Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Forces generated by cable faults
- Thread starter raewyn
- Start date
I am certain that I cannot provide the direct answer you seek as I am not an expert in this matter. However, I will respond with what I believe is a suggestion to help you find the answer.
At a minimum, I think you need to know the system capacity for supplying fault current at the point of interest as well as the arrangement of the conductors, the size of the conductors, and the spacing between them. I assume that you are describing the arrangement of the conductors when you say that they are in "trefoil" as being in an equilateral or triangular arrangement, so that part of the question may be answered.
At a minimum, I think you need to know the system capacity for supplying fault current at the point of interest as well as the arrangement of the conductors, the size of the conductors, and the spacing between them. I assume that you are describing the arrangement of the conductors when you say that they are in "trefoil" as being in an equilateral or triangular arrangement, so that part of the question may be answered.
If the forces you are talking about are the magnetic fields generated under fault conditions, that can be calculated, but (as rhatcher points out) it depends on the available fault current at the point of the fault.
Since the magnetic field is directly proportional to the current flowing in the conductor, the force genrated will also be directly related to the fault current. It will also be related to the distance between the conductors (for trefoil arrangement, this will be the radius of the conductors).
For two conductors, the calculation is relatively easy, but the calculation for three conductors gets more complicated if you consider a fault that may be on two conductors and not on the third. The current and polarity will not be the same in all three conductors, so you could get some H-field cancelling, which would change the resultant of all forces involved. That just means you get to outline the results for a couple of different fault situations, though.
Some places you might want to look for background on this are:
Since the magnetic field is directly proportional to the current flowing in the conductor, the force genrated will also be directly related to the fault current. It will also be related to the distance between the conductors (for trefoil arrangement, this will be the radius of the conductors).
For two conductors, the calculation is relatively easy, but the calculation for three conductors gets more complicated if you consider a fault that may be on two conductors and not on the third. The current and polarity will not be the same in all three conductors, so you could get some H-field cancelling, which would change the resultant of all forces involved. That just means you get to outline the results for a couple of different fault situations, though.
Some places you might want to look for background on this are:
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: The following could be considered:
1. Two straight conductors
2. Forces on unit length of each conductor
Conductor 1 has:
F1 force toward conductor 2 depending on a direction of current flow in conductors
I1 current flowing in conductor 1
Conductor 2 has:
F2 force toward conductor 1 depending on a direction of current flow in conductors
I2 current flowing in conductor 2
3. Relationship between forces for two conductors:
F1 = F2 = mu x I1 x I2 / (2 x pi x r)
where
mu is mu relative x muo, permeability
muo is permeability for free space 4 x pi x 10**(-7) in Newton/Ampere**2
pi = 3.14
r is a distance between two conductor electrical (if near) or geometrical (if far) centers
If currents flow in the opposite directions, then F1 and F2 will be repulsive
If currents flow in the same directions, then F1 and F2 will be attractive
The trefoil configuration will have to have all forces properly geometrically located and oriented (and principle of the superposition may be considered since there is no need to consider any nonlinear phenomena). They will be dependent on short circuit currents in individual conductors. Now, since there are various types of faults, there will be various types of forces within your trefoil conductor configuration. 132kV cables usually have a small harmonic content. Therefore, normal fault current calculation results may be applied, i.e symmetrical and close & latch.
1. Two straight conductors
2. Forces on unit length of each conductor
Conductor 1 has:
F1 force toward conductor 2 depending on a direction of current flow in conductors
I1 current flowing in conductor 1
Conductor 2 has:
F2 force toward conductor 1 depending on a direction of current flow in conductors
I2 current flowing in conductor 2
3. Relationship between forces for two conductors:
F1 = F2 = mu x I1 x I2 / (2 x pi x r)
where
mu is mu relative x muo, permeability
muo is permeability for free space 4 x pi x 10**(-7) in Newton/Ampere**2
pi = 3.14
r is a distance between two conductor electrical (if near) or geometrical (if far) centers
If currents flow in the opposite directions, then F1 and F2 will be repulsive
If currents flow in the same directions, then F1 and F2 will be attractive
The trefoil configuration will have to have all forces properly geometrically located and oriented (and principle of the superposition may be considered since there is no need to consider any nonlinear phenomena). They will be dependent on short circuit currents in individual conductors. Now, since there are various types of faults, there will be various types of forces within your trefoil conductor configuration. 132kV cables usually have a small harmonic content. Therefore, normal fault current calculation results may be applied, i.e symmetrical and close & latch.
- Status
Similar threads
- Question
- Locked
- Question