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Example: Jacobian Matrix and Power flow Solution by Newton-Raphson Determine the dimension of the jacobian matrix for the power system in ex:6.9. Also calculate ΔP 2 (0) in step(1) and Ji 24 (0) In step(2) of the first Newton Raphson iteration, Assume zero initial phase angle and 1.0 p.u. initial voltage magnitudes (except V 3 =1.05)

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Solution: Since N=5 buses NO. of eqns=2(N-1)=8 Jacobian matrix will have dimension 8х8 There is one voltage controlled bus, bus 3 Therefore V 3 and Q 3 could be eliminated.: J(i) is reduced to 7x7 matrix ΔP 2 = P 2 – P 2 (x) = P 2 – V 2 (0) {Y 21 V 1 cos[δ 2 (0)-δ 1 (0)-θ 21 ] + Y 22 V 2 cos [-θ 22 ] + Y 23 V 3 cos [δ 2 (0)-δ 3 (0)-θ 23 ] + Y 24 V 4 cos [δ 2 (0)- δ 4 (0)-θ 24 ] + Y 25 V 5 cos [δ 2 (0)-δ 5 (0)-θ 25 ]

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ΔP = [ *1.0*cos(84.624) *1.0*cos( ) *1.0*cos( )] = p.u. For step(2) and J i J i 24 (0) = V 2 (0)Y 24 V 4 (0)sin[δ 2 (0)-δ 4 -θ 24 ] = 1.0* *1.0*sin[ ] = PU See EX 6-11 Power World Simulator

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Example: Using the power flow system in Ex 6.9, Determine the acceptable generation range at bus3, keeping each line and transformer loaded at a below 100% of its MVA. Computer Simulate

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6.8 SPARSITY TECHNIQUES - Sparse matrix has only a few non zero elements. - Newton Raphson power flow program employ sparse matrix techniques to reduce computer and time requirement. These techniques include compact storge of Y bus and J(i).

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Consider the following matrix: One method for compact storage of S consists of four vectors DIAG=[ ] OFFDIAG=[ ] Col=[ ] Row=[ ]

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6.9 Fast Decoupled Power Flow - Contingencies are a major concern in Power system operations. For example, operating personnel need to know what power flow changes will occur due to a particular generator outage a transmission line outage. - Contingency information when obtained in real time,can be used to Anticipate problems caused by such outages and can be used to develop operating stategies to overcome problems. - Fast power flow algorithms have been developed to give power flow solutions in second or less.

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These algorithms a k based on the following simplification of the jacobian matrix. * Neglect J 2 (i) and J 3 (i) This will reduce the problem ( power flow ) to : J 1 (i)Δδ(i)=ΔP(i) J 4 (i)ΔV(i)=ΔQ(i) * Further reduction in computer time can be obtained, for ex Assume V k V n 1 p.u. and δ k =δ n Then J 1 and J 4 constant matrices this means that J 1 and J 4 do not have to be Recalculated during successive iterations.

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Chapter 7 SYMMETRICAL FAULTS

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7.1 Series R-L Circuit Transients Consider the series R-L circuit shown: The switch is closed at t=0. This represent a shot circuit across the circuit. For simplicity assume here fault impedance. This is usually called solid or bolted fault.

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To get the current after closing the switch

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- The total fault current I(t) is called the asymmetrical fault current. - The ac component is called the steady state fault current. - The dc component is called the offset current or the complementary current. This component decays with time constant. - The rms value of the ac component is (A).

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- The rms value of i(t) is give by: (A) In terms of the time constant and frequency w, This can be written in the following from

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Example: A bolted short circuit occurs in the series R-L circuit shown before with V=20KV, X=8, R=0.8 and with the maximum dc offset current. The circuit breaker opens 3 cycles after fault inception. Determine: a) The rms ac fault current. b) The rms momentary current at τ =0.5 cycle which pass through the breaker before it opens. c) The asymmetrical fault current the breaker interrupts.

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