Heading work on TM-25-6 power transformer

Contents

Contents

Transformer Course Design Assignment

1 Design features and determination of main dimensions of the transformer active part

1.1 Calculation of basic electrical values

1.2 Selection of material and design of magnetic system

1.3 Selection of material and design of windings

1.4 Determination of dimensions of main insulation of windings

1.5 Determination of rod diameter and winding height

2 Calculation of LV and HV windings

2.1 General Information

2.2 Calculation of LV windings

2.2.1 General calculations for any type of winding

2.2.2 Calculation of multilayer cylindrical winding from wire

rectangular section

2.3 Calculation of HV windings

2.3.1 General calculations for all winding types

2.3.2 Calculation of multilayer cylindrical winding from rectangular wire

3 Calculation of short circuit parameters

3.1 Determination of short circuit losses

3.2 Calculation of short circuit voltage

3.3 Determination of mechanical forces in windings and winding heating at short circuit

4. Transformer Magnetic System Calculation

4.1 Determination of magnetic system dimensions and mass

4.2 Determination of idling losses

4.3 Determination of transformer idling current

5 Transformer Thermal Calculation

5.1 Overview of Thermal Calculation

5.2 Thermal calculation of windings

5.3 Heat calculation of transformer tank

5.4 Calculation of winding and oil temperature exceedances

Conclusion

Bibliographic list

Task

for a course project in the discipline "Electric Machines"

Perform calculation and structural development of the transformer with the following data:

1. Transformer type................................................ TM 25/6

2. Number of phases..................................................................................................... 3

3. Mains voltage frequency................................................................................................. 50 Hz.

4. Rated power......................................................................................................................... 25kVA

5. HV winding rated voltage............................ 6 kV.

6. Nominal voltage of LV winding........................... 0.4 kV.

7. Winding connection diagrams and group.............................. Y/YH -0.

8. Cooling system - with natural circulation of oil and air (natural oil).

9. Operating mode - long-term load.

10. Unit..................................................................................................................................................................................................... External.

Transformer parameters

1. Short-circuit voltage................................................................................. 4,5 %.

2. Short circuit losses............................................................................................................................................................. 0.5 kW.

3. Idle current................................................................. 3.2%.

4. Idle losses..................................................... 0.125 kW.

. conclusion

In the course design, a TM25 type power transformer was calculated. The developed transformer has a core magnetic conductor, assembled from sheets of cold-rolled electrical steel of the brand.

Aluminium is used as conductor material in windings. Design of windings: ONN - cylindrical single and double-layer of rectangular wire; OVN - cylindrical multilayer of round wire.

Design value of short circuit losses is less than required by GOST. The calculated value of losses of idling of Rkh is more than preset value that does not exceed a tolerance of 5%. The resulting loss values can be considered satisfactory, since the transformer parameters depending on them are within the permissible limits:

The calculated idling current Ix in the calculations turned out to be almost the same as in the task. In this regard, the transformer will be more efficient during operation. The design value of short circuit voltage Uc was the same as in the task, which does not exceed 5%.

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