Download Tutorial question electric machine and more Exams Electric Machines in PDF only on Docsity!
Find the self GMD for each conductor configuration shown in Figure 1. [(a)GMR =
7036 r; (b)GMR = 1. 6921 r]
Figure 1:
Solution:
(a)
GMR = 3
GMR 1 × GMR 2 × GMR 3
GMR 1 = GMR 3 =
- 7788 r × 2 r × 4 r = 1. 84 r
GMR 2 =
- 7788 r × 2 r × 2 r = 1. 4604 r
GMR =
- 84 r × 1. 4604 r × 1. 84 r = 1. 7036 r
(b) GMR 1 = GMR 3 =
- 7788 r × 2 r × 2 r × 2 r = 1. 5798 r
GMR 2 = GMR 4 =
4
- 7788 r × 2 r × 2 r ×
12 r = 1. 8124 r
GMR =
5798 r × 1. 5798 r × 1. 8124 r × 1. 8124 r = 1. 6921 r
Determine the inductance of a 3 phase line operating at 50 Hz and conductors arranged
as given in figure 2. The conductor diameter is 0.8 cm. [L = 1. 294 mH/km]
Figure 2:
Solution: GMR = 0. 7788 × 0 .004 = 0. 0031152 m
GMD =
6
(1. 6 × 1 .6)(1. 6 × 3 .2)(1. 6 × 3 .2) = 2. 0158 m
L = 2 × 10 −^7 ln
= 1. 294 mH/km
- Determine the inductance of a single phase transmission line consisting of three conduc-
tors of 2.5 mm radii in the ‘go’ conductor and 5 mm radii in the return conductor. The configuration of line is as shown in figure 3.[(a)L = 1. 42 mH/km; (b)L = 1. 485 mH/km]
Figure 3:
Solution:
(a)
GMRA = 3
GMRa × GMRb × GMRc
GMRa = GMRc =
- 7788 × 0. 0025 × 6 × 12 = 0. 51947 m
GMRb = 3
- 7788 × 0. 0025 × 6 × 6 = 0. 4123 m
GMRA =
3
- 51947 × 0. 4123 × 0 .51947 = 0. 4809 m
GMRB =
4
(0. 7788 × 5 × 6)(0. 7788 × 5 × 6) = 0. 1528 m
GMDA = GMDB =
9 × 10. 81 × 9 × 10. 81 × 10. 81 × 15 = 10. 74 m
LA = 2 × 10
− 7 ln
= 0. 62 mH/km
Figure 4:
GMRc = 4
r
′ cDcc′^ ×^ r
′ c′^ Dc′c = 4
- 7788 × 0. 01266 × 10. 965 × 0. 7788 × 0. 01266 × 10 .965 = 0. 3288 m
GMRb = 4
r
′ bDbb′^ ×^ r
′ b′^ Db′b =
- 7788 × 0. 01266 × 9 × 0. 7788 × 0. 01266 × 9 = 0. 3288 m
GMRA =
- 3288 × 0. 2978 × 0 .3288 = 0. 3181 m
GMD =
3
GMDab × GMDbc × GMDca
GMDab =
4
DabDab′^ Da′bDa′b′^ =
- 069 × 9. 168 × 9. 168 × 4 .069 = 6. 1077 m
GMDbc = 4
DbcDbc′ Db′cDb′c′ = 4
- 069 × 9. 168 × 9. 168 × 4 .069 = 6. 1077 m
GMDca = 4
DcaDca′ Dc′aDc′a′ =
8 × 7. 5 × 7. 5 × 8 = 7. 7459 m
GMD = 6. 610 m
L = 0. 606 mH/km
- Determine the inductance per km per phase of a double circuit 3 phase line arranged as
shown in figure 5. The radius of each conductor is 15 mm. [L = 0. 51035 mH/km/ph]
Figure 5:
Solution:
GMRA =
3
GMRa × GMRb × GMRc
GMRa = 4
r ′ aDaa′^ ×^ r
′ a′^ Da′a = 4
- 7788 × 0. 015 × (1. 75 × 3) × 0. 7788 × 0. 015 × (1. 75 × 3) = 0. 2476 m
Since GMRa = GMRb = GMRc, GMR = GMRa = 0. 2476 m.
GMDab = 4
DabDab′ Da′bDa′b′ = 4
- 75 × 7 × 3. 5 × 1 .75 = 2. 9431 m
GMDbc =
4
DbcDbc′^ Db′cDb′c′^ =
- 75 × 7 × 3. 5 × 1 .75 = 2. 9431 m
GMDca = 4
DcaDca′ Dc′aDc′a′ =
- 5 × 1. 75 × 8. 75 × 3 .5 = 3. 7 m
GMD =
- 9431 × 2. 9431 × 3 .7 = 3. 1766 m
L = 0. 51035 mH/km/ph
- Find the inductive reactance per phase in Ω/m and the capacitive reactance to neutral
in Ω − m for a three phase line that has equilaterally spaced conductors of ACSR ‘Dove. The distance between lines of various phases is 3.048m and the operating frequency is 60 Hz. From standard table, Dove GMR = 0.00954024 m, Dia of Conductor = 0. m. [XL = 4347. 86 × 10 − 7 Ω/m/ph, XC = 0. 000265266 × 10 12 Ω − m/ph]
Solution: Since conductors are spaced equilaterally,
GMD = 3. 048 m
GMR = 0. 00954 m
L = 2 × 10
− 7 ln
= 11. 5334 × 10
− 7 H/m/ph
XL = 2 × 3. 1415 × 60 × 11. 5334 × 10
− 7 = 4347. 86 × 10 − 7 Ω/m/ph
For calculation of C, we have to use radius of conductor as GMR.
C =
2 × π × ǫ 0
ln
= 10 × 10
− 12 F/m
XC =
2 × 3. 1415 × 60 × 10 × 10 −^12
= 0. 000265266 × 10
12 Ω − m/ph
XL = 2 × 3. 1415 × 60 × 6. 131 × 10
− 7 = 2311 × 10 − 7 Ω/m/ph
For calculation of capacitance of the line, r=0.008636 m.
GMRa = GMRc =
- 008636 × 8 .2 = 0. 266 m
GMRb =
- 008636 × 6 .4 = 0. 2350 m
GMR =
- 266 × 0. 2350 × 0 .266 = 0. 255 m
C =
2 × π × ǫ 0
ln
= 18. 797 × 10
− 12 F/m
XC =
2 × 3. 1415 × 60 × 18. 797 × 10 −^12
= 0. 000141117 × 10
12 Ω − m/ph
- Each conductor of the bundled conductor line shown in figure. is ACSR, ’pheasant’.
Find (a) the inductive reactance in ohms per km per phase for d=45 cm. (b) the per unit series reactance of the line if its length is 160 km and the base is 100 MVA, 345 kV. (c) the capacitive reactance to neutral of the line in ohm-km/phase. From standard table, GMR=0.0466 ft (convert this into meters), outside diameter of each conductor is 1.382 inches (convert this into meters) and system frequency=60 Hz. [(a)XL = 0 .365Ω/km/phase; (b)Xp.u = 0. 049 p.u; (c)Xcn = 0. 22589 × 106 Ω − km/ph to neutral]
Figure 7:
Solution:
(a) GMR of each conductor = 0. 0466 × 0 .3048 = 0. 014204 m
GMRa = GMRb = GMRc =
- 014204 × 0 .45 = 0. 08 m
GMD =
3
Dab × Dbc × Dca =
8 × 8 × 16 = 10. 08 m
XL = 2 × π × 60 × 2 × 10 − 7 × ln
= 0.365Ω/km/phase
(b)
Baseimpedance, Z =
(basekV )^2
baseMV A
(345 × 103 )^2
100 × 106
Xp.u =
Xactual
Zbase
0. 365 × 160
= 0. 049 p.u
(c)
r =
1. 382 × 2. 54
2 × 100
= 0. 01755 m
GMRa = GMRb = GMRc =
- 01755 × 0 .45 = 0. 0889 m
GMD =
3
Dab × Dbc × Dca =
8 × 8 × 16 = 10. 08 m
Cn =
2 πǫo
ln
GMD
GMR
2 × π × 8. 842 × 10 −^12
ln
= 11. 743 × 10 −^12 F/m
Xcn =
2 πf C
12 × 10 − 3
2 × π × 60 × 11. 743
= 0. 22589 × 10
6 Ω − km/ph to neutral
- A 3 phase 60 Hz transmission line has its conductors arranged in a triangular formation
so that two of the distances between conductors are 25 ft and the third is 42 ft. The conductors are ACSR Osprey. Determine the capacitance to neutral in microfarads per mile and the capacitive reactance to neutral in Ω-miles. If the line is 150 miles long, find the capacitance to neutral and capacitive reactance of the line. From standard table, GMR=0.0284 ft (convert this into meters), outside diameter of each conductor is 0. inches (convert this into meters). [Cn = 13. 34 × 10 −^9 F/mi; Xc =
- 1988 × 106 Ω − mi; Cn,T otal = 2. 001 μF ; Xc,T otal = 1325Ω]
Solution: GMD = 3
25 × 42 × 25 = 29. 72 f t
r = 0. 879 /2 = 0. 4395 inches
Cn =
2 πǫo
ln
GMD
r
2 × π × 8. 842 × 10 −^12
ln
29. 72 × 12
= 8. 293 × 10
− 9 F/km
= 8. 293 × 10
− 9 × 1. 609 F/mi = 13. 34 × 10 − 9 F/mi
Xc =
2 πf C
2 × π × 60 × 13. 34
= 0. 1988 × 10
6 Ω − mi
Cn,T otal = 150 × 13. 34 × 10 − 9 = 2. 001 μF
Xc,T otal =
0. 1988 × 10
6
