ece 333 renewable energy systems lecture 3:basic circuits, complex power prof. tom overbye dept. of...
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ECE 333 Renewable Energy Systems
Lecture 3:Basic Circuits, Complex Power
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
Announcements
• Be reading Chapters 1 and 2 from the book• Be reading Chapter 3 from the book• Homework 1 is 1.1, 1.11, 2.6, 2.8, 2.14. It will be
covered by the first in-class quiz on Thursday Jan 29• As mentioned in lecture 2, your two lowest
quiz/homework scores will be dropped
2
Engineering Insight: Modeling
• Engineers use models to represent the systems we study
• Guiding motto: “All models are wrong but some are useful” George Box, 1979
• The engineering challenge, which can be quite difficult sometimes, is to know the limits of the underlying models.
3
• Ideal Voltage Source
• Ideal Current Source
Basic Electric Circuits
sv
+
-
i
+
-
si
Load
Load
si
i
i
v
sv
v
4
Example – Power to Incandescent Lamp
• Find R if the lamp draws 60W at 12 V
• Find the current, I• What is P if vs doubles and R stays the same? 240W
12sv V+
-
i
Load
125
2.4v
i AR
2vP v i
R
60P W
2 2122.4
60v
RP
5
Equivalent Resistance for Resistors in Series and Parallel
• Resistors in series – voltage divides, current is the same
v
+
-
1R
2R
NR
i
1 2EQ NR R R R
+
-
v
i
nodevoltages
6
Equivalent Resistance for Resistors in Series and Parallel
• Resistors in parallel – current divides, voltage is the same
1 2
11 1 1
...EQ
N
R
R R R
Simplification for 2 resistors
1 2
1 2EQ
R RR
R R
+
-
i
i
1R 2R NRv
branch currents
v
7
Voltage and Current Dividers
i1R
2R
+
-
v+
-outv
1 2EQ
v vi
R R R
2outv i R
2
1 2out
Rv v
R R
i
1R2R
v+
-2i1i 2
2
vi
R
1 2
1 2EQ
R Rv i R i
R R
12
1 2
Ri i
R R
Voltage Divider
Current Divider
8
Wire Resistance
• For dc systems wire resistance is key; for high voltage ac often the inductance (reactance) or capacitance (susceptance) are limiting
• Resistance causes 1) losses (i2R) and 2) voltage drop (vi)• Need to consider wire resistance in both directions
9
AC: Phase Angles
• Angles need to be measured with respect to a reference - depends on where we define t=0
• When comparing signals, we define t=0 once and measure every other signal with respect to that reference
• Choice of reference is arbitrary – the relative phase shift is what matters
• Relative phase shift between signals is independent of where we define t=0
10
Example: Phase Angle Reference
• Pick the bottom wave as the reference
1 sin4
v V t
2 sin 0v V t
1 sin 0v V t
2 sin4
v V t
1 2 4
1 2 4
• Or pick the top as the reference- it does not matter!
11
Important Properties: RMS
• RMS = root of the mean of the square• RMS for a periodic waveform
• RMS for a sinusoid (derive this for homework)
21( )
o
o
t T
RMSt
V v tT
let ( ) cos( )pv t V t
2p
RMS
VV
T period
12
In 333 we are mostly only concerned withsinusoidals
Important Properties:Instantaneous Power
• Instantaneous power into a load
p(t)= ( ) ( )v t i t( )v t
+
-
( )i t
( )= cos( )
( )= cos( )
p V
p I
v t V t
i t I t
( )= ( ) ( )p t v t i t
( )= cos cos 22p p
V I V I
V Ip t t
“Load sign convention” with current and power into load
positive
1cos cos cos cos
2
Identity
13
Important Properties: Average Power
• Average power is found from
• Find the average power into the load (derive this for homework)
( )= cos cos 22p p
V I V I
V Ip t t
1( )
o
o
t T
t
P p t dtT
T period
P= cos2p p
V I
V I P= cosRMS RMS V IV I or
14
Important Properties:Real Power
• P is called the Real Power
• cos(θV-θI) is called the Power Factor (pf)
• We’ll review phasors and then come back to these definitions…
P= cosRMS RMS V IV I
P= Re{VI*}
15
Review of Phasors
• Phasors are used in electrical engineering (power systems) to represent sinusoids of the same frequency
• A quick derivation…
2 f ( ) cos( )pA t A t
1cos( )
2jx jxx e e
cos( )2
j t j tAA t e e
Identity
Ap denotes the peak value of A(t)
16
Review of Phasors
• Use Euler’s Identity
• Written in phasor notation as
cos sinjxe x j x Identity
( ) cos( )
( ) Re
p
j t jp
A t A t
A t A e e
cos Re jxx e
or jRMS RMSA A e A A Tilde denotes a phasor
or jA A e A A Other, simplified notation
Regardless of what notation you use, it helps to be consistent.
Note, a convention- the amplitude used here is the RMS value, not the peak value as used in some other classes!
17
Why Phasors?
• Simplifies calculations– Turns derivatives and integrals into algebraic equations
– Makes it easier to solve AC circuits
dA j A
dt
R( )
R i (t)= Rv tR
L( )
L (t)= Ldi tv L
dt
C( )
C (t)= Cdv ti C
dt
=V
RI
=Lj IV Vj L
I
I=Cj V 1VI j C
LjX j L
1cjX j
C
18
Why Phasors: RLC Circuit
R j L
1j C
R L
C
( ) cosv t V t V V
1( ) ( ) ( )
div t Ri t L i t dt
dt C
I( )i t
1V RI j LI I
j C
Solve for the current- which circuit do you prefer?
+
-
+
-
19
RLC Circuit Example
( ) 2 100cos 30v t t
3LX L
2 f
60Hzf
2 24 3 5Z 1 3
tan 36.94Z
100 30
20 6.95 36.9
VI
Z
( ) 2 20cos( 6.9 )i t t 20
Complex Power
VV= RMSV
II= RMSI
Asterisk denotes complex conjugate
*
*
VI
VI cos sin
RMS RMS V I
RMS RMS V I RMS RMS V I
V I
V I jV I
S
Apparent
power
P
Real Power
Q
Reactive Power S = P+jQ
SQ
P(θV-θI)
Power triangle
21
Apparent, Real, Reactive Power
• P = real power (W, kW, MW)• Q = reactive power (VAr, kVAr, MVAr)• S = apparent power (VA, kVA, MVA)• Power factor angle• Power factor
*
*
*
VI
VI
VI cos sin
RMS RMS V I
RMS RMS V I RMS RMS V I
S P jQ
V I
V I jV I
V I cos( )pf
22
Apparent, Real, Reactive Power
• Remember ELI the ICE man
ELI ICEInductive loads
I lags V (or E)
Capacitive loads
I leads V (or E)
S Q
P(θV-θI) P
QS(θV-θI)
Q and θ positive Q and θ negative
(producing Q)
“Load sign convention” – current and power into load are assumed positive
23
Apparent, Real, Reactive Power
• Relationships between P, Q, and S can be derived from the power triangle just introduced
• Example: A load draws 100 kW with leading pf of 0.85. What are the power factor angle, Q, and S?
cos
sin
P S
Q S
-1cos 0.85 31.8
100 kW117.6 kVA
0.85Q=117.6 kVA sin( 31.8 ) 62.0 kVAr
S
24
Conservation of Power
• Kirchhoff’s voltage and current laws (KVL and KCL)– Sum of voltage drops around a loop must be zero– Sum of currents into a node must be zero
• Conservation of power follows – Sum of real power into every node must equal zero– Sum of reactive power into every node must equal zero
25
Conservation of Power Example
* 100 30 20 6.9 2000 36.9S VI 36.9 0.8 laggingpf
Resistor, consumed power
Inductor, consumed power
* 4 20 6.9 20 6.9 1600 R R RS V I 2 1600 WR RP I R
* 3 20 6.9 20 6.9 1200 L L LS V I j j 2 1200 VArL L LQ I X
0 VArRQ
0 WRP 26
Power Consumption in Devices
• Resistors only consume real power
• Inductors only consume reactive power
• Capacitors only produce reactive power
2R RP I R
2L L LQ I X LX L
2C C CQ I X
1CX
C
2R
RV
PR
2L
LL
VQ
X
2C
CC
VQ
X
27
Example
40000 0400 0 Amps
100 0I
40000 0 5 40 400 0
42000 16000 44.9 20.8 kV
V j
V j
* 44.9 20.8 400 0
17.98 20.8 MVA 16.8 6.4 MVA
S VI
S j
Solve for the total power delivered by the source
28
Reactive Power Compensation
• Reactive compensation is used extensively by utilities
• Capacitors are used to correct the power factor• This allows reactive power to be supplied locally• Supplying reactive power locally leads to decreased
line current, which results in– Decrease line losses– Ability to use smaller wires– Less voltage drop across the line
29
Power Factor Correction Example
• Assume we have a 100 kVA load with pf = 0.8 lagging, and would like to correct the pf to 0.95 lagging
S 80 60 kVAj 1cos (0.8) 36.9 We have:
We want: 1cos (0.95) 18.2desired
S Qdes.=?
P=8018.2
.tan(18.2 ) desQP
. tan(18.2 )*40 26.3 kVArdesQ This requires a capacitance of:
60 26.3 33.7 kVArcapQ Q=60 Q=-33.7
PQdes=26.3
P
30
Distribution System Capacitors for Power Factor Correction
31