Transcript
Presented byBrenden Motte, Alissa Olson & Long Pham
AC Power Engineering Review
Why not just use DC?
New York Cityutility lines,
1980
• DC was first use for electricity transmission.• Then “The war of current.” NY, 1880s:
▫ AC Vs DC ~ Testla Vs Edison ~ Employee Vs Boss. ▫ Edison’s least favorite of Tesla’s “impractical” ideas was the concept of using alternating current (AC) to bring electricity to
the people. Edison insisted that his own direct current (DC) system was superior, in that it maintained a lower voltage from power station to consumer, and was, therefore, safer. “Direct current is like a river flowing peacefully to sea, while alternating current is like a torrent rushing violently over a precipice.” - Tom
▫ Tesla insisted that he could increase the efficiency of Edison’s prototypical dynamos (by changing from DC to AC generator). Edison promised him $50,000 if he succeeded. Tesla worked around the clock for several months and made a great deal of progress. When he demanded his reward, Edison claimed the offer was a joke, saying, “When you become a full-fledged American, you will appreciate an American joke.” Ever prideful, Tesla quit, and spent the next few months picking up odd jobs across New York City. Nikola Tesla: ditch digger.
▫ Tesla eventually raised enough money to found the Tesla Electric Light Company, where he developed several successful patents including AC generators, wires, transformers, lights, and a 100 horsepower AC motor. The most significant contribution to the early success of ac was his patenting of the poly-phase ac motor in 1888.
▫ Always more of a visionary than a businessman, Tesla ended up selling most of his patents (for the healthy but finite sum of $1 million) to George Westinghouse, an inventor, entrepreneur, and engineer who had himself been feuding with Edison for years. Their partnership, made the eventual popularizing of AC that much more bitter for Edison.
• AC electrical energy has been the most convenient form of energy to be generated, transmitted, and distributed.
AC vs DC
• While AC was perfectly adequate for the conditions of much of the 19th & 20th century, the needs of the 21st century are showing its limits.
• We are facing a revolution in the way our electricity is produced and used. More and more electricity is being generated from renewable sources of energy in remote areas: hydropower plants (mountains far from urban centers), wind farms (people tend not to live in windy areas), offshore wind farms (have higher capacity factor, better alignment with peak demand), etc. DC is the only technology that allows power to be transmitted economically over very long distances, and DC is the type of power produced by photovoltaic panels.
• On the consumer side, more and more equipment runs on DC: computers, cell phones, LED lights, CFLs, high efficiency motor drives which found in new HVAC systems, industrial, etc. ABB estimates the savings from using DC instead of AC in buildings could be in the order of 10 to 20 percent.
DC vs AC
Reference: http://www.abb.com/cawp/seitp202/c646c16ae1512f8ec1257934004fa545.aspx
•Part 1: AC Circuits Analysis▫Steady-state sinusoidal response, Impedance Model.
•Part 2: AC Power Analysis▫Power in AC circuits, Power Factor, Power factor corrections, Poly-Phase
Circuits.
•Part 3: Elements of the AC Power Systems▫Basics elements comprising the AC grid: Power Generator, Transformers,
Capacitor Banks, Transmission circuits, etc.
AC Power: Today’s topics
•Complex Number: A Quick Review•Resistor, capacitor, Inductor: A practical explanation•Wave form of a signal & characteristic of Sinusoids signal▫Period, frequency, radian frequency (or angular frequency), phase leading
& lagging, etc.•Sinusoidal response of RC Network: usual approach•Phasor domain analysis or Frequency domain analysis•Sinusoidal response of RC Network: Impedance model•Using Impedance Model to solve AC circuits•Examples
Part 1: AC Circuits
Rectangular form Polar form Exponential form
Complex Number: A Quick Review
a
b
Im
Re
0
|A|
A
Note: Most calculator can do complex arithmetic & convert between Rectangular form to/from polar form
Complex NumberExample:• Evaluate the following complex
numbers into rectangular form:a.
b.
Solution:a. –15.5 +
j13.67b. 8.293 + j2.2]605j4)1j2)([(5 o
oo
3010j43
403j510
•Solution:•Problem:
Using complex number Solving Trigonometric nightmare
1 20cos 45 Vv t t
2 10sin 60 Vv t t
1 2Find ?sv v v
1 20 45 V V
2 10 30 V V
1 2
20 45 10 3014.14 j14.14 8.660 j523.06 j19.14
29.97 39.7 V
sV V V
𝒗𝑺 (𝒕 )=𝟐𝟗 .𝟗𝟕𝒄𝒐𝒔 (𝝎𝒕−𝟑𝟗 .𝟕° )𝑽 Complex number doesn’t make life more complicated but more simple!
Why can we do this?𝒆± 𝒋𝝋=cos𝝋 ± 𝒋 sin𝝋
Resistors• Purely resistive loads are almost non-
existent in AC networks. AC network loads consist primarily out of inductive and to lesser extent capacitive loads, both in combination with resistive loads. It is therefore VERY important to understand the characteristics of capacitors and inductors in an AC environment.
• Inductors and capacitors are energy storage elements. The difference lies in how and the type of energy that is stored by each.
Inductors• When current flows, a magnetic field is created. • Energy provided by the current is stored in the
magnetic field. The stronger the current or higher the number of coils (higher inductance), the greater the stored amount of energy will be.
• Inductor does however have a limit as to the amount of energy it can store and the rate at which it can store the energy. (~Saturated)
• Energy stored in inductor’s magnetic field can be retrieved.
• Inductance is represented by: L and measured in Henry: (H)
It is clear that an inductor stores energy in the form of a magnetic field created by current flowing through the inductor coil.
Inductors
Large 50 MVAR three-phase iron-core loading inductor at German utility substation
Inductors characteristics• When you pushing a car, it will slowly increase it’s velocity (~kinetic energy).
Similarly, when you apply a voltage across an inductor, it will slowly increase it’s current (~magnetic energy). Inductor’s current cannot change instantly just as car’s velocity. (voltage can change instantly)
• When you stop pushing, the car will continue to run. Similarly, when you stop applying voltage, the inductor’s current will continue to flow.
• What happens if something block the car from running? What happens if something block the inductor’s current from flowing?
• The car will try its best to run until it can’t run anymore; all the kinetic energy stored in the car will dissipate to the blocker. A big force will created by the car’s effort to try to maintain it’s velocity.
• Similarly, the inductor will try its best to flow the current until it can’t run anymore; all the magnetic energy store in the field will dissipate to the blocker. A big voltage is created by the inductor in its effort to try to maintain it’s current.
• Mathematically, the relation between inductor’s current & voltage is the same as the relation between car’s velocity & force:
𝒗=𝑳 𝑑𝒊𝑑𝑡 𝑭=𝒎 𝑑𝒗
𝑑𝑡
Capacitors• When apply a voltage to a capacitor, a electrical field
is created between the capacitor’s plates.• Energy provided by a voltage difference is stored in
the electric field. The higher the voltage or bigger capacitor (higher capacitance), the stronger the electric field and more energy is stored inside the capacitor.
• Capacitor does however have a limit as to the amount of energy it can store.
• Energy stored in capacitor’s magnetic field can be retrieved.
• Capacitance is represented by: C and measured in Farad: (F)
It is clear that a capacitor stores energy in the form of a electric field created by potential difference across the capacitor plates.
Capacitors
𝒊=𝑪 𝑑 𝒗𝑑𝑡
• A waveform is the shape or form of a signal against time, physical medium or an abstract representation.
• Common periodic waveforms are: Sine, Square, Triangle, etc.
Example:▫ Electrocardiogram: 1Hz▫ Main power: 50Hz▫ Audio signal: 20Hz▫ Wifi signal: 2.4Ghz▫ GPS signal L1 band: 1575.42Mhz▫ GLONASS L1 signal: 1602Mhz
• Sinusoids’ important because signals can be represented as a sum of sinusoids. Response to sinusoids of various frequencies -- aka frequency response -- tells s a lot about the system.
Waveform of a signal
• General form:
Sinusoidal signal: 𝒗 (𝒕 )=𝑽𝒎 𝒔𝒊𝒏 (𝝎𝒕 )
is plotted vs -> period:
is plotted vs -> period:
▫ : radian frequency or angular frequency (rads) = ▫ : frequency (Hz)▫ : signal’s period (s)▫ : Amplitude▫ : period (s)▫ : phase angle (rads)
Sinusoidal response of RC Network
{𝒗 𝑰 (𝒕 )=𝑽 𝒊𝒄𝒐𝒔 (𝝎𝒕 ) 𝒇𝒐𝒓 𝒕 ≥𝟎(𝑽 𝒊𝒓𝒆𝒂𝒍 )𝒗 𝑰 (𝒕 )=𝟎 𝒇𝒐𝒓 𝒕<𝟎𝒗 (𝟎 )=𝟎 𝒇𝒐𝒓 𝒕=𝟎
Given input voltage source is sinusoidal. Find the voltage across the capacitor:
Let’s try to solve a typical problem using what we’ve known in DC circuit analysis. (KVL, KCL, etc.)
Current flow out of node i equal 0:
Usual approach
𝑣−𝑣𝐼
𝑹 +𝑪 𝑑𝑣𝑑𝑡 =0
𝑹𝑪 𝑑𝑣𝑑𝑡 +𝑣=𝑣𝐼
𝑹𝑪 𝑑𝑣𝑑𝑡 +𝑣=𝑽 𝑖𝑐𝑜𝑠𝜔𝑡
That was easy!
Method of homogeneous and particular solutions:
1. Find the homogeneous solution. - also called the natural response
2. Find the particular solution. - also called the force response
3. The total solution is the sum of the particular and homogeneous solutions. Use the initial conditions to solve for the remaining constants.
How to solve DE𝑹𝑪 𝑑𝑣
𝑑𝑡 +𝑣=𝑽 𝑖𝑐𝑜𝑠𝜔𝑡
{ 𝒕 ≥𝟎 𝒗 (𝟎 )=𝟎
Solving differential equations is actually quite easy –commonly guesswork and applying patterns
𝒗=𝒗𝑷+𝒗 𝑯
Note: The homogeneous equation is derived from the original nonhomogeneous equation by setting the driving function, , to zero. The homogeneous solution is also called the natural response of the circuit because it depends only on the internal energy storage properties of the circuit and not on the external inputs.
1. Find the homogeneous solution:
𝑹𝑪 𝑑𝑣𝑑𝑡 +𝑣=0
We assume the solution of the form: (Homogeneous solution for any linear constant-coefficient ODE is always of this form.)
𝐴𝑒𝑠𝑡+𝐴𝑒𝑠𝑡
𝑅𝐶 =0⟹𝑠=− 1𝑅𝐶
𝑣𝐻=𝐴𝑒−𝑡 /𝑅𝐶
•First try: •Second try: •Third try:
2. Find the particular solution: (guesswork!)
𝑹𝑪 𝑑𝑣𝑑𝑡 +𝑣=𝑽 𝑖𝑐𝑜𝑠𝜔𝑡
It WORK!But trig. Nightmare!
(is there an easier way to find Vp?)
3. The total solution:
𝑹𝑪 𝑑𝑣𝑑𝑡 +𝑣=𝑽 𝑖𝑐𝑜𝑠𝜔𝑡
¿𝑽 𝑖
√𝟏+(𝜔𝑹𝑪 )𝟐cos (𝜔𝑡+𝜑 )+ 𝐴𝑒−𝑡 /𝑅𝐶
𝒗=𝒗𝑷+𝒗𝑯
Given
𝑨=𝑽 𝑖
√𝟏+(𝜔𝑹𝑪 )𝟐cos (𝜑 )
𝒘𝒉𝒆𝒓𝒆 φ= tan−𝟏 (−𝜔 𝑹𝑪 )
Sinusoidal Steady State (SSS)
𝒗=𝑽 𝑖
√𝟏+(𝜔 𝑹𝑪)𝟐cos (𝜔𝑡+𝜑 )+𝐴𝑒−𝑡 / 𝑅𝐶 { φ=tan−𝟏 (−𝜔𝑹𝑪 )
𝑨=𝑽 𝑖
√𝟏+(𝜔𝑹𝑪 )𝟐cos (𝜑 )
In AC Power, we are usually interested only in the particular solution for sinusoids, i.e. after transients have died.
0
All information about SSS response is contained in Is there a simpler way to get ?
When as
Phasor Analysis“I have found the equation that will enable us to transmit electricity through alternating current over thousands of miles. I have reduced it to a simple problem in algebra.”Charles Proteus Steinmetz, 1893
The use of complex numbers to solve ac circuit problems was first done by German-Austrian mathematician and electrical engineer Charles Proteus Steinmetz in a paper presented in 1893. He is noted also for the laws of hysteresis and for his work in manufactured lighting. Steinmetz was born in Breslau, Germany, the son of a government railway worker. He was deformed from birth and lost his mother when he was 1 year old, but this did not keep him from becoming a scientific genius. Just as his work on hysteresis later attracted the attention of the scientific community, his political activities while he was at the University at Breslau attracted the police. He was forced to flee the country just as he had finished the work for his doctorate, which he never received. He did electrical research in the United States, primarily with the General Electric Company. His paper on complex numbers revolutionized the analysis of ac circuits, although it was said at the time that no one but Steinmetz understood the method. In 1897 he also published the first book to reduce ac calculations to a science.
• A phase vector, or phasor is a complex number that represents the amplitude and phase of a sinusoidal signal whose amplitude, frequency, and phase are time-invariant.
• Frequency is common to all signal in our analysis. Phasor allow this common to be factored out, leaving just the amplitude and phase features. The result is that trigonometry reduces to algebra, and linear differential equations become algebraic ones.
• Phasor is a complex number. So it can be represent in one of the three forms of complex number: Rectangular, Polar, Exponential.
Phasor: A useful way to think about sine wave
A phasor can be considered a vector rotating about the origin in a complex plane
The sum of phasors as addition of rotating vectors
Time Domain Vs Phasor Domain representation of sinusoidal signals
• is a time function• (frequency domain) is a complex number. • Phasor domain (or frequency domain) analysis is very useful not only in AC circuits
analysis but in other fields as well. MATLAB Simulink provides a variety of simulation methods on: Time domain, Discrete domain, Phasor domain.
Note: Phasor analysis applies only when frequency is constant; when it is applied to two or more signals only if they have the same frequency.
Time domain Phasor domain
𝝎
Example: Convert SS between Time Domain & Phasor Domain
Time domain Phasor domain
𝑉=3 .2∠30 °=3 .2𝑒( 𝑗 𝜋/6 )𝑉
𝐼=1 .5∠ (−120 ° ) 𝐴¿1 .5 cos (𝜔𝑡+60 ° −180 ° )
𝑉=0 .1∠𝜋=0 .1𝑒( 𝑗 𝜋 )𝑉
𝐼=0 .18∠120 ° 𝐴𝑖 (𝑡 )=0 .18 cos (𝜔𝑡+120 ° ) 𝐴
𝐼=1+ 𝑗1¿√2∠ 45 °𝑖 (𝑡 )=√2 cos (𝜔𝑡+45 ° )
𝜔𝝎
Sinusoidal Signals: Arithmetic ComparisonsTime domain
𝑣( )= −𝑡 𝑋𝑐𝑜𝑠𝜔𝑡 𝑌𝑠𝑖𝑛𝜔𝑡 Phasor domain𝑉= +𝑋 𝑗𝑌 Adding & Scaling trigonometry Adding & scaling complex
numbers
Derivative and Integral of sin & cos Multiply & Divide by
Example: Solving DE using Phasor Domain
Answer: i(t) = 4.642cos(2t + 143.2o) A
)752cos(50384 tdtdiidti
Normal Guesswork + Trig. nightmare
- Convert to Phasor domain - Solve algebraic equation - Convert back to Time
domain
Find i(t):
Voltage / Current relationshipsTime domain Phasor domain
R
L
CComplicated! Simple ohm Law:
The Impedance Model in Phasor domain
Voltage & Current in Phasor domain follow simple ohm Law:
This allow us to generalize Z in phasor domain as the of R, L, C. impedance
Back to the RC example…Time domain Phasor domain
Calculator
DONE!
𝜔
𝜔
ExampleTime domain Phasor domain
𝜔
Another example of Sinusoidal Steady state response (SSS)Remember, we want only the Steady-state response to sinusoid: SSS SSS: Sinusoidal Steady-state response
Calculator
Solving AC Circuits The Easy Way
•The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms Ω.
where R = Re(Z) is the resistance and X = Im(Z) is the reactance. Positive X is for L and negative X is for C.
•The admittance Y is the reciprocal of impedance, measured in siemens (S).
Impedance and Admittance
VI
ZY
1
jXRZIV
IV
IVZ ziv
M
M
iM
vM
||)(
Series Impedances
Parallel Impedance
Wye – Delta conversion
Impedance example
Impedance example
Impedance example
Basic circuit example
Basic example
Thévenin and Norton equivalent circuits apply in AC analysis• Equivalent voltage/current is complex and frequency dependent
AC Equivalent Circuits
Load+V–
I
Source
VT(jω)+–
ZT
Load
+
V
–
I
IN(jω)
ZN Load
+
V
–
I
Norton EquivalentThévenin Equivalent
Thevenin and Norton Transformation
Thevenin transform
Norton transform
1. Remove the load (open circuit at load terminal)2. Zero all independent sources
Voltage sources short circuit (v = 0) Current sources open circuit (i = 0)
3. Compute equivalent impedance across load terminals (with load removed)
Computation of Thévenin and Norton Impedances:
NB: same procedure as equivalent resistance
ZL
Z1
Vs(jω)
+–
Z3
Z2
Z4
a
b
Z1 Z3
Z2
Z4
a
b
ZT
1. Remove the load (open circuit at load terminals)2. Define the open-circuit voltage (Voc) across the load terminals3. Chose a network analysis method to find Voc
node, mesh, superposition, etc.4. Thévenin voltage VT = Voc
Computing Thévenin voltage:
Z1
Vs(jω)
+–
Z3
Z2
Z4
a
b
Z1
Vs(jω)
+–
Z3
Z2
Z4
a
b
+VT
–
1. Replace the load with a short circuit2. Define the short-circuit current (Isc) across the load terminals3. Chose a network analysis method to find Isc
node, mesh, superposition, etc.4. Norton current IN = Isc
Computing Norton current:
Z1
Vs(jω)
+–
Z3
Z2
Z4
a
b
Z1
Vs(jω)
+–
Z3
Z2
Z4
a
b
IN
• Example: find the Thévenin equivalent• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
Rs
vs(t) +–~ RL
L
C +vL
–
• Example: find the Thévenin equivalent• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
Rs
vs(t) +–~ RL
L
C +vL
–
1. Note frequencies of AC sources
Only one AC source: ω = 103 rad/s
• Example: find the Thévenin equivalent• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
Rs
vs(t) +–~ RL
L
C +vL
–
1. Note frequencies of AC sources2. Convert to phasor domain
Zs
ZLD+–~
ZL
ZCVs(jω)
• Example: find the Thévenin equivalent• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF 1. Note frequencies of AC sources
2. Convert to phasor domain3. Find ZT
• Remove load & zero sources
Zs
ZLD+–~
ZL
ZCVs(jω)
Zs
ZL
ZC
9182.033.82414.6550
1
)/1()()/1)((
||
2
jLCLjR
CjLjCjLjR
ZZZZ
S
S
LCST
• Example: find the Thévenin equivalent• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF 1. Note frequencies of AC sources
2. Convert to phasor domain3. Find ZT
• Remove load & zero sources
4. Find VT(jω)• Remove load
Zs
ZLD+–~
ZL
ZCVs(jω) NB: Since no current flows in the
circuit once the load is removed:
ST VV
9182.033.82 TZ
Zs
+–~
ZL
ZCVs(jω)
+VT(jω)
–
• Example: find the Thévenin equivalent• ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF
9182.033.82 TZST VV
Zs
ZLD
+–~
ZCVs(jω)
ZT
+–~
VT(jω)
ZLD
ZL
Thevenin and Norton Equivalent Circuits
Thevenin and Norton Equivalent Circuits
Thevenin and Norton Equivalent Circuits
08 2 8 2 8.246 14.04
1 1 2 2 3 1 3.162 18.43j j
j j j
I
6 2(1 ) 8 2OC j j V
2 0 A 6 0 V AC
-j1
+
-
j2
I0
2 0 A 6 0 V AC
-j1
+
- VOC
+
-
-j1
1 1TH j Z
8 2j AC
-j1
+
-
j2
I00 2.608 32.47 I
Obtain current Io using Norton’s theorem at terminals a-b.
Obtain the Thevenin equivalent circuit atterminals a-b:
Source Transformation
Source Transformation
Mesh Analysis
Nodal Analysis
•Usual procedures for DC circuits apply.•However, phasor transformation must be carefully carried out if the
circuit has sources operating at different frequencies
•A different phasor circuit for each source frequency because impedance is a frequency-dependent quantity.
•For sources of different frequencies, the total response must be added in the time domain
•DO NOT ADD INDIVIDUAL RESPONSES IN THE PHASOR DOMAIN IF THE SOURCES HAVE DIFFERENT FREQUENCIES.
Super position for AC Circuits
SuperpositionFind I0.
Superposition
Circuit exampleTime domain Phasor domain
Circuit example
End of part 1
Thank you!
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