Skip to main content

TRANSFORMER ON NO LOAD

For an ideal transformer, we have assumed that there are no core losses and copper losses. For practical transformers, these two losses cannot be neglected. At no-load condition, the primary current is not fully reactive and it supplies (i) iron loss in the core, that is, hysteresis loss and eddy current loss and (ii) very small amount of copper loss in the primary. There is copper loss in the secondary because it is an open circuit. The no-load current lags behind V1 by an angle θ0, which is less than 90° (around 80°–85°). The no-load input power is given by
W0=V1I0cosθ0 (1.8)
where cosθ0 is the no-load power factor. Figure 1.23 shows the no-load phasor diagram of a practical transformer. From Figure 1.23, the no-load primary current (I0) has the following two components:
  • One component of I0, that is Iw = I0 cosθ0 is in phase with V1. Since Iw supplies the iron loss and primary copper loss at no load, it is known as active or working or iron loss component.
  • The other component of I0 that is, = I0 sinθ0 is in quadrature with V1. It is known as magnetizing component. Its function is to sustain the alternating flux in the core and it is wattless.
    From Figure 1.23, we have
    images
    and images
The following points are most important:
  • The no-load primary current is 1–5 per cent of full-load current.
  • Since the permeability of the core varies with the instantaneous value of exciting or magnetizing current, the waveform of exciting or magnetizing current is not truly sinusoidal.
  • Since I0 is very small, the no-load copper loss is negligible. Hence, no-load input is practically equal to the iron loss in the transformer.
  • Since core loss is solely responsible for shifting the current vector I0, the angle θ0 is known as hysteresis angle of advance.
images
Figure 1.23 Phasor Diagram at No-Load
Figure 1.24 shows the equivalent circuit of the transformer at no load.
images
Figure 1.24 Equivalent Circuit of Transformer at No-load
Example 1.5 The no-load current of a 4,400/440 V, sinlge-phase, 50 Hz transformer is0.04. It consumes power 80 W at no load when supply is given to LV side and HV side is kept open. Calculate the following:
(i) Power factor of no-load current.
(ii) Iron loss component of current.
(iii) Magnetizing component of current.

Solution
W0 W, I0 =0.04 A, V1 = 4,400 V
  1. Since W0 = V1I0cosθ0
    images
    The no-load power factor is 0.454 (lagging).
  2. IW = I0 cosθ0 =0.04 × 0.454 = 0.0187 A
  3. images
    ∴ Iμ = I0 sinθ0 = 0.04 × 0.891 = 0.0356

Labels:

Comments

Post a Comment

Comment Policy
We’re eager to see your comment. However, Please Keep in mind that all comments are moderated manually by our human reviewers according to our comment policy, and all the links are nofollow. Using Keywords in the name field area is forbidden. Let’s enjoy a personal and evocative conversation.

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
4 comments
Read more

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
Post a Comment
Read more

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
Post a Comment
Read more