Question

For the power system given in Problem 7.12, a three-phase short circuit occurs at bus 2, where the Prefault voltage is$\mathbf{525}\mathbf{KV}$. Prefault load current is neglected. Determine the (a) Thévenin equivalent at the fault, (b) subtransient fault current in per unit and in$\mathbf{kA}\mathbf{}\mathbf{rms}$, and (c) contributions to the fault from lines 1–2, 2–3, and 2–4.

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The}\mathrm{Thévenin}\mathrm{equivalent}\mathrm{fault}\mathrm{is}0.197\mathrm{p}.\mathrm{u}. \\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{sub}\mathrm{transient}\mathrm{fault}\mathrm{current}\mathrm{in}\mathrm{per}\mathrm{unit}\mathrm{and}\mathrm{kA}\mathrm{are}-\mathrm{j}5.33\mathrm{p}.\mathrm{u}.\mathrm{and}-\mathrm{j}6.18\mathrm{kA}. \\ \left(\mathrm{c}\right)\mathrm{The}\mathrm{current}\mathrm{contributed}\mathrm{from}1-2\mathrm{line},2-3\mathrm{and}2-4\mathrm{line}\mathrm{is}-\mathrm{j}1.46\mathrm{A},-\mathrm{j}2.60\mathrm{A}\mathrm{and}-\mathrm{j}2.60\mathrm{A}. \end{gathered}$$

Step-by-step solution

Write the given data.

$$\begin{gathered} \mathrm{From}\mathrm{reference}\mathrm{problem}7.1. \\ \mathrm{Consider}\mathrm{the}\mathrm{following}\mathrm{data}: \\ \mathrm{GeneratorG}1:\mathrm{Power}\mathrm{rating}:500\mathrm{MVA} \\ \mathrm{Voltage}\mathrm{rating}:13.8\mathrm{kV} \\ \mathrm{Reactance}:0.2\mathrm{p}.\mathrm{u}. \\ \mathrm{GeneratorG}2:\mathrm{Power}\mathrm{rating}:750\mathrm{MVA} \\ \mathrm{Voltage}\mathrm{rating}:18\mathrm{kV} \\ \mathrm{Reactance}:0.18\mathrm{p}.\mathrm{u}. \\ \mathrm{GeneratorG}3:\mathrm{Power}\mathrm{rating}:1000\mathrm{MVA} \\ \mathrm{Voltage}\mathrm{rating}:20\mathrm{kV} \\ \mathrm{Reactance}:0.17\mathrm{p}.\mathrm{u}. \\ \mathrm{TransformerT}1:\mathrm{Power}\mathrm{rating}:500\mathrm{MVA} \\ \mathrm{Voltage}\mathrm{rating}:13.8∆/500\mathrm{YkV} \\ \mathrm{Reactance}:0.12\mathrm{p}.\mathrm{u}. \\ \mathrm{TransformerT}2:\mathrm{Power}\mathrm{rating}:750\mathrm{MVA} \\ \mathrm{Voltage}\mathrm{rating}:18∆/500\mathrm{YkV} \\ \mathrm{Reactance}:0.10\mathrm{p}.\mathrm{u}. \\ \mathrm{TransformerT}3:\mathrm{Power}\mathrm{rating}:1000\mathrm{MVA} \end{gathered}$$

$$\begin{gathered} \mathrm{Voltage}\mathrm{rating}:20∆/500\mathrm{YkV} \\ \mathrm{Reactance}:0.10\mathrm{p}.\mathrm{u}. \\ \mathrm{Prefault}\mathrm{Voltage}:525\mathrm{kV} \\ \mathrm{Base}:\mathrm{Power}\mathrm{rating}:1000\mathrm{MVA} \\ \mathrm{For}\mathrm{G}3\mathrm{Voltage}\mathrm{rating}:20\mathrm{kV} \end{gathered}$$

Determine the required formula.

$$\begin{gathered} \mathrm{Here},{\mathbf{V}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{voltage}\mathrm{at}\mathrm{the}\mathrm{fault}\mathrm{and}{\mathit{X}}_{\mathbf{thp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{Thévenin}\mathrm{reactance}. \\ {\mathit{l}}_{\mathbf{F}}\mathbf{=}{\mathit{l}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}\mathbf{4}}\mathbf{\times }{\mathit{l}}_{\mathbf{F}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\dots \dots \left(3\right) \\ \mathrm{Here},{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathrm{is}\mathrm{base}\mathrm{current}\mathrm{and}{\mathit{l}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{fault}\mathrm{current}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{current}\mathrm{contributed}\mathrm{from}1-2,2-3\mathrm{and}2-4\mathrm{line}. \\ {\mathit{l}}_{\mathbf{12}\mathit{F}}\mathbf{=}{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathbf{\times }{\mathit{l}}_{\mathbf{12}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{}\dots \dots \left(4\right) \\ \mathrm{Here},{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathrm{is}\mathrm{base}\mathrm{current}\mathrm{and}{\mathit{l}}_{\mathbf{12}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{fault}\mathrm{current}. \\ {\mathit{l}}_{\mathbf{23}\mathbf{F}}\mathbf{=}{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathbf{\times }{\mathit{l}}_{\mathbf{21}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}.......\left(5\right) \\ \mathrm{Here},{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathrm{is}\mathrm{base}\mathrm{current}\mathrm{and}{\mathit{l}}_{\mathbf{21}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{fault}\mathrm{current}. \\ {\mathit{l}}_{\mathbf{24}\mathbf{F}}\mathbf{=}{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathbf{\times }{\mathit{l}}_{\mathbf{24}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}.......\left(6\right) \\ \mathrm{Here},{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathrm{is}\mathrm{base}\mathrm{current}\mathrm{and}{{\mathit{l}}_{\mathbf{24}}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{fault}\mathrm{current}. \end{gathered}$$Write the formula of Thévenin reactance at equivalent fault

localid="1655365033445" ${\mathbf{X}}_{\mathbf{t}\mathbf{h}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{=}\mathbf{\left(}\begin{array}{c}{\mathbf{X}}_{\mathbf{l}\mathbf{i}\mathbf{n}\mathbf{e}\mathbf{12}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{+}{\mathbf{X}}_{\mathbf{T}\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{n}\mathbf{e}\mathbf{w}}\\ \mathbf{+}{\mathbf{X}}_{\mathbf{G}\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{n}\mathbf{e}\mathbf{w}}\end{array}\mathbf{\right)}\mathbf{\square }\mathbf{\left(}\begin{array}{c}{\left(X_{line24p.u.}+X_{12p.u.new}+X_{G2p.u.new}\right)}^{\mathbf{\square }}\\ \left(X_{line23p.u.}+X_{13p.u.new}+X_{G3p.u.new}\right)\end{array}\mathbf{\right)}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

localid="1655364988381" $$\begin{gathered} \mathrm{Here},{\mathbf{X}}_{\mathbf{l}\mathbf{i}\mathbf{n}\mathbf{e}\mathbf{12}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{line}\mathrm{reactance},{\mathbf{X}}_{\mathbf{T}\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{n}\mathbf{e}\mathbf{w}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of}\mathrm{transformer} \\ \mathbf{T}\mathbf{1}\mathrm{according}\mathrm{to}\mathrm{new}\mathrm{base},{\mathbf{X}}_{\mathbf{G}\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{n}\mathbf{e}\mathbf{w}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of}\mathrm{generator}\mathbf{G}\mathbf{1}\mathrm{according} \\ \mathrm{to}\mathrm{new}\mathrm{base},{\mathbf{X}}_{\mathbf{l}\mathbf{i}\mathbf{n}\mathbf{e}\mathbf{24}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{line}\mathrm{reactance},{\mathbf{X}}_{\mathbf{12}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{n}\mathbf{e}\mathbf{w}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}, \\ {\mathbf{X}}_{\mathbf{G}\mathbf{2}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{n}\mathbf{e}\mathbf{w}\mathbf{}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of}\mathrm{generator}\mathbf{G}\mathbf{2}\mathrm{according}\mathrm{to}\mathrm{new}\mathrm{base},\mathbf{}{\mathbf{X}}_{\mathbf{l}\mathbf{i}\mathbf{n}\mathbf{e}\mathbf{23}}\mathrm{is}\mathrm{line} \\ \mathrm{reactance},{\mathbf{X}}_{\mathbf{T}\mathbf{3}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{n}\mathbf{e}\mathbf{w}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of}\mathrm{transformer}\mathbf{T}\mathbf{3}\mathbf{}\mathrm{according}\mathrm{to}\mathrm{new}\mathrm{base} \\ \mathrm{and}{\mathbf{X}}_{\mathbf{G}\mathbf{3}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{n}\mathbf{e}\mathbf{w}\mathbf{}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of}\mathrm{generator}\mathbf{}\mathbf{G}\mathbf{3}\mathrm{according}\mathrm{to}\mathrm{new}\mathrm{base}. \end{gathered}$$

Write the formula of sub transient fault current in per unit and$\mathrm{kA}$.

$$\begin{gathered} {\mathit{l}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{=}\frac{{\mathbf{V}}_{{}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}}}{{\mathbf{X}}_{{}_{\mathbf{thp}\mathbf{.}\mathbf{u}\mathbf{.}}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \\ \mathrm{Here},{\mathbf{V}}_{{}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{voltage}\mathrm{at}\mathrm{the}\mathrm{fault}\mathrm{and}{\mathit{X}}_{{}_{\mathbf{thp}\mathbf{.}\mathbf{u}\mathbf{.}}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{Thévenin}\mathrm{reactance}. \\ {\mathit{l}}_{\mathbf{F}}\mathbf{=}{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathbf{\times }{\mathit{l}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{}\mathbf{\left(}\mathbf{3}\mathbf{\right)} \end{gathered}$$

$$\begin{gathered} \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{current}\mathrm{contributed}\mathrm{from}1-2,2-3\mathrm{and}2-4\mathrm{line}. \\ {\mathit{l}}_{\mathbf{12}\mathbf{F}}\mathbf{=}{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathbf{\times }{\mathit{l}}_{\mathbf{12}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{}\mathbf{\left(}\mathbf{4}\mathbf{\right)} \\ \mathrm{Here},{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathrm{is}\mathrm{base}\mathrm{current}\mathrm{and}{\mathit{l}}_{\mathbf{12}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{fault}\mathrm{current}\mathbf{.}\mathbf{} \\ {\mathit{l}}_{\mathbf{23}\mathbf{F}}\mathbf{=}{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathbf{\times }{\mathit{l}}_{\mathbf{21}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{}\mathbf{\left(}\mathbf{5}\mathbf{\right)} \\ \mathrm{Here},\mathbf{}{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathrm{is}\mathrm{base}\mathrm{current}\mathrm{and}\mathit{}{\mathit{l}}_{\mathbf{21}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{}\mathbf{}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{fault}\mathrm{current}.\mathbf{} \\ {\mathit{l}}_{\mathbf{24}\mathbf{F}}\mathbf{=}{\mathbf{l}}_{\mathbf{base}\mathbf{4}}\mathbf{\times }{\mathbf{l}}_{\mathbf{24}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{}\mathbf{\left(}\mathbf{6}\mathbf{\right)} \\ \mathrm{Here},\mathbf{}{\mathit{l}}_{\mathbf{base}\mathbf{4}}\mathrm{is}\mathrm{base}\mathrm{current}\mathrm{and}{\mathit{l}}_{\mathbf{24}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{fault}\mathrm{current}. \end{gathered}$$

Determine Thévenin equivalent fault.

(a)

First determine the per unit reactance of generator $\mathrm{G}3$.

$\frac{X_{G3p.u.new}}{X_{G3p.u.}}=\frac{S_{base}}{S_{G3rated}}\times {\left(\frac{V_{G3rated}}{V_{baseG3}}\right)}^2......\left(7\right)$

$$\begin{gathered} \mathrm{Here},X_{G3\mathrm{p}.\mathrm{u}.\mathrm{new}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of}\mathrm{generator}G3,X_{G3\mathrm{p}.\mathrm{u}.}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of} \\ \mathrm{the}\mathrm{rated}\mathrm{generator},S_{base}\mathrm{is}\mathrm{given}\mathrm{power}\mathrm{base},{\mathrm{S}}_{\mathrm{G}3\mathrm{rated}}\mathrm{is}\mathrm{rated}\mathrm{power}\mathrm{as}\mathrm{base},V_{\mathrm{G}3\mathrm{rated}} \\ \mathrm{is}\mathrm{rated}\mathrm{voltage}\mathrm{base}\mathrm{and}{V}_{\mathrm{baseG}3}\mathrm{is}\mathrm{new}\mathrm{voltage}\mathrm{base}\mathrm{for}\mathrm{G}3. \end{gathered}$$

localid="1655369663047" $$\begin{gathered} \mathrm{Substitute}0.17\mathrm{for}{X}_{\mathrm{G}3\mathrm{p}.\mathrm{u}.},1000\mathrm{for}{S}_{base},1000\mathrm{for}{S}_{\mathrm{G}3\mathrm{rated}},20\mathrm{for}{V}_{\mathrm{G}3\mathrm{rated}}\mathrm{and}20\mathrm{for} \\ {V}_{baseG3}\mathrm{into}\mathrm{equation}\left(7\right). \\ {\mathrm{X}}_{{}_{\mathrm{G}3\mathrm{p}.\mathrm{u}.\mathrm{new}}}=0.17\left(\frac{1000}{1000}\right){\left(\frac{20}{20}\right)}^2 \\ =0.17\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine the per unit reactance of generator $\mathrm{G}2$.

localid="1655369445631" $\frac{X_{\mathrm{G}2\mathrm{p}.\mathrm{u}.\mathrm{new}}}{X_{G2\mathrm{p}.\mathrm{u}.}}=\frac{S_{base}}{S_{\mathrm{G}2\mathrm{rated}}}\times {\left(\frac{V_{\mathrm{G}2\mathrm{rated}}}{V_{\mathrm{baseG}2}}\right)}^2......\left(8\right)$

$$\begin{gathered} \mathrm{Here},X_{\mathrm{G}2\mathrm{p}.\mathrm{u}.\mathrm{new}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of}\mathrm{generator}\mathrm{G}2,X_{\mathrm{G}2\mathrm{p}.\mathrm{u}.}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of} \\ \mathrm{the}\mathrm{rated}\mathrm{generator},S_{base}\mathrm{is}\mathrm{given}\mathrm{power}\mathrm{base},{\mathrm{S}}_{\mathrm{G}2\mathrm{rated}}\mathrm{is}\mathrm{rated}\mathrm{power}\mathrm{as}\mathrm{base},V_{\mathrm{G}2\mathrm{rated}} \\ \mathrm{is}\mathrm{rated}\mathrm{voltage}\mathrm{base}\mathrm{and}{V}_{\mathrm{baseG}2}\mathrm{is}\mathrm{new}\mathrm{voltage}\mathrm{base}\mathrm{for}\mathrm{G}2. \end{gathered}$$

localid="1655370313948" $$\begin{gathered} \mathrm{Substitute}0.18\mathrm{for}{X}_{\mathrm{G}2\mathrm{p}.\mathrm{u}.},1000\mathrm{for}{S}_{\mathrm{base}},750\mathrm{for}{S}_{\mathrm{G}2\mathrm{rated}},18\mathrm{for}{V}_{\mathrm{G}2\mathrm{rated}}\mathrm{and}18\mathrm{for} \\ {V}_{\mathrm{baseG}2}\mathrm{into}\mathrm{equation}\left(8\right). \\ {\mathrm{X}}_{{}_{\mathrm{G}2\mathrm{p}.\mathrm{u}.\mathrm{new}}}=0.18\left(\frac{1000}{750}\right){\left(\frac{18}{18}\right)}^2 \\ =0.24\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine the per unit reactance of generator$\mathrm{G}1$.

$\frac{X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}}}{X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.}}=\frac{S_{base}}{S_{\mathrm{G}1\mathrm{rated}}}\times {\left(\frac{V_{\mathrm{G}1\mathrm{rated}}}{V_{\mathrm{baseG}1}}\right)}^2......\left(9\right)$

$$\begin{gathered} \mathrm{Here},X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of}\mathrm{generator}\mathrm{G}2,X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of} \\ \mathrm{the}\mathrm{rated}\mathrm{generator},S_{base}\mathrm{is}\mathrm{given}\mathrm{power}\mathrm{base},S_{\mathrm{G}1\mathrm{rated}}\mathrm{is}\mathrm{rated}\mathrm{power}\mathrm{as}\mathrm{base},V_{\mathrm{G}1\mathrm{rated}} \\ \mathrm{is}\mathrm{rated}\mathrm{voltage}\mathrm{base}\mathrm{and}{V}_{\mathrm{baseG}1}\mathrm{is}\mathrm{new}\mathrm{voltage}\mathrm{base}\mathrm{for}\mathrm{G}1. \end{gathered}$$

localid="1655371441329" $$\begin{gathered} \mathrm{Substitute}0.2\mathrm{for}{X}_{\mathrm{G}1\mathrm{p}.\mathrm{u}.},1000\mathrm{for}{S}_{\mathrm{base}},500\mathrm{for}{S}_{\mathrm{G}1\mathrm{rated}},13.8\mathrm{for}{V}_{\mathrm{G}1\mathrm{rated}}\mathrm{and}13.8\mathrm{for} \\ {V}_{\mathrm{baseG}1}\mathrm{into}\mathrm{equation}\left(9\right). \\ {\mathrm{X}}_{{}_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}}}=0.2\left(\frac{1000}{500}\right){\left(\frac{13.8}{13.8}\right)}^2 \\ =0.4\mathrm{p}.\mathrm{u}. \end{gathered}$$

For transformer $\mathrm{T}1$determine the new per unit reactance of transformer according to new base.

$\frac{X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.\mathrm{new}}}{X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.}}=\frac{S_{\mathrm{base}}}{S_{\mathrm{T}1\mathrm{rated}}}\times {\left(\frac{V_{\mathrm{T}1\mathrm{rated}}}{V_{\mathrm{baseT}1}}\right)}^2......\left(10\right)$

$$\begin{gathered} \mathrm{Here},X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.\mathrm{new}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of}\mathrm{transformer}\mathrm{T}1\mathrm{according}\mathrm{to}\mathrm{new}\mathrm{base},X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.}\mathrm{is} \\ \mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{on}\mathrm{the}\mathrm{rated}\mathrm{generator},S_{\mathrm{base}}\mathrm{is}\mathrm{given}\mathrm{power}\mathrm{base},S_{\mathrm{T}1\mathrm{rated}}\mathrm{is}\mathrm{rated}\mathrm{power} \\ \mathrm{as}\mathrm{base},V_{\mathrm{T}1\mathrm{rated}}\mathrm{is}\mathrm{rated}\mathrm{voltage}\mathrm{base}\mathrm{and}{V}_{\mathrm{baseT}1}\mathrm{is}\mathrm{new}\mathrm{voltage}\mathrm{base}\mathrm{for}\mathrm{T}1. \end{gathered}$$

localid="1655372564208" $$\begin{gathered} \mathrm{Substitute}0.12\mathrm{for}{X}_{\mathrm{T}1\mathrm{p}.\mathrm{u}.},1000\mathrm{for}{S}_{\mathrm{base}},500\mathrm{for}{S}_{\mathrm{T}1\mathrm{rated}},13.8\mathrm{for}{V}_{\mathrm{T}1\mathrm{rated}}\mathrm{and}13.8\mathrm{for} \\ {V}_{\mathrm{baseT}1}\mathrm{into}\mathrm{equation}\left(10\right). \\ {\mathrm{X}}_{\mathrm{T}1\mathrm{p}.\mathrm{u}.\mathrm{new}}=0.12\left(\frac{1000}{500}\right){\left(\frac{13.8}{13.8}\right)}^2 \\ =0.24\mathrm{p}.\mathrm{u}. \end{gathered}$$

For transformer$\mathrm{T}2$determine the new per unit reactance of transformer according to new base.

$\frac{X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}}}{X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.}}=\frac{S_{\mathrm{base}}}{S_{\mathrm{T}2\mathrm{rated}}}\times {\left(\frac{V_{\mathrm{T}2\mathrm{rated}}}{V_{\mathrm{baseT}2}}\right)}^2......\left(11\right)$

$$\begin{gathered} \mathrm{Here},X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of}\mathrm{transformer}\mathrm{T}2\mathrm{according}\mathrm{to}\mathrm{new}\mathrm{base},X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.}\mathrm{is} \\ \mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{on}\mathrm{the}\mathrm{rated}\mathrm{generator},S_{\mathrm{base}}\mathrm{is}\mathrm{given}\mathrm{power}\mathrm{base},S_{\mathrm{T}2\mathrm{rated}}\mathrm{is}\mathrm{rated}\mathrm{power} \\ \mathrm{as}\mathrm{base},V_{\mathrm{T}2\mathrm{rated}}\mathrm{is}\mathrm{rated}\mathrm{voltage}\mathrm{base}\mathrm{and}{V}_{\mathrm{baseT}2}\mathrm{is}\mathrm{new}\mathrm{voltage}\mathrm{base}\mathrm{for}\mathrm{T}2. \end{gathered}$$

$$\begin{gathered} \mathrm{Substitute}0.10\mathrm{for}{X}_{\mathrm{T}2\mathrm{p}.\mathrm{u}.},1000\mathrm{for}{S}_{\mathrm{base}},750\mathrm{for}{S}_{\mathrm{T}2\mathrm{rated}},18\mathrm{for}{V}_{\mathrm{T}2\mathrm{rated}}\mathrm{and}18\mathrm{for} \\ {V}_{\mathrm{baseT}2}\mathrm{into}\mathrm{equation}\left(11\right). \\ {\mathrm{X}}_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}}=0.10\left(\frac{1000}{750}\right){\left(\frac{18}{18}\right)}^2 \\ =0.13\mathrm{p}.\mathrm{u}. \end{gathered}$$

For transformer $\mathrm{T}3$determine the new per unit reactance of transformer according to new base.

$\frac{X_{\mathrm{T}3\mathrm{p}.\mathrm{u}.\mathrm{new}}}{X_{\mathrm{T}3\mathrm{p}.\mathrm{u}.}}=\frac{S_{\mathrm{base}}}{S_{\mathrm{T}3\mathrm{rated}}}\times {\left(\frac{V_{\mathrm{T}3\mathrm{rated}}}{V_{\mathrm{baseT}3}}\right)}^2......\left(12\right)$

$$\begin{gathered} \mathrm{Here},X_{\mathrm{T}3\mathrm{p}.\mathrm{u}.\mathrm{new}}\mathrm{is}\mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{of}\mathrm{transformer}\mathrm{T}3\mathrm{according}\mathrm{to}\mathrm{new}\mathrm{base},X_{\mathrm{T}3\mathrm{p}.\mathrm{u}.}\mathrm{is} \\ \mathrm{per}\mathrm{unit}\mathrm{reactance}\mathrm{on}\mathrm{the}\mathrm{rated}\mathrm{generator},S_{\mathrm{base}}\mathrm{is}\mathrm{given}\mathrm{power}\mathrm{base},S_{\mathrm{T}3\mathrm{rated}}\mathrm{is}\mathrm{rated}\mathrm{power} \\ \mathrm{as}\mathrm{base},V_{\mathrm{T}3\mathrm{rated}}\mathrm{is}\mathrm{rated}\mathrm{voltage}\mathrm{base}\mathrm{and}{V}_{\mathrm{baseT}3}\mathrm{is}\mathrm{new}\mathrm{voltage}\mathrm{base}\mathrm{for}\mathrm{T}3. \end{gathered}$$

$$\begin{gathered} \mathrm{Substitute}0.10\mathrm{for}{X}_{\mathrm{T}3\mathrm{p}.\mathrm{u}.},1000\mathrm{for}{S}_{\mathrm{base}},1000\mathrm{for}{S}_{\mathrm{T}3\mathrm{rated}},20\mathrm{for}{V}_{\mathrm{T}3\mathrm{rated}}\mathrm{and}20\mathrm{for} \\ {V}_{\mathrm{baseT}3}\mathrm{into}\mathrm{equation}\left(12\right). \\ {\mathrm{X}}_{\mathrm{T}3\mathrm{p}.\mathrm{u}.\mathrm{new}}=0.10\left(\frac{1000}{1000}\right){\left(\frac{20}{20}\right)}^2 \\ =0.13\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine the base impedance for the transmission lines.

$Z_{base}=\frac{V_{base}^2}{S_{base}}\dots \dots \left(13\right)$

$$\begin{gathered} \mathrm{Here},V_{base}^2\mathrm{is}\mathrm{base}\mathrm{voltage}\mathrm{and}{S}_{\mathit{base}}\mathrm{is}\mathrm{given}\mathrm{power}\mathrm{base}. \\ \mathrm{Substitute}500\times {10}^3\mathrm{for}{V}_{base}^2\mathrm{and}1000\times {10}^6\mathrm{for}{S}_{base}\mathrm{into}\mathrm{equation}\left(13\right). \\ {\mathrm{Z}}_{\mathrm{base}}=\frac{{\left(500\times {10}^3\right)}^2}{1000\times {10}^6} \\ =250\mathrm{\Omega } \end{gathered}$$

Determine the per unit reactance of the line.

$X_{linep.u.}=\frac{X_{line}}{Z_{base}}\dots \dots \left(14\right)$

$$\begin{gathered} \mathrm{Here},X_{\mathrm{line}}\mathrm{is}\mathrm{base}\mathrm{voltage}\mathrm{and}{Z}_{base}\mathrm{is}\mathrm{given}\mathrm{power}\mathrm{base}. \\ \mathrm{Substitute}50\mathrm{for}{\mathrm{X}}_{\mathrm{line}}\mathrm{and}250\mathrm{for}{Z}_{base}\mathrm{into}\mathrm{equation}\left(14\right). \\ {X}_{\mathrm{linep}.\mathrm{u}.}=\frac{50}{250} \\ =0.2\mathrm{p}.\mathrm{u}. \\ \mathrm{Solve}\mathrm{further}\mathrm{as}, \\ {\mathrm{X}}_{\mathrm{linep}.\mathrm{u}.12={\mathrm{X}}_{\mathrm{linep}.\mathrm{u}.23}} \\ ={\mathrm{X}}_{\mathrm{linep}.\mathrm{u}.24} \\ =0.2\mathrm{p}.\mathrm{u}. \end{gathered}$$

Here, fault occurs at bus 2.

Thévenin reactance at the fault will be the equivalent reactance of the circuit with respect to the fault point and ground.

Determine the per unit Thévenin reactance at the fault point.$$\begin{gathered} \mathrm{Substitute}0.2\mathrm{for}{X}_{\mathit{line}\mathit{12}\mathit{p}\mathit{.}\mathit{u}\mathit{.}},0.24\mathrm{for}{X}_{\mathrm{T}1\mathrm{p}.\mathrm{u}.\mathrm{new}}\mathit{},0.4\mathrm{for}{X}_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}},0.2\mathrm{for}{X}_{line\mathit{24}\mathrm{p}\mathit{.}\mathrm{u}\mathit{.}},0.13 \\ \mathrm{for}{X}_{12\mathrm{p}.\mathrm{u}.\mathrm{new}},0.24\mathrm{for}{X}_{\mathrm{G}2\mathrm{p}.\mathrm{u}.\mathrm{new}},0.2\mathrm{for}{X}_{\mathrm{line}23},0.1\mathrm{for}{X}_{\mathrm{T}3\mathrm{p}.\mathrm{u}.\mathrm{new}}\mathrm{and}0.17\mathrm{for}{X}_{\mathrm{G}3\mathrm{p}.\mathrm{u}.\mathrm{new}} \\ \mathrm{into}\mathrm{equation}\left(1\right). \end{gathered}$$

$$\begin{gathered} X_{\mathrm{thp}.\mathrm{u}.}=\left(0.2+0.24+0.4\right)\square \left(0.2+0.13+0.24\right)\square \left(0.2+0.1+0.17\right) \\ =0.84\square 0.57\square 0.47 \\ =0.197\mathrm{p}.\mathrm{u} \end{gathered}$$

Determine the sub transient fault current in per unit andkA.

(b)

First determine the per unit fault voltage.

$$\begin{gathered} V_{\mathrm{Fp}.\mathrm{u}.}=\frac{V_{\mathrm{F}}}{V_{\mathrm{base}4}}\dots \dots \left(15\right) \\ \mathrm{Here},V_{\mathrm{F}}\mathrm{is}\mathrm{fault}\mathrm{voltage}\mathrm{and}{\mathrm{V}}_{\mathrm{base}4}\mathrm{is}\mathrm{base}\mathrm{voltage}\mathrm{for}\mathrm{the}\mathrm{zone}4 \\ \mathrm{Substitute}525\mathrm{for}{V}_{\mathrm{F}}\mathrm{and}500\mathrm{for}{V}_{\mathrm{base}4}\mathrm{into}\mathrm{equation}\left(15\right). \\ {V}_{\mathrm{Fp}.\mathrm{u}.}=\frac{525}{500} \\ =1.05\mathrm{p}.\mathrm{u}. \end{gathered}$$

For the actual current, determine the base current.

$$\begin{gathered} l_{\mathrm{base}4}=\frac{S_{\mathrm{base}}}{V_{\mathrm{baseL}L}}\dots \dots \left(16\right) \\ \mathrm{Here},S_{\mathrm{base}}\mathrm{is}\mathrm{given}\mathrm{power}\mathrm{base},V_{\mathrm{baseL}L}\mathrm{is}\mathrm{base}\mathrm{line}\mathrm{to}\mathrm{line}\mathrm{voltage}. \\ \mathrm{Substitute}1000\times {10}^6\mathrm{for}{S}_{\mathrm{base}}\mathrm{and}\sqrt{3}\left(500\times {10}^3\right)\mathrm{for}{V}_{\mathrm{baseLL}}\mathrm{into}\mathrm{equation}\left(16\right). \\ {l}_{\mathrm{base}4}=\frac{1000\times {10}^6}{\sqrt{3}{V}_{\mathrm{baseLN}}} \\ =\frac{1000\times {10}^6}{\sqrt{3}\left(500\times {10}^3\right)} \\ =1.16\mathrm{kA} \end{gathered}$$

Determine the per unit current at the fault

$$\begin{gathered} \mathrm{Substitute}1.05\mathrm{for}{V}_{\mathrm{Fp}.\mathrm{u}.}\mathrm{and}j0.197\mathrm{for}{X}_{\mathrm{thp}.\mathrm{u}.}\mathrm{into}\mathrm{equation}\left(2\right). \\ {l}_{\mathrm{Fp}.\mathrm{u}.}=\frac{1.05}{j0.197} \\ =-j5.33\mathrm{p}.\mathrm{u}. \end{gathered}$$

Now determine the actual current.

$$\begin{gathered} \mathrm{Substitute}1.16\mathrm{for}{l}_{\mathrm{base}4}\mathrm{and}-\mathrm{j}5.33\times {10}^3\mathrm{for}{l}_{fp.u.}\mathrm{into}\mathrm{equation}\left(3\right). \\ {l}_{\mathrm{F}}=\left(1.16\right)\left(-\mathrm{j}5.33\times {10}^3\right) \\ =-j6.18\mathrm{kA} \end{gathered}$$

Determine the current contributed from 1-2 line, 2-3 and 2-4 line.

$$\begin{gathered} \mathrm{Substitute}1.16\mathrm{for}{l}_{\mathrm{base}4}\mathrm{and}-j5.33\mathrm{for}{l}_{\mathrm{Fp}.\mathrm{u}.}\mathrm{into}\mathrm{equation}\left(6\right). \\ {l}_{24\mathrm{F}}=\left(1.16\right)\left(-j5.33\right)\left(\frac{\left[\begin{array}{c}{\left(X_{\mathrm{G}1}+X_{\mathrm{T}1}+X_{12}\right)}^{\square }\\ \left(X_{\mathrm{T}3}+X_{\mathrm{G}3}+X_{23}\right)\end{array}\right]}{\left(X_{\mathrm{G}1}+X_{\mathrm{T}1}+X_{12}\right)\square \left[\begin{array}{c}\left({\mathrm{X}}_{\mathrm{T}3}+{\mathrm{X}}_{\mathrm{G}3}+{\mathrm{X}}_{23}\right)\\ +\left(X_{\mathrm{T}2}+X_{\mathrm{G}2}+X_{24}\right)\end{array}\right]}\right) \\ =\left(1.16\right)\left(-j5.33\right)\left(\frac{\left[\begin{array}{c}{\left(0.4+0.24+0.2\right)}^{\square }\\ 0.1+0.17+0.2\end{array}\right]}{\left(0.4+0.24+0.2\right)\square \left[\begin{array}{c}\left(0.1+0.17+0.2\right)\\ +\left(0.1+0.17+0.2\right)\end{array}\right]}\right) \\ =\left(1.16\right)\left(-J5.33\right)\left(\frac{0.30}{0.30+0.57}\right) \\ =-J2.13\mathrm{A} \end{gathered}$$(c)

Determine the current contributed from 1-2 line by using the current division rule between the line 1-2 and others as,

$$\begin{gathered} \mathrm{Substitute}1.16\mathrm{for}{l}_{base4}\mathrm{and}-j5.33\mathrm{for}{l}_{\mathrm{Fp}.\mathrm{u}.}\mathrm{into}\mathrm{equation}\left(4\right). \\ {l}_{12\mathrm{F}}=\left(1.16\right)\left(-j5.33\right)\left(\frac{\left[\begin{array}{c}{\left(X_{\mathrm{T}3}+X_{\mathrm{G}3}+X_{23}\right)}^{\square }\\ \left(X_{12}+X_{\mathrm{G}2}+X_{24}\right)\end{array}\right]}{\left(X_{\mathrm{T}3}+X_{\mathrm{G}3}+X_{23}\right)\square \left[\begin{array}{c}\left(X_{\mathrm{T}2}+X_{\mathrm{G}2}+X_{24}\right)\\ +\left(X_{G1}+X_{\mathrm{T}1}+X_{12}\right)\end{array}\right]}\right) \\ =\left(1.16\right)\left(-j5.33\right)\left(\frac{\left[\begin{array}{c}{\left(0.1+0.17+0.2\right)}^{\square }\\ 0.13+0.24+0.2\end{array}\right]}{\left(0.1+0.17+0.2\right)\square \left[\begin{array}{c}\left(0.13+0.24+0.2\right)\\ +\left(0.4+0.24+0.2\right)\end{array}\right]}\right) \\ =\left(1.16\right)\left(-J5.33\right)\left(\frac{0.26}{0.26+0.84}\right) \\ =-J1.46\mathrm{A} \end{gathered}$$

Determine the current contributed from 2-3 line by using the current division rule between the line 2-3 and others as,

$$\begin{gathered} \mathrm{Substitute}1.16\mathrm{for}{l}_{\mathrm{base}4}\mathrm{and}-j5.33\mathrm{for}{l}_{\mathrm{Fp}.\mathrm{u}.}\mathrm{into}\mathrm{equation}\left(5\right). \\ {l}_{23\mathrm{F}}=\left(1.16\right)\left(-j5.33\right)\left(\frac{\left[\begin{array}{c}{\left(X_{\mathrm{G}1}+X_{\mathrm{T}1}+X_{12}\right)}^{\square }\\ \left(X_{\mathrm{T}2}+X_{\mathrm{G}2}+X_{24}\right)\end{array}\right]}{\left(X_{\mathrm{G}1}+X_{\mathrm{T}1}+X_{12}\right)\square \left[\begin{array}{c}\left(X_{\mathrm{T}2}+X_{\mathrm{G}2}+X_{24}\right)\\ +\left(X_{\mathrm{T}3}+X_{\mathrm{G}3}+X_{23}\right)\end{array}\right]}\right) \\ =\left(1.16\right)\left(-j5.33\right)\left(\frac{\left[\begin{array}{c}{\left(0.4+0.24+0.2\right)}^{\square }\\ 0.13+0.24+0.2\end{array}\right]}{\left(0.4+0.24+0.2\right)\square \left[\begin{array}{c}\left(0.13+0.24+0.2\right)\\ +\left(0.1+0.17+0.2\right)\end{array}\right]}\right) \\ =\left(1.16\right)\left(-J5.33\right)\left(\frac{0.34}{0.34+0.47}\right) \\ =-J2.60\mathrm{A} \end{gathered}$$

Determine the current contributed from 2-4 line by using the current division rule between the line 2-4 and others as,

$$\begin{gathered} \mathrm{Substitute}1.16\mathrm{for}{l}_{\mathrm{base}4}\mathrm{and}-j5.33\mathrm{for}{l}_{\mathrm{Fp}.\mathrm{u}.}\mathrm{into}\mathrm{equation}\left(6\right). \\ {l}_{24\mathrm{F}}=\left(1.16\right)\left(-j5.33\right)\left(\frac{\left[\begin{array}{c}{\left(X_{\mathrm{G}1}+X_{\mathrm{T}1}+X_{12}\right)}^{\square }\\ \left(X_{\mathrm{T}3}+X_{\mathrm{G}3}+X_{23}\right)\end{array}\right]}{\left(X_{\mathrm{G}1}+X_{\mathrm{T}1}+X_{12}\right)\square \left[\begin{array}{c}\left({\mathrm{X}}_{\mathrm{T}3}+{\mathrm{X}}_{\mathrm{G}3}+{\mathrm{X}}_{23}\right)\\ +\left(X_{\mathrm{T}2}+X_{\mathrm{G}2}+X_{24}\right)\end{array}\right]}\right) \\ =\left(1.16\right)\left(-j5.33\right)\left(\frac{\left[\begin{array}{c}{\left(0.4+0.24+0.2\right)}^{\square }\\ 0.1+0.17+0.2\end{array}\right]}{\left(0.4+0.24+0.2\right)\square \left[\begin{array}{c}\left(0.1+0.17+0.2\right)\\ +\left(0.1+0.17+0.2\right)\end{array}\right]}\right) \\ =\left(1.16\right)\left(-J5.33\right)\left(\frac{0.30}{0.30+0.57}\right) \\ =-J2.13\mathrm{A} \end{gathered}$$