Load flow and short circuit studies in ETAP, explained

The short answer

A load flow study calculates the steady-state voltage at every bus and the real and reactive power in every branch with the plant running normally. A short circuit study calculates the current that flows when a fault appears, at every bus, in the first cycle and at the instant the breaker contacts part. The first says whether the design works; the second says whether it survives failure. Between them they decide breaker breaking capacity, cable sizes and relay settings.

What each study computes

Load flow solves the network for one operating case. In go generation, utility voltage, transformer taps and connected load; out come bus voltage, branch flows in kW and kvar, current, losses and power factor at the supply point. Run it for several cases: all loads on, future load, one transformer out with the bus coupler closed, generator islanded. The worst case decides the design.

Short circuit solves the network for a bolted fault at each bus in turn. Under IEC 60909 the outputs are the initial symmetrical current Ik", the peak current ip, the breaking current Ib and the steady-state current Ik. The ANSI method gives momentary and interrupting duties instead. Pick one and stay in it; mixing IEC and ANSI numbers in one report is a visible mistake.

The data you must collect first

The model is only as good as the nameplates behind it.

Two of these are routinely wrong. The utility fault level gets quoted from an old letter, and transformer percentage impedance gets taken from the specification rather than the routine test certificate, where the measured value sits.

Reading the results

Bus voltage. Compare every bus against the project design basis, not a rule of thumb. A common band is 95 to 105 per cent of nominal in steady state, with a lower transient limit during the largest motor start. Motor terminal voltage is bounded separately by IEC 60034-1.

Transformer and cable loading. Both are reported as a percentage of rating. A transformer above roughly 80 per cent continuous leaves no room for future load or for losing its twin; a cable above 100 per cent of its derated capacity is a fire, not a warning.

Fault duty. Compare Ik" at each bus against the breaker's rated breaking capacity and the switchboard's short-time withstand current. Check line-to-ground as well as three-phase: on a solidly earthed star winding close to the transformer, where zero-sequence reactance is below positive-sequence reactance, the line-to-ground current can exceed the three-phase value. Reporting only the three-phase number is how an under-rated earth path gets built.

Worked example: a 1600 kVA transformer

Transformer 1600 kVA, 11 kV / 415 V, Dyn11, declared impedance 6 per cent.

Full load current on the 415 V side:

I = S / (root 3 x V) = 1 600 000 / (1.732 x 415) = 1 600 000 / 718.8 = 2 226 A

So the incomer sits on a 2500 A air circuit breaker frame.

Symmetrical fault current at the LV terminals, treating the 11 kV source as infinite:

Isc = 2 226 / 0.06 = 37 100 A, about 37.1 kA

Now add a real source. If the utility declares 500 MVA at 11 kV, source impedance on the 1600 kVA base is 1600 / 500 000 = 0.0032 per unit. Total 0.0632 per unit, so Isc = 2 226 / 0.0632 = 35.2 kA. At low voltage the transformer dominates; the utility shaved five per cent.

Then add the motors, which feed a fault for the first few cycles at roughly four to six times their own full load current. With 1000 kVA of motors connected, that is 1 000 000 / 718.8 = 1 391 A, and at five times, about 7 kA more first-cycle current, taking the duty towards 42 kA.

A 36 kA device does not survive that. The selection is a 50 kA breaking capacity ACB, with the switchboard's rated short-time withstand current matched to it under IEC 61439-1, and the peak making duty checked too: IEC 60947-2 applies a multiplier to breaking capacity, 2.1 in the 20 to 50 kA band, so 50 kA implies about 105 kA peak.

One more sum explains the margin. IEC 60076-1 allows a tolerance on declared impedance, so a transformer sold as 6 per cent may test lower. At 5.4 per cent, Isc = 2 226 / 0.054 = 41.2 kA before any motor contribution. Model the test certificate, not the purchase order.

What the report is for

Three sign-offs depend on it. Switchgear selection takes the fault duty and the continuous current. Relay settings take the maximum fault current for the instantaneous element and the minimum fault current for earth fault pickup, so each device grades against the one downstream. Cable sizing takes the derated ampacity from load flow and the withstand current from the fault study. A study nobody converts into those three was an expensive way to produce a PDF.

The mistakes people make

• Modelling nameplate load instead of running load, which inflates voltage drop and hides the real problem.
• One operating case. Worst bus voltage and worst fault duty rarely occur together: bus coupler closed is usually the worst fault case, single transformer running the worst voltage case.
• Ignoring motor contribution, which at an LV motor control centre is not a rounding error.
• Leaving taps at nominal when the utility runs high or low, then wondering why site measurements disagree with the study.

What to learn next

Carry the same network into protection coordination: plot relay, fuse and breaker curves on one time-current graph and check each device clears before the one upstream. Arc flash reuses that fault model plus the clearing times, and motor starting reuses the load flow. Related reading: the electrical design engineer roadmap and DOL vs star-delta vs soft starter vs VFD.