Power Losses. b. Field winding copper losses Losses due to the shunt field (i sh 2 R sh. ) or series field winding (i s 2 R s

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Dayna Barton
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1 Power Losses The various losses inside a generator can be subdivided according to: 1. copper losses a. armature copper losses = i a 2 R a Where R is the resistance of the armature, interpoles and series field windings etc. this constitutes around 30-40% of full load losses. b. Field winding copper losses Losses due to the shunt field (i sh 2 R sh ) or series field winding (i s 2 R s ). this constitutes around 20-30% of full load losses. c. Brushes and contact losses Usually included in the armature losses.

2 Power Losses 2. Magnetic losses a. Hysteresis loss (W h ) Loss that is due to the reversal of magnetization of the armature core as it passes under the North and South pole alternately. The losses depends on the volume and grade of the iron, maximum value of flux density, and frequency of magnetic reversal. W h =nb 1.6 fv watts Where: n = Steinmetz hysteresis coefficient B = maximum flux density V = volume of the iron core (m 3 ) F = frequency

3 Power Losses 2. Magnetic losses a. Eddy Current loss (W e ) As the armature rotates, it also cuts the magnetic field, therefore a small emf is induced in the body of the core. Though the voltage is rather small, but it produces a large current because of the small resistance offered by the iron core. This current is known as the eddy current. W h =kb 2 f 2 t 2 V 2 watts Where: t = thickness of laminations

4 Power Losses 2. Magnetic losses a. Eddy Current loss (W e ) To counter act the effects of eddy current losses, instead of one solid continuous body, the core is made up of thin laminations stacked and riveted at right angles to the path of eddy currents. These laminations are insulated by thin coat of varnish. Both these losses constitute around 20-30% of the total losses.

5 Power Losses 3. Mechanical losses a. Frictional losses at the bearings and brushes b. Air resistance losses and windage Both these losses constitute around 20-30% of the total losses. Usually the magnetic losses and mechanical losses are collectively known as Stray Power Loss.

6 Power Losses The various losses of a DC generator can be summarized below:

7 Power Stages Input power (P in ) = output of the driving engine (ie. Turbine, ICE, motor) Power Generated (P g ) power generated in the armature Output power (P o ) also referred as the power delivered to the load

8 Generator Efficiency 1. Mechanical efficiency, η m η m = P g P i x100 Power generated in the armature Mechanical power supplied 2. Electrical efficiency, η e η e = P out P g x100 Power delivered to the load Power generated in the armature 3. Commercial or Overall efficiency, η η= P out P i x 100 Power delivered to the load Mechanical power supplied

9 Generator Efficiency P in = P o + P losses = P o + P e + P m (SPL) P g = P o + P e P in = P g + P m

10 Generator Efficiency Variable losses = Constant losses

11 Example Problem A long-shunt dynamo running at 1000 r.p.m. supplies 22 kw at a terminal voltage of 220 V. The resistances of armature, shunt field and the series field are 0.05, 110 and 0.06 Ω respectively. The overall efficiency at the above load is 88%. Find (a) Cu losses (b) iron and friction losses (c) the torque exerted by the prime mover (d) commercial, mechanical and electrical efficiencies (e) maximum efficiency. Answer: W, 1,415.5 W, N-m

12 Example Problem A shunt generator has a F.L. current of 196 A at 220 V. The stray losses are 720 W and the shunt field coil resistance is 55 Ω. If it has a F.L. efficiency of 88%, find the armature resistance. Also, find the load current corresponding to maximum efficiency.

EE6352-ELECTRICAL ENGINEERING AND INSTRUMENTATION UNIT I D.C. MACHINES PART A

EE6352-ELECTRICAL ENGINEERING AND INSTRUMENTATION 1. What is prime mover? UNIT I D.C. MACHINES PART A The basic source of mechanical power which drives the armature of the generator is called prime mover.