TRANSFORMER TANK

The factors such as weight, stray load losses and minimum cost are kept in mind while selecting the material for transformer tank. The material must be capable to withstand stresses due to jacking and lifting and must have capacity to house cores, windings and internal connections giving adequate clearance between the windings and the walls. Rolled steel plates are used for tank bodies of most of the transformers. Small tanks are generally welded from steel plates. The larger tanks are assembled from boiler plates. Usage of aluminium for transformer tank reduces the weight and the stray magnetic losses. However, it increases the cost and also needs special attention for lifting to present stressing. The aluminium tanks are usually made of cast aluminium tank parts, which are mounted on a shallow mild steel tray and are arranged to carry the main lifting.

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Transformer multiple choice questions part 1

Hello Engineer's Q.[1] A transformer transforms (a) frequency (b) voltage (c) current (d) voltage and current Ans : D Q.[2] Which of the following is not a basic element of a transformer ? (a) core (b) primary winding (c) secondary winding (d) mutual flux. Ans : D Q.[3] In an ideal transformer, (a) windings have no resistance (b) core has no losses (c) core has infinite permeability (d) all of the above. Ans : D Q.[4] The main purpose of using core in a transformer is to (a) decrease iron losses (b) prevent eddy current loss (c) eliminate magnetic hysteresis (d) decrease reluctance of the common magnetic circuit. Ans :D Q.[5] Transformer cores are laminated in order to (a) simplify its construction (b) minimize eddy current loss (c) reduce cost (d) reduce hysteresis loss. Ans : B Q.[6] A transformer having 1000 primary turns is connected to a 250-V a.c. supply. For a secondary voltage of 400 V, the number of secondary turns should be (a) 1600 (b) 250 (c) 400 (d) 1250 A

Condition for Maximum Power Developed In Synchronous Motor

The value of δ for which the mechanical power developed is maximum can be obtained as, Note : Thus when R a is negligible, θ = 90 o for maximum power developed. The corresponding torque is called pull out torque. 1.1 The Value of Maximum Power Developed The value of maximum power developed can be obtained by substituting θ = δ in the equation of P m . When R a is negligible, θ = 90 o and cos (θ) = 0 hence, . . . R a = Z s cosθ and X s = Z s sinθ Substituting cosθ = R a /Z s in equation (6b) we get, Solving the above quadratic in E b we get, As E b is completely dependent on excitation, the equation (8) gives the excitation limits for any load for a synchronous motor. If the excitation exceeds this limit, the motor falls out of step. 1.2 Condition for Excitation When Motor Develops ( P m ) R max Let us find excitation condition for maximum power developed. The excitation

Effect of Slip on Rotor Parameters : Part2

Effect of Slip on Rotor Parameters 2. Effect of Slip on Magnitude of Rotor Induced E.M.F We have seen that when rotor is standstill, s = 1, relative speed is maximum and maximum e.m.f. gets induced in the rotor. Let this e.m.f. be, E 2 = Rotor induced e.m.f. per phase on standstill condition As rotor gains speed, the relative speed between rotor and rotating magnetic field decreases and hence induced e.m.f. in rotor also decreases as it is proportional to the relative speed N s - N. Let this e.m.f. be, E 2r = Rotor induced e.m.f. per phase in running condition Now E 2r α N s while E 2r α N s - N Dividing the two proportionality equations, E 2r /E 2 = ( N s - N)/N s but (N s - N)/N = slip s E 2r /E 2 = s E 2r = s E 2 The magnitude of the induced e.m.f in the rotor also reduces by slip times the

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