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| Subject - Single phase Short Circuit Force calculations? | |
| dslad |
Does anyone know the formula for calculating the force generated during a single phase short circuit condition if the asymetrical short circuit current and distance between conductors is known? Thanks! Don |
| Ryan_J | What type of force? Amperage? Calories? |
| JimmyDee | quote:I think I'll wait for Scott Vickrey to have a go at this one.  Just a tad over my head.Jim |
| iwire | I have no clue how to figure that.
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| Scott Vickrey | Wow what a question. I'm sure there is such a formula and the transmition types might be able to answer your question. I have to say I don't know. |
| Ryan_J | I don't know the formula, but I know where a calculator for it is. http://www.bussman.com/apen/arcflash |
| jackleung | I think he's asking about the magnetic force (either repulsive or attractive force between 2 conductors) generated from the short circuit current. As I know, this is quite important for the busbar inside switchboard. The force generated from short circuit current can bend the busbar and cause further damage! Please clarify your question. If you are really asking this force, the answer is around those equation we learn from college. i.e. F=B*I*l, F=B*v*q...etc. I forgot the detail, but someone must be able to quote the equation correctly here! |
| dslad | Hello, Folks, Thanks for your inputs. I am looking for the formula to calculate the magnetic force between two conductors due to a short circuit, when the fault current and distance between conductors is known. I'm primarily interested in 60hz AC. |
| mistermike | Found a formula in a dusty old reference book. F/l=((2*10E7)*I1*I2)/S where: S = spacing. Reference point not specified, so I assume center-to-center. I1 = current in first wire. I2 = current in second wire. 10E7 = 10 to the 7th power (can't figure out how to do superscripts). F/l = Force per unit length. S and l are same units. F in newtons. For AC systems, multiply the formula result by 2 if you use RMS values for current. Above formula is for wires. I also have a formula for bus bars if needed. |
| Scott Vickrey | Thanks for chiming in with a solution Mike. We are glad to have you here. I hope you will continue to share your resourcefulness. This is the beauty of open forum sites like this and the other similar great sites. If it is known someone has the answer and is willing to take the time to share it. Now when someone else searches for this information they are more likely to find it and when they do they have the option to further the topic. Like a living document. Available to all. |
| RalphChristie | Hi Some info regarding short-circuit forces on busbars at: http://www.cda.org.uk/megab2/elecapps/pub22/sec8.htm#Maximum%20Permissible%20Stress go to maximum permissible stress You can also try IEEE Std 106-1998 (on page 16) (IEEE Guide for Design of Substation Rigid-Bus Structures) Hope it helps and not to late Regards Ralph |
| lucky1122 | force calculations? |
| lucky1122 | Since the magnetic field of the wire is based on the amount of current flowing through it ,this would be the first step in any calculations that would be necessary to determine the destructive power caused by a short or fault. When equipment is not properly applied within proper withstand ratings severe damage has been known to be caused by faults having values in excess of the rating of the equipment. |
| lucky1122 | please except my humble apology for the math error in the previous posting to this issue . The correct proceedure is 150KVAX1000/208x 1.73=416.85A. (416.85A x 100/2)= 20842.59rms sym amperes . I do apologize for making this error in the math when I posted a reply it simply has been a while since Ive worked out the math and simply forgot.Simply dividing by the impedance will give you the same answer . there is much more to this formula obviously and it can be found(bulletin EDP-1) |
| lucky1122 | just remove the 1.73 from the equation and you have the single phase version. |