Short circuit forces - Direction? 4

[sunny81]

I am trying to design a bus brace for a 3 phase busbar syatem. I was able to calculate the forces acting on the brace from the information I found on this forum. I am trying to apply this load on to the brace and do some analysis using pro/mechanica.I am having difficulty understanding the direction of these forces on the three phases. If my calculated force is 1000lbs, would this force be acting on each of the three phases depending on the direction of current.
This is the first time I am working on something like this, Please help me understand.

Thanks

 

[jghrist]

The forces on the bus are sinusoidal. When the instantaneous current in two conductors is in the same direction, the two conductors are attracted to each other. During a Ø-Ø fault, the currents are in opposite directions and the conductors would repel each other.

During a 3Ø fault, the forces (for symmetrical current) in the center conductor would average zero and would peak in opposite directions at different points in the current cycle. The average direction of forces on the center conductor during an asymmetrical fault (with dc offset) would depend on the point of initiation of the fault with respect to the voltage waveform. The forces in the outside conductors would be away from the center conductor.

 

[electricpete]

Sounds like a great response. For 3-phase symmetrical fault, I would think the average force on outer conductors feeding the fault would be attractive (because 120 degrees-apart currents are in opposite direction for 2/3 of the cycle), right?

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[desertfox]

Hi sunny81

This site might help:-

http://www.copperinfo.co.uk/busbars/pub22-copper-for-busbars/sec6.htm#Electromagnetic

Stresses

desertfox

 

[jghrist]

For 3-phase symmetrical fault, I would think the average force on outer conductors feeding the fault would be attractive (because 120 degrees-apart currents are in opposite direction for 2/3 of the cycle), right?

When the currents are in the opposite direction, the forces are repulsive.

 

[electricpete]

Yes, you’re right – I had that backwards for a moment.
Now let’s circle back and revisit the phase-to-phase fault. In that case currents are flowing opposite direction as well, so shouldn’t they be repulsive?


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[electricpete]

Never mind - that's exactly what you said.

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[sunny81]

Thanks everyone, great information.

Now, would it be safe to assume for my pro/mechanica analysis for a symmetrical 3phase fault, that the brace would be seeing a force of say 1000lbs on A and 1000lbs on C phase going in the opposite directions, cause B phase forces would average out to zero.

 

[jghrist]

Now, would it be safe to assume for my pro/mechanica analysis for a symmetrical 3phase fault, that the brace would be seeing a force of say 1000lbs on A and 1000lbs on C phase going in the opposite directions, cause B phase forces would average out to zero.

IEEE Std 605, IEEE Guide for Design of Substation Rigid-Bus Structures considers peak force during an asymmetrical fault, not average during a symmetrical fault. The force on the middle phase during a 3Ø fault is 0.866 times the force on the buses during a Ø-Ø fault of the same magnitude. The force on the outside phases is 0.808 times the Ø-Ø fault force.

 

[desertfox]

Hi sunny81

If you click on the link I gave you in my earlier post, then click on electromagnetic stresses it will give you formula's for calculating the forces on the busbars for your situation.

desertfox

 

[waross]

Anytime during the cycle that the current is zero in a bus, the force on that bus will be zero.
In practice the force on the outer bus bars will be outward and the center bus will tend to swing back and forth between them.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jack6238]

Well, this is a great question and let's start with the fundamentals. The question is where is the greatest force depends on the bus configuration. Surely this makes sense.

For example a bus that is confiruged is a horizontal plane like ABC does not create the same force as a bus that has a configuration of an equalitratel plane( an equal lateral traiangle). The equal-lateral triangle has equal force on each conductor (that is the design parameter) wher a ABC arrangement has the maximum force onthe middle conductor.

So what is the force on conductors in the same plane?

Here are the considerations:

1.You need to determine the section modulus of the conductors. For rectangular conductors the section modulus is S= b* d^2/6 where d is the width and b is the height.(Reference Mechanic of Materials ,Fairman & Cutshall, John Wiley & Sons, page 113.
2.Is the bus Aluminum or Copper. The bus may have the same section modulus but as you know, aluminum may have a lower yield strength than Copper and this makes it a weaker system.
3. For a buses in the same plane, the maximum force is on the phase B conductor and this force occurs 45 degrees after the short circuit occurs in phase B as the current passes through a current zero value. All othe short circut occurrences produce less force.
4. The Force is given by:
F(lbs/In^2= 37'5* I^2*10^6/(10^7*D) where D distance between the center to center bus buses).
6.Once you know the section modulus, the short circuit in symmetrical amps, the distance between buses , you can easily determine the required distance between bracing requirements.

Obviously this is not a calculation for rookies.

Jack

 

[waross]

the maximum force is on the phase B conductor and this force occurs 45 degrees after the short circuit occurs in phase B as the current passes through a current zero value. All othe short circut occurrences produce less force.

Maximum force at zero current?
Did I read this properly?
I thought that the repelling force was greatest between adjacent conductors when the currents were equal. That is when the sum of the squares of the currents will be greatest.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[desertfox]

Hi waross

No you read jacks post correctly.

Obviously my link to short circuit forces on busbars must be invisible other than to me.

desertfox

 

[jack6238]

Mr. Bill,

I should have been more specific about when the maximum force occurs---and thank you for noting I was not clear on that specific point. Actually there were several points which needed additional explation.

Concerning when the maximum force occurs.

The maximum force occurs when the short circuit occurs 45 degrees after the zero current in Phase B. The maximum force will then occurs 1/2 cycle after the instant of the short circuit. Any other combination of circumstances will produce a smaller value.

Stanley Killian, he was a Developmet and Research Engineer for Delta Star Electric Company (Chicago), describes this in greater detail in "Forces Due to Short Circuit Currents" Electrical World, December 12, 1942.

Also concerning the Bracing Requirements

Often, or sometimes, the bus bars are not in the same plane. That is they are in a staggered plane or they contain bends. Not an easy solution to this. Here is what is happening. When the forces are not in the same plane, the forces causes the buses to want to "twist" or rotate so the bus bars will need to resist this motion. I solve this by calculating the bracing distance for buses in the same plane and adding a brace in the middle of the calculation. For example, Rather than 12 " I would add one at 6 inches.

Another item concerning bracing.

To really calculate the bracing distances, you need to know the exact yield strength (PSI) of the bus bar material. Most calculation are for copper. Aluminum alloy are now almost equal to the yield of Copper. The calculations are made for copper and then adjusted for AL. For example if the yield strength of aluminum is 6000 PSI and copper is 15,000 PSI, then the calculations are adjusted to {6000/15000}^0.5 =0.633 of the distance for copper material. So you calculate 10 inch bracing for copper and for Al the bracing requirement is 10*.633 = 6.33 inches. It all depends on the yield strength for the material used in your calculations.

I Hope this helps and many thanks to Bill for asking that question. I was in a hurry and should have taken the time to fully explain. Sorry!

Also, there appears to be significant interest in this subject as the insurance companies for industrials ask for short circuit and flash hazard calculations. I have helped several solve these problems and avoid a replacement or rebuild.

The basic engineering for this topic can be found in the work by C.H. Van Aspen of Hydro Electro Power Comission of Ontario -1922 and also H.B Dwight.

Jack

 

[ScottyUK]

desertfox,

I can see it.

The content of the web page was once published as a hardcover book by the CDA called 'Copper for Busbars', and some of the diagrams have been scanned in which explains the indifferent quality of the images. The book is well worth tracking down if only for the images of contemporary equipment. My personal favourite is the 40,000A 250V DC circuit breaker, as in 40,000A thermal rating, not breaking capacity. It is the size of a small truck.


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If we learn from our mistakes I'm getting a great education!

 

[desertfox]

Hi ScottyUK

Yes I have a copy of that very book and I thought it was still available, it is well worth having a copy I agree.

Here's a quote on the max force's due to a single phase and three phase fault:-



Maximum stresses

When a busbar system is running normally the interphase forces are normally very small with the static weight of the busbars being the dominant component. Under short-circuit conditions this is very often not the case as the current rises to a peak of some thirty times its normal value, falling after a few cycles to ten times its initial value. These high transitory currents create large mechanical forces not only in the busbars themselves but also in their supporting system. This means that the support insulators and their associated steelwork must be designed to withstand these high loads as well as their normal structural requirements such as wind, ice, seismic and static loads.

The peak or fully asymmetrical short circuit current is dependent on the power factor (cos f) of the busbar system and its associated connected electrical plant. The value is obtained by multiplying the r.m.s. symmetrical current by the appropriate factor given in Balanced three-phase short-circuit stresses.

If the power factor of the system is not known then a factor of 2.55 will normally be close to the actual system value especially where generation is concerned. Note that the theoretical maximum for this factor is 2Ö2 or 2.828 where cos f = 0. These peak values reduce exponentially and after approximately 10 cycles the factor falls to 1.0, i.e., the symmetrical r.m.s. short circuit current. The peak forces therefore normally occur in the first two cycles (0.04 s) as shown in Figure 13.

In the case of a completely asymmetrical current wave, the forces will be applied with a frequency equal to that of the supply frequency and with a double frequency as the wave becomes symmetrical. Therefore in the case of a 50 Hz supply these forces have frequencies of 50 or 100 Hz.

The maximum stresses to which a bus structure is likely to be subjected would occur during a short-circuit on a single-phase busbar system in which the line short-circuit currents are displaced by 180°.

In a three-phase system a short-circuit between two phases is almost identical to the single-phase case and although the phase currents are normally displaced by 120°, under short-circuit conditions the phase currents of the two phases are almost 180° out of phase. The effect of the third phase can be neglected.

In a balanced three-phase short-circuit, the resultant forces on any one of the three phases is less than in the single-phase case and is dependent on the relative physical positions of the three phases.

In the case of a single-phase short-circuit, the forces produced are unidirectional and are therefore more severe than those due to a three-phase short-circuit, which alternate in direction.

The short-circuit forces have to be absorbed first by the conductor. The conductor therefore must have an adequate proof strength to carry these forces without permanent distortion. Copper satisfies this requirement as it has high strength compared with other conductor materials (Table 2). Because of the high strength of copper, the insulators can be more widely spaced than is possible with lower-