Question

For the line in Problems 5.14 and 5.38, determine (a) the practical line load ability in MW, assuming ${\mathbf{V}}_{\mathbf{s}}\mathbf{=}\mathbf{1}$ per unit,role="math" localid="1655187951347" ${\mathbf{V}}_{\mathbf{R}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{95}$ per unit, and${\mathbf{d}}_{\mathbf{max}}\mathbf{=}{\mathbf{35}}^{\mathbf{0}}$ ; part (b) the full-load current at $\mathbf{0}\mathbf{.}\mathbf{99}$ pf. leading, based on the above practical line load ability; (c) the exact receiving-end voltage for the full-load current in part (b); and (d) the percent voltage regulation. For this line, is load ability determined by the thermal limit, the voltage-drop limit, or steady-state stability?

Short answer

(a) The practical line load ability is $921.25\mathrm{MW}$.

(b) The full load current at receiving end at$0.99$ power factor is $1.131\angle 8.{109}^0\mathrm{kA}$.

(c) The receiving end voltage for the full load current is $528.6\mathrm{kV}$.

(d) The voltage regulation is$16.1\%$ .

The ratio of voltage at full load and base voltage is greater than $0.95$. Therefore, laudability determined by steady state stability.

Step-by-step solution

Write the given data by the question.

The length of the transmission line,$\mathrm{l}=500\mathrm{km}$

The positive sequence series impedance,$\mathrm{Z}=0.03+\mathrm{j}0.35\frac{\mathrm{\Omega }}{\mathrm{km}}$

The positive sequence shunt admittance,$\mathrm{y}=\mathrm{j}4.4\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}$

Per unit sending end voltage,${\mathrm{V}}_{\mathrm{S}}=1\mathrm{pu}$

Per unit receiving end voltage,${\mathrm{V}}_{\mathrm{R}}=0.95\mathrm{pu}$

The sending ending base voltage,${\mathrm{V}}_{\mathrm{s}.\mathrm{Base}}=500\mathrm{kV}$

Receiving end power factor, $\mathrm{pf}=1$

Maximum phase angle, ${\mathrm{\delta }}_{\max }={35}^0$

Determine the formulas to calculate the load ability, full load current, receiving end voltage and percentage voltage regulation.

The equation to calculate the propagation constant is given as follows,

$\mathbf{\gamma l}\mathbf{=}\mathbf{l}\sqrt{\mathbf{zy}}$ …… (1)

The equation to calculate the characteristic impedance is given as follows.

${\mathbf{Z}}_{\mathbf{c}}\mathbf{=}\sqrt{\frac{\mathbf{z}}{\mathbf{y}}}$ …… (2)

The equation to calculate the ABCD parameter of the transmission line is given as follows.

role="math" localid="1655189664686" $$\begin{gathered} \mathbf{A}\mathbf{=}\mathbf{D} \\ \mathbf{A}\mathbf{=}\mathbf{cosh}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}\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{}....\left(3a\right) \\ \mathbf{B}\mathbf{=}{\mathbf{Z}}_{\mathbf{c}}\mathbf{sinh}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}\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{}.....\left(3b\right) \\ \mathbf{C}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{sinh}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}\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{}......\left(3\mathrm{c}\right) \end{gathered}$$

The equation to calculate the per unit reactance is given as follows.

${\mathbf{X}}_{\mathbf{pu}}\mathbf{=}\frac{{\mathbf{X}}_{\mathbf{actual}}}{{\mathbf{X}}_{\mathbf{Base}}}$ …… (4)

The equation to calculate the sending end voltage is given as follows.

role="math" localid="1655189909068" ${\mathbf{V}}_{\mathbf{S}}\mathbf{=}{\mathbf{V}}_{\mathbf{s}\mathbf{p}\mathbf{u}}{\mathbf{V}}_{\mathbf{S}\mathbf{.}\mathbf{B}\mathbf{a}\mathbf{s}\mathbf{e}}$ …… (5)

The equation to calculate the receiving end voltage is given as follows.

${\mathbf{V}}_{\mathbf{R}}\mathbf{=}{\mathbf{V}}_{\mathbf{Rpu}}{\mathbf{V}}_{\mathbf{S}\mathbf{.}\mathbf{Base}}$ …… (6)

The equation to calculate the practical line load ability is given as follows.

${\mathbf{P}}_{\mathbf{R}\mathbf{.}\mathbf{max}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{R}}{\mathbf{V}}_{\mathbf{S}}}{{\mathbf{Z}}^{\mathbf{\text{'}}}}\mathbf{cos}\left({\theta }_z-\delta \right)$ …… (7)

The equation to calculate the full load receiving current is given as follows.

${\mathbf{l}}_{\mathbf{R}\mathbf{.}\mathbf{FL}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{R}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{R}}\mathbf{(}\mathbf{p}\mathbf{.}\mathbf{f}\mathbf{)}}\mathbf{\angle }\mathbf{-}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{p}\mathbf{.}\mathbf{f}\mathbf{)}$ …… (8)

The equation to calculate the exact receiving end voltage at full load is given as follows.

role="math" localid="1655190514011" ${\mathbf{V}}_s\mathbf{=}{\mathbf{AV}}_{\mathbf{R}}\mathbf{+}{\mathbf{BI}}_{\mathbf{R}\mathbf{.}\mathbf{FL}}$ …… (9)

The equation to calculate the no load receiving voltage is given as follows.

${\mathbf{V}}_{\mathbf{s}}\mathbf{=}{\mathbf{AV}}_{\mathbf{R}\mathbf{.}\mathbf{NL}}\mathbf{+}{\mathbf{BI}}_{\mathbf{R}\mathbf{.}\mathbf{NL}}$ …… (10)

The equation to calculate the voltage regulation is given as follows.

$\mathbf{VR}\mathbf{\%}\mathbf{=}\left(\frac{V_{R.\mathrm{NL}}-V_{R.\mathrm{FL}}}{V_{R.\mathrm{FL}}}\right)\mathbf{\times }\mathbf{100}$ …… (11)

Calculate the load ability of the line.

(a)

Calculate the propagation constant.

Substitute $500\mathrm{km}$for l ,$0.03+\mathrm{j}0.35\frac{\mathrm{\Omega }}{\mathrm{km}}$for z and $\mathrm{j}4.4\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}$for y into equation (1).

$$\begin{gathered} \mathbb{y}=500\sqrt{(0.03+\mathrm{j}0.35)(\mathrm{j}4.4\times {10}^{-6})} \\ \mathbb{y}=500\sqrt{1.54\times {10}^{-6}\angle 175.1^0} \\ \mathbb{y}=500\times 1.24\times {10}^{-3}\angle 87.{55}^0 \\ \mathbb{y}=0.620\angle 87.{55}^0 \end{gathered}$$

Convert into cartesian form.

$\mathbb{y}=0.0265+\mathrm{j}0.619\mathrm{pu}$

Calculate the characteristics impedance.

Substitute $0.03+\mathrm{j}0.35\frac{\mathrm{\Omega }}{\mathrm{km}}$for Z and$\mathrm{j}4.4\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}$ for y into equation (2).

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{C}}=\sqrt{\frac{0.03+\mathrm{j}0.35}{\mathrm{j}4.4\times {10}^{-6}}} \\ {\mathrm{Z}}_{\mathrm{C}}=\sqrt{79.84\times {10}^3\angle -4.{89}^0} \\ {\mathrm{Z}}_{\mathrm{C}}=282.6\angle -2.{45}^0 \end{gathered}$$

Calculate the value ofrole="math" localid="1655356708504" $\sinh \left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)$ ,

role="math" localid="1655357044815" $$\begin{gathered} \sinh \left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)=\frac{{\mathrm{\theta }}^{0.0266+\mathrm{j}0.619}-{\mathrm{\theta }}^{-0.0265-\mathrm{j}0.619}}{2} \\ \sinh \left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)=\frac{1.026\angle 0.619-0.9738\angle -6.19}{2} \\ \sinh \left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)=\frac{1.162\angle 1.534}{2}\mathrm{redian} \\ \sinh \left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)=0.581\angle 87.{89}^0 \end{gathered}$$

Calculate $\left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)$,

$$\begin{gathered} \cosh \left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)=\frac{{\mathrm{e}}^{0.0266+\mathrm{j}0.619}+{\mathrm{e}}^{-0.0265-\mathrm{j}0.619}}{2} \\ \cosh \left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)=\frac{1.026\angle 0.619+0.9738\angle -6.19}{2} \\ \cosh \left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)=\frac{1.63\angle 0.019}{2}\mathrm{redian} \\ \cosh \left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)=0.815\angle 1.{089}^0 \end{gathered}$$

Calculate the ABCD parameter of the transmission line.

Calculate the A and D parameter,

Substitute $0.815\angle 1.{089}^0$for cosh$\left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)$into equation (3a).

$$\begin{gathered} \mathrm{A}=\mathrm{D} \\ \mathrm{A}=0.815\angle 1.{089}^0\mathrm{pu} \end{gathered}$$

Calculate B parameter,

Substitute $282.6\angle -2.{45}^0$for ${\mathrm{Z}}_{\mathrm{C}}$and $0.581\angle 87.{89}^0$ for$\sinh \left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)$ into equation (3b).

$$\begin{gathered} \mathrm{B}={\mathrm{Z}}^{\text{'}} \\ \mathrm{B}=282.6\angle 2.{45}^0(0.581\angle 87.{89}^0) \\ \mathrm{B}=164.19\angle 85.{44}^0\mathrm{\Omega } \end{gathered}$$

Calculate the C parameter,

Substitute $282.6\angle -2.{45}^0$for${\mathrm{Z}}_{\mathrm{C}}$ and $0.581\angle 87.{89}^0$ for$\sinh \left(\stackrel{\rightharpoonup }{\mathbb{y}}\mathrm{l}\right)$ into equation (3c).

$$\begin{gathered} \mathrm{C}=\frac{0.581\angle 87.{89}^0}{282.6\angle -2.{45}^0} \\ \mathrm{C}=2.056\angle 90.{34}^0\mathrm{mS} \end{gathered}$$

Calculate the sending end voltage.

Substitute 500 kV for ${\mathrm{V}}_{\mathrm{L},\mathrm{Base}}$and I for${\mathrm{V}}_{\mathrm{s}.\mathrm{pu}}$ into equation (5).

$$\begin{gathered} {\mathrm{V}}_{\mathrm{S}}=1\times 500 \\ {\mathrm{V}}_{\mathrm{S}}=500\mathrm{kV} \end{gathered}$$

Calculate the receiving end voltage.

Substitute 500 kV forrole="math" localid="1655358825111" ${\mathrm{V}}_{\mathrm{L},\mathrm{Base}}$ and 0.95 for${\mathrm{V}}_{\mathrm{R}.\mathrm{pu}}$ into equation (6).

role="math" localid="1655360173514" $$\begin{gathered} {\mathrm{V}}_{\mathrm{R}}=(0.95)\left(500\right) \\ {\mathrm{V}}_{\mathrm{R}}=475\mathrm{kV} \end{gathered}$$

Calculate the practical line load ability.

Substitute 500 kV for ${\mathrm{V}}_{\mathrm{s}}$, 475 kV for ${\mathrm{V}}_{\mathrm{R}}$, $85.{44}^0$for ${\mathrm{\theta }}_{\mathrm{z}}$and ${36}^0$for $\mathrm{\delta }$into equation (7).

$$\begin{gathered} {\mathrm{P}}_{\mathrm{R},\max }=\left(\frac{500\times {10}^3\times 475\times {10}^3}{164.19}\right)\cos {\left(85.44-35\right)}^0 \\ {\mathrm{P}}_{\mathrm{R},\max }=\left(\frac{2.375\times {10}^{11}}{164.19}\right)\cos 50.{44}^0 \\ {\mathrm{P}}_{\mathrm{R},\max }=921.25\mathrm{MW} \end{gathered}$$

Hence the practical line load ability is$921.25\mathrm{MW}$ .

Calculate the full receiving end current at 0.99 power factor leading.

(b)

Calculate the full load receiving end current.

Substitute 921.25 MW for ${\mathrm{P}}_{\mathrm{R}}$, $475\mathrm{kV}$for ${\mathrm{V}}_{\mathrm{R}}$and $0.99$for pf into equation (8).

$$\begin{gathered} {\mathrm{I}}_{\mathrm{R}.\mathrm{FL}}=\frac{921.252}{\sqrt{3}\times 475\times {10}^3(0.99)}\angle {\cos }^{-1}(0.99) \\ {\mathrm{I}}_{\mathrm{R}.\mathrm{FL}}=\frac{921.252}{814.49\times {10}^3}\angle -8.{109}^0 \\ {\mathrm{I}}_{\mathrm{R}.\mathrm{FL}}=1.131\angle 8.{109}^0\mathrm{kA} \end{gathered}$$

Hence the full load current at receiving end at 0.99 power factor is $1.131\angle 8.{109}^0\mathrm{kA}$.

Calculate the per phase exact receiving voltage for full load current.

(c)

Calculate full load receiving end voltage.

Substitute $0.815\angle 1.089°\mathrm{pu}$for A , localid="1655362361110" $164.19\angle 85.44°$for B ,$1.131\angle 8.109°\mathrm{kA}$for${\mathrm{I}}_{\mathrm{R},\mathrm{FL}}$into equation (9).

localid="1655360864228" $$\begin{gathered} \frac{500\times {10}^3}{\sqrt{3}}\angle 35°=\left[\right(0.815\angle 1.089°\left)\right({\mathrm{V}}_{\mathrm{RFL}}\angle 0°)+(164.19\angle 85.44°)\times (1.131\angle 8.109°\mathrm{kA}\left)\right] \\ 288.67\times {10}^3\angle 35°(0.815\angle 1.089°)({\mathrm{V}}_{\mathrm{RFL}}\angle 0°)+(185.689\times {10}^3\angle 93.95°) \\ (0.815\angle 1.089°)({\mathrm{V}}_{\mathrm{RFL}}\angle 0°)=288.67\times {10}^3\angle 35°-185.689\times {10}^3\angle 93.95 \\ (0.815\angle 1.089°)({\mathrm{V}}_{\mathrm{RFL}}\angle 0°)=248.716\times {10}^3\angle 4.556° \end{gathered}$$

Solve further as,

localid="1655360516090" $$\begin{gathered} (0.815\angle 1.089°)(┤)=248.716\times {10}^3\angle 4.556 \\ {\mathrm{V}}_{\mathrm{RFL}}\angle 0°\frac{(248.716\times {10}^3\angle 4.556)}{(0.815\angle 1.089°)} \\ {\mathrm{V}}_{\mathrm{RFL}}\angle 0=305.17\angle {10}^3\angle -5.67°\mathrm{V} \end{gathered}$$

Calculate the line to line receiving end voltage.

localid="1655360334577" $$\begin{gathered} {\mathrm{V}}_{\mathrm{R}}=305.17\times {10}^3\times \sqrt{3} \\ {\mathrm{V}}_{\mathrm{R}}=528.6\mathrm{kV} \end{gathered}$$

Hence the receiving end voltage for the full load current is$528.6\mathrm{kV}$.

Calculate the volage regulation.

(d)

At no load, the receiving end current,${\mathrm{l}}_{\mathrm{R}}=0$

Substitute 0 for ${\mathrm{I}}_{\mathrm{R}}$into equation (10).

$$\begin{gathered} {\mathrm{V}}_{\mathrm{S}}={\mathrm{AV}}_{\mathrm{R},\mathrm{NL}}+\mathrm{B}\left(0\right) \\ {\mathrm{V}}_{\mathrm{R},\mathrm{NL}}\frac{{\mathrm{V}}_{\mathrm{S}}}{\mathrm{A}} \end{gathered}$$

Calculate the no load receiving end voltage.

Substitute $0.815\angle 1.{089}^0\mathrm{pu}$for role="math" localid="1655361995675" $\mathrm{A}$and role="math" localid="1655361941041" $500\mathrm{kV}$ for ${\mathrm{V}}_{\mathrm{S}}$in above equation.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{R}.\mathrm{NL}}=\frac{500\times {10}^3}{0.815} \\ {\mathrm{V}}_{\mathrm{R},\mathrm{NL}}=613.5\mathrm{kV} \end{gathered}$$

Calculate the voltage regulation.

Substitute$613.5\mathrm{kV}$ for ${\mathrm{V}}_{\mathrm{R},\mathrm{NL}}$and$528.6\mathrm{kV}$ for${\mathrm{V}}_{\mathrm{R},\mathrm{FL}}$ into equation (11).

$$\begin{gathered} \mathrm{VR}\%=\left(\frac{613.5-528.6}{528.6}\right)\times 100 \\ \mathrm{VR}\%=16.1\% \end{gathered}$$

Hence the voltage regulation is 16.1%.

Calculate the ratio of voltage at full load and base voltage.

$$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{R},\mathrm{FL}}}{{\mathrm{V}}_{\mathrm{S}}}=\frac{525.6}{500} \\ \frac{{\mathrm{V}}_{\mathrm{R},\mathrm{FL}}}{{\mathrm{V}}_{\mathrm{S}}}=1.0572 \\ 1.0572>0.95 \end{gathered}$$

The ratio of voltage at full load and base voltage is greater than 0.95. Therefore, laudability determined by steady state stability.