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POWER (LOAD)
  FLOW STUDY
              INTRODUCTION
           BASIC TECHNIQUES
              TYPE OF BUSES
                Y BUS MATRIX
  POWER SYSTEM COMPONENTS
      BUS ADMITTANCE MATRIX



      POWER SYSTEM 2 - EPO 643   1
INTRODUCTION
   Power (Load) flow study is the analysis of a
    power system in normal steady-state
    operation
   This study will determine:
     Voltages
     Currents                   In a power system under a
                                 given set of load conditions
     Real power
     Reactive power

    Why we need load flow study?

                         POWER SYSTEM 2 - EPO 643               2
   The power flow problem was originally
    motivated within planning environments
    where engineers considered different network
    configurations necessary to serve an
    expected future load.
   Later, it became an operational problem as
    operators and operating engineers were
    required to monitor the real-time status of the
    network in terms of voltage magnitudes and
    circuit flows.
                     POWER SYSTEM 2 - EPO 643     3
   A power flow solution procedure is a numerical
    method that is employed to solve the power flow
    problem.
   A power flow program is a computer code that
    implements a power flow solution procedure.
   The power flow solution contains the voltages and
    angles at all buses, and from this information, we
    may compute the real and reactive generation and
    load levels at all buses and the real and reactive
    flows across all circuits.
                      POWER SYSTEM 2 - EPO 643       4
Terminology
   The above terminology is often used with the
    word “load” substituted for “power,” i.e., load
    flow problem, load flow solution procedure,
    load flow program, and load flow solution.
   However, the former terminology is preferred
    as one normally does not think of “load” as
    something that “flows.”


                     POWER SYSTEM 2 - EPO 643     5
Power system components
   Generator
   Transmission Lines
   Load




                     POWER SYSTEM 2 - EPO 643              6
                                                Figure 1
Generator
   Generators have maximum and minimum real and
    reactive power capabilities.
   Maximum reactive power capability:
       maximum reactive power that the generator may produce
        when operating with a lagging power factor.
   minimum reactive power capability:
       maximum reactive power the generator may absorb when
        operating with a leading power factor.
   These limitations are a function of the real power
    output of the generator,
       as the real power increases, the reactive power limitations
        move closer to zero.
                           POWER SYSTEM 2 - EPO 643               7
Figure 2
POWER SYSTEM 2 - EPO 643   8
   Figure 2 illustrates several important elements of the
    power flow problem.
   First, identify each bus depending on whether
    generation and/or load is connected to it.
   A bus may have
       generation only (buses B1, B2, and B3),
       load only (buses B5, B7, and B9),
       neither generation or load (buses B4, B6, and B8).
       both generation and load (leads us to define “bus
        injection”)

                        POWER SYSTEM 2 - EPO 643         9
Basic Technique for Load Flow
Studies
   In a load flow study, assumptions are made
    about:
       Voltage at a bus or                        For each bus
                                                   in the system
       Power being supplied to the bus




                        POWER SYSTEM 2 - EPO 643                   10
Types of Buses
   For each bus, there are four possible
    variables that characterize the buses
    electrical condition.
   The four variables are
       real and reactive power injection, Pi and Qi,
       voltage magnitude and angle, |Vi| and δi ,
        respectively



                         POWER SYSTEM 2 - EPO 643       11
Types of Buses (cont..)
   Generation Bus
       Also called the P-V bus or voltage-controlled buses
       Voltage magnitude |Vi| and real power Pi are specified
       Able to specify (and therefore to know) the voltage
        magnitude of this bus.
       Most generator buses fall into this category, independent of
        whether it also has load

   Load Bus
       Also called the P-Q bus
       Real power Pi and Qi are specified
       All load buses fall into this category, including buses that
        have not either load or generation.
                           POWER SYSTEM 2 - EPO 643               12
   Slack or Swing Bus
       Known as reference bus
       Voltage magnitude |Vi| and phase angle δi are
        specified
       There is only one swing bus, and it can be
        designated by the engineer to be any generator
        bus in the system.
       This generator “swings” to compensate for the
        network losses, or, one may say that it “takes up
        the slack.”

                        POWER SYSTEM 2 - EPO 643            13
Bus types        Quantities             Unknown
                 specified              values
Generator Bus |Vi| , Pi                 Q i , δi


Load Bus         Pi , Qi                |Vi| , δi


Slack Bus        |Vi| , δi              Pi , Qi



             POWER SYSTEM 2 - EPO 643               14
Bus injection
    An injection is the power (P or Q), that is being
    injected into or withdrawn from a bus by an element
    having its other terminal (in the per-phase
    equivalent circuit) connected to ground. Such an
    element would be either a generator or a load.
   Positive injection is defined as one where power is
    flowing from the element into the bus.
   Negative injection is then when power is flowing
    from the bus, into the element.


                      POWER SYSTEM 2 - EPO 643        15
Bus injection (cont..)
   Generators normally have positive real power
    injections, although they may also be assigned
    negative real power injections when they are
    operating as a motor.
   Generators may have either positive or negative
    reactive power injections:
       positive if the generator is operating lagging and delivering
        reactive power to the bus,
       negative if the generator is operating leading and
        absorbing reactive power from the bus, and
       zero if the generator is operating at unity power factor.

                            POWER SYSTEM 2 - EPO 643               16
Loads
   Loads normally                               Pk=100
                                                 Qk=30
                                                                           Pk= - 40
                                                                           Qk= -20
    have negative
    real and reactive
    power injections.                              (a)                        (b)

   Figure 3: Illustration of (a)
                                                         Pk=100+(-40)=60
    positive injection, (b)                              Qk=30+(-20)=10
    negative injection, and (c)
    net injection

                                                                (c)

                             POWER SYSTEM 2 - EPO 643                                 17
   Figure 3 illustrates the net injection as the algebraic
    sum when a bus has both load and generation;
   In this case, the net injection for both real and
    reactive power is positive (into the bus).
   Thus, the net real power injection is Pk=Pgk-Pdk, and
    the net reactive power injection is Qk=Qgk-Qdk.
   We may also refer to the net complex power
    injection as Sk=Sgk-Sdk, where Sk=Pk+jQk.


                        POWER SYSTEM 2 - EPO 643              18
Power Flow solution
   Most common and important tool in power
    system analysis
       also known as the “Load Flow” solution
       used for planning and controlling a system
       assumptions: balanced condition and single phase analysis
   The utility wants to know the voltage profile
        the nodal voltages for a given load and generation
        schedule
   From the load flow solution –
       the voltage magnitude and phase angle at each bus could
        be determined and hence the active and reactive power
        flow in each line could be calculated

                           POWER SYSTEM 2 - EPO 643               19
   The currents and powers are expressed as
    going into the bus
       for generation the powers are positive
       for loads the powers are negative
       the scheduled power is the sum of the generation
        and load powers




                        POWER SYSTEM 2 - EPO 643       20
The Bus Admittance Matrix
   The matrix equation for relating the nodal
    voltages to the currents that flow into and out
    of a network using the admittance values of
    circuit branches is given by :-




                     POWER SYSTEM 2 - EPO 643     21
Forming the Admittance Matrix

              1              y13               3               4
         I1                                              y34
                                  2                                 I4
                  y12                 y23
                             I2                     I3
    y1                                                         y4
                        y2               y3




                             POWER SYSTEM 2 - EPO 643                    22
From Kirchoff’s Current Law (KCL) –
 the current injections be equal to the sum of
  the currents flowing out of the bus and into
  the lines connecting the bus to other buses,or
  to the ground.
 Therefore, recalling Ohm’s Law, I=V/Z=VY,
  the current injected into bus 1 may be written
  as:
         I1=(V1-V2)y12 + (V1-V3)y13 + V1y1
                  POWER SYSTEM 2 - EPO 643     23
   I2 = (V2 - V1)(y21) + V2 y2 + (V2 – V3)y23

   I1= V1( y1 + y12 + y13) + V2(-y12) + V3(-y13)

   I2= V1(-y21) + V2( y2 + y21 + y23) + V3(-y23)

   I3= V1(-y31) + V2(-y32) + V3( y3 + y31 + y32+ y34) +
        V4(-y34)

   I4= V3(-y43) + V4( y4 + y43)
                          POWER SYSTEM 2 - EPO 643         24
Admittance Matrix


  ( y1 + y12 + y13 )         - y12                  - y13                  0 
         - y 21      ( y 2 + y 21 + y 23 )          - y 23                 0 
                                                                                  
         - y31               - y32         ( y3 + y31 + y32 + y34 )      - y34 
                                                                                  
           0                   0                    - y 43          ( y 4 + y 43 )




                                POWER SYSTEM 2 - EPO 643                               25
Matrix Equation


  I1   ( y1 + y12 + y13 )         - y12                  - y13                0   V1 
 I            - y 21      ( y 2 + y 21 + y 23 )          - y 23               0   2 V 
  2 = 
  I3          - y31               - y32         ( y3 + y31 + y32 + y34 ) - y34   V3 
                                                                                       
  I4            0                   0                    - y 43         ( y 4 + y 43 )  V4 




                                     POWER SYSTEM 2 - EPO 643                                       26
Y-Bus Matrix Building Rules
   The matrix is symmetric, i.e., Yij=Yji.
   A diagonal element Yii = Self Admittance
       is obtained as the sum of admittances for all branches
        connected to bus i, including the shunt branch
                                          N
                      Yii = y i +        ∑y
                                      k =1, k ≠ i
                                                    ik



   The off-diagonal elements are the negative of the
    admittances connecting buses i and j, i.e., Yij=-yji =
    mutual admittance.
                           POWER SYSTEM 2 - EPO 643              27
   E.g. for a 4 bus system

              Y11 Y12                    Y13     Y14 
              Y    Y22                   Y23         
                                                  Y24 
           Y =  21

              Y31 Y32                    Y33     Y34 
                                                     
              Y41 Y42                    Y43     Y44 

                       POWER SYSTEM 2 - EPO 643           28
The power flow equations
   The net complex power injection into a bus
         Sk=Sgk-Sdk
          Sk=VkIk*
          Vk=| Vk|∠δk
          Ik =Σ | Ykj|∠θkj | Vj|∠δj
          Ik =Σ | Ykj|| Vj|∠ (θkj + δj)
          Ik* =Σ | Ykj|| Vj|∠ -(θkj + δj)
                       POWER SYSTEM 2 - EPO 643   29
Sk=VkIk*
Sk= | Vk|∠δk x Σ | Ykj|| Vj|∠ -(θkj + δj)
Sk= Σ | Ykj| | Vk|| Vj|∠ (δk - δj - θkj )
Pk= Σ | Ykj| | Vk|| Vj|cos (δk - δj - θkj )
Qk= Σ | Ykj| | Vk|| Vj|sin (δk - δj - θkj )



                        POWER SYSTEM 2 - EPO 643   30

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load flow 1

  • 1. POWER (LOAD) FLOW STUDY INTRODUCTION BASIC TECHNIQUES TYPE OF BUSES Y BUS MATRIX POWER SYSTEM COMPONENTS BUS ADMITTANCE MATRIX POWER SYSTEM 2 - EPO 643 1
  • 2. INTRODUCTION  Power (Load) flow study is the analysis of a power system in normal steady-state operation  This study will determine:  Voltages  Currents In a power system under a given set of load conditions  Real power  Reactive power Why we need load flow study? POWER SYSTEM 2 - EPO 643 2
  • 3. The power flow problem was originally motivated within planning environments where engineers considered different network configurations necessary to serve an expected future load.  Later, it became an operational problem as operators and operating engineers were required to monitor the real-time status of the network in terms of voltage magnitudes and circuit flows. POWER SYSTEM 2 - EPO 643 3
  • 4. A power flow solution procedure is a numerical method that is employed to solve the power flow problem.  A power flow program is a computer code that implements a power flow solution procedure.  The power flow solution contains the voltages and angles at all buses, and from this information, we may compute the real and reactive generation and load levels at all buses and the real and reactive flows across all circuits. POWER SYSTEM 2 - EPO 643 4
  • 5. Terminology  The above terminology is often used with the word “load” substituted for “power,” i.e., load flow problem, load flow solution procedure, load flow program, and load flow solution.  However, the former terminology is preferred as one normally does not think of “load” as something that “flows.” POWER SYSTEM 2 - EPO 643 5
  • 6. Power system components  Generator  Transmission Lines  Load POWER SYSTEM 2 - EPO 643 6 Figure 1
  • 7. Generator  Generators have maximum and minimum real and reactive power capabilities.  Maximum reactive power capability:  maximum reactive power that the generator may produce when operating with a lagging power factor.  minimum reactive power capability:  maximum reactive power the generator may absorb when operating with a leading power factor.  These limitations are a function of the real power output of the generator,  as the real power increases, the reactive power limitations move closer to zero. POWER SYSTEM 2 - EPO 643 7
  • 8. Figure 2 POWER SYSTEM 2 - EPO 643 8
  • 9. Figure 2 illustrates several important elements of the power flow problem.  First, identify each bus depending on whether generation and/or load is connected to it.  A bus may have  generation only (buses B1, B2, and B3),  load only (buses B5, B7, and B9),  neither generation or load (buses B4, B6, and B8).  both generation and load (leads us to define “bus injection”) POWER SYSTEM 2 - EPO 643 9
  • 10. Basic Technique for Load Flow Studies  In a load flow study, assumptions are made about:  Voltage at a bus or For each bus in the system  Power being supplied to the bus POWER SYSTEM 2 - EPO 643 10
  • 11. Types of Buses  For each bus, there are four possible variables that characterize the buses electrical condition.  The four variables are  real and reactive power injection, Pi and Qi,  voltage magnitude and angle, |Vi| and δi , respectively POWER SYSTEM 2 - EPO 643 11
  • 12. Types of Buses (cont..)  Generation Bus  Also called the P-V bus or voltage-controlled buses  Voltage magnitude |Vi| and real power Pi are specified  Able to specify (and therefore to know) the voltage magnitude of this bus.  Most generator buses fall into this category, independent of whether it also has load  Load Bus  Also called the P-Q bus  Real power Pi and Qi are specified  All load buses fall into this category, including buses that have not either load or generation. POWER SYSTEM 2 - EPO 643 12
  • 13. Slack or Swing Bus  Known as reference bus  Voltage magnitude |Vi| and phase angle δi are specified  There is only one swing bus, and it can be designated by the engineer to be any generator bus in the system.  This generator “swings” to compensate for the network losses, or, one may say that it “takes up the slack.” POWER SYSTEM 2 - EPO 643 13
  • 14. Bus types Quantities Unknown specified values Generator Bus |Vi| , Pi Q i , δi Load Bus Pi , Qi |Vi| , δi Slack Bus |Vi| , δi Pi , Qi POWER SYSTEM 2 - EPO 643 14
  • 15. Bus injection  An injection is the power (P or Q), that is being injected into or withdrawn from a bus by an element having its other terminal (in the per-phase equivalent circuit) connected to ground. Such an element would be either a generator or a load.  Positive injection is defined as one where power is flowing from the element into the bus.  Negative injection is then when power is flowing from the bus, into the element. POWER SYSTEM 2 - EPO 643 15
  • 16. Bus injection (cont..)  Generators normally have positive real power injections, although they may also be assigned negative real power injections when they are operating as a motor.  Generators may have either positive or negative reactive power injections:  positive if the generator is operating lagging and delivering reactive power to the bus,  negative if the generator is operating leading and absorbing reactive power from the bus, and  zero if the generator is operating at unity power factor. POWER SYSTEM 2 - EPO 643 16
  • 17. Loads  Loads normally Pk=100 Qk=30 Pk= - 40 Qk= -20 have negative real and reactive power injections. (a) (b)  Figure 3: Illustration of (a) Pk=100+(-40)=60 positive injection, (b) Qk=30+(-20)=10 negative injection, and (c) net injection (c) POWER SYSTEM 2 - EPO 643 17
  • 18. Figure 3 illustrates the net injection as the algebraic sum when a bus has both load and generation;  In this case, the net injection for both real and reactive power is positive (into the bus).  Thus, the net real power injection is Pk=Pgk-Pdk, and the net reactive power injection is Qk=Qgk-Qdk.  We may also refer to the net complex power injection as Sk=Sgk-Sdk, where Sk=Pk+jQk. POWER SYSTEM 2 - EPO 643 18
  • 19. Power Flow solution  Most common and important tool in power system analysis  also known as the “Load Flow” solution  used for planning and controlling a system  assumptions: balanced condition and single phase analysis  The utility wants to know the voltage profile  the nodal voltages for a given load and generation schedule  From the load flow solution –  the voltage magnitude and phase angle at each bus could be determined and hence the active and reactive power flow in each line could be calculated POWER SYSTEM 2 - EPO 643 19
  • 20. The currents and powers are expressed as going into the bus  for generation the powers are positive  for loads the powers are negative  the scheduled power is the sum of the generation and load powers POWER SYSTEM 2 - EPO 643 20
  • 21. The Bus Admittance Matrix  The matrix equation for relating the nodal voltages to the currents that flow into and out of a network using the admittance values of circuit branches is given by :- POWER SYSTEM 2 - EPO 643 21
  • 22. Forming the Admittance Matrix 1 y13 3 4 I1 y34 2 I4 y12 y23 I2 I3 y1 y4 y2 y3 POWER SYSTEM 2 - EPO 643 22
  • 23. From Kirchoff’s Current Law (KCL) –  the current injections be equal to the sum of the currents flowing out of the bus and into the lines connecting the bus to other buses,or to the ground.  Therefore, recalling Ohm’s Law, I=V/Z=VY, the current injected into bus 1 may be written as: I1=(V1-V2)y12 + (V1-V3)y13 + V1y1 POWER SYSTEM 2 - EPO 643 23
  • 24. I2 = (V2 - V1)(y21) + V2 y2 + (V2 – V3)y23  I1= V1( y1 + y12 + y13) + V2(-y12) + V3(-y13)  I2= V1(-y21) + V2( y2 + y21 + y23) + V3(-y23)  I3= V1(-y31) + V2(-y32) + V3( y3 + y31 + y32+ y34) + V4(-y34)  I4= V3(-y43) + V4( y4 + y43) POWER SYSTEM 2 - EPO 643 24
  • 25. Admittance Matrix  ( y1 + y12 + y13 ) - y12 - y13 0   - y 21 ( y 2 + y 21 + y 23 ) - y 23 0     - y31 - y32 ( y3 + y31 + y32 + y34 ) - y34     0 0 - y 43 ( y 4 + y 43 ) POWER SYSTEM 2 - EPO 643 25
  • 26. Matrix Equation  I1   ( y1 + y12 + y13 ) - y12 - y13 0   V1  I   - y 21 ( y 2 + y 21 + y 23 ) - y 23 0   2 V   2 =   I3   - y31 - y32 ( y3 + y31 + y32 + y34 ) - y34   V3        I4   0 0 - y 43 ( y 4 + y 43 )  V4  POWER SYSTEM 2 - EPO 643 26
  • 27. Y-Bus Matrix Building Rules  The matrix is symmetric, i.e., Yij=Yji.  A diagonal element Yii = Self Admittance  is obtained as the sum of admittances for all branches connected to bus i, including the shunt branch N Yii = y i + ∑y k =1, k ≠ i ik  The off-diagonal elements are the negative of the admittances connecting buses i and j, i.e., Yij=-yji = mutual admittance. POWER SYSTEM 2 - EPO 643 27
  • 28. E.g. for a 4 bus system Y11 Y12 Y13 Y14  Y Y22 Y23  Y24  Y = 21 Y31 Y32 Y33 Y34    Y41 Y42 Y43 Y44  POWER SYSTEM 2 - EPO 643 28
  • 29. The power flow equations  The net complex power injection into a bus Sk=Sgk-Sdk Sk=VkIk* Vk=| Vk|∠δk Ik =Σ | Ykj|∠θkj | Vj|∠δj Ik =Σ | Ykj|| Vj|∠ (θkj + δj) Ik* =Σ | Ykj|| Vj|∠ -(θkj + δj) POWER SYSTEM 2 - EPO 643 29
  • 30. Sk=VkIk* Sk= | Vk|∠δk x Σ | Ykj|| Vj|∠ -(θkj + δj) Sk= Σ | Ykj| | Vk|| Vj|∠ (δk - δj - θkj ) Pk= Σ | Ykj| | Vk|| Vj|cos (δk - δj - θkj ) Qk= Σ | Ykj| | Vk|| Vj|sin (δk - δj - θkj ) POWER SYSTEM 2 - EPO 643 30