RTUEE / EC / EEEYr 2019 · Sem 72019

Q3Power System Analysis

Question

16 marks

2. (a) Discuss the analysis of short circuit on a loaded synchronous machine and draw models for computing subtransient and transient current. [8]

(b) Fig shows a system having 4 alternators each rated at 11 kV, 50 MVA and each having a subtransient reactance of 15%. Find - [8]

  • (i) Fault level for a fault on one of the feeders (near the bus) with zero value of reactance X.
  • (ii) The reactance of the current limiting reactor X to limit the fault level to 800 MVA for a fault on the far bus (alternators G1, G2 on the near bus; G3, G4 on the far bus, connected via reactor X).

Answer

(a) Short Circuit on a Loaded Synchronous Machine - Subtransient and Transient Models

When a three-phase short circuit occurs suddenly at the terminals of a synchronous machine that was previously carrying load, the resulting fault current does not jump instantly to its final steady-state value but instead exhibits a characteristic decaying AC envelope, conventionally divided into three successive time periods: the subtransient period (the first few cycles, typically 0-2 cycles, during which the armature reaction flux is very effectively opposed by induced currents in the rotor's damper windings and, in a solid rotor, by induced eddy currents in the rotor iron itself, giving the machine its lowest effective reactance and hence its highest fault current), the transient period (extending from roughly 2 cycles out to several cycles, during which the damper-winding currents have decayed away but the field winding's own induced currents continue to oppose the armature reaction flux, giving an intermediate effective reactance and current level), and the steady-state period (after all transient rotor currents have fully decayed, leaving only the synchronous reactance to limit the fault current at its lowest level).

Synchronous Machine Fault Current Envelopesubtransient (I")transient (I')steady-state (I)

For computing the subtransient fault current, the machine is represented by its internal subtransient EMF E'' behind its subtransient reactance X''d (and, if resistance/asymmetry matters, subtransient reactance in the quadrature axis X''q as well) - E'' is computed from the machine's pre-fault loaded operating condition (pre-fault terminal voltage, load current, and power factor), since the subtransient internal EMF, unlike the simpler synchronous EMF used for steady-state analysis, must properly account for the machine's actual loaded condition immediately before the fault occurred, typically via a phasor diagram construction: E'' = V + I*(Ra + jX''d), where V is the pre-fault terminal voltage and I is the pre-fault load current.

Similarly, for the transient fault current, the machine is modeled by its transient EMF E' behind its transient reactance X'd, with E' computed from the same pre-fault loaded condition using X'd in place of X''d: E' = V + I*(Ra + jX'd). The transient fault current magnitude is then given by an analogous expression using E' and X'd, and the steady-state fault current is given by the synchronous EMF E (computed similarly using the synchronous reactance Xd) divided by Xd.

It is important to note that because the machine was loaded before the fault, E'', E', and E (the subtransient, transient, and synchronous internal EMFs respectively) are generally all different in magnitude from each other and from the simple no-load terminal voltage, since each is computed using a different effective reactance combined with the same actual pre-fault load current - this loaded-machine short-circuit analysis is therefore meaningfully more involved than the simpler unloaded-machine case (where the internal EMF simply equals the pre-fault terminal voltage for all three reactance stages, since with zero pre-fault current there is no voltage drop term to add), and correctly modeling this loaded pre-fault condition is essential for accurate fault-current prediction in any real power system, where generators are essentially always supplying some load immediately before a fault occurs.

(b)(i) Fault Level with X=0

Each of the 4 alternators (G1, G2 on the near bus; G3, G4 on the far bus) has subtransient reactance X''=15% (0.15 pu) on its own 50 MVA, 11 kV base. If the current-limiting reactor X between the two buses is set to zero, the two buses become effectively a single common bus, and all 4 alternators appear in parallel as seen from the fault point (a fault on one of the feeders near this now-single bus):

So with X=0, the fault level for a fault near the bus is approximately 1333.33 MVA.

(b)(ii) Reactance X to Limit Fault Level to 800 MVA

For a fault on the near bus (where G1 and G2 are directly connected), the local contribution comes from G1 and G2 in parallel (each 0.15 pu on the common 50 MVA base), giving a local branch reactance of Xlocal = 0.15/2 = 0.075 pu, contributing a local fault MVA of 50/0.075 = 666.67 MVA. The remote contribution comes from G3 and G4 (also in parallel, giving 0.075 pu at their own bus) in series with the current-limiting reactor X, this remote branch being in parallel with the local branch as seen from the fault point.

So the required current-limiting reactor reactance is X = 0.30 p.u. on the 50 MVA, 11 kV base. Converting to ohms using base impedance Zbase = (11)^2/50 = 2.42 ohm:

This result can be cross-checked: with X=0.30 pu, the remote branch (G3, G4 plus reactor) has total reactance 0.075+0.30=0.375 pu, giving a remote fault MVA contribution of 50/0.375=133.33 MVA; adding this to the local contribution of 666.67 MVA gives 666.67+133.33=800 MVA exactly, confirming the reactor value satisfies the specified fault-level limitation correctly.

It is worth noting that the fault-level results obtained here (1333.33 MVA with the reactor bypassed, reduced to 800 MVA with the properly sized 0.30 pu reactor) illustrate the fundamental engineering purpose of a current-limiting (bus-tie) reactor in power system design: as generating capacity is added to a system over time, the available fault level at existing buses tends to rise correspondingly (since more parallel generation sources mean a lower combined Thevenin fault impedance), potentially exceeding the interrupting capability of already-installed circuit breakers and switchgear - rather than replacing this expensive existing switchgear with higher-interrupting-capacity equipment throughout the system, inserting a current-limiting reactor between generation groups (exactly as calculated in this problem) is often a more economical way to keep fault levels within the existing equipment's rated interrupting capability as new generation is added.

It is also worth relating the loaded-machine short-circuit modeling approach described in part (a) to its practical application in this specific fault-level calculation: the fault-level calculation performed in part (b) implicitly assumes each alternator's subtransient EMF magnitude is simply 1.0 per unit (as if each machine were unloaded immediately before the fault), a simplification appropriate for a standard fault-level/switchgear-rating study where a conservative, standardized assumption is normally used rather than a specific, momentarily-varying actual pre-fault loading condition - a more detailed transient stability or protection coordination study, by contrast, would need to use the full loaded-machine E'' calculation described in part (a), using each machine's actual pre-fault terminal voltage and current, to obtain a more precise instantaneous fault current prediction for that specific operating condition.

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