Consider the simple circuit of Figure 1 in which a load impedance of r is connected to a voltage source. The no load voltage of r

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1 1 Intductin t Pe Unit Calculatins Cnside the simple cicuit f Figue 1 in which a lad impedance f L 60 + j70 Ω Ω is cnnected t a vltage suce. The n lad vltage f the suce is E The intenal esistance f the suce is j10 Ω. The ttal impedance f the cicuit is T 60 + j80 Ω. The lad cuent thugh the cicuit is E IL 10A j80 T The appaent pwe pduced by the suce is E IL ,000A 53 1 j10ω E 1000 L 60+ j70ω Figue 1 Basic Cicuit: Example f p.u. calculatins In this cicuit, we will designate the magnitude f each f these values as the efeence value f that quantity, i.e. the efeence value will be ne unit f that quantity. That is, 1000 is equal t ne unit f vltage ( 1000 ) 100 Ω is equal t ne unit f esistance ( 100 Ω) 10 A is equal t ne unit f cuent (I 10 A) 10,000 A is equal t ne unit f pwe ( 10,000 A) These basic units ae called the values. Theefe, the values f u cicuit ae 1000, 100 Ω, 10 A, and 10 ka. 1 I L is the cnjugate f the lad cuent. In this case I L 10A + 53

2 Nw we can efe any value f any quantity in the cicuit t these values. F example, the magnitude f the vltage acss the lad is L 9. L E T In tems f u chsen vltage, the lad vltage is the ati f the actual vltage and L 9 the chsen vltage, i.e f the chsen vltage. We say 1000 that the vltage is 0.9 pe unit f vlts. The designatin is 0.9 p.u. (). We can als say that the lad vltage is 9.% f the vltage. imilaly, in tems f u impedance, the lad impedance is L p.u.( Ω). 100 Just like the magnitude f the lad impedance, we can expess any hmic value in tems f the impedance. the pe unit esistance f the lad is R R L p.u.( Ω) X L 70 The pe unit eactance f the lad is X 0.7 p.u.( Ω) 100 Ntice hee that bth, the esistance and the eactance ae cmpaed t impedance. The pwe dissipated by the esistance is P RI W. Expessed as pe unit value n u f 10,000 A, the dissipated pwe is P P ,000 p.u. 0.6p.u.(W)

3 3 The eactive pwe t the lad is Q L X L I A (ind). Expessed as pe unit value n u f 10,000 A, the lad eactive pwe is Q Q L Lp.u 0.7 p.u.(a) (ind) imilaly as in handling the pe unit values f hmic quantities, bth eal pwe and eactive pwe ae cmpaed t the appaent pwe, usually called MA. All values ae nly magnitude. They ae nt assciated with any angle. The pe unit values, hweve, ae phass. The phase angles f the cuents and vltages and the pwe fact f the cicuit ae nt affected by the cnvesin t pe unit values. The pe unit values can be witten as phass, and all calculatins that can be dne with the actual quantities can be als dne with pe unit values. F example, in u cicuit, the pe unit lad impedance and the pe unit lad cuent ae: p.u.( Ω) and I 1 53 p.u.(a) L Thus the vltage acss the lad is I p.u.() In geneal, the pe unit value is the ati f the actual value and the value f the same quantity. actual value pe unit value value Manufactues give impedance f equipment in pecent n wn. The pecent value is the pe unit value multiplied by 100: % p.u. 100%. The expessin wn means that the vltage is the ated vltage f the equipment, and the pwe is the ated appaent pwe (in A) f the equipment. The cuent and the impedance ae calculated fm the vltage and the A:

4 4 I and The pecent impedance is then in Ω in Ω actual % 100% Geneats and Pe Unit ystem A geneat ated 1000 A and 00 has intenal impedance f j10 Ω. Figue Basic Cicuit: Example f p.u. calculatins f geneats The atings f the geneat ae chsen as the values. In this example, I 1000 A A Ω 1000 The geneat impedance is stamped n the nameplate tgethe with the the atings. This geneat has impedance f j5%. This means that the pe unit impedance is % j5 p.u. j0.5 p.u. ( Ω) The actual impedance in Ω is actual p.u. j j10 Ω

5 5 Example: The geneat abve is sht cicuited at its teminals. Find the sht cicuit cuent and the sht cicuit pwe deliveed by the geneat in p.u., in %, and in the actual units. lutin: Figue 3 Example: ht Cicuit n Geneat Teminals E G p.u. 1 Isc p.u. j4.0 p.u.(a) sc % I j0.5 p.u. I sc p.u. 100% j400 % * sc p.u. E G p.u. Isc p.u. 1 ( j4) j4.0 p.u. (A) sc % sc p.u. 100% j400 % The actual values calculated fm the pe unit values ae: I sc sc I sc p.u. sc p.u. I j j0.0 A j j4000 A Tansfmes and Pe Unit ystem A tansfme is ated 000 A, 00/400, and has an intenal impedance f j4.0 Ω as seen fm the lw vltage side.

6 6 (a) (b) Figue 4 Basic Cicuit: Example f Pe Unit Calculatins f a Tansfme (a) Tansfme Impedance Refeed t the Lw ltage ide (b) Tansfme Impedance Refeed t the High ltage ide The intenal impedance f the tansfme as seen fm the high vltage side is H 400 L j4.0 j16. 0 Ω 1 00 The ated values f pwe and vltage ae used as the s f the calculatins. It means that the vltage is diffeent n each side f the tansfme. Cmpaisn f the s and the pe unit value n bth sides f the tansfme is in Table 1

7 7 I lw vltage side 000 A 00 10A high vltage side 000 A 400 5A p.u. L 0Ω j4.0 0 j0.p.u( Ω) L 80Ω j16 80 j0.p.u( Ω) Table 1 Bases f tansfme lw side and high side Ntice in Table 1 that the tansfme pe unit impedance is the same, egadless f t which side f the tansfme it is efeed. In the tansfme equivalent cicuit the diffeent vltage levels disappea and the tansfme equivalent cicuit is educed t a single impedance: p.u. Figue 5 Equivalent Cicuit f a Tansfme in Pe Unit Analysis Thee Phase ystem and Pe Unit Calculatins In thee phase systems the ttal appaent pwe is given by 3φ 3 LLI L whee LL is the line-t-line vltage and I L is the line cuent.

8 8 imilaly, φ whee 3 3 L LI L 3φ is the chsen thee phase pwe, usually the ated pwe f the equipment, the chsen pwe used f all system calculatin by the pwe utility LL is the chsen line-t-line vltage, usually the ated vltage f the equipment the nminal vltage level f a pat f the pwe system I L is the line cuent, usually calculated fm the pwe and the vltage. Δ and Y Cnnectin in Pe Unit Calculatins I L I Δ Δ Y Figue 6 - Δ - Y Cnvesin The values ae elated thugh the same elatinships as the actual quantities: LL 3 LN LN Δ 3 3 I I I Δ L 3 L Y Y The pe unit values f the Δ cnnected impedances and the Y cnnected impedances ae Δ p.u. Δ Δ Y p.u. Y Y

9 9 Fm that it is easy t shw that the pe unit value f the Δ cnnectin is the same as the pe unit value f the Y cnnectin. 3 Δ Y Y Δ p.u. Δ 3Y Y Y p.u. Change f Base When pieces f equipment with vaius diffeent atings ae cnnected t a system, it is necessay t cnvet thei impedances t a pe unit value expessed n the same. The that we ae cnveting fm will be dented by subscipt M, the we ae cnveting t will be dented by subscipt N. The impedance f the s M and N ae, espectively, M M M N N N The pe unit impedances n the s M and N ae, espectively M p.u. M N p.u. N whee is the actual hmic value f the impedance f the equipment. It fllws that M p.u. M N p.u. N ubstituting f the impedances we get

10 10 M p.u. M M N p.u. N N N p.u. M p.u. N M M N Using the MA and k ntatin, N p.u. M p.u. MA MA N M (k (k M N ) )