Homework #1, EE 553, Fall, 2012, Dr.

McCalley

SOLUTIONS

1. Problem 4.1a.
2. Before working this problem, you may review page 102 of your W&W text (note that in class, we use
V to denote voltages whereas your W&W text uses E to denote voltages).
Observe that Jacobian derivative expressions given by equations T7.47 and T7.53 in your notes are
almost the same. Also observe that Jacobian derivative equations T7.49 and T7.51 in your notes are
almost the same. These two sets of equations are given below.

Pj ( x)
 V j Vk G jk sin( j   k )  B jk cos( j   k ) 
P
J jk  (T7.47)
 k
Q j ( x)
J jk   V j G jk sin( j   k )  B jk cos( j   k ) 
QV
(T7.53)
 Vk
Q j ( x)
  V j Vk G jk cos( j   k )  B jk sin( j   k ) 
Q
J jk  (T7.49)
 k
Pj ( x)
  V j G jk cos( j   k )  B jk sin( j   k ) 
PV
J jk (T7.51)
 Vk

a. What is the difference between the right-hand-side expression of T7.47 and that of T7.53?
Solution: T7.47 has an extra |Vj|.
What is the difference between the right-hand-side expression of T7.49 and that of T7.51?
Q j ( x)
b. Modify (T7.53) to revise the Jacobian element to J jk
QV
 Vk
 Vk

Solution:
Q j ( x)
J jk QV ,new  Vk
 Vk

 Vk V j G jk sin( j   k )  B jk cos( j   k ) 
Pj ( x)
c. Modify (T7.51) to revise the Jacobian element to J jk
PV
  Vk
 Vk

Solution:
Pj ( x)
J jk PV ,new  Vk
 Vk

 Vk V j G jk cos( j   k )  B jk sin( j   k ) 

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d. Express the relations in (T7.42e) below to account for the two modifications made above. This
requires that you express the new Jacobian submatricies J PV,new and JQV,new in terms of its revised
elements (from parts (a) and (b) above). It also requires that you modify the elements of the
voltage-related part of solution vector ∆|V|new.

J P θ  J PV |V |  P
(T7.42e)
J Q θ  J QV |V | Q
Solution:
Define:
 P2 ( x) P2 ( x) P2 ( x) 
 V N g 1 VN g 2  VN
 V N g 1  VN g 2  VN 
 
 P3 ( x) P3 ( x) P3 ( x) 
 V N 1 VN g 2  VN 
J PV ,new
 g  V N g 1  VN g 2  VN 
 
     
V PN ( x) PN ( x) P3 N ( x) 
VN g 2  VN
 N g 1  V N g 1  VN g 2  VN 
 
 Q N g 1 ( x) Q N g 1 ( x) Q N g 1 ( x) 
 V N g 1 VN g 2  VN 
  V N g 1  VN g 2  VN 
 
 Q N g  2 ( x) Q N g  2 ( x) Q N g  2 ( x) 
V VN g 2  VN
J QV ,new  g N  1
 V N g 1  VN g 2  VN 
 
     
 Q N ( x) Q N ( x) Q3 N ( x) 
 V N g 1 VN g 2  VN 
  V N g 1  VN g 2  VN 
 
T
new
 |V N g 1 | |V N g  2 | |V N | 
V   
 |V N g 1 | |V N g 2 | |V N | 

Then
J P θ  J PV ,new V
new
  P
J Q θ  J QV ,new V
new
  Q

This can be expressed in matrix form as

 J P J PV ,new   θ   P 
 Q   new     Q 
 J J QV ,new   V   

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 e. What is the advantage of this new formulation? Hint: Express and compare JjkPθ to JjkQV,new and
express and compare JjkQθ to JjkPV,new.

Solution:
To see the advantage, one must observe:
Pj ( x)
 V j Vk G jk sin( j   k )  B jk cos( j   k ) 
P
J jk  (T7.47)
 k
Q j ( x)
J jk QV ,new  Vk
 Vk

 Vk V j G jk sin( j   k )  B jk cos( j   k )  (T7.53)

Q j ( x)
  V j Vk G jk cos( j   k )  B jk sin( j   k ) 
Q
J jk  (T7.49)
 k
Pj ( x)
J jk PV ,new  Vk
 Vk

 Vk V j G jk cos( j   k )  B jk sin( j   k )  (T7.51)

Thus, we see that

J jk P  J jk QV ,new
Q
  J jk
PV , new
J jk
and

Therefore, the benefit is in terms of computation (only have to compute one element for each pair
of the above) and in storage (only have to store one element for each pair of the above).

3. A transformer with an off-nominal tap ratio t connects two buses as shown in Fig. 1. The transformer
admittance is y. An equivalent representation of the Fig. 1 transformer, used for power flow
calculations, is shown in Fig. 2. For each model, give the elements of the admittance matrix Y, where
I=YV, I=[I1 I2]T, and V=[V1 V2]T. From your results, express the admittances of Fig. 2 in terms of
tap t and transformer admittance y. Doing so will allow us to model the off-nominal turns transformer
of Fig. 1 in the standard π-equivalent model of Fig. 2 so that we can represent the tap changer in our
power flow algorithm.
Transformer
I1 I2
V1 y V2
t
Bus 1 Bus 2

Fig. 1
I1 I2
V1 yL V2

Bus 1 ys1 ys2 Bus 2

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 Fig. 2

Solution:
From Fig. 1,

 V 1t  V 2 y  I 1  t 2 yV 1  tyV 2
I1
t
I 2  V 2  tV 1 y  I 2  yV 2  tyV 1
In matrix form, we have
 I1  t 2 y  ty  V 1 
I     
 2   ty y  V 2 
From Fig. 2,
I 1  V 1  V 2 y L  V 1 y s1  I 1  V 1  y L  y s1   V 2 y L

I 2  V 2  V 1 y L  V 2 y s 2  I 2  V 2 y L  y sq  V 1 y L 
In matrix form, we have
 I 1   y L  y s1  y L  V 1 
I     y y L  y s 2  V 2 
 2  L

To model the off-nominal turns transformer of Fig. 1 in the standard y-equivalent model of Fig. 2 (so that we
can represent the tap changer in our power flow studies), we require:
 Elements in row 1, column 2, and in row 2, column 1, to be equal:
 ty   y L  y L  ty (eq *)
 Elements in row 1, column 1 to be equal:
y L  y s1  t 2 y  y s1  t 2 y  y L
But by (eq *), we have:
y s1  t 2 y  ty  t (t  1) y
 Elements in row 2, column 2 to be equal:
y L  ys2  y  ys2  y  y L
Again, by (eq *), we have:
y s 2  y  ty  (1  t ) y

In summary, when the program data indicates a non-unity tap transformer, the program should
model it as a standard pi-equivalent circuit (as in Fig. 2) with numerical values according to:
y L  ty
y s1  t (t  1) y
y s 2  (1  t ) y

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4. Consider that the transformer of Fig. 1 is a tap-changing-under-load (TCUL) transformer, regulating
bus 1, and that buses 1 and 2 are interconnected to a larger system. Bus 2 is a type PQ bus. Which
parameters associated with the bus 1 to bus 2 subsystem would be included in the state vector ∆x used
to solve the equation J∆x=-f(x) for one iteration of the Newton-Raphson power follow solution? That
is, what are the unknowns for this subsystem? Note that the matrices J and f(x) are the Jacobian and
the mismatch vector, respectively.

Solution: Since the TCUL transformer is regulating bus 1, the voltage magnitude at bus 1 is fixed and
not a variable. This regulation is performed by modifying the tap, t, and so it is a variable, along with
both voltage angles and the bus voltage magnitude at bus 2. Therefore the variables are:
V2, t, θ1, and θ2.

5. Consider the two-bus system of Fig. 3, where the tap t is used to regulate the voltage at bus 2 to be
equal to 1.0. pu.
Transformer
V2=V2∟θ2

V1=1∟0º y V2
=
t
Bus 1 Bus 2

P2+jQ2= - (Pd+jQd)

Fig. 3
To solve this system by Newton-Raphson, we need to develop the equation J∆x=-f(x). For this
system, specify J, ∆x, and f(x) in terms of the unknowns in the problem and the admittance y=|y|∟γ.

Solution: From Problem 3, Fig. 3 can be redrawn as below:

I1 I2
V1 ty V2

Bus 1 t(t-1)y (t-t)y Bus 2

The Y-bus is therefore:

Y11 Y12   | Y11 |  11 | Y12 |  12   t 2 y  ty   t 2 | y |   t | y |  

Y bus      
Y21 Y22  | Y21 |  21 | Y22 |  22   ty y   t | y |  | y |  

We have two unknowns: θ2 and t, so that the solution vector is x=[ θ2 t ]T.
And so we need two equations.
These equations will be the real and reactive power flow equations at bus 2, which we write below.

P2 ( x)  V2 | Y21 | V1 cos( 2   21) | Y22 | V2 cos( 22 )

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Q2 ( x)  V2 | Y21 | V1 sin( 2   21) | Y22 | V2 sin( 22 )

With V1 and V2 specified to be 1.0, and with |Y21|=-t|y| and |Y22|=|y|, we have:
P2 ( x)  t | y | cos( 2   ) | y | cos( )
Q2 ( x)   t | y | sin( 2   ) | y | sin( )
And so the equations to solve are, in vector form, as follows:
 f1 ( x )   P2 ( x )  P2   t | y | cos( 2   ) | y | cos( )  P2  0
f ( x)      
 f 2 ( x ) Q2 ( x )  Q2    t | y | sin( 2   ) | y | sin( )  Q2  0 ,
with the solution vector given as:
 2 
x 
t 
Then the Jacobian is written as:
 f 1 f1 
 t   t | y | sin( 2   )  | y | cos( 2   )
J 2    t | y | cos(   )  | y | sin(   ) 
 f 2 f 2   2 2 
  2 t 
Define the mismatch vector as
 P   P2 ( x )  P2 
Q   Q ( x )  Q 
   2 2

Then we can write:

 P2 ( x )  P2 
J x    
Q2 ( x )  Q2 
Or
    P2 ( x )  P2 
J  2    
 t  Q 2 ( x )  Q 2 
And this will allow us to perform the Newton-Raphson method.