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ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION 1 ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 3 : PER UNIT SYSTEM - SOLUTION Question 1. A single Phase 50kVA, 2400/240V, 60Hz distribution transformer is used as a step down transformer at the load end of a 2400V feeder whose series impedance is (1.0 + j2.0) ohms. The equivalent series impedance of the transformer is (1.0 + j2.5) ohms referred to the high voltage (i.e. primary) side. The transformer is delivering rated total power at 0.8 power factor lagging, and at rated secondary voltage. Neglecting the transformer excitation current, determine: (a) The voltage at the transformer primary terminals, (b) The voltage at the sending end of the feeder, (c) The real and reactive power delivered to the sending end of the feeder. Work in the Per Unit System, using the transformer ratings as base quantities. Answer : First Determine the base quantities. Sbase = 50kVA Vbase1 = 2400V Vbase2 = 240V Therefore: A V SI base base base 833.202400 000,50 1 1 === A V SI base base base 333.208240 000,50 2 2 === Ω=== 2.115 8333.20 2400 1 1 1 base base base I VZ Ω=== 152.1 333.208 240 2 2 2 base base base I VZ So the per unit impedances become: upjj Z Z Z base eq eqpu .0217.000868.02.115 5.21 1 _1 _1 += + == upjj Z ZZ base line puline .01736.000868.02.115 21 1 _ += + == Now the load power is given by: ( ) 01 9.36508.0cos50 ∠=∠= − kVAkVASload Or: ( ) 01 9.36..0.18.0cos50 ∠=∠= − upkVASload So the load current is given by: ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION 2 0 0 0 _ _ 9.36..0.1 00.1 9.36..0.1 −∠= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∠ ∠ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ∗∗ upup V SI puload load puload The load current is then: 09.36333.208 −∠= AIload (a) The voltage at the transformer primary terminals is then given by: eqpupuloadpupu ZIVV _1__2_1 += ( )( )0217.000868.09.360.100.1 00 _1 jV pu +−∠+∠= upV pu .69.002.1 0 _1 ∠= The transformer primary voltage is then: 0 1 69.02448 ∠= VV (b) The supply voltage is given by: ( )pulineeqpupuloadpupuS ZZIVV __1__2_ ++= ( )( )03906.0001736.09.360.100.1 00 _ jV puS +−∠+∠= upV puS .15.1037.1 0 _ ∠= The supply voltage is then: 015.12489 ∠= VVS (c) The supply real and reactive power is then given by: ∗ = loadSS IVS ( )( )00 9.360.115.1037.1 ∠∠=SS upjupSS .6387.0..8169.002.38037.1 0 +=∠= So the real and reactive power are: kWP 845.40= kVarQ 936.31= ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION 3 Question 2. Three zones of a single phase distribution level circuit are identified in figure 1. The zones are connected by transformers T1 and T2, whose ratings are also shown. Using base values of 3MVA and 11kV in zone 1, draw the per unit circuit and determine the per-unit impedances and the per- unit source voltage. Then calculate the load current both in per-unit and in amperes. Transformer winding resistances and shunt admittance branches are neglected. Zone 1 Zone 2 Zone 3 T1 T2 2MVA3MVA 11 kV/6.6 k V 7.2 kV /3.3 kV Xeq = 0.1 p.u. Xeq = 0.12 p.u. Xline = j0.2 Ω Xload = j2.9 Ω Rload = 5.2 ΩVs = 13 kV Figure 1 : Three Zone Distribution System for question 2. Answer : Choose Bases: Sbase = 3MVA Vbase1 = 11kV Vbase2 = 6.6kV ( )( ) kV kV kVkVVbase 025.32.7 3.36.6 3 == This requires: A V SI base base base 727.272 1 1 == A V SI base base base 545.454 2 2 == A V SI base base base 736.991 3 3 == Similarly: Ω== 333.40 2 1 1 base base base S VZ Ω== 520.14 2 2 2 base base base S VZ Ω== 050.3 2 3 3 base base base S VZ So the per unit impedance values are: ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION 4 ( ) ..9508.0075.1 3 _ upj Z ZZ base load puload +== ..1.01_ upjZ puTeq = ..01377.0 2 _ upj Z ZZ base line puline == ..2142.0 2 3 6.6 2.712.0 22 2 _22_ upMVA MVA kV kV S S V VZZ rate base base rate rateTpuTeq =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = The per unit supply voltage is then: ..182.1 11 13 _ up kV kVV pus == The per unit equivalent circuit is then given by: Zone 1 Zone 2 Zone 3 XT1eq p.u. = j0.1 p.u. XT2eq = j0.2142 p.u. Xline p.u. = j0.01377pu Xload p.u. = j0.9508 p.u Rload p.u. = 1.705 p.u. Vs = 1.182 p.u. The load current is then given by: puloadpuTpulinepuT pus puload ZXXX V I __2__1 _ _ +++ = ( ) 0 00 _ 87.36131.2 0182.1 705.19508.001377.02142.01.0 0182.1 ∠ ∠ = ++++ ∠ = jI puload ..87.365546.0 0 _ upI puload −∠= So the load current is : 087.360.550 −∠= AIload ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION 5 Question 3. A balanced Y-connected voltage source with Eab = 480∠0° V is applied to a balanced ∆ load with Z∆ = 30∠40° ohms. The line impedance between the source and the load is ZL = 1∠85° p.u. for each phase. Calculate the per-unit and actual current in phase a of the line using Sbase3φ = 100kVA and VbaseLL = 600V. Answer: Define the base quantities as: kVASbase 1003 =φ kVASbase 333.331 =φ VVbaseLL 600= VVVbaseLN 412.3463 600 == Ω=== 6.3 1 2 3 2 φφ base baseLN base baseLL base S V S VZ A V S I baseLL base base 225.963 3 == φ So : 0 0 _ 0..8.0 600 0480 ∠=∠= upE puab and 0 _ 30..8.0 −∠= upE pua 0 0 _ 40..333.8 6.3 4030 ∠=∠=∆ upZ pu 0_ _ 40..7778.2 3 ∠== ∆ up Z Z pupuY 0 0 _ 85..2778.0 6.3 851 ∠=∠= upZ puline So the total impedance seen by the source is: 0 ___ 78.43..9807.2 ∠=+= upZZZ pulinepuYputot Therefore the supply current is given by: ..78.732684.0 78.43..9807.2 30..8.0 0 0 0 _ _ _ up up up Z V I putot pua pua −∠=∠ −∠ == So the actual load current is given by: 078.7383.25 −∠= AIa ELEC 4100 TUTORIAL THREE : PER UNIT SYSTEM - SOLUTION 6 Question 4. A balanced Y-connected voltage source with Eag = 277∠0° V is applied to a balanced Y load in parallel with a balanced ∆ load, where ZY = 30 + j10 ohms and Z∆ = 45 – j25 ohms. The Y load is solidly grounded. Using base values of Sbase1φ = 10kVA and VbaseLN = 277 V, calculate the source current Ia in per-unit and in amperes. Answer : Define the base quantities as: kVASbase 303 =φ kVASbase 101 =φ VVbaseLN 277= VVV baseLNbaseLL 77.4793 == Ω== 6729.7 3 3 2 φbase baseLL base S VZ A V S I baseLL base base 101.363 3 == φ So the per unit impedance values are given by: 0 _ 055.29..709.6 6729.7 2545 −∠=−=∆ up jZ pu 0 _ 435.18..121.4 6729.7 1030 ∠=+= upjZ puY Now the delta load can be converted to an equivalent star load as: 0_ _ 055.29..2364.2 3 −∠== ∆∆ up Z Z puputoY The total per-phase impedance is then given by: puYputoYputot ZZZ ___ //∆= 0 __ __ _ 74.125705.1 −∠= + = ∆ ∆ puYputoY puYputoY putot ZZ ZZ Z The per unit source current is then given by: 0 _ _ _ 74.126367.0 ∠== putot puag pua Z E I So the source current is : 074.1299.22 ∠= AIa ELEC 4100 TUTORIAL FOUR : THREE PHASE TRANSFORMERS 1 ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 4 : TRANSFORMERS SOLUTIONS. Question 1. a1 a2 b2 c1b1 c2 A1 B1 C1 A2 B2 C2 a3 a4 b4 c3b3 c4 A1A2 B1 B2 C1 C2 a1a2 b1 b2 c1 c2 a3a4 b3 b4 c3 c4 A1 B1 C1 a1 b1 c1 a3b3 c3 (a) a1 a2 b2 c1b1 c2 A1 B1 C1 A2 B2 C2 a3 a4 b4 c3b3 c4 a5 a6 b6 c5b5 c6 A1A2 B1 B2 C1 C2 a1a2 b1 b2 c1 c2 a3a4 b3 b4 a5a6 b5 b6 c5 c6 c4 c3 A1 B1 C1 a1b1 c1 c3 c4b6 b5 a3a4c6 c5 b4 b3 a5a6 (b) ELEC 4100 TUTORIAL FOUR : THREE PHASE TRANSFORMERS 2 a1 a2 b2 c1b1 c2 A1 B1 C1 A2 B2 C2 a3 a4 b4 c3b3 c4 a5 a6 b6 c5b5 c6 A1A2 B1 B2 C1 C2 a1a2 b1 b2 c1 c2 a3a4 b3 b4 a5a6 b5 b6 c5 c6 c4 c3 A1 B1 C1 a1 b1 c1 a3a4 b5 b6 b3 b4 c5 c6 c4 c3 a5 a6 (c) ELEC 4100 TUTORIAL FOUR : THREE PHASE TRANSFORMERS 3 Question 2. Consider the single line diagram of the power system shown below. The equipment ratings are as follows: • Generator 1 : 750MVA, 18kV, Xeq = 0.2 p.u. • Generator 2 : 750 MVA, 18kV, Xeq = 0.2 p.u • Synchronous Motor 3 : 1500MVA, 20kV, Xeq = 0.2 p.u. • 3 Phase Transformers, T1 to T4 : 750MVA, 500kV Y/20kV ∆, Xeq = 0.1 p.u. • 3 Phase Transformer T5 : 1500MVA, 500kV Y/20kV Y, Xeq = 0.1 p.u. Neglecting winding resistances, transformer phase shifts, and the excitation phenomena, draw the equivalent per unit reactance diagram. Use a base of 100MVA and 500kV for the 40 Ω transmission line. Determine all per unit reactance’s. T1 1 2 3 Bus 1 Bus 2 Bus 3 j40 ohm j25 ohm j25 ohm T2 T3 T4 T5 Answer: The equivalent per phase, per unit circuit diagram is shown below: Bus 1 Bus 2 Bus 3 XT1 XT3 XT2 XT4 XT5 Xline1 Xline2 Xline3 XG2 XM3 XG1 EG2 EM3 EG1 j0.0133pu j0.0133pu j0.0133puj0.0133pu j0.00666pu j0.01333pu j0.01pu j0.01pu j0.016pu j0.0216pu j0.0216pu ELEC 4100 TUTORIAL FOUR : THREE PHASE TRANSFORMERS 4 The impedance values in the circuit diagram are calculated as will be detailed below: MVASbase 100= kVV HVbase 500_ = Transmission line zones kVV LVbase 20_ = Generator zones ( ) ( ) Ω=== 2500 100 500 22 _ _ MVA kV S V Z base HVbase LVbase ( ) kAkV MVA V SI HVbase base LVbase 887.2203 100 3 _ _ === So the generator per unit impedances are: ..0216.0 750 100 20 182.0 2 1 upMVA MVA kV kVXG =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ..0216.0 750 100 20 182.0 2 2 upMVA MVA kV kVXG =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ..01333.0 1500 1002.03 upMVA MVAX M =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = The transformer per unit impedances are: ..01333.0 750 1001.04321 upMVA MVAXXXX TTTT =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ==== ..00666.0 1500 1001.05 upMVA MVAXT =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = The transmission line per unit impedances are: ..016.0 2500 40 1 upXline =Ω Ω = ..01.0 2500 25 32 upXX lineline =Ω Ω == Question 3. For the power system discussed in question 2, consider the case where the motor absorbs 1200MW at 0.8p.f. leading with the Bus 3 voltage at 18kV. Determine the Bus 1 and Bus 2 voltages in kV. Assume that generators 1 and 2 deliver equal real powers and equal reactive powers. Also assume a balanced three-phase system with positive-sequence sources. Answer : The bus 3 voltage is given by: ..09.0 20 018 00 3 upkV kVV pu ∠= ∠ = ELEC 4100 TUTORIAL FOUR : THREE PHASE TRANSFORMERS 5 The motor current is then: ( )( ) ( ) ( )( ) 01 3 87.3611.488.0183 1200 ..cos ..3 ∠==∠ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − kA kV MWfpfpV PI LL The motor current in per unit: ..87.3667.16 887.2 87.3611.48 00 _3 upkA kAI pu ∠= ∠ = Due to symmetry: ( )2_3_3_5_3_3_2_1 2 linepuTpupuTpupupupu XX I XIVVV +++== ( )( ) ( )01333.001.0 2 87.3667.16 00666.087.3667.1609.0 0 00 _2_1 jj jVV pupu + ∠ + ∠+∠== ..83.187572.0 0 _2_1 upVV pupu ∠== So the bus 1 and bus 2 voltages are: 0 21 83.1814.15 ∠== kVVV ELEC 4100 TUTORIAL FIVE : LOAD FLOW - SOLUTION 1 ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 5 : LOAD FLOW – SOLUTION. Question 1. Answer : (a) For sinusoidal time varying voltage and current waveforms, define: ( ) ( ) tjexVtxv ω=, ( ) ( ) tjexItxi ω=, Then substituting into the partial differential equations gives: ( ) ( ) ( )[ ] ( ) tjtjtj exzIexLIjxrI dx xdV e ωωω ω −=−−= ( ) ( ) ( )[ ] ( ) tjtjtj exyVexCVjxGV dx xdI e ωωω ω −=−−= These expressions can be simplified as: ( ) ( )xzI dx xdV −= ( ) ( )xyV dx xdI −= Differentiating with respect to x: ( ) ( ) ( )xzyV dx xdI z dx xVd =−=2 2 ( ) ( ) ( )xzyI dx xdV y dx xId =−=2 2 These expressions are separate, second order linear differential equations involving one spatial variable only. (b) From the π-section model it can be shown that: ( ) 1 2 Y YVIVV sssr − −= Rearranging: 11 21 Y I Y YVV ssr −⎥ ⎦ ⎤ ⎢ ⎣ ⎡ += Comparing this expression with the general transmission line solutions provided gives: ( )dZY C γsinh 1 1 = Using this value and again comparing with the general transmission line solutions gives: ( )[ ] ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − = 2 tanh1 sinh 1cosh 2 d ZdZ d Y CC γ γ γ ELEC 4100 TUTORIAL FIVE : LOAD FLOW - SOLUTION 2 Now : ( ) ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − −−−=−−= 1 2 3232 Y VYIVYVYIVYVYII sssssrssr ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ++− ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ += 1 2 23 1 3 11 Y YYYV Y YII ssr Equating this expression with the general transmission line solutions gives: ( )[ ] ( )[ ]( ) 213 2tanh 1 sinh 1cosh1cosh Yd ZdZ ddYY CC = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − =−= γ γ γγ (b) The π-section model of a transmission line is used in load flow analysis since load flow is interested only in the steady state characteristics at the voltage buses in the network. Hence it is only necessary to consider the behaviour at the terminating ends of the transmission line, not in the middle of the line. A lumped element model is sufficient to provide this information. Furthermore the full distributed model is used to predict dynamic characteristics along the line, but since this information is irrelevant for load flow it is not necessary to utilise the full model. Question 2. Answer : (a) The diagonal elements are the self admittances at each node of the network, and are the sum of all admittances connected to that node. The off-diagonal elements are the mutual admittances between two nodes of a network, and are the negative values of the admittances linking the two nodes in question. The YBUS matrix is square since the network consists of N buses, and for the ith bus there are N-1 potential mutual connections, and 1 self admittance – hence the matrix is square. The matrix is symmetric since the mutual connections between buses i and k are the same as the connections between buses k and i. The matrix is sparse since in power systems there is generally a low level of connectivity between the nodes, with couplings only between a few adjacent couplings. Hence the bulk of the mutual couplings are zero, and so the matrix is sparse. (b) The complex conjugate of the apparent power at bus i can be written as: iii IVS ∗∗ = k n k ikiiii VyVjQPS ∑ = ∗∗ =−= 1 Where the yik are the elements of the admittance bus. (c) The apparent power is given by: k n k ikiiii VyVjQPS ∑ = ∗∗ =−= 1 iiik n ik k ik i ii VyVy V jQP += − ∑ ≠ = ∗ 1 ELEC 4100 TUTORIAL FIVE : LOAD FLOW - SOLUTION 3 Rearranging gives: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − = ∑ ≠ = ∗ k n ik k ik i ii ii i VyV jQP y V 1 1 So the Gauss implementation of a voltage calculation is: ( ) ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − = ∑ ≠ = ∗ + p k n ik k ikp i p i p i ii p i Vy V jQP y V 1 1 1 The Gauss-Seidel implementation of a voltage calculation is: ( ) ⎥⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ −− − = ∑∑ += + − = ∗ + p k n ik ik p k i k ikp i p i p i ii p i VyVy V jQP y V 1 1 1 1 1 1 Since the Gauss-Seidel method uses the most recently available iteration data it generally shows a faster convergence rate than the Gauss method. Furthermore the Gauss method must store the pth and (p+1)th bus data, whereas the Gauss-Seidel discards the previous data as soon as the new data has become available. This results in a memory allocation and storage requirement advantage for the Gauss-Seidel approach, and furthermore programming the Gauss-Seidel method is simpler. (d) The apparent power is given by: k n k ikiiii VyVjQPS ∑ = ∗∗ =−= 1 Define: iii VV δ∠= ikikik yy γ∠= Then : ( )ikikki n k ikiii VVyjQPS γδδ +−∠=−= ∑ = ∗ 1 Or: ( ) ( )ikkiki n k ikikikki n k iki VVyVVyP γδδγδδ −−=+−= ∑∑ == coscos 11 ( ) ( )ikkiki n k ikikikki n k iki VVyVVyQ γδδγδδ −−=+−−= ∑∑ == sinsin 11 ELEC 4100 TUTORIAL FIVE : LOAD FLOW - SOLUTION 4 (e) Define the power flow mismatches at bus i as: ( ) ( )ikkiki n k ikiikkiki n k ikLiGiii VVyPVVyPPPf γδδγδδ −−−=−−−−=∆= ∑∑ == coscos 11 ( ) ( )ikkiki n k ikiikkiki n k ikLiGiii VVyQVVyQQQg γδδγδδ −−−=−−−−=∆= ∑∑ == sinsin 11 So applying the Newton-Raphson method: ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ∆ ∆ + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ∆ ∆ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − + + p p p p p p p p p p VVQ P V gg V ff VV δδ δ δδδ 1 1 1 Alternatively this can be expressed as: ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ∆ ∆ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ∆ ∆ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ −= ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ∆ ∆ p p pp pp p p p p VJJ JJ V V gg V ff Q P δδ δ δ 43 21 Where pP∆ are the real power mismatches at all PQ and PV buses, pQ∆ are the reactive power mismatches at all PQ buses, pδ∆ are the voltage angle corrections for all PQ and PV buses, and pV∆ are the voltage magnitude corrections for all PQ buses. The Jacobian matrix is defined by: ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − = n nnn n n p fff fff fff J δδδ δδδ δδδ L MMM L L 32 3 3 3 2 3 2 3 2 2 2 1 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − = ++ ++ ++ n n m n m n nmm nmm p V f V f V f V f V f V f V f V f V f J L MMM L L 21 3 2 3 1 3 2 2 2 1 2 2 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − = +++ +++ n nnn n mmm n mmm p ggg ggg ggg J δδδ δδδ δδδ L MMM L L 32 2 3 2 2 2 1 3 1 2 1 3 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ −∂ ∂ −∂ ∂ − = ++ + + + + + + + + + + n n m n m n n m m m m m n m m m m m p V g V g V g V g V g V g V g V g V g J L MMM L L 21 2 2 2 1 2 1 2 1 1 1 4 The Swing bus is bus 1, while buses 2 to m are the PV buses, and buses m+1 to n are the PQ buses. The 8 derivatives in the Jacobian matrices are given by: ( ) kiVVyf ikkikiik k i ≠−−= ∂ ∂ − ,sin γδδδ ( )ikkiki n ik k ik i i VVyf γδδδ −−−=∂ ∂ − ∑ ≠ = sin 1 ELEC 4100 TUTORIAL FIVE : LOAD FLOW - SOLUTION 5 ( ) kiVy V f ikkiiik k i ≠−−= ∂ ∂ − ,cos γδδ ( ) ( )ikiiiikkik n k ik i i VyVy V f γγδδ coscos 1 +−−= ∂ ∂ − ∑ = ( ) ikVVyg ikkikiik k i ≠−−−= ∂ ∂ − ,cos γδδδ ( )ikkiki n ik k ik i i VVyg γδδδ −−=∂ ∂ − ∑ ≠ = cos 1 ( ) ikVy V g ikkiiik k i ≠−−= ∂ ∂ − ,sin γδδ ( ) ( )iiiiiikkik n k ik i i VyVy V g γγδδ sinsin 1 −−−= ∂ ∂ − ∑ = (f) Since the Newton Raphson method uses a first order Taylor series approximation of the non-linear power flow equations to iteratively find a solution, it has a much faster convergence rate than the Gauss or Gauss Seidel methods. These latter methods are limited by the sparsity of the admittance bus matrix, which limits the rate that corrective terms can propagate through the solution. (g) The Swing Bus is needed to condition the YBUS admittance matrix so as to make solutions to the power flow problem possible. Without conditioning it may be possible to have many solutions to the load flow problem which satisfy the constraints. Hence by fixing one bus with respect to earth potential one of these many solution cases is selected. The Swing Bus also serves the purpose of carrying the slack or net power from the rest of the network. ELEC 4100 TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS 1 ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 7 : SYMMETRICAL COMPONENTS - SOLUTIONS Question 1. Determine the symmetrical components of the following line currents : (a) Ia = 5∠900, Ib = 5∠3400, Ic = 5∠2000, and (b) Ia = 50, Ia = j50, Ic = 0. Answer (a) The symmetrical component currents are given by: ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 200 340 90 2 2 2 2 2 1 0 5 5 5 1 1 111 3 1 1 1 111 3 1 j j j c b a e e e aa aa I I I aa aa I I I ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 0 0 0 00 00 200 340 90 120120 120120 2 1 0 5 5 5 1 1 111 3 1 j j j jj jj e e e ee ee I I I ( ) ( ) ( ) ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ++ ++ ++ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − 000 000 000 4022090 8010090 20034090 2 1 0 3 5 3 5 3 5 jjj jjj jjj eee eee eee I I I ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − + + = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ Aj Aj Aj I I I 4760.0 9490.4 5266.0 2 1 0 (b) The symmetrical component currents are given by: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 50 50 1 1 111 3 1 1 1 111 3 1 090 2 2 2 2 2 1 0 j c b a e aa aa I I I aa aa I I I ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 0 50 50 1 1 111 3 1 0 00 00 90 120120 120120 2 1 0 j jj jj e ee ee I I I ( ) ( ) ( ) ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ + + + = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − 0 0 0 30 210 90 2 1 0 13 50 13 50 13 50 j j j e e e I I I ( ) ( ) ( ) ⎥⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − + = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ Aj Aj Aj I I I 333.8100.31 333.8233.2 667.16667.16 2 1 0 ELEC 4100 TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS 2 Question 2. One line of a three phase generator is open circuited, while the other two are short-circuited to ground. The line currents are Ia = 0, Ib = 1000A∠900, and Ic = 1000A∠-300. Find the symmetrical components of these currents. Also find the current into ground. Answer The symmetrical component currents are given by: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − 0 0 30 90 2 2 2 2 2 1 0 1000 1000 0 1 1 111 3 1 1 1 111 3 1 j j c b a e e aa aa I I I aa aa I I I ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −− − 0 0 00 00 30 90 120120 120120 2 1 0 1000 1000 0 1 1 111 3 1 j j jj jj e e ee ee I I I ( ) ( ) ( ) ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ + + + = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − 00 00 00 9030 150210 3090 2 1 0 3 1000 3 1000 3 1000 jj jj jj ee ee ee I I I ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ −∠ ∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 2 1 0 303.333 1507.666 303.333 A A A I I I The ground current is the sum of the b and c phase currents and is given by: ( ) ( ) 03090 2 1 0 3010005008661000 00 ∠=+=+= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ =+= − AAjee I I I III jjcbgnd Question 3. Given the line to ground voltages Vag = 280V∠00, Vbg = 290V∠-1300, and Vcg = 260V∠1100, calculate (a) the sequence components of the line to ground voltages, denoted VLg0, VLg1, and VLg2. (b) the line to line voltages Vab, Vbc, Vca. (c) The sequence components of the line to line voltages VLL0, VLL1, and VLL2. Also verify the following general relation : VLL0 = 0, 011 303 ∠= LLgLL VV , and 0 21 303 −∠= LLgLL VV . Answer (a) The symmetrical components of the line to ground voltages are given by: ELEC 4100 TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS 3 ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − 0 0 0 110 130 0 2 2 2 2 2 1 0 260 290 280 1 1 111 3 1 1 1 111 3 1 j j j cg bg ag Lg Lg Lg e e e aa aa V V V aa aa V V V ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − 0 0 0 00 00 110 130 0 120120 120120 2 1 0 260 290 280 1 1 111 3 1 j j j jj jj Lg Lg Lg e e e ee ee V V V ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −∠ −∠ ∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 2 1 0 43.7987.24 63.673.275 11.7855.7 V V V V V V Lg Lg Lg (b) The line to line voltages are calculated according to: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ −∠ ∠ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 0 0 0 0110 110130 1300 49.146491.442 80.101550.476 47.25613.516 280260 260290 290280 00 00 00 V V V ee ee ee VV VV VV V V V jj jj jj agcg cgbg bgag ca bc ab (c) The symmetrical components of the line to line voltages are given by: ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − 0 0 0 49.146 80.101 47.25 2 2 2 2 2 1 0 491.442 550.476 613.516 1 1 111 3 1 1 1 111 3 1 j j j cg bg ag LL LL LL e e e aa aa V V V aa aa V V V ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ ∠= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 2 1 0 43.4907.43 37.2357.477 0 V V V V V V LL LL LL So : 00 0 0 1 1 303307321.1 63.673.275 37.2357.477 ∠=∠= −∠ ∠ = V V V V Lg LL 00 0 0 2 2 303307321.1 43.7987.24 43.4907.43 −∠=−∠= −∠ ∠ = V V V V Lg LL Question 4. The voltages given in question 3 are applied to a balanced Y load consisting of (12+j16) ohms per phase. The load neutral is solidly grounded. Draw the sequence networks and calculate I0, I1, and I2, the sequence components of the line currents. Then calculate the line currents Ia, Ib, and Ic from the sequence components, and compare with the line currents calculated directly from the network equations. ELEC 4100 TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS 4 Answer The sequence networks are shown below: IL1 VLg1 12+j16 Ω Positive Sequence Negative Sequence Zero Sequence IL2 IL0 VLg2 VLg0 12+j16 Ω 12+j16 Ω The three sequence currents can be calculated as: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ −∠ ∠ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ + −∠ + −∠ + ∠ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 0 0 0 2 2 1 1 0 0 2 1 0 30.26243.1 76.59787.13 98.24378.0 1612 43.7987.24 1612 63.673.275 1612 11.7855.7 A A A j V j V j V Z V Z V Z V I I I Lg Lg Lg L L L The line currents are then given by: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ −∠ ∠ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 0 0 0 120120 120120 2 1 0 2 2 30.26243.1 76.59787.13 98.24378.0 1 1 111 1 1 111 00 00 A A A ee ee I I I aa aa I I I jj jj L L L c b a ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ ∠ −∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 87.560.13 87.1765.14 13.530.14 A A A I I I c b a From the network equations directly: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ ∠ −∠ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ + ∠ + −∠ + ∠ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 0 0 0 87.560.13 87.1765.14 13.530.14 1612 110260 1612 130290 1612 0280 A A A j V j V j V Z V Z V Z V I I I cg bg ag c b a This matches the result calculated using the symmetrical component model. ELEC 4100 TUTORIAL SEVEN : SYMMETRICAL COMPONENTS - SOLUTIONS 5 Question 5. As shown in figure 1, a balanced three-phase, positive sequence source with VAB = 480V∠00 is applied to an unbalanced ∆ load. Note that one leg of the ∆ is open. Determine (a) the load currents IAB and IBC. (b) the line currents IA, IB, IC, which feed the ∆ load. (c) the zero, positive, and negative sequence components of the line currents. I a Ib I c E a E c Eb (18+j10)Ω (18+j10)Ω V ab= 480V 00 Ibc I ab Figure 1: Network for Question 5. Answer (a) The load currents are given by: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −∠ −∠ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ + −∠ + ∠ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 05.149311.23 05.29311.23 0 1018 120480 1018 0480 0 0 0 0 0 A A j j Z V Z V I I I bc ab ca bc ab (b) The line currents are given by: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ −∠ −∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − −= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 95.30311.23 05.179376.40 05.29311.23 A A A I II I I I I bc abbc ab c b a (c) The sequence currents are given by: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ −∠ −∠ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 2 2 2 2 2 1 0 95.30311.23 05.179376.40 05.29311.23 1 1 111 3 1 1 1 111 3 1 A A A aa aa I I I aa aa I I I c b a ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ −∠= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 2 1 0 95.60459.13 055.59917.26 0 A A A I I I ELEC 4100 TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION 1 ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 8 : THREE PHASE FAULTS - SOLUTION Question 1. Equipment ratings for the 4-bus system shown in figure 1 are as follows: • Generator G1 : 500MVA, 13.8kV, X’ = 0.20 p.u. • Generator G2 : 750MVA, 18.0kV, X’ = 0.18 p.u. • Generator G3 : 1000MVA, 20.0kV, X’ = 0.17p.u. • Transformer T1 : 500MVA, 13.8kV delta/500kV star, X = 0.12 p.u. • Transformer T2 : 750MVA, 18kV delta/500kV star, X = 0.10 p.u. • Transformer T3 : 1000MVA, 20kV delta/500kV star, X = 0.10 p.u. • Each transmission line : X = 50 ohms. A three phase short circuit occurs at bus 1, where the pre-fault voltage is 525kV. Pre-fault load current is negligible. Draw the positive sequence reactance diagram in per unit on a 1000MVA base, 20kV base in the zone of generator G3. Determine: (a) The Thevenin reactance in per unit at the fault : [0.2670] (b) The transient fault current in per unit and kA : [-j3.933, -j4.541kA] (c) Contributions to the fault current from G1 and from line 1-2. [-j1.896, -j2.647kA] 1 2 3 Bus 1 Bus 3 Bus 4 j50 ohm j50 ohm T3 T3 T2 Bus 2 j50 ohm Figure 1 : Four Bus Power System. Answer: The positive sequence per unit network is shown below. The per unit values are determined as follows: MVASbase 1000= kVVbase 203 = Zone of Generator 3. kVkV kV kVVbase 5002020 500 4 == Zone of Transmission lines. kVkV kV kVVbase 18500500 18 2 == Zone of Generator 2. ELEC 4100 TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION 2 kVkV kV kVVbase 8.13500500 8.13 1 == Zone of Generator 1. ( ) ( ) Ω=== 250 1000 500 224 4 MVA kV S VZ base base base ( ) kAkV MVA V SI base base base 155.15003 1000 3 4 4 === Bus 1 Bus 4 XT1 XT3 X24 X12 X23 XG3 XG2 XG1 EG3 EG2 EG1 j0.24 pu j0.1pu j0.2pu j0.24pu j0.2pu j0.2puj0.4pu j0.17pu Bus 3Bus 2 XT2j0.133pu So applying these base values to the generators: ( ) ..4.0 500 10002.01 upXG =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ( ) ..24.0 750 100018.02 upXG =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ..17.03 upXG = Similarly for the transformers: ( ) ..24.0 500 100012.01 upXT =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ( ) ..1333.0 750 10001.02 upXT =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ..1.03 upXT = For the transmission lines: ..2.0 250 50 242312 upXXX ==== ELEC 4100 TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION 3 Part (a) The Thevenin equivalent impedance of the network when viewed from voltage bus 1 is: ( ) ( ) ( )[ ]332322241211 //// GTGTTGTh XXXXXXXXXX ++++++= ( ) ( ) ( )[ ]24.01333.02.0//17.01.02.02.0//4.024.0 jjjjjjjjjXTh ++++++= ( ) ( )4583.0//64.0 jjXTh = ..2670.0 upjXTh = Part (b) The pre-fault voltage, neglecting pre-fault currents is: ..005.1 500 0525 00 up kV kVVF ∠= ∠ = So the fault current is: ..933.3 2670.0 ..005.1 0 upjj up Z VI Th F F −= ∠ == kAjIF 541.4−= Part (c) Using the current divider rule: ..641.1 64.04583.0 4583.0 1 upjjj jII FG −= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = kAjIG 896.11 −= ..292.2 64.04583.0 64.0 2 upjjj jII FG −= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = kAjIG 647.21 −= ELEC 4100 TUTORIAL EIGHT : THREE PHASE FAULTS - SOLUTION 4 Question 2. For the above described power system, consider the case where a balanced 3-phase short circuit occurs at bus 2 where the pre-fault voltage is 525kV (neglect the pre-fault current). Determine – (a) The Thevenin equivalent impedance of the network viewed from the fault location : [0.1975 p.u.] (b) The fault current in per unit and in kA [-j5.3155 p.u., -j6.138kA] (c) The contribution to the fault from lines 1-2, 2-3 and 2-4. [-j1.44, -j2.58, -j2.21 kA] Answer: Part (a) For faults on bus 2, the Thevenin equivalent impedance is given by: ( ) ( ) ( )332322241211 //// GTGTTGTh XXXXXXXXXX ++++++= ( ) ( ) ( )24.01333.02.0//17.01.02.0//2.04.024.0 jjjjjjjjjXTh ++++++= ( ) ( ) ( )5733.0//47.0//84.0 jjjXTh = ..1975.0 upjXTh = Part (b) The pre-fault voltage, neglecting pre-fault currents is: ..005.1 500 0525 00 up kV kVVF ∠= ∠ = So the fault current is: ..3155.5 1975.0 ..005.1 0 upjj up Z VI Th F F −= ∠ == kAjIF 1379.6−= Part (c) The contribution to the fault from line 12 is given by: kAjupjjI 443.1..25.184.0 005.1 0 12 −=− ∠ = kAjupjjI 580.2..234.247.0 005.1 0 23 −=− ∠ = kAjupjjI 115.2..8315.15733.0 005.1 0 24 −=− ∠ = ELEC 4100 TUTORIAL NINE : FAULT STUDIES 1 ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 9 : FAULT STUDIES Question 1. The single-line diagram and equipment ratings of a three phase electrical system are given below. The inductor connected to the neutral of generator 3 has a reactance of 0.05 p.u. using the ratings of generator 3 as a base. Draw the positive, negative and zero sequence network diagrams for the system using a 1000MVA base, and a 765kV base in the zone of line 1-2. Neglect the effects of ∆-Y transformer phase shifts. Line 1 - 3 3 1 2 Bus 1 Bus 3 T1 T3 T2 Line 1 - 2 Line 2 - 3 Bus 2 4 T4 Transformers: • T1 : 1000MVA, 15 kV ∆ / 765 kV Y , X = 0.1 p.u. • T2 : 1000MVA, 15 kV ∆ / 765 kV Y , X = 0.1 p.u. • T3 : 500MVA, 15 kV ∆ / 765 kV Y , X = 0.12 p.u. • T4 : 750MVA, 15 kV ∆ / 765 kV Y , X = 0.11 p.u. Transmission Lines : • 1-2 : 765 kV, X1 = 50 Ω, X0 = 150 Ω. • 1-3 : 765 kV, X1 = 40 Ω, X0 = 100 Ω. • 2-3 : 765 kV, X1 = 40 Ω, X0 = 100 Ω. Synchronous Generators : • G1 : 1000MVA, 15 kV, X1 = X2 = 0.18 p.u., X0 = 0.07 p.u. • G2 : 1000MVA, 15 kV, X1 = X2 = 0.20 p.u., X0 = 0.10 p.u. • G3 : 500MVA, 13.8 kV, X1 = X2 = 0.15 p.u., X0 = 0.05 p.u. • G4 : 750MVA, 13.8 kV, X1 = 0.30 p.u. X2 = 0.40 p.u., X0 = 0.10 p.u. Answer: The three sequence networks for the system are shown below. The per unit impedance values are calculated as follows: MVASbase 1000= kVVbaseHV 765= Zone of Transmission Lines. kVVbaseLV 15= Zone Generators. ELEC 4100 TUTORIAL NINE : FAULT STUDIES 2 ( ) ( ) Ω=== 23.585 1000 765 22 MVA kV S VZ base baseHV baseHV ( ) kAkV MVA V SI baseHV base baseHV 7547.07653 1000 3 === The per unit sequence impedances of the generators are then given by: ..18.01_1 upXG = ..18.02_1 upXG = ..07.00_1 upXG = ..20.01_2 upXG = ..20.02_2 upXG = ..10.00_2 upXG = ( ) ..2539.0 500 1000 15 8.1315.0 2 1_3 upXG =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ( ) ..2539.0 500 1000 15 8.1315.0 2 2_3 upXG =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ( ) ( ) ..3385.0..2539.0..08464.0 500 1000 15 8.1305.03 500 1000 15 8.1305.0 22 0_2 upupup XG =+= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ( ) ..3386.0 750 1000 15 8.133.0 2 1_4 upXG =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ( ) ..4514.0 750 1000 15 8.1340.0 2 2_4 upXG =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ( ) ( ) ..1129.0750 15 8.131.0 2 0_4 upXG =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = The per unit sequence impedances of the transformers are then given by: ..1.01 upXT = ..1.02 upXT = ..24.0 500 1000 15 1512.0 2 3 upXT =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ..1467.0 750 1000 15 1511.0 2 4 upXT =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = The per unit sequence impedances of the transmission lines are then given by: ..08544.0 23.585 50 2_121_12 upXX === ..2563.023.585 150 0_12 upX == ..06835.0 23.585 40 2_131_13 upXX === ..1709.023.585 100 0_13 upX == ..06835.0 23.585 40 2_231_23 upXX === ..1709.023.585 100 0_23 upX == ELEC 4100 TUTORIAL NINE : FAULT STUDIES 3 Bus 1 XT1 XT3 X12_1 X23_1 XG3_1 XG2_1 XG1_1 EG3 EG2 EG1 j0.1 pu j0.24pu j0.20pu j0.06835pu j0.18pu j0.254pu Bus 3 Bus 2 XT2j0.1pu X13_1 j0.08544 pu EG4 XG4_1 j0.339pu XT4j0.147pu j0.06835pu Positive Sequence Network. Bus 1 XT2 XT3 X12_2 X23_2 XG3_2 XG2_2 XG1_2 j0.1 pu j0.24pu j0.20pu j0.06835pu j0.18pu j0.254pu Bus 3 Bus 2 XT2j0.1pu X13_2 j0.08544 pu XG4_2 j0.451pu XT4j0.147pu j0.06835pu Negative Sequence Network. Bus 1 XT1 XT3 X12_0 X23_0 XG3_0 XG2_0 XG1_0 j0.1 pu j0.24pu j0.10pu j0.1709pu j0.07pu j0.339pu Bus 3 Bus 2 XT2j0.1pu X13_0 j0.2563 pu XG4_0 j0.113pu XT4j0.147pu j0.1709pu Zero Sequence Network. ELEC 4100 TUTORIAL NINE : FAULT STUDIES 4 Question 2. Faults at bus 1 in question 1 are of interest. Determine the Thevenin equivalent impedance of each sequence network as viewed from the fault bus. The pre-fault voltage is 1.0 p.u. Pre-fault load currents and ∆-Y transformer phase shifts are neglected. Answer: The first step towards obtaining the Thevenin equivalent networks for the sequence networks above is to simplify the networks using a Y-∆ transformation. Recall that the Y-∆ transformation is of the form: ZB ZA ZC ZCA ZBC ZAB CABCAB CAAB A ZZZ ZZZ ++ = C ACCBBA AB Z ZZZZZZZ ++= CABCAB BCAB B ZZZ ZZZ ++ = A ACCBBA BC Z ZZZZZZZ ++= CABCAB BCCA C ZZZ ZZZ ++ = B ACCBBA CA Z ZZZZZZZ ++= So the three sequence networks can be simplified to the form: Bus 1 EG3 EG2 EG1 j0.4939pu j0.30pu j0.06835pu j0.28pu Bus 3 Bus 2 j0.08544 pu EG4 j0.4860pu j0.06835pu Bus 1 EG3 EG2 EG1 j0.4939pu j0.7605pu j0.06835pu j0.28pu Bus 3 j0.1733 pu EG4 j0.4860pu j0.6083pu Positive Sequence. Bus 1 j0.4939pu j0.30pu j0.06835pu j0.28pu Bus 3 Bus 2 j0.08544 pu j0.5981pu j0.06835pu Bus 1 j0.4939pu j0.7605pu j0.06835pu j0.28pu Bus 3 j0.1733 pu j0.6083pu j0.5981pu Negative Sequence. ELEC 4100 TUTORIAL NINE : FAULT STUDIES 5 Bus 1 j0.1 pu j0.1709pu j0.07pu j0.339pu Bus 3 Bus 2 j0.1pu j0.2563 pu j0.09116puj0.1709pu Bus 1 j0.1 pu j0.1709pu j0.07pu j0.339pu Bus 3 j0.5063pu j0.09116puj0.8652pu j0.3376pu Zero Sequence. So from these simplified networks, the Thevenin equivalent impedances can be derived looking in at bus 1, as: ( ) { } { }( ) _ 1 0.28 // 0.7605 // 0.06835 // 0.1733 0.4939 // 0.4860 // 0.6083THZ j j j j j j j⎡ ⎤= + ⎣ ⎦ _1 0.1069THZ j= And: ( ) { } { }( ) _ 2 0.28 // 0.7605 // 0.06835 // 0.1733 0.4939 // 0.5981// 0.6083THZ j j j j j j j⎡ ⎤= + ⎣ ⎦ _ 2 0.1097THZ j= And: ( ) { } { }( ) _ 0 0.1 // 0.5063// 0.1709 // 0.8652 0.3376 // 0.09116THZ j j j j j j⎡ ⎤= + ⎣ ⎦ _ 0 0.0601THZ j= Question 3. For a bolted three phase fault, the fault current is given by: 0 2 0I I= = , 0 0 1 _1 1 0 9.355 . . 90 0.1069 F TH VI p u Z j ∠ = = = ∠ − Similarly : 1 9.355 . .a b cI I I I p u= = = = So in ampere: 1 7.06a b cI I I I kA= = = = ELEC 4100 TUTORIAL NINE : FAULT STUDIES 6 Question 4. For a single line to ground fault the sequence networks are connected in series as: I1 ZTh1 VF I2 I0 V1 V2 V0 ZTh2 ZTh0 Positive Sequence Negative Sequence Zero Sequence Hence the sequence currents are: 0 0 0 1 2 _1 _ 2 _ 0 1 0 3.614 . . 90 0.2767 F TH TH TH VI I I p u Z Z Z j ∠ = = = = = ∠ − + + And: 0b cI I= = 0 13 10.84 . . 90aI I p u= = ∠ − In amperes. 08.183 . 90aI kA= ∠ − The sequence voltages are: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − 0 0 0 0 90 90 90 0 2 1 0 2 1 0 2_ 1_ 0_ 2 1 0 614.3 614.3 614.3 1097.000 01069.00 000601.0 0 0.1 0 00 00 00 0 0 j j j j TH TH TH F e e e j j j e V V V I I I Z Z Z V V V V ELEC 4100 TUTORIAL NINE : FAULT STUDIES 7 ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 0 0 0 2 1 0 3965.0 6137.0 2172.0 j j j e e e V V V Hence the phase to ground voltages are given by: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 0 0 0 00 00 0 0 0 120120 120120 2 1 0 2 2 3965.0 6137.0 2172.0 1 1 111 1 1 111 j j j jj jj c b a e e e ee ee V V V aa aa V V V ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −+− −+−= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 00 00 120120 120120 3965.06137.02172.0 3965.06137.02172.0 0 jj jj c b a ee ee V V V ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠− −∠= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ +− −−= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 4.110..9336.0 4.110..9336.0 0 8749.03258.0 8749.03258.0 0 up up j j V V V c b a Question 5. The fault impedance in per unit is: 30 0.0513 585.23F Z = = The connection of the sequence networks is then as shown below: I1 ZTh1 VF I2 I0 V1 V2 V0 ZTh2 ZTh0 Positive Sequence Negative Sequence Zero Sequence 3ZF ELEC 4100 TUTORIAL NINE : FAULT STUDIES 8 So for a single line to ground fault through this impedance: 0 0 0 1 2 _1 _ 2 _ 0 1 0 2.322 . . 90 3 0.4306 F TH TH TH F VI I I p u Z Z Z Z j ∠ = = = = = ∠ − + + + And: 0b cI I= = 0 13 6.967 . . 90aI I p u= = ∠ − In amperes. 05.258 . 90aI kA= ∠ − The sequence voltages are: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − 0 0 0 0 90 90 90 0 2 1 0 2 1 0 2_ 1_ 0_ 2 1 0 322.2 322.2 322.2 1097.000 01069.00 000601.0 0 0.1 0 00 00 00 0 0 j j j j TH TH TH F e e e j j j e V V V I I I Z Z Z V V V V ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 0 0 0 2 1 0 2547.0 7518.0 1396.0 j j j e e e V V V Hence the phase to ground voltages are given by: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 0 0 0 00 00 0 0 0 120120 120120 2 1 0 2 2 2547.0 7518.0 1396.0 1 1 111 1 1 111 j j j jj jj c b a e e e ee ee V V V aa aa V V V ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −+− −+−= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 00 00 0 120120 120120 0 2547.07518.01396.0 2547.07518.01396.0 3575.0 jj jj j c b a ee ee e V V V ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠− −∠ ∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ +− −−= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 00 114..9542.0 114..9542.0 0..3575.0 8717.03882.0 8717.03882.0 3575.0 0 up up up j j e V V V j c b a ELEC 4100 TUTORIAL NINE : FAULT STUDIES 9 Question 6. For a bolted line to line fault the sequence networks are connected as shown below: I1 ZTh1 VF I2 I0 V1 V2 V0 ZTh2 ZTh0 Positive Sequence Negative Sequence Zero Sequence The sequence currents are therefore: 0 0I = 0 0 1 2 _1 _ 2 1 0 4.617 . . 90 0.2166 F TH TH VI I p u Z Z j ∠ = − = = = ∠ − + And: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − − 0 0 00 00 90 90 120120 120120 2 1 0 2 2 617.4 617.4 0 1 1 111 1 1 111 j j jj jj c b a e e ee ee I I I aa aa I I I ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ..997.7 ..997.7 0 up up I I I c b a In amperes: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ kA kA I I I c b a 035.6 035.6 0 The sequence voltages are: ELEC 4100 TUTORIAL NINE : FAULT STUDIES 10 ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 0 00 90 900 2 1 0 2 1 0 2_ 1_ 0_ 2 1 0 617.4 617.4 0 1097.000 01069.00 000601.0 0 0.1 0 00 00 00 0 0 j jj TH TH TH F e e j j j e V V V I I I Z Z Z V V V V ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 5064.0 5064.0 0 5064.0 4936.01 0 2 1 0 V V V Hence the phase to ground voltages are given by: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 5064.0 5064.0 0 1 1 111 1 1 111 00 00 120120 120120 2 1 0 2 2 jj jj c b a ee ee V V V aa aa V V V ( ) ( )⎥⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − ∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 00 00 120120 120120 0 5064.0 5064.0 0..0128.1 jj jj c b a ee ee up V V V ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ −∠ ∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 180..5064.0 180..5064.0 0..0128.1 up up up V V V c b a ELEC 4100 TUTORIAL NINE : FAULT STUDIES 11 Question 7. Recall that for a bolted double line to ground fault, the sequence networks are connected together as shown below: I1 ZTh1 VF I2 I0 V1 V2 V0 ZTh2 ZTh0 Positive Sequence Negative Sequence Zero Sequence Therefore the positive sequence fault current is given by: 0601.0//1097.01069.0 0.1 // 0_2_1_ 1 jjjZZZ VI THTHTH F + = + = 0 0 1 90..862.61457.0 00.1 −∠=∠= upjI The negative and zero sequence currents can be determined by current divider rule, as: 1097.00601.0 0601.090862.6 0 2_0_ 0_ 12 jj j ZZ Z II THTH TH + −∠−= + −= 0 2 90..429.2 ∠= upI Similarly: 1097.00601.0 1097.090862.6 0 2_0_ 2_ 10 jj j ZZ Z II THTH TH + −∠−= + −= 0 0 90..433.4 ∠= upI The phase currents are therefore: ELEC 4100 TUTORIAL NINE : FAULT STUDIES 12 ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − 0 0 0 90 90 90 2 2 429.2 862.6 433.4 1 1 111 j j j c b a e e e aa aa I I I ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −+− −+− −+− = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − −− 00 000 120120 12012090 429.2862.6433.4 429.2862.6433.4 429.2862.6433.4 jj jjj c b a ee eee I I I ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ +− −−= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −− 0 0 0 000 6.39 4.140 6.129 4.2309090 ..44.10 ..44.10 0 44.10 44.10 0 046.865.6 046.865.6 0 j j j jjj c b a eup eup e ee j je I I I So in amperes: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 6.39 4.140 879.7 879.7 0 j j c b a ekA ekA I I I The sequence voltages are: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −∠− −∠ −∠− ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 0 2 1 0 2 1 0 2_ 1_ 0_ 2 1 0 90429.2 90862.6 90433.4 1097.000 01069.00 000601.0 0 0.1 0 00 00 00 0 0 0 j j j e V V V I I I Z Z Z V V V V j TH TH TH F ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ ∠ ∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ − ∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 0 0 2 1 0 0..2664.0 0..2664.0 0..2664.0 02664.0 7335.01 02664.0 up up up V V V The phase voltages are then: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − 2664.0 2664.0 2664.0 1 1 111 1 1 111 00 00 120120 120120 2 1 0 2 2 jj jj c b a ee ee V V V aa aa V V V ELEC 4100 TUTORIAL NINE : FAULT STUDIES 13 ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0..7993.0 0up V V V c b a Question 8. The phase voltages and phase fault currents in the above cases are: Three Phase Fault. 1 7.06a b cI I I I kA= = = = Single Line to Ground Fault. 08.183 . 90aI kA= ∠ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠− −∠= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 4.110..9336.0 4.110..9336.0 0 up up V V V c b a Single Line to Ground Fault Through an Impedance. 0 13 6.967 . . 90aI I p u= = ∠ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠− −∠ ∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 114..9542.0 114..9542.0 0..3575.0 up up up V V V c b a Line to Line Fault. ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ..997.7 ..997.7 0 up up I I I c b a ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ −∠ ∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0 180..5064.0 180..5064.0 0..0128.1 up up up V V V c b a Double Line to Ground Fault. ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ ∠= ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 6.39..44.10 4.140..44.10 0 up up I I I c b a ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∠ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 0 0 0..7993.0 0up V V V c b a The following observations apply: • The Double Line to Ground Fault leads to the worst case fault current. • Line to Line voltages lead to an increase in the un-faulted phase voltage. ELEC 4100 TUTORIAL NINE : FAULT STUDIES 14 Question 9. Now since: ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 2 1 0 2 2 1 1 111 I I I aa aa I I I c b a Then: [ ] 0210 =++= IIIIa : Then via KCL, the three sequence currents must form a node. And: [ ] 0'''' 210 =++= IIIIa : Then via KCL, the three sequence currents must form a node. Similarly: ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 3 3 3 0 0 1 1 111 3 1 1 1 111 3 1 ' ' ' ' 2 2 ' ' ' 2 2 '22 '11 '00 aa aa aa aa cc bb aa V V VV aa aa V V V aa aa V V V The three sequence voltages are therefore equal in magnitude: I2 V00' I0 I1 V11' V22' I2' I0' I1' ELEC 4100 TUTORIAL TEN : TRANSIENT STABILITY SOLUTIONS 1 ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 10 : TRANSIENT STABILITY - SOLUTIONS Question 1. The pre-fault electrical power delivered to the infinite bus is given by: δsin _ _ EQLtd prefaulte XXX EV P ++ = ∞ ( )( ) ( ) ( ) δsin 2.01.0//2.01.03.0 0.128.1 _ +++ = prefaulte P δsin462.2 _ = prefaulte P During the fault, a star-delta transform can be applied to simplify the circuit as shown below: X L12 X L13 X' d X T1 X EQ The X EQ is given by: ( ) ( ) 12 13 13 12 13 d t L d t L L L EQ X X X X X X X X X X + + + + = ( ) ( ) ( )0.3 0.1 0.2 0.3 0.1 0.1 0.1 0.2 0.1 EQ X + + + + = 1.4 EQ X = So the faulted electrical power delivered to the infinite bus is given by: _ sin e fault EQ EV P X δ ∞ = ( ) ( ) _ 1.28 1.0 sin 1.4 e fault P δ= _ 0.9143 sin e fault P δ= The post-fault electrical power delivered to the infinite bus is given by: _ _ sin e postfault d t L EQ EV P X X X δ ∞ = + + ( ) ( ) _ 1.28 1.0 sin 0.3 0.1 0.2 e postfault P δ= + + ELEC 4100 TUTORIAL TEN : TRANSIENT STABILITY SOLUTIONS 2 _ 2.133sin e prefault P δ= To apply the equal area criterion, it is necessary to determine the maximum possible swing angle, and the initial operating angle. This is done as follows: _ 0 2.462 sin 1.0 e prefault m P Pδ= = = 1 0 1.0 sin 2.462 δ − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 0 0 23.96δ = Similarly: ( )0 _ 2.133sin 180 1.0 e prefault m m P Pδ= − = = 0 1 1.0 180 sin 2.133 m δ − ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ 0 152 m δ = Now apply the equal area criterion. ( ) ( ) 1 0 1 1.0 0.9143 sin 2.133sin 1.0 m d d δ δ δ δ δ δ δ δ− = −∫ ∫ ( ) ( ) 0 0 1 2.133cos 0.9143cos 2.133 0.9143 cos m m δ δ δ δ δ− + − = − 1 0.4835 1.219cosδ− = 1 1 0.4835 cos 1.219 δ − −⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 0 1 113.4δ = This is the critical clearing angle. Question 2. Again apply the equal angle criterion, but with: 0 1 51.2δ = So the equal area criterion requires that: ( ) ( ) max1 0 1 1.0 0.9143sin 2.133sin 1.0d d δ δ δ δ δ δ δ δ− = −∫ ∫ ( ) max 0 1 0 max 1 0.9143cos 0.9143cos 2.133cos 2.133cosδ δ δ δ δ δ− + − = − + ( ) max max 0 1 0 2.133cos 2.133 0.9143 cos 0.9143cosδ δ δ δ δ+ = + − + max max 2.133cos 2.017δ δ+ = ELEC 4100 TUTORIAL TEN : TRANSIENT STABILITY SOLUTIONS 3 This is a non-linear equation, which can be solved iteratively using the Newton-Raphson method. Recall that the NR method approximates a solution using the local gradient of the function as: ( ) ( ) 1 ' p p p p y f x x x f x + ⎡ ⎤−⎣ ⎦= + So: ( ) 1 1 max max max max max 1 2.133sin 2.017 2.133cos p p p p p δ δ δ δ δ − + ⎡ ⎤= + − − −⎣ ⎦ Solving Iteratively: 1 max 1.0δ = , 2 max 1.1704δ = , 3 max 1.1546δ = , 4 max 1.1545δ = , 5 max 1.1545δ = , 6 max 1.1545δ = , 7 max 1.1545δ = 8 max 1.1545δ = So after 8 iterations the solution is: 0 max 66.15δ = Since this rotor angle is well below the maximum possible stable rotor angle, the generator can remain synchronised to the infinite bus. Question 3. Substituting in the system parameters. 2 2 1.0 0.03183 0.01 2.462sin d d dt dt δ δ δ= + + Or: 2 2 31.42 0.3142 77.35sin d d dt dt δ δ δ= + + Now consider small deviations in the rotor angle. If the rotor angle changes from 0 δ to 0 δ δ+ ∆ , then: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 sin sin cos cos sin sin cos δ δ δ δ δ δ δ δ δ ∆ + ∆ = ∆ + ∆ = + So: 2 0 0 0 2 31.42 0.3142 77.35sin d d dt dt δ δ δ= + + Changes to: ( ) ( ) ( ) ( ) 2 0 0 0 0 2 31.42 0.3142 77.35 sin cos d d dt dt δ δ δ δ δ δ δ ∆ ∆ ∆ + + = + + +⎡ ⎤⎣ ⎦ The difference becomes: ELEC 4100 TUTORIAL TEN : TRANSIENT STABILITY SOLUTIONS 4 ( ) 2 0 2 0 0.3142 77.35cos d d dt dt δ δ δ δ ∆ ∆ ∆ = + + Taking the Laplace transform: ( ) ( )2 0 0 0.3142 77.35coss s sδ δ ∆ ⎡ ⎤= + +⎣ ⎦ Solving for the roots of the quadratic equation: ( ) 0 1 2 0.3142 0.0987 309.4cos , 2 s s δ− ± − = For the system to be stable, it is necessary that: ( ) 0 0.3142 0.0987 309.4cos 0δ− + − < ( ) 0 0.0987 309.4cos 0.3142δ− < ( ) 0 0.0987 309.4cos 0.0987δ− < ( ) 0 309.4cos 0δ > For ( ) 0 cos 0δ > , it is necessary to have 0 0 0 90 90δ− < < + . This is the small signal stability constraint on the system. ELEC 4100 TUTORIAL ELEVEN : PROTECTION SOLUTIONS 1 ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 11 : PROTECTION - SOLUTIONS Question 1. The input current to a Westinghouse CO-8 relay is 10A. Determine the relay operating time for the following current tap settings (CTS) and time dial settings (TDS). (a) CTS = 1.0, TDS = 0.5. (b) CTS = 2.0, TDS = 1.5. (c) CTS = 2.0, TDS = 7. (d) CTS = 3.0, TDS = 7. (e) CTS = 12.0, TDS = 1. Answer: From the inverse time curves : (a) Time to operate : 0.1s. (b) Time to operate : 0.55s. (c) Time to operate : 3.0s. (d) Time to operate : 5.2s. (e) The breaker can not operate – the current is less than the pick-up current. Question 2. For the system shown in figure 1, directional over-current relays are used at breakers B12, B21, B23, B32, B34 and B43. Over-current relays alone are used at B1 and B4. (a) For a fault at P1, which breakers do not operate? Which breakers should be coordinated? Repeat (a) for a fault at (b) P2, (c) P3. (d) Explain how the system is protected against bus faults. Bus 1 B1 B12 Bus 2 L1 L2 Bus 3 Bus 4 L3 L4 B21 B23 B32 B34 B43 B4P1P2P3 Figure 1 Answer : (a) For a fault at P1, only B34 and B43 should operate. If B34 fails to operate, then B23, B12 and B1 would operate as a backup. So B23, B12 and B1 must coordinate with B34 in the sequence (B34 – B23 – B12 – B1). If B43 fails to operate, B4 would operate as a backup, so B4 must coordinate with B43 in the sequence (B43 – B4). (b) For a fault at P2, only B23 and B32 should operate. As backup protection, B12 and B1 should coordinate with B23 in the sequence (B23 – B12 – B1), and B43 and B4 should coordinate with B32 in the sequence (B32 – B43 – B4). ELEC 4100 TUTORIAL ELEVEN : PROTECTION SOLUTIONS 2 (c) For a fault at P3, only B12 and B21 should operate. As backup protection, B1 should coordinate with B12 in the sequence (B12 – B1), and B32, B43 and B4 should coordinate with B21 in the sequence (B21 – B32 – B43 – B4). (d) Fault at Bus 1 : Breakers B1 and B21 should open. Fault at Bus 2 : Breakers B12 and B32 should open. Fault at Bus 3 : Breakers B23 and B43 should open. Fault at Bus 4 : Breakers B34 and B4 should open. Question 3. (a) Draw the protective zones for the power system shown in figure 2. (b) Which circuit breakers should open for a fault at (i) P1, (ii) P2, (iii) P3? (c) For case (i), if circuit breaker B21a failed to operate, which circuit breakers would open as back-up? Bus 1 Bus 2 B1 Bus 3 P1 Bus 4B12a B12b B13 B31 B3 B46 B32 B32 B21b B21a B24a B42a B24b B42b P2 P3 Figure 2 Answer : (a) The figure below shows the protective zones of the system in figure 2. Bus 1 Bus 2 B1 Bus 3 P1 Bus 4 B12a B12b B13 B31 B3 B46 B32 B32 B21b B21a B24a B42a B24b B42b P2 P3Zone 1 Zone 2 Zone 3 Zone 4 Zone 5 Zone 6 Zone 7 Zone 8 Zone 9 Zone 10 Zone 11 Zone 12 Zone 13 ELEC 4100 TUTORIAL ELEVEN : PROTECTION SOLUTIONS 3 (b i) For a fault at P1, breakers in Zone 3 should operate – i.e. B12a and B21a. (b ii) For a fault at P2, breakers in Zone 9 should operate – i.e. B21a, B21b, B23, B24a and B24b. (b iii) For a fault at P3, breakers in both Zone 6 and Zone 9 should operate – i.e. B21a, B21b, B23, B32, B24a and B24b. (c) If in case b(i), B21a did not operate, then back-up protection would be achieved by opening the breakers in Zone 9 – i.e. B21b, B23, B24a and B24b. Question 4. Three-zone mho relays are used for transmission line protection of the power system shown in figure 3. Positive sequence line impedances are given as follows: • Line 1-2 : Z12_1 = (6+j60) Ω • Line 2-3 : Z23_1 = (5+j50) Ω • Line 2-4 : Z24_1 = (4+j40) Ω Rated voltage for the high voltage buses is 500kV. Assume a 1500:5 CT ratio and a 4500:1 VT ratio at B12. (a) Determine the zone 1, zone 2 and zone 3 settings Zr1, Zr2 and Zr3 for the mho relay at B12 if zone 1 is set for 80% reach of line 1-2, zone 2 is set for 120% reach of line 1-2, and zone 3 is set to cover 120% of adjacent lines. (b) Maximum current for line 1-2 under emergency loading conditions is 1400A at 0.9 p.f. lagging. Verify that B12 does not trip during emergency loading conditions. 1 2 3 Bus 1 Bus 3 Bus 4 line 1-2 Bus 2 line 2-3 line 2-4 B1 B4 B3B12 B21 B23 B32 B24 B42 Figure 3. Answer: Part (a): The impedance seen by the mho relay at B12 is: ( ) ( ) 4500 /1' ' ' 1500 / 5 1 ' 15 15 LNLN L L LN L VVZ I I V ZZ I = = ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ Set the B12 relay zone 1 Zr1 setting for 80% reach of line 1-2, as: ELEC 4100 TUTORIAL ELEVEN : PROTECTION SOLUTIONS 4 ( )1 6 600.8 0.32 3.215r jZ j+⎡ ⎤= = + Ω ⎢ ⎥ ⎣ ⎦ secondary Set the B12 relay zone 2 Zr2 setting for 120% reach of line 1-2, as: ( )1 6 601.2 0.48 4.815r jZ j+⎡ ⎤= = + Ω ⎢ ⎥ ⎣ ⎦ secondary Set the B12 relay zone 3 Zr3 setting for 100% reach of line 1-2, and 120% reach of line 2-3, as: ( )1 6 60 5 501.2 0.8 8.015 15r j jZ j+ +⎡ ⎤ ⎡ ⎤= + = + Ω ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ secondary Part (b) The secondary impedance viewed by B12 during emergency loading is: 0 0 1 500 0 ˆ ' 13 ˆ ' 13.7 25.8 ˆ 1.4 cos 0.9 15' LN L VZ I − ∠ = = = ∠ Ω ∠ − This value is well in excess of the zone 3 impedance setting, so the impedance seen by B12 during emergency loading will not trip the three zone mho relay. Question 5. Line impedances for the power system shown in figure 4 are Z12 = Z23 = (3+j40) Ω, and Z24 = (6+j80)Ω. Reach for the zone 3 B12 impedance relay is set for 100% of line 1-2 plus 120% of line 2-4. (a) for a bolted three phase fault at bus 4, show that the apparent primary impedance “seen” by the B12 relay is: 32 12 24 24 12 apparent IZ Z Z ZI ⎛ ⎞ = + + ⎜ ⎟ ⎝ ⎠ Where ( )32 12I I is the line 2-3 to line 1-2 fault current ratio. (b) if 32 12 0.2I I > , does the B12 relay see the fault at bus 4? NOTE: This problem illustrates the “infeed effect”. Fault currents from line 2-3 can cause the zone 3 B12 relay to under-reach. As such, remote backup of line 2-4 at B12 is ineffective. 1 3 Bus 1 Bus 3 line 1-2 Bus 2 line 2-3 B1 B3B12 B21 B23 B32 B24 I12 I32 Bus 4 I24B24 B42 Figure 4. ELEC 4100 TUTORIAL ELEVEN : PROTECTION SOLUTIONS 5 Answer : Part (a) For the bolted three phase fault at bus 4 the system of figure 4 can be reduced to: Z12 Z23 Z24V2 I24 I12 I32 V1 V3 The primary impedance seen by the B12 relay is then: 1 1 2 2 2 12 12 12 12 12 ' ' ' ' ' ' ' ' ' apparent V V V V VZ Z I I I I − = = + = + ( )24 12 32 12 12 ' ' ' apparent Z I I Z Z I + = + 32 12 24 24 12 apparent IZ Z Z Z I ⎛ ⎞ = + + ⎜ ⎟ ⎝ ⎠ Part (b): The apparent secondary impedance seen by B12 for the bolted three phase fault at bus 4 is: ( )' apparent apparent V I Z Z N N = Where NV and NI are the turns ratios of the potential and current transformers for B12. ( ) ( ) ( ) ( )( ) ( ) 12 24 32 12 32 121 3 40 6 80 1 'apparent V I V I Z Z I I j j I I Z N N N N + + + + + + = = ( )( ) ( )( ) ( ) 32 12 32 129 6 120 80 'apparent V I I I j I I Z N N + + + = Also the zone 3 relay for B12 is set for 100% reach of line 1-2, and 120% of line 2-4. So: ( ) ( ) ( ) ( )3 3 40 1.2 6 80 10.2 136 r V I V I j j jZ N N N N + + + + = = Ω Comparing this expression with the case for the balanced three phase fault when 32 12 0.2I I > shows that: ( ) 10.2 136 ' apparent V I jZ N N + > So the apparent impedance for the three phase fault is greater than the zone 3 set point, and B12 will not trip in this case. Remote backup in this case is not effective. ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS 1 ELEC 4100 ELECTRICAL ENERGY SYSTEMS TUTORIAL 12 : TRANSMISSION LINES – SOLUTIONS Question 1. A single phase transmission line of 1p.u. length with distributed parameters R, L, C and G has a step voltage E applied to the sending end of the line. The general solutions to the transmission line partial differential equations is given by: ( ) 1 2, x xV x s k e k eγ γ−= + and ( ) 1 2 0 1 , x xI x s k e k e Z γ γ− = −  Where: ( )( )R sL G sCγ = + + ( ) ( )0 R sL Z G sC + = + a) For the case where the receiving end of the transmission line is short circuited determine the constants k1 and k2, and derive simplified expressions for the voltage and current. b) For the case where the receiving end of the transmission line is open circuited determine the constants k1 and k2, and derive simplified expressions for the voltage and current. Answer : Part (a) The boundary conditions are : ( )0,V s E= , ( )1, 0V s = Substituting these constraints into the general solutions gives: ( ) 1 20,V s k k E= + = ( ) 1 21, 0V s k e k eγ γ−= + = From the second equation: 2 1 2k k e γ= − Substituting this expression into the first equation gives: ( )22 1k e Eγ− = or ( ) ( ) ( )2 2 2sinh1 E e E e E k e e e γ γ γ γ γ γ − − − = = = − − − So: ( ) 2 1 2 2sinh e E k k e γ γ γ = − = The voltage expression for the transmission line is then given by: ( ) ( ) ( ) ( )1 1 , 2sinh x xE V x s e e γ γ γ − − − = −  ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS 2 ( ) ( )( ) ( ) sinh 1 , sinh x V x s E γ γ − = The current expression for the transmission line is then given by: ( ) ( ) ( ) ( )1 1 0 1 , 2sinh x xE I x s e e Z γ γ γ − − − = +  ( ) ( )( ) ( )0 cosh 1 , sinh xE I x s Z γ γ − = Part (b) The boundary conditions are : ( )0,V s E= , ( )1, 0I s = Substituting these constraints into the above expressions for the voltage and current reveals: ( ) 1 20,V s k k E= + = ( ) 1 2 0 1 1, 0I s k e k e Z γ γ− = − =  From the second expression involving the current at the receiving end of the line: 2 1 2k k e γ= Substituting into the expression for the voltage at the sending end of the line: ( )22 1k e Eγ+ = or ( )2 21 2cosh E e E e E k e e e γ γ γ γ γ γ − − − = = = + + And so: ( ) 2 1 21 2cosh e E e E e E k e e e γ γ γ γ γ γ γ− = = = + + Hence: ( ) ( ) ( ) , 2cosh 2cosh x xe E e E V x s e e γ γ γ γ γ γ − −= + ( ) ( )( ) ( ) cosh 1 , cosh x V x s E γ γ − = Similarly ( ) ( ) ( )0 1 , 2cosh 2cosh x xe E e EI x s e e Z γ γ γ γ γ γ − −   = −    ( ) ( )( ) ( )0 sinh 1 , cosh xE I x s Z γ γ − = ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS 3 Question 2. A single phase lossless transmission line of 150m length has an inductance and shunt capacitance per unit length of L = 1µH/m and C = 11.111 pF/m. The line is terminated by a 600Ω resistance. The transmission line is struck by lightning through an effective 100Ω impedance at the sending end of the line, creating a surge voltage of 30kV peak and 50µs duration. a) Determine the characteristic impedance, travelling wave propagation velocity and the one-way transit time for the transmission line. b) Draw the equivalent circuit of the transmission line under the surge voltage conditions, and calculate the reflection coefficients at each end of the transmission line. c) Plot the voltages at the sending and receiving ends of the line for the first 5µs. Answer: Part (a) The characteristic impedance is given by: 6 0 12 1 10 300 11.111 10 L Z C − − × = = = Ω × The travelling wave propagation velocity is : ( )( ) 8 1 0 12 6 1 1 3 10 11.111 10 1 10 v ms LC − − − = = = × × × This is the speed of light in free space. The one-way transit time for the transmission line is then: ( ) ( ) 7 8 1 0 150 5 10 0.5 3 10 md s s v ms τ µ− − = = = × = × Part (b) v(0,t) v(d,t) 30kV Z0 = 300 Ω τ = 0.5µs ZL= 600 Ω 100Ω The reflection coefficients at the sending and receiving ends of the line are then: 0 0 600 300 1 600 300 3 L R L Z Z Z Z − − Γ = = = + + 0 0 100 300 1 100 300 2 S L S Z Z Z Z − − Γ = = = − + + The initial surge voltage on the line is then: ( )(0,0) 30 300 100 300 22.5v kV kV= Ω Ω+ Ω =   Part (c): The Bewley lattice diagram can be developed as shown below: ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS 4 τ = 0.5µs 2τ = 1.0µs 3τ = 1.5µs 4τ = 2.0µs 5τ = 2.5µs 6τ = 3.0µs 7τ = 3.5µs 22.5 kV x = 0 x = d 8τ = 4.0µs 9τ = 4.5µs 10τ = 5.0µs 7.5 kV -3.750 kV -1.25 kV 0.625 kV 0.2083 kV -0.1042 kV -0.0347 kV 0.0174 kV 0.0058 kV -0.0029 kV From the lattice diagram the sending and receiving end voltages can be developed as shown below: v(d,t) v(0,t) t t 30kV 25kV 0.5µs 1.0µs 1.5µs 2.0µs 2.5µs 3.0µs 3.5µs 4.0µs 4.5µs 5.0µs 0.5µs 1.0µs 1.5µs 2.0µs 2.5µs 3.0µs 3.5µs 4.0µs 4.5µs 5.0µs 25.83kV 25.69kV 25.72kV 25.63kV 25.73kV 25.71kV26.25kV22.5kV ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS 5 Question 3. Figure 1 below shows a single phase lossless transmission line composed of two different sections of underground cable. The first section has a characteristic impedance 100Ω and a one-way propagation time of 0.1ms, while the second section has a characteristic impedance 400Ω and a one-way propagation time of 0.1ms. A surge voltage of 20kV is applied to the line through a 100Ω impedance, and the line is terminated with a 800Ω load impedance. Plot the voltages at the transmission line junction, and the sending and receiving ends of the total line for the first 0.6ms. Z2=400Ω 100Ω v(0,t) v(d1+d2,t)20kV Z1=100Ω τ = 0.1msτ = 0.1ms ZL=800Ωv(d1,t) ZS Figure 1 : Single Phase lossless transmission line. Answer: The reflection coefficient at the sending end of the transmission line is: 1 1 100 100 0 100 100 S S S Z Z Z Z − − Γ = = = + + The reflection and refraction coefficients on the sending side of the junction are: 2 1 12 2 1 400 100 3 400 100 5 Z Z Z Z − − Γ = = = + + ( ) 2 12 1 2 2 4002 8 400 100 5 Z Z Z β = = = + + The reflection and refraction coefficients on the receiving side of the junction are: 1 2 21 2 1 100 400 3 400 100 5 Z Z Z Z − − Γ = = = − + + ( ) 1 21 1 2 2 1002 2 400 100 5 Z Z Z β = = = + + The reflection coefficient at the receiving end of the transmission line is: 2 2 800 400 1 800 400 3 L R L Z Z Z Z − − Γ = = = + + The surge voltage entering the line is given by: ( ) 1 1 100 0,0 20 10 100 100S Z v E kV kV Z Z = = = + + The Bewley lattice diagram can now be developed as shown below: ELEC 4100 TUTORIAL TWELVE : TRANSMISSION LINES - SOLUTIONS 6 0.1ms 10kV 16kV6kV 5.33kV 2.133kV -3.2kV -1.067kV -0.427kV 0.640kV 0.213kV x = 0 x = d1+d2x = d1 0.2ms 0.3ms 0.4ms 0.5ms 0.6ms 0.7ms From this lattice diagram it is possible to plot the three relevant voltage profiles as shown below: 10kV v(d1+d2,t) v(d1,t) v(0,t) t t t τ 2τ 3τ 4τ 5τ 6τ 7τ τ 2τ 3τ 4τ 5τ 6τ 7τ τ 2τ 3τ 4τ 5τ 6τ 7τ 16kV 21.33kV 16kV 18.13kV 17.06kV 18.13kV 17.7kV 17.71kV 17.92kV ELEC4100_Tutorial_1_Three_Phase_CCTS_Soln ELEC4100_Tutorial_2_Three_Phase_CCTS_Part2_Soln ELEC4100_Tutorial_3_Per_Unit_System_Soln ELEC4100_Tutorial_4_Transformers_Soln ELEC4100_Tutorial_5_Load_FLow_Soln ELEC4100_Tutorial_7_Symmetrical_Components_Soln ELEC4100_Tutorial_8_Three_Phase_Faults_Soln ELEC4100_Tutorial_9_Fault_Studies_Soln ELEC4100_Tutorial_10_Transient_Stability_Soln ELEC4100_Tutorial_11_Protection_Soln ELEC4100_Tutorial_12_Tranmission_Lines_Soln /ColorImageDict > /JPEG2000ColorACSImageDict > /JPEG2000ColorImageDict > /AntiAliasGrayImages false /DownsampleGrayImages true /GrayImageDownsampleType /Bicubic /GrayImageResolution 300 /GrayImageDepth -1 /GrayImageDownsampleThreshold 1.50000 /EncodeGrayImages 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