Power Systems Test 8

Concept:

Short circuit MVA = 3V_base I_sc

For three-phase fault, short circuit current, \({I_{sc}} = \frac{{{V_{base}}}}{{{X_{eq}}}}\)

Calculation:

Short circuit MVA = 4000 MVA

Base voltage, V_base =220 kV

Now, short circuit MVA = 3V_base I_sc

Also, I_sc = fault current

\({I_{sc}} = \frac{{{V_{base}}}}{{{X_{1\;eq}}}}\)

\( \Rightarrow 3{V_{base}}\frac{{{V_{base}}}}{{{X_{1eq}}}} = 4000\)

\({X_{1eq}} = \frac{{3 \times V_{base}^2}}{{4000}}\)

\({X_{1eq}} = \frac{{3 \times {{\left( {\frac{{220}}{{\sqrt 3 }}} \right)}^2}}}{{4000}}\)

Since the star point is isolated from ground LLG fault is just like LL fault.

\({I_b} = - {I_c} = \frac{{ - j\;\sqrt 3 \times 1}}{{j\;0.35 + j\;0.25}} = - 2.887\;pu\)

Voltage (V_R) = 6.6 kV

MVA rating = 20 MVA

X^’’_d = 10% = 0.1

X^’_d = 20% = 0.2

X_d = 100% = 1

Sustained short circuit current through breaker (I_s) is the steady state fault current.

During steady state,

\({I_s} = \frac{{{V_{pu}}}}{{{X_{{d_{pu}}}}}}\) 

\(= \frac{1}{1} = 1\;pu\) 

Rated current \({I_R} = \frac{{\left( {VA} \right)}}{{\sqrt 3 \times V}}\)

\(= \frac{{20 \times {{10}^6}}}{{\sqrt 3 \times 6.6 \times {{10}^3}}}\) 

= 1749.5 A

X^’’ = X₁ = X₂ = 0.15 pu

X₀ = 0.05 pu

Sub transient line current for LG fault,

\({I_{f\left( {L - G} \right)}} = \frac{3}{{X_1 + {X_2} + {X_0}}}\)

Sub transient line current for a symmetrical three phase fault

\({I_{f\left( {3 - \phi } \right)}} = \frac{1}{{X''}}\)

I_f(L-G) = I_f(3-ϕ)

\(\Rightarrow \frac{3}{{X'' + {X_2} + {X_0} + 3{X_n}}} = \frac{1}{{0.15}}\)

⇒ X_n = 0.033 pu

Voltage (E) = 2200 V

3-phase fault current = 127 A

\({I_{f\left( {3 - \phi } \right)}} = \frac{{{E_{ph}}}}{{{X_1}}}\) 

\(\Rightarrow 127 = \frac{{\frac{{2200}}{{\sqrt 3 }}}}{{{X_1}}}\) 

⇒ X₁ = 10 Ω

L-L fault current = 129.5 A

\({I_{f\left( {L - L} \right)}} = \frac{{\sqrt 3 {E_{ph}}}}{{{X_1} + {X_2}}}\) 

\(\Rightarrow 129.5 = \frac{{\sqrt 3 \times \frac{{2200}}{{\sqrt 3 }}}}{{10 + {X_2}}}\) 

⇒ X₂ = 7 Ω

L-G fault current = 190 A

\({I_{f\left( {L - G} \right)}} = \frac{{3{E_{ph}}}}{{{X_1} + {X_2} + {X_0}}}\) 

\(\Rightarrow 190 = \frac{{3 \times \frac{{2200}}{{\sqrt 3 }}}}{{10 + 7 + {X_0}}}\) 

Concept:

The fault current in a single line to ground fault is,

\({I_f} = \frac{{3 \times {E_{R1}}}}{{{X_1} + {X_2} + {X_0} + 3{X_n}}}\)

Short-circuit capacity = I_f × Base MVA

X₁ is the positive sequence reactance

X₂ is the negative sequence reactance

X₀ is the zero-sequence reactance

X_n is the ground reactance

Calculation:

Short circuit capacity = 500 MvA,

Neutral reactance, X_n = 0.05 pu

Zero sequence reactance, X₀ = 25% of positive sequence reactance

Base MVA = 100 MVA

3ϕ fault current, \({I_f} = \frac{{{E_{R1}}}}{{X_d^{''}}} \times 100 = \frac{1}{{{X_d}}}pu\)

Short circuit MVA = 500

\( \Rightarrow \frac{1}{{X_d^{''}}} \times 100 = 500\)

\( \Rightarrow X_d^{''} = 0.2\;pu\)

In turbo generator,

Positive sequence reactance, \(X_d^{''}\) = Negative sequence reactance, X₂

i.e. \(X_d^{''} = {X_1} = {X_2} = 0.2\;pu\)

Zero sequence reactance, X₀ = 25% of \(X_d^{''}\)

= 0.25 × 0.2

= 0.05 pu

For the LG fault,

\({X_{f\left( {LG} \right)}} = \frac{{3 \times {E_{R1}}}}{{{X_1} + {X_2} + {X_0} + 3{X_n}}}\)

\( = \frac{{3 \times 1\;\;}}{{0.2 + 0.2 + 0.05 + 3 \times 0.05}} = 5\;pu\)

Short-circuited capacity for a line to ground fault = I_f(LG) × Base MVA