3 3 ( ) 277 V ( ) 429 V ⎞ P A V E = = = max X S 1.5 & 238 kW (d) Since E A must be decreased linearly with frequency, the maximum value at 300 r/min would be 15 Hz ( ) 429 V 129 V = = E A ,300 50 Hz (e) If the applied voltage ⎞ V is derated by the same amount as E A , then V ⎞ = (15/50)(277) = 83.1 V. Also, note that be P X S = (15/50)(1.5 & ) = 0.45 & . The maximum power that the motor could supply would 3 3 ( ) 83.1 V ( 129 V ) ⎞ A V E = = = max X S 0.45 & 71.5 kW (f) As we can see by comparing the results of (c) and (e) , the power-handling capability of the synchronous motor varies linearly with the speed of the motor. 6-7. A 208-V Y-connected synchronous motor is drawing 40 A at unity power factor from a 208-V power system. The field current flowing under these conditions is 2.7 A. Its synchronous reactance is 0.8 & . Assume a linear open-circuit characteristic. (a) Find the torque angle ™ . (b) How much field current would be required to make the motor operate at 0.8 PF leading? (c) What is the new torque angle in part (b) ? S OLUTION (a) The phase voltage of this motor is V ⎞ = 120 V, and the armature current is Therefore, the internal generated voltage is I A 40 0 A = ° . = E V I I A A ⎞ A S A R jX 120 0 V ( ) 0.8 ( 40 0 A ) E A = ° j & ° E A 124 14.9 = V ° The torque angle ™ of this machine is –14.9 ° . (b) A phasor diagram of the motor operating at a power factor of 0.78 leading is shown below. P I A 2 I A 1 ⎞ V jX I S A } P EE A 1 A 2 Since the power supplied by the motor is constant, the quantity I A cos ⎝ , which is directly proportional to power, must be constant. Therefore, ( ) ( )( ) I A2 0.8 = 40 A 1.00 159 I A2 50 36. = 87 A ° The internal generated voltage required to produce this current would be = E V I I 2 2A A ⎞ 2A S A R jX ( )( ) E A 2 120 0 V = ° j 0.8 50 & 36.8 7 A ° E A2 147.5 12.5 = V ° The internal generated voltage E A is directly proportional to the field flux, and we have assumed in this problem that the flux is directly proportional to the field current. Therefore, the required field current is E A 2 147 V ( ) 2 = = .7 A 3.20 A = 2 1 F F I I E 1 A 124 V (c) The new torque angle ™ of this machine is –12.5 ° . 6-8. A synchronous machine has a synchronous reactance of 2.0 & per phase and an armature resistance of 0.4 & per phase. If E A =460 -8 ° V and ⎞ V = 480 0 ° V, is this machine a motor or a generator? How much power P is this machine consuming from or supplying to the electrical system? How much reactive power Q is this machine consuming from or supplying to the electrical system? S OLUTION This machine is a motor, consuming power from the power system, because E A is lagging ⎞ V . It is also consuming reactive power, because A cos E V ™ < ⎞ . The current flowing in this machine is ⎞ A V E I 480 0 V ° 4 60 8 V ° 33.6 9.6 A = = = ° A 0.4 2.0 + + & A S R jX j Therefore the real power consumed by this motor is 3 cos 3 ( ) 480 V ( 3 = = 3.6 A ) cos ( 9.6 ) 47.7 kW ° = ⎞ A P V I ⎝ and the reactive power consumed by this motor is 3 sin 3 ( ) 480 V ( 3 = = 3.6 A ) sin ( 9.6 ) 8.07 kVA ° R = ⎞ ⎝ A Q V I 6-9. Figure P6-2 shows a synchronous motor phasor diagram for a motor operating at a leading power factor with no R A . For this motor, the torque angle is given by tan = ™ X I co S A s ⎝ V X ⎞ + si S A I n ⎝ ™ =tan -1 X I co S A s ⎝ V X ⎞ + si S A I n ⎝ Derive an equation for the torque angle of the synchronous motor if the armature resistance is included . 160 S OLUTION The phasor diagram with the armature resistance considered is shown below. I A X I ⎞ V S A ⎝ ™ sin ⎝ } cos ⎝ I X jX I S A S A ⎝ ⎝ R I A A E A cos ⎝ R I Therefore, ™ A A ⎝ ⎝ tan = X I co S A s + A R I si A n ⎝ ⎝ + V X ⎞ si S A n I A c A R o I s ™ ⎝ ⎝ = tan 1 X I co S A s + A R I sin A ⎝ ⎝ + V X ⎞ sin S A I A R c A o I s 6-10. A 480-V 375-kVA 0.8-PF-lagging Y-connected synchronous generator has a synchronous reactance of 0.4 & and a negligible armature resistance. This generator is supplying power to a 480-V 80-kW 0.8-PF- leading Y-connected synchronous motor with a synchronous reactance of 1.1 & and a negligible armature resistance. The synchronous generator is adjusted to have a terminal voltage of 480 V when the motor is drawing the rated power at unity power factor. (a) Calculate the magnitudes and angles of E A for both machines. (b) If the flux of the motor is increased by 10 percent, what happens to the terminal voltage of the power system? What is its new value? (c) What is the power factor of the motor after the increase in motor flux? S OLUTION (a) The motor is operating at rated power and unity power factor, so the current flowing in the motor is P I I = = = 80 kW 96.2 A = ,m ,mA L ( ) 3 PF 3 ( ) 480 V 1.0 T V so I A,m 96.2 0 A = ° . This machine is Y-connected, so the phase voltage is V ⎞ = 480 / 3 = 277 V. The internal generated voltage of the motor is 161 = E V I ,m ⎞ ,m jX ,mA S A ( )( ) E ,m A E ,mA 277 0 V = ° j 1.1 297 20. = 9 V ° 96 & .2 0 A ° This same current comes from the generator, so the internal generated voltage of the generator is = + E V I ,g ⎞ ,g jX ,g A S A ( )( ) E ,gA 277 0 V = ° j 0.4 + 96 & .2 0 A ° E A ,g 280 7.9 V = ° j 0.4 & I A,g I A,m j 1.1 & + + + E A,g - V ⎞ ,g V ⎞ ,m + E A,m - - - E A,g A jX I I A I V ⎞ S,g A ⎞ V jX I S,m A E A,m Generator Motor (b) The power supplied by the generator to the motor will be constant as the field current of the motor is varied. The 10% increase in flux will raise the internal generated voltage of the motor to (1.1)(297 V) = 327 V. To make finding the new conditions easier, we will make the angle of the phasor E , A g the reference during the following calculations . The resulting phasor diagram is shown below. E I A,g A g ™ jX I m ™ S,g A ⎞ V jX I Then by Kirchhoff’s Voltage Law, E A,m S,m A = + E E , ,A g A m + I j X X ,S g , ( ) S m A 162 or A I = , ,A g A m E E + j X X , , ( ) S g S m Note that this combined phasor diagram looks just like the diagram of a synchronous motor, so we can apply the power equation for synchronous motors to this system. , ,A g A m E E P = 3 sin © + , ,S g S m X X ™ ™ = + . From this equation, where © g m ( ) X X + P ( ) & ( ) , , 1 1 S g S m sin 1.5 80 kW 25.9 sin © ( )( = = ) = ° Therefore, 3 , , A g A m E E E E 3 280V 327 V ° ° , , A g A m 280 0 V 327 = = 25.9 V 95.7 5.7 A = ° A I + & , , ( ) S g j X S X m 1. j 5 The phase voltage of the system would be = V E I = ° ( )( & ) ° = ° ⎞ , , A g jX S g A 280 0 V j 0.4 95.7 5.7 A 286 7.6 V If we make ⎞ V the reference (as we usually do), these voltages and currents become: E , A g 280 7.6 V = ° V ⎞ 286 0 V = ° E , A m 327 18. = 3 V ° I A 95.7 13.3 = A ° The new terminal voltage is ( ) 3 286 V 495 V T V = = , so the system voltage has increased . (c) The power factor of the motor is now -18.3 ° implies an impedance angle of 18.3 ° . Note: The reactive power in the motor is now ( )( = = PF cos ( ) 13.3 = 0.9 ° 73 = leading , since a current angle of ) ( ) ° = motor 3 ⎞ A sin Q V I ⎝ 3 286 V 95.7 A sin 13.3 18.9 kVAR The motor is now supplying 18.9 kVAR to the system. Note that an increase in machine flux has increased the reactive power supplied by the motor and also raised the terminal voltage of the system . This is consistent with what we learned about reactive power sharing in Chapter 5. 6-11. A 480-V, 100-kW, 50-Hz, four-pole, Y-connected synchronous motor has a rated power factor of 0.85 leading. At full load, the efficiency is 91 percent. The armature resistance is 0.08 & , and the synchronous reactance is 1.0 & . Find the following quantities for this machine when it is operating at full load: (a) Output torque (b) Input power (c) m n (d) E A 163 . ,mA S A ( )( ) E ,m A E ,mA 277 0 V = ° j 1.1 297 20. = 9 V ° 96 & .2 0 A ° This same current comes from the generator,. ) 13.3 = 0 .9 ° 73 = leading , since a current angle of ) ( ) ° = motor 3 ⎞ A sin Q V I ⎝ 3 286 V 95 .7 A sin 13.3 18 .9 kVAR The. consuming from or supplying to the electrical system? How much reactive power Q is this machine consuming from or supplying to the electrical system? S OLUTION This