2 ( ) = = 2 ( ) & = CU 3 A A P I 3 R 79.4 A 0.22 4.16 kW (f) co P nv at rated conditions is = = = conv IN C P P U 83 P .8 kW 4.16 kW 79.6 kW (g) If E A is decreased by 10%, the new value if E A = (0.9)(603 V) = 543 V. To simplify this part of the problem, we will ignore R A . Then the quantity E A sin ™ will be constant as E A changes. Therefore, sin E 1 A sin sin 603 V ™ ™ sin ( ) 1 1 = = 19.5 21.8 ° = ° 2 1 Therefore, I E A2 ⎞ A V E 440 0 V ° 543 V 5 43 2 1.8 ° 70.5 17.7 A = = = ° A 3.0 jX S j and the reactive power supplied by the motor to the power system will be 3 sin 3 ( ) 440 V ( 7 = = 0.5 A ) sin ( 17.7 ) 28.3 kVA ° R = ⎞ ⎝ A Q V I 6-16. Answer the following questions about the machine of Problem 6-15. (a) If E A = 430 13.5 ° V and ⎞ V = 440 0 ° V, is this machine consuming real power from or supplying real power to the power system? Is it consuming reactive power from or supplying reactive power to the power system? (b) Calculate the real power P and reactive power Q supplied or consumed by the machine under the conditions in part (a) . Is the machine operating within its ratings under these circumstances? (c) If E A = 470 -12 ° V and ⎞ V = 440 0 ° V, is this machine consuming real power from or supplying real power to the power system? Is it consuming reactive power from or supplying reactive power to the power system? (d) Calculate the real power P and reactive power Q supplied or consumed by the machine under the conditions in part (c) . Is the machine operating within its ratings under these circumstances? S OLUTION (a) This machine is a generator supplying real power to the power system, because E A is ahead of ⎞ V . It is consuming reactive power because A cos E V ™ < ⎞ . (b) This machine is acting as a generator, and the current flow in these conditions is ° ° A ⎞ E V 430 13.5 440 0 V 34.2 16.5 A = = = ° A I 0.22 3.0 + + A S R jX j The real power supplied by this machine is 3 cos 3 ( ) 440 V ( 3 = = 4.2 A ) cos ( 16.5 ) 43. 3 kW ° = ⎞ A P V I ⎝ The reactive power supplied by this machine is 3 sin 3 ( ) 440 V ( 3 = = 4.2 A ) sin ( 16.5 ) 12 .8 kV ° A = R ⎞ ⎝ A Q V I 169 (c) This machine is a motor consuming real power from the power system, because E A is behind ⎞ V . It is supplying reactive power because A cos E V ™ > ⎞ . (d) This machine is acting as a motor, and the current flow in these conditions is ⎞ A V E I 440 0 V ° 4 70 12 ° 33.1 15.6 A = = = ° A 0.22 3.0 + + A S R jX j The real power consumed by this machine is 3 cos 3 ( ) 440 V ( 3 = = 3.1 A ) cos ( 15.6 ) 42.1 kW ° = ⎞ A P V I ⎝ The reactive power supplied by this machine is 3 sin 3 ( ) 440 V ( 3 = = 3.1 A ) sin ( 15.6 ) 11.7 kV ° A = R + ⎞ ⎝ A Q V I 170 . decreased by 10%, the new value if E A = (0.9)(603 V) = 543 V. To simplify this part of the problem, we will ignore R A . Then the quantity E A sin ™ will be constant. power P and reactive power Q supplied or consumed by the machine under the conditions in part (a) . Is the machine operating within its ratings under these circumstances? (c) . power P and reactive power Q supplied or consumed by the machine under the conditions in part (c) . Is the machine operating within its ratings under these circumstances? S OLUTION