Electricity + Control August 2015

TRANSFORMERS + SUBSTATIONS

For all the efficiency calculations a per-phase load voltage V 2 of 220<0° is used. All calculations performed for the per-phase circuit. The load current I 2 is given by the following calculation:

s

3

θ

I 2 = rated% ·

(19)

V 2

where θ is given by the inverse cosines of the power factor, in this case 30°. The formulas will be shown completely for the 60 % rated load. The turns ratio a for the Y-Y configuration is 220/220 = 1.

At 60 % the per-phase load current in the primary winding is:

Figure 5: The exact equivalent circuit of the transformer.

s

3

θ

I 2 = rated% ·

(20)

V 2

The core-loss resistance and the magnetising reactance is much bigger than the winding resistances and the leakage reactance. The reactance of the low voltage and high voltage is as expected since the transformer has a one-to-one ratio and a Y-Y configuration was used. The practical test was done on a 1 kVA, 380/380 V transformer. The following measurements were taken down during the laboratory test (see Table 4 ):

The induced EMF in the secondary winding is: E 2 = V 2 + [ I 2 ( R ] L + jX L )

(21)

The induced voltage in the primary winding is given by: E 1 = a E 2 30

(22)

The current in the primary winding of the transformer is given by:

Table 4: The laboratory readings. Parameters

Open circuit

Short circuit

I 2

I p =

30

(23)

a

Voltage (V)

120

20

Current (A)

0,05

1,5

The per-phase source current is thus given by:

Power 1 (W)

4

12,5

1 ( + ) R c jX 1 1

Power 2 (W)

2

15

I 2 = I p

+ E

(24)

Total Power (W)

16

27,5

m

where R c – is resistance of the core

As seen in Table 4 , the short-circuit test was only done at a rated volt- age and current and care was taken to not pass the rated current of the transformer. As expected, there is a great current at a low voltage. The efficiency of the transformer was calculated at a rated load of 60 % to 90 % in 5 % increases and can be seen in Table 5 . Table 5: Transformer efficiency at different rated loads. Rated load (%) Efficiency (%) 60 88,160 65 88,720 70 89,188 75 89,580 80 89,909 85 90,186 90 90.419

The per-phase voltage supplied by the source: V 1 = E 1 + [ I 1 ( R ] H + jX H )

(25)

The power input is calculated as follows: P out = R e [ V 2 I 2 ]

(26)

Lastly, the efficiency calculated by means of equation:

 =  ( P out ) ( P in )

(27)

The calculated data for the transformer is enclosed in Table 3 .

Table 3: Transformer parameters. Parameters

Impedance ( Ω )

2 450

R c

j 1707,32

X m

• Many plants are served by a supply system that includes on-site transformers and substations. • Transformer efficiency is an increasingly important pa- rameter to understand, and the efficiency improves with loading (within the specified operating range). • Open and short circuit tests can be effectively used to determine the machine parameter, and hence its efficiency.

2,04

R H

2,04

R L

j 3,27

X H

j 3,27

take note

X L

Electricity+Control August ‘15

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