HAKAN ALTIN ELEKTRİK ve ELEKTRONİK MÜHENDİSİ

*In the calculation of reactance values: HV underground cables are single and laid underground at 7 cm distance, and in overhead lines, flatcross arm and dm = 1.1 m are taken.

Upstream Network :Z_{Upstream
Network}=100/Sk pu , X_{Upstream
Network}=0,98.Z_{Upstream
Network
,}R_{Upstream
Network}=0,1.X_{Upstream
Network}

-Three Phase Short Circuit Calculations :

X₁(1)=X_s+X_tr(1)+X_h1(1) =0,0681+0,1462+0,0011=0,2154 pu

R₁(1)=R_s+R_tr(1)+R_h1(1) =0,0069+0+0,0004=0,0073 pu

Z₁(1)=R₁(1)+j X₁(1) =0,0073+ j 0,2154=0,2155 pu

I_k3''=6,803 kA

-Phase Ground short circuit calculations

X₀(1)=X_s0+X_tr0(1)+X_h0(1) =0,1363+0,1462+0,0044=0,2869 pu

R₀(1)=R_s0+R_tr0(1)+3.Rn+R_h0(1) =0,0069+0+9,2214+0,0036=9,2319 pu

Z₀(1)=R₀(1)+j X₀(1) =9,2319+ j 0,2869=9,2364 pu

Z₀(1)/Z₁(1) =9,2364/0,2155=42,86 > 1 the phase-to-ground short circuit current is greater than the two phase-to-ground fault currents.

I_k1t''= I_k3''_x 3 ₌ 6,803 _x 3 ₌0,455 kA

2₊ Z₀ 2+ 42,86

Z₁

Three phase short circuit current is greater than phase-to-ground short circuit current

-Peak Asymmetrical Current Calculations

IEC 60909 standardına göre tepe akımı:

k₁=1,02+0,98 e^(-3R/x)=1,91

I_p1= k.√2 . I_k3'' =18,376 kA

-Thermal Short Circuit Current

At a point far from the generator I_k3''/I_k3=1 and n=1

m₌ 1 [ e^{4.f.Tk.ln(K-1)}-1]= 0,106

2.f.Tk.Ln(K-1)

I_th=I_k3''.√(m+n) = 6,803

Breaker Power : S_ksc=√3.Un.1,1.I_k3''₌√3.36.1,1.6,803= 467 MVA

-Three Phase Short Circuit Calculations :

X₁(2)=X_s+X_tr(1)+X_h1(1)+X_h1(2)=0,0681+0,1462+0,0011+0,0686=0,284 pu

R₁(2)=R_s+R_tr(1)+R_h1(1)+R_h1(2) =0,0069+0+0,0004+0,0271=0,0344 pu

Z₁(2)=R₁(2)+j X₁(2) =0,0344+ j 0,284=0,2861 pu

I_k3''=5,124 kA

-Phase Ground short circuit calculations

X₀(2)=X_s0+X_tr0(1)+X_h0(1)+X_h0(2) =0,1363+0,1462+0,0044+0,2744=0,5613 pu

R₀(2)=R_s0+R_tr0(1)+3.Rn+R_h0(2) =0,0069+0+9,2214+0,0036+0,2168=9,4487 pu

Z₀(2)=R₀(2)+j X₀(2) =9,4487+ j 0,5613=9,4654 pu

Z₀(2)/Z₁(2) =9,4654/0,2861=33,08 > 1 the phase-to-ground short circuit current is greater than the two phase-to-ground fault currents.

I_k1t''= I_k3''_x 3 ₌ 5,124 _x 3 ₌0,438 kA

2₊ Z₀ 2+ 33,08

Z₁

Three phase short circuit current is greater than phase-to-ground short circuit current

-Peak Asymmetrical Current Calculations

IEC 60909 standardına göre tepe akımı:

k₂=1,02+0,98 e^(-3R/x)=1,7

I_p2= k.√2 . I_k3'' =12,319 kA

-Thermal Short Circuit Current

At a point far from the generator I_k3''/I_k3=1 and n=1

m₌ 1 [ e^{4.f.Tk.ln(K-1)}-1]= 0,028

2.f.Tk.Ln(K-1)

I_th=I_k3''.√(m+n) = 5,124

Breaker Power: S_ksc=√3.Un.1,1.I_k3''₌√3.36.1,1.5,124= 351 MVA

-Three Phase Short Circuit Calculations :

X₁(3)=X_s+X_tr(1)+X_h1(1)+X_h1(2)+X_h1(3)=0,0681+0,1462+0,0011+0,0686+0,0365=0,3205 pu

R₁(3)=R_s+R_tr(1)+R_h1(1)+R_h1(2)+R_h1(3) =0,0069+0+0,0004+0,0271+0,0144=0,0488 pu

Z₁(3)=R₁(3)+j X₁(3) =0,0488+ j 0,3205=0,3242 pu

I_k3''=4,522 kA

-Phase Ground short circuit calculations

X₀(3)=X_s0+X_tr0(1)+X_h0(1)+X_h0(2)+X_h0(3)=0,1363+0,1462+0,0044+0,2744+0,146=0,7073 pu

R₀(3)=R_s0+R_tr0(1)+3.Rn+R_h0(1)+R_h0(2) +R_h0(3) =0,0069+0+9,2214+0,0036+0,2168+0,1152=9,5639 pu

Z₀(3)=R₀(3)+j X₀(3) =9,5639+ j 0,7073=9,59 pu

Z₀(3)/Z₁(3) =9,59/0,3242=29,58 > 1 the phase-to-ground short circuit current is greater than the two phase-to-ground fault currents.

I_k1t''= I_k3''_x 3 ₌ 4,522 _x 3 ₌0,43 kA

2₊ Z₀ 2+ 29,58

Z₁

Three phase short circuit current is greater than phase-to-ground short circuit current

-Peak Asymmetrical Current Calculations

Peak current according to IEC 60909 standard:

k₃=1,02+0,98 e^(-3R/x)=1,64

I_p3= k.√2 . I_k3'' =10,488 kA

-Thermal Short Circuit Current

At a point far from the generator I_k3''/I_k3=1 and n=1

m₌ 1 [ e^{4.f.Tk.ln(K-1)}-1]= 0,0224

2.f.Tk.Ln(K-1)

I_th=I_k3''.√(m+n) = 4,522

Breaker Power : S_ksc=√3.Un.1,1.I_k3''₌√3.36.1,1.4,522= 310 MVA

-Three Phase Short Circuit Calculations :

X₁(4)=X_s+X_tr(1)+X_h1(1)+X_h1(2)+X_h1(3)+X_h1(4)=0,0681+0,1462+0,0011+0,0686+0,0365++0,0269=0,3474 pu

R₁(4)=R_s+R_tr(1)+R_h1(1)+R_h1(2)+R_h1(3)+R_h1(4) =0,0069+0+0,0004+0,0271+0,0144+0,0106=0,0594 pu

Z₁(4)=R₁(4)+j X₁(4) =0,0594+ j 0,3474=0,3524 pu

I_k3''=4,16 kA

-Phase Ground short circuit calculations

X₀(4)=X_s0+X_tr0(1)+X_h0(1)+X_h0(2)+X_h0(3)+X_h0(4)=0,1363+0,1462+0,0044+0,2744+0,146+0,1076=0,8149 pu

R₀(4)=R_s0+R_tr0(1)+3.Rn+R_h0(1)+R_h0(2) +R_h0(3)+R_h0(4) =0,0069+0+9,2214+0,0036+0,2168+0,1152+0,0848=9,6487 pu

Z₀(4)=R₀(4)+j X₀(4) =9,6487+ j 0,8149=9,6831 pu

Z₀(3)/Z₁(3) =9,6831/0,3524=27,48 > 1 the phase-to-ground short circuit current is greater than the two phase-to-ground fault currents.

I_k1t''= I_k3''_x 3 ₌ 4,522 _x 3 ₌0,423 kA

2₊ Z₀ 2+ 27,48

Z₁

Three phase short circuit current is greater than phase-to-ground short circuit current

-Peak Asymmetrical Current Calculations

Peak current according to IEC 60909 standard:

k₄=1,02+0,98 e^(-3R/x)=1,61

I_p4= k.√2 . I_k3'' =9,472 kA

-Thermal Short Circuit Current

At a point far from the generator I_k3''/I_k3=1 and n=1

m₌ 1 [ e^{4.f.Tk.ln(K-1)}-1]= 0,0202

2.f.Tk.Ln(K-1)

I_th=I_k3''.√(m+n) = 4,16

Breaker Power : S_ksc=√3.Un.1,1.I_k3''₌√3.36.1,1.4,16= 285 MVA

-Three Phase Short Circuit Calculations :

X₁(5)=X_s+X_tr(1)+X_h1(1)+X_h1(5)=0,0681+0,1462+0,0011+0,192=0,4074 pu

R₁(5)=R_s+R_tr(1)+R_h1(1)+R_h1(5) =0,0069+0+0,0004+0,075=0,0823 pu

Z₁(5)=R₁(5)+j X₁(5) =0,0823+ j 0,4074=0,4156 pu

I_k3''=3,527 kA

-Phase Ground short circuit calculations

X₀(5)=X_s0+X_tr0(1)+X_h0(1)+X_h0(5)=0,1363+0,1462+0,0044+0,6144=0,9013 pu

R₀(5)=R_s0+R_tr0(1)+3.Rn+R_h0(1)+R_h0(5) =0,0069+0+9,2214+0,0036+0,225=9,4569 pu

Z₀(5)=R₀(5)+j X₀(5) =9,4569+ j 0,9013=9,4998 pu

Z₀(5)/Z₁(5) =9,4998/0,4156=22,86 > 1 the phase-to-ground short circuit current is greater than the two phase-to-ground fault currents.

I_k1t''= I_k3''_x 3 ₌ 4,522 _x 3 ₌0,426 kA

2₊ Z₀ 2+ 22,86

Z₁

Three phase short circuit current is greater than phase-to-ground short circuit current

-Peak Asymmetrical Current Calculations

Peak current according to IEC 60909 standard:

k₅=1,02+0,98 e^(-3R/x)=1,55

I_p5= k.√2 . I_k3'' =7,731 kA

-Thermal Short Circuit Current

At a point far from the generator I_k3''/I_k3=1 and n=1

m₌ 1 [ e^{4.f.Tk.ln(K-1)}-1]= 0,0167

2.f.Tk.Ln(K-1)

I_th=I_k3''.√(m+n) = 3,527

Breaker Power: S_ksc=√3.Un.1,1.I_k3''₌√3.36.1,1.3,527= 242 MVA

-Three Phase Short Circuit Calculations :

X₁(6)=X_s+X_tr(1)+X_h1(1)+X_h1(5)+X_h1(6)=0,0681+0,1462+0,0011+0,192+0,1407=0,5481 pu

R₁(6)=R_s+R_tr(1)+R_h1(1)+R_h1(5)+R_h1(6) =0,0069+0+0,0004+0,075+0,055=0,1373 pu

Z₁(6)=R₁(6)+j X₁(6) =0,1373+ j 0,5481=0,565 pu

I_k3''=2,595 kA

-Phase Ground short circuit calculations

X₀(6)=X_s0+X_tr0(1)+X_h0(1)+X_h0(5)+X_h0(6)=0,1363+0,1462+0,0044+0,6144+0,4502=1,3515 pu

R₀(6)=R_s0+R_tr0(1)+3.Rn+R_h0(1)+R_h0(5)+R_h0(6) =0,0069+0+9,2214+0,0036+0,225+0,205=9,6619 pu

Z₀(6)=R₀(6)+j X₀(6) =9,6619+ j 1,3515=9,756 pu

Z₀(6)/Z₁(6) =9,756/0,565=17,27 > 1 the phase-to-ground short circuit current is greater than the two phase-to-ground fault currents.

I_k1t''= I_k3''_x 3 ₌ 4,522 _x 3 ₌0,404 kA

2₊ Z₀ 2+ 17,27

Z₁

Three phase short circuit current is greater than phase-to-ground short circuit current

-Peak Asymmetrical Current Calculations

Peak current according to IEC 60909 standard:

k₅=1,02+0,98 e^(-3R/x)=1,48

I_p5= k.√2 . I_k3'' =5,431 kA

-Thermal Short Circuit Current

At a point far from the generator I_k3''/I_k3=1 and n=1

m₌ 1 [ e^{4.f.Tk.ln(K-1)}-1]= 0,0136

2.f.Tk.Ln(K-1)

I_th=I_k3''.√(m+n) = 2,595

Breaker Power : S_ksc=√3.Un.1,1.I_k3''₌√3.36.1,1.2,595= 178 MVA

-Three Phase Short Circuit Calculations :

X₁(7)=X_s+X_tr(1)+X_h1(1)+X_h1(5)+X_h1(6)+X_h1(7)=0,0681+0,1462+0,0011+0,192+0,1407+0,0831=0,6312 pu

R₁(7)=R_s+R_tr(1)+R_h1(1)+R_h1(5)+R_h1(6)+R_h1(7)=0,0069+0+0,0004+0,075+0,055+0,0325=0,1698 pu

Z₁(7)=R₁(7)+j X₁(7) =0,1698+ j 0,6312=0,6536 pu

I_k3''=2,243 kA

-Phase Ground short circuit calculations

X₀(7)=X_s0+X_tr0(1)+X_h0(1)+X_h0(5)+X_h0(6)+X_h0(7)=0,1363+0,1462+0,0044+0,6144+0,4502+0,2659=1,6174 pu

R₀(7)=R_s0+R_tr0(1)+3.Rn+R_h0(1)+R_h0(5)+R_h0(6)+R_h0(7) =0,0069+0+9,2214+0,0036+0,225+0,205++0,1825=9,8444 pu

Z₀(7)=R₀(7)+j X₀(7) =9,8444+ j 1,6174=9,9764 pu

Z₀(7)/Z₁(7) =9,9764/0,6536=15,26 > 1 the phase-to-ground short circuit current is greater than the two phase-to-ground fault currents.

I_k1t''= I_k3''_x 3 ₌ 4,522 _x 3 ₌0,39 kA

2₊ Z₀ 2+ 15,26

Z₁

Three phase short circuit current is greater than phase-to-ground short circuit current

-Peak Asymmetrical Current Calculations

Peak current according to IEC 60909 standard:

k₅=1,02+0,98 e^(-3R/x)=1,46

I_p5= k.√2 . I_k3'' =4,631 kA

-Thermal Short Circuit Current

At a point far from the generator I_k3''/I_k3=1 and n=1

m₌ 1 [ e^{4.f.Tk.ln(K-1)}-1]= 0,0129

2.f.Tk.Ln(K-1)

I_th=I_k3''.√(m+n) = 2,595

Breaker Power: S_ksc=√3.Un.1,1.I_k3''₌√3.36.1,1.2,243= 154 MVA

-Three Phase Short Circuit Calculations :

X₁(8)=X_s+X_tr(1)+X_h1(1)+X_h1(2)+X_h1(3)+X_h1(4)+X_str(1)=0,0681+0,1462+0,0011+0,0686+0,0365+0,0269+3,75=4,0974 pu

R₁(8)=R_s+R_tr(1)+R_h1(1)+R_h1(2)+R_h1(3)+R_h1(4)+R_h1(5)+R_str(1) =0,0069+0+0,0004+0,0271+0,0144+0,0106+0=0,0594 pu

Z₁(8)=R₁(8)+j X₁(8) =0,0594+ j 4,0974=4,0978 pu

I_k3''=0,358 kA

LV Side Three Phase Short Circuit : I_k3AG''= I_k3''. V_p/V_s = 28,19 kA

-Phase Ground short circuit calculations

X₀(8)=X_s0+X_tr0(1)+X_h0(1)+X_h0(2)+X_h0(3)+X_h0(4)+X_str10(1)=0,1363+0,1462+0,0044+0,2744+0,146+0,1076+3,75=4,5649 pu

R₀(8)=R_s0+R_tr0(1)+3.Rn+R_h0(1)+R_h0(2) +R_h0(3)+R_h0(4)+R_str10(1) =0,0069+0+9,2214+0,0036+0,2168+0,1152+0,0848+0=9,6487 pu

Z₀(8)=R₀(8)+j X₀(8) =9,6487+ j 4,5649=10,6741 pu

Z₀(8)/Z₁(8) =10,6741/4,0978=2,6 > 1 the phase-to-ground short circuit current is greater than the two phase-to-ground fault currents.

I_k1t''= I_k3''_x 3 ₌ 0,358 _x 3 ₌0,233 kA

2₊ Z₀ 2+ 2,6

Z₁

Three phase short circuit current is greater than phase-to-ground short circuit current

LV Side Phase Ground Short Circuit : I_k1fAG''= I_k1f''. V_p/V_s = 18,35 kA

-Three Phase Short Circuit Calculations :

X₁(9)=X_s+X_tr(1)+X_h1(1)+X_h1(5)+X_h1(6)+X_h1(7)+X_str(1)=0,0681+0,1462+0,0011+0,192+0,1407+0,0831+3,75=4,3812 pu

R₁(9)=R_s+R_tr(1)+R_h1(1)+R_h1(5)+R_h1(6)+R_h1(7)+R_str(1)=0,0069+0+0,0004+0,075+0,055+0,0325+0=0,1698 pu

Z₁(9)=R₁(9)+j X₁(9) =0,1698+ j 4,3812=4,3845 pu

I_k3''=0,334 kA

LV Side Three Phase Short Circuit : I_k3AG''= I_k3''. V_p/V_s = 26,3 kA

-Phase Ground short circuit calculations

X₀(9)=X_s0+X_tr0(1)+X_h0(1)+X_h0(5)+X_h0(6)+X_h0(7)+X_str0(1)=0,1363+0,1462+0,0044+0,6144+0,4502+0,2659+3,75=5,3674 pu

R₀(9)=R_s0+R_tr0(1)+3.Rn+R_h0(1)+R_h0(5)+R_h0(6)+R_h0(7)+R_str0(1) =0,0069+0+9,2214+0,0036+0,225+0,205+0,1825+0=9,8444 pu

Z₀(9)=R₀(9)+j X₀(9) =9,8444+ j 5,3674=11,2125 pu

Z₀(9)/Z₁(7) =11,2125/4,3845=2,56 > 1 the phase-to-ground short circuit current is greater than the two phase-to-ground fault currents.

I_k1t''= I_k3''_x 3 ₌ 0,334 _x 3 ₌0,22 kA

2₊ Z₀ 2+ 2,56

Z₁

LV Side Phase Ground Short Circuit: I_k1fAG''= I_k1f''. V_p/V_s = 17,32 kA