Question

Recalculate the sub transient current through the breaker in Problem 7.6 if the generator is initially delivering rated MVA at 0.8 pf lagging and at rated terminal voltage.

Short answer

The sub transient current flowing through breaker is 6.876 kA.

Step-by-step solution

Write the given data from the question.

Base MVA of generator,$S_{base}=1000\mathrm{MVA}$

Base voltage of generator,$V_{base}=20\mathrm{kV}$

Base MVA of transformer rating;localid="1655438951931" $S_{base}=1000\mathrm{MVA}$,

Base voltage of transformer,localid="1655438955397" $V_{base}=20\mathrm{kV}∆/345\mathrm{kV}\mathrm{Y}$

Generator reactances:

localid="1655438959709" $$\begin{gathered} X_d^{\text{'}\text{'}}=0.17\mathrm{pu} \\ {X}_d^{\text{'}}=0.30\mathrm{pu} \\ {X}_d=1.5\mathrm{pu} \end{gathered}$$

Time constant for generator;

localid="1655438964322" $$\begin{gathered} T_d^{\text{'}\text{'}}=0.05\mathrm{s} \\ {T}_d^{\text{'}}=1.0\mathrm{s} \\ {T}_A=0.1\mathrm{s} \end{gathered}$$

Transformer series reactance localid="1655438970849" $X_{tr}=0.1\mathrm{pu}$.

The voltage base value,localid="1655438976009" $V_{base}=345\mathrm{kV}$

Power base value,localid="1655438981399" $S_{base}=1000\mathrm{MVA}$

Power factor,localid="1655438986697" $pf=0.8$

The rated voltage in per unit,localid="1655438991351" $V=1\angle 0°$

Determine the formula to calculate the sub transient current through the breaker.

The equation to calculate base current on transmission line is given as follows.

${\mathbf{I}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{base}}}$ …… (1)

The equation to calculate the pre fault current is given as follows.

$\mathbf{I}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{{\mathbf{V}}_{\mathbf{base}}}\mathbf{D}\mathbf{‐}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{pf}\mathbf{\right)}$ …… (2)

The equation to calculate generator voltage is given as follows.

$\mathbf{E}{\mathbf{\text{'}}}_{\mathbf{g}}\mathbf{=}\mathbf{V}\mathbf{+}\mathbf{jX}{\mathbf{\text{'}}}_{\mathbf{d}}\mathbf{I}$ …… (3)

The equation to calculate the Thevenin’s reactance is given as follows.

${\mathbf{X}}_{\mathbf{Th}}\mathbf{=}{\mathbf{X}}_{\mathbf{d}}^{\mathbf{\text{'}\text{'}}}\mathbf{+}{\mathbf{X}}_{\mathbf{tr}}$ …… (4)

The equation to calculate the sub transient current through the breaker is given as follows.

${\mathbf{I}}_{\mathbf{pu}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{F}}}{{\mathbf{X}}_{\mathbf{Th}}}$ …… (5)

The equation to calculate the sub transient current through the breaker in kA is given as follows.

$\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{base}}{\mathbf{I}}_{\mathbf{pu}}$ …… (6)

Calculate the sub transient current through the circuit breaker.

Calculate the base current,

Substitute 1000 MVA for $S_{base}$and 345 kV for $V_{base}$into equation (1).

$$\begin{gathered} I_{base}=\frac{1000\times {10}^6}{\sqrt{3}\times 345\times {10}^3} \\ {I}_{base}=1.673\mathrm{kA} \end{gathered}$$

The equivalent circuit before the fault is shown below.

Calculate the pre fault current.

Substitute 1.0 pu for $S_{base}$, 1.0 pu for $V_{base}$and 0.8 for$pf$into equation (2).

$$\begin{gathered} I=\frac{1}{1}\angle {\cos }^{-1}\left(0.8\right) \\ I=1.0\angle 36.87° \end{gathered}$$

Calculate the generated voltage.

Substitute 1.0 pu for $V_{base}$, 0.17 for $X_d^{\text{'}\text{'}}$and $1\angle 36.87°\mathrm{A}$for $I$into equation (3).

localid="1655439174737" $$\begin{gathered} E_g^{\text{'}}=1+j0.17\left(1\angle 36.87°\right) \\ {E}_g^{\text{'}}=1+0.17\angle 53.13° \\ {E}_g^{\text{'}}=1.11\angle 7.035°\mathrm{pu} \end{gathered}$$

In case of three phase fault at transmission line the pre fault voltage and generated voltages are equals.

localid="1655439179699" $V_F=1.11\angle 7.035°\mathrm{A}$

Calculate the equivalent reactance.

Substitute 0.17 pu for localid="1655439184849" $X_d^{\text{'}\text{'}}$and 0.1pu localid="1655439189741" $X_{tr}$for into equation (4).

localid="1655439194578" $$\begin{gathered} X_{Th}=j0.17+j0.1 \\ {X}_{Th}=j0.27\mathrm{pu} \end{gathered}$$

Calculate the sub transient current flowing through breaker.

Substitute 1.11 V for localid="1655439200718" $V_F$and 0.27 pu for localid="1655439206445" $X_{Th}$into equation (5).

localid="1655439212943" $$\begin{gathered} I_{pu}=\frac{1.11}{0.27} \\ {I}_{pu}=4.11\mathrm{pu} \end{gathered}$$

Calculate the sub transient current flowing through breaker in kA.

Substitute 4.11 pu for localid="1655439218821" $I_{pu}$and 1.673 A for localid="1655439228539" $I_{base}$into equation (6).

localid="1655439235323" $$\begin{gathered} I=1.673\times 4.11 \\ I=6.876\mathrm{kA} \end{gathered}$$

Hence, the sub transient current flowing through breaker is 6.876 kA .