8/6/2019 Test Ovvn a Single Face ]

http://slidepdf.com/reader/full/test-ovvn-a-single-face- 1/4

TEST ON A SINGLE PHASE TRANSFORMER

Name of Instructor: Miss K R S Dematagolla Name : N.D.AmarasingheIndex No : 020017Field : Elect.Eng. (Level-2)Date of Per. : 18/09/2003

Date of Sub. : 25/09/2003

8/6/2019 Test Ovvn a Single Face ]

http://slidepdf.com/reader/full/test-ovvn-a-single-face- 2/4

CALCULATIONS

Rated currentHV side = 4000/400 = 10 ALV side = 4000/230 = 17.39 A

Equivalent circuit parameters

1. Core loss resistance r c = V rated2 /P

R c = 210 2 /40R c = 1102.5 Ω

2. Magnetizing reactance (x m) x m = V rated /Im

= 210/.78= 269.2 Ω

3. Primary winding resistance (r1) r 1 = P/I rated2

= 65/10 2 = 0.65 Ω

4. Primary winding leakage reactance (x1)V in

2 = V x2 + V r

2

V12 = I rated

2 (r1 2 +x1 2)8.4 2 = 10 2 (0.65 2 + x1 2)x1 = 0.532Ω

turns ratio = a 2 = 400/230 = 1.74

r 2eq = r 1/ a2 x2eq=x1/ a 2

= 0.65 Ω/ 1.74 2 = 0.532 Ω / 1.74 2

= 0.214 Ω = 0.175 Ω

Voltage Regulation = 0.65*10*1+269.2*10*08.4

= 0.77

Efficiency = 8.4*10*1 * 100

8.4*10*1 + 102

*0.65 + 65= 39.2%

RESULTS

8/6/2019 Test Ovvn a Single Face ]

http://slidepdf.com/reader/full/test-ovvn-a-single-face- 3/4

Voltage Vs w attmeter readings

0

5

10

15

20

25

30

35

40

45

0 50 100 150 200 250

V

W

Current Vs w attmeter readings

0

5

10

15

20

25

30

35

40

45

0 0.2 0.4 0.6 0.8 1

A

W

8/6/2019 Test Ovvn a Single Face ]

http://slidepdf.com/reader/full/test-ovvn-a-single-face- 4/4

DISCUSSION

.

A transformer, operated on AC voltage can be categorized into two main categories, namely, step uptransformers and step down transformers. When the device is supplied with the AC voltage to the

primary winding, a higher or a lower voltage depending on the category of the transformer will be givenout in the secondary winding which is connected through a magnetic core.

When it comes to an Engineering point of view consideration of losses of an electrical machine is of great importance due to reasons such as determination of efficiency which influences the operating cost,determination of heating of the machine which places an upper limit for output which can be obtainedwithout any deterioration of the insulation, etc.

Considering a transformer, losses which occur can be stated as,• Copper loss in the windings, which vary the current• Hysterisis and eddy current losses in the laminated iron-core constant in the neighbourhood

of a normal operating voltage• Dielectric losses in the insulation material (appreciable only in case of high voltage

transformer)• Leakage flux in the magnetic circuit

When current is increased power generated due to resistance in increased. Therefore when it comes to power transmission in an electrical system it might be necessary for the power to be transmitted tohundreds of kilometres. So if a high current is used for the transmission the power loss along the lines ishigh as shown in the above equation. To minimize this power loss a higher voltage rather than a higher current is needed. Since it is not easy to generate a high voltage at generating stations itself, the role of the transformer is brought in. I.e. the step up transformers, which have the capability of producing ahigher voltage and a lower current output than the input, are used. These transformers output voltages ashigh as 33,000 volts (even higher than that) for long distance transmission.

When it comes down to distributing power to the consumer again the transformers are brought in as thevoltages have to be brought down to the usage levels such as 230 V, 400 V etc. For this purpose stepdown transformers are used which produce lower voltages than the input. Sometimes this is done insteps requiring the usage of the transformer even more.

So considering the above points which mainly involve transformers’ role in power transmission could be mentioned as one of the most important aspects in a electrical power system. In conclusion thetransformer plays a very big role in an electrical power system.