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Symmetrical components
- 1. Lec3:Symmetrical components Prepared by : Eng/Zaki Mohammed Al Shadaddi
- 2. Symmetrical components Type of fault: - Symmetrical fault - unsymmetrical fault (L-L-L) (L-L,L-L-G,L-G) (value of V&I will change and value of V&I will change remain in balance condition) • For balance system and symmetrical component , if we want to find the electrical quantities (V&I) of system , we will use per phase model of system • For unbalance system we cant use per phase model of system , we will write equation for each phase which will take more time , so we will use symmetrical components
- 3. Symmetrical components set of vector / component Unbalanced ͢͢ balance condition N phase = (n-1)set of n-phase + one set of n phase For unbalanced system balance phasor with different rotation Ex :3Ф,3wire/4wire Phase = 2 set of 3-phase + one set of cophasial phasor -3 Unbalance condition balance phasor ↓ ↓ - positive sequence zero sequence component - negative sequence component
- 4. Symmetrical components 3-phase unbalance system(a ,b ,c ) represented Positive sequence 1 Negative sequence 2 Zero sequence 0
- 6. Symmetrical components As a 3-phase unbalance condition 3 voltage variable 3 current variable Due to this there will be 9 variable , so we will use operator a which reduce the variable from 9 to 3
- 9. Symmetrical components Unbalance voltage and current in term of symmetrical components
- 10. Symmetrical components Unbalance voltage and current in term of symmetrical components
- 13. Symmetrical components (i) 𝐼120 = [𝐴]−1[𝐼 𝑎𝑏𝑐 ] 𝐼 𝑎1 𝐼 𝑎2 𝐼 𝑎0 = 1 3 1 𝛼 𝛼2 1 𝛼2 𝛼 1 1 1 𝐼 𝑎 𝐼 𝑏 𝐼 𝑐 𝐼 𝑎1= 1 3 (𝐼 𝑎+ 𝛼 𝐼 𝑏+𝛼2 𝐼𝑐)= 1 3 (100∟30 +1 ∟120*50∟300 +1 ∟240*30∟180)= 57.98 ∟43.3 A 𝐼 𝑎2= 1 3 (𝐼 𝑎+ 𝛼2 𝐼 𝑏+𝛼𝐼𝑐 ) = 1 3 (100∟30 + 1 ∟240*50∟300 + 1 ∟120*30∟180 ) = 18.97 ∟24.96 A 𝐼 𝑎0= 1 3 (𝐼 𝑎+ 𝐼 𝑏+𝐼𝑐) = 1 3 (100∟30 +50∟300 +30∟180 ) =27.29 ∟4.69 A
- 14. • (ii) current in neutral circuit :- 𝐼 𝑛=𝐼 𝑎+ 𝐼 𝑏+𝐼𝑐 = 𝐼 𝑎1+𝐼 𝑎2 +𝐼 𝑎0 + 𝐼 𝑏1+ 𝐼 𝑏2 + 𝐼 𝑏0 + 𝐼𝑐1+ 𝐼𝑐2 + 𝐼𝑐0 = 𝐼 𝑎0 + 𝐼 𝑏0 + 𝐼𝑐0 = 3 𝐼 𝑎0 = 3*27.29 ∟4.69 A = 81.877 ∟4.69A
- 15. Symmetrical components • Relationship between line and phase voltage of sequence components :-
- 17. Symmetrical components = − = − 𝛼 =(1−𝛼) = 3 ∟-30 = − =0 Conculation :- 1) +ve line voltage ( ) = 3 times of +ve phase voltage ( ) and lead by 30 degree . 2) −ve line voltage ( ) = 3 times of -ve phase voltage ( ) and lag by 30 degree . 3) No zero sequence component exist for line voltage .
- 18. Symmetrical components Relationship between line and phase current(delta) of symmetrical component: • For balance condition I= 3 𝐼𝑝ℎ ∠ − 30 • Line current lag phase current by 30(deg) • Abnormal Condition • Line current • Ia= 𝐼𝑎1 + 𝐼𝑎2 + 𝐼𝑎0 • Ib= 𝐼𝑏1 + 𝐼𝑏2 + 𝐼𝑏0 • Ic= 𝐼𝑐1 + 𝐼𝑐2 + 𝐼𝑐0
- 21. Symmetrical components Conculation:for eq(1) +ve sequence:- Line current (Ia1)=√3 time of +ve phase current (Iab1) and lag by30 -ve sequence:- (Ia2)=√3 time of –ve phase current (Iab2)and lead by 30 No zero sequence component in the line current
- 29. Symmetrical components • 1-symmetrical/sequence component of line current? • 2-symmetrical/sequence component of phase current? • 3- if the value of R=10 then find the unbalanced phase/line voltage? • 4-symmetrical component of line /phase voltage ?
- 32. Symmetrical components Iab=Iab1+Iab2+Iab0=8.3299∠-6.9 A Ibc=aᶺ2Iab1+aIab2+Iab0=10.542∠138.41 A Ica=aIab1+𝑎ᶺ2Iab2+Iab0= 6.0092∠-93.66 A Step4 :- unbalance phase voltage Vab= Iab ∗ Zab =83.299∠-6.9 v vbc= Ibc*Zbc =105.42∠138.41 v Vca= Ica ∗ Zca=60.098∠-93.66 v Step5 :- sequence component of phase /line voltage Step5 :-sequence(symmetrical)component of phase /line voltage :- 𝑉𝑎𝑏1 𝑉𝑎𝑏2 𝑉𝑎𝑏0 =1/3 1 𝑎 𝑎ᶺ2 1 𝑎ᶺ2 𝑎 1 1 1 𝑉𝑎𝑏 𝑉𝑏𝑐 𝑉𝑐𝑎 𝑉𝑎𝑏1=1/3[𝑉𝑎𝑏+a𝑉𝑏𝑐+𝑎ᶺ2𝑉𝑏𝑐]=26.931 ∠-81.8 v 𝑉𝑎𝑏2=1/3[𝑉𝑎𝑏+𝑎ᶺ2𝑉𝑏𝑐+𝑎𝑉𝑐 𝑎]=80.597 ∠11.922 v 𝑉𝑎𝑏0=1/3[𝑉𝑎𝑏 + 𝑉𝑏𝑐+𝑉𝑐 𝑎]=4.709*10^3 ∠-12.49 v
- 33. Symmetrical components Expression of complex power in term of symmetrical component :- Suppose 3 phase system : S=P+jQ = V 𝐼∗ 𝑆3𝑝ℎ = 3 𝑉𝑝ℎ 𝐼∗ 𝑝ℎ for balanced condition = 3𝑉𝐿 𝐼∗ 𝐿 For unbalance condition :- 𝑆3𝑝ℎ = 𝑉𝑎 𝐼 𝑎 ∗ + 𝑉𝑏 𝐼 𝑏 ∗ + 𝑉𝑐 𝐼𝑐 ∗ = [ 𝑉𝑎 𝑉𝑏 𝑉𝑐 ] 𝐼 𝑎 ∗ 𝐼 𝑏 ∗ 𝐼𝑐 ∗ = 𝑉𝑎 𝑉𝑏 𝑉𝑐 𝑇 𝐼 𝑎 ∗ 𝐼 𝑏 ∗ 𝐼𝑐 ∗ = [ 𝑉𝑎 𝑏𝑐] 𝑇 [ 𝐼 𝑎 𝑏𝑐]∗
- 35. Symmetrical components Ex :- star configuration with line voltage =100∟0 v , =80.8 ∟-121.44 v , =90∟130 v =10Ώ find the power exponded in the resistance Sol :- method (1):- 𝑆3𝑝ℎ = 𝑉𝑎 𝐼 𝑎 ∗ + 𝑉𝑏 𝐼 𝑏 ∗ + 𝑉𝑐 𝐼𝑐 ∗ For last example as we found , , 𝑆3𝑝ℎ= P + j Q = P+j0 as a resistance load 𝑆3𝑝ℎ= 327.664+275.41504+212.3366 =815.59 w method (2):- 𝑆3𝑝ℎ = 3*[ 𝑉𝑎1 𝐼 𝑎1 ∗ + 𝑉𝑎2 𝐼 𝑎2 ∗ + 𝑉𝑎0 𝐼 𝑎0 ∗ ]=815.59w method (3) :- P= 𝑃𝑎+ 𝑃𝑏+ 𝑃𝑐= 𝑉𝑎𝑛 2 𝑅 𝑎 + 𝑉 𝑏𝑛 2 𝑅 𝑏 + 𝑉𝑐𝑛 2 𝑅 𝑐 method (4):-𝑃3𝑝ℎ = 𝐼 𝑎 2 𝑅 𝑎 + 𝐼 𝑏 2 𝑅 𝑏 + 𝐼𝑐 2 𝑅 𝑐
- 36. Symmetrical components 26 Ex :- = 0.9 pu, = -0.2 pu, = -0.1 pu Base voltage is 11 kv (L.L) , find 1) in pu and kv 2) L.L in pu and kv Sol :- = + + = 0.9 – 0.2 – 0.1 = 0.6 pu = * = 0.6 * 11 3 = 3.811 kv = - = - - - = - 𝛼2 - 𝛼 - = 0.6 - 1 ∟240 *0.9 +1 ∟120*0.2 + 0.1 = 0.6 – 1.054 ∟-115.28 = 1.418 ∟42.226 pu = - = ( - =1.418 ∟42.226 * 11 3 = 9.0062 kv ∟42.23
- 37. Symmetrical components • Sequence impedance of transmission line By kvl at node a: 𝑉𝑎 − 𝑉𝑎′= 𝐼 𝑎* 𝑍 𝑠 + 𝐼 𝑛 * 𝑍 𝑛 + 𝐼 𝑏* 𝑍 𝑚+ 𝐼𝑐* 𝑍 𝑚 = 𝐼 𝑎[ 𝑍 𝑛+ 𝑍 𝑠]+ 𝐼 𝑏[ 𝑍 𝑚+ 𝑍 𝑛]+ 𝐼𝑐[ 𝑍 𝑚+ 𝑍 𝑛]→(1) By kvl at node b: 𝑉𝑏 − 𝑉𝑏′= 𝐼 𝑏* 𝑍 𝑠 + 𝐼 𝑛* 𝑍 𝑛+ 𝐼 𝑎* 𝑍 𝑚+ 𝐼𝑐* 𝑍 𝑚 = 𝐼 𝑎[ 𝑍 𝑚+ 𝑍 𝑛]+ 𝐼 𝑏[ 𝑍 𝑚 + 𝑍 𝑠]+ 𝐼 𝐶 [ 𝑍 𝑚+ 𝑍 𝑛]→(2)