Idea Transcript
Bus Admittance Matrix
Chapter 6: Power Flow
• Let
Network Matrices Network Solutions Power Flow Equations Newton-Raphson Method Fast Decoupled Method 6 Power Flow
Notes on Power System Analysis
I = vector of currents injected into nodes V = vector of node voltages Ybus = bus admittance matrix Then: I = Ybus V
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Example
• Ybus is symmetric unless the circuit has phase shifters or active devices • Diagonal term Yii is the sum of all admittances connected to bus i • Off-diagonal term Yij is the negative of the sum of all admittances directly connecting bus i to bus j
I = YV Notes on Power System Analysis
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Network Solution
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• Triangular matrices L and U are obtained so that Ybus = L U
– For small systems, simply invert the Ybus matrix: V = Ybus-1 I = Zbus 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 Notes on Power System Analysis
Notes on Power System Analysis
Triangular factorization
• Given current injections, find node voltages
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Constructing Ybus
I V I = 1 V = 1 I 2 V2 I1 Y11 Y12 V1 I = Y 2 21 Y22 V2 6 Power Flow
Notes on Power System Analysis
– The triangular factors L and U are saved to be used on other calculations – Note that L is a lower triangular matrix and U is an upper triangular matrix 5
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Triangular factorization
Triangular factorization
• Forward substitution gives V' by solution of set of (lower) triangular equations that can be solved sequentially starting at 1 • Back substitution gives V by solution of a set of (upper) triangular equations that are solved sequentially starting at n
• Once the factors are known, solve: – forward substitution I = L V' – back substitution V' = U V
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Kron Reduction
Kron reduction
• If one bus has current injection of zero, that bus can be eliminated:
• The Kron reduction method is equivalent to performing a general wye-delta conversion to eliminate a node in the network
I1 Y11 Y12 I = Y 2 21 Y22 0 Y31 Y32 Yij( new ) = Yij − 6 Power Flow
Yik Ykj Ykk
Y13 V1 Y23 V2 Y33 V3
– It applies to any node that has a zero current injection – It can be generalized to give a partial inverse (see Mathcad worksheet)
i, j = 1, 2, L , n i, j ≠ k
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Bus Impedance Matrix • V = Zbus I, so Zbus = Ybus • The bus impedance matrix can be formed by inverting the bus admittance matrix or by the building it one bus at a time
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Power Flow Analysis
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Notes on Power System Analysis
Power Flow Equations Power Flow Problem Several Iterative Solutions Newton-Raphson and Decoupled Methods Control of Power Flow 11
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Power Flow Equations
Example
• Complex power injection: V1 SD1
SG1
V2
SG2
V3
Si = SGi – SDi = generation – load Si = Σk Sik where k = 1, … , n and i = 1, … , n
SG3 SD3
• Current injection (phasor) V4 6 Power Flow
SD4
V5
Ii = IGi – IDi = Σk Iik = Σk (Yik Vk)
SD5
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• Break the complex power flow equation into real and imaginary parts:
Ii = IGi – IDi = Σk Iik = Σk (Yik Vk) Si = Vi Ii* = Vi Σk (Yik* Vk*) Si = Σk |Vi| |Vk| ejδik (Gik – j Bik)
Si = Pi + j Qi = Σk |Vi| |Vk| ejδik (Gik – j Bik) gives Pi = Σk |Vi| |Vk| [Gik cos(δik) + Bik sin(δik)] and Qi = Σk |Vi| |Vk| [Gik sin(δik) - Bik cos(δik)]
Vk = |Vk| ejδk δik = δi – δk Yik = Gik + jBik Notes on Power System Analysis
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Power Flow Problem – At most generator buses, the active power PG 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 approximation is that the load active and reactive power demand PD and QD are known. – At one generator bus, leave the active power as a variable (to make up system losses). Notes on Power System Analysis
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Power flow problem: bus type
• Assumptions:
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Power flow equations
Power Flow Equations
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– The slack bus: the one generator bus that has a variable PG. Bus number 1 – PV buses: those having known P and |V| (mostly all other generator buses). buses number 2, ... , m – PQ buses: those having known P and Q (mostly load buses, but also fixed generators). buses number m+1, ... , n 6 Power Flow
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Power flow problem
• Remarks: Seek an iterative solution
• Statement: – Given: (|V1|, δ1), (P2, |V2|),...,(Pm, |Vm|) (Pm+1, Qm+1), ..., (Pn, Qn) – Find: (P1, Q1), (Q2, δ2), ..., (Qm, δm), (|Vm+1|, δm+1), ..., (|Vn|, δn)
• Note: P1 and Q1, ..., Qm are calculated once all voltages and phase angles are known 6 Power Flow
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– The equations are nonlinear algebraic equations – There are n-1 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
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Iterative Solutions • Remarks: Seek an iterative solution – The equations are nonlinear algebraic equations – Multiple solutions are possible – Solutions may fail to exist – Numerical method may fail – Well-designed system will usually have only one realistic solution 6 Power Flow
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• After rearranging, the power flow equation gives a form that can be used for a Gauss or Gauss-Seidel iterative solution
x h(x) x0 x1
x
1 Vi = Yii
Note slow convergence, depending on shape of h(x) 6 Power Flow
Notes on Power System Analysis
Solve x = h(x) by using initial guess x0 to compute x1 = h(x0), then iterate xp+1=h(xp) where p is iteration Vector case, use Gauss-Seidel: x1p+1=h(x1p, x2p , x3p, ..., xnp) x2p+1=h(x1p+1, x2p , x3p, ..., xnp) x3p+1=h(x1p+1, x2p+1 , x3p, ..., xnp) ... 6 Power Flow Notes on Power System Analysis 22
Gauss-Seidel power flow
Gauss iteration (scalar case) h(x1) h(x0)
• Gauss iteration:
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n S* i* − ∑ Yik Vk Vi k =1 k ≠i Notes on Power System Analysis
i = 2, 3,L , n
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Gauss-Seidel power flow
Newton-Raphson
• Results of the Gauss-Seidel power flow:
• Newton’s method (scalar): Equations in this form: f(x) = 0 where x0 is an initial guess Linearize: f(xp + ∆x) = f(xp) + f’(xp) ∆x ≈ 0 Solve for ∆x = - f '(xp)-1 f(xp)
– 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
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Newton-Raphson
Newton’s method (scalar)
• Newton-Raphson (vector): Vector function of vector: f(x) = 0 where x0 = [x10,..., xn0]T is initial guess Linearize: f(xp+∆x) = f(xp) + J(xp) ∆x ≈ 0 where J(xp) is the Jacobian matrix:
a matrix of partial derivatives Jij(xp) = ∂fi/∂xj evaluated at xp p = iteration number 6 Power Flow
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Newton-Raphson
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• Define a vector of unknowns x = [δ2, δ3,..., δn, |Vm+1|, |Vm+2|,..., |Vn|]T
Linearize: f(xp+∆x) = f(xp) + J(xp) ∆x ≈ 0 Solve for ∆x = - J(xp)-1 f(xp) J(xp) is the Jacobian matrix:
• Net power injections computed Pi(x) = Σk [|Vi| |Vk| [Gik cos(δi-δk) + Bik sin(δi-δk)] Qi(x) = Σk [|Vi| |Vk| [Gik sin(δi-δk) - Bik cos(δi-δk)]
Jij(xp) = ∂fi/∂xj evaluated at xp. Notes on Power System Analysis
Notes on Power System Analysis
Newton Raphson Power Flow
• Newton-Raphson (vector):
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Newton Raphson Power Flow • Power mismatches (= specified value – computed value) – Solution is obtained when mismatches go to zero
∆P(x) = [P2 – P2(x), ... , Pn – Pn(x)]T ∆Q(x) = [Qm+1–Qm+1(x), ... , Qn–Qn(x)]T
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• The solution will be achieved when the mismatches are driven to zero • Solve the following linear equation for the updates ∆δ and ∆|V| : J11 J12 ∆δ ∆P J = 21 J 22 ∆ | V | ∆Q
•After each iteration gives ∆δ and ∆|V|: δp+1 = δp + ∆δ and |V|p+1 = |V|p + ∆|V| mismatch vector is updated for next iteration 6 Power Flow
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• J21 is a matrix of ∂Q(x)/∂δ :
• J11 is a matrix of ∂P(x)/∂δ : ∂Pi/∂δj =|Vi||Vj|[Gijsin(δi-δj)-Bijcos(δi-δj) ∂Pi/∂δi = - Qi - Bii | Vi|2
∂Qi/∂δj= -|Vi||Vj|[Gijcos(δi-δj) + Bij sin(δi-δj)]
∂Qi/∂δi = Pi - Gii | Vi|2
• J12 is a matrix of ∂P(x)/∂|V| : ∂Pi/∂|V|j =|Vi| [Gijcos(δi-δj) + Bijsin(δi-δj) ∂Pi/∂|V|i = Pi /|Vi| + Gii |Vi|
• J22 is a matrix of ∂Q(x)/∂|V| : ∂Qi/∂|V|j=|Vi|[Gijsin(δi-δj) - Bij cos(δi-δj)]
∂Qi/∂|V|i = Qi /|Vi| - Bii |Vi| 6 Power Flow
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Jacobian elements
Decoupled Power Flow
• Elements of J12 are relatively small
• Neglect the relatively small terms in the Jacobian, but calculate the power mismatches exactly
• Elements of J21 are relatively small • Elements of J11 and J22 are relatively large
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J11 ∆δ = ∆P and J22 ∆|V| = ∆Q
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Fast Decoupled Method
Decoupled Power Flow
-B ∆δ = ∆P' and -B ∆|V| = ∆Q' where ∆Pi' = ∆Pi/|Vi| and ∆Qi' = ∆Qi/|Vi| The approximations do not affect the mismatches. If the solution converges, we get the exact solution This may take more iterations but less computation time than the NewtonRaphson.
• Fast decoupled method makes constant Jacobian approximation |Vi| ≈ 1 pu and δi > Gij
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Control Implications
Notes on Power System Analysis
Control Implications
• P primarily depends on δ , change active power flow by controlling angles
• Q primarily depends on V, change reactive power flow by controlling voltage magnitudes
– change Pg which drives the local angle ahead – or operate phase shifting transformers
– change voltage regulator settings – or operate tap changers on transformers
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Large Systems • Large systems typically exhibit a sparse Ybus matrix • A sparse matrix is best stored as a linked list – only the non-zero elements of the matrix are stored in the linked list – pointers are used to indicate the position of the next non-zero entry 6 Power Flow
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