Question

Using per-unit calculations rework Problem 5.18 to determine the sending-end voltage and current.

Short answer

(a)The value of sending-end voltage of the line is ${}^{\mathbf{230}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{kV}}$.

(b)The value of sending-end current of the line is${}^{\mathbf{332}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{A}}$ .

Step-by-step solution

Given data.

As using reference problem 5.18.

The given load at receiving-end of line at unity power factor is 125 MW.

The given load at receiving-end of line is 215 kV.

The given frequency of line is 60-Hz.

The given length of the line${}^{\mathbf{l}\mathbf{=}\mathbf{230}\mathbf{}\mathbf{mile}}$ .

The given series impedance of line isrole="math" localid="1655379865722" ${}^{\mathit{Z}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8431}\mathbf{\angle }\mathbf{79}\mathbf{.}{\mathbf{04}}^{\mathbf{°}}}\frac{\mathbf{\Omega }}{\mathbf{mi}}$.

The given shunt admittance of line is${}^{\mathbf{y}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{105}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{\angle }{\mathbf{90}}^{\mathbf{°}}\frac{\mathbf{S}}{\mathbf{mi}}}$.

Determine the required formula.

Write the formula of sending-end voltage of the line.

…… (1)

Here, ${}^{{\mathbf{V}}_{{\mathbf{R}}_{\mathbf{p}\mathbf{u}}}}$is receiving-end per unit voltage, ${}^{{\mathit{l}}_{{\mathit{R}}_{\mathit{p}\mathit{u}}}}$is receiving-end current, ${}^{{\mathbf{Z}}_{\mathbf{c}}}$is characteristics impedance and ${}^{\mathit{\gamma }\mathit{l}}$is dimensionless quantity.

Write the formula of sending-end current of the line.

${\mathbf{l}}_{\mathbf{S}}\mathbf{=}{\mathbf{l}}_{{\mathbf{R}}_{\mathbf{pu}}\mathbf{}}\mathbf{cosh}\left(\mathrm{gl}\right)\mathbf{+}\frac{{\mathbf{V}}_{{\mathbf{R}}_{\mathbf{pu}}}}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{sinh}\left(\mathrm{gl}\right)$ …… (2)

Here, ${}^{{\mathbf{V}}_{{\mathbf{R}}_{\mathbf{p}\mathbf{u}}}}$is receiving-end voltage, ${}^{{\mathit{l}}_{{\mathit{R}}_{\mathit{p}\mathit{u}}}}$is receiving-end current, ${}^{{\mathbf{Z}}_{\mathbf{c}}}$is characteristics impedance and role="math" localid="1655380081055" ${}^{\mathit{\gamma }\mathit{l}}$is dimensionless quantity.

Determine the sending-end voltage.

Assume a base of 125 MVA and 215 kV,

Determine the base impedance.

${\mathit{Z}}_{\mathbf{B}\mathbf{A}\mathbf{S}\mathbf{E}}\mathbf{=}\frac{{\left(K{V}_{LL}\right)}^{\mathbf{2}}}{\mathbf{M}\mathbf{V}{\mathbf{A}}_{\mathbf{3}\mathbf{\varphi }}}\left(ohms\right)$ …… (3)

Here, ${}^{\mathit{K}{\mathit{V}}_{\mathit{L}\mathit{L}}}$is line-to-line receiving-end voltage and localid="1655380577570" ${}^{\mathit{M}\mathit{V}{\mathit{A}}_{\mathit{3}\mathit{\varphi }}}$is receiving-end load at unity power factor.

Substitute ${}^{\mathbf{215}}$for ${}^{\mathit{K}{\mathit{V}}_{\mathit{L}\mathit{L}}}$and ${}^{\mathbf{125}}$for ${}^{\mathit{M}\mathit{V}{\mathit{A}}_{\mathit{3}\mathit{\varphi }}}$into equation (3).

$$\begin{gathered} {\mathit{Z}}_{\mathbf{B}\mathbf{A}\mathbf{S}\mathbf{E}}\mathbf{=}\frac{{\left(215\right)}^{\mathbf{2}}}{\mathbf{125}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{370}\mathbf{}\mathit{\Omega } \end{gathered}$$

Determine the base current.

${\mathit{l}}_{\mathbf{B}\mathbf{A}\mathbf{S}\mathbf{E}}\mathbf{=}\frac{\mathbf{M}\mathbf{V}{\mathbf{A}}_{\mathbf{3}\mathbf{\varphi }}}{\sqrt{\mathbf{3}}\mathbf{\times }\mathbf{K}{\mathbf{V}}_{\mathbf{L}\mathbf{L}}\mathbf{\times }\mathbf{1000}}\left(amps\right)$ …… (4)

Here, ${}^{\mathit{K}{\mathit{V}}_{\mathit{L}\mathit{L}}}$is line-to-line receiving-end voltage and ${}^{\mathit{M}\mathit{V}{\mathit{A}}_{\mathit{3}\mathit{\varphi }}}$is receiving-end load at unity power factor.

Substitute ${}^{\mathbf{125}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}}$for ${}^{\mathit{M}\mathit{V}{\mathit{A}}_{\mathit{3}\mathit{\varphi }}}$and ${}^{\mathbf{215}}$for ${}^{\mathit{K}{\mathit{V}}_{\mathit{L}\mathit{L}}}$into equation (4).

$$\begin{gathered} {\mathit{l}}_{\mathbf{B}\mathbf{A}\mathbf{S}\mathbf{E}\mathbf{=}}\frac{\mathbf{125}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}}{\sqrt{\mathbf{3}\mathbf{\times }\mathbf{215}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{335}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{A} \end{gathered}$$

Determine the characteristic impedance.

${\mathit{Z}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{z}}{\mathbf{y}}$ …… (5)

Here, ${}^{\mathbf{z}}$ is the impedance and ${}^{\mathbf{y}}$is the admittance of the conductor.

Substitute ${}^{\mathbf{0}\mathbf{.}\mathbf{8431}\mathbf{\angle }\mathbf{79}\mathbf{.}{\mathbf{04}}^{\mathbf{°}}}$for ${}^{\mathbf{z}}$and ${}^{\mathbf{5}\mathbf{.}\mathbf{105}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}}$for ${}^{\mathbf{y}}$into equation (5).

$$\begin{gathered} {\mathit{Z}}_{\mathbf{c}}\mathbf{=}\sqrt{\frac{\mathbf{0}\mathbf{.}{\mathbf{8431}}^{\mathbf{°}}}{\mathbf{5}\mathbf{.}\mathbf{105}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}}}\mathbf{\angle }\left(79.{04}^{°}-{90}^{°}\right)\mathbf{/}\mathbf{2} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{406}\mathbf{.}\mathbf{4}\mathbf{\angle }\mathbf{-}\mathbf{5}\mathbf{.}{\mathbf{48}}^{\mathbf{°}}\mathit{\Omega } \end{gathered}$$

Now, determine the characteristic impedance in per unit (Pu).

${\mathit{Z}}_{\mathbf{P}\mathbf{U}}\mathbf{=}\frac{\mathbf{Z}\left(Inohms\right)}{{\mathbf{Z}}_{\mathbf{B}\mathbf{A}\mathbf{S}\mathbf{E}}}$ …… (6)

Here ${}^{\mathbf{Z}}$is series impedance and ${}^{{\mathbf{Z}}_{\mathbf{BASE}}}$is base impedance.

Substitute ${}^{\mathbf{406}\mathbf{.}\mathbf{4}\mathbf{\angle }\mathbf{-}\mathbf{5}\mathbf{.}{\mathbf{48}}^{\mathbf{°}}}$for localid="1655381682434" ${}^{\mathbf{Z}\mathbf{(}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{o}\mathbf{h}\mathbf{m}\mathbf{s}\mathbf{)}}$and ${}^{\mathbf{370}}$for ${}^{{\mathbf{Z}}_{\mathbf{BASE}}}$into equation (6).

$$\begin{gathered} {\mathit{Z}}_{\mathbf{P}\mathbf{U}}\mathbf{=}\frac{\mathbf{406}\mathbf{.}\mathbf{4}\mathbf{\angle }\mathbf{-}\mathbf{5}\mathbf{.}{\mathbf{48}}^{\mathbf{°}}}{\mathbf{370}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{098}\mathbf{\angle }\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{48}\mathbf{}\mathit{p}\mathit{u} \end{gathered}$$

Determine the line to neutral receiving end voltage.

$$\begin{gathered} {\mathit{V}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{215}}{\sqrt{\mathbf{3}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{124}\mathbf{.}\mathbf{13}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{}\mathbf{kV}\mathbf{/}\mathbf{ph} \end{gathered}$$

Now, the line to neutral receiving end voltage in per unit (Pu)

${\mathit{V}}_{{\mathbf{R}}_{\mathbf{P}\mathbf{U}}\mathbf{=}}\frac{{\mathbf{V}}_{\mathbf{R}}\left(InkV/ph\right)}{{\mathbf{V}}_{{\mathbf{R}}_{\mathbf{B}\mathbf{A}\mathbf{S}\mathbf{E}}}}$ …… (7)

Here, ${}^{{\mathbf{V}}_{\mathbf{R}}\mathbf{(}\mathbf{In}\mathbf{}\mathbf{kV}\mathbf{/}\mathbf{ph}\mathbf{)}}$is line to neutral receiving-end voltage and ${}^{{\mathbf{V}}_{{\mathbf{R}}_{\mathbf{B}\mathbf{A}\mathbf{S}\mathbf{E}}}}$is line to neutral receiving-end base voltage.

Substitute${}^{\mathbf{124}\mathbf{.}\mathbf{13}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}}$for ${}^{{\mathbf{V}}_{\mathbf{R}}\mathbf{(}\mathbf{In}\mathbf{}\mathbf{kV}\mathbf{/}\mathbf{ph}\mathbf{)}}$and ${}^{\mathbf{124}\mathbf{.}\mathbf{13}}$for${}^{{\mathbf{V}}_{{\mathbf{R}}_{\mathbf{B}\mathbf{A}\mathbf{S}\mathbf{E}}}}$into equation (7).

$$\begin{gathered} {\mathit{V}}_{{\mathbf{R}}_{\mathbf{P}\mathbf{U}}}\mathbf{=}\frac{\mathbf{124}\mathbf{.}\mathbf{13}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}}{\mathbf{124}\mathbf{.}\mathbf{13}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{pu} \end{gathered}$$

Determine the receiving end current.

$\mathbf{=}\frac{\mathbf{S}}{\mathbf{3}{\mathbf{V}}_{\mathbf{R}}}\mathbf{\angle }\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{PF}$ …… (8)

Here, ${}^{\mathbf{S}}$is complex power, ${}^{{\mathbf{V}}_{\mathbf{R}}}$is receiving-end voltage and ${}^{\mathbf{P}\mathbf{F}}$is power factor.

Substitute ${}^{\mathbf{125}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}}$for${}^{\mathbf{S}}$,${}^{\mathbf{215}}$for ${}^{{\mathbf{V}}_{\mathbf{R}}}$and ${}^{\mathbf{0}}$for PF into equation (8).

$$\begin{gathered} {\mathit{l}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{125}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}}{\sqrt{\mathbf{3}}\mathbf{\times }\mathbf{215}}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{335}\mathbf{.}\mathbf{7}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{}\mathbf{A} \end{gathered}$$

Now, determine the receiving end current in per unit (Pu)

.

${\mathit{l}}_{{\mathbf{R}}_{\mathbf{P}\mathbf{U}}}\mathbf{=}\frac{{\mathbf{l}}_{\mathbf{R}}\mathbf{}\left(Inamps\right)}{{\mathbf{l}}_{{\mathbf{R}}_{\mathbf{B}\mathbf{A}\mathbf{S}\mathbf{E}}}}$ …… (9)

Here,localid="1655382902072" ${}^{{\mathit{l}}_{\mathit{R}}\mathit{}\mathit{(}\mathit{I}\mathit{n}\mathit{}\mathit{a}\mathit{m}\mathit{p}\mathit{s}\mathit{)}}$ is receiving-end current and ${}^{{\mathbf{l}}_{{\mathbf{R}}_{\mathbf{B}\mathbf{A}\mathbf{S}\mathbf{E}}}}$is receiving-end base current.

Substitute ${}^{\mathbf{335}\mathbf{.}\mathbf{7}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}}$for ${}^{{\mathit{l}}_{\mathit{R}}\mathit{}\mathit{(}\mathit{I}\mathit{n}\mathit{}\mathit{a}\mathit{m}\mathit{p}\mathit{s}\mathit{)}}$and for ${}^{{\mathbf{l}}_{{\mathbf{R}}_{\mathbf{BASE}}}}$into equation (9).

$$\begin{gathered} {\mathit{l}}_{{\mathbf{R}}_{\mathbf{P}\mathbf{U}}}\mathbf{=}\frac{\mathbf{335}\mathbf{.}\mathbf{7}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}}{\mathbf{335}\mathbf{.}\mathbf{7}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{\angle }{\mathbf{0}}^{\mathbf{°}}\mathbf{}\mathbf{pu} \end{gathered}$$

Determine the value of ${}^{\mathbf{A}\mathbf{B}\mathbf{C}\mathbf{D}}$parameters.

Determine the value of${}^{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{h}\left(\gamma l\right)}$.

$$\begin{gathered} \mathit{c}\mathit{o}\mathit{s}\mathit{h}\left(\gamma l\right)\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\left(0.0456+j0.475\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{0}\mathbf{.}\mathbf{0456}}\mathbf{}\mathbf{\angle }\mathbf{27}\mathbf{.}{\mathbf{22}}^{\mathbf{°}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{0456}}\mathbf{}\mathbf{\angle }\mathbf{-}\mathbf{27}\mathbf{.}{\mathbf{22}}^{\mathbf{°}}}{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8904}\mathbf{\angle }\mathbf{1}\mathbf{.}{\mathbf{34}}^{\mathbf{°}}\mathbf{}\mathbf{A} \end{gathered}$$

Determine the value of. ${}^{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{h}\left(\gamma l\right)}$

$$\begin{gathered} \sin \left(\mathrm{\gamma l}\right)=\sinh (0.0456+\mathrm{j}0.475) \\ =\frac{{\mathrm{e}}^{0.0456}\angle 27.22°-{\mathrm{e}}^{-0.0456}\angle -27.22°}{2} \\ =0.4597\angle 84.93° \\ \mathrm{Now},\mathrm{determine}\mathrm{the}\mathrm{sending}-\mathrm{end}\mathrm{voltage}\mathrm{of}\mathrm{line}. \\ \mathrm{Substitute}1\angle 0°\mathrm{for}{\mathrm{V}}_{\mathrm{Rpu}},0.8904\angle 1.34°\mathrm{for}\cosh \left(\mathrm{\gamma l}\right),1\angle 0°\mathrm{for}{\mathrm{l}}_{\mathrm{Rpu}},1.098\angle -5.48° \\ \mathrm{for}{\mathrm{Z}}_{\mathrm{c}}\mathrm{nnd}0..4597\angle 84.93°\mathrm{for}\sinh \left(\mathrm{\gamma l}\right)\mathrm{into}\mathrm{equation}\left(1\right). \\ {\mathrm{V}}_{\mathrm{s}}=1\angle 0°\times 0.8904\angle 1.34°)+(1\angle 0°\times 1.098\angle -5.48°\times 0.4597\angle 84.93°) \\ =1.1102\angle 27.75°\mathrm{pu} \\ \mathrm{Now},\mathrm{determine}\mathrm{the}\mathrm{line}\mathrm{to}\mathrm{line}\mathrm{sending}\mathrm{end}\mathrm{voltage}. \\ {\mathrm{V}}_{\mathrm{S}.\mathrm{LL}}=\left(1.1102\right)\times 215 \\ =238.7\mathrm{kV} \end{gathered}$$

Determine the sending-end current.

$$\begin{gathered} \mathrm{Substitute}1\angle 0°\mathrm{for}{\mathrm{V}}_{\mathrm{Rpu}},0.8904\angle 1.34°\mathrm{for}\cosh \left(\mathrm{\gamma l}\right),1\angle 0°\mathrm{for}{\mathrm{l}}_{\mathrm{Rpu}},1.098\angle -5.48° \\ \mathrm{for}{\mathrm{Z}}_{\mathrm{c}}\mathrm{nnd}0..4597\angle 84.93°\mathrm{for}\sinh \left(\mathrm{\gamma l}\right)\mathrm{into}\mathrm{equation}\left(1\right). \\ {\mathrm{l}}_{\mathrm{s}}=1\angle 0°\times 0.8904\angle 1.34°)+\frac{(1.0\angle 0°}{1.098\angle -5.48°}\times 0.4597\angle 84.93°) \\ =0.99\angle 26.35°\mathrm{pu} \\ \mathrm{Now},\mathrm{determine}\mathrm{the}\mathrm{line}\mathrm{to}\mathrm{line}\mathrm{sending}\mathrm{end}\mathrm{current}. \\ {\mathrm{l}}_{\mathrm{S}.\mathrm{LL}}=0.99\times 335.7° \\ =332.3\mathrm{A} \\ \mathrm{Therefore},\mathrm{the}\mathrm{sending}\mathrm{end}\mathrm{current}\mathrm{is}332.3\mathrm{A} \end{gathered}$$