22-Elec-B7 Power Systems Engineering · December 2017
Question 1 of 7: Long transmission line — ABCD constants, sending-end conditions and power-factor correction
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2017 — 16-Elec-B7 Power
Systems Engineering, 3 hours, open book, any non-communicating calculator permitted.
The paper prints seven problems; the rubric states that any five questions constitute a
complete paper and that all questions are of equal value, so only the first
five in the answer book are marked. All seven are worked here, because this set is a study
resource rather than a sitting. Sub-part point values are reproduced from the paper.
Reference texts. This exam code follows M. E. El-Hawary,
Electrical Power Systems: Design and Analysis (IEEE Press) — the notation used
below is his: the long-line ABCD constants of Problem 1, the round-rotor power-angle
relations of Problem 2, the cantilever transformer equivalent of Problem 3, the
power-flow injection equations of Problem 4, the network-reduction fault analysis of
Problems 5 and 6, and the equal-area criterion of Problem 7. Corroborating
references: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and
Design, 6th ed. (ch. 5 transmission lines, ch. 6 power flow,
ch. 7–9 symmetrical components and unsymmetrical faults, ch. 11 transient
stability); J. J. Grainger and W. D. Stevenson, Power System Analysis; and
S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (ch. 2
transformers, ch. 4–5 synchronous machines). Canadian practice references for the
protection and grounding discussion: CSA C22.1 Canadian Electrical Code, Part I
and the relevant IEEE C37 and C57 series.
Question 1: Long transmission line — ABCD constants, sending-end conditions and
power-factor correction (25 points)
Given. A three-phase transmission line rated 220 kV and 200 MVA,
described by its total series impedance and total shunt admittance, together
with one receiving-end operating point and a target sending-end power factor.
Given data
Quantity
Symbol
Value
Rated line-to-line voltage
$V_{\text{rated}}$
220 kV
Rated three-phase apparent power
$S_{\text{rated}}$
200 MVA
Total series impedance
$Z$
$17 + j125\ \Omega$
Total shunt admittance
$Y$
$j6\times10^{-4}\ \text{S}$
Receiving-end line-to-line voltage
$V_{R,LL}$
205 kV
Receiving-end load
$S_R$
175 MVA at 0.8 pf lagging
Target sending-end power factor
$\cos\phi_S'$
0.85 lagging
Find. (a) the constants $A$, $B$ and $C$ of the exact (distributed)
long-line model; (b) the sending-end voltage together with the active and reactive power
delivered into the line at the sending end; (c) the three-phase reactive rating of a shunt
capacitor bank that raises the sending-end power factor to 0.85 lagging.
Figure 1.1 — The 220 kV line as a two-port. Because the whole line is represented by one pair of terminal relations, every quantity asked for in parts (b) and (c) follows from the four constants found in part (a).
Approach. Form the propagation constant and characteristic impedance from
the lumped $Z$ and $Y$, evaluate the hyperbolic ABCD constants, push the given receiving-end
phasors through them to get the sending-end voltage and current, then take the reactive
difference between the resulting sending-end power factor and the target as the capacitor
rating.
Form the propagation constant and the characteristic impedance. For the
exact model the whole line is characterised by
$$\gamma l=\sqrt{ZY},\qquad Z_c=\sqrt{Z/Y}$$
Substituting the given totals,
$$ZY=(17+j125)(j6\times10^{-4})=-0.0750+j0.01020$$
so that
$$\gamma l=0.018580+j0.274491=0.275119\,\angle\,86.128^\circ$$
and
$$Z_c=\sqrt{\frac{17+j125}{j6\times10^{-4}}}=457.485-j30.966=458.53\,\angle\,-3.872^\circ\ \Omega$$
The real part of $\gamma l$ is the total attenuation in nepers and the imaginary part is the
total electrical length in radians; 0.274491 rad is about 86.128 degrees of line, i.e. a
genuinely long line for which the nominal-$\pi$ approximation is no longer safe.
Evaluate the ABCD constants. The distributed-parameter two-port relations
are $A=D=\cosh\gamma l$, $B=Z_c\sinh\gamma l$ and $C=\sinh\gamma l/Z_c$. Evaluating the
hyperbolic functions of the complex argument found above,
$$\boxed{\begin{aligned}
A=D&=0.96274\,\angle\,0.2997^\circ\\
B&=124.580\,\angle\,82.353^\circ\ \Omega\\
C&=5.9253\times10^{-4}\,\angle\,90.098^\circ\ \text{S}
\end{aligned}}$$
In rectangular form $A=0.96273+j0.005036$ and $B=16.577+j123.472\ \Omega$. Two free checks
confirm the arithmetic. First, reciprocity requires $AD-BC=1$, which the numbers satisfy to
better than one part in $10^{15}$. Second, $C$ must be almost a pure positive susceptance,
because the shunt path of a transmission line is capacitive; its angle of 90.098 degrees is
within a tenth of a degree of $+90^\circ$, as it should be for a line whose series resistance
is small compared with its series reactance.
Write the receiving-end phasors. Working per phase and taking the
receiving-end voltage as reference,
$$V_R=\frac{205\ \text{kV}}{\sqrt3}=118.357\ \text{kV}\,\angle\,0^\circ$$
$$\begin{aligned}
I_R&=\frac{S_R}{\sqrt3\,V_{R,LL}}\angle-\cos^{-1}(0.8)
=\frac{175\times10^{6}}{\sqrt3\,(205\times10^{3})}\angle-36.870^\circ\\
&=492.86\,\angle\,-36.870^\circ\ \text{A}
\end{aligned}$$
that is $I_R=394.29-j295.72$ A. The lagging power factor puts the current behind the
voltage, which is the sign convention that makes the reactive power positive into the load.
Push the operating point through the two-port. Substituting into
$V_S=AV_R+BI_R$ gives
$$V_S=156.994+j44.377\ \text{kV}=163.146\ \text{kV}\,\angle\,15.784^\circ$$
per phase, so the sending-end line-to-line voltage is
$$\boxed{V_{S,LL}=\sqrt3\,(163.146)=282.58\ \text{kV}}$$
and from $I_S=CV_R+AI_R$,
$$I_S=380.96-j212.58=436.26\,\angle\,-29.162^\circ\ \text{A}$$
The shunt charging current has rotated the current phasor forward by about eight degrees
relative to the load current, and it has reduced the magnitude from 492.86 A to
436.26 A — the line is supplying part of the load's reactive demand from its own
capacitance.
Take the sending-end complex power. With three phases,
$$S_S=3V_SI_S^{*}=151.13+j150.84\ \text{MVA}$$
so the active and reactive powers entering the line are
$$\boxed{P_S=151.13\ \text{MW},\qquad Q_S=150.84\ \text{MVAr}}$$
The corresponding power factor is
$$\cos\phi_S=\frac{P_S}{|S_S|}=\frac{151.13}{213.52}=0.7078\ \text{lagging}
\qquad(\phi_S=44.95^\circ)$$
As a check on the whole calculation, the load absorbs $P_R=140.0$ MW, so the line loss
is $151.13-140.0=11.13$ MW — 7.4 per cent of the power entering the
line, which is the right order for a circuit whose series resistance is 17 Ω
carrying roughly 490 A.
Size the shunt capacitor bank. A shunt bank at the sending bus changes no
real power, so the active power stays at $P_S$ and only the reactive component is reduced:
$$Q_C=P_S\left[\tan\phi_S-\tan\phi_S'\right]
=151.13\left[0.99811-0.61974\right]$$
which gives
$$\boxed{Q_C=57.18\ \text{MVAr (three-phase)}}$$
Confirming, the reactive power drawn from the source falls to
$150.84-57.18=93.66$ MVAr, and
$\cos\phi_S'=151.13/\sqrt{151.13^2+93.66^2}=0.850$ lagging as required. In practice such a
bank would be split into switched steps so that the correction can follow the load.
Final results — Question 1
Quantity
Symbol
Value
Characteristic impedance
$Z_c$
458.53 ∠ -3.872° Ω
Propagation constant (total)
$\gamma l$
0.275119 ∠ 86.128°
Two-port constant
$A=D$
0.96274 ∠ 0.2997°
Two-port constant
$B$
124.580 ∠ 82.353° Ω
Two-port constant
$C$
5.9253 × 10−4 ∠ 90.098° S
Sending-end voltage (line-to-line)
$V_{S,LL}$
282.58 kV
Sending-end current
$I_S$
436.26 ∠ -29.162° A
Sending-end active power
$P_S$
151.13 MW
Sending-end reactive power
$Q_S$
150.84 MVAr
Sending-end power factor
$\cos\phi_S$
0.7078 lagging
Capacitor bank for 0.85 lagging
$Q_C$
57.18 MVAr
Check: two features of this answer look wrong and are not. First, the
sending-end voltage of 282.58 kV is 37.8 per cent above the 205 kV receiving
end and well above the 220 kV rating. That is what the stated data produce: the surge
impedance loading of this line is $V^2/|Z_c| = 105.6$ MW, and the question loads it with
140.0 MW at only 0.8 power factor, so the series reactive drop
$I(R\cos\phi+X\sin\phi)$ dominates. A real 220 kV circuit would carry series or shunt
compensation, or would be operated closer to unity power factor; the numbers here are
illustrative of the method, not of practice. Second, part (c) does not say where
the bank is connected. It is taken here to be at the sending bus, which is the only reading
that leaves parts (a) and (b) intact and makes the answer a single line of algebra.
A bank installed at the receiving end would also raise the sending-end power factor but would
change $I_R$, hence $V_S$, hence $P_S$, and would have to be solved iteratively.
