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Transcript
ECE 320 DC Machines Problems Lesson 26 1, Questions 8.2, 8.3, 8.4, 8.5, and 8.6 on page 552. 8.2 How can the speed of a shunt dc motor be controlled? Explain in detail. A shunt DC motor speed can be controlled by one of three methods: 1. Adjusting the field resistance: Increasing the field resistance decreases the field current. Field weakening increases the steady state speed. 2. Adjusting the terminal voltage adjusts the field current and the armature current. This adjusts torque which leads to a corresponding adjustment in steady state speed. 3. Inserting a resistance in series with the armature makes the machine more sensitive to load, decreasing its steady state speed if it is under load. This method is not popular because a resistance in series with the main energy-carrying current wastes a lot of energy. Details on these methods are explained on pages 479-491 of the textbook. 8.3 What is the practical difference between a separately excited and a shunt dc motor? The main difference is the way that the field is connnected to its source of energy. In the shunt machine, the same source serves both armature and field; in the separately excited machine, a separate source serves each machine winding. The practical effect is less complicated control of speed with greater flexibility in the separately excited machine. 8.4 What effect does armature reaction have on the torque-speed characteristic of a shunt dc motor? Can the effects of armature reaction be serious? What can be done to remedy this problem? Armature reaction shifts the neutral plane of the machine and reduces the magnetic flux within the field; in other words, it weakens the field. Shifting of the neutral plane causes commutation under load to occur when voltage remains on the commutator. This leads to sparking and arcing at the brushes; in severe cases, it can lead to flashover between commutator segments. The field weakening aspect of armature reaction gives greater than expected speed for a given terminal condition of voltage and current. In severe cases, this may cause runaway in a manner similar to what occurs with poorly designed differential compounding. See pages 433-444 of the textbook for details. 8.5 What are the desirable characteristics of the permanent magnets in a PMDC machine? Permanent magnet dc machines have no field circuit. There is no requirement for a source of energy for the field, saving the expense of a separate excitation or the slower control of shunt or series machines. Consequently, PMDC machines tend to be smaller than comparably rated wound field motors There is little energy loss in the field circuit of a PMDC machine. These advantages tend to be more significant in smaller machines. 8.6 What are the principal characteristics of a series dc motor? What are its uses? The series connection of the field winding leads to a torque-speed characteristic that has the speed proportional to the reciprocal of the square of the torque, shifted a little lower by the effect of the winding resistances. This leads to an enormous starting torque and a need to always have some load applied to prevent overspeed. Winding resistances are quite small. A series dc machine has an advantage where high starting torque is needed. Applications include starter motors, elevator motors, and traction motors in locomotives. 2. Problems 8.1, 8.2 and 8.3 on page 553 of the textbook. ILrated 110 A VT 240 V NF 2700 NSE 14 RA 0.19 ohm RF 75 ohm RS 0.02 ohm Radj = 100 to 400 ohm ωrated 1800 RPM rad ωrated 188.496 sec RPM 2π Prated 30 hp 60 Hz Protational 3550 W In problems 8.1 through 8.7, assume that the motor described above can be connected in shunt. The equivalent circuit of the shunt motor is shown in Figure P8-2. 8.1 If the resistor Radj is adjusted to 175 ohms, what is the rotational speed of the motor at no-load conditions? Radj 175 ohm ωfl ωrated No load means no armature current. IAnl 0 A VAnl VT VAnl 240 V Find the generated voltage at no load. With no current, it's the same at the terminal voltage EAnl VAnl RA IAnl EAnl 240 V Calculate the field current in the machine VT IF IF 0.96 A RF Radj From Figure P8-1, at 1800 RPM, this field current gives a generated voltage of 241V. EA0 241 V ω0 1800 RPM Calculate the speed that corresponds to the same field current (same flux) but 240V=EA. EA0 K Φ ω0 EAnl rad = ωnl ω0 ωnl 187.713 ωnl 1793 RPM EAnl K Φ ωnl EA0 sec 8.2 Assuming no armature reaction, what is the speed of themotor at full load? What is the speed regulation of the motor? We have given the terminal full load current. There is no indication that the field has changed. ILrated 110 A IF 0.96 A By the current law, we can find the armature current. IAfl ILrated IF IAfl 109.04 A Use a loop equation to find the induced voltage. EAfl VT IAfl RA EAfl 219.3 V From Figure P8-1, at 1800 RPM, this field current gives a generated voltage of 271V. EA0 271 V ω0 1800 RPM Calculate the speed that corresponds to the same field current (same flux) but 218.3V=EA. EA0 K Φ ω0 EAfl rad = ωfl ω0 ωfl 152.523 ωfl 1456 RPM EAfl K Φ ωfl EA0 sec ωnl ωfl SpeedRegulation SpeedRegulation 23.1 % ωfl 8.3 If the motor is operating at full load and if its variable resistance R adj is increased to 250Ω what is the new speed of the motor? Compare the full load speed with Radj=175Ω to the full load speed with Radj=250Ω Assume no armature reaction. When we change Radj to 250Ω, the field current changes. VT IF3 RF Radj Radj 250 ohm IF3 0.738 A Armature current remains the same at rated conditions because ratings are based upon the heating that occurs in response to armature current and field current levels. Therefore, induced voltage remains the same. IA3 IAfl IA3 109.04 A EA3 VT IA3 RA EA3 219.3 V From Figure P8-1, at 1800 RPM, this field current gives a generated voltage of 215V EA0 215 V ω0 1800 RPM Calculate the speed that corresponds to the same field current (same flux) but 218.3V=EA. EA0 EA3 = K Φ ω0 K Φ ω3 EA3 ω3 ω0 EA0 rad ω3 192.25 sec ω3 1836 RPM This speed is somewhat higher than the speed under the same load at a lower field resistance. Field weakening increases speed for the same load. 3. Problems 8.4 and 8.6 on page 553 of the textbook. 8.4 Assume that the motor is operating at full load and that the variable resistor R adj is again 175Ω. If the armature reaction is 2000 A*turns at full load, what is the speed of the motor? How does it compare to the results of Problem 8-2? Radj 175 Ω turns 1 AR 2000 A turns We have the same field current as we began with, but this will be reduced by the armature reaction. The armature reaction is related to the field current by the number of field turns: VT IF RF Radj IF 0.96 A AR IFeff IF NF IFeff 0.219 A Looking up the nominal induced voltage from the magnetization curve in Figure P8-1 on page 554: EA0 84V ω0 1800 RPM The armature current is: IA4 ILrated IFeff IA4 109.781 A Calculating the induced voltage for the load and armature current specified: EA4 VT IA4 RA EA4 219.142 V Compare this to the nominal induced voltage at nominal speed and we can find the operating speed. Armature reaction has the effect of weakening the field. EA4 K φ ω4 EA4 ω0 Field weakening speeds up = ω4 ω4 4696 RPM EA0 K φ ω0 EA0 the machine. In this case, the numbers are big. 8.6 What is the starting current of thes machine if it is started by connecting it directly to the power supply V T? How does this starting current compare to the full load current of the motor? The starting current for a line start is found by setting the induced voltage to zero and using the remaining armature loop to find the current. VT 3 IAstart IAstart 1.263 10 A RA To get the line current, add in the field current. The field current is negligible under these circumstances. ILstart IAstart IF 3 ILstart 1.264 10 A This is more than ten times the 110A full load machine current. A current this large, particularly if the motor is started under load (which makes it quite slow to come up to speed), is likely to damage the motor. For this reason, there is a practical limit of a few horsepower for machines to be line-started. Larger machines must have some form of starter circuit for safe starting. 4. Problem 8.7 on page 554 of the textbook. 8.7 Plot the torque-speed characteristic of this motor assuming no armature reaction and again assuming an armature reaction of 1200 A-turns. Restate the given. VT 240 V RF 75 Ω Radj 175 Ω RA 0.40 Ω IL0 110 A NF 2700 Describe the magnetization curve. Here, I just took a set of points that give a reasonable piecewise linearization of the curve. ω0 1800 RPM Far0 1200 A RPM 2 π 60 Hz T EAx ( 18 50 82 113 140 153 184 203 220 233 243 253 261 267 273 ) V T IFx ( 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 ) A Define the armature reaction as linear with terminal current,as the textbook does in its example. IL FAR IL Far0 Far0 IL0 Set up the equations. Use circuit analysis to find I A, EA, and IF. VT IA IL IL RF Radj EA IL VT IA IL RA VT FAR IL Far0 IF IL Far0 NF RF Radj Enter the magnetization curve and use linear interpolation to get the nominal value of EA0. EA0 IL Far0 linterp IFx EAx IF IL Far0 Calculate the speed for this case. EA IL ω87 IL Far0 ω0 EA0 IL Far0 Calculate the torque for this case. EA IL IA IL τind87 IL Far0 ω87 IL Far0 240 220 200 ω 87 IL 1200 A ω 87 IL 0 A 180 160 0 50 τ ind87 IL 0 A τ ind87 IL 1200 A The machine exhibits instability when armature reaction is considered. 100 5. Problem 8.8 on page 554 of the textbook. 8.8 For Problems 8-8 and 8-9, the shunt dc motor is reconnected separately excited,as shown in Figure P8-3. It has a fixed field voltage V F of 240V and an armature voltage V A that can be varied from 120V to 240V. What is the no load speed of this separately excited motor when R adj=175ohms and (a) V A=120V, (b) VA=180V, and (c) VA=240V? VF 240 V 120 VA 180 V 240 Radj 175 ohm Find the field current. VF IF IF 0.96 A RF Radj From Figure P8-1, at 1800 RPM, this field current gives a generated voltage of 241V. EA0 241 V ω0 1800 RPM At no load, the armature voltage is equal to the generated voltage; no load means no armature current. IA 0 A 120 EAnl 180 V 240 EAnl VA IA RA Calculate the speed that corresponds to the same field current (same flux). EA0 EAnl = K Φ ω0 K Φ ωnl EAnl ωnl ω0 EA0 93.857 rad ωnl 140.785 sec 187.713 896 ωnl 1344 RPM 1793 Check: the third operating point here is the same as the operating conditions in Problem 8.1. 6. Problem 8.20 on page 559 of the textbook. 8.20 An automatic starter circuit is to be designed for a shunt motor rated at 20hp, 240V, and 75A. The armature resistance is 0.12Ω and the shunt field resistance is 40Ω. The motor is to start with no more than 250% of its rated armature current and as soon as the current falls to rated value, a starting resistor stage is to be cut out. How many stages of starting resistance are need and how big should each one be? PR 20 hp VAR 240 V ILR 75 A RA 0.12 Ω VAR IFR RF IFR 6 A IAR ILR IFR RF 40 Ω IX 250% IAR 69 A To limit the starting current to 250% of this rated value, IAPk IX IAR IAPk 172.5 A Starting resistance needed to limit the current to the peak value specified is found as follows: VAR RAstart1 RAstart1 1.391 Ω IAPk We calculate the expected number of steps and rounding up: RA log RAstart1 n ceil 3 log IAR IAPk Subtracting the armature resistance gives us the necessary value of added start resistance. Rstart1 RAstart1 RA Rstart1 1.271 Ω When the machine accelerates, it eventually reaches the rated current value, at an induced voltage value of EA1 VT IAR RA Rstart1 EA1 144 V We now reduce the starting resistance. Again, the limit is the specified peak current. RAstart2 VAR EA1 RAstart2 0.557 Ω IAPk Subtracting the armature resistance to get the necessary value of added resistance. Rstart2 RAstart2 RA Rstart2 0.437 Ω When the machine accelerates, it eventually reaches the rated current value, at an induced voltage value of EA2 VT IAR RA Rstart2 EA2 201.6 V We find the next value of starting resistance using this induced voltage and the specified peak current. RAstart3 VAR EA2 RAstart3 0.223 Ω IAPk Subtracting the armature resistance to get the necessary value of added resistance. Rstart3 RAstart3 RA Rstart3 0.103 Ω When the machine accelerates, it eventually reaches the rated current value, at an induced voltage value of EA3 VT IAR RA Rstart3 EA3 224.64 V We find the next value of starting resistance using this induced voltage and the specified peak current. RAstart4 VAR EA3 IAPk RAstart4 0.089 Ω This is less than the armature resistance, so no further additional resistance is needed to complete the starting process. There are only three stages of armature resistance needed for starting. Rstart1 1.271 Ω Rstart2 0.437 Ω Rstart3 0.103 Ω Thd individual starting resistances are found by evaluating each successive stage of the starting process. Rstart1 = R1 R2 R3 Rstart2 = R2 R3 Solving these for the starting resistances, R3 Rstart3 R3 0.103 Ω R2 Rstart2 R3 R2 0.334 Ω R1 Rstart1 R2 R3 R1 0.835 Ω Rstart3 = R3 7. Problem 8.14a on page 556 of the textbook; do only the 100% case. 8.14a A 20-hp, 240V, 76A, 900RPM series motor has a field winding of 33 turns per pole. Its armature resistance is 0.09Ω and its field resisitance is 0.06Ω. The magnetization curve expressed in terms of magnetomotive force versus EA at 900 RPM is given by the following table EA (V) F (A*turns) 95 500 150 1000 188 1500 212 2000 229 2500 243 3000 Armature reaction is negligible for this machine. Computer the motor's torque, speed, and output power at 100% of full load armature current. Neglect rotational losses. Pout 20 hp VT 240 V IA 76 A ωfl 900 RPM T NF 33 RA 0.09 Ω RF 0.06 Ω T EA ( 95 150 188 212 229 234 ) V F ( 500 1000 1500 2000 2500 3000 ) A Set up the magnetization curve for interpolation. M cs cspline F EA Calculate the generated voltage and the MMF under the conditions given. EAx VT IA RA RF EAx 228.6 V F0 NF IA F0 2508 A Perform the interpolation to get the normalized generated voltage at the nominal speed of ω0 900 RPM EA0 interp M cs F EA F0 229.197 V Calculate the speed that corresponds to the same field current (same flux) but at the calculated E A. Then initiate a collection of values for later plotting. EA0 EAx = K Φ ω0 K Φ ωx EAx ωx ω0 EA0 ωx 897.656 RPM rad ωm ωx ( 94.002 ) sec Output power is Pconv EAx IA Pconv 17.374 kW Pconv 23.298 hp which is close to the nominal value Torque is τind Pconv ωx Collect torque values for plotting: τind 184.821 N m Check τind ωx 17.374 kW τm τind ( 184.821 ) N m IA 76 A 0.33 25.08 A EAx 236.238 V F0 NF IA Repeat the calculations for 33% load current: EAx VT IA RA RF F0 827.6 A EA0 interp M cs F EA F0 133.018 V EA0 EAx = K Φ ω0 EAx ωx ω0 EA0 K Φ ωx Pconv EAx IA τind ω0 900 RPM ωx 1598.4 RPM 94 rad ωm stack ωm ωx 167.4 sec Pconv 7.945 hp Pconv 5.925 kW Pconv τind 35.397 N m ωx τm stack τm τind Check Repeat the calculations for 67% load current: EAx VT IA RA RF τind ωx 5.925 kW IA 76 A 0.67 50.92 A EAx 232.362 V F0 NF IA EA0 interp M cs F EA F0 197.842 V EA0 EAx = K Φ ω0 Pconv EAx IA τind ωx 1057 RPM 94 rad ωm stack ωm ωx 167.4 sec Pconv 15.867 hp 110.7 Pconv 11.832 kW Pconv τind 106.89 N m ωx F0 1680.4 A ω0 900 RPM EAx ωx ω0 EA0 K Φ ωx 184.821 N m 35.397 184.821 τm stack τm τind 35.397 N m 106.89 τind ωx 11.832 kW Check Repeat the calculations for 133% load current: IA 76 A 1.33 101.08 A EAx VT IA RA RF EAx 224.838 V F0 NF IA EA0 interp M cs F EA F0 226.396 V EA0 EAx = K Φ ω0 K Φ ωx Pconv EAx IA τind Pconv ωx EAx ωx ω0 EA0 Pconv 22.727 kW ω0 900 RPM 94 ωx 893.8 RPM 167.4 rad ωm stack ωm ωx 110.7 sec Pconv 30.477 hp 93.6 τind 242.808 N m Check F0 3335.6 A τind ωx 22.727 kW 184.821 35.397 τm stack τm τind N m 106.89 242.808 Plot the torque vs speed curve. 300 200 τm 100 0 80 100 120 140 ωm 160 180