= V I = Bus Admittance Matrix. Chapter 6: Power Flow. Constructing Ybus. Example. Network Solution. Triangular factorization. Let

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1 Chapter 6: Power Flow Network Matrices Network Solutions Newton-Raphson Method Fast Decoupled Method Bus Admittance Matri Let I = vector of currents injected into nodes V = vector of node voltages Y bus = bus admittance matri Then: I = Y bus V 6 Power Flow Notes on Power System Analysis 6 Power Flow Notes on Power System Analysis I I = I I I Eample Y = Y I = Y V V V = V Y Y 6 Power Flow Notes on Power System Analysis 3 V V Constructing Ybus Ybus is symmetric unless the circuit has phase shifters or active devices Diagonal term Y ii is the sum of all admittances connected to bus i Off-diagonal term Y ij is the negative of the sum of all admittances directly connecting bus i to bus j 6 Power Flow Notes on Power System Analysis 4 Network Solution Given current injections, find node voltages For small systems, simply invert the Ybus matri: V = Y bus - I = Z bus I For medium, solve the set of linear equations using Gaussian elimination For large systems, the Gaussian elimination is best done by triangular (LU) factorization 6 Power Flow Notes on Power System Analysis 5 Triangular factorization Triangular matrices L and U are obtained so that Y bus = L U The triangular factors L and U are saved to be used on other calculations Note that L is a lower triangular matri and U is an upper triangular matri 6 Power Flow Notes on Power System Analysis 6

2 Triangular factorization Once the factors are known, solve: forward substitution I = L V' back substitution V' = U V Triangular factorization Forward substitution gives V' by solution of set of (lower) triangular equations that can be solved sequentially starting at Back substitution gives V by solution of a set of (upper) triangular equations that are solved sequentially starting at n 6 Power Flow Notes on Power System Analysis 7 6 Power Flow Notes on Power System Analysis 8 Kron Reduction If one bus has current injection of zero, that bus can be eliminated: I Y Y Y3 V = I Y Y Y3 V 0 Y3 Y3 Y33 V3 (new) Yik Ykj Yij = Yij i, j =,, L, n Ykk i, j k 6 Power Flow Notes on Power System Analysis 9 Kron reduction The Kron reduction method is equivalent to performing a general wye-delta conversion to eliminate a node in the network It applies to any node that has a zero current injection It can be generalized to give a partial inverse (see Mathcad worksheet) 6 Power Flow Notes on Power System Analysis 0 Bus Impedance Matri V = Z bus I, so Z bus = Y bus - The bus impedance matri can be formed by inverting the bus admittance matri or by the building it one bus at a time Power Flow Analysis Power Flow Problem Several Iterative Solutions Newton-Raphson and Decoupled Methods Control of Power Flow 6 Power Flow Notes on Power System Analysis 6 Power Flow Notes on Power System Analysis

3 S G Eample S G V V V 3 S D V 4 V 5 S D4 S D5 S G3 S D3 Comple power injection: S i = S Gi S Di = generation load S i = Σ k S ik where k =,, n and i =,, n Current injection (phasor) I i = I Gi I Di = Σ k I ik = Σ k (Y ik V k ) 6 Power Flow Notes on Power System Analysis 3 6 Power Flow Notes on Power System Analysis 4 I i = I Gi I Di = Σ k I ik = Σ k (Y ik V k ) S i = V i I i * = V i Σ k (Y ik * V k *) S i = Σ k V i V k e jδ ik (G ik j B ik ) V k = V k e jδ k δ ik = δ i δ k Y ik = G ik + jb ik 6 Power Flow Notes on Power System Analysis 5 Power flow equations Break the comple power flow equation into real and imaginary parts: S i = P i + j Q i = Σ k V i V k e jδ ik (G ik j B ik ) gives P i = Σ k V i V k [G ik cos(δ ik ) + B ik sin(δ ik )] and Q i = Σ k V i V k [G ik sin(δ ik ) - B ik cos(δ ik )] 6 Power Flow Notes on Power System Analysis 6 Power Flow Problem Assumptions: At most generator buses, the active power P G is controlled (by speed governor) and the voltage magnitude is controlled (by the voltage regulator). Treat these as known. At most load buses, a reasonable approimation is that the load active and reactive power demand P D and Q D are known. At one generator bus, leave the active power as a variable (to make up system losses). Power flow problem: bus type The slack bus: the one generator bus that has a variable P G. Bus number PV buses: those having known P and V (mostly all other generator buses). buses number,..., m PQ buses: those having known P and Q (mostly load buses, but also fied generators). buses number m+,..., n 6 Power Flow Notes on Power System Analysis 7 6 Power Flow Notes on Power System Analysis 8

4 Power flow problem Statement: Given: ( V, δ ), (P, V ),...,(P m, V m ) (P m+, Q m+ ),..., (P n, Q n ) Find: (P, Q ), (Q, δ ),..., (Q m, δ m ), ( V m+, δ m+ ),..., ( V n, δ n ) Note: P and Q,..., Q m are calculated once all voltages and phase angles are known Remarks: Seek an iterative solution The equations are nonlinear algebraic equations There are n- unknown phase angles (all but the slack bus, which is given as 0) There are n-m unknown voltage magnitudes (all but the PV and slack buses, which total m) 6 Power Flow Notes on Power System Analysis 9 6 Power Flow Notes on Power System Analysis 0 Remarks: Seek an iterative solution The equations are nonlinear algebraic equations Multiple solutions are possible Solutions may fail to eist Numerical method may fail Well-designed system will usually have only one realistic solution 6 Power Flow Notes on Power System Analysis Iterative Solutions Gauss iteration: Solve = h() by using initial guess 0 to compute = h( 0 ), then iterate p+ =h( p ) where p is iteration Vector case, use Gauss-Seidel: p+ =h( p, p, 3p,..., np ) p+ =h( p+, p, 3p,..., np ) p+ 3 =h( p+, p+, 3p,..., np )... 6 Power Flow Notes on Power System Analysis Gauss iteration (scalar case) h( ) h( 0 ) 0 6 Power Flow Notes on Power System Analysis 3 h() Note slow convergence, depending on shape of h() Gauss-Seidel power flow After rearranging, the power flow equation gives a form that can be used for a Gauss or Gauss-Seidel iterative solution * n S V i i = YikVk Y * ii Vi k k = i i =,3, L, n 6 Power Flow Notes on Power System Analysis 4

5 Gauss-Seidel power flow Results of the Gauss-Seidel power flow: acceptable for small systems, but it takes many iterations to converge number of iterations for convergence increases as the number of buses in the system increase not suitable for medium and large sizes of systems 6 Power Flow Notes on Power System Analysis 5 Newton-Raphson Newton s method (scalar): Equations in this form: f() = 0 where 0 is an initial guess Linearize: f( p + ) = f( p )+ f ( p ) 0 Solve for = - f '( p ) - f( p ) 6 Power Flow Notes on Power System Analysis 6 Newton s method (scalar) 6 Power Flow Notes on Power System Analysis 7 Newton-Raphson Newton-Raphson (vector): Vector function of vector: f() = 0 where 0 = [ 0,..., n0 ] T is initial guess Linearize: f( p + ) = f( p )+ J( p ) 0 where J( p ) is the Jacobian matri: a matri of partial derivatives J ij ( p )= f i / j evaluated at p p = iteration number 6 Power Flow Notes on Power System Analysis 8 Newton-Raphson Newton-Raphson (vector): Linearize: f( p + ) = f( p )+ J( p ) 0 Solve for = - J( p ) - f( p ) J( p ) is the Jacobian matri: J ij ( p )= f i / j evaluated at p. Newton Raphson Power Flow Define a vector of unknowns = [δ, δ 3,..., δ n, V m+, V m+,..., V n ] T Net power injections computed P i () = Σ k [ V i V k [G ik cos(δ i -δ k ) +B ik sin(δ i -δ k )] Q i () = Σ k [ V i V k [G ik sin(δ i -δ k ) -B ik cos(δ i -δ k )] 6 Power Flow Notes on Power System Analysis 9 6 Power Flow Notes on Power System Analysis 30

6 Newton Raphson Power Flow Power mismatches (= specified value computed value) Solution is obtained when mismatches go to zero P() = [P P (),..., P n P n ()] T Q() = [Q m+ Q m+ (),..., Q n Q n ()] T The solution will be achieved when the mismatches are driven to zero Solve the following linear equation for the updates δ and V : J J J δ P = J V Q After each iteration gives δ and V : δ p+ = δ p + δ and V p+ = V p + V mismatch vector is updated for net iteration 6 Power Flow Notes on Power System Analysis 3 6 Power Flow Notes on Power System Analysis 3 J is a matri of P()/ δ : P i / δ j = V i V j [G ij sin(δ i -δ j )-B ij cos(δ i -δ j ) P i / δ i = - Q i -B ii V i J is a matri of P()/ V : P i / V j = V i [G ij cos(δ i -δ j )+ B ij sin(δ i -δ j ) P i / V i = P i / V i + G ii V i J is a matri of Q()/ δ : Q i / δ j =- V i V j [G ij cos(δ i -δ j ) +B ij sin(δ i -δ j )] Q i / δ i = P i -G ii V i J is a matri of Q()/ V : Q i / V j = V i [G ij sin(δ i -δ j ) -B ij cos(δ i -δ j )] Q i / V i = Q i / V i -B ii V i 6 Power Flow Notes on Power System Analysis 33 6 Power Flow Notes on Power System Analysis 34 Jacobian elements Elementsof J are relatively small Elements of J are relatively small Elements of J and J are relatively large Decoupled Power Flow Neglect the relatively small terms in the Jacobian, but calculate the power mismatches eactly J δ = P and J V = Q 6 Power Flow Notes on Power System Analysis 35 6 Power Flow Notes on Power System Analysis 36

7 Decoupled Power Flow Fast decoupled method makes constant Jacobian approimation V i pu and δ i << rad Ybus = G + j B B ij >> G ij Fast Decoupled Method -B δ = P' and -B V = Q' where P i ' = P i / V i and Q i ' = Q i / V i The approimations do not affect the mismatches. If the solution converges, we get the eact solution This may take more iterations but less computation time than the Newton- Raphson. 6 Power Flow Notes on Power System Analysis 37 6 Power Flow Notes on Power System Analysis 38 Control Implications P primarily depends on δ, change active power flow by controlling angles change Pg which drives the local angle ahead or operate phase shifting transformers Control Implications Q primarily depends on V, change reactive power flow by controlling voltage magnitudes change voltage regulator settings or operate tap changers on transformers 6 Power Flow Notes on Power System Analysis 39 6 Power Flow Notes on Power System Analysis 40 Large Systems Large systems typically ehibit a sparse Ybus matri A sparse matri is best stored as a linked list only the non-zero elements of the matri are stored in the linked list pointers are used to indicate the position of the net non-zero entry 6 Power Flow Notes on Power System Analysis 4