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Topic Summary: Created On: 14 July 2012 10:19 AM Status: Post and Reply |
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 16 July 2012 03:04 PM
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Here we go some smarta$$ bounced this to top without revealing themselves. 
I have to take credit for being the only one to answer the OP's question without putting them through the usual technical hoops that are created on this forum, Peacock,  you're very sad!
Regards |
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16 July 2012 08:06 PM
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Feathers and blood every where, not a pretty sight; good job the OP is not looking.
------------------------- http://www.niceic.biz |
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18 July 2012 02:16 PM
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As opposed to In / Correction factors But that assumes you already know It (ie you have a cable size already) In (or Ib)/corrections is the more usal design approach as it gives you a number you then select against from cable tabulated ratings. Of course - if we are being picky, it would be Ib in this case, as the circuit is a motor circuit and will almost certainly have seperate overload protection - so we are really only designing for short circuit conditions regards OMS ------------------------- Failure is always an option |
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20 July 2012 11:56 AM
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As opposed to In / Correction factors But that assumes you already know It (ie you have a cable size already) Of course we both know that, I was merely pointing out the error in the example given by R&R, he did ask. 
In (or Ib)/corrections is the more usal design approach as it gives you a number you then select against from cable tabulated ratings. I would have thought volt drop would be a better starting point. Of course - if we are being picky, it would be Ib in this case, as the circuit is a motor circuit and will almost certainly have seperate overload protection - so we are really only designing for short circuit conditions I did point out that there was more information needed. 
regards OMS ------------------------- http://www.niceic.biz |
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20 July 2012 12:18 PM
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Well, many cables could comply for voltage drop - the first consideration has to be CCC in my opinion. You need a minimum conductor size for CCC, from there you test for volt drop - which may increase a cable size - and then you have a cable to test for EFLI (which again may increase a cable size and thermal constraints (earth fault and short circuit) whaich again may increase a cable size - ie you are always going from the minimum needed - increasing because of a further consideration doesn't then negate your previous calculation. Regards OMS ------------------------- Failure is always an option |
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 20 July 2012 12:24 PM
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Good afternonn, I am new member in this forum and it's my first participation that I hope will not be the last! I'm a mechanical ingeenier and I'm looking for the complete and correct manner or formula to determine and calculate a cable section in a three phase circuit. It will be very helpful to give a detailed calculation with detailed terms in such circumstances (I am not known in electrical calculation, but I have to make it during a present project): Motor power: 8kW, voltage: 415 V, Nominal current: I= 22 A cable length = 50m ambiant temperature: 50°C thank you in advance Walid Hi Walid, If you do your volt drop calculation (mV per amp per metre x length of circuit x design current /1000) for an 8KW motor, the nominal current rating of 22amps will be OK for a 6mm Steel wire armoured cable, but it is likely that the starting current (likely to be around 32amps given that this is an 8KW motor) will result in excessive voltage drop. I would move up to a 10mm Steel wire armoured supply cable. |
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20 July 2012 12:35 PM
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The starting current would be a lot more than 32A david - try something like 132A Keep in mind that you could allow the voltage drop to fall by say 20% on starting anyway (motor pull out torque requirements not withstanding) OMS ------------------------- Failure is always an option |
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20 July 2012 12:42 PM
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The starting current would be a lot more than 32A david - try something like 132A Keep in mind that you could allow the voltage drop to fall by say 20% on starting anyway (motor pull out torque requirements not withstanding) OMS Why would a figure of 132amps be relevant to the starting characteristics of a Motor labelled '8KW and with a nominal current rating of 22amps'? |
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20 July 2012 12:43 PM
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Fair enough if you only design small installations, but the over-riding factor determining most commercial/industrial circuits is their length.
A volt drop calculation is a quick way to start the calculation, but then each to his/her own. ------------------------- http://www.niceic.biz |
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20 July 2012 12:45 PM
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OMG - try a google for motors, DOL and starting current - take a look for factors of around 6 x FLC OMS ------------------------- Failure is always an option |
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20 July 2012 12:49 PM
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Why would a figure of 132amps be relevant to the starting characteristics of a Motor labelled '8KW and with a nominal current rating of 22amps'? The motor starting current can be anything upto 10 times its FLC dependant upon its type, function and starting methods. ------------------------- http://www.niceic.biz |
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20 July 2012 12:49 PM
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Fair enough if you only design small installations, but the over-riding factor determining most commercial/industrial circuits is their length. A volt drop calculation is a quick way to start the calculation, but then each to his/her own. LoL - OK So you guess a cable size to evaluate volt drop and then go from there ? - you then find that your guess is less than It x corrections I've done this just a few times now, Keith, in some reasonable sized systems - Trust me, the starting point is Ib and installation method , and then CCC - everything else flows from there because, without exception, that's the smallest the cable could ever be Regards OMS ------------------------- Failure is always an option |
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20 July 2012 12:55 PM
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KJ, or can I call you Scotty,
The length of the circuit is considered within the volt drop calculation kid. OMS, Though a figure of 6x full load current my well be reached momentarily it is irrelevant to the starting calculations, as this motor is clearly labelled and the crucial figure is "8KW" (with with a nominal current of 22amps)! |
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20 July 2012 01:02 PM
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Why would a figure of 132amps be relevant to the starting characteristics of a Motor labelled '8KW and with a nominal current rating of 22amps'? The motor starting current can be anything upto 10 times its FLC dependant upon its type, function and starting methods. I don't doubt what you have to say Scotty, but this motor is labelled with the relevant information for calculating cable size and overcurrent protection. |
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 20 July 2012 03:46 PM
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david911,
The inrush current for the motor will tend toward that of LRC at the instant of start, thus the current will be quite a bit more than that given by the 8kW & 22A FLC, as this will be the rating at nominal speed, less acceptable slip. As you are so keen on experiments, give it a try with a suitable power analyser, or even a fast acting clamp meter will do. Also, we have no idea of the mechanical inertia connected to the motor shaft, this could have a significant affect on ramp up times if the starter is of the star delta type this will affect the transition time. We don't know the number of starts per hour, which again could affect circuit considerations, especially if the motor has a high mechanical load. Oh, and david911, please, you seem to pick on every little thing that others post in here as being inaccurate, one for you, kilowatts, is kW not KW!
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20 July 2012 04:33 PM
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david911, The inrush current for the motor will tend toward that of LRC at the instant of start, thus the current will be quite a bit more than that given by the 8kW & 22A FLC, as this will be the rating at nominal speed, less acceptable slip. As you are so keen on experiments, give it a try with a suitable power analyser, or even a fast acting clamp meter will do. Also, we have no idea of the mechanical inertia connected to the motor shaft, this could have a significant affect on ramp up times if the starter is of the star delta type this will affect the transition time. We don't know the number of starts per hour, which again could affect circuit considerations, especially if the motor has a high mechanical load. Oh, and david911, please, you seem to pick on every little thing that others post in here as being inaccurate, one for you, kilowatts, is kW not KW! Paul, Read the earlier entries please. I wouldn't be in the slightest bit supprised if current peaked at 6 or even 10 times full load current, but the crucial information is on the motor (8Kw). |
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20 July 2012 04:40 PM
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Simply that the absolute peak is made irrelevant by the figure of 8Kw supplied with the motor. It's kW!, don't you pay attention? For your further edification, google for "circle diagram electric motor"; study one and how it is derived, then attempt to make a sensible contribution to this post, or just desist from making another fool of yourself! 
Regards |
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20 July 2012 04:45 PM
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Here we go again - so which bit of the post is technically meaningless then David - just because you don't undertsand it doesn't change the physics you know OMS "if you think 130 +A through a Type C MCB rated at say 20amps then you don't understand my point" - is technically meaningless, unless you would like to explain (without using your stock response of "if you don't know you must be an idiot"). OK - as you know, a 20MCB type C has a tripping characteristic of between 5 and 10 x In, so taking the lower value of instantaneous tripping and using a simple crest factor of 1.414 we could say that the breaker will trip instantaneously if it sees a current greater than: In x 1.414 x 5 = 7.07In For a 20A MCB that's about 141 A If the motor start current is in the range of 6 - 10 x FLC (lets call it 8 x) then we have 22A x 8 = 176A - way more than the instantaneous trip point of the MCB at about 141A Does that help you understand the comment ? "you know nothing about electrical engineering" & "I thought you had at least a bit of basic knowledge" - is offensive. you appear to have a very short and selective memory david "trust me that starting current will influence volt drop (and such pull out torque) along with tripping of CPD" - is irrelevant as the required information for cable size and overcurrent protection are already given, nobody has claimed that starting current will not influence volt drop or the tripping of the CPD. Simply that the absolute peak is made irrelevant by the figure of 8Kw supplied with the motor. How irrelevant ? - I've just shown you above just how relevant it actually is - particularly if that start current occurs frequently, as cables are unable to cool fully from the starting current before the next start occurs - think lift motors for example OMS ------------------------- Failure is always an option |
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20 July 2012 04:54 PM
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Simply that the absolute peak is made irrelevant by the figure of 8Kw supplied with the motor. It's kW!, don't you pay attention? For your further edification, google for "circle diagram electric motor"; study one and how it is derived, then attempt to make a sensible contribution to this post, or just desist from making another fool of yourself!
Regards Calm down son, this is a simple calculation to provide a cable size for an 8Kw motor, 22amp nominal current motor. One of your buddies recommended a 6mm SWA, I just pointed out that with the available information a 10mm SWA would be a better idea. |
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20 July 2012 05:10 PM
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Here we go again - so which bit of the post is technically meaningless then David - just because you don't undertsand it doesn't change the physics you know OMS "if you think 130 +A through a Type C MCB rated at say 20amps then you don't understand my point" - is technically meaningless, unless you would like to explain (without using your stock response of "if you don't know you must be an idiot"). OK - as you know, a 20MCB type C has a tripping characteristic of between 5 and 10 x In, so taking the lower value of instantaneous tripping and using a simple crest factor of 1.414 we could say that the breaker will trip instantaneously if it sees a current greater than: In x 1.414 x 5 = 7.07In For a 20A MCB that's about 141 A If the motor start current is in the range of 6 - 10 x FLC (lets call it 8 x) then we have 22A x 8 = 176A - way more than the instantaneous trip point of the MCB at about 141A Does that help you understand the comment ? "you know nothing about electrical engineering" & "I thought you had at least a bit of basic knowledge" - is offensive. you appear to have a very short and selective memory david "trust me that starting current will influence volt drop (and such pull out torque) along with tripping of CPD" - is irrelevant as the required information for cable size and overcurrent protection are already given, nobody has claimed that starting current will not influence volt drop or the tripping of the CPD. Simply that the absolute peak is made irrelevant by the figure of 8Kw supplied with the motor. How irrelevant ? - I've just shown you above just how relevant it actually is - particularly if that start current occurs frequently, as cables are unable to cool fully from the starting current before the next start occurs - think lift motors for example OMS The guy was asking for help sizing a cable! The phrase "if you think 130+A through a Type C MCB rated at say 20amps then you don't understand my point" is meaningless! A 20amp Type C MCB has a 5 second disconnection time of 200amps (this will therefore cope with peak starting current); but an I2 value only approaching 30amps. Therefore if the 8Kw motor is under extreme load for a long period of time, the 20amp breaker may well fail. |
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