Skip to main content

Load Characteristics of D.C. Series Generator

Consider a series generator shown in the Fig. 1
       In case of series generator,
        Ia = Ise =  IL
       As load current increases,  Ise increases. The flux Φ is directly proportional to. So flux also increases. The induced e.m.f. E is proportional to  Ise flux hence induced e.m.f. also increases. Thus the characteristics of E against i.e. internal characteristics is of increasing nature. As increases Ia, armature reaction increases but its effect is negligible compared to increase in E. But for high load current, saturation occurs and flux remains constant. In such case, due to the armature reaction E starts decreasing as shown by dotted line in the Fig. 2.
Fig. 1 Characteristics of d.c. series motor
       Now as IL = Ia  increases, thus the drop Ia  (Ra  +Rse) increases.
      Vt = E - Ia  (Ra  +Rse)
      Thus the external characteristics is also of rising nature as E increases but it will be below internal characteristics due to drop Ia  (Ra  +Rse). This is shown in the Fig. 2.
       In self excited series generator, open circuit characteristics can not be obtained. In open circuit, Ia  = IL = 0 hence induced e.m.f. is zero. Thus open circuit characteristics is possible only by separately exciting the field winding. It is also shown in the Fig. 2.
       In practice when there is no load IL = 0, then there exists certain induced e.m.f. due to residual flux retained by the field winding.
Note : Hence for zero load current, there is a voltage present and the internal and external characteristics do not pass through origin as shown in the Fig. 2.
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