Question

A 60-Hz, 100-mile, three-phase overhead transmission line, constructed of ACSR conductors, has a series impedance of ${}^{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1826}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{784}\mathbf{)}\mathbf{}\mathbf{\Omega }\mathbf{/}\mathbf{mi}}$per phase and a shunt capacitive reactance-to-neutral of${}^{\mathbf{185}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{\angle }\mathbf{-}{\mathbf{90}}^{\mathbf{°}}\mathbf{}\mathbf{\Omega }\mathbf{‐}\mathbf{mi}\mathbf{}}\mathbf{}$ per phase. Using the nominal ${}^{\mathbf{\pi }}$ circuit for a medium-length transmission line, (a) determine the total series impedance and shunt admittance of the line; (b) compute the voltage, the current, and the real and reactive power at the sending end if the load at the receiving end draws 200 MVA at unity power factor and at a line-to-line voltage of 230 kV; and (c) find the percent voltage regulation of the line.

Short answer

(a)The total series impedance and shunt admittance of the transmission line are ${}^{\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{\angle }\mathbf{76}\mathbf{.}{\mathbf{89}}^{\mathbf{°}}\mathbf{}\mathbf{\Omega }\mathbf{}\mathbf{per}\mathbf{}\mathbf{phase}}$and ${}^{\mathbf{0}\mathbf{.}\mathbf{5391}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{\angle }{\mathbf{90}}^{\mathbf{°}}\mathbf{}\mathbf{S}\mathbf{}\mathbf{per}\mathbf{}\mathbf{phase}}$.

(b)The voltage, current, real and reactive power at the sending end of the transmission line are ${}^{\mathbf{144}\mathbf{.}\mathbf{79}\mathbf{\angle }\mathbf{16}\mathbf{.}{\mathbf{04}}^{\mathbf{°}}\mathbf{}\mathbf{kV}}$, ${}^{\mathbf{496}\mathbf{.}\mathbf{7}\mathbf{\angle }\mathbf{8}\mathbf{.}{\mathbf{5}}^{\mathbf{°}}\mathbf{}\mathbf{A}}$, ${}^{\mathbf{213}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{MW}}$and ${}^{\mathbf{28}\mathbf{.}\mathbf{32}\mathbf{}\mathbf{MVAR}}$.

(c) The percent voltage regulation of the line is ${}^{\mathbf{9}\mathbf{\%}}$

Step-by-step solution

Given data.

The frequency of the transmission line is of 60 Hz frequency.

The series impedance of transmission line${}^{\left(0.1826+j0.784\right)\mathbf{}\mathbf{\Omega }\mathbf{/}\mathbf{mi}}$

The shunt capacitive reactance-to-neutral${}^{\mathbf{185}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{\angle }\mathbf{-}{\mathbf{90}}^{\mathbf{°}}\mathbf{}\mathbf{\Omega }\mathbf{‐}\mathbf{mi}\mathbf{}}\mathbf{}$ per phase.

The load at the receiving end is${}^{\mathbf{200}\mathbf{}\mathbf{MVA}}$ , line to line voltage is ${}^{\mathbf{230}\mathbf{}\mathbf{V}}$ .

The power factor${}^{\mathbf{pf}\mathbf{=}\mathbf{1}\left(\mathrm{unity}\right)}$ .

The line to line voltage is${}^{{\mathbf{V}}_{\mathbf{R}\mathbf{L}\mathbf{L}}\mathbf{=}\mathbf{230}\mathbf{}\mathbf{k}\mathbf{V}}$ .

Consider the nominal  circuit.

Consider the diagram for the nominal ${}^{\mathbf{\pi }}$network.

Figure: network

Write the formula for the voltage regulation.

$\mathbf{VR}\mathbf{\%}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}\mathbf{-}{\mathbf{V}}_{\mathbf{R}}}{{\mathbf{V}}_{\mathbf{R}}}\mathbf{\times }\mathbf{100}\mathbf{\%}$

Here, ${}^{{\mathbf{V}}_{\mathbf{S}}}$is the sending end voltage and ${}^{{\mathbf{V}}_{\mathbf{R}}}$is the receiving end voltage.

Write the formula for the impedance current.

${\mathbf{I}}_{\mathbf{Z}}\mathbf{+}{\mathbf{I}}_{\mathbf{R}}\mathbf{+}{\mathbf{V}}_{\mathbf{R}}\mathbf{\left(}\frac{\mathbf{Y}}{\mathbf{2}}\mathbf{\right)}$

Here, ${}^{{\mathbf{I}}_{\mathbf{Z}}}$is impedance current, ${}^{{\mathbf{I}}_{\mathbf{R}}}$is the receiving end current, ${}^{{\mathbf{V}}_{\mathbf{R}}}$is the receiving-end voltage and ${}^{\mathbf{Y}}$is the admittance.

Determine the total series impedance and shunt admittance of the line.

Determine the total impedance of the transmission line.

$\mathit{Z}\mathbf{=}\mathit{z}\mathit{l}$

Here,${}^{\mathbf{z}}$ is series impedance and ${}^{\mathbf{l}}$is length of line.

Substitute ${}^{\mathbf{0}\mathbf{.}\mathbf{1826}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{784}}$for ${}^{\mathbf{z}}$and ${}^{\mathbf{100}}$for ${}^{\mathbf{l}}$into equation.

$$\begin{gathered} \mathit{Z}\mathbf{=}\left(0.1826+j0.784\right)\left(100\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{80}\mathbf{.}\mathbf{5}\mathbf{\angle }\mathbf{76}\mathbf{.}{\mathbf{89}}^{\mathbf{°}}\mathbf{}\mathit{\Omega }\mathbf{}\mathbf{per}\mathbf{}\mathbf{phase} \end{gathered}$$

Determine the shunt admittance of the transmission line.

$\mathit{Y}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{X}}_{\mathbf{c}}\mathbf{\angle }\mathbf{-}{\mathbf{90}}^{\mathbf{°}}}$

Here,${}^{\mathbf{Y}}$ is admittance and ${}^{{\mathbf{X}}_{\mathbf{c}}}$is shunt capacitance reactance of the line.

Substitute ${}^{\mathbf{\left(}\frac{\mathbf{185}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}}{\mathbf{100}}\mathbf{\right)}}$for ${}^{{\mathbf{X}}_{\mathbf{c}}}$into equation.

$$\begin{gathered} \mathit{Y}\mathbf{=}\frac{\mathbf{1}}{\left({\displaystyle \frac{185.5\times {10}^3}{100}}\right)\mathbf{\angle }{\mathbf{90}}^{\mathbf{°}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5391}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{\angle }{\mathbf{90}}^{\mathbf{°}}\mathbf{}\mathbf{S}\mathbf{}\mathbf{per}\mathbf{}\mathbf{phase} \end{gathered}$$

Determine the voltage, current, real and reactive power at the sending end of the transmission line.

Determine the line to neutral receiving end voltage.

$$\begin{gathered} {\mathit{V}}_{\mathit{R}\mathit{L}\mathit{N}}\mathbf{=}\frac{\mathbf{230}}{\sqrt{\mathbf{3}}}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{132}\mathbf{.}\mathbf{8}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{kV} \end{gathered}$$

Determine the receiving end current (line to line).

${\mathit{I}}_{\mathbf{R}\mathbf{L}\mathbf{L}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{R}}}{\sqrt{\mathbf{3}\mathbf{\times }{\mathbf{V}}_{\mathbf{R}\mathbf{L}\mathbf{L}}}}$

Here, localid="1655349931569" ${}^{{\mathbf{I}}_{\mathbf{RLL}}}$is the receiving end line current, ${}^{{\mathbf{S}}_{\mathbf{R}}}$is receiving-end power and ${}^{{\mathbf{V}}_{\mathbf{R}\mathbf{L}\mathbf{L}}}$is receiving-end voltage (line to line).

Substitute ${}^{\mathbf{200}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}}$for ${}^{{\mathbf{S}}_{\mathbf{R}}}$and ${}^{\mathbf{230}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}}$for ${}^{{\mathbf{V}}_{\mathbf{R}\mathbf{L}\mathbf{L}}}$into equation.

$$\begin{gathered} {\mathit{I}}_{\mathbf{R}\mathbf{L}\mathbf{L}}\mathbf{=}\frac{\mathbf{200}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}}{\sqrt{\mathbf{3}\mathbf{\times }\mathbf{230}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}}}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{502}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{}\mathbf{A} \end{gathered}$$

Apply Kirchhoff current law in nominal ${}^{\mathbf{\pi }}$network shown above.

${\mathit{I}}_{\mathbf{Z}}\mathbf{=}{\mathit{I}}_{\mathbf{R}}\mathbf{=}{\mathit{V}}_{\mathbf{R}}\left(\frac{Y}{2}\right)$

Here, ${}^{{\mathbf{I}}_{\mathbf{Z}}}$is impedance current, ${}^{{\mathbf{I}}_{\mathbf{R}}}$is the receiving end current,${}^{{\mathbf{V}}_{\mathbf{R}}}$is the receiving-end voltage and ${}^{\mathbf{Y}}$is the admittance.

Substitute ${}^{\mathbf{502}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}}$for localid="1655354224584" ${}^{{\mathbf{I}}_{\mathbf{R}}}$, ${}^{\mathbf{132}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}}$ for ${}^{{\mathbf{V}}_{\mathbf{R}}}$and ${}^{\mathbf{0}\mathbf{.}\mathbf{5391}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{\angle }{\mathbf{90}}^{\mathbf{°}}}$for ${}^{\mathbf{Y}}$into equation.

$$\begin{gathered} {\mathit{I}}_{\mathit{Z}}\mathbf{=}\mathbf{502}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{+}\left(132.8\times {10}^3\angle 0^{°}\right)\left(\frac{0.5391\times {10}^{-3}\angle {90}^{°}}{2}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{502}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{+}\left(132.9\times {10}^3\angle 0^{°}\right)\left(0.27\times {10}^{-3}\angle {90}^{°}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{502}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{+}\mathbf{35}\mathbf{.}\mathbf{856}\mathbf{\angle }{\mathbf{90}}^{\mathbf{°}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{503}\mathbf{\angle }\mathbf{4}\mathbf{.}{\mathbf{09}}^{\mathbf{°}}\mathbf{}\mathbf{A} \end{gathered}$$

Apply Kirchhoff voltage law in nominal ${}^{\mathbf{\pi }}$network shown above.

${\mathit{V}}_{\mathbf{S}}\mathbf{=}{\mathit{V}}_{\mathbf{R}}\mathbf{+}{\mathit{I}}_{\mathbf{Z}}\mathit{Z}$

Here,${}^{{\mathbf{V}}_{\mathbf{R}}}$is the receiving-end voltage,${}^{{\mathbf{I}}_{\mathbf{Z}}}$is impedance current and ${}^{\mathbf{Z}}$is total series impedance.

Substitute ${}^{\mathbf{132}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}}$for ${}^{{\mathbf{V}}_{\mathbf{R}}}$, ${}^{\mathbf{503}\mathbf{.}\mathbf{3}\mathbf{\angle }\mathbf{4}\mathbf{.}{\mathbf{09}}^{\mathbf{°}}}$for localid="1655352633286" ${}^{{\mathbf{I}}_{\mathbf{Z}}}$and ${}^{\mathbf{80}\mathbf{.}\mathbf{5}\mathbf{\angle }\mathbf{76}\mathbf{.}{\mathbf{89}}^{\mathbf{°}}}$for ${}^{\mathbf{Z}}$into equation.

localid="1655356943488" $$\begin{gathered} {\mathit{V}}_{\mathbf{S}}\mathbf{=}\mathbf{(}\mathbf{132}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{503}\mathbf{.}\mathbf{3}\mathbf{\angle }\mathbf{4}\mathbf{.}{\mathbf{09}}^{\mathbf{°}}\mathbf{)}\mathbf{(}\mathbf{80}\mathbf{.}\mathbf{5}\mathbf{\angle }\mathbf{76}\mathbf{.}{\mathbf{89}}^{\mathbf{°}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{132}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{+}\mathbf{6352}\mathbf{+}\mathit{J}\mathbf{40014}\mathbf{.}\mathbf{6} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{139152}\mathbf{+}\mathbf{}\mathit{J}\mathbf{40014}\mathbf{.}\mathbf{6} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{144}\mathbf{.}\mathbf{79}\mathbf{\angle }\mathbf{16}\mathbf{.}{\mathbf{04}}^{\mathbf{°}\mathbf{}}\mathit{k}\mathit{V} \end{gathered}$$

Apply Kirchhoff current law in nominal ${}^{\mathbf{\pi }}$network shown above.

Determine the sending-end current.

${\mathit{I}}_{\mathbf{S}}\mathbf{=}{\mathit{I}}_{\mathbf{Z}}\mathbf{+}{\mathit{V}}_{\mathbf{S}}\left(\frac{Y}{2}\right)$

Here, ${}^{{\mathbf{I}}_{\mathbf{Z}}}$is impedance current, ${}^{{\mathbf{S}}_{\mathbf{R}}}$is the sending end voltage and ${}^{\mathbf{Y}}$is the admittance.

Substitute${}^{\mathbf{502}\mathbf{+}\mathbf{J}\mathbf{35}\mathbf{.}\mathbf{86}}$for localid="1655354202522" ${}^{{\mathbf{I}}_{\mathbf{Z}}}$, ${}^{\mathbf{144}\mathbf{.}\mathbf{79}\mathbf{\angle }\mathbf{16}\mathbf{.}{\mathbf{04}}^{\mathbf{°}}}$ for ${}^{{\mathbf{V}}_{\mathbf{S}}}$and ${}^{\mathbf{0}\mathbf{.}\mathbf{5391}\mathbf{\angle }{\mathbf{90}}^{\mathbf{°}}}$for ${}^{\mathbf{Y}}$into equation.

$$\begin{gathered} {\mathit{I}}_{\mathbf{S}}\mathbf{=}\left(502+J35.86\right)\mathbf{+}\left(144.79\angle 16.{04}^{°}\right)\left(\frac{0.5391\angle {90}^{°}}{2}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{502}\mathbf{+}\mathit{J}\mathbf{35}\mathbf{.}\mathbf{86}\mathbf{-}\mathbf{10}\mathbf{.}\mathbf{78}\mathbf{+}\mathit{J}\mathbf{37}\mathbf{.}\mathbf{8} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{491}\mathbf{.}\mathbf{2}\mathbf{+}\mathit{J}\mathbf{73}\mathbf{.}\mathbf{46} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{496}\mathbf{.}\mathbf{7}\mathbf{\angle }\mathbf{8}\mathbf{.}{\mathbf{5}}^{\mathbf{°}}\mathbf{}\mathbf{A} \end{gathered}$$

Determine the formula for 3 phase sending end power.

${\mathit{S}}_{\mathbf{S}\mathbf{3}\mathbf{\varphi }}\mathbf{=}\mathbf{3}{\mathit{V}}_{\mathbf{S}}{\mathit{I}}_{\mathbf{S}}^{\mathbf{*}}$

Here, ${}^{{\mathit{S}}_{\mathbf{S}\mathbf{3}\mathbf{\varphi }}}$is sending-end power, ${}^{{\mathbf{V}}_{\mathbf{S}}}$is the sending end voltage and ${}^{{\mathbf{I}}_{\mathbf{S}}^{\mathbf{*}}}$is real power of transmission line.

Substitute ${}^{\mathbf{144}\mathbf{.}\mathbf{79}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{\angle }\mathbf{16}\mathbf{.}{\mathbf{04}}^{\mathbf{°}}}$for ${}^{{\mathbf{V}}_{\mathbf{S}}}$and ${}^{\mathbf{496}\mathbf{.}\mathbf{7}\mathbf{\angle }\mathbf{8}\mathbf{.}{\mathbf{5}}^{\mathbf{°}}}$for ${}^{{\mathbf{I}}_{\mathbf{S}}^{\mathbf{*}}}$into equation.

$$\begin{gathered} {\mathbf{S}}_{\mathbf{S}\mathbf{3}\mathbf{\varphi }}\mathbf{=}\mathbf{3}\left(144.79\times {10}^3\angle 16.{04}^{°}\right)\left(496.7\angle 8.5^{°}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\left(71.3+J9.44\right)\mathbf{\times }{\mathbf{10}}^{\mathbf{6}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(213.9+J28.31\right)\mathbf{\times }{\mathbf{10}}^{\mathbf{6}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{213}\mathbf{.}\mathbf{9}\mathbf{}\mathit{M}\mathit{W}\mathbf{+}\mathit{J}\mathbf{28}\mathbf{.}\mathbf{31}\mathbf{}\mathbf{MVAR} \end{gathered}$$

Since,

$\mathbf{complex}\mathbf{}\mathbf{power}\mathbf{}\mathbf{\left(}\mathbf{S}\mathbf{\right)}\mathbf{=}\mathbf{real}\mathbf{}\mathbf{power}\mathbf{}\mathbf{\left(}\mathbf{P}\mathbf{\right)}\mathbf{+}\mathbf{jreactive}\mathbf{}\mathbf{power}\mathbf{\left(}\mathbf{Q}\mathbf{\right)}$

Compare the values $\mathbf{213}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{MW}\mathbf{+}\mathbf{}\mathit{J}\mathbf{28}\mathbf{.}\mathbf{31}\mathbf{}\mathbf{MVAR}$from equation.

localid="1655350424446" $$\begin{gathered} {\mathbf{P}}_{\mathbf{S}\mathbf{3}\mathbf{\varphi }}\mathbf{=}\mathbf{213}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{MW} \\ {\mathbf{Q}}_{\mathbf{S}\mathbf{3}\mathbf{\varphi }}\mathbf{=}\mathbf{28}\mathbf{.}\mathbf{32}\mathbf{}\mathbf{MVAR} \end{gathered}$$

Therefore, the values of sending end real and reactive powers are

$\mathbf{213}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{MW}\mathbf{,}\mathbf{}\mathbf{28}\mathbf{.}\mathbf{32}\mathbf{}\mathbf{MVAR}$

Determine the percent voltage regulation of the transmission line.

(c)

Write the formula for percent voltage regulation.

$\mathbf{VR}\mathbf{\%}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}\mathbf{-}{\mathbf{V}}_{\mathbf{R}}}{{\mathbf{V}}_{\mathbf{R}}}\mathbf{\times }\mathbf{100}\mathbf{\%}$

Here, ${}^{{\mathbf{V}}_{\mathbf{S}}}$is sending-end voltage and ${}^{{\mathbf{V}}_{\mathbf{R}}}$is receiving-end voltage.

Substitute ${}^{\mathbf{144}\mathbf{.}\mathbf{79}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}}$for ${}^{{\mathbf{V}}_{\mathbf{S}}}$and ${}^{\mathbf{132}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}}$for ${}^{{\mathbf{V}}_{\mathbf{R}}}$ into equation.

$$\begin{gathered} \mathbf{VR}\mathbf{\%}\mathbf{=}\frac{\mathbf{144}\mathbf{.}\mathbf{79}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{-}\mathbf{132}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}}{\mathbf{132}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{09}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{9}\mathbf{\%} \end{gathered}$$