EQUIVALENT REACTANCE

MAGNETIC LEAKAGE

Till now we have assumed that all the flux linked with the primary also links with the secondary. But in practice, the permeability of the core of the transformer is finite. All the flux linked with the primary do not link with the secondary. As shown in Figure 1.29(a), Φ_L1 and Φ_L2 induce emf e_L1 and e_L2 in primary and secondary windings respectively. Therefore, in effect, we can consider it as an equivalent inductive coil in phase with the winding shown in Figure 1.29(b).

EQUIVALENT REACTANCE

If X₂ be transferred to primary, let its referred value be X₂^′. We have

I₂²X₂=I₁²X₂^′

i.e.,

If X₁ be transferred to secondary, let the referred value be X₁^′ We have

I₁²X₁=I₂²X₁^′

i.e.,

The total reactance referred to as primary (X₀₁) is X_l + a²X₂ and that of referred to as secondary (X₀₂) is X₂+

. The total reactance is known as equivalent reactance. It is denoted by

X₀₁=X₁+a²X₂ (1.18)

Comments

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

Electrical World

Search This Blog