Skip to main content

CALCULATION FOR VOLTAGE REGULATION

The voltage regulation up is expressed mathematically by
images
Positive sign is for lagging power factor and negative sign is for leading power factor.

1.31.1 Zero Voltage Regulation

For lagging power factor and unity power factor, 0V2 > V2. Therefore, we get positive voltage regulation. For leading power factor, V2 starts increasing. At a certain leading power factor, 0V2 = V2 and hence regulation becomes zero. If the load power factor is further increased, 0V2 becomes less than V2 and hence regulation becomes negative. For zero voltage regulation, we have
0V2V2 = 0
i.e.,     I2(R02cosθX02sinθ)=0
     images
     images
Equation (1.56) shows the leading power factor at which voltage regulation becomes zero.

1.31.2 Condition for Maximum Voltage Regulation

Maximum voltage regulation can be obtained for lagging power factor. For maximum voltage regula-tion, we have
images
i.e.,      –R02sinθ+X02cosθ=0
i.e., images
This is satisfied only when the power factor of the load is lagging. The regulation is maximum when the load power factor angle is equal to the impedance angle of the transformer.
i.e.,      images
Equation (1.58) shows the power factor of the load at which voltage regulation is maximum.
Figure 1.41 shows the variation of regulation with the power factor of the load. Figure 1.41 also shows that regulation becomes maximum when the power factor of the load is lagging and it becomes zero when the power factor is leading.

Comments

Popular posts from this blog

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