Electric Machines I DC Machines DC Generators Dr. Firas Obeidat 1
Table of contents 1 Construction of Simple Loop Generator 2 Working of Simple Loop Generator 3 Types of DC Generators 4 The Terminal Characteristic of a Separately Excited DC Generator 5 The Terminal Characteristic of a Self Excited Shunt DC Generator 6 The Terminal Characteristic of a Self Excited Series DC Generator 7 The Terminal Characteristic of Cumulatively Compound DC Generator 8 E.M.F. Equation of DC Generator 9 Total Loss in a DC Generator 10 Power Stages and Efficiency 11 Voltage Regulation 12 Uses of DC Generators 2
Construction of Simple Loop Generator A single turn rectangular copper ABCD rotating about its own axis in a magnetic field provided by either permanent magnet or electromagnet. The two ends of the coil are joined to slip ring a and b which are insulated from each other and from the central shaft. Two collecting brushes press against the slip rings; their function is to collect the current induced in the coil and to convey it to external load resistance. The rotating coil is called the armature. 3
Working of Simple Loop Generator Imagine the coil to be rotating in clockwise direction. As the coil assumes successive positions in the field, the flux linked with it changes. An emf is induced in it which is proportional to the rate of change of flux linkages (e=ndϕ/dt). When the plane of coil is in position 1, then flux linked with the coil is maximum but rate of change of flux linkage is minimum. Hence, there is no induced emf in the coil. As the coil continues rotating further, the rate of change of flux linkages (and hence induced emf in it) increases, till position 3 is reached where θ=90 o. The coil plane is horizontal (parallel to the lines of flux). The flux linked with the coil is minimum but rate of change of flux linkage is maximum. Hence, maximum emf is induced in the coil when in this position. In the second half revolution, the direction of the current flow is DCMLBA. Which is just the reverse of the previous direction of flow. 4
Working of Simple Loop Generator In the next quarter revolution (from 90 o to 180 o ), the flux linked with the coil gradually increases but the rate of change of flux linkages decreases. Hence the induced emf decreases gradually till in position 5 of the coil, it reduces to zero value. In the first half revolution of the coil, no emf is induced in it when in position 1, maximum when in position 3 and no emf when in position 5. In this half revolution, the direction of the current flow is ABMLCD. The current through the load resistor R flows from M to L during the first half revolution of the coil. In the next half revolution (from 180 o to 360 o ), the variations in the magnitude of emf are similar to those in the first half revolution. Its value is maximum when the coil is in position 7 and minimum when it in position 1. 5
Working of Simple Loop Generator The current which is obtained from such a simple generator reverses its direction after every half revolution, this current is known as alternating current. To make the flow of current unidirectional in the external circuit, the slip rings are replaced by split rings. In the first half revolution segment a is connected to brush 1 and segment b is connected to brush 2, while in the second half revolution segment b is connected to brush 1 and segment a is connected to brush 2. In this case the current will flow in the resistor from M to L in the two halves of revolution. The resulting current is unidirectional but not continuous like pure direct current. 6
Types of DC Generators Generators are usually classified according to the way in which their fields are excited A. Separately Excited Generators: are those whose field magnets are energized from an independent external source of DC current. B. Self Excited Generators: are those whose field magnets are energized by current produced by the generators themselves. There are three types of self excited generators named according to the manner in which their field coils are connected to the armature. i. Shunt Wound: the field windings are connected across or in parallel with the armature conductors and have the full voltage of the generator applied across them. ii. Series Wound: the field windings are joined in series with the armature conductors iii.compound Wound: it is a combination of a few series and a few shunt windings and can be either shortshunt or longshunt. In compound generator, the shunt field is stronger than the series field. When series field aids the shunt field, generator is said to be commutativelycompound. In series field oppose the shunt field, the generator is said to be differentially compounded. 7
Types of DC Generators Separately Excited Generators Shunt Wound Generators Series Wound Generators Short Shunt Generators Long Shunt Generators 8
Types of DC Generators 9
The Terminal Characteristic of a Separately Excited DC Generator For Separately Excited DC Generator I A = I L V T = E A I A R A V F = I F R F E A = kϕω m + V F I F R F L F I A + R A E A I L + V T The terminal voltage can be controlled by: 1. Change the speed of rotation: If ω increases, then E A =kϕω m increases, so V T = E A I A R A increases as well. 2. Change the field current. If R F is decreased. then the field current increases (V F = I F R F ). Therefore, the flux in the machine increases. As the flux rises, E A =kϕω m must rise too, so V T = E A I A R A increases. Where I A : is the armature current I L : is the load current E A : is the internal generated voltage V T : is the terminal voltage I F : is the field current V F : is the field voltage R A : is the armature winding resistance R F : is the field winding resistance ϕ: is the flux ω m : is the rotor angular speed 10
The Terminal Characteristic of a Self Excited Shunt DC Generator For Self Excited Shunt DC Generator I A = I F + I L V T = E A I A R A V T = I F R F E A = kϕω m I A + R A E A R F L F I F I L + V T The terminal voltage can be controlled by: 1. Change the speed of rotation: If ω increases, then E A =kϕω m increases, so V T = E A I A R A increases as well. 2. Change the field current. If R F is decreased. then the field current increases (V F = I F R F ). Therefore, the flux in the machine increases. As the flux rises, E A =kϕω m must rise too, so V T = E A I A R A increases. 11
The Terminal Characteristic of a Self Excited Series DC Generator For Self Excited Series DC Generator I A = I s = I L V T = E A I A (R A +R s ) E A = kϕω m I A + R A E A I s R s L s I L + V T At no load, there is no field current, so V T is reduced to a small level given by the residual flux in the machine. As the load increases, the field current rises, so E A rises rapidly The I A (R A + R s ) drop goes up too, but at first the increase in E A goes up more rapidly than the I A (R A + R s ) drop rises, so V T increases. After a while, the machine approaches saturation, and E A becomes almost constant. At that point, the resistive drop is the predominant effect, and V T starts to fall. 12
The Terminal Characteristic of Cumulatively Compound DC Generator For Long Shunt Cumulatively Compound DC Generator I A = I F + I L V T = E A I A (R A +R s ) V T = I F R F E A = kϕω m I A + R A E A For Short Shunt Cumulatively Compound DC Generator I A = I F + I L V T = E A I A R A I L R s E A = kϕω m + R A E A R s L s R F The terminal voltage Cumulatively Compound DC Generator can be controlled by: 1. Change the speed of rotation: If ω increases, then E A =kϕω m increases, so V T = E A I A R A increases as well. 2. Change the field current. If R F is decreased. then the field current increases (V F = I F R F ). Therefore, the flux in the machine increases. As the flux rises, E A =kϕ ω m must rise too, so V T = E A I A R A increases. I A R F L F I F R s L F I F L s I L I L + V T + V T 13
Examples Example: A shunt DC generator delivers 450A at 230V and the resistance of the shunt field and armature are 50Ω and 0.3 Ω respectively. Calculate emf. I f = 230 50 = 4.6A I A = I F + I L = 4.6 + 450 = 454.6A E A = V T + I A R A = 230 + 454.6 0.3 = 243.6V I A I L =450A + R A R + F E A I F L F VT=230V Example: A long shunt compound DC generator delivers a load current of 50A at 500V and has armature, series field and shunt field resistances of 0.05Ω, 0.03Ω and 250Ω respectively. Calculate the generated voltage and the armature current. Allow 1V per brush for contact drop. I F = 500 250 = 2A I A = I F + I L = 2 + 50 = 52A Voltage drop across series winding=i A R s =52 0.03=1.56V Armature voltage drop=i A R A =52 0.05=2.6V Drop at brushes=2 1=2V E A = V T + I A R A + series drop + brushes drop = 500 + 2.6 + 1.56 + 2 = 506.16V I A + R A E A R s L s R F L F I F I L =50A + 14 VT=500V
Examples Example: A short shunt compound DC generator delivers a load current of 30A at 220V and has armature, series field and shunt field resistances of 0.05Ω, 0.3Ω and 200Ω respectively. Calculate the induced emf and the armature current. Allow 1V per brush for contact drop. Voltage drop across series winding=i L R s =30 0.3=9V Voltage across shunt winding=220 + 9=229V I F = 229 200 = 1.145A Armature voltage drop=i A R A = 31.145 0.05 = 1.56V Drop at brushes=2 1=2V I A + R A E A R F L F I F R s L s I L =30A + VT=220V E A = V T + I A R A + series drop + brushes drop = 220 + 9 + 1.56 + 2 = 232.56V 15
Examples (i) (ii) Example: A long shunt compound DC generator delivers a load current of 150A at 230V and has armature, series field and shunt field resistances of 0.032Ω, 0.015Ω and 92Ω respectively. Calculate (i) induced emf (ii) total power generated and (iii) distribution of this power. I F = 230 92 = 2.5A I A = I F + I L = 2.5 + 150 = 152.5A Voltage drop across series winding= I A R s =152.5 0.015=2.2875V I A + R A E A Armature voltage drop=i A R A =152.5 0.032=4.88V E A = V T + I A R A + I A R s = 230 + 2.2875 + 4.88 = 237.1675V R s L s R F Total power generated by the armature=e A I A =237.1675 152.5=36168.04375W (iii) Power lost in armature=i A 2 R A =152.5 2 0.032=744.2W Power dissipated in shunt winding=v T I F =230 2.5=575W Power dissipated in series winding=i A 2 R s =152.5 2 0.015=348.84375W Power delivered to the load=v T I L =230 150=34500W L F I F I L =150A + Total power generated by the armature=744.2 + 575 + 348.843 + 34500=36168.04375W VT=230V 16
E.M.F. Equation of DC Generator Let ϕ: flux/pole in weber. Z: total number of armature conductors Z=number of slots number of conductors/slot A: number of parallel paths in armature N: armature rotation in rpm E: emf induced in any parallel path in armature Generated emf E A =emf generated in any one of the parallel paths Average emf generated/conductor=dϕ/dt volt Flux cut/conductor in one revolution dϕ=ϕp Wb Number of revolutions /second=n/60 Time for one revolution dt=60/n second E.M.F. generated/conductor= dϕ/dt= ϕpn/60 volt 17
E.M.F. Equation of DC Generator For simplex wavewound generator Number of parallel paths=2 Number of conductors (in series) in one path=z/2 E. M. F. generated/path(e A ) = φpn 60 Z 2 = φpzn 120 volt For simplex lapwound generator Number of parallel paths=p Number of conductors (in series) in one path=z/p E. M. F. generated/path(e A ) = φpn 60 Z P = φzn 60 volt In general E A = φzn 60 P where A volt A=2 for simplex wavewinding A=P for simplex lapwinding E A = 1 2π 2πN 60 φz P A = ZP 2πA φω m volt For a given DC machine Z,P and A are constant E A = kφω m volt Where ω m = 2πN 60 Where k = ZP 2πA 18
E.M.F. Equation of DC Generator Example: A four pole generator, having wave wound armature winding has 51 slots, each slot containing 20 conductor. What will be the voltage generated in the machine when driven at 1500 rpm assuming the flux per pole to be 7mWb? E A = φzn 60 P A = 7 10 3 51 20 1500 4 60 2 = 357volt Example: An 8 pole Dc generator has 500 armature conductors, and a useful flux of 0.05Wb per pole. what will be the emf generated if it is lapconnected and runs at 1200 rpm? What must be the speed at which it is to be driven produce the same emf if it is wavewound? With lapwound, P=A=8 E A = φzn 60 P A = 0.05 500 1200 60 8 8 = 500volt With wavewound, P=8, A=2 E A = φzn 60 P A = 0.05 500 N 60 8 = 500 N = 300rpm 2 19
E.M.F. Equation of DC Generator Example: A four pole lapconnected armature of a DC shunt generator is required to supply the loads connected in parallel: (a) 5kW Geyser at 250 V and (b) 2.5kW lighting load also at 250V. The generator has an armature resistance 0.2Ω and a field resistance of 250Ω. The armature has 120 conductors in the slots and runs at 1000 rpm. Allowing 1V per brush for contact drops, find (1) Flux per pole, (2) armature current per parallel path (1) With lapwound, P=A=4 5000 + 2500 I L = 250 I F = 250 250 = 1A = 30A I A = I L + I F = 30 + 1 = 31A E A = V T + I A R A + brushes drop = 250 + 31 0.2 + 2 1 = 258.2V I A + E A R A =0.2Ω R F =250Ω L F I F I L + VT=250V 5kW Geyser 2.5kW lighting (2) E A = φzn 60 P A = φ 120 1000 60 4 4 Armature current per parallel path=31/4=7.75a = 258.2volt φ = 129.1mWb 20
E.M.F. Equation of DC Generator Example: A separately excited DC generator, when running at 1000 rpm supplied 200A at 125V. What will be the load current when the speed drops to 800 rpm if I F is unchanged? Given that the armature resistance 0.04Ω and brush drop 2V. E A1 = V T1 + I A R A + brushes drop = 125 + 200 0.04 + 2 = 135V N A1 = 1000rpm E A2 = E A1 N A2 N A1 = 135 800 1000 = 108V R load = 125 200 = 0.625Ω + V F I F R F L F + R A E A1 I L =200A + VT1=125V E A2 = V T2 + I A2 R A + brushes drop V T2 = I A2 R load 108 = I A2 0.625 + I A2 0.04 + 2 I A2 = 108 2 0.625 + 0.04 = 159.4A + V F I F R F L F + R A E A2 I L =159.4A + VT2=99.6V V T2 = I A2 R load = 159.4 0.625 = 99.6V 21
Total Loss in a DC Generator (A) Copper Losses (i) Armature copper losses=i a2 R a This loss is about 3040% of full load losses. (ii) Field copper loss: In case of shunt generator, field copper losses=i F2 R F In case of shunt generator, field copper losses=i L2 R s This loss is about 2030% of full load losses. (iii) The loss due to brush contact resistance. (B) Magnetic (Iron or Core) Losses (i) Hysteresis Loss, W h B max 1.6 f (ii) Eddy Current Loss, W e B max 2 f 2 These losses are practically constant for shunt and compound wound generators, because in their case, field current is approximately constant. This loss is about 2030% of full load losses. (C) Mechanical Losses (i) Friction Loss at bearing and commutator. (ii) Air Friction or Windage Loss of rotating armature This loss is about 1020% of full load losses. 22
Total Losses Total Loss in a DC Generator Armature Cu Loss Copper Losses Shunt Cu Loss Series Cu Loss Iron Losses Mechanical Losses Hysteresis Loss Eddy Current Loss Friction Loss Air Friction or Windage Loss Stray Losses Iron and mechanical losses are collectively known as Stray (Rotational) losses. Constant or Standing Losses Field Cu losses is constant for shunt and compound generators. Stray losses and shunt Cu loss are constant in their case. These losses are together known as Constant or Standing Losses (W c ). 23
Power Stages and Efficiency Mechanical Efficiency η m = Total watts generated in armature Mechanical power supplied 100% = E A I A Output of driving engine 100% Electrical Efficiency η e = Watts available in load ciruit Total watts generated in armature 100% = VI L E A I A 100% Overall or Commercial Efficiency η c = η m η e = Watts available in load ciruit Mechanical power supplied 100% = VI L Output of driving engine 100% 24
Power Stages and Efficiency Example: A shunt generator delivers 195A at terminal voltage of 250V. The armature resistance and shunt field resistance are 0.02Ω and 50Ω respectively. The iron and friction losses equal 950W. Find (a) emf generated (b) Cu losses (c) output of the prime motor (d) commercial, mechanical and electrical efficiencies. (a) I f = 250 50 = 5A I A = I F + I L = 5 + 195 = 200A (b) E A = V T + I A R A = 250 + 200 0.02 = 254V Armature Cu loss = I A 2 R A = 200 2 0.02 = 800W Shunt Cu loss = I f 2 R f = 5 2 50 = 1250W (c) Total Cu loss = 800 + 1250 = 2050W Stray losses=950w Total losses=950+2050=3000w Generator output = VI L = 250 195 = 48750W Output of the prime motor = Generator input 25
Power Stages and Efficiency Generator input = Generator output + total losses = 48750 + 3000 = 51750W Output of the prime motor = 51750W (c) Generated electrical power(e A I A ) = Generator input stray loss Generated electrical power(e A I A ) = 51750 950 = 50800W η m = E A I A 50800 100% = 100% = 98.2% Output of driving engine 51750 η e = VI L = 48750 100% = 95.9% E A I A 50800 η c = VI L 48750 100% = 100% = 94.2% Output of driving engine 51750 26
Power Stages and Efficiency Example: A shunt generator has a full load current of 196 A at 220V. The stray lassos are 720W and the shunt field coil resistance is 55Ω. If it has full load efficiency of 88%, find the armature resistance. Generator output = VI L = 220 196 = 43120W η e = VI L E A I A 100% = 88% Total losses = 49000 43120 = 5880W I f = 220 55 = 4A E A I A = 43120 0.88 = 49000W Shunt Cu loss = I f V = 4 220 = 880W Constant losses=shunt Cu losses+stray losses=880+720=1600w Total losses=armature losses + Constant losses=i A 2 R A +1600=5880 I A 2 R A = 5880 1600 = 4280W I A = I L + I f = 195 + 4 = 199A R A = 4280 199 2 = 0.108Ω 27
Voltage Regulation The voltage regulation (VR) is defined as the difference between the noload terminal voltage (V NL ) to full load terminal voltage (V FL ) and is expressed as a percentage of full load terminal voltage. It is therefore can be expressed as, Voltage Regulation VR = V NL V FL V FL 100% = E A V FL V FL 100% Example: A 4pole shunt DC generator is delivering 20A to a load of 10Ω. If the armature resistance is 0.5 Ω and the shunt field resistance is 50 Ω, calculate the induced emf and the efficiency of the machine. Allow a drop of 1V per brush. Terminal Voltage = I L R = 20 10 = 200V I f = 200 50 = 4A I A = I L + I f = 20 + 4 = 24A I A R A = 24 0.5 = 12V E A = I A R A + V + brush drop = 12 + 200 + 2 = 214V η e = VI L E A I A 100% = VR = E A V FL V FL 100% = 200 20 100% = 77.9% 214 24 214 200 200 100% = 7% 28
Uses of DC Generators Shunt Generators Shunt generators with field regulators are used for ordinary lighting and power supply purposes. They are also used for charging batteries because their terminal voltages are almost constant. Series Generators Series generators are used as boosters in a certain types of distribution systems particularly in railway service. Compound Generators The cumulatively compound generator is the most used DC generator because its external characteristics can be adjusted for compensating the voltage drop in the line resistance. Cumulatively compound generators are used for motor driving which require DC supply at constant voltage, for lamp loads and for heavy power service such as electric railways. The differential compound DC generator has an external characteristic similar to that of shunt generator but with large demagnetization armature reaction. Differential compound DC generators re widely used in arc welding where larger voltage drop is desirable with increase in current. 29
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