[email protected] engr-43_lec-09-2_complex_power.ppt 1 bruce mayer, pe engineering-43:...
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[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PERegistered Electrical & Mechanical Engineer
Engineering 43
Chp 9 [5-7]Chp 9 [5-7]Complex Complex PowerPower
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Outline – AC SS Power cont.Outline – AC SS Power cont. Effective or RMS Values
• Heating Value for Sinusoidal Signals
Power Factor• A Measure Of The Angle Between Current
And Voltage Phasors within a Load
Power Factor Correction• Improve Power Transfer To a Load By
“Aligning” Phasors
Single Phase Three-Wire Circuits• Typical HouseHold Power Distribution
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Outline – AC Steady State PowerOutline – AC Steady State Power
Instantaneous Power Concept• For The Special Case Of Steady State
Sinusoidal Signals
Average Power Concept• Power Absorbed Or Supplied During in
Integer Number of Complete Cycles
Maximum Average Power Transfer• When The Circuit Is In Sinusoidal Steady
State
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Power FactorPower Factor Consider A Complex
Current Thru a Complex Impedance Load
The Current and Load-Voltage Phasors (Vectors) Can Be Plotted on the Complex Plane
By Ohm & Euler
in the Electrical Power Industry θZ is the Power Factor Angle, or Simply the Phase Angle
LZiMI
vMV
iv
VIz
ivz
izv
or
IZV
ZIV
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Power Factor contPower Factor cont The Phase Angle Can
Be Positive or Negative Depending on the Nature of the Load
Typical Industrial Case is the INDUCTIVE Load• Large Electric Motors are
Essentially Inductors
Now Recall The General Power Eqn
Measuring the Load with an AC DMM yields• Vrms
• Irms
Zrmsrmsivrmsrms
ZMMivMM
IVIVP
IVIVP
cos)cos(
cos2
1)cos(
2
1
0MVV
e)(capacitiv
leadscurrent 090 z
)(inductive
lagscurrent 900 z
ZZ
ZVI
0
V is the BaseLine
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Power Factor cont.2Power Factor cont.2 The Product of the
DMM Measurements is the APPARENT Power
The Apparent Power is NOT the Actual Power, and is thus NOT stated in Watts.• Apparent Power
Units = VA or kVA
Now Define the Power Factor for the Load
Some Load Types
rmsrmsapparent IVP pfIVP
P
Ppf
rmsrmsactual
zivapparent
actual
and
cos)cos(
inductive pure
inductiveor lagging
resistive
capacitiveor leading
capacitive pure
90
900
0
10
01
09010
900
z
z
z
pf
pf
pf
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
pf – Why do We Care?pf – Why do We Care? Consider this case
• Vrms = 460 V
• Irms = 200A• pf = 1.5%
Then• Papparent = 92kVA
• Pactual =1.4 kW
This Load requires The Same Power as a Hair Dryer
However, Despite the low power levels, The WIRES and CIRCUIT BREAKERS that feed this small Load must be Sized for 200A!• The Wires would be
nearly an INCH in Diameter
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt8
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example Example Power Factor Power Factor The Local Power
Company Services this Large Industrial Load
Find Irms by Pwr Factor
Then the I2R Loses in the 100 mΩ line
Improving the pf to 94%
1.0
kW100080V4
Power company
rmsrms
rmsrms
VpfPI
pfIVP
22
22 1
pfV
RPRIP
rms
linelinermslosses
234.4
)(707.0
1
480
1.010)707.0(
22
10
kW
WpfPlosses
13.134.4
)(94.0
1
480
1.010)94.0(
22
10
kW
WpfPlosses
kWkWPsaved 77.334.487.0
I lags V
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt9
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example - Power Factor contExample - Power Factor cont For This Ckt The
Effect of the Power Factor on Line Losses
1.0
kW100080V4
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt10
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complex PowerComplex Power Consider a general Ckt
with an Impedance Ld Mathematically
For this Situation Define the Complex Power for the Load:
Converting to Rectangular Notation
*rmsrmsIVS
Ziv
ivrmsrms
irmsvrms
irmsvrms
IV
IV
IV
:recall
*
S
S
S
)sin()cos( ivrmsrmsivrmsrms IVjIVS
PActive Power
Q
Reactive Power
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt11
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complex Power contComplex Power cont Thus S in Shorthand Alternatively,
Reconsider the General Sinusoidal Circuit
S & Q are NOT Actual Power, and Thus all Terms are given Non-Watt Units• S→ Volt-Amps (VA)
• Q → Volt-Amps, Reactive (VAR)
P is Actual Power and hence has Units of W
First: U vs. Urms
jQP S
rmsrms
Mrms
M
U
UU
U
U
U
2and
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt12
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complex Power cont.2Complex Power cont.2 Now in the General Ckt
By Ohm’s Law
In the Last Expression Equate the REAL and Imaginary Parts
And Again by Ohm
iviv
rmsrms
ivrms
rms
irms
vrms
rmsrms
jZZ
j
soZIV
I
V
I
V
sincos
ImRe
and
ZZ
Z
Z
IVZ
Z
Z
iv
iv
Z
Z
Imsin
Recos
ZVI
Z
V
Z
V
rmsrms
irms
iv
vrmsrms
rmsrms
So
I
ZVI
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt13
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complex Power cont.3Complex Power cont.3 And by Complex Power
Definition
Using the Previous Results for P
Similarly for Q
ivrmsrms
ivrmsrms
IVQ
IVP
jQP
sinIm
cosRe
then
S
S
S
RIIP
IZ
V
ZIVP
rmsrms
rmsrms
rmsrms
22 Re
Re
ReRe
Z
Z
ZS
XIIQ rmsrms22 Im Z
So Finally the Alternative Expression for S
ZS 2rmsI
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt14
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complex Power TriangleComplex Power Triangle The Expressions for S
Plotting S in the Complex Plane
From The Complex Power “Triangle” Observe
ZS
S2rmsI
jQP
P
Qiv tan
Note also That Complex Power is CONSERVED
kkrmsktot I ZSS 2,
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt15
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example - Complex PowerExample - Complex Power For the Circuit At Right
• Zline =0.09 Ω + j0.3 Ω
• Pload = 20 kW
• Vload = 2200°
• pf = 80%, lagging
• f = 60 Hz → ω = 377s-1
Lagging pf → Inductive
From the Actual Power
Thusinductive
capacitive
pfSP iv )cos(||Re SS
kVAkW
pf
PSL 25
8.0
20
And Q from Pwr Triangle
kVAR15222 QPSQ L
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt16
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example - Complex Power contExample - Complex Power cont Then SL
Recall the S Mathematical Definition
Alternatively
87.36251520 kVAkVAjLS
*LLL IVS
Note also that [U*]* = U In the S Definition,
Isolating the Load Current and then Conjugating Both Sides
)(86.3664.113
0220
87.3625**
A
V
kVA
L
L
LL
I
V
SI
)(18.6891.90
220
000,15000,20*
Aj
j
L
L
I
I
Lagging
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt17
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example - Complex Pwr cont.2Example - Complex Pwr cont.2 Now Determine VS
Then VS Then The Phase Angle
To find the Src Power Factor, Draw the I & V Phasor Diagram
)(220)18.6891.90)(3.009.0(
0220)3.009.0(
Vjj
j
S
LS
LlineS
V
IV
VVV
86.453.249
14.2163.248
rmsS
S
V
j
V
V
86.4
86.36
SV
LI
7464.0
72.41coscos
also and
Load Inductive I Leads V
72.4186.3686.4
pf
pf iv
iv
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt18
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example - Complex Power kVARExample - Complex Power kVAR For the Circuit At
Right, Determine• Real And Reactive
Power losses in the Ln
• Real And Reactive Power at the Source
Lagging pf → Inductive
From the Actual Power
Thusinductive
capacitive
kVAkW
pf
PSL 62.47
84.0
40
And by S Definition
laggingpf
kW
84.0
40
1.0 25.0j
pfSP iv )cos(||Re SS
rmsL
LL A
V
SI )(45.216* VIS
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt19
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example - Complex kVAR cont.Example - Complex kVAR cont. Also from the S
Relation
Now the PowerFactor Angle Then for Line Loses
Quantitatively
laggingpf
kW
84.0
40
1.0 25.0j
)(839,25|||| 22 VARPLL SQ
86.3284.0acosiv
pf = cos(θv − θi); hence
2
** )(
Lline
LLlinelinelineline
IZ
IIZIVS
VAj
j
line
line
117134685
)45.216)(25.01.0( 2
S
S
I Lagging V
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt20
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example - Complex kVAR cont.2Example - Complex kVAR cont.2 Find Power Supplied
by Conservation of Complex Power
Then to Summarize the Answer• Pline = 4.685 kW
• Qline = 11.713 kVAR
• PS = 44.685 kW
• QS = 37.552 kVAR
laggingpf
kW
84.0
40
1.0 25.0j
kVA
kVAj
j
jjSup
04.4037.58
552.37685.44
839.25713.1140685.4
839.2540713.11685.4S
LoadlineSupplied SSS
In this Case
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt21
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Power Factor CorrectionPower Factor Correction As Noted Earlier, Most
Industrial Electrical Power Loads are Inductive• The Inductive
Component is Typically Associated with Motors
The Motor-Related Lagging Power Factor Can Result in Large Line Losses
The Line-Losses can Be Reduced by Power Factor Correction
To Arrive at the Power Factor Correction Strategy Consider A Schematic of a typical Industrial Load
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt22
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Power Factor Correction cont.Power Factor Correction cont. Prior to The Addition of
the Capacitor
For The Capacitive Load
)cos(
||
oldold
oldoldoldoldold
pf
SjQPS
22
2
C
Im
90
LL
CCC
LC
LLLC
L
CVC
CVIQ
CVI
CVCj
Z
VZ
VI
After Addition of the Capacitor
)cos(
||
newnew
newnew
Coldold
Coldnew
pf
jQjQP
S
SSS
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt23
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Power Factor Correction cont.2Power Factor Correction cont.2 Find θnew
• Cap is a Purely REACTIVE Load
The Vector Plot Below Shows Power Factor Correction Strategy
old
new
old
Coldnewtan
P
Q
P
Use Trig ID to find QC to give desired θnew
2tan1
1cos
1cos
1
new2
old
Cold
P
QL
QC
QL-QC
P
Qnew
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt24
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Trig ID DigressionTrig ID Digression
Start with the ID
2tan1
1cos
Solve for tan
old
Coldnewtan
P
Recall tanθnew
1cos
1
new2
old
Cold
P
Or
new
new2
new2
new2
new2
new2
new2
old
Cold
cos
cos1
cos
cos1
cos
cos
cos
1
P
But: cosθnew = pfnew
new
new
pf
pf 2
new
1tan
1cos
1tan
22
Substituting
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt25
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example – pf CorrectionExample – pf Correction Kayak Centrifugal
Injection-Molding Power Analysis• Improve Power Factor
to 95%
Find Sold
Now Qold
Adding A Cap DoesNOT Change P• Use Trig ID to Find Tan(new)
And by S Relation
laggingpf
VkW rmsL
8.0
0220,50
Roto-molding
process
pfSSP iv )cos(Re S
kVAkW
pf
PSold 5.62
80.0
50
)(5.37|||| 22 kVARPoldold SQ
329.01
tan95.0cos2
new
newnewnewnew pf
pf
P
Q
kVARPQP
Qnew
new 43.16329.0
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt26
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example – pf Correction contExample – pf Correction cont Then the Needed QC
Recall The Expression for QC
laggingpf
VkW rmsL
8.0
0220,50
Roto-molding
process
kVAQ
QQQ
C
newoldC
07.21
43.165.37
CVQ LL 2CC |||| IV
Then C from QC
FFC
V
QC
L
C
1155)(001155.0
)220()602(
1007.212
3
2
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt27
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
WhiteBoardWhiteBoard Work Work
Let’s Work (w/ 2-changes)
Problem 9.81• Determine at the input
SOURCE – Voltage & Current
– Complex Power
– Δθ & pf
19V220 rms
60 kVA
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt28
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt29
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt30
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt31
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt32
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt33
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis