ELK322E POWER TRANSMISSION SYSTEMS
ITU ELECTRICAL ENGINEERING
May 24, 2021
Ö. USTA & R. ÇAĞLAR
Name & Family Name:
TOTAL
Q1 Q2
Student Number:
Signature:
Instructions:
+This EXAM PAPER consists of 2 questions.
+This is an open-book examination. You are not permitted to take any assistance of others to solve the problems.
+Your solutions should be methodical. You must write the unit of each physical quantity or variable.
+You will be severely penalized, if your solutions are illegible.
+Please use your own statements while answering the problems.
4 Please add a copy of your ID in every page of your exam booklet.
MIDTERM EXAM 2
PROBLEM-1 [60 points]: A 60Hz 3-phase transmission line is 175mi long. It has a total series impedance
of (35+j140) ohm and a shunt admittnce of 930 × 10−6 ∠90° S. It delivers 40 MW at 220 kV, with 90%
power factor lagging. Find the voltage and power at the sending end using (a) the short line model, (b) the
nominal pi-model and (c) the long line model.
PROBLEM 2 [40 points]: A three-phase transposed line is composed of one ACSR 1,431,000cmil, 47/7
Bobolink conductor per phase with flat horizontal spacing of 11 meters as shown in Figure 1a. The
conductors have a diameter of 3.625cm and a GMR of 1.439cm. The line is to be replaced by a threeconductor bundle of ACSR 477,000cmil, 26/7 Hawk conductors having the same cross-sectional area of
aluminum as the single-conductor line. The conductors have a diameter of 2.1793cm and a GMR of
0.8839cm. The new line will also have a flat horizontal configuration, but it is to be operated at a higher
voltage and therefore the phase spacing is increased to 14m as measured from the center of the bundles. The
spacing between the conductors in the bundle is 45cm as shown in Figure 1b.
Determine:
(a) The percentage change in the inductance.
(b) The percentage change in the power transmitted.
(c) The percentage change in the capacitance.
(d) The change in the charging current (Line to line voltage is 400kV).
Figure 1a.
L  2 x10 7 ln
Deq
Ds
𝑥𝐿 = 𝑗𝜔𝐿
( H / m)
Figure 1b.
Dsb  Ds  d
Deq  Dm  ( Dab  Dbc  Dca )1 / 3
Cn 
2
( F / m) to neutral
ln ( Deq / r )