Skip to main content

Load Characteristics of D.C. Compound Generator

The characteristics depends on whether generator is cumulatively compound or differentially compound generator. In cumulatively compound, Φ2r = Φ2r + Φ2r. As load current increases,  Ia increases hence  Ise  also increases producing more flux. But as  Ia increases, the various voltage drops and armature reaction drop also increases. Hence there is drop in the terminal voltage.
       If drop in Vt due to increasing  IL is more dominating than increase in Vt due to increase in flux then generator is called under compounded and its characteristics is dropping in nature, as shown in the Fig. 1.
Fig. 1 Characteristics of compound generator
       If drop in Vt due to armature reaction and other drops is much less than increase in Vt due to increase in flux then generator is called over compound and its characteristics is rising in nature, as shown in the Fig. 1. If the effects of the two are such that on full load current Vt is same as no load induced e.m.f. i.e. the effects are neutralizing each other on full then then generator is called flat compounded or level compound. Its characteristics is shown in the Fig. 1.
       In differentially compound, ΦT = Φsh ~ Φse.. The net flux is difference between the two. As IL increases, Φsh  is almost constant but Φse increases rapidly. Hence the resultant flux ΦT reduces. There is drop due to armature resistance, series field resistance, armature reaction due to which terminal voltage drops further. Thus we get the characteristics of such differentially compound generator as shown in the Fig. 1.
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