Deck 4: Transmission Lines: Steady-State Operation

Question 1

A $30-\mathrm{km}, 34.5-\mathrm{kV}, 60-\mathrm{Hz}$ three-phase line has a positive-sequence series impedance $z=0.19+j 0.34 \Omega / \mathrm{km}$ . The load at the receiving end absorbs 10 MVA at $33 \mathrm{kV}$ . Assuming a short line, calculate: (a) the $A B C D$ parameters, (b) the sending-end voltage for a load power factor of 0.9 lagging, (c) the sending-end voltage for a load power factor of 0.9 leading.

Answer 1

(a) $\bar{A}=\bar{D}=1.0 \angle 0^{\circ} \mathrm{pu} ; \bar{C}=0.0 \mathrm{~S}$
$\bar{B}=\bar{Ƶ}=(0.19+j 0.34)(30)=11.685 \angle 60.8^{\circ} \Omega$
(b) $\bar{V}_{R}=(33 / \sqrt{3}) \angle 0^{\circ}=19.05 \angle 0^{\circ} \mathrm{kV}_{\mathrm{LN}}$
$\begin{aligned}\bar{I}_{R} & =\frac{S_{R}}{\sqrt{3} V_{R L-L}} \angle-\cos ^{-1}(p f)=\frac{10}{\sqrt{3}(33)} \angle-\cos ^{-1} 0.9 \\& =0.1750 \angle-25.84^{\circ} \mathrm{kA} \\\bar{V}_{S} & =\bar{A} \bar{V}_{R}+\bar{B} \bar{I}_{R}=1.0(19.05)+\left(11.685 \angle 60.8^{\circ}\right)\left(0.175 \angle-25.84^{\circ}\right) \\& =19.05+2.045 \angle 34.96^{\circ}=20.73+j 1.172 \\& =20.76 \angle 2.22^{\circ} \mathrm{kV}_{\mathrm{LN}} ; V_{S}=20.76 \sqrt{3}=35.96 \mathrm{kV}_{\mathrm{LL}}\end{aligned}$
(c) $\bar{I}_{R}=0.175 \angle 25.84^{\circ} \mathrm{kA}$
$\begin{aligned}\bar{V}_{S} & =1.0(19.05)+\left(11.685 \angle 60.8^{\circ}\right)\left(0.175 \angle 25.84^{\circ}\right) \\& =19.05+2.044 \angle 86.64^{\circ} \\& =19.17+j 2.04 \\& =19.28 \angle 4.07^{\circ} \mathrm{kV}_{\mathrm{LN}} \\V_{S} & =19.28 \sqrt{3}=33.39 \mathrm{kV}_{\mathrm{LL}}\end{aligned}$

Question 2

A $150-\mathrm{km}, 230-\mathrm{kV}, 60-\mathrm{Hz}$ three-phase line has a positive-sequence series impedance $z=0.08+j 0.48 \Omega / \mathrm{km}$ and a positive-sequence shunt admittance $y=j 3.33 \times 10^{-6} \mathrm{~S} /$ $\mathrm{km}$ . At full load, the line delivers $250 \mathrm{MW}$ at 0.99 p.f. lagging and at $220 \mathrm{kV}$ . Using the nominal $\pi$ circuit, calculate: (a) the $A B C D$ parameters, (b) the sending-end voltage and current, and (c) the percent voltage regulation.

Answer 2

$$\begin{array}{l}\text { (a) } \begin{array}{l}\bar{A}=\bar{D}=1+\frac{\bar{Y} \bar{Ƶ}}{2}=1+\frac{1}{2}\left(3.33 \times 10^{-6} \times 150 \angle 90^{\circ}\right)(.08+j .48)(150) \\\begin{aligned}\bar{A}=\bar{D} & =1+\frac{1}{2}\left(4.995 \times 10^{-4} \angle 90^{\circ}\right)\left(72.99 \angle 80.54^{\circ}\right) \\& =1+0.01823 \angle 170.54^{\circ}=0.9820+j 0.002997 \\& =0.9820 \angle 0.175^{\circ} \text { per unit }\end{aligned} \\\begin{aligned}\bar{B}= & \bar{Ƶ}=72.99 \angle 80.54^{\circ} \Omega \\\bar{C}= & \bar{Y}\left(1+\frac{\bar{Y} \bar{Ƶ}}{4}\right)=4.995 \times 10^{-4} \angle 90^{\circ}\left(1+.009115 \angle 170.54^{\circ}\right)\end{aligned} \\\begin{aligned}\bar{C} & =\left(4.995 \times 10^{-4} \angle 90^{\circ}\right)(0.991+j 0.00150) \\& =\left(4.995 \times 10^{-4} \angle 90^{\circ}\right)\left(0.991 \angle 0.0867^{\circ}\right)=4.950 \times 10^{-4} \angle 90.09^{\circ}\end{aligned}\end{array}\end{array}$$
(b) $\bar{V}_{R}=\frac{220}{\sqrt{3}} \angle 0^{\circ}=127.02 \angle 0^{\circ} \mathrm{kV} \mathrm{V}_{\mathrm{LN}}$
$$\begin{array}{l}\begin{aligned}\bar{I}_{R}= & \frac{P_{R} \angle-\cos ^{-1}(P . F .)}{\sqrt{3} V_{R L L}(P . F .)}=\frac{250 \angle-\cos ^{-1} .99}{\sqrt{3}(220)(0.99)}=0.6627 \angle-8.11^{\circ} \\\bar{V}_{S}= & \bar{A} \bar{V}_{R}+\bar{B} \bar{I}_{R}=\left(0.9820 \angle 0.175^{\circ}\right)\left(127.02 \angle 0^{\circ}\right)+\left(72.99 \angle 80.54^{\circ}\right) \\& \quad \times\left(.6627 \angle-8.11^{\circ}\right) \\\bar{V}_{S}= & 124.73 \angle 0.175^{\circ}+48.37 \angle 72.43^{\circ} \\\bar{V}_{S}= & 139.33+j 46.49=146.9 \angle 18.45^{\circ} \mathrm{kV} \mathrm{LN} \\V_{S}= & 146.9 \sqrt{3}= \underline{ 254.4 } \mathrm{kV} \mathrm{LI} \\\bar{I}_{S}= & \bar{C} \bar{V}_{R}+\overline{D I}_{R}=\left(4.95 \times 10^{-4} \angle 90.09^{\circ}\right)(127.02)+ \\& \quad\left(.9820 \angle 175^{\circ}\right)\left(.6627 \angle-8.6^{\circ}\right) \\= & 0.06287 \angle 90.09^{\circ}+0.6508 \angle-7.935^{\circ} \\\bar{I}_{S}= & 0.6445-j 0.02697= \underline{\underline{ 0.6750}} \angle-2.396^{\circ} \mathrm{kA}\end{aligned}\end{array}$$
(c)
$$\begin{aligned}V_{R V Z} & =\frac{V_{S}}{A}=\frac{254.4}{0.9820}=259.1 \mathrm{kV} \mathrm{~V}_{\mathrm{LL}} \\\% V_{r} R_{r} & =\frac{V_{R V Z}-V_{R F L}}{V_{R F L}} \times 100=\frac{259.1-220}{220} \times 100=\underline{\underline{17.8}} \%\end{aligned}$$

Question 3

Rework Test Bank Problem 5.2 in per-unit using 100-MVA (three-phase) and 230-kV(line-to-line) base values. Calculate (a) the per-unitABCDparameters, (b) the per-unitsending-end voltage and current, and (c) the percent voltage regulation.

Answer 3

$$\begin{array}{l}Z_{\text {base }}=\frac{V_{\text {base }}^{2}}{S_{\text {base }}}=\frac{(230)^{2}}{100}=529 \Omega \\Y_{\text {base }}=1 / Ƶ_{\text {base }}=1.890 \times 10^{-3} \end{array}$$
(a)
$$\begin{array}{l}\bar{A}_{p u}=\bar{D}_{p u}=0.9820 \angle 0.175^{\circ} \text { per unit } \\\bar{B}_{p u}=\frac{\bar{B}}{Z_{\text {base }}}=\frac{72.99}{529} \angle 80.54^{\circ}=0.1380 \angle 80.54^{\circ} \text { per unit } \\\bar{C}_{p u}=\frac{\bar{C}}{Y_{\text {base }}}=\frac{4.950 \times 10^{-4}}{1.890 \times 10^{-3}} \angle 90.09^{\circ}=0.2619 \angle 90.09^{\circ} \text { per unit }\end{array}$$



$$\begin{array}{l} \bar{V}_{S P U}= \bar{A}_{P U} \bar{V}_{R P U}+\bar{B}_{P U} \bar{I}_{R P U}=\left(0.982 \angle 0.175^{\circ}\right)\left(0.9565 \angle 0^{\circ}\right)+ \\\quad\left(0.1380 \angle 80.54^{\circ}\right)\left(2.64^{\circ} \angle-8.11^{\circ}\right) \\= 0.9393 \angle 0.175^{\circ}+0.3643 \angle 72.43^{\circ}=1.049+j 0.3501 \\\bar{V}_{S P U}= \underline{\overline{1.106} } \angle 18.45^{\circ} \text { per unit }\\\bar{I}_{S P U}=\bar{C}_{P U} \bar{V}_{R P U}+\bar{D}_{P U} \bar{I}_{R P U}=\left(0.2619 \angle 90.09^{\circ}\right)(0.9565)+ \\\quad\left(0.982 \angle 0.175^{\circ}\right)\left(2.64 \angle-8.11^{\circ}\right) \\\bar{I}_{S P U}= \underline{\underline{2.569}} \angle-2.398^{\circ} \text { per unit }\end{array}$$
(c)
$$\begin{array}{l}V_{R N L_{p u}}=V_{S p u} / A_{p u}=1.106 / 0.982=1.1263 \text { per unit } \\\% V . R .=\frac{V_{R N L_{p u}}-V_{R F L p u}}{V_{R F L p u}} \times 100=\frac{1.126-.9565}{.9565} \times 100=\underline{\underline{17.8}} \%\end{array}$$

Question 4

\underline{\underline{0.6507}

-Evaluate $\cosh (\gamma l)$ and $\tanh (\gamma l / 2)$ for $\gamma l=0.45 \angle \underline{ 87 }^{\circ}$ per unit.

Question 5

A $500-\mathrm{km}, 500-\mathrm{kV}, 60-\mathrm{Hz}$ uncompensated three-phase line has a positive-sequence series impedance $z=0.03+j 0.35 \Omega / \mathrm{km}$ and a positive-sequence shunt admittance $y=j 4.4 \times 10^{-6} \mathrm{~S} / \mathrm{km}$ . Calculate: (a) $Z_{c}$ , (b) $(\gamma l)$ , and (c) the exact $A B C D$ parameters for this line.

Question 6

At full load the line in Test Bank Problem 5.5 delivers 1000 MW at unity power factor and at 480 kV. Calculate (a) thesending-end voltage, (b) the sending-end current, (c) the sending-end power factor, (d) the full-loadline losses, and (e) the percent voltage regulation.

Question 7

etermine the equivalent π circuit for the line in TestBank Problem 5.5 and compare it with the nominal π circuit.

Question 8

A $320-\mathrm{km} 500-\mathrm{kV}, 60-\mathrm{Hz}$ three-phase uncompensated line has a positive-sequence series reactance $x=0.34 \Omega / \mathrm{km}$ and a positive-sequence shunt admittance $y=j 4.5 \times$ $10^{-6} \mathrm{~S} / \mathrm{km}$ . Neglecting losses, calculate: (a) $\mathrm{Z}_{c}$ , (b) $(\gamma l)$ , (c) the $A B C D$ parameters, (d) the wavelength $\lambda$ of the line, in kilometers, and (e) the surge impedance loading in MW.

Question 9

Determine the equivalent π circuit for the line in Test Bank Problem 5.8.

Question 10

Rated line voltage is appliedto the sending end of the line in Test Bank Problem 5.8.Calculate the receiving-end voltage when thereceiving end is terminated by (a) an open circuit, (b) the surge impedance of the line,and (c) one-half of thesurge impedance. (d)Also calculate the theoretical maximum realpower that the line can deliver when rated voltage is applied toboth ends of the line.

Question 11

The line in Test Bank Problem 5.5 has three ACSR 1113-kcmil conductors per chase.Calculate the theoretical maximum real power that this line can deliver and compare withthe thermal limit of the line. Assume V_S-V_R=1.0 per unit and unity power factor at thereceiving end.

Question 12

Repeat Test Bank Problems 5.5 and 5.11 if the line length is (a) $200 \mathrm{~km}$ and (b) $550 \mathrm{~km}$ .

Question 13

For the line in Test Bank Problems 5.5 and 5.11 , determine (a) the practical line loadability in MW, assuming $V_{S}=1.0$ per unit, $V_{R} \approx 0.95$ per unit, and $\delta_{\max }=35^{\circ}$ ; (b) the full-load current at 0.99 leading power factor, based on the above practical line loadability; (c) the exact receiving-end voltage for the full-load current in (b) above; and (d) the percent voltage regulation. For this line, is loadability determined by the thermal limit, the drop limit, or stead-state stability?

Question 14

Determine the practical line loadability in MWand in per unit of SIL for the line in Test Bank Problem 5.5 if the length is(a) 200 km and (b) 600 km. Assume V_S = 1.0 per unit, V_R =0.95 per unit , $\delta_{\max }=35^{\circ}$ and 0.99 leading power factor at the receiving end.