Fault_calculationsFault calculations 1. Per Unit System Revision The per unit quantity = • • actual quantity base value of quantity The actual quantity and base value of quantity are in actual units, therefore the per unit value is dimensionless. The base value is always a real number, therefore the angle of the per unit quantity is the same as for the angle of the actual quantity. In a power system, balanced systems can be solved on a per phase basis after converting ∆ load impedances to Y impedances. The base voltage (Vbase) and complex power (Sbase) are selected. The other base quantities for a three phase Y system can be found from these as: V base , line = V line (the nominal line-line voltage of the HV or LV side of the transformer as appropriate. Note Vbase,line and Vbase,phase vary with location) Vbase , phase = V line 3 S base = Pbase = Q base S base I base = (where the value is chosen and remains the same throughout the system) = 3 V base ,line S base 3Vbase , phase (This comes from ST = √3 VLIL = 3VpIp) (For a Y system Ibase=Ibase,phase=Ibase,line) Z base = V base, phase I base = 2 V base , line S base = Rbase = X base The per unit system allows for complex systems involving many transformers to be analysed more simply, in particular the calculation of symmetrical three phase faults. The calculation of fault level on a system allows for the switch gear to be properly sized. The approach used to solve three phase symmetrical faults is as follows: • • • • • • • Use the per unit system Neglect normal load currents Assume all generators are in phase and operate at rated voltage (to allow all generators to be replaced by a single unit) Neglect system resistance (X/R ≈ 20 for a generator, X/R ≈ 10 for a transformer, X/R ≈ 8 to 3 for HV lines) Assume system is symmetrical so calculations can be performed on a per phase basis Draw reactance diagram and reduce to a single reactance Current and voltage obtained by retracing steps of network reduction Note: 1. Current limiting reactors can be introduced to limit fault current 2. Circuit breakers are rated in MVA (product of fault current and rated voltage) 3. A specified fault infeed is represented by an equivalent generator and source reactance However, it is sometimes necessary to consider other types of faults as shown below and to do this an additional system must be used. Earth fault EEE301 Phase to phase fault Double earth fault 1 Broken conductor Fault_calculations 2.). However any 3phase unbalanced system can be broken down into three balanced systems known as the positive. ±120° phase displacement and are positive sequenced (this is sometimes subscripted as 1 in literature and sometimes as + ). but displaced by ± 120°. Vc+ Vb− Vc0 Vb0 Va0 = V0 Va+ = V+ Vb+ Va− = V− Vc− Where the voltage in phase a. The negative sequence components are made up of three voltage (or current) phasors all equal in magnitude. The values in the other phases are the same magnitude. negative and zero sequence systems. ±120° phase displacement and are negative sequenced (this is sometimes subscripted as 2 in literature and sometimes as . we can refer to V0 . Vb = Vb0 + Vb+ + Vb− the voltage in phase c. The zero sequence components consist of three voltage (or current) phasors with equal magnitude and zero phase displacement (subscripted 0) The positive sequence components are made up of three voltage (or current) phasors all equal in magnitude. In an unbalanced system the currents and voltages are unequal and not displaced by 120°. Unbalanced systems In a 3-phase system it is easier to analyse a balanced system.and use the transform “a” where a = 1∠120° = -1/2 + j√3/2 and a 3 = 1∠240° and a = 1 The phase equations can therefore be written as : Va = V0 + V+ + V− 2 Vb = V0 + a V+ + aV− 2 Vc = V0 + aV+ + a V− EEE301 (Equation set 1) 2 . current and impedance in one phase. After analysis the original system can be rebuilt using superposition. Va = Va0 + Va+ + Va− the voltage in phase b. Vc = Vc0 + Vc+ + Vc− as shown below: Va+ Va0 Va− Vb Vb+ Vb− Vc+ Vc Vc0 Vb0 Va a phase Vc− b phase c phase 2 For simplicity. since it is necessary to only calculate the voltage. V+ and V. (This can be extended for an N-phase system which can be broken into N symmetrical components) Each balanced system can be analysed in turn using volts/phase and current/phase. In a balanced system or in a 3-phase system with no neutral such as a ∆ or an ungrounded Y system there is no neutral current and therefore no zero sequence components. because the sum of the three balanced phasors is zero. while that of ‘a’ is 120°. EEE301 3 . Note: generators normally generate balanced positive sequence voltages and supply balanced positive sequence currents.Fault_calculations The complex number ‘a’ is very similar to the complex number ‘j’. even under a 3-phase symmetrical fault. the neutral current In is the sum of the line currents In = Ia + Ib + Ic By comparison with the equations for sequence components In = 3I0 The neutral current is three times the zero sequence current. Similar equations can be deduced for the phase current as: Ia = I0 + I+ + I− 2 Ib = I0 + a I+ + aI− 2 Ic = I0 + aI+ + a I− (Equation set 3) And the sequence currents as: I0 = 1/3 (Ia + Ib + Ic) 2 I+ = 1/3 (Ia + aIb + a Ic) 2 I− = 1/3 (Ia + a Ib + aIc) (Equation set 4) In a 3-phase Y connected system with a neutral. The above three equations can be written in terms of V0.V+ and V− as: V0 = 1/3 (Va + Vb + Vc) 2 V+ = 1/3 (Va + aVb + a Vc) 2 V− = 1/3 (Va + a Vb + aVc) (Equation set 2) This shows there is no zero sequence voltage in a balanced 3-phase. The only difference being that the angle of ‘j’ is 90°. The positive and negative sequence components are normally fairly straightforward. Z+ = Z − For a generator: All impedances are different. Similarly. it possible for the impedance presented to positive sequence currents to be different from that presented to negative or zero sequence currents. However. Positive.5 times Z+. when there are transformers in a system. or a five limb core type transformer. changes its direction). The per-unit positive and negative impedances are identical for all non-rotating equipment. If for some reason the system becomes unbalanced.limb type 4 . or 3 separate single phase units. The zero sequence impedance (if present) depends on the type of transformer and the neutral impedance. Usually the positive sequence component is much greater than negative sequence component keeping the machine running in the positive direction. then if the neutral impedance is very small compared to the positive and negative impedances (for example if it is solidly earthed) then Z0 can usually be assumed to be equal to the positive and negative impedances. The impedance presented to the fundamental is often different to the impedance presented to the third and fifth harmonic systems. Transformers There are two aspects associated with transformers in an unbalanced system. As a typical guide: For non rotating pieces of equipment Z+ = Z − For transmission lines: Z0 = 3 to 4 times Z+. The negative sequence system is particularly interesting. After analysis they are reconstructed into the original system. the currents and voltages set up a rotating flux in the positive direction. Z+ = Z − For underground cables: Z0 = 1.5 to 2. In harmonic analysis a complex waveform containing. Firstly. say fundamental. a flux wave rotating in the opposite direction will be set up (interchanging two phases of an induction motor. third and fifth harmonics is broken down into three harmonic components and analysed as pure sine waves. there is the impedance and secondly the phase shifts. If the transformer is a shell type 3-phase transformer.Fault_calculations 3. negative and zero sequence impdeances In some respects the idea of symmetrical components is similar to that of harmonic analysis. Z+ = Z − = Z0 Shell type EEE301 5. In a balanced induction machine system. it is important to get the zero sequence components correct since these are more complicated. However. even a small negative sequence current can cause the machine to overheat because the rotor conductors are travelling through the negative sequence flux at approximately twice synchronous speed. Fault_calculations But for a 3 limb core type X0 « X+ since the zero sequence fluxes of each phase are in phase in each limb and have no return path in the iron core (there is a small amount of leakage flux contributing to the zero sequence impedance) 3. Star windings – no neutral winding If we look closely at all the phases of a star connected system and remember that each phase has the same zero sequence current (with no phase shift).limb type φ0 φ0 φ0 The per-unit positive and negative impedances do not depend on the winding connection. This depends on the circuit configuration. However the base voltages do depend on winding connection. In order to determine the zero sequence impedance it is necessary to determine where the zero sequence currents can flow. then summing the currents at the star point gives : I0 I0 I0 I0 + I0 + I0 = 0 3I0 = 0 I0 = 0 THERE ARE NO ZERO SEQUENCE CURRENTS IN AN UNEARTHED STAR WINDING Star windings – earthed If we look closely at all the phases of an earthed star connected system and remember that each phase has the same zero sequence current (with no phase shift). ∆-Y or ∆-∆ are the same. Y-∆. that is the per-unit impedances of a transformer connected Y-Y. then summing the currents at the star point gives : I0 I0 I0 In I0 + I0 + I0 = In 3I0 = In POSSIBLE ZERO SEQUENCE CURRENTS IN AN EARTHED STAR WINDING – depends on circuit and other windings EEE301 5 . Consquently from consideration of the ampere turns : Ip0 Np = Is0 Ns Is0 = 0 and In = 0 This can be easily shown to be the case for all unearthed star connected two winding transformers. then summing the currents at point P gives : Ia0 Ib0 P I0 I0 Ic0 I0 .star (earthed) transformer : Ip0 Is0 Ip0 Is0 Ip0 Is0 In Under all circumstances Ip0 is zero. For example. We will limit our work to two windings.Fault_calculations Delta windings If we look closely at all the phases of a delta connected system and remember that each phase has the same zero sequence current (with no phase shift). 2-winding transformers All transformers are a combination of two or more windings. In our transformers the fundamental rule is that the AMPERE TURNS ON THE PRIMARY MUST EQUAL THE AMPERE TURNS ON THE SECONDARY.I0 + Ia0 = 0 Ia0 = 0 I0 = any value I0 ZERO SEQUENCE CURRENTS CAN CIRCULATE WITHIN A DELTA WINDING BUT NO ZERO SEQUENCE CURRENT MAY ENTER OR LEAVE THE WINDING. if we were to look at the zero sequence current in the windings of a star (unearthed). EEE301 6 . zero sequence current possible Xpu 4. No zero sequence current Xpu 6. zero sequence current in Y winding and internal circulating currents in ∆ winding possible EEE301 7 . No zero sequence current Xpu 3.Fault_calculations SCHEMATIC EQUIVALENT CIRCUIT I0 = 0 Xpu 1. No zero sequence current Xpu 2. No zero sequence current Xpu 5. compensating current can flow in the secondary windings and therefore the zero sequence impedance is as for the positive and negative impedance assuming the primary circuit is complete. for a fault on the secondary side of the transformer. Example 3 – expanded to show the zero sequence currents The impedance to the zero phase sequence is the impedance of the generator plus the impedance of the transformer: Xgenerator Xtransformer Z0 Example 6 – expanded to show zero sequence currents EEE301 8 . This can be shown more clearly using another diagram. the zero phase sequence reactance looks like an open circuit and therefore no current can flow. For a fault in the primary windings. Lets assume a single phase to earth fault.Fault_calculations In example 6. Step 2 – Move the phasors next to their appropriate terminals. the positive sequence currents and voltages (on the Y. Step 1 – Draw the phasor diagram of the positive (or negative – depending on the sequence being considered) sequence voltages which are applied to the primary Y (or ∆ .V. For a Y-∆ transformer. R Phase shifts There is no voltage or current phase shift in a Y-Y or a ∆-∆ transformer. Observe where the labelling lies in relation to the primary winding. the H. side by 30°. side) lead the corresponding values on the L. The phase shift can be worked out by referring to the figure and steps below. Note in our example that VAB leads Vab by 30° EEE301 9 .depending on the transformer being considered) winding.V. quantities lag by 30°. Step 4 – Label the lines by inspection. It can be seen that the neutral current is three times the zero sequence phase current. In order to represent the earthing resistor in a phase equivalent circuit it is necessary to think of it as three resistors of value 3R in parallel (this gives the same voltage drop namely INR ) 3R R I0 3I0 3R I0 3R I0 3I0 Earthing resistor.Fault_calculations The impedance to the zero phase sequence is the impedance of the transformer : Xtransformer Z0 Note the zero sequence impedance network may not be the same as the positive and negative sequence networks. H.V. So draw a line parallel to the phasors in step 2 next to the secondary terminals. For a Y-∆ or ∆-Y transformer a phase shift may need to be included in the per-unit positive and negative sequence networks. Earthing Resistors Neutral earthing resistors are used to help limit the fault current. While for negative sequence. Step 5 – Bring the lines together to complete the new phasor diagram. Step 3 – For each phase the voltage induced in the secondary ∆ (or Y) windings is in phase with the voltage in the primary windings. Fault_calculations (Step 1). (Step 5) C c VAB 30° A N a b Vab B A a B b C c N EEE301 10 . I+ represents the positive sequence current and V+ represents the positive sequence voltage at the fault. E+ represents the positive sequence supply voltage. Analysis of unbalanced faults In any unbalanced system the three sequence components can be treated individually. Z+ I+ V+ E+ E+ = V+ + I+Z+ Similarly for the negative and zero sequence diagrams. Z− I− E− Z0 E0 E. In the positive sequence diagram.= V− + I −Z − V− I0 E0 = V0 + I0Z0 V0 Normally the voltages being generated by a supply system are balanced and therefore contain only positive components.Fault_calculations 4. The three equations simplify to: E+ = V+ + I+ Z+ V− = − I− Z− V0 = − I0 Z0 EEE301 11 . Fault_calculations Single phase to earth fault Now examine a real 3-phase system when a single earth fault occurs on phase A C B A Any load connected to the system is ignored.and E0) to give: E+ Z+ Z− Z0 I+ V− V+ Solving for I+ gives: I+ = E+ Z+ + Z− + Z0 IA can then be calculated from equation set 3 as: IA = 3I+ Two phase to earth fault A B C EEE301 12 V0 . For a single earth fault. the boundary conditions can be written down: IB = IC = 0 and VA = 0 From equation set 4 the following can be written: I0 = 1/3 IA I+ = 1/3 IA I− = 1/3 IA That is I0 = I+ = I − Similarly from equation set 2 the voltages can be written as: V0 = 1/3 ( Vb + Vc) 2 V+ = 1/3 (aVb + a Vc) 2 V− = 1/3 ( a Vb + aVc) 2 Adding the voltage equations together and noting that 1 + a + a = 0 gives: V0 + V+ + V− = 0 This suggests that the three circuits can be connected together in series (ignoring E. Fault_calculations Setting the boundary conditions gives : IA = 0 .IC VB = VC Using these equations gives I+ = -I− and V+ = V− E+ = V+ + I+ Z+ = V− + I+ Z+ = -I− Z − + I+Z+ = I+ Z − + I+Z+ hence I+ = EEE301 E+ Z+ + Z− 13 - . VB = VC = 0 Solving for the equations in sets 2 and 4 gives : I+ + I− + I0 = 0 V+ = V− = V0 This suggests a parallel connection E+ Z+ I+ V+ Z− Z0 Where I+ can be found as : I + = −( I − + I 0 ) = I− I0 Z + Z− V− V0 + = V+ 0 Z− Z0 Z0Z− Substituting for E+ and re-arranging gives E+ = I+ I + = E+ Z0Z− + I+ Z+ Z0 + Z− Z0 + Z− Z0 Z− + Z0 Z+ + Z− Z+ Phase to Phase fault A B C There is no earth current therefore no I0. The boundary conditions are: IA = 0 IB = . IYA IXB = .VY+ = VA/3 VX− .= -IY− IX0 = -IY0 VX+ .VYA = VA VXB = VYB VXC = VYC Two broken conductors EEE301 VB VC IA = 0 VB = 0 VC = 0 V+ = V− I+ = -I − IXA = .IYC VXA.IYB IXC = .IYA IXB = IYB = 0 IXC = IYC = 0 VXA= VYA VXB .VY0 = VA/3 Crossed phases IYA IYB IYC IXA VXA VYA IXB VXB VYB IXC VXC VYC IXA = .VYC = VC IB One broken conductor IA IXC VA IC Phase-to-phase fault IA = 0 VB = VC IA IYA IYB IYC IX+ + IX− + IX0 = 0 IY+ + IY− + IY0 = 0 IX+ = -IY+ IX. Earth fault Double earth fault IA VA IB VB IC VC IB = 0 IC = 0 VA = 0 I+ = I − = I0 V+ + V− + V0 = 0 IB IC V+ = V− = V0 I+ + I − + I0 = 0 IXA VXA IXB VXB V VYB B VXC VC VYC VYA VXA VYA VA IXA VB VC IXB VXB VYB IXC VXC VYC VA IXA = 0 IYA = 0 IXB = .= VA/3 VX0 .VY.IYB VXA= VYA VXB = VYC VXC = VYB 14 IYA IYB IYC .Fault_calculations A summary of fault boundary conditions The diagram below shows the more common faults and the boundary conditions that are applicable.VYB = VB VXC .IYC IXC = .