Electrical Engineering Mathematics

Question

David · Accepted Answer

Solution: 
Question 1 :
For XLPE insulated cable 
Length = 5000 m 
 Internal radius r= 10.8/2 = 5.4 mm = 0.54 cm 
External radius R = (10.8+3.4)/2= 7.1 mm = 0. 71 cm 
Capacitance per meter  
2
loge
C
R
r


      faraday 
0 r    
Putting the value of constants and converting to base 10  we get following 
equation : 
Capacitance in micro farad per kilo meter 
10
0.0242
log
rC
R
r


 
C= 0.4886 uF per kilometer 
Hence total length = 5 km 
C = 2.443 uF 
Question 1A
V= 11 kV  
I = V/Xc 
1
1/ 1302
2
Xc c ohm
fC


    
I= 8.45 amp 
Assuming 11 kV is line voltage and 8.45 amp is line current 
Total charging kVAr = 1.732* 11 kv* 8.45 amp = 161 kVAr 
Question 2 
A single core cable for use on a nominal 6.6kV three phase system has a conductor 
20 mm in diameter and the diameter at the internal surface of the Leads heat h is 
40mm.The relative permittivity of the insulation is 3.5 and the loss angle is 7 
minutes (0.002radians). Calculate the insulation resistance and the insulation 
power loss per km. 
Solution 
First calculating the capacitance per kilo meter 
0
2
loge
r
C
R
r

  



Capacitance in micro farad per kilo meter 
10
0.0242
log
rC
R
r


 
Given  
relative permittivity of the insulation =  3.5 
R= 20 mm 
And conductor radius = 10 mm 
So  
10
0.0242*3.5
0.281 /
20
log
10
C uF km 
 
So for 1 km length of cable C= 0.281 uF 
Operating voltage = 6.6 kV = 6600 V 
Loss angle = 7 minute = 0.

Electrical Engineering Mathematics

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