NPTEL Online Certification Course

<Power Electronics Applications in Power Systems>

<Assignment 04>: Detailed Solution
Indian Institute of Technology Guwahati

For answering Q.1-Q.10, please find the following text:

A symmetrical lossless, long transmission line with following parameters: L=0.857 mH/km, C
= 15.4 nF/km, Line length = 850 km, f = 50 Hz, Voltage=735 kV. We need to design a
compensator to hold the mid-point voltage to 1.05 p.u. irrespective of loading.
Q1. What is the voltage at mid-point at no load without compensation?
(a) 1.09 p.u.
(b) 1.13 p.u.
(c) 1.15 p.u.
(d) 1.07 p.u.
Solution:

𝑍𝑐 = √𝐿⁄𝐶 = √0.857⁄15.4 × 103 = 235.90 Ω

𝛽 = 𝜔√𝐿 × 𝐶 = 2𝜋 × 50 × √0.857 × 15.4 × 10−6 = 1.141 × 10−3 𝑟𝑎𝑑/𝑘𝑚

𝛽𝑙 = 0.9698 𝑟𝑎𝑑 = 55.56°
𝛽𝑙
= 27.78°
2

𝑐𝑜𝑠𝛿 ⁄2 1 1
(𝑉𝑚 )𝑝.𝑢. = = 𝑐𝑜𝑠𝛽𝑙⁄2 = 𝑐𝑜𝑠27.78° = 1.13 𝑝. 𝑢.
𝑐𝑜𝑠𝛽𝑙⁄2

Q2. What is the expression for active power of uncompensated line?

(a) 2297.7sin 𝛿
(b) 2991.3sin 𝛿
(c) 2776.7𝐬𝐢𝐧 𝜹
(d) 2527.5sin 𝛿
Solution:
2
𝑉𝐿−𝐿 sin 𝛿 7352 ×sin 𝛿
P= = = 2776.77 sin 𝛿 𝑀𝑊
𝑍𝑐 sin 𝛽𝑙 235.9×sin 55.56°

Q3. Determine 𝑸𝒗 = 𝟐𝑸𝒎 i.e. the amount of reactive power compensation required to hold the
midpoint voltage at 1.05 p.u.
𝜹
(a) 9585.49 – 10318.28𝐜𝐨𝐬 𝟐 MVAr
𝛿
(b) 4792.74 – 5159.14 cos MVAr
2
 NPTEL Online Certification Course
< Power Electronics Applications in Power Systems >
<Assignment 04>: Detailed Solution
Indian Institute of Technology Guwahati
𝛿
(c) 8271.41 – 9317.16cos 2 MVAr
𝛿
(d) 9725.48 – 11358.28cos 2 MVAr

Solution:
𝛽𝑙
2 𝑐𝑜𝑠 −𝑉𝑉 𝑐𝑜𝑠 𝛿
𝑉𝑚𝑐 𝑚𝑐
2 2
𝑄𝑣 = 2 × [ 𝛽𝑙 ]
𝑧𝑐 𝑠𝑖𝑛
2

𝛿
(1.05×735)2 𝑐𝑜𝑠27.78°−(735×1.05×735)𝑐𝑜𝑠
2
𝑄𝑣 = 2 × [ ]
235.90×𝑠𝑖𝑛27.78°

𝛿
𝑄𝑣 = (9585.49 − 10318.28𝑐𝑜𝑠 2) 𝑀𝑉𝐴𝑟

Q4. What is the expression for active power of compensated line?

𝛿
(a) 5209.81× sin 2
𝛿
(b) 42219.78× sin 2
𝜹
(c) 5159.14× 𝐬𝐢𝐧
𝟐
𝛿
(d) 4318.59× sin 2

Solution:
𝑉𝐿−𝐿 ×(𝑉𝑚𝑐 )𝐿−𝐿 𝛿 735×(1.05×735) 𝛿 𝛿
Compensated line active power = 𝛽𝑙 × sin 2 = 235.9×sin(27.78°) × sin 2 = 5159.14 sin 2
𝑍𝑐 ×sin
2

Q5. What is the range of mid-point compensator?

(a) [−𝟕𝟑𝟐. 𝟕𝟗, 𝟗𝟓𝟖𝟓. 𝟒𝟗] MVAr
(b) [−366.4, 4792.74] MVAr
(c) [−1045.75, 8271.41] MVAr
(d) [−1632.8, 9725.48] MVAr
Solution:
𝐴𝑡 𝛿 = 0°, (𝑄𝑣 )𝑚𝑖𝑛 = 9585.49 − 10318.28 = −732.79 𝑀𝑉𝐴𝑟
𝐴𝑡 𝛿 = 180°, (𝑄𝑣 )𝑚𝑎𝑥 = 9585.49𝑀𝑉
 NPTEL Online Certification Course
< Power Electronics Applications in Power Systems >
<Assignment 04>: Detailed Solution
Indian Institute of Technology Guwahati

Q6. Let us keep the size of the compensator to a reasonably practical range i.e. -650 MVAr
to +400 MVAr. What is the range of angle of loading for which the compensator can provide
required compensation to keep mid-point voltage constant?
(a) 17.39° ≤ 𝛿 ≤ 59°
(b) 14.52° ≤ 𝜹 ≤ 𝟓𝟒. 𝟐𝟎°
(c) 19.54° ≤ 𝛿 ≤ 58.60°
(d) 15.43° ≤ 𝛿 ≤ 55.43°
Solution:
𝛿
𝑄𝑣 = 9585.49 – 10318.28 cos 2
𝛿
−650 = 9585.49 – 10318.28 cos 2

𝛿 = 14.45°
𝛿
400 = 9585.49 – 10318.28 cos 2

𝛿 = 54.20°
Q7. The compensator operated in fixed reactor mode of operation. What is the value of fixed
reactance?
(a) 715.19Ω
(b) 971.34Ω
(c) 872.51 Ω
(d) 916.30 Ω
Solution:
(1.05×735)2
𝑄𝐿 = 650 = 𝑋𝐿

(1.05×735)2
𝑋𝐿 = = 916.30 Ω
650

Q8. The compensator operated in fixed capacitor mode of operation. What is the value of
fixed capacitor?
(a) 1359.1 Ω
(b) 1241.4 Ω
 NPTEL Online Certification Course
<Power Electronics Applications in Power Systems>
<Assignment 04>: Detailed Solution
Indian Institute of Technology Guwahati

(c) 𝟏𝟒𝟖𝟖. 𝟗 Ω
(d) 1537.5 Ω
Solution:
(1.05×735)2
𝑄𝑐 = 400 = 𝑋𝑐

(1.05×735)2
𝑋𝑐 = = 1488.9 Ω
400

Q9. When the compensator will operate as fixed capacitive mode. What would be the
expression for mid-point voltage?
(a) 𝟖𝟔𝟔. 𝟗 𝐜𝐨𝐬 𝜹⁄𝟐 𝒌𝑽
(b) 887.5 cos 𝛿 ⁄2 𝑘𝑉
(c) 941.0 cos 𝛿 ⁄2 𝑘𝑉
(d) 916.2 cos 𝛿 ⁄2 𝑘𝑉
Solution:
cos𝛿 ⁄2 cos𝛿 ⁄2
𝑉𝑚𝑐 = 𝛽𝑙 𝑍 𝛽𝑙 = 235.9 𝑝. 𝑢. = 1.179 cos 𝛿 ⁄2 𝑝. 𝑢. =
cos − 𝑐 sin cos 27.78−
2×1488.9
sin 27.78
2 2𝑋𝑐 2
866.9 cos 𝛿 ⁄2 𝑘𝑉
Q10. When the compensator will operate as fixed capacitive mode. What would be the
expression for power flow?
(e) 𝟐𝟖𝟗𝟕. 𝟔𝟏 𝐬𝐢𝐧 𝜹
(f) 2516.26 sin 𝛿
(g) 2687.49 sin 𝛿
(h) 2758.34 sin 𝛿
Solution:
𝑉𝑠 ×𝑉𝑚𝑐 735×866.9 ×sin 𝛿
𝑃𝑐𝑜𝑚𝑝 = 𝛽𝑙 = 2×235.9×sin 27.78° = 2897.61 sin 𝛿
𝑍𝑐 sin
2
 NPTEL Online Certification Course
<Power Electronics Applications in Power Systems>
<Assignment 04>: Detailed Solution
Indian Institute of Technology Guwahati
 NPTEL Online Certification Course
<Power Electronics Applications in Power Systems>
<Assignment 04>: Detailed Solution
Indian Institute of Technology Guwahati
 NPTEL Online Certification Course
<Power Electronics Applications in Power System>
<Assignment 04>: Detailed Solution
Indian Institute of Technology Guwahati
 NPTEL Online Certification Course
<Power Electronics Applications in Power System>
<Assignment 04>: Detailed Solution
Indian Institute of Technology Guwahati
 NPTEL Online Certification Course
<COURSE NAME>
<Assignment Number>: Detailed Solution
Indian Institute of Technology Guwahati
 NPTEL Online Certification Course
<COURSE NAME>
<Assignment Number>: Detailed Solution
Indian Institute of Technology Guwahati
 NPTEL Online Certification Course
<COURSE NAME>
<Assignment Number>: Detailed Solution
Indian Institute of Technology Guwahati
 NPTEL Online Certification Course
<COURSE NAME>
<Assignment Number>: Detailed Solution
Indian Institute of Technology Guwahati
 NPTEL Online Certification Course
<COURSE NAME>
<Assignment Number>: Detailed Solution
Indian Institute of Technology Guwahati