Question

Nodal equations $\mathbf{I}\mathbf{=}{\mathbf{Y}}_{\mathbf{bus}}\mathbf{}\mathbf{V}$are a set of linear equations

analogous tolocalid="1655199005807" $\mathbf{y}\mathbf{=}\mathbf{Ax}$.

(a) True

(b) False

Short answer

The correct answer is (a) true.

Step-by-step solution

Determine the formula of Nodal equations.            

Write the formula of nodal equation for a given set of linear equation $\mathbf{Y}\mathbf{=}\mathbf{Ax}$.

$\mathbf{I}\mathbf{=}{\mathbf{Y}}_{\mathbf{bus}}\mathbf{V}$ …… (1)

Here, $\mathit{I}$is the current source vector, $\mathit{V}$is the voltage source vector and ${\mathbf{Y}}_{\mathbf{bus}}$is

bus matrix.

Determine the correct answer.

Using, ${\mathrm{Y}}_{\mathrm{bus}}$the nodal equations for a power system network are written as

$\mathrm{I}={\mathrm{Y}}_{\mathrm{bus}}\mathrm{V}$ …… (2)

Here, $\mathit{I}$is the N vector of source currents injected into each bus and V is the

voltage source vector and ${\mathrm{Y}}_{\mathrm{bus}}$is bus matrix.

For bus k , the kthequation (2) of current/.

$I_k=\sum _{n-1}^N{Y}_{kn}{V}_n$ …… (3)

Here, $Y_{kn}$is rectangular coordinate and $V_n$is N vector of voltage source.

The complex power delivered to bus k .

$S_k=P_k+j{Q}_k=V_k{I}_k^{*}$ …… (4)

Here,$V_k$is N vector of voltage source, and $I_k^{*}$is N vector of current source.

Substitute ${\left[\sum _{n-1}^N{Y}_{kn}{V}_n\right]}^{*}$for$i_k^{*}$ into equation (4).

${\mathrm{P}}_{\mathrm{k}}+j{\mathrm{Q}}_{\mathrm{k}}={\mathrm{V}}_{\mathrm{k}}{\left[\sum _{n-1}^N{Y}_{kn}{V}_n\right]}^{*}k=1,2,..,N$ …… (5)

Here,$Y_{kn}$is rectangular coordinate,$V_k$is N vector of voltage source,$V_n$is

voltage source.

Substitute following notation $V_n{e}^{j\delta n}$for $V_n$and $V_n{e}^{j\delta n}$for$V_{kn}$into equation (5)

${\mathrm{P}}_{\mathrm{k}}+{\mathrm{jQ}}_{\mathrm{k}}={\mathrm{V}}_{\mathrm{k}}\sum _{\mathrm{n}-1}^{\mathrm{n}}{\mathrm{Y}}_{\mathrm{kn}}{\mathrm{V}}_{\mathrm{n}}{\mathrm{e}}^{\mathrm{j}\left({\mathrm{\delta }}_{\mathrm{k}}-{\mathrm{\delta }}_{\mathrm{n}}-{\mathrm{\delta }}_{\mathrm{kn}}\right)}$ …… (6)

Here, $Y_{kn}$is rectangular coordinate, $V_k$is N vector of voltage source, $Y_n$is voltage source and $e^{J\left({\mathrm{\delta }}_{\mathrm{k}}-{\mathrm{\delta }}_{\mathrm{n}}-{\mathrm{\delta }}_{\mathrm{kn}}\right)}$is exponential form.

Taking the real and imaginary parts of (6), the power balance equations are written as either

${\mathrm{P}}_{\mathrm{k}}={\mathrm{V}}_{\mathrm{k}}\sum _{\mathrm{n}-1}^{\mathrm{n}}{\mathrm{Y}}_{\mathrm{kn}}{\mathrm{V}}_{\mathrm{n}}\cos \left({\mathrm{\delta }}_{\mathrm{k}}-{\mathrm{\delta }}_{\mathrm{n}}-{\mathrm{\delta }}_{\mathrm{kn}}\right)$

Here, is real power, is rectangular coordinate, is vector of voltage source and is voltage source

$Q_{\mathrm{k}}={\mathrm{V}}_{\mathrm{k}}\sum _{\mathrm{n}-1}^{\mathrm{n}}{\mathrm{Y}}_{\mathrm{kn}}{\mathrm{V}}_{\mathrm{n}}\sin \left({\mathrm{\delta }}_{\mathrm{k}}-{\mathrm{\delta }}_{\mathrm{n}}-{\mathrm{\delta }}_{\mathrm{kn}}\right)\mathrm{k}=1,2..,\mathrm{N}$

Here, $Q_k$is reactive power, $Y_{kn}$is rectangular coordinate, $V_k$is N vector of

voltage source and $V_n$is voltage source

Or when the is expressed in rectangular coordinates as,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{k}}={\mathrm{V}}_{\mathrm{k}}\sum _{\mathrm{n}-1}^{\mathrm{n}}{V}_n\left[G_{kn}\cos \left({\delta }_k-{\delta }_n\right)+B_{kn}\sin \left({\delta }_k-{\delta }_n\right)\right] \\ {Q}_{\mathrm{k}}={\mathrm{V}}_{\mathrm{k}}\sum V_n\left[G_{kn}\sin \left({\delta }_k-{\delta }_n\right)-B_{kn}\cos \left({\delta }_k-{\delta }_n\right)\right]K=1,2..,N \end{gathered}$$

Hence, nodal gauss equations are a set of linear equations analogous to

y = Ax.

Therefore, the correct answer is (a) true.