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Need a 8000A-10,OOO amp magnetic field? help? 1
- Thread starter drax
- Start date
- Moderator
- #21
Respectfully ScottyUK I notice on another thread you have suggested that voltages in the range of 1000 Volts may be applied to the secondary of a CT. At 277 volts the OP achieved 900 amps. At 1000 he should have 3250 amps. That is almost halfway to the 7000 amps he needs. Reduced impedance or two CT's in series and he has it. (Yes series not parallel. His current is being limited by the available voltage, not the impedance.)
Respectfully
Respectfully
Hi waross,
The 1000V kneepoint is high but not out of the question for a Class X CT. They are designed specifically to have a very high kneepoint voltage - essentially lots of high grade iron in the core to give a low mag current and low core flux. Common CTs have a much lower kneepoint.
Strictly it is not voltage than we are short of, but magnetic flux. Increasing the flux is achieved by circulating more current, which is achieved by applying more voltage, up to the point where the core saturates. The only options to avoid saturation are: bigger core, or more turns, or higher frequency. The idea of series CTs is interesting - essentially it gives the effect of a bigger core. The easiest way to picture it is that for a given core at a given frequency, the volts/turn is a constant. If the volts/turn is not large enough to circulate rated current, the core will saturate.
The big distribution transformer I suggested using would give about 27V on open circuit rather than 1V or 2V - they are fairly easy to rent over here in the UK. I'm pretty sure such a transformer would withstand a 200% overload for 90 seconds - the thermal mass is huge. The big unknown is the mechanical integrity of the winding and leadouts - they are designed to withstand massive short duration faults, not smaller sustained faults. This is essentially a fault-level test we are discussing.
You have made a very good suggestion: "Let's look at what he does have". Even the big primary injection set is going to take more power than the 277V supply to his lab can handle.
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I don't suffer from insanity. I enjoy it...
The 1000V kneepoint is high but not out of the question for a Class X CT. They are designed specifically to have a very high kneepoint voltage - essentially lots of high grade iron in the core to give a low mag current and low core flux. Common CTs have a much lower kneepoint.
Strictly it is not voltage than we are short of, but magnetic flux. Increasing the flux is achieved by circulating more current, which is achieved by applying more voltage, up to the point where the core saturates. The only options to avoid saturation are: bigger core, or more turns, or higher frequency. The idea of series CTs is interesting - essentially it gives the effect of a bigger core. The easiest way to picture it is that for a given core at a given frequency, the volts/turn is a constant. If the volts/turn is not large enough to circulate rated current, the core will saturate.
The big distribution transformer I suggested using would give about 27V on open circuit rather than 1V or 2V - they are fairly easy to rent over here in the UK. I'm pretty sure such a transformer would withstand a 200% overload for 90 seconds - the thermal mass is huge. The big unknown is the mechanical integrity of the winding and leadouts - they are designed to withstand massive short duration faults, not smaller sustained faults. This is essentially a fault-level test we are discussing.
You have made a very good suggestion: "Let's look at what he does have". Even the big primary injection set is going to take more power than the 277V supply to his lab can handle.
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- Moderator
- #23
Hi ScottyUK
Thanks for the response.
I found out years ago working on series lighting circuits that when using constant current transformers all the normal circuit characteristics were turned upside down. The only thing that you could depend on was ohms law.
This situation has many similarities.
The CTs are capable of the current, low voltage is the problem. If we can increase the voltage and reduce the voltage required we have a good chance of doing it.
I have a question, ScottyUK, about saturation tests.
Am I right in assuming that the test current is in excess of 5 amps (or rated secondary current) at the knee of the saturation curve?
If so we may be able to increase the voltage significantly above 277 volts.
If the original poster will let us know what he was using for hardware when he pushed 900 amps through the bus we can start making valid suggestions. If the circuit was cable connected and if it is possible to change to bolted bus connections we are well on the way.
Another point, with one bus CT the current will travel the length of one bus bar and return on another. With 3 CTs in star the voltage required from each CT is about half. That should translate into double current right there.
I understand that the original poster has several 10,000 amp CT's available. The problem of course is the very low voltage generated when these are used in reverse.
I think we can address this issue on two fronts. First, by using the bus bars directly instead of connecting with cables we can reduce the circuit impedance.
In response to my own last post, if you have a big transformer in the yard you can abuse it, but the rental agency will frown on you renting a transformer with the intention of forcing 200% of rated current through it for however short a time.
Respectfully
Thanks for the response.
I found out years ago working on series lighting circuits that when using constant current transformers all the normal circuit characteristics were turned upside down. The only thing that you could depend on was ohms law.
This situation has many similarities.
The CTs are capable of the current, low voltage is the problem. If we can increase the voltage and reduce the voltage required we have a good chance of doing it.
I have a question, ScottyUK, about saturation tests.
Am I right in assuming that the test current is in excess of 5 amps (or rated secondary current) at the knee of the saturation curve?
If so we may be able to increase the voltage significantly above 277 volts.
If the original poster will let us know what he was using for hardware when he pushed 900 amps through the bus we can start making valid suggestions. If the circuit was cable connected and if it is possible to change to bolted bus connections we are well on the way.
Another point, with one bus CT the current will travel the length of one bus bar and return on another. With 3 CTs in star the voltage required from each CT is about half. That should translate into double current right there.
I understand that the original poster has several 10,000 amp CT's available. The problem of course is the very low voltage generated when these are used in reverse.
I think we can address this issue on two fronts. First, by using the bus bars directly instead of connecting with cables we can reduce the circuit impedance.
In response to my own last post, if you have a big transformer in the yard you can abuse it, but the rental agency will frown on you renting a transformer with the intention of forcing 200% of rated current through it for however short a time.
Respectfully
kptx,
Very easy to say, much less easy to do! Care to design the core?
Waross,
At the kneepoint the CT will hopefully only be drawing a few tens of mA, maybe a couple of hundred at the absolute outside on a horrible cheap CT. Once past the kneepoint the current starts rising at an exponential rate (well, maybe not truly exponential but bloody quickly: the mathematical analysis is gonna be complicated!). You may not be able to raise the voltage more than 10% above the kneepoint before all the rated current is used for magnetising current.
Only if you tell them!
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I don't suffer from insanity. I enjoy it...
Very easy to say, much less easy to do! Care to design the core?
Waross,
At the kneepoint the CT will hopefully only be drawing a few tens of mA, maybe a couple of hundred at the absolute outside on a horrible cheap CT. Once past the kneepoint the current starts rising at an exponential rate (well, maybe not truly exponential but bloody quickly: the mathematical analysis is gonna be complicated!). You may not be able to raise the voltage more than 10% above the kneepoint before all the rated current is used for magnetising current.
Only if you tell them!
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rcw retired EE
Electrical
Find a company in your area that tests low voltage circuit breakers using primary injection. Look for a NETA company. Typical breaker test sets put out up 50,000 amps. A new or different power supply from the utility would nto be needed, since most test sets operate on 240 V or 480 V power (or equivalent.)
The actual power requirement is relatively small, as long as the voltage drop in the loop of cable or bus is reasonable, say 5 volts or less. (50 kA @ 5 V = 250 kVA).
A test set with standard CT will never make it. The impedance of the primary circuit changes by the square of the turns ratio when viewed from the CT secondary. With just 0.5 ohms of impedance in your high current circuit and a 10,000:5 CT (2000/1 ratio) the secondary impedance will be 2 meg ohms.
A 480 VAC supply could push about 0.25 milliamps into a 2 megohm impedance and get about 1/2 amp output on the high side of the 1000/5 CT.
I would be concerned about generating a "10,000 amp" field for a test. Without a gauss meter to measure the field, how do you know what field strength actually is applied to the device under test? The field strength is very dependent on the conductor configuration around the device.
For example, you could get high current by using a coax confuguration using two aluminum pipes, one inside the other, and welded or bolted together at one end. This low impedance circuit could easily carry "10,000 Amps" for 90 seconds. But the external magnetic field would be close to nil, since current in the two pipes would cancel each other. My point is make sure the test set up gives your client what they need, not what they say.
The actual power requirement is relatively small, as long as the voltage drop in the loop of cable or bus is reasonable, say 5 volts or less. (50 kA @ 5 V = 250 kVA).
A test set with standard CT will never make it. The impedance of the primary circuit changes by the square of the turns ratio when viewed from the CT secondary. With just 0.5 ohms of impedance in your high current circuit and a 10,000:5 CT (2000/1 ratio) the secondary impedance will be 2 meg ohms.
A 480 VAC supply could push about 0.25 milliamps into a 2 megohm impedance and get about 1/2 amp output on the high side of the 1000/5 CT.
I would be concerned about generating a "10,000 amp" field for a test. Without a gauss meter to measure the field, how do you know what field strength actually is applied to the device under test? The field strength is very dependent on the conductor configuration around the device.
For example, you could get high current by using a coax confuguration using two aluminum pipes, one inside the other, and welded or bolted together at one end. This low impedance circuit could easily carry "10,000 Amps" for 90 seconds. But the external magnetic field would be close to nil, since current in the two pipes would cancel each other. My point is make sure the test set up gives your client what they need, not what they say.
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1
- #28
CarlPugh
Electrical
- Jul 16, 2002
- 142
Get a junk transformer with a good 115 volt or 230 volt primary on the inside. Cut the secondary off. (don't disassemble the transformer) Loop one turn of bus bar through the core. Using a variable transformer, slowly increase the voltage to the junk transformer primary.
If you don't have enough current add turns.
I have done this and it works great.
If you don't have enough current add turns.
I have done this and it works great.
- Moderator
- #29
Hi CarlPugh
I like it.
In the original test the popster got 900 amps at 277 volts.
That would be a primary current of 900A/10,000A x 5A = 0.45A.
At 277 volts that would be about 125 VA. We need about 8 times the current so that would be 64 times the VA.
That makes our requirement 8 KVA. To allow a safety margin for the extra bus bar, I would suggest looking for a transformer in the range of 15KVA to 25 KVA.
For testing the mechanical strength of busbar mounting you should use three transformers and three phase power.
For testing the magnetic effects around the busbars, whatever makes the customer happy. Hopefully you can get by with single phase testing.
I would expect that single phase currents will be the "worst case" for adjacent magnetic effects. You will not get any field cancelling from the other phases. That's the argument that I would use to try and avoid the extra expense of three phase testing. Likewise testing with the current out on one bar and back on an adjacent bar will probably subject those two bars to greater mechanical stress than a three phase current. The currents will be in phase rather than 120 deg. displaced.
Bear in mind that a momentary outage in the source current during the test may cause a transient inrush that can easily apply enough force to blow the busbars out of the enclosure.
Doing a 7000 amp test on bus bars with a mechanical capability of less than 30,000 or 50,000 amps would make me nervous.
If you are using more than one transformer in any configuration, be very carful to avoid open circuit primaries.
I hope you have the customer sign off on the testing method in case the test destroys the equipment.
respectfully
I like it.
In the original test the popster got 900 amps at 277 volts.
That would be a primary current of 900A/10,000A x 5A = 0.45A.
At 277 volts that would be about 125 VA. We need about 8 times the current so that would be 64 times the VA.
That makes our requirement 8 KVA. To allow a safety margin for the extra bus bar, I would suggest looking for a transformer in the range of 15KVA to 25 KVA.
For testing the mechanical strength of busbar mounting you should use three transformers and three phase power.
For testing the magnetic effects around the busbars, whatever makes the customer happy. Hopefully you can get by with single phase testing.
I would expect that single phase currents will be the "worst case" for adjacent magnetic effects. You will not get any field cancelling from the other phases. That's the argument that I would use to try and avoid the extra expense of three phase testing. Likewise testing with the current out on one bar and back on an adjacent bar will probably subject those two bars to greater mechanical stress than a three phase current. The currents will be in phase rather than 120 deg. displaced.
Bear in mind that a momentary outage in the source current during the test may cause a transient inrush that can easily apply enough force to blow the busbars out of the enclosure.
Doing a 7000 amp test on bus bars with a mechanical capability of less than 30,000 or 50,000 amps would make me nervous.
If you are using more than one transformer in any configuration, be very carful to avoid open circuit primaries.
I hope you have the customer sign off on the testing method in case the test destroys the equipment.
respectfully
waross,
What is the difference between Carl's vandalised power transformer and a large CT driven in 'reverse'? Both are just iron cores wound with copper windings. Whether they are used in a CT or VT mode is down to the way they are applied. The transformer laws apply to all transformers - the significant one here being the impedance transformation being proportional to the square of the turns ratio. rcwilson has already mentioned this.
rcwilson,
I'm curious where you can easily get hold of a 50kA breaker test set - they are certainly not common over in the UK. The Programma unit I mentioned earlier is the largest I've seen. I'd be interested to have a look at one if you know of a manufacurer & model.
----------------------------------
I don't suffer from insanity. I enjoy it...
What is the difference between Carl's vandalised power transformer and a large CT driven in 'reverse'? Both are just iron cores wound with copper windings. Whether they are used in a CT or VT mode is down to the way they are applied. The transformer laws apply to all transformers - the significant one here being the impedance transformation being proportional to the square of the turns ratio. rcwilson has already mentioned this.
rcwilson,
I'm curious where you can easily get hold of a 50kA breaker test set - they are certainly not common over in the UK. The Programma unit I mentioned earlier is the largest I've seen. I'd be interested to have a look at one if you know of a manufacurer & model.
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- Moderator
- #31
Hi ScottyUK
I agree with you 100%
Electrically, there is probably not much difference.
The main difference is possibly in availability and in the ease or difficulty of adding an extra turn or two on the high current side. Depending on what's on hand, the choice could go either way. We don't know if the OP's CTs are bar type or window type.
Do you concur with my KVA suggestions, SkottyUK?
respectfully
I agree with you 100%
Electrically, there is probably not much difference.
The main difference is possibly in availability and in the ease or difficulty of adding an extra turn or two on the high current side. Depending on what's on hand, the choice could go either way. We don't know if the OP's CTs are bar type or window type.
Do you concur with my KVA suggestions, SkottyUK?
respectfully
Hi waross,
Your arithmetic seems to make sense, but assuming the 'secondary' - the high current path - remains a constant impedance it will take 8x 277V i.e. 2.2kV primary voltage to drive the full 10kA through the secondary. The core does not have enough cross-section to allow application of such a voltage assuming the frequency is maintained constant.
The relationship between flux and voltage is proportional if we assume constant frequency. A ferrous core can only operate at a certain flux density prior to saturation, so if the flux doubles the magnetic cross section of the core must also double to avoid saturation. Note that 'flux' and 'flux density' are absolutely not interchangeable terms.
The point at which saturation occurs in the CT in question is a bit unknown - if the kneepoint is 400V then saturation is probably around 450V - 500V. It varies with different CT types. With that assumption for saturation, the core is probably about one-fifth of the required cross-section. If the larger core were available it would need an HV supply to drive the required current.
It will be interesting to try reducing the number of 'primary' - low curent - turns on the available CT so that saturation occurs just above operating voltage. This would make maximum use of the core by running it at rated flux density, and reduce the 'secondary' impedance as seen from the 'primary' side, causing more current to flow in both. I have to go to work now so I haven't time. Maybe later!
----------------------------------
I don't suffer from insanity. I enjoy it...
Your arithmetic seems to make sense, but assuming the 'secondary' - the high current path - remains a constant impedance it will take 8x 277V i.e. 2.2kV primary voltage to drive the full 10kA through the secondary. The core does not have enough cross-section to allow application of such a voltage assuming the frequency is maintained constant.
The relationship between flux and voltage is proportional if we assume constant frequency. A ferrous core can only operate at a certain flux density prior to saturation, so if the flux doubles the magnetic cross section of the core must also double to avoid saturation. Note that 'flux' and 'flux density' are absolutely not interchangeable terms.
The point at which saturation occurs in the CT in question is a bit unknown - if the kneepoint is 400V then saturation is probably around 450V - 500V. It varies with different CT types. With that assumption for saturation, the core is probably about one-fifth of the required cross-section. If the larger core were available it would need an HV supply to drive the required current.
It will be interesting to try reducing the number of 'primary' - low curent - turns on the available CT so that saturation occurs just above operating voltage. This would make maximum use of the core by running it at rated flux density, and reduce the 'secondary' impedance as seen from the 'primary' side, causing more current to flow in both. I have to go to work now so I haven't time. Maybe later!
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- Moderator
- #33
Hello ScottyUK
I think that we are pretty much in agreement. With a 400 volt rated CT, we may be able to get to twice 277V. That would push the 900 amps achieved by the OP up to 1800 amps.
I am sure that our secondary current is limited by the available secondary voltage so two CTs in series will push the current from there up to 3600 amps. We don't yet know how the OP has connected the CTs to the bus bars. WE don't know if the CTs are bar type or window type. If he has been using cables, and if it is possible to change cable connections to bus bar connections, he may be able to cut the impedance of the secondary circuit in half, at which point he will have 7200 amps. The OP was trying for 7000 amps.
I think we are safe in saying that it is theoretically possible to get 7,000 or 10,000 amps out of the high current side of a 10,000:5 CT but the extremely low voltage makes it quite dificult to arrange a circuit with low enough impedance.
Respectfully
I think that we are pretty much in agreement. With a 400 volt rated CT, we may be able to get to twice 277V. That would push the 900 amps achieved by the OP up to 1800 amps.
I am sure that our secondary current is limited by the available secondary voltage so two CTs in series will push the current from there up to 3600 amps. We don't yet know how the OP has connected the CTs to the bus bars. WE don't know if the CTs are bar type or window type. If he has been using cables, and if it is possible to change cable connections to bus bar connections, he may be able to cut the impedance of the secondary circuit in half, at which point he will have 7200 amps. The OP was trying for 7000 amps.
I think we are safe in saying that it is theoretically possible to get 7,000 or 10,000 amps out of the high current side of a 10,000:5 CT but the extremely low voltage makes it quite dificult to arrange a circuit with low enough impedance.
Respectfully
davidbeach
Electrical
waross said:
If the 5A side of the 10,000:5 CT is fed at 5A and limited to 600V - reasonable numbers - the impedance on the 10,000A side of the CT can not be allowed to exceed 1.2 micro ohms.
I'm not sure you can even come within an order of magnitude or two of 1.2 micro ohms. At 10,000A, that circuit will have a total voltage drop of 12 mV (millivolts).
It is probably worth pointing out that a CT fed from a mains power source doesn't exhibit normal CT behaviour. It is effectively just a normal power transformer with a large step-down ratio. It might prove beneficial to draw out the circuit using the classic transformer model and refer all the parameters to the high current side: source impedance, CT leakage reactance, CT winding resistance, load resistance, etc. The problems should then become apparent.
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I don't suffer from insanity. I enjoy it...
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