF correction is a straightforward engineering
design
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Power factor correction example

• 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
  planned that will increase power demand 125kW.
  How many kVAR of capacitance should be added to
  avoid purchasing additional 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 to Real/PF 885/(0.8) 1106kVA
  gt 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
  planned that will increase power demand 125kW.
  How many kVAR of capacitance should be added to
  avoid purchasing additional substation
  transformer capacity?
• For substation to handle the growth, power factor
  must improve to 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 capacitor bank 664 466 198 kVAR
• Specify a gt 200 kVAR cap bank at industrial
  customers service entrance switchgear

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3-wire single phase residential wiring

• Figure 2.11 page 68 is helpful
• Figure 2.12 is not as clear
• Panel box is best

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3-phase systems

• Benefits
• Generators and motors work more efficiently in
  torque transfer and have higher stability (they
  run smoother 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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3? current
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3? neutral current
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New homework

• HW 2 due next Monday
• will be posted on web
• 2.1, 2.2, 2.3, 2.6, 2.7, 2.8, 2.10