Power System Representation and Equations. A one-line diagram of a simple power system

Transcription

1 Power ystem epresenttion nd Equtions Lod B Lod A Bus Bus A oneline digrm of simple power system Oil or liquid iruit reker otting mhine Twowinding power trnsformer Wye onnetion, neutrl ground PerPhse, Per Unit ystem se se se Φ,se L,se L,se se L,se ( ) Φ,se tul vlue Quntity in per unit se vlue of quntity As we hve seen in Chpter, mny trnsformers nd mhines hve their internl impednes speified s per unit resistnes nd retnes, using the voltge nd pprent power rtings of the devie itself s the se quntity. f these impednes re expressed in per unit to se other thn the one seleted s se for power system, we must onvert the impednes to per unit on the se. This onversion ould e done y using the originl se impedne to onvert the impednes k into ohms,

2 nd then using the power system se impedne to onvert the vlue on ohms to per unit on the se. Alterntively, we my omine the two steps into single eqution: Per unit per unit Exmple A simple power system onsisting of one synhronous genertor nd one synhronous motor onneted y two trnsformers nd trnsmission line is shown in the following Figure. Develop perphse, per unit equivlent system for this power system using se pprent power of 00 MA nd se line voltge t genertor G of.8 k. egion egion egion L 5Ω 75Ω Lod A G rting: 00 MA.8 k 0. pu 0.9 pu T rtings: 00 MA.8/0 k 0.0 pu 0.05 pu T rtings: 50 MA 0/4.4 k 0.0 pu 0.05 pu M rting 50 MA.8 k 0. pu. pu olution: se,.8 k egion 0 k egion se, se, 0 k.8 k 4.4 k egion se, se,. k 0 k The orresponding se impednes in eh region re: se, se, ( ) (.8 k) L,se φφse 00 MA ( ) ( 0 k) L,se φφse.904 Ω Ω 00 MA egion egion

3 se, ( ) L,se (. k) φφse 00 MA.74 Ω egion G,pu G,pu T,pu T,pu 0.per unit 0.9 per unit 0.0 per unit 0.05 per unit line, system line, system 5 Ω 0.4 per unit Ω 75 Ω 0.60 per unit Ω Per unit per unit T,pu T,pu ( 0.0) 4.4 k 00 MA 0.8 per unit. k 50 MA 4.4 k. k 00 MA 50 MA ( 0.05) 0.9 per unit Per unit M,pu M,pu ( 0.0) per unit.8 k 00 MA 0.9 per unit. k 50 MA.8 k 00 MA. k 50 MA ( 0.05).405 per unit

4 0.0 j0.05 pu 0. j0.9 pu 0.4 j0.6 pu 0.0 j0.9 pu 0.9 j.405 pu G M Perphse, per unit equivlent iruit of the simple power system. Writing Node Equtions for Equivlent Ciruit One the perphse, per unit equivlent iruit of power system is reted, it my e used to find the voltges, urrents, nd powers present t vrious points in power system. The most ommon tehnique used to solve suh iruits is nodl nlysis. n nodl nlysis, we use Kirhhoff s urrent lw equtions to determine the voltges t eh node (eh us) in the power system, nd then using the resulting voltges to lulte the urrents nd power flows t vrious points in the power system, nd then use the resulting voltges to lulte the urrents nd power flows t vrious points in the system.consider the following simple threephse power system ontining three usses onneted y three trnsmission lines. The system inludes genertor onneted to us, lod onneted to us, nd motor onneted to us.

5 Line T T G M Line Line T T 6 T 4 T 5 Lod A simple threephse power system

6 um of urrents out of node sum of urrents into the node Apply KCL on node ( ) ( ) d Apply KCL on node ( ) ( ) e Apply KCL on node ( ) ( ) f errnge these equtions to ollet the terms in eh voltge ( ) ( ) ( ) _ f e d The ove eqution n e expressed in mtrix form f e d This eqution is of the form us Where us is the self dmittne of system. d e n f

7 There re mny wys for solving systems of simultneous liner equtions, suh s sustitution, Gussin elimintion, LU ftoriztion, nd so forth! For us, MATLAB hs very effiient eqution solvers uilt diretly into it! f system of n simultneous liner equtions in n unknowns n e expressed in the form: Ax Where A is n n mtrix, nd is nelement olumn vetor. t my e expressed s Where A is the inverse of mtrix! x A