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WEEK 07: TRANSFOMERS: CALCULATIONS & TEST DATA Sections: Volt-Turns | Density | Insulation | Stampings | Window Volt-Amperes, V-A = 4.44f x F x TI x 10-8 But the Volts per turn, Vt = E/T = EI/TI Thus Vt² = 4.44 f F x 10-8 x EI/TI = k x (fF/TI) x EI Thus:
Sections: Volt-Turns | Density | Insulation | Stampings | Window
Usually, permissible current densities are limited by heating. The resistivity of copper is usually at 1.0W per CM in 60°C and 1.05W per CM in 75°C. Note that 1 cu. in. of copper is equal to 4/p x 106 CM, thus, the watts lost per cu. in. would depend on the current density. Then, Watts lost per cu. in. = D² x 105 / [(4/p) x 106] = 0.825D² x 10-6 but 1 cu. in. of copper = 0.32 lb., thus Watts lost per lb. of copper = 2.57D² x 10-6. Sections: Volt-Turns | Density | Insulation | Stampings | Window
Insulations should be mechanically strong, high dielectric strength and not soluble in oil. Usual types include: cotton tape, empire cloth, micanite, press board and horn paper. When used in dry oil, up to 50,000 volts alternating between metal disks 0.5 in in diameter, there should be 0.2 in. separation. Note: 25 parts of water per million by volume will lower dielectric strength of approximately 50%. The oil ducts should be at least 3/16 to 5/16 of an inch. The creepage distance (under oil) must be approximately 0.25 + 0.07kV. Generally, the insualtion thickness should be about 0.2 + 0.35 kV, where kV is the rms value of kilovolts of the transformer. Increase this up by 25% for coils that are rectangular. For 69kV or higher, end turns should be insulated from breakdown due to surge. 2.5% of the end windings should be with special insulations up to 10,000 volts. Add ¼% for each increase of 10 volts, and add 15% for 220,000 V transformers. |