EE320 Homework # 9 Synchronous Machines SOLUTION Problem 1: A 4-pole, 60Hz, Y-connected, 3-phase generator has a regulated terminal line-to-neutral voltage of Vas and = 260Vr ms ∠0 , a synchronous reactance of 0.06Ω , a stator resistance of 0.003Ω , Lsf = 0.02H . The balanced 3-phase load draws 2MW at a lagging 0.8 power factor. a. b. c. d. e. Compute the 3-phase complex power of the load (magnitude and angle) Compute the generator phase current (magnitude and angle; hint: the angle of the current should be negative for the load to have a lagging power factor) Find the required excitation voltage ( E a ) Calculate the required field current Calculate the required speed of the prime mover Problem 2: Given the same machine as in problem 13.1. If the excitation voltage is E a = 360Vrms∠14 and the field current is adjusted to maintain the same terminal voltage as in Problem 13.1, a. b. c. Find the new generator phase current Calculate the 3-phase real power consumed by the load and identify the new power factor Calculate the generator efficiency (consider stator winding losses as the only losses) Problem 3: A 3-phase, Y-connected, 6-pole, 60Hz generator has resistance), and Lsf = 0.2H . The machine operates at a power angle of 31.48 with a field current of 100A. The line-to-neutral terminal voltage is 2400Vrms. a. b. c. d. e. X S = 1Ω , negligible rS (stator Find the speed of the prime mover in rad/sec Determine the excitation voltage Find the line current Find the power consumed by the load and its power factor Find the developed torque E£3d-0 + LoAD ... w :E 0 0 K s-= o.oG ..n._ f.� ::: 0. Q() 1 .D­ 'S = �- !;MVA Q: :- l lof (o.�): '3b.�? f\ � � a. N\ 'v\} Lsf = 0. 0d- 1-J p34 t� '?JC,, i7 l. 5 AA.VAR -;;...�Kio' /_3b.'67 0 3 ( -:2,0 Lo I T¥s -::: � :l.o� [3,.�7 � � �- Q ° J A 3���: L- ����7<� -::a. (;' Q,, -> - ' :::: � J a. 5 ( � 'x_ S) - lQ.s (rs) - � � = 0 Fa_ :: (i�o� l1G.�7) (Jo.Ob) +(3:to�[-3(8'7')( 0,00'3) 7 tc:,.. - 4-1 o. 7 [�I, I O V - �olo 0 Pro\,f� I J) Fitvt> ( COk'T '()) a lF l;t Ii We - {"�(4-10 .?') ( o.oi') ( d-n ,c') ! :: 7 7.03 Al EF 3)..o � o. o� S"l. o. 001 ..c1.. (/J (/J (/J WWW a:a:a: ggg� <( <( <( a: U') IO IO LL I I I I (/J ��� ti:iwww wwww :c:c:c:c (/J Cl) Cl) (/J g888 I I I I ... (II (II l:f:l � t,, t,, (c._ - j0.0(; (fc..s J ���9 C') C') C') C') - Off)3 (L_s) (.) f�r\J fscL /' � Cc>� loS�S :::: o � '31(> � -= 3 � � cp . C-J �5 = \:)o7b._::� :4�:.:;��] :E 0 b) t=-\ (\,� - ---J,.. � 5 I� ct. Cl.� - 3 ( �'--0 L 0 ) (d-o7C:..'-t L l.fd.., i�i) ) ° 1� d-. 0 ::: "3 ) Ta,, l C · r�f\ :: �ovt '=-: 00:.') P, f\ - :: Ps"''- s "6 • <J t:. \JJ I · ''i N\. \V + 3 �' � f:.W -:::. \ , � 3 fl\ \I\J I 'f r,I. w_ lh � I. 1 / .�3 IJAIJ} ::. 1(, , Z . J _n_ �) w t=: rv� w� = t-we =-%- 1- ::e 0 (..) lJ .,_ = b) F�nJ w t(\ ___:::. E °'" lf:1 -= L�Jw• - (dffbo) � l)_S-.7 � (o.a.J<;;o')(d-:rT,o)::: S-331.� lt:: � ��-�--L ��· �0-�] 3 ......, � � 0 '3 \r�� 1<:\_; == '3 ( �lfoo Lo ) {ss1 r;, 7 !._s?� 6 °') �VA ::. �6-. 31 L'37, ''f 'S3d) � O © .. r,)o / oSS2 � 1111 :I 0 Is =-o / (.) _..... p3cf; _ ';;)_Q, 0� '{IO " l �S,7 '?;1,ev = 5 \fo.., 'f .._ S: ,J W M 'X s -..... "- 5 ( d--LfOO )( D 3 I. 5) S� V\ ( 1 I, Lf � u}