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simple network with power capacitors.

7.5.1 Simple network with no power capacitors

Figure 7.2 shows a circuit representation of a simple industrial or distribution network

containing a harmonic source In with no power capacitors. Lsc is the system short circuit

equivalent inductance made up mainly of the leakage reactance of the supply transformer

but might also include the effects of some loads such as large motors.

Lsc

Vn

In

I

Fig. 7.2 Simple network with no power capacitors

In the absence of power capacitors, the level of harmonic voltage distortion Vn is limited

and proportional to the size of the non-linear load (i.e. the magnitude of In) and system

equivalent short-circuit impedance Zn at the harmonic frequency under consideration. This

is given by:

S7: Supply Harmonics

118 EEE8044: Fundamentals

Zn = R s+ jωnLsc = R s + jn(2πfs)Lsc = R s + jn Xsc (7.3)

where R s is system resistance, which may be significant and must not be completely

ignored, (R s + jωnLsc) is the system short circuit impedance as a function of frequency ωn

= n(2πfs).

Once Zn is calculated, the voltage distortion Vn at each harmonic frequency may be

calculated from:

Vn = In Zn (7.4)

And finally, the total harmonic distortion (THD) may be calculated from Equation 7.2.

S7: Supply Harmonics

119 EEE8044: Fundamentals

Worked Example 7.1

You have been asked to investigate if a paper mill is in compliance with ER G5/4. It is

supplied from an 80 MVA, 33/11 kV transformer with 4% reactance and 1% resistance.

The site demand on the utility is 9 MVA. The load is a 6-pulse, three-phase bridge diode

rectifier.

In order to do this you need to calculate for each harmonic, the rms value of current (In),

the circuit impedance (Zn ) and the voltage (Vn ), then sum the THD.

Note: For this type of repetitive calculation it is a good idea to use a spreadsheet

Base impedance = (base Voltage)2/base VA

Base impedance at the 11 kV side of the transformer is given by

(11×103)2/(80×106) = 1.5125 Ω.

Supply short circuit reactance at 50 Hz, Xsc, is thus given by

(1.5125)(0.04) = 0.0605 Ω.

Supply resistance Rs is given by (1.5125)(0.01) = 0.0151 Ω.

The three-phase rectifier load is 9 MVA. This means that the rms rectifier line current

IL, is given by

11 10 3

9 10 3

3

6

×

×

= 472.4 A

The rectifier dc current Id is given by IL/0.8165 = 578.6 A

For a three-phase diode bridge rectifier, the ac current waveform will contain harmonics of

the order (6k ± 1), where k = 1, 2, 3, etc.

The rms value of the nth current harmonic In is given by

n

0.78 Id

.

The circuit impedance to the nth harmonic Zn = Rs + j n Xsc.

Thus, for the 5th harmonic:

I5 = 90.25 A, Z5 = 0.3029 Ω, and the rms 5th harmonic voltage distortion V5 = I5 Z5 = 27.3

V

For the 7th harmonic, I7 = 64.5 A, Z7 = 0.4238 Ω, and V7 = 27.3 V

S7: Supply Harmonics

120 EEE8044: Fundamentals

For the 11th harmonic, I11 = 41 A, Z11 = 0.6656 Ω and V11 = 27.3 V

For the 13th harmonic, I13 = 34.7 A, Z13 = 0.7866 Ω and V13 = 27.3 V