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 5 6 7 • The Terminal Characteristic of a Separately Excited DC Generator • The Terminal Characteristic of a Self Excited Shunt DC Generator • The Terminal Characteristic of a Self Excited Series DC Generator • 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 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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=90o. 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 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Working of Simple Loop Generator In the next quarter revolution (from 90o to 180o), 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 180o to 360o), 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 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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 short-shunt or long-shunt. 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 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Types of DC Generators Separately Excited Generators Short Shunt Generators Shunt Wound Generators Series Wound Generators Long Shunt Generators 8 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Types of DC Generators 9 Dr. Firas Obeidat Faculty of Engineering Philadelphia University The Terminal Characteristic of a Separately Excited DC Generator For Separately Excited DC Generator 𝐼𝐴 = 𝐼𝐿 IF + 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 VF 𝑉𝐹 = 𝐼𝐹 𝑅𝐹 - 𝐸𝐴 = 𝑘ϕ𝜔𝑚 The terminal voltage can be controlled by: 1. Change the speed of rotation: If 𝜔 increases, then 𝐸𝐴=𝑘ϕ𝜔𝑚 increases, so 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases as well. 2. Change the field current. If RF is decreased. then the field current increases (𝑉𝐹 = 𝐼𝐹 𝑅𝐹 ). Therefore, the flux in the machine increases. As the flux rises, 𝐸𝐴=𝑘ϕ𝜔𝑚 must rise too, so 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases. IA RF LF IL + + RA VT EA - Where IA: is the armature current IL: is the load current EA: is the internal generated voltage VT: is the terminal voltage IF: is the field current VF: is the field voltage RA: is the armature winding resistance RF: is the field winding resistance ϕ: is the flux 𝜔m: is the rotor angular speed 10 Dr. Firas Obeidat Faculty of Engineering Philadelphia University The Terminal Characteristic of a Self Excited Shunt DC Generator For Self Excited Shunt DC Generator 𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 IA + RA RF IF EA LF - 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 𝑉𝑇 = 𝐼𝐹 𝑅𝐹 IL + VT - 𝐸𝐴 = 𝑘ϕ𝜔𝑚 The terminal voltage can be controlled by: 1. Change the speed of rotation: If 𝜔 increases, then 𝐸𝐴=𝑘ϕ𝜔𝑚 increases, so 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases as well. 2. Change the field current. If RF is decreased. then the field current increases (𝑉𝐹 = 𝐼𝐹 𝑅𝐹 ). Therefore, the flux in the machine increases. As the flux rises, 𝐸𝐴=𝑘ϕ𝜔𝑚 must rise too, so 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases. 11 Dr. Firas Obeidat Faculty of Engineering Philadelphia University The Terminal Characteristic of a Self Excited Series DC Generator For Self Excited Series DC Generator 𝐼𝐴 = 𝐼𝑠 = 𝐼𝐿 IA + RA EA - 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 (𝑅𝐴 +𝑅𝑠 ) 𝐸𝐴 = 𝑘ϕ𝜔𝑚 Is Rs IL Ls + VT - At no load, there is no field current, so VT is reduced to a small level given by the residual flux in the machine. As the load increases, the field current rises, so EA rises rapidly The IA(RA+ Rs) drop goes up too, but at first the increase in EA goes up more rapidly than the IA(RA+ Rs) drop rises, so VT increases. After a while, the machine approaches saturation, and EA becomes almost constant. At that point, the resistive drop is the predominant effect, and VT starts to fall. 12 Dr. Firas Obeidat Faculty of Engineering Philadelphia University The Terminal Characteristic of Cumulatively Compound DC Generator For Long Shunt Cumulatively Compound DC Generator IA 𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 + RA EA - 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 (𝑅𝐴 +𝑅𝑠 ) 𝑉𝑇 = 𝐼𝐹 𝑅𝐹 𝐸𝐴 = 𝑘ϕ𝜔𝑚 IL Rs Ls + RF IF LF VT - For Short Shunt Cumulatively Compound DC Generator 𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 IA 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 −𝐼𝐿 𝑅𝑠 + RA RF IF EA LF - 𝐸𝐴 = 𝑘ϕ𝜔𝑚 IL Rs Ls + VT - The terminal voltage Cumulatively Compound DC Generator can be controlled by: 1. Change the speed of rotation: If 𝜔 increases, then 𝐸𝐴=𝑘ϕ𝜔𝑚 increases, so 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases as well. 2. Change the field current. If RF is decreased. then the field current increases (𝑉𝐹 = 𝐼𝐹 𝑅𝐹 ). Therefore, the flux in the machine increases. As the flux rises, 𝐸𝐴=𝑘ϕ 𝜔𝑚 must rise too, so 𝑉𝑇 = 𝐸𝐴 − 𝐼𝐴 𝑅𝐴 increases. 13 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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. IA 𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 4.6 + 450 = 454.6𝐴 𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 = 230 + 454.6 × 0.3 = 243.6V + RA RF IF EA LF - IL=450A + - VT=230V 230 𝐼𝑓 = = 4.6𝐴 50 VT=500V 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. 500 𝐼𝐹 = = 2𝐴 250 IA IL=50A + 𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 2 + 50 = 52𝐴 + RA Rs Ls RF IF Voltage drop across series winding=𝐼A𝑅s=52×0.03=1.56V EA LF Armature voltage drop=𝐼A𝑅A=52×0.05=2.6V Drop at brushes=2×1=2V 𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 + 𝑠𝑒𝑟𝑖𝑒𝑠 𝑑𝑟𝑜𝑝 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 500 + 2.6 + 1.56 + 2 = 506.16V 14 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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=𝐼L𝑅s =30 ×0.3=9V Voltage across shunt winding=220 + 9=229V Drop at brushes=2×1=2V IA + RA RF IF EA LF - Rs IL=30A + Ls - VT=220V 229 𝐼𝐹 = = 1.145𝐴 200 Armature voltage drop=𝐼𝐴 𝑅𝐴 = 31.145 × 0.05 = 1.56V 𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 + 𝑠𝑒𝑟𝑖𝑒𝑠 𝑑𝑟𝑜𝑝 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 220 + 9 + 1.56 + 2 = 232.56V 15 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Examples VT=230V 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. IA IL=150A 230 (i ) 𝐼 = + = 2.5𝐴 𝐹 L R R R s A s F 92 + IF EA 𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 2.5 + 150 = 152.5𝐴 LF Voltage drop across series winding= 𝐼A𝑅s=152.5 ×0.015=2.2875V Armature voltage drop=𝐼A𝑅A=152.5 ×0.032=4.88V 𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 + 𝐼A𝑅s = 230 + 2.2875 + 4.88 = 237.1675V (ii) Total power generated by the armature=𝐸𝐴 𝐼𝐴 =237.1675×152.5=36168.04375W (iii) Power lost in armature=𝐼𝐴 2 𝑅𝐴 =152.52 ×0.032=744.2W Power dissipated in shunt winding=𝑉𝑇 𝐼𝐹 =230×2.5=575W Power dissipated in series winding=𝐼𝐴 2 𝑅𝑠 =152.52 ×0.015=348.84375W Power delivered to the load=𝑉𝑇 𝐼𝐿 =230×150=34500W Total power generated by the armature=744.2 + 575 + 348.843 + 34500=36168.04375W 16 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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 EA=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 Dr. Firas Obeidat Faculty of Engineering Philadelphia University E.M.F. Equation of DC Generator For simplex wave-wound generator Number of parallel paths=2 Number of conductors (in series) in one path=Z/2 𝜙𝑃𝑁 𝑍 𝜙𝑃𝑍𝑁 × = 𝑣𝑜𝑙𝑡 60 2 120 For simplex lap-wound generator 𝐸. 𝑀. 𝐹. 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑/𝑝𝑎𝑡ℎ(𝐸𝐴 ) = Number of parallel paths=P Number of conductors (in series) in one path=Z/P 𝜙𝑃𝑁 𝑍 𝜙𝑍𝑁 𝐸. 𝑀. 𝐹. 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑/𝑝𝑎𝑡ℎ(𝐸𝐴 ) = × = 𝑣𝑜𝑙𝑡 60 𝑃 60 In general where 𝜙𝑍𝑁 𝑃 A=2 for simplex wave-winding 𝐸𝐴 = × 𝑣𝑜𝑙𝑡 60 𝐴 A=P for simplex lap-winding 1 2π𝑁 𝑃 𝑍𝑃 2π𝑁 Where 𝜔𝑚 = 60 𝐸𝐴 = × × 𝜙𝑍 × = 𝜙𝜔𝑚 𝑣𝑜𝑙𝑡 2π 60 𝐴 2π𝐴 For a given DC machine Z,P and A are constant 𝑍𝑃 Where 𝑘 = 2π𝐴 𝐸𝐴 = 𝑘𝜙𝜔𝑚 𝑣𝑜𝑙𝑡 18 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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? 𝜙𝑍𝑁 𝑃 7 × 10−3 × 51 × 20 × 1500 4 𝐸𝐴 = × = × = 357𝑣𝑜𝑙𝑡 60 𝐴 60 2 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 lap-connected and runs at 1200 rpm? What must be the speed at which it is to be driven produce the same emf if it is wave-wound? With lap-wound, P=A=8 𝜙𝑍𝑁 𝑃 0.05 × 500 × 1200 8 𝐸𝐴 = × = × = 500𝑣𝑜𝑙𝑡 60 𝐴 60 8 With wave-wound, P=8, A=2 𝜙𝑍𝑁 𝑃 0.05 × 500 × 𝑁 8 𝐸𝐴 = × = × = 500 60 𝐴 60 2 → 𝑁 = 300𝑟𝑝𝑚 19 Dr. Firas Obeidat Faculty of Engineering Philadelphia University E.M.F. Equation of DC Generator Example: A four pole lap-connected 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) IA With lap-wound, P=A=4 RF=250Ω IF LF - 5kW Geyser 2.5kW lighting + EA - IL + VT=250V 5000 + 2500 = 30𝐴 250 250 𝐼𝐹 = = 1𝐴 250 𝐼𝐿 = RA=0.2Ω 𝐼𝐴 = 𝐼𝐿 + 𝐼𝐹 = 30 + 1 = 31𝐴 𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 250 + 31 × 0.2 + 2 × 1 = 258.2V 𝐸𝐴 = 𝜙𝑍𝑁 𝑃 𝜙 × 120 × 1000 4 × = × = 258.2𝑣𝑜𝑙𝑡 60 𝐴 60 4 → 𝜙 = 129.1𝑚𝑊𝑏 (2) Armature current per parallel path=31/4=7.75A 20 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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 IF is unchanged? Given that the armature resistance 0.04Ω and brush drop 2V. 𝐸𝐴1 = 𝑉𝑇1 + 𝐼𝐴 𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 = 125 + 200 × 0.04 + 2 = 135V 𝑁𝐴1 = 1000𝑟𝑝𝑚 𝑁𝐴2 800 = 𝐸𝐴1 = 135 = 108𝑉 𝑁𝐴1 1000 𝑅𝑙𝑜𝑎𝑑 125 = = 0.625Ω 200 + RF VF LF + EA1 - RA - 𝐸𝐴2 = 𝑉𝑇2 + 𝐼𝐴2 𝑅𝐴 + 𝑏𝑟𝑢𝑠ℎ𝑒𝑠 𝑑𝑟𝑜𝑝 108 = 𝐼𝐴2 × 0.625 + 𝐼𝐴2 × 0.04 + 2 108 − 2 𝐼𝐴2 = = 159.4A 0.625 + 0.04 𝑉𝑇2 = 𝐼𝐴2 𝑅𝑙𝑜𝑎𝑑 = 159.4 × 0.625 = 99.6𝑉 IL=159.4A VF - RF LF + EA2 - RA + VT2=99.6V 𝑉𝑇2 = 𝐼𝐴2 𝑅𝑙𝑜𝑎𝑑 IF + + VT1=125V 𝐸𝐴2 IL=200A IF - 21 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Total Loss in a DC Generator (A) Copper Losses (i) Armature copper losses=Ia2Ra This loss is about 30-40% of full load losses. (ii) Field copper loss: In case of shunt generator, field copper losses=IF2RF In case of shunt generator, field copper losses=IL2Rs This loss is about 20-30% of full load losses. (iii) The loss due to brush contact resistance. (B) Magnetic (Iron or Core) Losses (i) Hysteresis Loss, 𝑾𝒉 ∝ 𝑩𝒎𝒂𝒙 𝟏.𝟔 𝒇 (ii) Eddy Current Loss, 𝑾𝒆 ∝ 𝑩𝒎𝒂𝒙 𝟐 𝒇𝟐 These losses are practically constant for shunt and compound wound generators, because in their case, field current is approximately constant. This loss is about 20-30% 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 10-20% of full load losses. 22 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Total Loss in a DC Generator Armature Cu Loss Total Losses Copper Losses Shunt Cu Loss Series Cu Loss Hysteresis Loss Iron Losses Eddy Current Loss Friction Loss Mechanical Losses 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 (Wc). 23 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Power Stages and Efficiency Mechanical Efficiency 𝜂𝑚 = 𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝐸𝐴 𝐼𝐴 × 100% = × 100% 𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒 Electrical Efficiency 𝜂𝑒 = 𝑊𝑎𝑡𝑡𝑠 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑢𝑖𝑡 𝑉𝐼𝐿 × 100% = × 100% 𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝐸𝐴 𝐼𝐴 Overall or Commercial Efficiency 𝜂𝑐 = 𝜂 𝑚 × 𝜂𝑒 = 𝑊𝑎𝑡𝑡𝑠 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑢𝑖𝑡 𝑉𝐼𝐿 × 100% = × 100% 𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒 24 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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) 250 𝐼𝑓 = = 5𝐴 50 𝐼𝐴 = 𝐼𝐹 + 𝐼𝐿 = 5 + 195 = 200𝐴 𝐸𝐴 = 𝑉𝑇 + 𝐼𝐴 𝑅𝐴 = 250 + 200 × 0.02 = 254V (b) 𝐴𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝐶𝑢 𝑙𝑜𝑠𝑠 = 𝐼𝐴 2 𝑅𝐴 = 2002 × 0.02 = 800𝑊 𝑆ℎ𝑢𝑛𝑡 𝐶𝑢 𝑙𝑜𝑠𝑠 = 𝐼𝑓 2 𝑅𝑓 = 52 × 50 = 1250𝑊 𝑇𝑜𝑡𝑎𝑙 𝐶𝑢 𝑙𝑜𝑠𝑠 = 800 + 1250 = 2050𝑊 (c) Stray losses=950W Total losses=950+2050=3000W 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑉𝐼𝐿 = 250 × 195 = 48750𝑊 𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑒 𝑚𝑜𝑡𝑜𝑟 = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡 25 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Power Stages and Efficiency 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡 = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑡𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 48750 + 3000 = 51750𝑊 𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑒 𝑚𝑜𝑡𝑜𝑟 = 51750𝑊 (c) 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟(𝐸𝐴 𝐼𝐴 ) = 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡 − 𝑠𝑡𝑟𝑎𝑦 𝑙𝑜𝑠𝑠 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟(𝐸𝐴 𝐼𝐴 ) = 51750 − 950 = 50800𝑊 𝐸𝐴 𝐼𝐴 50800 × 100% = × 100% = 98.2% 𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒 51750 𝜂𝑚 = 𝑉𝐼𝐿 48750 𝜂𝑒 = = × 100% = 95.9% 𝐸𝐴 𝐼𝐴 50800 𝜂𝑐 = 𝑉𝐼𝐿 48750 × 100% = × 100% = 94.2% 𝑂𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑒𝑛𝑔𝑖𝑛𝑒 51750 26 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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 = 𝑉𝐼𝐿 = 220 × 196 = 43120𝑊 𝑉𝐼𝐿 𝜂𝑒 = × 100% = 88% → 𝐸𝐴 𝐼𝐴 = 43120 ÷ 0.88 = 49000𝑊 𝐸𝐴 𝐼𝐴 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 49000 − 43120 = 5880𝑊 𝐼𝑓 = 220 ÷ 55 = 4𝐴 𝑆ℎ𝑢𝑛𝑡 𝐶𝑢 𝑙𝑜𝑠𝑠 = 𝐼𝑓 𝑉 = 4 × 220 = 880𝑊 Constant losses=Shunt Cu losses+stray losses=880+720=1600W Total losses=Armature losses + Constant losses=𝐼𝐴 2 𝑅𝐴 +1600=5880 𝐼𝐴 2 𝑅𝐴 = 5880 − 1600 = 4280𝑊 𝐼𝐴 = 𝐼𝐿 + 𝐼𝑓 = 195 + 4 = 199𝐴 𝑅𝐴 = 4280 ÷ 1992 = 0.108Ω 27 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Voltage Regulation The voltage regulation (VR) is defined as the difference between the no-load terminal voltage (VNL) to full load terminal voltage (VFL) and is expressed as a percentage of full load terminal voltage. It is therefore can be expressed as, 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑉𝑅 = 𝑉𝑁𝐿 − 𝑉𝐹𝐿 𝐸𝐴 − 𝑉𝐹𝐿 × 100% = × 100% 𝑉𝐹𝐿 𝑉𝐹𝐿 Example: A 4-pole 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 = 𝐼𝐿 𝑅 = 20 × 10 = 200𝑉 𝐼𝑓 = 200 ÷ 50 = 4𝐴 𝐼𝐴 = 𝐼𝐿 + 𝐼𝑓 = 20 + 4 = 24𝐴 𝐼𝐴 𝑅𝐴 = 24 × 0.5 = 12𝑉 𝐸𝐴 = 𝐼𝐴 𝑅𝐴 + 𝑉 + 𝑏𝑟𝑢𝑠ℎ 𝑑𝑟𝑜𝑝 = 12 + 200 + 2 = 214𝑉 𝑉𝐼𝐿 200 × 20 × 100% = × 100% = 77.9% 𝐸𝐴 𝐼𝐴 214 × 24 𝐸𝐴 − 𝑉𝐹𝐿 214 − 200 𝑉𝑅 = × 100% = × 100% = 7% 𝑉𝐹𝐿 200 𝜂𝑒 = Dr. Firas Obeidat Faculty of Engineering Philadelphia University 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 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 30