9/16/2011
Load Models
Simplified Power System Modeling
• Balanced three phase systems can be analyzed using
per phase analysis
• A “per unit” normalization is simplify the analysis of
systems with different voltage levels.
• To provide an introduction to power flow analysis we
need models for the different system devices:
– Transformers and Transmission lines, generators and loads
• Transformers and transmission lines are modeled as a
series impedances
Energy Systems Research Laboratory, FIU
Generator Models
• Engineering models depend upon application
• Generators are usually synchronous machines
• For generators we will use two different models:
– a steady-state
y
model,, treating
g the ggenerator as a constant
power source operating at a fixed voltage; this model will be
used for power flow and economic analysis
– This model works fairly well for type 3 and type 4 wind
turbines
– Other models include treating as constant real power with a
fixed power factor.
Energy Systems Research Laboratory, FIU
Per Unit Conversion Procedure, 1
1. Pick a 1 VA base for the entire system, SB
2. Pick a voltage base for each different voltage level,
VB. Voltage bases are related by transformer turns
ratios. Voltages are line to neutral.
3 Calculate
3.
C l l the
h impedance
i
d
base,
b
ZB= (VB)2/SB
4. Calculate the current base, IB = VB/ZB
5. Convert actual values to per unit
Note, per unit conversion on affects magnitudes, not the angles. Also, per unit
quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)
• Ultimate goal is to supply loads with electricity at
constant frequency and voltage
• Electrical characteristics of individual loads matter,
but usually they can only be estimated
– actual loads are constantly changing,
changing consisting of a large
number of individual devices
– only limited network observability of load characteristics
• Aggregate models are typically used for analysis
• Two common models
– constant power: Si = Pi + jQi
– constant impedance: Si = |V|2 / Zi
Energy Systems Research Laboratory, FIU
Per Unit Calculations
• A key problem in analyzing power systems is the
large number of transformers.
– It would be very difficult to continually have to refer
impedances to the different sides of the transformers
• Thi
This problem
bl is
i avoided
id d by
b a normalization
li i off all
ll
variables.
• This normalization is known as per unit analysis.
quantity in per unit 
actual quantity
base value of quantity
Energy Systems Research Laboratory, FIU
Three Phase Per Unit
Procedure is very similar to 1 except we use a 3VA base, and use line to line
voltage bases
1. Pick a 3 VA base for the entire system, S B3
2. Pick a voltage base for each different voltage level,
VB. Voltages are line to line.
3. Calculate the impedance base
ZB 
VB2, LL
S B3

( 3 VB , LN ) 2
3S 1B

VB2, LN
S 1B
Exactly the same impedance bases as with single phase!
Energy Systems Research Laboratory, FIU
Energy Systems Research Laboratory, FIU
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9/16/2011
Three Phase Per Unit, cont'd
4. Calculate the current base, IB
I3B 
S B3
3 S 1B
S 1B


 I1B
3 VB , LL
3 3 VB , LN VB , LN
Exactly the same current bases as with single phase!
5. Convert actual values to per unit
Energy Systems Research Laboratory, FIU
Bus Admittance Matrix or Ybus
• First step in solving the power flow is to create what
is known as the bus admittance matrix, often call the
Ybus.
gives the relationships
p between all the bus
• The Ybus g
current injections, I, and all the bus voltages, V,
I = Ybus V
• The Ybus is developed by applying KCL at each bus in
the system to relate the bus current injections, the bus
voltages, and the branch impedances and admittances
Energy Systems Research Laboratory, FIU
Power Flow Analysis
• When analyzing power systems we know neither
the complex bus voltages nor the complex current
injections
• Rather, we know the complex
p ppower being
g
consumed by the load, and the power being injected
by the generators plus their voltage magnitudes
• Therefore we can not directly use the Ybus equations,
but rather must use the power balance equations
Energy Systems Research Laboratory, FIU
Power Flow Analysis
• We now have the necessary models to start to
develop the power system analysis tools
• The most common power system analysis tool is the
power flow (also known sometimes as the load flow)
– power flow determines how the power flows in a network
– also used to determine all bus voltages and all currents
– because of constant power models, power flow is a
nonlinear analysis technique
– power flow is a steady-state analysis tool
Energy Systems Research Laboratory, FIU
Ybus General Form
•The diagonal terms, Yii, are the self admittance
terms, equal to the sum of the admittances of all
devices incident to bus i.
•The off-diagonal terms, Yij, are equal to the negative
of the sum of the admittances joining the two buses.
•With large systems Ybus is a sparse matrix (that is,
most entries are zero)
•Shunt terms, such as with the  line model, only
affect the diagonal terms.
Energy Systems Research Laboratory, FIU
Power Flow Slack Bus
• We can not arbitrarily specify S at all buses because
total generation must equal total load + total losses
• We also need an angle reference bus.
• To solve these problems we define one bus as the
"slack" bus. This bus has a fixed voltage magnitude
and angle, and a varying real/reactive power
injection.
• A “slack bus” does not exist in the real power system.
Energy Systems Research Laboratory, FIU
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9/16/2011
Power Balance Equations
From KCL we know at each bus i in an n bus system
the current injection, I i , must be equal to the current
that flows into the network
Ii  I Gi  I Di 
Pi 
n
 YikVk
k 1
Energy Systems Research Laboratory, FIU
Newton-Raphson Method (scalar)
f ( xˆ )  0
(v)
, define
 Vi Vk
(cos ik  j sin  ik )(Gik  jBik )
Qi 
n
 Vi Vk (Gik cos ik  Bik sin ik )  PGi  PDi
k 1
n
 Vi Vk (Gik sin  ik  Bik cos ik )  QGi  QDi
k 1
Newton-Raphson Method, cont’d
3. Approximate f ( xˆ ) by neglecting all terms
f ( xˆ )  0  f ( x( v ) ) 
x ( v )  xˆ - x ( v )
2. Represent f ( xˆ ) by a Taylor series about f ( x )
f ( xˆ )  f ( x ( v ) ) 
df ( x ( v ) ) ( v )
x 
dx
1 d 2 f ( x( v ) )
x ( v )
2 dx 2


2
 higher order terms
Energy Systems Research Laboratory, FIU
df ( x (v ) ) (v )
x
dx
4. Use this linear approximation to solve for x( v )
1
 df ( x( v ) ) 
(v)
x ( v )   
 f (x )
 dx 
5. Solve for a new estimate of x̂
x (v 1)  x ( v )  x (v )
Energy Systems Research Laboratory, FIU
Newton-Raphson Example
Use Newton-Raphson to solve f ( x)  x 2 - 2  0
The equation we must iteratively solve is
1
x ( v )
k 1
except the first two
1. For each guess of xˆ , x
x
k 1
Energy Systems Research Laboratory, FIU
G en eral fo rm o f p ro b lem : F in d an x su ch th at
(v)
n
Resolving into the real and imaginary parts
k 1
The network power injection is then Si  Vi I i*

n
k 1
 Iik
n
Si  Pi  jQi  Vi  Yik*Vk*   Vi Vk e jik (Gik  jBik )

n
Since I = Ybus V we also know
Ii  I Gi  I Di 
Real Power Balance Equations
 df ( x( v ) ) 
(v )
 
 f (x )
d
 dx

1
   ( v )  (( x ( v ) ) 2 - 2)
2x 
Newton-Raphson Example, cont’d
1
x( v 1)  x( v )   ( v )  (( x ( v ) ) 2 - 2)
2x 
Guess x (0)  1. Iteratively solving we get
v
x(v)
f ( x(v ) )
x ( v )
0.5
0
1
1
1
1.5
0.25
0.08333
3
x (v 1)  x ( v )  x ( v )
2
1.41667
6.953  10
1
x (v 1)  x ( v )   ( v )  (( x ( v ) ) 2 - 2)
2x 
3
1.41422
6.024  106
Energy Systems Research Laboratory, FIU
2.454  103
Energy Systems Research Laboratory, FIU
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Newton-Raphson Comments
• When close to the solution the error decreases quite
quickly -- method has quadratic convergence
• f(x(v)) is known as the mismatch, which we would
like to drive to zero
• Stopping criteria is when f(x(v))  < 
• Results are dependent upon the initial guess. What if
we had guessed x(0) = 0, or x (0) = -1?
• A solution’s region of attraction (ROA) is the set of
initial guesses that converge to the particular solution.
The ROA is often hard to determine
Energy Systems Research Laboratory, FIU
Multi-Variable Newton-Raphson
Next we generalize to the case where x is an ndimension vector, and f (x) is an n-dimension function
 x1 
x 
2
x   

x 
 n
Again define the solution xˆ so f (xˆ )  0 and
x  xˆ  x
Energy Systems Research Laboratory, FIU
Multi-Variable Case, cont’d
The Taylor series expansion is written for each fi (x)
f1 (xˆ )  f1 (x) 
f1 (x)
f (x)
x1  1 x2  
x1
x2
f1 (x)
g
order terms
xn  higher
xn

f n (xˆ )  f n (x) 
f n (x)
f (x)
x1  n
x2  
x1
x2
f n (x)
xn  higher order terms
xn
Energy Systems Research Laboratory, FIU
Jacobian Matrix
The n by n matrix of partial derivatives is known
as the Jacobian matrix, J (x)
 f1 (x)
 x
1

 f 2 (x)
J (x)   x1

 
 f (x)
 n
 x1
f1 (x)
f1 (x) 

x2
xn 

f 2 (x)
f 2 (x) 

x2
xn 



 
f n (x)
f n (x) 


x2
xn 
Energy Systems Research Laboratory, FIU
 f1 (x) 
 f ( x) 
2

f ( x)  
  
 f ( x) 
 n 
Multi-Variable Case, cont’d
This can be written more compactly in matrix form
 f1 (x)
 x
1
 f1 (x)  
 f (x)   f 2 (x)
2
   x1
f (xˆ )  
   
 f ( x)   
 n  
f (x)
 n
 x1
f1 (x)
f1 (x) 

x2
xn 
  x1 
f 2 (x)
f 2 (x)  

x 
x2
xn   2 
  


 

 xn 
f n (x)
f n (x) 


x2
xn 
 higher order terms
Energy Systems Research Laboratory, FIU
Multi-Variable N-R Procedure
Derivation of N-R method is similar to the scalar case
f (xˆ )  f (x)  J (x)x  higher order terms
f (xˆ )  0  f (x)  J (x)x
x
  J (x) 1 f (x)
x(v 1)  x(v )  x( v )
x(v 1)  x(v )  J (x( v ) ) 1 f (x( v ) )
Iterate until f (x( v ) )  
Energy Systems Research Laboratory, FIU
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9/16/2011
Multi-Variable Example
x 
Solve for x =  1  such that f (x)  0 where
x2 
f1 (x)  2 x12  x22  8  0
f 2 (x)  x12  x22  x1 x2  4  0
First symbolically determine the Jacobian
 f1 (x)
 x
1
J (x) = 
f
(

 2 x)
 x1
f1 (x) 
x2 

f 2 (x) 
x2 
specified tolerance 
0.1556 
f (x )  

0.0900 
If  = 0.2 then we would be done. Otherwise we'd
continue iterating.
(2)
Energy Systems Research Laboratory, FIU
Newton-Raphson Power Flow
In the Newton-Raphson power flow we use Newton's
method to determine the voltage magnitude and angle
at each bus in the power system.
We need to solve the power balance equations
 Vi
Vk (Gik cos ik  Bik sin  ik )  PGi  PDi
 Vi
Vk (Gik sin  ik  Bik cos ik )  QGi  QDi
k 1
Energy Systems Research Laboratory, FIU
1
  f1 (x) 
x1  2 x2   f 2 (x) 
1
 
1
2 x2
1  4 2   5
 2.1
  
  



1  3 1  3
1.3 
Energy Systems Research Laboratory, FIU
1
Qi 
2 x2 
x1  2 x2 
1
 2.1 8.40 2.60   2.51
1.8284 
 
x(2)     




1.3  5.50 0.50  1.45 
1.2122 
Each iteration we check f (x) to see if it is below our
k 1
n
 x1 
 4 x1
 x     2 x  x
 2
 1 2
x(1)
Multi-variable Example, cont’d
Pi 
 4 x1
J (x) = 
 2 x1  x2
Then
Arbitrarily guess x(0)
Energy Systems Research Laboratory, FIU
n
Multi-variable Example, cont’d
Real Power Balance Equations
n
n
k 1
k 1
Si  Pi  jQi  Vi  Yik*Vk*   Vi Vk e jik (Gik  jBik )

n
 Vi Vk
k 1
(cos ik  j sin  ik )(Gik  jBik )
Resolving into the real and imaginary parts
Pi 
Qi 
n
 Vi Vk (Gik cos ik  Bik sin ik )  PGi  PDi
k 1
n
 Vi Vk (Gik sin ik  Bik cos ik )  QGi  QDi
k 1
Energy Systems Research Laboratory, FIU
Power Flow Variables
Assume the slack bus is the first bus (with a fixed
voltage angle/magnitude). We then need to determine
the voltage angle/magnitude at the other buses.
 2 
  


 n 
x  
V 
 2
  


 Vn 
 P2 (x)  PG 2  PD 2 





 Pn (x)  PGn  PDn 
f ( x)  
Q (x)  QG 2  QD 2 
 2








x
Q
(
)
Q
Q
 n
Gn
Dn 
Energy Systems Research Laboratory, FIU
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N-R Power Flow Solution
The power flow is solved using the same procedure
discussed last time:
Set v  0; make an initial guess of x, x( v )
While f (x )   Do
(v)
x( v 1)  x(v )  J (x( v ) ) 1 f (x(v ) )
v
 v 1
End While
Energy Systems Research Laboratory, FIU
Power Flow Jacobian Matrix
The most difficult part of the algorithm is determining
and inverting the n by n Jacobian matrix, J (x)
 f1 (x)
 x
1

 f 2 (x)
J (x)   x1

 
 f (x)
 n
 x1
f1 (x)
f1 (x) 

x2
xn 

f 2 (x)
f 2 (x) 

x2
xn 



 
f n (x)
f n (x) 


x2
xn 
Energy Systems Research Laboratory, FIU
Power Flow Jacobian Matrix, cont’d
Jacobian elements are calculated by differentiating
each function, fi (x), with respect to each variable.
Two Bus Newton-Raphson Example
For the two bus power system shown below, use the
Newton-Raphson power flow to determine the
voltage magnitude and angle at bus two. Assume
that bus one is the slack and SBase = 100 MVA.
For example, if fi (x) is the bus i real power equation
fi ( x) 
fi ( x)

 i
n
 Vi Vk (Gik cosik  Bik sin ik )  PGi  PDi
k 1
n
One
 Vi Vk (Gik sin ik  Bik cosik )
k 1
k i
fi ( x)
 Vi V j (Gik sin  ik  Bik cos ik ) ( j  i )
 j
Energy Systems Research Laboratory, FIU
Two Bus Example, cont’d
General power balance equations
Pi 
Qi 
n
 Vi Vk (Gik cosik  Bik sin ik )  PGi  PDi
k 1
n
i ik  Bik cos ik )  QGi  QDi
 Vi Vk (Gik sin
k 1
Bus two power balance equations
V2 V1 (10sin  2 )  2.0  0
V2 V1 ( 10cos  2 )  V2 (10)  1.0  0
2
Energy Systems Research Laboratory, FIU
Line Z = 0.1j
0 1j
1.000 pu
Two
0 MW
0 MVR
1.000 pu
200 MW
100 MVR
2 
  j10 j10 
x   
Ybus  

 j10  j10 
 V2 
Energy Systems Research Laboratory, FIU
Two Bus Example, cont’d
P2 (x)  V2 (10sin  2 )  2.0  0
Q2 (x)  V2 (10cos 2 )  V2 (10)  1.0  0
2
Now calculate the power flow Jacobian
 P2 (x)
 
2
J ( x)  
 Q 2 (x)
 
2

P2 (x) 
V 2 

Q2 (x) 
 V 2 
10 V2 cos 2
 
10 V2 sin  2
10sin  2

10 cos 2  20 V2 
Energy Systems Research Laboratory, FIU
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9/16/2011
Two Bus Example, First Iteration
0 
Set v  0, guess x(0)   
1 
Calculate
V2 (10sin  2 )  2.0


 2.0 
f(x(0) )  
   
2
1.0 
 V2 (10 cos 2 )  V2 (10)  1.0 
10 V2 cos 2
J (x(0) )  
10 V2 sin  2
10sin  2

10 0 
 

10 cos 2  20 V2 
 0 10 
1
 0.2 
 

 0.9 
0  10 0   2.0 
Solve x(1)     
  
1   0 10  1.0 
Energy Systems Research Laboratory, FIU
Once the voltage angle and magnitude at bus 2 are known we can
calculate all the other system values, such as the line flows and the
generator reactive power output
One
-200.0 MW
-100.0 MVR
Line Z = 0.1j
1.000 pu
Two
0.855 pu -13.522 Deg
200.0 MW
168.3 MVR
200 MW
100 MVR
Energy Systems Research Laboratory, FIU
Three Bus PV Case Example
 P2 ( x)  PG 2  PD 2 
f ( x)   P3 ( x)  PG 3  PD3   0


 Q2 (x)  QD 2 
Line Z = 0.1j
0 1j
0.941 pu
One
170.0 MW
68.2 MVR
1.000 pu
Line Z = 0.1j
Three
Two
Line Z = 0.1j
1.000 pu
30 MW
63 MVR
Energy Systems Research Laboratory, FIU
1
 0.2   8.82 1.986  0.212 
 0.233
x(2)  

 
   0.8586 
 0.9   1.788 8.199  0.279 


0.0145

0.236




x (3)  
f(x (2) )  


0.0190 
 0.8554 
0.0000906 
f(x (3) )  

 0.0001175 
Done!
V2  0.8554  13.52
PV Buses
• Since the voltage magnitude at PV buses is fixed
there is no need to explicitly include these voltages
in x or write the reactive power balance equations
– the reactive power output of the generator varies to
maintain
i t i the
th fixed
fi d terminal
t
i l voltage
lt
(within
( ithi limits)
li it )
– optionally these variations/equations can be included by
just writing the explicit voltage constraint for the
generator bus
|Vi | – Vi setpoint = 0
Energy Systems Research Laboratory, FIU
For this three bus case we have
 2 
x   3 
 
 V2 
0.9(10sin(0.2))  2.0

 0.212 
f(x (1) )  


2





0.9(
10cos(
0.2))
0.9
10
1.0

 0.279 
 8.82 1.986 
J (x(1) )  

 1.788 8.199 
Energy Systems Research Laboratory, FIU
Two Bus Solved Values
200.0 MW
168.3 MVR
Two Bus Example, Next Iterations
-7.469 Deg
200 MW
100 MVR
Solving Large Power Systems
• The most difficult computational task is inverting the
Jacobian matrix
– inverting a full matrix is an order n3 operation, meaning
the amount of computation increases with the cube of the
size size
– this amount of computation can be decreased substantially
by recognizing that since the Ybus is a sparse matrix, the
Jacobian is also a sparse matrix
– using sparse matrix methods results in a computational
order of about n1.5.
– this is a substantial savings when solving systems with
tens of thousands of buses
Energy Systems Research Laboratory, FIU
7
9/16/2011
“DC” Power Flow
Five Bus Power Flow Example
• The “DC” power flow makes some approximations to
the power balance equations to simplify the problem
5
400 MVA
15/345 kV
345 kV
100 mi
Line 1
400 MVA
15 kV
• This makes the power flow a linear set of equations,
-1
which can be solved directly = B P
• While the DC power flow is approximate, it is widely
used to get a feel for power system MW flows.
Energy Systems Research Laboratory, FIU
T2
800 MVA
4 345/15 kV
Line 3
345 kV
50 mi
Line 2
– completely ignore reactive power, assume all the voltages
are always 1.0 per unit, ignore line conductance, assumes
angles
l across the
th lines
li
are small.
ll
– Line flow is then approximated as (i – j)/Xij
T1
1
3
520 MVA
800 MVA
15 kV
40 Mvar 80 MW
345 kV
200 mi
2
280 Mvar
800 MW
Single-line diagram
Energy Systems Research Laboratory, FIU
37 Bus Power Flow Example
DC Power Flow Example
Metropolis Light and Power Electric Design Case 2
• For the system shown in the previous slide with bus 1 as
the system slack and with the below B matrix (100 MVA
base) numbered from buses 2 to 5, determine the bus
angles using the DC power flow approximation the and
the flow on the line between bus 1 and 5,
5 which has a pu
X of 0.02.
SLA CK3 4 5
A
MVA
A
MVA
2 2 0 MW
5 2 Mvar
RA Y 3 4 5
1 .0 3 pu
slack
System Losses: 10.70 MW
1 .0 2 pu
T IM3 4 5
A
A
MVA
MVA
A
SLA CK1 3 8
MVA
RA Y1 3 8
1 .0 2 pu
A
A
1 .0 3 pu
A
MVA
MVA
33 MW
1 3 M var
T IM1 3 8
1 .0 0 pu
A
A
MVA
3 9 MW
1 3 Mvar
H A NNA H 6 9
60 MW
1 9 M var
1 .0 0 pu
A
5 8 MW
4 0 Mvar
MVA
1 .0 0 pu
2 0 MW
1 2 Mvar
UIUC6 9
1 .0 0 pu
1 2 .8 Mvar
KYLE69
A
1 .0 0 pu
A
A
MVA
1 .0 2 pu
H O MER6 9
1 .0 1 pu
2 0 MW
3 M var
1 .0 0 pu
A
BLT 1 3 8
1 .0 0 pu
A
MVA
0 .0 M var
1 .0 0 pu
A
2 0 MW
2 8 Mvar
PA T T EN6 9
MVA
MVA
1 4 MW
4 5 MW
0 Mvar
WEBER6 9
2 2 MW
1 5 Mvar
MVA
1 0 MW
5 Mvar
RO GER6 9
2 Mvar
1 4 MW
3 Mvar
A
MVA
1 .0 2 pu
LA UF1 3 8
1 .0 1 pu
A
A
1 .0 0 pu
1 .0 1 pu
2 3 MW
6 Mvar
A
MVA
1 .0 0 pu
1 .0 1 pu
MVA
A
MVA
LA UF6 9
1 .0 2 pu
A
MVA
A
7 .3 Mvar
SA VO Y 6 9
BUCKY1 3 8
1 .0 2 pu
A
JO 1 3 8
3 8 MW
3 Mvar
MVA
A
MVA
1 .0 1 pu
A
A
MVA
1 5 MW
5 Mvar
55 MW
2 5 M var
A
MVA
MVA
1 .0 2 pu
SH IMKO 6 9
7 .4 Mvar
A
MVA
MVA
36 MW
1 0 M var
A
60 MW
1 2 M var
MVA
14 MW
4 Mvar
MVA
BLT 6 9
1 .0 1 pu
H A LE6 9
MVA
A
LYNN1 3 8
1 3 M var
16 MW
-1 4 Mvar
A
MVA
A
MVA
A
A
BO B6 9
56 MW
124 MW
4 5 M var
A
MVA
MVA
25 MW
3 6 M var
A M A NDA 6 9
MVA
A
MVA
MVA
MVA
25 MW
1 0 M var
B O B1 3 8
A
MVA
A
MVA
A
1 .0 1 pu
DEM A R6 9
A
0 .9 9 pu
WO LEN6 9
4 .9 Mvar
2 8 .9 M var
MVA
A
MVA
1 .0 1 pu
12 MW
3 M var
1 4 .2 M var
0 .9 9 pu
1 3 Mvar
FERNA 6 9
PET E6 9
A
12 MW
5 M var
A
MVA
A
MVA
A
MVA
H ISKY 6 9
3 7 MW
1 7 MW
3 Mvar
MVA
GRO SS6 9
1 .0 1 pu
A
MVA
MO RO 1 3 8
RA Y 6 9
1 .0 2 pu
PA I6 9
1 .0 1 pu
T IM6 9
A
MVA
A
1 .0 3 pu
MVA
1 8 MW
5 Mvar
A
MVA
MVA
1 .0 2 pu
2 3 MW
7 Mvar
A
MVA
MVA
A
1 5 .9 Mvar
JO 3 4 5
A
SA VO Y 1 3 8
1 5 0 MW
0 Mvar
MVA
A
MVA
MVA
1 5 0 MW
0 Mvar
A
P15 = (1 – 5)/X15 = 0.072/0.02 = 3.6 pu = 360 MW
Energy Systems Research Laboratory, FIU
Good Power System Operation
MVA
1 .0 2 pu
A
1 .0 3 pu
MVA
Energy Systems Research Laboratory, FIU
Looking at the Impact of Line Outages
Metropolis Light and Power Electric Design Case 2
• Good power system operation requires that there be no reliability
violations for either the current condition or in the event of
statistically likely contingencies
A
MVA
slack
T IM 3 4 5
A
MVA
A
SLA CK1 3 8
MVA
RA Y1 3 8
1 .0 3 pu
A
MVA
MVA
33 MW
1 3 M var
T IM1 3 8
1 .0 1 pu
A
A
MVA
1 .0 2 pu
1 6 .0 M var
3 9 MW
1 3 Mvar
H A NNA H6 9
60 MW
1 9 M var
0 .9 0 pu
FERNA 6 9
P ET E6 9
DEM A R6 9
KYLE69
A
1 .0 0 pu
A
20 MW
1 2 M var
UIUC6 9
1 .0 0 pu
1 2 .8 Mvar
A
MVA
25 MW
3 6 M var
A M A NDA 6 9
5 6 MW
124 MW
4 5 M var
A
SH IM KO 6 9
7 .3 Mvar
A
MVA
A
A
P A T T EN6 9
2 2 MW
1 5 Mvar
10 MW
5 M var
MVA
LA UF1 3 8
BUCKY1 3 8
ROGER6 9
2 M var
A
MVA
SA V OY6 9
1 .0 2 pu
A
3 8 MW
9 M var
JO 1 3 8
MVA
A
MVA
14 MW
1 4 MW
3 M var
1 .0 1 pu
1 .0 0 pu
MVA
MVA
45 MW
0 M var
WEBER6 9
1 .0 0 pu
23 MW
6 Mvar
A
80%
1 .0 1 pu
MVA
1 .0 0 pu
MVA
LA UF6 9
1 .0 1 pu
A
MVA
15 MW
5 Mvar
A
MVA
A
7 .2 M var
1 .0 0 pu
1 .0 2 pu
MVA
55 MW
3 2 M var
A
MVA
3 6 MW
1 0 Mvar
MVA
A
0 .9 9 pu
A
BLT 6 9
MVA
60 MW
1 2 M var
20 MW
4 0 M var
MVA
A
MVA
1 .0 1 pu
H A LE6 9
MVA
A
0 .0 M var
LYNN1 3 8
14 MW
4 M var
A
BLT 1 3 8
1 .0 0 pu
MVA
A
135%
20 MW
3 M var
0 .9 4 pu
A
MVA
1 3 Mvar
16 MW
-1 4 M var
A
MVA
A
1 .0 1 pu
MVA
BO B6 9
1 .0 2 pu
A
MVA
A
MVA
MVA
MVA
MVA
A
MVA
25 MW
1 0 M var
BOB1 3 8
MVA
A
0 .9 0 pu
MV A
WOLEN6 9
A
1 .0 1 pu
2 8 .9 Mvar
MVA
A
110%
HO MER6 9
1 3 M var
1 .0 1 pu
A
MVA
1 .0 0 pu
A
MVA
12 MW
3 M var
4 .9 M var
58 MW
4 0 M var
1 1 .6 M var
37 MW
17 MW
3 M var
A
MVA
MVA
A
12 MW
5 M var
RA Y6 9
1 .0 2 pu
A
GROSS6 9
A
H ISKY6 9
MO RO1 3 8
1 .0 3 pu
MVA
1 .0 1 pu
A
MVA
A
MVA
18 MW
5 M var
P A I6 9
1 .0 1 pu
T IM6 9
2 3 MW
7 M var
A
MVA
MVA
A
MVA
A
• North American Electric Reliability Corporation now has legal
authority to enforce reliability standards (and there are now lots of
them). See
http://www.nerc.com for details (click on Standards)
A
MVA
1 .0 2 pu
A
A
1 .0 0 pu
2 2 7 MW
4 3 Mvar
RA Y3 4 5
1 .0 3 pu
System Losses: 17.61 MW
MVA
• Reliability requires as a minimum that there be no transmission
line/transformer limit violations and that bus voltages be within acceptable
limits (perhaps 0.95
0 95 to 1.08)
1 08)
• Example contingencies are the loss of any single device. This is known as n1 reliability.
SLA CK3 4 5
A
MVA
1 .0 2 pu
1 .0 1 pu
A
MVA
SA V OY1 3 8
JO3 4 5
A
150 MW
4 Mvar
MVA
A
MVA
150 MW
4 Mvar
A
MVA
1 .0 2 pu
A
1 .0 3 pu
MVA
Opening one line (Tim69-Hannah69) causes an overload. This would not be
allowed (i.e., we can’t operate this way when line is in.
Energy Systems Research Laboratory, FIU
Energy Systems Research Laboratory, FIU
8
9/16/2011
Contingency Analysis
Generation Changes and The Slack Bus
Contingency
analysis provides
an automatic
way of looking
at all the
statistically
i i ll
likely
contingencies. In
this example the
contingency set
Is all the single
line/transformer
outages
Energy Systems Research Laboratory, FIU
• The power flow is a steady-state analysis tool, so
the assumption is total load plus losses is always
equal to total generation
• Generation mismatch is made up at the slack bus
• When doing generation change power flow studies
one always needs to be cognizant of where the
generation is being made up
• Common options include system slack, distributed across
multiple generators by participation factors or by
economics
Energy Systems Research Laboratory, FIU
Generation Change Example 1
Generation Change Example 2
SLA CK3 4 5
A
SLA CK3 4 5
A
MVA
A
MVA
A
MVA
MVA
0 .0 0 pu
sla ck
0 .0 0 pu
T IM 3 4 5
A
MVA
MVA
SLA CK1 3 8
T IM1 3 8
0 MW
0 M var
A
A
T IM6 9
MVA
0 .0 0 pu
0 MW
0 Mvar
RA Y6 9
A
A
HISKY6 9
MVA
A
0 MW
0 M var
0 .0 0 pu
DEMA R6 9
MVA
0 MW
0 Mvar
HA NNA H6 9
0 MW
0 M var
-0 .2 Mvar
A
MVA
0 MW
0 M var
0 .0 0 pu
0 MW
0 M var
A M A NDA 6 9
0 MW
0 M var
MVA
A
-0 .0 1 pu
0 .0 0 pu
MVA
A
0 MW
0 Mvar
MVA
MVA
0 .0 Mvar
0 MW
0 M var
0 .0 0 pu
0 MW
0 Mvar
A
0 MW
4 M var
0 .0 0 pu
0 .0 0 pu
BUCKY1 3 8
1 9 MW
5 1 M var
A
A
MVA
0 MW
0 Mvar
A
A
MVA
A
0 .0 0 pu
MVA
0 .0 0 pu
0 .0 0 pu
0 MW
0 Mvar
A
P A T T EN6 9
0 MW
0 M var
0 .0 0 pu
BUCKY1 3 8
0 .0 0 pu
4 2 MW
-1 4 M var
JO1 3 8
MVA
0 .0 0 pu
A
MVA
SA VOY 1 3 8
JO3 4 5
A
0 MW
0 Mvar
MVA
A
MVA
0 MW
0 Mvar
A
MVA
0 .0 0 pu
0 .0 0 pu
A
A
MVA
SA VOY 6 9
A
A
0 MW
2 Mvar
0 .0 0 pu
RO GER6 9
0 M var
0 MW
0 M var
0 .0 0 pu
MVA
0 MW
2 Mvar
MVA
A
MVA
MVA
0 MW
0 MW
0 M var
WEBER6 9
0 MW
0 M var
MVA
LA UF1 3 8
MVA
A
MVA
0 .0 0 pu
MVA
A
0 .0 Mvar
A
0 .0 0 pu
JO3 4 5
A
SA VOY 1 3 8
MVA
0 .0 0 pu
SHIM KO6 9
-0 .1 Mvar
MVA
MVA
MVA
JO1 3 8
0 MW
3 M var
LYNN1 3 8
0 MW
0 M var
A
BLT 6 9
-0 .0 1 pu
0 MW
0 Mvar
LA UF6 9
A
99 MW
-2 0 Mvar
0 .0 0 pu
A
0 .0 0 pu
A
HA LE6 9
MVA
0 .0 0 pu
A
MVA
SA VOY 6 9
MVA
A
MVA
A
0 .0 M var
MVA
RO GER6 9
0 M var
0 MW
0 M var
0 .0 0 pu
-0 .0 1 pu
MVA
0 MW
0 MW
0 M var
0 MW
0 M var
0 M var
A
MVA
A
MVA
MVA
0 MW
0 Mvar
MVA
A
P A T T EN6 9
WEBER6 9
0 MW
0 M var
MVA
LA UF1 3 8
0 MW
0 MW
0 Mvar
B LT 1 3 8
-0 .0 3 pu
MVA
A
0 .0 0 pu
0 .0 0 pu
0 MW
0 Mvar
A
MVA
MVA
-1 5 7 M W
-4 5 M var
A
A
MVA
A
A
A
MVA
LA UF6 9
0 .0 0 pu
BOB6 9
MVA
A
0 .0 0 pu
MVA
0 .0 0 pu
0 .0 M var
0 MW
0 M var
A
A
MVA
A
0 .0 0 pu
-0 .1 Mvar
MVA
A M A NDA 6 9
0 MW
0 M var
A
MVA
0 MW
0 Mvar
A
UIUC6 9
MVA
0 .0 0 pu
A
MVA
MVA
0 MW
5 1 M var
A
MVA
MVA
0 MW
0 M var
A
MVA
MVA
HA LE6 9
A
BLT 6 9
MVA
0 .0 0 pu
A
MVA
0 .0 0 pu
HO MER6 9
MVA
MVA
A
A
A
SHIM KO6 9
0 .0 M var
MVA
MVA
0 .0 0 pu
A
0 MW
0 M var
0 .0 0 pu
-0 .2 Mvar
A
A
MVA
A
HO MER6 9
DEMA R6 9
0 MW
0 Mvar
-0 .1 Mvar
-0 .0 0 3 pu
0 MW
0 M var
MVA
A
A
WO LEN6 9
A
BO B1 3 8
-0 .0 3 pu
P ET E6 9
MVA
LYNN1 3 8
A
B LT 1 3 8
-0 .0 3 pu
0 Mvar
0 .0 0 pu
0 .0 0 pu
A
0 MW
0 M var
0 MW
0 Mvar
A
MVA
A
MVA
0 MW
0 M var
HA NNA H6 9
0 MW
0 M var
MVA
-1 5 7 M W
-4 5 M var
A
MVA
MVA
MVA
MVA
0 .0 0 pu
0 MW
0 MW
0 M var
FERNA 6 9
A
0 MW
0 Mvar
BOB6 9
0 .0 0 pu
A
0 .0 0 pu
-0 .1 Mvar
A
A
A
-0 .0 0 2 pu
RA Y6 9
A
MVA
MVA
0 .0 M var
MVA
MVA
UIUC6 9
-0 .1 Mvar
A
MVA
MVA
0 .0 0 pu
A
MVA
GROSS6 9
0 .0 0 pu
A
0 .0 0 pu
A
0 MW
0 Mvar
P A I6 9
HISKY6 9
MVA
BO B1 3 8
-0 .0 3 pu
P ET E6 9
MVA
0 .0 0 pu
MVA
0 MW
0 M var
A
-0 .0 1 pu
A
0 .0 0 pu
MVA
A
MVA
A
A
0 MW
0 Mvar
-0 .1 Mvar
0 .0 0 pu
A
0 MW
0 M var
MO RO1 3 8
WO LEN6 9
0 .0 0 pu
0 MW
0 M var
MVA
-0 .1 Mvar
MVA
A
A
MVA
FERNA 6 9
MVA
A
T IM6 9
0 .0 0 pu
0 Mvar
MVA
A
MVA
A
MVA
RA Y1 3 8
MVA
0 MW
0 M var
MVA
MVA
0 MW
0 MW
0 M var
MVA
GROSS6 9
0 .0 0 pu
A
MVA
-0 .0 1 pu
A
P A I6 9
0 .0 0 pu
A
0 MW
0 M var
MO RO1 3 8
MVA
A
A
-0 .1 Mvar
MVA
MVA
0 .0 0 pu
A
MVA
A
SLA CK1 3 8
A
T IM1 3 8
0 .0 0 pu
MVA
MVA
A
-0 .0 1 pu
A
MVA
0 .0 0 pu
A
0 .0 0 pu
sla ck
T IM 3 4 5
MVA
RA Y1 3 8
A
0 MW
3 7 Mvar
RA Y3 4 5
A
0 .0 0 pu
A
-0 .0 1 pu
A
MVA
0 .0 0 pu
1 6 2 MW
3 5 Mvar
RA Y3 4 5
A
A
0 .0 0 pu
MVA
MVA
Display shows “Difference Flows” between original 37 bus case, and case with
a BLT138 generation outage; note all the power change is picked up at the slack
Energy Systems Research Laboratory, FIU
Display repeats previous case except now the change in generation is picked
up by other generators using a participation factor approach
Energy Systems Research Laboratory, FIU
What does the Future Hold for Wind?
Sitting New Wind Generation Example
MVA
1.02 pu
System Losses:
1.02 pu
RAY345
sl
8.73 MW
TIM345
A
A
M VA
M VA
A
SLACK138
1.01 pu
A
A
MVA
MVA
M VA
RAY138
1.03 pu
A
TIM138
1.00 pu
M VA
A
1.02 pu
MVA
1.02 pu
A
A
MVA
MVA
1.02 pu
PAI69
1.01 pu
TIM69
1.01 pu
A
M VA
A
MORO138
GROSS69
A
FERNA69
M VA
1.00 pu
12 MW
3 Mvar
M VA
HISKY69
PETE69
M VA
A
M VA
1.01 pu
M VA
0.99 pu
12 MW
5 Mvar
1.00 pu
A
M VA
A
A
A
M VA
RAY69
A
M VA
HANNAH69
DEMAR69
Wind69
BOB138
A
A
MVA
1.00 pu
UIUC69
1.00 pu
50 MW
20 MW
12 Mvar
MVA
BOB69
1.01 pu
A
M VA
1.00 pu
A
A
MVA
MVA
56 MW
0 MW
0 Mvar
13 Mvar
A
A
A
M VA
A
HOMER69
0.99 pu
MVA
-2 Mvar
AMANDA69
1.00 pu
1.00 pu
MVA
SHIMKO69
M VA
M VA
1.01 pu
A
A
BLT138
A
A
MVA
A
BLT69
MVA
MVA
HALE69
M VA
A
1.01 pu
A
MVA
A
MVA
A
1.01 pu
A
MVA
MVA
M VA
1.00 pu
A
1.00 pu
A
M VA
LAUF69
1.02 pu
A
A
Energy Systems Research Laboratory, FIU
PATTEN69
MVA
14 MW
1.01 pu
WEBER69
2 Mvar
A
ROGER69
• Wind has experienced rapid growth over the last decade;
but with a large drop in installations expected for 2010.
• Nuclear and renewal generation are running into strong
“head winds”
causedd bby low
l
natural gas prices.
• Expiring tax credits is a
continual concern.
• State renewable
portfolio standards will
help with growth.
Energy Systems Research Laboratory, FIU
9
9/16/2011
Hurricanes Impact Energy Prices
Hurricanes Impact Energy Prices
• Many oil refineries
and natural gas
pipelines off the coast
• Need to be shut down
and evacuated
• Takes time to get the
systems back up and
running afterwards
Hurricane Ike, Sept. 2008
http://tonto.eia.doe.gov/oog/special/hurricanes/
gustav_091308.html
http://www.eia.doe.gov/emeu/steo/pub/special/pdf/2008_sp_03.pdf
Energy Systems Research Laboratory, FIU
Energy Systems Research Laboratory, FIU
What does the Future Hold for Wind?
Distributed Generation (DG)
Vestas to Cut 3000 Jobs (14%) (10/26/10)
Vestas Stock Price Over Last Five Years
• Small-scale, up to about 50 MW
• Includes renewable and non-renewable
sources
• May
M bbe iisolated
l t d from
f
the
th grid
id or gridid
connected
• Usually near the end user
Closed
at 176 on
Wednesday and
169.4 on
Thursday!
Long-term the outlook for wind is probably good. On 9/1/10 EIA reduced their 2010 forecast
for US capacity
additions to 4.3 GW in 2010 and 6.5 GW in 2011; the totals for 2007 were 5.2 GW, 2008 8.4
GW and 2009 10.0GW
Energy Systems Research Laboratory, FIU
Energy Systems Research Laboratory, FIU
Pluggable Hybrid Electric Vehicles
(PHEVs) as Distributed Generation
Integrated Generation, Transmission,
Buildings, Vehicles
•
Can charge at night when electricity is
cheap
Renewables
Grid
kWh
kWh
PHEV
Smart
meters
Vehicle-to-Grid
Heat
N. Gas
Energy Systems Research Laboratory, FIU
kWh
Combined
Heat and
Power (CHP)
Source: Masters
Source: http://www.popularmechanics.com/automotive/new_cars/4215489.html
• Can provide services
back to the grid
Energy Systems Research Laboratory, FIU
Source: www.calcars.org
10
9/16/2011
DG Technologies, Excepting Solar
San Francisco charging stations, 2009
•
•
•
•
•
•
http://en.wikipedia.org/wiki/File:Charging_stations_in_SF_City_Hall_02_2009_02.jpg
Energy Systems Research Laboratory, FIU
Energy Systems Research Laboratory, FIU
Terminology
Reasons for Distributed Generation
•
•
•
•
•
Microturbines
Reciprocating Internal Combustion Engines
Biomass
y
Micro-Hydro
Fuel Cells (we’ll be skipping)
Concentrated solar power is talked about in Chapter 4,
but we’ll skip until after we cover Chapter 7.
• Cogeneration and Combined Heat and Power (CHP)
Good for remote locations
Renewable resources
Reduced emissions
Can use the waste heat
Can sell power back to the grid
– capturing and using waste heat while generating electricity
• When fuel is burned one product is water; if water vapor
exits stack then its energy is lost (about 1060 Btu per
ppound of water vapor)
p )
• Heat of Combustion for fuels
– Higher Heating Value (HHV) – gross heat, accounts for latent
heat in water vapor
– Lower Heating Value (LHV) – net heat, assumes latent heat in
water vapor is not recovered
– Both are used - Conversion factors (LHV/HHV) in Table 4.2
Energy Systems Research Laboratory, FIU
Energy Systems Research Laboratory, FIU
HHV and LHV Efficiency
Microturbines
• Find LHV efficiency or HHV efficiency from
the heat rate:
HHV( LHV ) 
3412 Btu/kWh
Heat Rate (Btu/kWh) HHV( LHV )
• Convert to get the
other
LHV
 efficiency:
HHV  LHV 
 HHV 
(3.16)
• Small natural gas turbines,
500 W to 100s kW
• Only one moving part
• Combined heat and power
80% CHP
• High overall efficiencyEfficiency
(4.1)
230 kW fuel
120 kW hot
water output
65 kW electrical
output
45 kW waste heat
Capstone 65 kW Microturbine
Note the LHV is less than the HHV
Source: http://www.capstoneturbine.com
Energy Systems Research Laboratory, FIU
Energy Systems Research Laboratory, FIU
11
9/16/2011
Microturbines
Microturbines and Renewable Energy
1. Incoming air is
compressed
2. Moves into cool side of
recuperator & is heated
3. Mixes with fuel in
combustion chamber
4. Expansion of hot gases
spins shaft
5. Exhaust leaves
Figure 4.1
Energy Systems Research Laboratory, FIU
Biomass – Current Conditions
• Biomass, including waster, currently provides about 4% of the US
total energy, a value that is projected to grow to about 7% by
2030.
• In 2008 the 3.85 quad of biomass was split between biofuels such
as ethanol (1.37 quad), waste such as landfill gas (0.436 quad)
and wood (2.04 quad).
– About 1/10 of the wood was used to produce electricity, primarily by the
paper industry; total generation capacity in US is about 11.3 GW
– We are not considering liquid fuels for non-electricity use in ECE 333;
ethanol usage was growing at 30% per year
• Little to no fuel cost for wood waste but little growth
Energy Systems Research Laboratory, FIU
Biomass – Future Possibilities for Electricity
• A full consideration of biomass is beyond our scope of since it
gets into agricultural economics issues
• Miscanthus can be harvested at rates of about 15 tons per acre in
Illinois Once established it does not need to be replanted.
Illinois.
replanted This
allows the energy potential of about 225 Mbtu per acre; income
depends on energy price, say $2/Mbtu = $450 per acre. For
comparison corn can yield up to 200 bushels per acre at say
$5.50/bushel = $1100 per acre
• Microturbines are not a renewable energy source since they
ultimately use natural gas as their fuel. But when used for
combined heat and electricity they can be quite efficient and emit
significantly less CO2 than coal.
• A possible scenario: If natural gas prices remain low,
commercial/industrial entities will increasingly go “off grid.”
Hence they will not be subject to state renewable portfolio
standards or utility taxes. This will mean renewable energy
subsidies will increasingly be born by residential customers.
Average residential price per therm is $0.57 in October in Illinois;
with a heat rate of 8.5 this gives electricity at 4.8 cents/kWh.
Energy Systems Research Laboratory, FIU
Biomass – Future Possibilities for
Electricity
• Newer crops are being considered for future biomass,
including various grasses (such as Miscanthus and
Switch Grass) along with algae
• Potential uses include both fuel to
create
t electricity
l t i it (primarily
( i
il the
th
grasses), and conversion to liquid
fuels such as ethanol.
– On campus 320 acres in the South
Farms are devoted to biofuel research
Energy Systems Research Laboratory, FIU
Cofiring
• Burn biomass and coal
• Modified conventional steam-cycle plants
• Allows use of biomass in plants with higher
efficiencies
ffi i i
• Reduces overall emissions
– Energy yield is about 225/15 = 15 Mbtu/ton which is similar to coal
– Not a native species so containment could be an issue
Energy Systems Research Laboratory, FIU
Energy Systems Research Laboratory, FIU
12
9/16/2011
Gas Turbines and Biomass
• Cannot run directly on biomass without
causing damage
• Gassify the fuel first and clean the gas before
combustion
• Coal-integrated gasifier/gas turbine
(CIG/GT) systems
• Biomass-integrated gasifier/gas turbine
(BIG/GT) systems
Energy Systems Research Laboratory, FIU
Biomass and Transportation
• A key issue associated with biomass is the transportation
costs – these grasses are quite bulky.
• A rough estimate of the cost per ton for transportation is
about $1 + 0.1 * round trip distance in miles +
h
harvesting
ti costs
t off about
b t $22 per ton.
t
So
S if the
th power
plant is 50 miles distant, total cost would be $33 per ton,
or about $33/ton/($15 Mbtu/ton) = $2.2 per Mbtu
• Total US corn planting is about 87 million acres, which
if planted in grass could yield about 20 quad of energy.
– Won’t meet all our needs but could play a major role
Energy Systems Research Laboratory, FIU
A Biomass plant
Energy Systems Research Laboratory, FIU
13