Ece421 Power System Analysis
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ECE421 POWER SYSTEM ANALYSIS
Homework #4
4.2 4.5(a) 4.5(b) 4.6 4.8(a) 4.8(b) 4.11 4.15 Total ECE42 12’ 9’ 9’ 20’ 10’ 10’ 20’ 10 100’ 1
4.2 12’
The dc resistance at 20 degree,
(3’)
The ac resistance at 60Hz,
(3’)
The ac resistance at 50 degree,
(3’)
The resistance of cable,
(3’)
4.5(a) 9’
(3’)
(3’)
(3’)
4.5(b) 9’
(3’)
(3’)
(3’)
4.6 20’ (0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(2’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(1’)
(2’)
(2’)
(2’)
(2’)
(2’)
4.8(a) 10’
(4’)
(6’) 4.8(b) 10’
(2’)
(4’)
(4’)
4.11 20’
For the a1b1c1,c2b2a2 configuration,
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’) (0.5’)
(0.5’)
(1’)
(1’)
Use gmd to verify the result, (1’)
Enter (1 or 2) -> 1
Enter spacing unit within quotes 'm' or 'ft'-> 'm'
Enter row vector [S11, S22, S33] = [16, 24, 17]
Enter row vector [H12, H23] = [10, 9]
Cond. size, bundle spacing unit: Enter 'cm' or 'in'-> 'cm'
Conductor diameter in cm = 4.4069
Geometric Mean Radius in cm = 1.7374
No. of bundled cond. (enter 1 for single cond.) = 2
Bundle spacing in cm = 45
GMD = 15.92670 m
GMRL = 1.47993 m GMRC = 1.57052 m
For the a1b1c1,a2b2c2 configuration,
(0.5’)
(0.5’)
(0.5’) (0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(0.5’)
(1’)
(1’)
Use gmd to verify the result, (1’)
Enter (1 or 2) -> 2
Enter spacing unit within quotes 'm' or 'ft'-> 'm'
Enter row vector [S11, S22, S33] = [16,24,17]
Enter row vector [H12, H23] = [10,9]
Cond. size, bundle spacing unit: Enter 'cm' or 'in'-> 'cm'
Conductor diameter in cm = 4.4069
Geometric Mean Radius in cm = 1.7374
No. of bundled cond. (enter 1 for single cond.) = 2
Bundle spacing in cm = 45
GMD = 17.08872 m GMRL = 1.28551 m GMRC = 1.36419 m
4.15 10’
(2’)
(2’)
(2’)
(2’)
The equivalent capacitance to neutral is,
(2’)