Experiment 4- Load test on single phase transformer. PDF

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Summary

It contains aim, apparatus, theory, procedure along with the tables, model calculations, graph and E-circuit for the given experiment....

Description

DEPT. OF ELECTRICAL & ELECTRONICS ENGINEERING SRM INSTITUTE OF SCIENCE AND TECHNOLOGY, Kattankulathur – 603 203

Title of Experiment

: 4. LOAD TEST ON SINGLE PHASE TRANSFORMER

Name of the candidate

: Vedant Kadam

Register Number

: RA2111003010241

Date of Experiment

: 17/12/2021

Sl. No. 1

Marks Split up Pre Lab questions

Maximum marks (50) 5

2

Preparation of observation

15

3

Execution of experiment

15

4

Calculation / Evaluation of Result

10

5

Post Lab questions Total

Marks obtained

5 50

Staff Signature

PRE LAB QUESTIONS

1. Explain the working principle of the transformer. A transformer works on the principle of mutual induction. Mutual induction is the phenomenon by which the amount of magnetic flux linked with a coil changes, an EMF induced in the neighboring coil.

2. What are the main parts of a transformer? 1.Iron core 2. Primary wounded coil of wire 3. Secondary wounded coil of wire 4. Slip rings 5. Armature coil 6. Commutator 7. Brushes 8. Yoke

3. What are the types of transformers? Based on construction there are two types of transformers: 1. Core type 2. Shell type Based on the working principle there are two types: 1. Step up transformer 2. Step down transformer.

4. What is the meaning of the KVA rating of a transformer? KVA stands for kilovolt-ampere and it is the rating normally used to rate a transformer. The size of the transformer is determined by the KVA of the loop.

5.What is the necessity of the load test for a transformer? The purpose of load test is to determine the parallel operation and calculate the efficiency, thermal stability and dynamic stability of the power transformer through measuring the short circuit loss and impedance voltage of the power transformer.

Experiment No. 4 Date : 17/12/2021

Load test on single phase transformer

Aim: To conduct the load test on the given single phase transformer for finding the efficiency and its regulation. Apparatus Required: S.NO

APPARATUS

RANGE

TYPE

QUANTITY

1.

Voltmeter

(0-150)V

MI

1

(0-300) V

MI

1

(0-10)A

MI

1

2.

Ammeter

(0-20) A

MI

1

3.

Wattmeter

150V,20A

UPF

1

300V,10A

UPF

1

4.

Auto transformer

240 V, 2.7 KVA,10A

Formula Used: 1.Percentage Regulation = (Vo2-V2) /Vo2*100 Where Vo2 = Secondary voltage on no load Vo = Secondary voltage at a particular load 2. Power factor = Pout/V2*I2 Where Pout = Secondary wattmeter readings in Watts V2 = Secondary voltage in Volts I2 = Secondary current in Amps

1

3. Percentage efficiency = Pout/Pin*100 Where Pout = Secondary wattmeter readings in Watts Pin = Primary wattmeter readings in Watts. Model Graph:

Pout Procedure 1.Connections are given as per the circuit diagram. 2.Verify whether the autotransformer is kept at zero voltage position. 3.By closing the DPST switch, 230V,1φ,50HZ AC supply is given to the transformer. 4.At no load, the readings from the meters are noted down. 5.The load is applied to the transformer in steps upto 125% of the rated value of the primary Current by using rheostatic load.. 6.The corresponding values from the meters are tabulated for different loads. 7.Then the load is removed gradually, auto transformer is brought to its minimum position and the supply is switched off. 8.From the recorded values, the regulation, power factor and efficiency are calculated.

TABULATION:

S. No

Primary Current I1 (A)

Secondary Voltage V2 (V)

Secondary Current I2 (A)

η %

1

Primary Voltage V1 (V) 230

1.05

229.2

0

0

2

230

1.9

228.2

1.38

72.06

3

230

3.12

227.1

2.05

65.08

4

230

4.25

224.67

3.23

74.23

5

230

5.6

223.3

4.55

78.88

E-circuit:

1 lamp connected:

2 lamps connected:

3 lamps connected:

4 lamps connected:

Model Calculation:

Result: Circuit successfully drawn and observation table, graphs duly filled and plotted. Functioning of a single phase transformer was understood.

POST LAB QUESTIONS 1. What will happen if a DC voltage is given to the transformer primary? A transformer cannot be operated on the DC source. If a rated DC voltage is applied to the primary coil of the transformer, the flux produced in the transformer core will not vary but remains constant in magnitude.

2. What are the losses in a transformer? Transformer losses are produced by the electrical current flowing in the coils and the magnetic field alternating in the core. The losses associated with the coils are called load losses while the losses produced in the core are called no load losses. 3. How can we minimize the core losses in a transformer? Core losses majorly include hysteresis loss and eddy current loss. Eddy current loss can be reduced by increasing the number of laminations. The laminations provide small gaps between the plates. As it is easier for magnetic flux to flow through iron than air or oil, stray flux that can cause core losses is minimized.

4. What is meant by eddy current losses? Eddy current loss is conductive I2R loss produced by circulating currents induced in response to AC flux linkage, flowing against the internal resistance of the core.

5. How can hysteresis loss be reduced? Hysteresis loss can be reduced by using material having least hysteresis loop area. Hence silicon steel or high grade steel is used for manufacturing of a transformer core as it is having less hysteresis loop area. Hysteresis losses can be reduced by increasing the number of laminations.

VEDANT KADAM (RA2111003010241)...