Electrical Engineering Problems: Motor and Generator Analysis, Study notes of Electronic Circuits Analysis

Download Electrical Engineering Problems: Motor and Generator Analysis and more Study notes Electronic Circuits Analysis in PDF only on Docsity! 85 PROBLEM SOLUTIONS: Chapter 7 Problem 7.1 part (a): ωm ∝ Va. part (b): ωm ∝ 1 If part (c): ωm will be constant. Problem 7.2 part (a): For constant terminal voltage, the product nIf (where n is the motor speed) is constant. Hence, since If ∝ 1/Rf Rf 1180 = Rf + 5 1250 and hence Rf = 84.2 Ω. part (b): 1380 r/min Problem 7.3 Check this part (a): ωm halved; Ia constant part (b): ωm halved; Ia doubled part (c): ωm halved; Ia halved part (d): ωm constant; Ia doubled part (e): ωm halved; Ia reduced by a factor of 4. Problem 7.4 part (a): Rated armature current = 25 kW/250-V = 100 A. part (b): At 1200 r/min, Ea can be determined directly from the magneti- zation curve of Fig. 7.27. The armature voltage can be calculated as Va = Ea + IaRa and the power output as Pout = VaIa. With Ia = 100 A If [A] Ea [V] Va [V] Pout [kW] 1.0 150 164 16.4 2.0 240 254 25.4 2.5 270 284 28.4 part (c): The solution proceeds as in part (b) but with the generated voltage equal to 900/1200 = 0.75 times that of part (b) If [A] Ea [V] Va [V] Pout [kW] 1.0 112 126 12.6 2.0 180 194 19.4 2.5 202 216 21.6 86 Problem 7.5 part (a): part (b): (i) (ii) 89 must equal 1.05 A and hence the shunt field resistance must be 219 Ω. Hence, under this loading condition, with a terminal voltage of 250 V, the armature voltage will be 250 + 61.5× 0.065 = 250.8 V, the shunt field current will equal 250.8/219 = 1.15 A and thus the armature current will equal 61.5 + 1.15 = 62.7 A. The generated voltage can now be calculated to be 250.8 + 62.7(0.175) = 286 V. The corresponding voltage on the 1195 r/min mag curve will be Ea = 286(1195/1139) = 285 V and hence the required net field ampere-turns is (using the MATLAB ‘spline()’ function) 1042 A-turns. The shunt-field ampere-turns is 1.15× 500 = 575 A-turns, the demagnetizing armature amp-turns are 204 A- turns and hence the required series turns are Ns = 1042− (575− 204) 61.5 = 10.6 ≈ 11 turns Problem 7.9 From the given data, the generated voltage at Ia = 90A and n(90) = 975 r/min is Ea(90) = Va − Ia(Ra +Rs) = 230− 90(0.11 + 0.08) = 212.9 V Similarly, the generated voltage at Ia = 30 A is Ea(30) = 230− 30(0.11 + 0.08) = 224.3 V Since Ea ∝ nΦ Ea(30) Ea(90) = ( n(30) n(90) )( Φ(30) Φ(90) ) Making use of the fact that Φ(30)/Φ(90) = 0.48, we can solve for n(30) n(30) = n(90) ( Ea(30) Ea(90) )( Φ(90) Φ(30) ) = 2140 r/min Problem 7.10 90 Problem 7.11 part (a): For constant field current, and hence constant field flux, constant torque corresponds to constant armature current. Thus for speeds up to 1200 r/min, the armature current will remain constant. For speeds above 1200 r/min, ignoring the voltage drop across the armature resistance, the motor speed will be inversely proportional to the field current (and hence the field flux). Thus the armature current will increase linearly with speed from its value at 1200 r/min. Note that as a practical matter, the armature current should be limited to its rated value, but that limitation is not considered in the plot below. part (b): In this case, the torque will remain constant as the speed is in- creased to 1200 r/min. However, as the field flux drops to increase the speed above 1200 r/min, it is not possible to increase the armature current as the field flux is reduced to increase the speed above 1200 r/min and hence the torque track the field flux and will decrease in inverse proportion to the change in speed above 1200 r/min. Problem 7.12 part (a): With constant terminal voltage and speed variation obtained by field current control, the field current (and hence the field flux) will be inversely proportional to the speed. Constant power operation (motor A) will then require 91 constant armature current. Constant torque (motor B) will require that the armature current variation be proportional to the motor speed. Thus motor A: Ia = 125 A motor B: Ia = 125(500/1800) = 34.7 A part (b): motor A: Ia = 125 A motor B: Ia = 125(1800/125) = 450 A part (c): Under armature voltage control and with constant field current, the speed will be proportional to the armature voltage. The generated voltage will be proportional to the speed. Constant-power operation (motor A) will re- quire aramture current that increases inversely with speed while constant torque operation (motor B) will require constant armature current. For the conditions of part (a): motor A: Ia = 125(1800/125) = 450 A motor B: Ia = 125 A For the conditions of part (b): motor A: Ia = 125(500/1800) = 34.7 A motor B: Ia = 125 A Problem 7.13 ωm = Ea KaΦd = Va − IaRa KaΦd Ia = T KaΦd Thus ωm = 1 KaΦd ( Va − TRa KaΦd ) The desired result can be obtained by taking the derivative of ωm with Φd dωm dΦd = 1 KaΦ2 d ( 2TRa KaΦd − Va ) = 1 KaΦ2 d (2IaRa − Va) = 1 KaΦ2 d (Va − 2Ea) From this we see that for Ea > 0.5Va, dωa/dΦd < 0 and for Ea < 0.5Va, dωa/dΦd > 0. Q.E.D. 94 Using MATLAB and its ‘spline()’ function to represent the fan character- istics, an iterative routine can be written to solve for the operating point (the intersection of the motor and fan characteristics). The result is that the motor will operate at a speed of 999 r/min and an output power of 8.39 kW. Problem 7.18 part (a): Assuming negligible voltage drop across the armature resistance at no load, the field current can be found from the 1300 r/min magnetization curve by setting Ea = 230 V. This can be most easily done using the MATLAB ‘spline()’ function. The result is If = 1.67 A. This corresponds to NfIf = 2500 a·turns of mmf. part (b): At rated load, Ea = Va−IaRa = 230−46.5×0.17 = 222.1 V. From the no-load, 1300 r/min magnetization curve, the corresponding field current is 1.50 A (again obtained using the MATLAB ‘spline()’ function). Thus, the effective armature reaction is Armature reaction = (1.67− 1.5) A× 1500 turns/pole = 251 A · turns/pole part (c): With the series field winding, Rtot = Ra + Rs = 0.208 Ω. Thus, under this condition, Ea = Va−IaRa = 220.3. This corresponds to a 1300 r/min generated voltage of 236.7 V and a corresponding field current (determined from the magnetization curve using the MATLAB ‘spline()’ function) of 1.84 A, corresponding to a total of 2755 A·turns. Thus, the required series field turns will be Ns = 2755− (2500− 251) 46.5 = 10.8 or, rounding to the nearest integer, Ns = 11 turns/pole. part (d): Now the effective field current will be Ieff = 2500− 251 + 20× 46.5 1500 = 2.12 A From the 1300 r/min magnetization curve, Ea = 246.1 V while the actual Ea = Va − RtotIa = 220.3 V. Hence the new speed is n = 1300 ( 220.3 246.1 ) = 1164 r/min Problem 7.19 part (a): At full load, 1185 r/min, with a field current of 0.554 A Ea = Va − IaRtot = 221.4 V where Rtot = 0.21 + 0.035 = 0.245 Ω. 95 An 1825 r/min magnetization curve can be obtained by multiplying 230 V by the ratio of 1185 r/min divided by the given speed for each of the points in the data table. A MATLAB ‘spline()’ fit can then be used to determine that this generated voltage corresponds to a field-current of 0.527 A . Thus, the armature reaction is (0.554− 0.527)2000 = 53.4 A·turns/pole. part (b): The full-load torque is T = EaIa ωm = 221.4× 34.2 1185(π/30) = 62.8 N ·m part (c): The maximum field current is 230/310 = 0.742 Ω. The effective field current under this condition will therefore be Ieff = 2000× 0.742− 160 2000 = 0.662 A From the 1185 r/min magnetization curve found in part (b), this corresponds to a generated voltage of 245 V. Thus, the corresponding torque will be T = EaIa ωm = 245× 65 1185(π/30) = 128 N ·m part (d): With the addition of 0.05 Ω, the total resistance in the armature circuit will now be Rtot = 0.295 Ω. The required generated voltage will thus be Ea = Va − IaRtot = 219.6 V This corresponds to 219.6(1185/1050) = 247.8 V on the 1185 r/min magnetiza- tion curve and a corresponding effective field current of 0.701 A. As can be seen from the data table, a no-load speed of 1200 r/min corre- sponds to a field current of 0.554 A. Thus the series-field A·turns must make up for the difference between that required and the actual field current as reduced by armature reaction. Ns = Nf(If,eff − If) + Armature reaction Ia = 2000(0.701− 0.554) + 53.4 35.2 = 9.8 turns Problem 7.20 part (a): From the demagnetization curve, we see that the shunt field current is 0.55 A since the no-load generated voltage must equal 230 V. The full-load generated voltage is Ea = Va − IaRa = 219.4 V and the corresponding field current (from the demagnetization curve obtained using the MATLAB ‘spline()’ function) is 0.487 A. Thus the demagnetization is equal to 2000(0.55− 0.487) = 127 A·turns. 96 part (b): The total effective armature resistance is now Rtot = 0.15+0.028 = 0.178 Ω. Thus, the full-load generated voltage will be Ea = Va − IaRtot = 217.4 V The net effective field current is now equal to 0.487+70.8(4/2000) = 0.628 A. The corresponding voltage at 1750 r/min (found from the magnetization curve using the MATLAB ‘spline()’ function) is 242.7 V and hence the full-load speed is n = 1750 ( 217.4 242.7 ) = 1567 r/min part (c): The effective field current under this condition will be Ieff = 0.55 + 125(4/2000)− 230/2000 = 0.685 A From the 1750 r/min magnetization curve (using the MATLAB ‘spline()’ func- tion), this corresponds to a generated voltage of 249 V. Thus, the corresponding torque will be T = EaIa ωm = 249× 125 1750(π/30) = 170 N ·m Problem 7.21 part (a): For a constant torque load, changing the armature resistance will not change the armature current and hence Ia = 60 A. part(b): Ea = Va − RaIa Thus, without the added 1.0Ω resistor, Ea = 216 V and with it Ea = 156 V. Thus, Speed ratio = 156 216 = 0.72 Problem 7.22 parts (a) and (b):