APPLICATION OF THE DOUBLING FACTOR TO FAULT CURRENT

11.7 APPLICATION OF THE DOUBLING FACTOR TO FAULT CURRENT

I frms ′′ FOUND IN 11.6

Now returning to the rms equations for I ′′ g ,I ′′ hm and I ′′ lm in sub-section 11.6 it can be seen that each of these currents can have different X-to-R ratios and will therefore decay at different rates. The peak fault current is,

I fpk ′′ = √ 2(DF g I g ′′ + DF hm I hm ′′ + DF lm I lm ′′ )

Where the doubling factors DF g , DF hm and DF lm are evaluated from the X-to-R ratios of each component using equation (11.5) or their nearest ratio given in Table 11.3 as I 1 (pu) or in Table H.1b.

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11.7.1 Worked Example

An LNG plant has an estimated load of 90 MW and is supplied by five 34.0 MVA generators. The main switchboard operates at 33 kV, and supplies DOL induction motors with unit transformers totalling 35 MW. There is a group of 10 DOL motors operating at 6.6 kV the total kW of which is 25 MW. There is a total of 5 MW of low voltage motors operating at 400 V. A large group of high voltage motors operate from variable speed power electronic rectifier-inverters. These consume

a total of 23 MW , and can be regarded as static loads in that they do not contribute fault currents to the main switchboard. There is a miscellaneous static load at 400 V totalling 2 MW.

Each generator has a sub-transient reactance of 0.13 pu. Each generator unit transformer is rated at 42.5 MVA and has a reactance of 0.08 pu. The marginal factors for MVA ratings are,

•K tg = 1.25 •K tm = 1.10

•K td = 1.3 The operating power factors of the loads are, • 0.87 for high voltage motors

• 0.85 for low voltage motors • 0.86 for high voltage static loads (motors) • 0.97 for low voltage static loads.

The MVA values for these loads are,

S hm1 = 35.0/0.87 = 40.23 MVA S hm2 = 25.0/0.87 = 28.74 MVA

S lm = 5.0/0.85 = 5.88 MVA S hs = 23.0/0.86 = 26.74 MVA

S ls = 2.0/0.97 = 2.06 MVA

The summations of active and reactive powers are,

P hm1 = 35.0 MW, P hm2 = 25.0 MVA P lm = 5.0 MW, P hs = 23.0 MVA P ls = 2.0

FAULT CALCULATIONS AND STABILITY STUDIES

Hence

P load =P hm1 +P hm2 +P lm +P hs +P ls = 90.0 MW

Q hm1 = 19.84 MVA r ,Q hm2 = 14.17 MVA r Q lm = 3.10 MVA r ,Q hs = 13.65 MVA r Q ls = 0.5 MVA r Q load = (Q hm1 +Q hm2 +Q lm +Q hs +Q ls ) 1.015 = 52.03

load = (P load 2 +Q load 2 ) = 103.96 MVA

The operating power factor PF load is,

P load

PF load =

The generator MVA is S gen which equals 37.5. Choose the base MVA to be S base = 100. Assume all five generators are operating when the three-phase zero impedance fault occurs. Calculate the rms symmetrical fault currents for the generators and each type of load.

a) The generators and unit transformers

I ′′

= 0.0 − j 9.639 pu

• The high voltage motors and unit transformers. These consist of two groups S hm1 and S hm2 , let their total be S hm .

= 0.033 + j 0.164 + j 0.0545

= 14.1184(0.033 − j 0.2185) = 0.4659 − j 3.0849 pu

• The high voltage variable speed drive motors. These can be ignored as sources of sub-transient current.

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• The low voltage motors and distribution transformers. There is one group of low voltage motors connected to various switchboards. Their total MVA is S lm . The transformer ratings also require the value of the total static MVA which is S ls .

= 0.054 + j 0.143 + j 0.0157

= 2.0924(0.054 − j 0.1587) = 0.1130 − j 0.3321 pu

• The total rms symmetrical sub-transient fault current. The total rms fault current I ′′ frms is,

I frms ′′ =I ′′ g +I ′′ hm +I ′′ lm = 0.0 − j 9.639 + 0.4659 − j 3.0849 + 0.1130 − j 0.3321

= 0.5789 − j 13.056 pu

The base current I base is,

S base

3V base √ = 3 × 33,000

= 1749.6 amps

Hence the total fault current in rms amps is,

I frms ′′ = 1749.6(0.5789 − j 13.056) = 1012.8 − j 22842.7 amps

The magnitude is,

|I ′′ frms | = 22,865 amps

Find the peak sub-transient fault current.

FAULT CALCULATIONS AND STABILITY STUDIES

The X-to-R ratios of the three symmetrical currents can be found from their real and imaginary parts, as shown in the table below:-

Current

Imaginary part

Real part

X-to-R ratio

I g ′′ 9.639

2.9389 The three ‘doubling factors’ are 2.0, 1.622 and 1.343 per unit. The magnitudes of the three

rms currents in amps are 16,864, 5458 and 614 respectively. Multiply each of these currents by 1.414 × doubling factor,

I gpk ′′ = 2.828 × 16,864 = 47,691 amps

I hmpk ′′ = 2.294 × 5458 = 12, 518 amps

I lmpk ′′ = 1.899 × 614 = 16 amps

The total of these currents is the peak asymmetrical sub-transient fault current I ′′ pk which is 61,375 amps. This is a conservative summation because it assumes that the three peaks occur at the same time. The fault making duty of the main switchboard must be greater than this value of current, i.e., choose a duty of at least 70,000 amps.

11.7.2 Breaking Duty Current

Modern switchboard circuit breakers are often able to clear a major fault current within 120 milliseconds, which is typically five or six cycles of the fundamental current. When these circuit breakers are used with generators, and switchboards that are fed by generators located only a short distance away, the decay of the sub-transient current merges with the decay of the transient current. Even at 120 milliseconds the current may have a substantial value. There are several ways of assessing the breaking duty current,

• Use the rigorous equations for a salient pole generator,

e T ′′ d cos(ωt + θ ) − ˆ V e Ta

e ′′ Ta cos(2ωt + θ ) (11.8)

2X d X ′′

See sub-sections 3.4 and 7.2.7 for an explanation of the variables and parameters. • Use the above equation but ignore the sub-transient terms, thereby leaving,

1 1 1 −t

i fa =ˆ V +

e T ′ d cos(ωt + θ )

• Use equation (11.5) and substitute suitable values for R and L the transient parameters of the generator.

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• Use a set of multiplying factors to modify the precalculated value of the rms symmetrical sub- transient current I ′′ f . Apply the factor at the given fault clearance time. (This factor functions in

a manner similar to the ‘doubling factor’ described in sub-sections 11.6 and 11.6.1.3.) Suitable values of the factor are given in clause 12.2.1.3 of IEC60909, equation (47) and Figure 16 therein.

Whichever method is used it is not usually necessary to include the contribution of fault current from induction motors, because such current will have decayed to almost zero at the fault clearance time. If there are large motors connected to the main switchboard then their contribution will be similar to a generator and should be included, see sub-section 7.2.7 and Reference 3 therein, and sub-section 11.8.5.