Pre-Fault Load Current | Power System | Electrical Engineering

The effects of pre-fault system loading have been neglected. Usually this is justified because fault currents are much larger than load currents. However, in some situations it may become necessary to consider the effect of load current.

The straight forward approach would be to connect load impedances and simply deal with a more elaborate network, but usually it is not done for two reasons:

1. All loads are not correctly modelled as simple impedances.

2. If effects for several different loads are to be considered, load impedances are required to be recalculated and a new fault study is to be made for each case.

The pre-fault line loadings are typically available as output data from a load flow study. The voltages at different points in the power network depart from unity and zero in magnitude and phase value due to line loadings. It is possible to calculate proper values for a given load condition and use these values in a fault study.

However, the same result can be obtained by Thevenin’s theorem, by which we determine the changes in currents in the branches and changes in the voltages at different buses of the network. By superposition of the pre-fault branch currents and changes in branch currents, post-fault branch currents are determined. Similarly the post-fault bus voltages can be determined.

If the load happens to be a synchronous motor, it continues to run due to its stored energy and supplies power to the fault. The motor circuit, under fault conditions, will have current equal to that of phasor sum of motor current during pre-fault condition and the fault current contribution of the motor.

The noteworthy points are that in case of a generator the directions of load current and fault current are same (both the currents flow out of generator terminals) while in case of a motor the directions of the two currents are opposite to each other because the load current flows into the motor terminals and the fault current flows out of the motor terminals.

The above facts will be illustrated by the following example:

Example:

A synchronous generator and a synchronous motor are each rated 30 MVA, 13.2 kV. Both have X” = 20 %. The line connecting the two has reactance of 10 % on the base of the machine ratings. The motor is receiving 20 MW at 0.8 pf leading and terminal voltage of 12.8 kV. If a symmetrical three phase short circuit occurs at the terminals of the motor, compute the sub-transient current in the generator, motor and fault.

Solution:

Single line diagram for the given system, pre-fault equivalent circuit and equivalent circuit during fault condition are shown in Figs. 4.26 (a), (b) and (c) respectively. All the reactances are given on a common base of 30 MVA.

Pre-fault voltage, V₀ = 12.8/13.2 = 0.9697 ∠0° pu

Load = 20 MW at 0.8 pf leading = 20/30 = 0.6667 pu, 0.8 pf leading

Pre-fault current, I₀ = load/V₀ x pf ∠cos^-1 pf

= 0.6667/0.9697 x 0.8 ∠cos^-1 0.8 = 0.8594 °∠36.87° pu

(a) By Direct Calculations:

Voltage behind subtransient reactance for the generator,

E”_g = V₀ – j I₀ X”_d = 0.9697 ∠0° + j 0.3 x 0.8594 ∠36.87°

= 0.9697 + j 0.2063 – 0.1547 = 0.815 + j 0.2063 pu

Voltage behind subtransient reactance for the motor,

E”_m = V₀ – j I₀ X”_d = 0.9697 ∠0° – j 0.2 x 0.8594 ∠36.87°

= 0.9697 + 0.1031 – j 0.1375 = (1.0728  – j 0.1375) pu

Under fault condition [Fig. 4.26 (c)]

I”_g = [E”_g/(j 0.2 + j 0.1)] = [(0.815 + j 0.2063)/j 0.3] = (0.6876 – j 2.717) pu

I”_m = [E”_m/j 0.2] = [(1.0728 – j 0.1375)/ j 0.2] = (0.6876 – j 5.364) pu

 

= – j 8.081 pu

Basic current, I_B = [(30 x 10³)/(√3 x 13.2)] = 1,312.16 A

So I”_g = 1,312.16 x (0.6876 – j 2.717) = (902.3 – j 3,565) A

= 3,677 ∠– 75.8° A Ans.

I”_m = 1,312.16 x (– 0.6876 – j 5.364) = (-902.3 – j 7,038) A

= 7,096 ∠– 97.3° A Ans.

I_f = 1,312.16 x (– j 8.081)= – j 10,603 A or 10, 603 ∠– 90° A Ans.

(b) By Thevenin’s and Superposition Theorems:

The circuit model for the system for computation of post fault condition is shown in Fig. 4.27. Thevenin’s reactance when viewed from fault terminals FG

= (j 0.2 + j 0.1) ll j 0.2

= [(j 0.3 x j 0.2)/(j 0.3 + j 0.2)] = j 0.12 pu

Total fault current, I_f = V₀/Thevenin’s reactance

= 0.9697/j 0.12 = – j 8.081 pu

Change in generator current due to fault,

ΔI_g = I_f x (j 0.2/j 0.5) = – j 8.081 x 0.4 = – j 3.32324 pu

Change in motor current due to fault, ΔI_m = – j 8.081 x (j 0.3/j 0.5) = – j 4.8486 pu

Generator sub-transient current, I”_g = I₀ + Δ I_g = 0.8594 ∠36.87° – j 3.2324

= (0.6876 – j 2.717) pu

Motor sub-transient current, I’m = – I”₀ + Δ I_m = – 0.8594 ∠36.87° – j 4.8486

= (– 0.6876 – j 5.364) pu

Which are the same, calculated already. This is as expected.

This method, indeed, is a powerful method for solution of large networks.