This document provides an overview of power systems concepts including: 1) It defines power and energy, and their units of measurement. It notes Germany's annual electric energy consumption and generation capacity. 2) It discusses the different types of electric generating technologies and the tradeoffs between their fixed and operating costs. It also explains how total average costs depend on these factors. 3) It outlines the main elements of power systems including generation, consumption, transmission, and distribution components. It provides brief descriptions of each.

1. Informatik 7
Rechnernetze und
Kommunikationssysteme
Basics of Power Systems
(Part 1)
21.10.2015
Dr.- Ing. Abdalkarim Awad

2. Power
 Power:
Instantaneous rate of consumption of energy, How hard
you work!
Power = voltage x current
Power Units:
Watts = amps times volts (W)
kW = 1 x 103 Watt
MW = 1 x 106 Watt
GW = 1 x 109 Watt
TW = 1x 1012 Watt
Installed generation capacity in Germany is about
174 GW ( about 2.1 kW per person)
1Dr.- Ing. Abdalkarim Awad

3. Energy
Energy:
Integration of power over time,
Energy is what people really want from a power system,
How much work you accomplish over time.
Energy Units:
Joule = 1 watt-second (J)
kWh = kilowatthour (3.6 x 106 J)
Btu (British thermal unit )= 1055 J
1 MBtu=0.292 MWh
Annual electric energy consumption in Germany is
about 617 billion kWh (about 7300 kWh per
person, which means on average we each use 0.8
kW of power continuously), Bayern (85 billion kWh)
2Dr.- Ing. Abdalkarim Awad

4. Energy Economics
• Electric generating technologies involve a tradeoff
between fixed costs (primarily capital costs to build
them) and operating costs:
– Nuclear, wind, and solar high fixed costs, but low operating
costs,
– Natural gas has low fixed costs but relatively high operating
costs (dependent upon fuel prices, CO2 certificate,…)
– Coal in between (although recent low natural gas prices has
meant that some coal plants have higher operating costs than
some natural gas).
• Total average costs depend on fixed costs,
operating costs, and capacity factor (ratio of
average power production to capacity).
3Dr.- Ing. Abdalkarim Awad

5. Power System
4Dr.- Ing. Abdalkarim Awad

6. Main Elements of Power Systems
Generation (Supply)
Consumption (Load or Demand)
Transmission System
Additional components include:
– distribution system: local reticulation of power
– control equipment: coordinate supply with load
Dr.- Ing. Abdalkarim Awad 5

7. Generation (Supply)
Traditional Sources
Gas
Oil
Nuclear
Coal
…
Renewable Energy Sources
Wind
Solar
Water
…
Capacity, Cost,
Carbon Emission
Dr.- Ing. Abdalkarim Awad 6

8. Demand
Power Consumption:
Industrial
Commercial
Residential
Demand Response
Controllable Load
Non‐Controllable
Dr.- Ing. Abdalkarim Awad 7

9. Power Transmission
Power Transmission:
High Voltage (HV) Transmission Lines
Several Hundred Miles
Do renewable energy sources have new requirements?
Die Südwest-Kuppelleitung „Thüringer Strombrücke“?
Dr.- Ing. Abdalkarim Awad 8

10. 9Dr.- Ing. Abdalkarim Awad

11. Power Distribution:
Medium Voltage (MV) Transmission Lines (< 50 kV)
Power Deliver to Load Locations
Interface with Consumers / Metering
Distribution Sub‐stations
Step‐Down Transformers
Distribution Transformers
Dr.- Ing. Abdalkarim Awad 10

12. 11
Basics
Dr.- Ing. Abdalkarim Awad

13. Review of Phasors
Goal of phasor analysis is to simplify the analysis of
constant frequency ac systems
v(t) = Vmax cos(wt + qv)
i(t) = Imax cos(wt + qI)
Root Mean Square (RMS) voltage of sinusoid
2 max
0
1
( )
2
T
V
v t dt
T

Dr.- Ing. Abdalkarim Awad 12

14. Phasor Representation
j
( )
Euler's Identity: e cos sin
Phasor notation is developed by rewriting
using Euler's identity
( ) 2 cos( )
( ) 2 Re
(Note: is the RMS voltage)
V
V
j t
j
v t V t
v t V e
V
q
w q
q q
w q

 
 
   
Dr.- Ing. Abdalkarim Awad 13

15. Phasor Representation, cont’d
The RMS, cosine-referenced voltage phasor is:
( ) Re 2
cos sin
cos sin
V
V
j
V
jj t
V V
I I
V V e V
v t Ve e
V V j V
I I j I
q
qw
q
q q
q q
  

 
 
(Note: Some texts use “boldface” type for
complex numbers, or “bars on the top”)
Dr.- Ing. Abdalkarim Awad 14

16. Advantages of Phasor Analysis
0
2 2
Resistor ( ) ( )
( )
Inductor ( )
1 1
Capacitor ( ) (0)
C
Z = Impedance
R = Resistance
X = Reactance
X
Z = =arctan( )
t
v t Ri t V RI
di t
v t L V j LI
dt
i t dt v V I
j C
R jX Z
R X
R
w
w


 
 
 
   


Device Time Analysis Phasor
(Note: Z is a
complex number but
not a phasor)
Dr.- Ing. Abdalkarim Awad 15

17. RL Circuit Example
Dr.- Ing. Abdalkarim Awad 16
)9.6cos(220)(
9.620
9.365
30100
9.36
534
3
4
i(t)Find
Hz50frequencyand
)30(100cos2v(t)ifcircuit,givenFor the
22











tti
Z
V
I
Z
LX
R
t
w

w
w
4Ω
9.55 mH

18. Complex Power
Dr.- Ing. Abdalkarim Awad 17
)2cos())(cos(2/1)(
))cos()(cos(2/1coscos
)cos()(
)cos()(
)()()(
maxmax
max
max
IvIv tIVtp
bababa
ItIti
vtVtv
titvtp
Power
qqwqq
qw
qw






19. Complex Power, cont’d
max max
0
max max
1
( ) [cos( ) cos(2 )]
2
1
( )
1
cos( )
2
cos( )
= =
V I V I
T
avg
V I
V I
V I
p t V I t
P p t dt
T
V I
V I
q q w q q
q q
q q
 q q
    

 
 


Power Factor
Average P
Angle
ower
Dr.- Ing. Abdalkarim Awad 18

20. Complex Power
 
*
cos( ) sin( )
P = Real Power (W, kW, MW)
Q = Reactive Power (var, kvar, Mvar)
S = Complex power (VA, kVA, MVA)
Power Factor (pf) = cos
If current leads voltage then pf is leading
If current
V I V I
V I
S V I j
P jQ
q q q q

  




lags voltage then pf is lagging
(Note: S is a complex number but not a phasor)
Dr.- Ing. Abdalkarim Awad 19
=θV-θI
Φ<0
Φ>0

21. Complex Power, cont’d
2
1
Relationships between real, reactive and complex power
cos
sin 1
Example: A load draws 100 kW with a leading pf of 0.85.
What are (power factor angle), Q and ?
-cos 0.85 31.8
100
0.
P S
Q S S pf
S
kW
S



 

   
   
 117.6 kVA
85
117.6sin( 31.8 ) 62.0 kVarQ

    
Dr.- Ing. Abdalkarim Awad 20

22. Conservation of Power
 At every node (bus) in the system
Sum of real power into node must equal zero
Sum of reactive power into node must equal zero
 This is a direct consequence of Kirchhoff’s current
law, which states that the total current into each
node must equal zero.
Conservation of power follows since S = VI*
Dr.- Ing. Abdalkarim Awad 21

23. Conservation of Power Example
Earlier we found
I = 20-6.9 amps
What is the power at the resistance
and at the inductance
*
*
R
2
R R
*
L
2
L L
100 30 20 6.9 2000 36.9 VA
36.9 pf = 0.8 lagging
S 4 20 6.9 20 6.9
P 1600 (Q 0)
S 3 20 6.9 20 6.9
Q 1200var (P 0)
R
L
S V I
V I
W I R
V I j
I X

        
 
       
  
       
  
Dr.- Ing. Abdalkarim Awad 22

24. Power Consumption in Devices
2
Resistor Resistor
2
Inductor Inductor L
2
Capacitor Capacitor C
Capa
Capacitor
Resistors only consume real power
P
Inductors only consume reactive power
Q
Capacitors only generate reactive power
1
Q
Q
C
I R
I X
I X X
C
V
w


  
 
2
citor
C
C
(Note-some define X negative)
X
Dr.- Ing. Abdalkarim Awad 23

25. Example
First solve
basic circuit
Dr.- Ing. Abdalkarim Awad 24
MWVIP
Z
V
I
L 160400*040000
0400
0100
040000
*





Power Consumed
in the load?

26. Example
*
40000 0
400 0 Amps
100 0
40000 0 (5 40) 400 0
42000 16000 44.9 20.8 kV
S 44.9k 20.8 400 0
17.98 20.8 MVA 16.8 6.4 MVA
V
I
V j
j
V I
j
 
   
  
      
    
     
    
First solve
basic circuit
Dr.- Ing. Abdalkarim Awad 25
How much loss?
V at source?
S at source?

27. Example, cont’d
Now add additional
reactive power load
and resolve
70.7 0.7 lagging
564 45 Amps
59.7 13.6 kV
S 33.7 58.6 MVA 17.6 28.8 MVA
LoadZ pf
I
V
j
  
   
  
    
Dr.- Ing. Abdalkarim Awad 26

28. Example, cont’d
70.7 0.7 lagging
564 45 Amps
59.7 13.6 kV
S 33.7 58.6 MVA 17.6 28.8 MVA
LoadZ pf
I
V
j
  
   
  
    
Dr.- Ing. Abdalkarim Awad 27
Power Consumed
in the load?
How much power
should be generated
at the source
How much Loss?

29. 59.7 kV
17.6 MW
28.8 MVR
40.0 kV
16.0 MW
16.0 MVR
17.6 MW 16.0 MW
-16.0 MVR28.8 MVR
Power System Notation
Power system components are usually shown as
“one-line diagrams.” Previous circuit redrawn
Arrows are
used to
show loads
Generators are
shown as circles
Transmission lines
are shown as a
single line
BusPowerWorld
simulator
Dr.- Ing. Abdalkarim Awad 28

30. Reactive Compensation
44.94 kV
16.8 MW
6.4 MVR
40.0 kV
16.0 MW
16.0 MVR
16.8 MW 16.0 MW
0.0 MVR6.4 MVR
16.0 MVR
Key idea of reactive compensation is to supply reactive
power locally. In the previous example this can
be done by adding a 16 Mvar capacitor at the load
Compensated circuit is identical to first example with
just real power loadDr.- Ing. Abdalkarim Awad 29

31. Reactive Compensation, cont’d
 Reactive compensation decreased the line flow
from 564 Amps to 400 Amps. This has advantages
Lines losses, which are equal to I2 R decrease
Lower current allows utility to use small wires, or alternatively,
supply more load over the same wires
Voltage drop on the line is less
 Reactive compensation is used extensively by
utilities
 Capacitors can be used to “correct” a load’s power
factor to an arbitrary value.
Dr.- Ing. Abdalkarim Awad 30

32. Power Factor Correction Example
1
1
desired
new cap
cap
Assume we have 100 kVA load with pf=0.8 lagging,
and would like to correct the pf to 0.95 lagging
80 60 kVA cos 0.8 36.9
PF of 0.95 requires cos 0.95 18.2
S 80 (60 Q )
60 - Q
ta
80
S j
j




    
  
  
 cap
cap
n18.2 60 Q 26.3 kvar
Q 33.7 kvar
   

Dr.- Ing. Abdalkarim Awad 31

33. Distribution System Capacitors
Dr.- Ing. Abdalkarim Awad 32

34. Balanced 3 Phase () Systems
 A balanced 3 phase () system has:
– three voltage sources with equal magnitude, but with an angle
shift of 120,
– equal loads on each phase,
– equal impedance on the lines connecting the generators to the
loads.
 Bulk power systems are almost exclusively 3.
 Single phase is used primarily only in low voltage,
low power settings, such as residential and some
commercial.
 Single phase transmission used for electric trains in
Europe.
33Dr.- Ing. Abdalkarim Awad

35. Balanced 3 -- Zero Neutral Current
34
* * * *
(1 0 1 1
3
Note: means voltage at point with respect to point .
n a b c
n
an a bn b cn c an a
xy
I I I I
V
I
Z
S V I V I V I V I
V x y
  
         
   
Dr.- Ing. Abdalkarim Awad

36. Three Phase - Wye Connection
 There are two ways to connect 3 systems:
– Wye (Y), and
– Delta ().
35
an
bn
cn
Wye Connection Voltages
V V
V V
V V



  
   
   
Dr.- Ing. Abdalkarim Awad

37. Wye Connection Line Voltages
36
Van
Vcn
Vbn
Vab
Vca
Vbc
-Vbn
(1 1 120
3 30
3 90
3 150
ab an bn
bc
ca
V V V V
V
V V
V V
 



       
   
   
   
Line to line
voltages are
also balanced.
(α = 0 in this case)
Dr.- Ing. Abdalkarim Awad

38. Wye Connection, cont’d
 We call the voltage across each element of a wye
connected device the “phase” voltage.
 We call the current through each element of a wye
connected device the “phase” current.
 Call the voltage across lines the “line-to-line” or just
the “line” voltage.
 Call the current through lines the “line” current.
37
6
*
3
3 1 30 3
3
j
Line Phase Phase
Line Phase
Phase Phase
V V V e
I I
S V I


   


Dr.- Ing. Abdalkarim Awad

39. Delta Connection
38
Ica
Ic
Iab
Ibc
Ia
Ib
*
3
For Delta connection,
voltages across elements
equals line voltages
For currents
3
3
a ab ca
ab
b bc ab
c ca bc
Phase Phase
I I I
I
I I I
I I I
S V I
 
   
 
 

Dr.- Ing. Abdalkarim Awad

40. Three Phase Example
Assume a -connected load, with each leg Z =
10020, is supplied from a 3 13.8 kV (L-L) source
39
13.8 0 kV
13.8 0 kV
13.8 0 kV
ab
bc
ca
V
V
V
  
   
  
13.8 0 kV
138 20 amps
138 140 amps 138 0 amps
ab
bc ca
I
I I
 
    
 
      
Dr.- Ing. Abdalkarim Awad

41. Three Phase Example, cont’d
40
*
138 20 138 0
239 50 amps
239 170 amps 239 0 amps
3 3 13.8 0 kV 138 amps
5.7 MVA
5.37 1.95 MVA
pf cos20 lagging
a ab ca
b c
ab ab
I I I
I I
S V I
j
        
   
      
       
 
 
   
Dr.- Ing. Abdalkarim Awad

42. Delta-Wye Transformation
41
Y
Line
phase
To simplify analysis of balanced 3 systems:
1) Δ-connected loads can be replaced by
1
Y-connected loads with
3
2) Δ-connected sources can be replaced by
Y-connected sources with
3 30
Z Z
V
V



 
Dr.- Ing. Abdalkarim Awad

43. Example (Losses)
5+j10 Ω
33 kV
2
1
4
186 22 24
2
1.5
3
186 22 24
Load Shifting
LoadMW
LoadMW
 Compute Energy Losses
Load MW(P) I=P/(√3*33*1000) Losses(kW)=I2R
1 17.5 1.53
1.5 26.3 3.45
2 35 6.13
3 52.5 13.78
4 70 24.5
Load
3 Phase 3 Phase
Dr.- Ing. Abdalkarim Awad 42

44. Example (Energy Loss)
5+j10 Ω
33 kV
2
1
4
186 22 24
2
1.5
3
186 22 24
Load Shifting
LoadMW
LoadMW
Load MW(P) I=P/(√3*33*1000) Losses(kW)=I2R
1 17.5 1.53
1.5 26.3 3.45
2 35 6.13
3 52.5 13.78
4 70 24.5
Loss/Line=8*1.53+12*6.13+4*24.5
=183.8 kWh Loss=3*183.8 kWh
Loss/Line=8*3.45+12*6.13+4*13.7
8=156.3 kWh  Loss=3*156.3 kWh
Load
 Compute Energy Loss in the Transmission Line
Dr.- Ing. Abdalkarim Awad 43
Savings
3*183.8-3*156.3
=82.5 kWh

45. Example (Costs)
5+j10 Ω
33 kV
2
1
4
186 22 24
2
1.5
3
186 22 24
Load Shifting
LoadMW
LoadMW
Time Cost (Euro/MWh)
0-8 20
8-12 30
12-16 40
16-22 50
22-0 20
Load
 Compute Energy Costs
Dr.- Ing. Abdalkarim Awad 44

46. Example (Costs)
5+j10 Ω
33 kV
2
1
4
186 22 24
2
1.5
3
186 22 24
Load Shifting
LoadMW
LoadMW
Time Cost (Euro/MWh)
0-8 20
8-12 30
12-16 40
16-22 50
22-0 20
Cost=1*8*20+2*2*20+2*4*30+2*4
*40+2*2*50+4*4*50=1800 Euro
Cost=1.5*8*20+2*2*20+2*4*30+2*4
*40+2*2*50+4*3*50=1680 Euro
Load
 Compute Energy Costs
Dr.- Ing. Abdalkarim Awad 45
Saving
=1800-1680
=120 Euro

47. 46
Economic Dispatch
And Optimal Power Flow
Dr.- Ing. Abdalkarim Awad

48. Economic dispatch
Objective: minimize the cost of generation
Constraints
Equality constraint: load generation balance
Inequality constraints: upper and lower limits on generating units
output
47
A B C
L
Dr.- Ing. Abdalkarim Awad

49. Limitations of economic dispatch
Generating units and loads are not all connected to
the same bus
The economic dispatch may result in unacceptable
flows or voltages in the network
48
Network
A
B C
D
Dr.- Ing. Abdalkarim Awad

50. Example of network limitation
LA=100 MW
LB=200 MW
Production of each Generator?
A
B
100$/MWh
50 $/MWh
Maximum flow on each line: 100MW
LBLA
Dr.- Ing. Abdalkarim Awad 49

51. Acceptable ED solution
50
A
B
LB = 200MWLA = 100 MW 100 MW
100 MW
The flows on the lines are below the limit
The economic dispatch solution is acceptable
The solution of this (trivial) economic dispatch is:
0 MW
300 MW
100$/MWh
50 $/MWh
Dr.- Ing. Abdalkarim Awad

52. Example of network limitation
LA=100 MW
LB=400 MW
Production of each Generator?
A
B
100$/MWh
50 $/MWh
Maximum flow on each line: 100MW
LBLA
Dr.- Ing. Abdalkarim Awad 51

53. Unacceptable ED solution
52
A
B
The solution of the economic dispatch is:
500 MW
0 MW
200 MW
200 MW
The resulting flows exceed their limit
The economic dispatch solution is not acceptable
LB = 400MWLA = 100 MW
100$/MWh50 $/MWh
Dr.- Ing. Abdalkarim Awad

54. Modified ED solution
53
In this simple case, the solution of the economic dispatch
can be modified easily to produce acceptable flows.
A
B
300 MW
200 MW
100 MW
100 MW
LB = 400MWLA = 100 MW
This could be done mathematically by adding the following
inequality constraint:
However, adding inequality constraints for each problem
is not practical in more complex situations
We need a more general approach
100$/MWh
50 $/MWh
Dr.- Ing. Abdalkarim Awad

55. Optimal Power Flow (OPF) - Overview
Optimization problem
Classical objective function
Minimize the cost of generation
Equality constraints
Power balance at each node - power flow equations
Inequality constraints
Network operating limits (line flows, voltages)
Limits on control variables
54Dr.- Ing. Abdalkarim Awad

56. Optimal Power Flow (OPF)
The goal of an optimal power flow (OPF) is to
determine the “best” way to instantaneously
operate a power system.
Usually “best” = minimizing operating cost.
OPF can incorporate and enforce transmission
limits, but we’ll introduce OPF initially ignoring
transmission limits
Dr.- Ing. Abdalkarim Awad 55

57. “Ideal” Power Market
 Ideal power market is analogous to a lake.
Generators supply energy to lake and loads remove
energy.
 Ideal power market has no transmission constraints
 Single marginal cost associated with enforcing
constraint that supply = demand
– buy from the least cost unit that is not at a limit
– this price is the marginal cost.
 This solution is identical to the economic dispatch
problem solution.
56Dr.- Ing. Abdalkarim Awad

58. Two Bus Example
Total Hourly Cost :
Bus A Bus B
300.0 MWMW
199.6 MWMW 400.4 MWMW
300.0 MWMW
8459 $/hr
Area Lambda : 13.02
AGC ON AGC ON
57Dr.- Ing. Abdalkarim Awad

59. Mathematical Formulation of Costs (C)
and incremental Cost (IC)
 Generator cost curves are usually not smooth.
However the curves can usually be adequately
approximated using piece-wise smooth, functions.
 Two representations predominate
– quadratic or cubic functions
– piecewise linear functions
 We'll assume a quadratic presentation
In order to minimize the total operating cost
IC1(PG1)=IC2(PG2)=…ICN(PGN)
2
( ) $/hr (fuel-cost)
( )
( ) 2 $/MWh
i Gi i Gi Gi
i Gi
i Gi Gi
Gi
C P P P
dC P
IC P P
dP
  
 
  
   =λ
Dr.- Ing. Abdalkarim Awad 58

60. Mathematical Formulation of Costs
For the two generator system
PGA+PGB=600
Economic dispatch IC(PGA)=IC(PGA)
Dr.- Ing. Abdalkarim Awad 59
Euro/MWh0.00298PG11.83)(PGIC
Euro/MWh0.00668PG11.69)(PGIC
Euro/hr0.00149PG11.83PG616.9)(PGC
Euro/hr0.00334PG11.69PG399.8)(PGC
BBB
AAA
2
BBBB
2
AAAA




600
8.39696.3350.00298PG11.83)(PGIC
17507.1490.00668PG11.69)(PGIC
BBB
AAA



BA
B
A
PGPG
PG
PG



61. Mathematical Formulation of Costs
λ =ICA(PGA)=13.02
λ = ICB(PGB)=13.02
PGA=200  CA(200)=2870.6
PGB=400 CB(400)=5587.3
Total cost 8457.9
Dr.- Ing. Abdalkarim Awad 60
400
200
02.13)6.3357.149/()6008.39691750(
600




B
A
BA
PG
PG
PGPG


62. Generator cost curves
The Problem can be written as:
MINIMIZE (CA(PGA)+CB(PGB))
Subject to
PGA+PGB=600
PGA>=0
PGB>=0
Dr.- Ing. Abdalkarim Awad 61

63. Generator cost curves
Dr.- Ing. Abdalkarim Awad 62
400,200,0
0.00483x+1.928x-8651.1
x)-00.00149(60x)-11.83(6009.616
0.00334x11.69x399.8
600,
600
0.00149PG11.83PG616.9
0.00334PG11.69PG399.8
Euro0.00149PG11.83PG616.9)(PGC
Euro0.00334PG11.69PG399.8)(PGC
2
2
2
2
BB
2
AA
2
BBBB
2
AAAA












BA
total
total
total
BA
BA
total
PGPGx
x
C
C
C
xPGxPG
PGPG
C

64. Optimal Power Flow (OPF)
 OPF functionally combines the power flow with
economic dispatch
 Minimize cost function, such as operating cost,
taking into account realistic equality and inequality
constraints
 Equality constraints:
– bus real and reactive power balance
– generator voltage setpoints
– area MW interchange
63Dr.- Ing. Abdalkarim Awad

65. OPF, cont’d
 Inequality constraints:
– transmission line/transformer/interface flow limits
– generator MW limits
– generator reactive power capability curves
– bus voltage magnitudes
 Available Controls:
– generator MW outputs
– transformer taps and phase angles
64Dr.- Ing. Abdalkarim Awad

66. OPF Solution Methods
 Non-linear approach using Newton’s method:
– handles marginal losses well, but is relatively slow and has
problems determining binding constraints
 Linear Programming (LP):
– fast and efficient in determining binding constraints, but can
have difficulty with marginal losses.
 Mixed Integer
 Difficult
 Battery, devices
65Dr.- Ing. Abdalkarim Awad

67. Bibliography
Book
Duncan Glover, Mulukutla Sarma, and Thomas Overbye, Power
System Analysis and Design, Thomson, Fifth Edition.
Power System Analysis by Thomas Overbye (ppt
slides).
PowerWorld tutrials
Power System Economics by Daniel Kirschen (ppt
Slides)
Communications and Control in Smart Grid by
Hamed Mohsenian-Rad (ppt slides)
Dr.- Ing. Abdalkarim Awad 66