Electric Machines I DC Machines - DC Motors Dr. Firas Obeidat 1 Table of Contents 1 • DC Motor Principle 2 • Types of DC Motors 3 • E.M.F. Equation of DC Motor 4 • Armature Torque of DC Motor 5 • Speed Regulation of DC Motor 6 • Total Losses in DC Motor 7 • Power Stages and Efficiency 8 2 Dr. Firas Obeidat Faculty of Engineering Philadelphia University DC Motor Principle ο Constructionally, there is no basic difference between DC generator and DC motor. The DC machine can be used as generator or as motor. ο When field magnets are excited in multipolar DC motor, and its armature conductors are supplied with current from the supply, they experience a force tending to rotate the armature. Armature conductors under N-pole are assumed to carry current downwards (crosses) and those under S-poles to carry current upwards (dots). ο By applying Fleming’s left hand rule, each conductor experiences a force F which tends to rotate the armature in anticlockwise direction. These forces collectively produce a driving torque which sets the armature rotating. 3 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Types of DC Motors 1- Separately Excited DC Motor IF + πΌπ΄ = πΌπΏ RF VF ππ = πΈπ΄ + πΌπ΄ π π΄ ππΉ = πΌπΉ π πΉ IA 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 Ο: is the flux πm: is the rotor angular speed πΈπ΄ = πΟππ 2- Shunt DC Motor πΌπΏ = πΌπ΄ + πΌπ ππ = πΈπ΄ + πΌπ΄ π π΄ IA ππΉ = πΌπΉ π πΉ + RA RF IF EA LF - πΈπ΄ = πΟππ IL + VT 4 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Types of DC Motors Speed control of separately excited and shunt DC motors Adjusting the field resistance RF • Increasing RF causes π°π = π½π» /πΉπ to decrease. • Decreasing IF decreases Ο. • Decreasing Ο, lowers πΈπ΄=π Οππ . • Decreasing EA increases π°π¨ = (π½π» − π¬π¨ )/πΉπ¨. • Increasing IA increases Tind=πΟIA , with the change in IA dominant over the change in flux. • Increasing Tind makes Tind>Tload and the speed π m increases. • Increasing πm increases πΈπ΄=πΟππ. • Increasing EA decreases Adjusting the terminal voltage applied to the armature • An increase in VT increases π°π¨ = (π½π» − π¬π¨ )/πΉπ¨ . • Increasing IA increases Tind=πΟIA • Increasing Tind makes Tind>Tload and the speed π m increases. • Increasing πm increases πΈπ΄=πΟππ. • Increasing EA decreases π°π¨ = (π½π» − π¬π¨ )/πΉπ¨ . • Decreasing IA decreases Tind until Tind=Tload at a higher speed ππ IA. • Decreasing IA decreases Tind until Tind=Tload at a higher speed ππ. Inserting a resistor in series with the armature circuit • resistor is inserted in series with the armature circuit, the effect is to drastically increase the slope of the motor's torquespeed characteristic, making it operate more slowly if loaded. • The insertion of a resistor is a very wasteful method of speed control, since the losses in the inserted resistor are very large. For this reason, it is rarely used. It will be found only in applications in which the motor spends almost all its time operating at full speed or in applications too inexpensive to justify a better form of speed control. 5 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Types of DC Motors 3- The Permanent Magnet DC Motor A permanent-magnet DC (PMDC) motor is a DC motor whose poles are made of permanent magnets. Permanent-magnet dc motors offer a number of benefits compared with shunt dc motors in some applications. Since these motors do not require an external field circuit, they do not have the field circuit copper losses associated with shunt dc motors. Because no field windings are required, they can be smaller than corresponding shunt DC motors. they are especially common in smaller fractional- and sub fractional-horsepower sizes. PMDC motors are generally less expensive, smaller in size, simpler, and higher efficiency than corresponding DC motors with separate electromagnetic fields. This makes them a good choice in many DC motor applications. The armatures of PMDC motors are essentially identical to the armatures of motors with separate field circuits, so their costs are similar too. However, the elimination of separate electromagnets on the stator reduces the size of the stator, the cost of the stator, and the losses in the field circuits. PMDC motors also have disadvantages. Permanent magnets cannot produce as high a flux density as an externally supplied shunt field, so a PMDC motor will have a lower induced torque Tind per ampere of armature current IA than a shunt motor of the same size and construction. In addition, PMDC motors run the risk of demagnetization. A permanent-magnet DC motor is basically the same machine as a shunt dc motor, except that the flux of a PMDC motor is fixed. Therefore, it is not possible to control the speed of a PMDC motor by varying the field current or flux. The only methods of speed control available for a PMDC motor are armature voltage control and armature resistance control. The techniques to analyze a PMDC motor are basically the same as the techniques to analyze a shunt dc motor with the field current held constant. 6 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Types of DC Motors 4- Series DC Motor IA πΌπ΄ = πΌπ = πΌπΏ + RA EA - ππ = πΈπ΄ + πΌπ΄ (π π΄ + π π ) Is Rs IL Ls + VT - πΈπ΄ = πΟππ Speed control of series DC motors 1) Change the terminal voltage of the motor. 2) Insertion of a series resistor into the motor circuit, but this technique is very wasteful of power and is used only for intermittent periods during the start-up of some motors. 7 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Types of DC Motors 5- Compound DC Motor For Long Shunt Cumulatively Compound DC Motor πΌπΏ = πΌπ΄ + πΌπΉ IA ππ = πΈπ΄ + πΌπ΄ (π π΄ +π π ) + EA - ππ = πΌπΉ π πΉ πΈπ΄ = πΟππ IL RA Rs Ls + RF IF LF VT - For Short Shunt Cumulatively Compound DC Motor πΌπΏ = πΌπ΄ + πΌπΉ IA + RA RF IF EA LF - ππ = πΈπ΄ + πΌπ΄ π π΄ +πΌπΏ π π πΈπ΄ = πΟππ IL Rs Ls + VT - Speed control of Cumulatively compound DC motors 1) Change the field resistance RF. 2) Change the armature voltage VA. 3) Change the armature resistance RA. 8 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Types of DC Motors Example: A 220V DC shunt machine has an armature resistance of 0.5Ω. If the full load armature current is 20A, find the induced emf when the machine acts as (i) generator (ii) motor. IA=20A (i) As generator πΈπ΄ = 220 + 20 × 0.5 = 230π + RA RF IF EA LF - - IA=20A + EA - ππ = πΈπ΄ + πΌπ΄ π π΄ πΈπ΄ = ππ − πΌπ΄ π π΄ πΈπ΄ = 220 − 20 × 0.5 = 210π RA RF IF LF IL + - VT=220V (ii) As motor VT=220V πΈπ΄ = ππ + πΌπ΄ π π΄ IL + 9 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Types of DC Motors Example: A 440V shunt DC motor has an armature resistance of 0.8Ω and field resistance of 200Ω. find the back emf when giving an output of 7.46kW at 85% efficiency. πππ’π‘ IA IL π= × 100% πππ πππ = 7.46π = 8.7765ππ 0.85 + VT + RA EA - 7.46π 85% = × 100% πππ RF IF LF - πππ = 8.7765π = ππ πΌπΏ = 440 × πΌπΏ 8.7765π = 19.9π΄ 440 440 πΌπΉ = = 2.2π΄ 200 πΌπΏ = πΌπ΄ = πΌπΏ − πΌπΉ = 19.9 − 2.2 = 17.7π΄ πΈπ΄ = ππ − πΌπ΄ × π π΄ = 440 − 17.7 × 0.8 = 425.84π 10 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Types of DC Motors Example: A 25kW, 250V DC shunt machine has an armature and field resistances of 0.06Ω and 100Ω respectively. Determine the total armature power developed when working (i) as generator delivering 25kW output and IA IL (ii) as motor taking 25kW input. (i) As generator → πΌπΏ = πππ’π‘ 25000 = = 100π΄ ππ 250 + VT πππ’π‘ = πΌπΏ ππ + RA RF IF EA LF - - 250 πΌπΉ = = 2.5π΄ πΌπ΄ = πΌπΏ + πΌπΉ = 100 + 2.5 = 102.5π΄ 100 πΈπ΄ = ππ + πΌπ΄ π π΄ = 250 + 0.06 × 102.5 = 256.15 ππ΄ = πΈπ΄ πΌπ΄ = 256.15 × 102.5 = 26255.375π (ii) As motor πππ = πΌπΏ ππ 250 πΌπΉ = = 2.5π΄ πΌπ΄ = πΌπΏ − πΌπΉ = 100 + 2.5 = 97.5π΄ 100 πΈπ΄ = ππ − πΌπ΄ π π΄ = 250 − 97.5 × 0.06 = 244.15V IA + EA - IL RA RF IF LF + VT πππ 25000 → πΌπΏ = = = 100π΄ ππ 250 - ππ΄ = πΈπ΄ πΌπ΄ = 244.15 × 97.5 = 23804.625π 11 Dr. Firas Obeidat Faculty of Engineering Philadelphia University E.M.F. Equation of DC Motor Let Ο: 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=dΟ/dt volt Flux cut/conductor in one revolution dΟ=ΟP Wb Number of revolutions /second=N/60 Time for one revolution dt=60/N second E.M.F. generated/conductor= dΟ/dt= ΟPN/60 volt 12 Dr. Firas Obeidat Faculty of Engineering Philadelphia University E.M.F. Equation of DC Motor For simplex wave-wound motor Number of parallel paths=2 Number of conductors (in series) in one path=Z/2 πΈ. π. πΉ. πππππππ‘ππ/πππ‘β(πΈπ΄ ) = πππ π ππππ × = π£πππ‘ 60 2 120 For simplex lap-wound motor Number of parallel paths=P Number of conductors (in series) in one path=Z/P πππ π πππ πΈ. π. πΉ. πππππππ‘ππ/πππ‘β(πΈπ΄ ) = × = π£πππ‘ 60 π 60 In general where πππ π A=2 for simplex wave-winding πΈπ΄ = × π£πππ‘ 60 π΄ A=P for simplex lap-winding 1 2ππ π ππ 2ππ Where ππ = 60 πΈπ΄ = × × ππ × = πππ π£πππ‘ 2π 60 π΄ 2ππ΄ For a given DC machine Z,P and A are constant ππ Where π = 2ππ΄ πΈπ΄ = ππππ π£πππ‘ 13 Dr. Firas Obeidat Faculty of Engineering Philadelphia University E.M.F. Equation of DC Motor Example: A 4-pole, 32 conductor, lap-wound DC shunt generator with terminal voltage of 200V delivering 12A to the load has RA=2 Ω and field circuit resistance of 200Ω. It is driven at 1000rpm. Calculate the flux per pole in the machine. If the machine has to be run as a motor with the same terminal voltage and drawing 5A from the mains, maintaining the same magnetic field, find the speed of the machine. IA=13A (i) As generator πππ π A=P × 60 π΄ 226 × 60 π= = 0.42375π€π 1000 × 32 πΈπ΄ = 226 = for lap-winding πΌπ΄ = πΌπΏ − πΌπΉ = 5 + 1 = 4π΄ πππ π πΈπ΄ = ππ − πΌπ΄ π π΄ = 200 + 4 × 2 = 192 = × π 60 π΄ 192 × 60 π= = 850πππ 0.42375 × 32 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Action as Generator IA=4A + EA - IL=5A + RA RF IF=1A LF - Action as Motor VT=200V (ii) As motor + EA - VT=200V 200 πΌπΉ = = 1π΄ πΌπ΄ = πΌπΏ + πΌπΉ = 12 + 1 = 13π΄ 200 πΈπ΄ = ππ + πΌπ΄ π π΄ = 200 + 13 × 2 = 226π IL=12A + RA RF IF=1A LF - 14 Armature and Shaft Torque of DC Motor Armature Torque of DC Motor Let Ta be the torque developed by the armature of a motor, then the power developed πππ€ππ πππ£ππππππ = ππ × 2ππ/60 π€ππ‘π‘ The electrical power converted into mechanical power in the armature=EAIA Equating the above two equations yields ππ × 2ππ = πΈπ΄ πΌπ΄ ππ = πΈπ΄ πΌπ΄ 60 πΈπ΄ πΌπ΄ πΈπ΄ πΌπ΄ = = 9.55 N. m 2ππ/60 2π π π πΈπ΄ = πππ π × π£πππ‘ 60 π΄ Or 9.55 π π ππ = πππΌπ΄ × = 0.159πππΌπ΄ × N. m 60 π΄ π΄ Shaft Torque of DC Motor ππ β πππ‘ππ ππ’π‘ππ’π‘ = 9.55 N. m π (ππ β − ππ ) is known as lost torque and is due iron and friction losses of the motor 15 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Armature Torque of DC Motor Example: A DC motor takes an armature current of 110A at 480V. The armature circuit resistance is 0.2Ω. The machine has 6 poles and the armature is lap-connected with 864 conductors. The flux per pole is 0.05wb. Calculate the speed and the gross torque developed by the armature. πΈπ΄ = ππ − πΌπ΄ π π΄ = 480 − 110 × 0.2 = 458V πππ π 0.05 × 864 × π × = = 458 60 π΄ 60 πΈπ΄ πΌπ΄ 458 × 110 ππ = 9.55 = 9.55 ≈ 756N. m π 636 πΈπ΄ = π = 636 πππ ππ = 0.159 × π × π × πΌπ΄ = 0.159 × 0.05 × 864 × 110 ≈ 756N. m Or Example: Determine armature torque and motor speed of 220V, 4-pole series motor with 800 conductors wave connected supplying a load by taking 45A from the mains. The flux per pole is 25mwb and its armature circuit resistance is 0.6Ω. π 4 ππ = 0.159πππΌπ΄ × = 0.159 × 0.025 × 800 × 45 × = 286.2N. m π΄ 2 πΈπ΄ = ππ − πΌπ΄ π π΄ = 220 − 45 × 0.6 = 193V πΈπ΄ = πππ π 0.025 × 800 × π 4 × = × = 193 60 π΄ 60 2 Dr. Firas Obeidat Faculty of Engineering Philadelphia University π = 579 πππ 16 Armature Torque of DC Motor Example: A 220V DC shunt motor runs at 500 rpm when the armature current is 50A. Calculate the speed if the torque id doubled. Given that RA=0.2Ω. In shunt DC motor the flux is constant. ππ1 π = 0.159πππΌπ΄1 × π΄ ππ1 πΌπ΄1 = 2ππ1 πΌπ΄2 → ππ2 = 2ππ1 = 0.159πππΌπ΄2 × 1 50 = 2 πΌπ΄2 π π΄ → πΌπ΄2 = 2 × 50 = 100π΄ πΈπ΄1 = ππ − πΌπ΄1 π π΄ = 220 − 50 × 0.2 = 210V πΈπ΄2 = ππ − πΌπ΄2 π π΄ = 220 − 100 × 0.2 = 200V πΈπ΄1 = πππ1 π × 60 π΄ πΈπ΄1 π1 = → πΈπ΄1 π2 πΈπ΄2 = 210 500 = 200 π2 πππ2 π × 60 π΄ → π2 = 500 × 200 = 476πππ 210 17 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Armature Torque of DC Motor Example: A 500V, 37.3kW, 1000rpm DC shunt motor has on full load an efficiency of 90%. Determine (i) full load line current (ii) full load armature torque (neglect iron and friction losses). (i) πππ‘ππ ππππ’π‘ = πππ’π‘ 37300 = = 41444π π 0.9 41444 πΉπ’ππ ππππ ππππ ππ’πππππ‘ = = 82.9π΄ 500 (ii) ππ = 9.55 ππ = 9.55 πΈπ΄ πΌπ΄ ππ’π‘ππ’π‘ ≈ 9.55 = N. m π π (neglect iron and friction losses) 37300 = 356N. m 1000 Example: Determine the torque established by the armature of a four-poles DC motor having 774 conductors, two paths in parallel, 24 milli-webers of pole-flux and the armature current is 50A. ππ = 0.159 × π × π × πΌπ΄ = 0.159 × 0.024 × 774 × 50 × 4/2 = 295.36N. m 18 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Speed Regulation of DC Motor The speed regulation refers to the change in speed of a motor with change in applied load torque, other conditions remaining constant. The Speed Regulation (SR) is defined as the change in speed when the load on the motor is reduced from rated value to zero, expressed as percent of the rated load speed. ππ = πππΏ − ππΉπΏ × 100% ππΉπΏ Where NNL: noload speed NFL: full load speed Example: A 4-pole series motor has 944 wave connected armature conductors. At a certain load, the flux per pole is 34.6 mWb and the total mechanical torque developed is 209 N.m. calculate the line current taken by the motor and the speed at which it will run with an applied voltage of 500V. Total motor resistance is 3 ohm. π 4 ππ = 209 = 0.159 × π × π × πΌπ΄ × = 0.159 × 0.0346 × 944 × πΌπ΄ × π΄ 2 πΌπ΄ = 20.1π΄ πΈπ΄ = ππ − πΌπ΄ π π΄ = 500 − 20.1 × 3 = 439.7V πππ π 0.0346 × 944 × π 4 πΈπ΄ = 439.7 = × = × 60 π΄ 60 2 Dr. Firas Obeidat Faculty of Engineering Philadelphia University → π = 403.8πππ 19 Speed Regulation of DC Motor Example: A 230V DC shunt motor has an armature resistance of 0.5Ω and field resistance of 115Ω. At no load, the speed is 1200rpm and the armature current 2.5A. On application of rated load, the speed drops to 1120 rpm. Determine the speed regulation, line current and power input when the motor delivers rated voltage. πππΏ − ππΉπΏ 1200 − 1120 ππ = × 100% = × 100% = 7.1% ππΉπΏ 1120 π1 = 1200πππ πΈπ΄1 = ππ − πΌπ΄1 π π΄ = 230 − 2.5 × 0.5 = 228.75V π2 = 1120πππ πΈπ΄2 = ππ − πΌπ΄2 π π΄ = 230 − πΌπ΄2 × 0.5 πΈπ΄1 π1 = πΈπ΄2 π2 → 228.75 1200 = πΈπ΄2 1120 πΈπ΄2 = 213.5 = 230 − πΌπ΄2 × 0.5 230 πΌπΏ = πΌπ΄2 + πΌπΉ = 33 + = 35π΄ 115 → πΈπ΄2 = → 1120 × 228.75 = 213.5π 1200 πΌπ΄2 = 230 − 213.5 = 33π΄ 0.5 πππ = πΌπΏ ππ = 35 × 230 = 8050W 20 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Total Loss in a DC Motor 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). 21 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Power Stages and Efficiency Mechanical Efficiency ππ = πππ‘ππ ππ’π‘ππ’π‘ ππ πππ‘π‘π πππ‘ππ ππ’π‘ππ’π‘ ππ πππ‘π‘π × 100% = × 100% π·πππ£πππ πππ€ππ ππ π΄ππππ‘π’ππ ππ πππ‘π‘π πΈπ΄ πΌπ΄ Electrical Efficiency ππ = π·πππ£πππ πππ€ππ ππ π΄ππππ‘π’ππ ππ πππ‘π‘π πΈπ΄ πΌπ΄ × 100% = × 100% πππ‘ππ πΌπππ’π‘ ππ πππ‘π‘π ππ πΌπΏ Overall or Commercial Efficiency ππ = ππ × ππ = πππ‘ππ ππ’π‘ππ’π‘ ππ πππ‘π‘π πππ‘ππ ππ’π‘ππ’π‘ ππ πππ‘π‘π × 100% = × 100% πππ‘ππ πΌπππ’π‘ ππ πππ‘π‘π ππ πΌπΏ 22 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Power Stages and Efficiency Example: The armature winding of a 4-pole, 250V DC shunt motor is lap connected. There are 120 slots, each slot containing 8 conductors. The flux per pole is 20mWb and current taken by the motor is 25A. The resistance of armature and field circuit are 0.1Ω and 125Ω respectively. If the rotational losses amount to be 810W. Find (i) gross torque (Ta) (ii) useful torque (Tsh) (iii) efficiency. (i) gross torque (Ta) ππ 250 πΌπΉ = = = 2π΄ π πΉ 125 πΌπ΄ = πΌπΏ − πΌπΉ = 25 − 2 = 23π΄ πΈπ΄ = ππ − πΌπ΄ π π΄ = 250 − 23 × 0.1 = 247.7π πΈπ΄ = 247.7 = ππ = 9.55 πππ π 0.02 × (120 × 8) × π 4 × = × 60 π΄ 60 4 → π = 773πππ πΈπ΄ πΌπ΄ 247.7 × 23 = 9.55 = 70.4N. m π 773 23 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Power Stages and Efficiency (ii) useful torque (Tsh) Driving power in armature = πΈπ΄ πΌπ΄ = 247.7 × 23 = 5697.1π Optput power = πΈπ΄ πΌπ΄ − πππ‘ππ‘πππππ πππ π ππ = 5697.1 − 810 = 4887.1π ππ β = 9.55 πππ‘ππ ππ’π‘ππ’π‘ 4887.1 = 9.55 = 60.4N. m π 773 (iii) efficiency ππ = πππ‘ππ ππ’π‘ππ’π‘ ππ πππ‘π‘π 4887.1 × 100% = × 100% = 85.8% πΈπ΄ πΌπ΄ 5697.1 ππ = πΈπ΄ πΌπ΄ 5697.1 × 100% = × 100% = 91% ππ πΌπΏ 250 × 25 ππ = 4887.1 × 100% = 78.2% 250 × 25 24 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Power Stages and Efficiency Example: A 20hp (14.92kW), 230V, 1150rpm, 4poles DC shunt motor has a total of 620 conductors arranged in two parallel paths and yielding an armature circuit resistance of 0.2Ω. When it delivers rated power at rated speed, it draws a line current of 74.8A and a field current of 3A. Calculate (i) the flux per pole (ii) armature torque (iii) the rotational losses (iv) total losses expressed as a percentage of power. (i) the flux per pole πΌπ΄ = πΌπΏ − πΌπΉ = 74.8 − 3 = 71.8π΄ πΈπ΄ = ππ − πΌπ΄ π π΄ = 230 − 71.8 × 0.2 = 215.64π πππ π π × 620 × 1150 4 × = × 60 π΄ 60 2 (ii) armature torque πΈπ΄ = 215.64 = → π = 9πππ πΈπ΄ πΌπ΄ 215.64 × 71.8 = 9.55 = 128.6N. m π 1150 (iii) the rotational losses ππ = 9.55 Rotational losses=EAIA – output power =215.64×71.8 - 14920= 562.952W (iv) total losses expressed as a percentage of power Total losses=input power (VTIL)-output power = 230×74.8 - 14920= 17204 - 14920= 2284W total losses expressed as a percentage of power=2284/17204=13.3% 25 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Power Stages and Efficiency Example: A 7.46kW, 250V shunt motor takes a line current of 5A when running light. Calculate the efficiency as a motor when delivering full load output, if the armature resistance are 0.5Ω and 250 Ω respectively. 1- When loaded lightly πππ = ππ πΌπΏ = 250 × 5 = 1250π ππ 250 πΌπ΄ = πΌπΏ − πΌπΉ = 5 − 1 = 4π΄ πΌπΉ = = = 1π΄ π πΉ 250 Field Cu loss = ππ πΌπΉ = 250 × 1 = 250W Armature Cu loss = πΌπ΄ 2 π π΄ = 42 × 0.5 = 8W Iron and friction losses=input power-Armature Cu loss-Filed Cu loss Iron and friction losses=1250-250-8=992W Iron and friction losses is constant 2- at full load ππ πΌπ΄ = πΌπ΄ 2 π π΄ + πΈπ΄ πΌπ΄ ππ πΌπ΄ = 250πΌπ΄ πΈπ΄ πΌπ΄ = ππ’π‘ππ’π‘ πππ€ππ + ππππ πππ π + πππ’π βππ πππ π = 7460 + 992 + 0 = 8452π 250πΌπ΄ = πΌπ΄ 2 × 0.5 + 8452 26 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Power Stages and Efficiency 250πΌπ΄ = πΌπ΄ 2 × 0.5 + 8452 Rearrange the above equation yields 0.5πΌπ΄ 2 − 250πΌπ΄ + 8452 = 0 −π ± π2 − 4ππ 250 ± (−250)2 −4 × 0.5 × 8452 πΌπ΄ = = = 36.5 ππ 463.5 2π 2 × 0.5 πΌπ΄ = 36.5π΄ πΌπΏ = πΌπ΄ + πΌπΉ = 36.5 − 1 = 37.5π΄ πππ = ππ πΌπΏ = 250 × 37.5 = 9375π πππ’π‘ = 7460π ππΉπΏ ππ’π‘ππ’π‘ πππ€ππ 7460 = × 100% = × 100% = 79.6% ππππ’π‘ πππ€ππ 9375 27 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Power Stages and Efficiency Example: A 6-pole, 500V, wave connected shunt motor has 1200 armature conductors and useful flux/pole of 20mWb. The armature and field resistances are 0.5Ω and 250Ω respectively. What will be the speed and torque developed by the motor when it draws 20A from the supply mains? If magnetic and mechanical losses amount to 900W, find (i) output in kW (ii) useful torque (Tsh) (iii) efficiency (ππ ) at this load. πΌπΉ = ππ 500 = = 2π΄ π πΉ 250 πΌπ΄ = πΌπΏ − πΌπΉ = 20 − 2 = 18π΄ πΈπ΄ = ππ − πΌπ΄ π π΄ = 500 − 18 × 0.5 = 491π πππ π 0.02 × 1200 × π 6 πΈπ΄ = 491 = × = × 60 π΄ 60 2 πΈπ΄ πΌπ΄ 491 × 18 ππ = 9.55 = 9.55 = 205.9N. m π 410 (i) output in kW → π = 410πππ Field Cu loss = ππ πΌπΉ = 500 × 2 = 1000W Armature Cu loss = πΌπ΄ 2 π π΄ = 182 × 0.5 = 162W Iron and friction losses=900W 28 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Power Stages and Efficiency Total losses=armature Cu loss + field Cu loss + iron and friction loss Total losses=162+1000+900=2062W Motor input power=VTIL=500×20=10000W Motor output power=motor input power-motor losses=10000-2062=7938W (ii) useful torque (Tsh) ππ β = 9.55 πππ‘ππ ππ’π‘ππ’π‘ 7938 = 9.55 = 184.9N. m π 410 (iii) Efficiency (ππ ) at this load ππ’π‘ππ’π‘ πππ€ππ 7938 ππ = × 100% = × 100% = 79.38% ππππ’π‘ πππ€ππ 10000 29 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Characteristics and Applications of DC Motors Type of Motor Shunt Motor Series Motor Cumulative Compound Motor Characteristics Applications Approximately constant speed Adjustable speed Medium starting torque (up to 1.5 F.L. torque) For driving constant speed line shafting Lathes Centrifugal pumps Machine tools Blowers and fans Reciprocating pumps Variable speed Adjustable variable speed High starting torque For traction work i.e. Electric locomotives Rapid transit systems Trolley, Cars etc Cranes and hoists conveyers Variable speed Adjustable variable speed High starting torque For intermittent high torque loads For shears and punches Elevators Conveyers Heavy planers Rolling mills, Ice machine, printing presses, Air compressors 30 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 31
