ansT1 (1)

ansT1 (1) - Module T1 Solution to problem 1 Two ways to...

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Module T1 Solution to problem 1 Two ways to work it. 1. Find pq P and ' pq P then ' pq pq loss P P P - = where ' pq P = qp P - 2. Find pq I and then R I P pq loss 2 3 = We will do #2. ( 29 ( 29 MW MW P j jX R V V I loss q q p p pq 3 61 . 0 183 . 0 30 45 3 26 . 27 45 150 30 0 06 . 65 0 . 6 51 . 63 10 2 3 × = = = ° = + ° - ° × = + - = θ #1: 666 . 2 = pq P 605 . 2 ' = pq P 061 . 0 / = φ loss P Solution to problem 2 ( 29 ( 29 ( 29 [ ] [ ] MW B V V P q p q p pq 885 . 2 10472 . 0 0067 . 0 10 06 . 65 10 51 . 63 3 3 = × × = - = ( 29 ( 29 [ ] [ ] MVAR V V B V Q q p p pq 6566 . 0 10 06 . 65 51 . 63 0067 . 0 10 51 . 63 3 3 - = × - × = - = ( 29 ( 29 MVAR B V Q c p sp 4034 . 0 0001 . 0 10 51 . 63 2 2 3 2 - = × - = - = So, MVAR Q Q Q sp pq 06 . 1 4034 . 0 6566 . 0 - = - - = + = MVAR Q MW P 18 . 3 06 . 1 3 , 655 . 8 885 . 2 3 - = × = = × = p q p q Q P , Q P , s p Q P Q
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Solution to problem 3 80 . 19 98 . 1 1 j jX R jB G - = + = - (a) P pq = ( 29 ( 29 q p q p q p q p p B V V G V V G V φ - + - - sin cos 2 = ) 10 sin( ) 80 . 19 )( 95 . 0 ( 15 . 1 ( ) 10 cos( ) 98 . 1 )( 95 . 0 )( 15 . 1 ( ) 98 . 1 ( 15 . 1 2 + - = 4.2445 So real power into line = 100(4.245) = 424.5 MW P' pq =-P pq = )] sin( ) cos( [ 2 p q p q p q p q q B V V G V V G V - + - - - = )] 10 sin( ) 8 . 19 )( 95 . 0 )( 15 . 1 ( ) 10 cos( ) 98 . 1 )( 15 . 1 )( 95 . 0 ( ) 98 . 1 ( 95 [. 2 - + - - - = 4.0996 So real power out of the line = 100(4.0996) = 409.96 MW (b) P loss = P pq -P' pq =424.5-409.96= 14.49 MW (c) F = (0.03 / kwhr)(14.49MW)(1000kW/MW) = $434.70 / hr (d) 5 . 424 96 . 409 = = in out P P γ = 96.6 % Solution to problem 4 (a) What are the typical voltage levels for bulk transmission lines? 138, 101, 230, 345, 500, 765
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This note was uploaded on 09/18/2011 for the course EE 303 taught by Professor Mccalley during the Spring '08 term at Iowa State.

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ansT1 (1) - Module T1 Solution to problem 1 Two ways to...

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