|
 SHEATH
LOSSES
The common practice of grounding cable shields in three
phase systems at multiple locations results in induced voltages
and circulating currents depending on the load currents and shield
impedance.
With individually
jacketed cables these currents can be eliminated by interrupting
the shield and grounding each section at only one point. When using
single point grounding, it is recommended that the voltage rise
at the opposite end from ground of each section be limited to approximately
120 volts, under normal operating conditions. Circulating shield
currents can also be reduced by cross bonding the shields to cancel
out the induced voltages that generate these currents. (For more
information on shield currents and cross bonding, refer to IEEE
Standard 575).
VOLTAGE
RISE IN OPEN SHEATHS
Eliminating
the problem of circulating sheath currents by operating the circuit
with grounding at one end results in a voltage being induced in
the sheath. This voltage rise is proportional to distance from the
ground point and the phase current. As covered under sheath losses
the voltage rise should be limited to approximately 120 volts under
normal operating conditions.
The following
equation may be used for approximating the voltage rise on an open
circuited sheath.
Where:
Vs = Voltage rise in open sheath (volts/1,000ft.)
Dm = Geometric mean distance between cables (inches)
Ds = Diameter of the shielding tape (inches)
lc = Current in the phase conductor (amps)
Note:
Ds (the diameter of the shielding tape) can be arrived at by adding
the value given in the table below to the diameter over the insulation
or outer Permashield.
Diameter
Over the
Insulation (Inches) |
Single
Permashield
Cables |
Double
and
Triple Permashield Cables |
| up
to 1.000 |
0.065 |
0.013 |
| 1.001
to 1.500 |
0.085 |
0.013 |
| 1.501
and up |
0.115 |
0.013 |
 |
 |
 |
1 Cable per duct, equilateral spacing Dm = S.
For 3-1/C or 3/C twisted in one duct. Dm = cable O.D. |
For
flat spaced configuration
Dm = 1.26S
|
For
rectangular configuration
Dm = 1.12S
|
The constants
below can be used with the following formula to calculate the voltage
rise in the sheath of 15kV, 25kV, and 35kV cables:
Where:
Vs = K x lc x 10-3
Volts/1,000 feet
Vs = Voltage rise in open sheath (Volts/1,000 ft.)
K = Constant from table below
lc = Phase Current (Amperes)
CONSTANTS
FOR USE IN CALCULATING VOLTAGE RISE ON OPEN
CIRCUITED SHIELDS
| |
15kV
Cable Configuration |
25kV
Cable Configuration |
35kV
Cable Configuration |
|
Conductor
Size
(AWG/kcmil)
|
| |
|
|
|
|
|
|
|
|
|
A |
B |
C |
A |
B |
C |
A |
B |
C |
| 2 |
23 |
77 |
74 |
-- |
-- |
-- |
-- |
-- |
-- |
| 1 |
23 |
76 |
73 |
22 |
71 |
68 |
-- |
-- |
-- |
| 1/0 |
23 |
75 |
72 |
22 |
70 |
67 |
21 |
66 |
63 |
| 2/0 |
22 |
73 |
71 |
21 |
69 |
66 |
21 |
65 |
62 |
| 3/0 |
22 |
72 |
69 |
21 |
68 |
65 |
21 |
64 |
61 |
| 4/0 |
22 |
72 |
68 |
21 |
67 |
64 |
20 |
63 |
60 |
| 250 |
21 |
69 |
67 |
21 |
65 |
63 |
20 |
62 |
59 |
| 350 |
21 |
67 |
64 |
21 |
64 |
61 |
20 |
60 |
58 |
| 500 |
21 |
65 |
62 |
20 |
61 |
59 |
20 |
58 |
56 |
| 750 |
21 |
61 |
58 |
20 |
58 |
55 |
20 |
56 |
53 |
| 1000 |
21 |
59 |
56 |
20 |
56 |
53 |
20 |
54 |
51 |
|
For
Configuration:
A: S = Cable O.D.
B: S = 8 Inches
C: S = 8 Inches
Example:
One circuit of three single 500 kcmil, 15kV cables twisted,
installed in a single duct carrying 495 amps.
K =21
Vs = 21 x 495 x 10-3
Vs = 10.4 Volts/1,000 feet
|
