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Line impedance query

me1899
Hi 


I am looking at a job where a supply to a replacement CT scanner has to be calculated.


The current scanner is fed by a cable with line impedance 0.06 at source end of the sub panel that feeds it.


The new scanner must have a line impedance no more than 0.06 so basically I can't use the way on the current panel.


The client engineer has said to run from the intake and provide a cable calc to prove the line impedance is OK for the new scanner feed (this is D and B).


I am using AMTECH and was wondering if there was a way to output line impedance or if I need to show a calculation another way?


The client engaged a firm to do line impedance tests, and their report shows values measured between L1-L2, L2-L3 and L1-L3, but looks like they did testing at the sub panel and not at the intake.


This is basically like PSCC right? So I can get PSCC from AMTECH, divide by root 3 and work out impedance from there?


Am I on the right track? Assistance would be much appreciated.


Regards


Anisur
  • 0
    As Dave Z and OMS have pointed out peak current short duration are going to be present.


    The equipment manufacturer should be able to give you a data sheet with maximum demand information, power factor, harmonics and minimum voltage drop. Plus any other relevant information. 


    I would want to know what was on the main LV panel the supply is being taken from to ensure the supply with existing loads can take the big inrush currents without impacting on the volt drop on the supply and the existing loads.
  • 0
    Thanks to all for the responses, looking to speak to the scanner man man tomorrow and pump him for info as well as try and dig about for more info from our site boys in relation to the existing.


    I will provide updates as to how things go.
  • Former Community Member
    0 Former Community Member
    Chris Pearson:

    I don't doubt for one minute the expertise of Mike, Dave, and OMS, but this all seems to be OTT. Just how are scanners put into a ship or a field hospital?

     


    Reasonably easily - the source impedance requirements aren't that difficult to achieve. Back in the day, even the sockets used for mobile x-ray had impedance limits - most of this stuff has been in HTM documentation for many decades.


    As the OP indicated, the CT will still work, just that you get ever increasing image degradation  - which is then exacerbated if you need to share those images over networks resulting in further degradation


    To get this into context, in a modern acute hospital of say 500 beds, it wouldn't be unreasonable to have a firm 4MVA supply capability (so essentially 8MVA depending on how the source transformers are configured for A+B primary and secondary supplies into the Healthcare estate)


    Most hospital electrical loads are disturbing to some extent, so sensible design is needed in terms of what you try and do. It's never that sensible to have your main diagnostic capability as far away from your supply as possible - usually you would design for the opposite. There are a few idiotic cases such as where a know nothing project manager decided that installing a new MRI scanner in a room directly above a set of transformers was perfectly reasonable - the magnetic field of the MRI sweeping through the transformers resulted in some interesting disturbance issues - ohh how we laughed.


    The above aside - 0.06 Ohm limit is only roughly 4% voltage drop in a cable - so it's hardly unreasonable assuming you have decent transformers in place, and noting this is a swap out rather than an additional scanner.


    Regards


    OMS


  • 0
    OMS


    Is this line impedance the impedance of one line conductor or 2 line conductors?


    I know the top man in the country for Medical Locations. I will see if I can get him to comment on this post.
  • 0
    LV Panel - Supply

    L1 - N = 6.19kA (For 3-phase condition √3 x 6.19kA = 10.72kA)

    Off on a tangent I know, but wouldn't the PSCC for a 3-phase (L-L-L) fault be more like 2x the L1-N value? (e.g. 12.38kA). I think √3 is only going to give you the L-L (two phase) fault value. (All presuming L and N impedances are similar).


       - Andy.
  • Former Community Member
    0 Former Community Member
    John Peckham:

    OMS


    Is this line impedance the impedance of one line conductor or 2 line conductors?


    I know the top man in the country for Medical Locations. I will see if I can get him to comment on this post. 


    Well, typically HTM 06-01 indicates that it is measured between one or more lines as required. Obviously, depending on the source of the equipment, manufacturers may state specific requirements (eg, some GE kit may be assuming a delta supply source)


    I can't specifically state that, John - read the instructions would be good advice.


    Regards


    OMS


     


  • 0
    I agree OMS he usual advice applies "if all else fails read the instructions"!


    Andy I think you are using the rule of thumb calculation from Appendix 14 of the good book. They will always come out on the very high side for safety.


    For a bolted 3 phase fault the maximum PFC would need to be measured with the installation cold. The PFC will be the open circuit phase voltage, so for a transformer taped to 433V divice by route 3 = 250V divided by the impedance of one winding added to the impedance of one line conductor. The calculated figure assumes the HV supply can supply unlimited current (which it cannot) and ignores any current limiting effect of any upstream circuit protection.


    To get a closer actual measurement in the field measure the fault current phase to phase with a decent loop tester and divide the reading by 0.87. 


    In this case I thing the requirement for very low supply impedance values is for supply voltage stability. I am thinking the voltage will be relatively stable for normal running current then short high current demands when the images are being taken will drop the voltage, depending on the supply impedance, leading to blurred images.


    If the running loads and peak load are known and the manufacturer gives voltage data then Amtech can do the rest as it will give supply impedances for the chosen cable and will calculate volt drop.
  • Former Community Member
    0 Former Community Member
    It's actually more about the power than the voltage per se. Typical rotating anodes in medical x ray sets draw in a huge amount of power very quickly and in a CT scan that is often very repetitive of the scan.


    To get an x ray you need an arc in a vacuum (so be careful about standing about in front of operational HV vacuum circuit breakers) - if the power delivery  and the voltage are unstable, you don't get the x ray you expect to get (ie poor image quality is down to an inability to supply the power and an inability to maintain the voltage (typically around 75kV).


    Just follow the instructions of the set manufacturer - he knows more about how his set operates in terms of stored energy and what parameters it's been  designed to. The market is hugely competitive in terms of the number of slices so it is very much a war in terms of the image quality output between manufacturers of medical x ray sets.


    No one would be measuring this with a cheap electrical meter - it's usually by calculation and verification at commissioning. I've seen a number of commissioning visits aborted because of supply line impedance issue. They can be solved by adding capacitance but that's fearsomely expensive to do when not required.


    regards


    OMS
  • 0
    The field solution involves one scanner in a container that arrives on a lorry and then with a big clockwork key it all telescopes out to make the  treatment room,
    This public domain link gives the flavour.  The video suggests it deploys in 3mins 11 seconds, actually it is more like a very  busy morning, but is still impressive (we borrowed one of these trick containers once for something else).


    For power there will be an adjacent container holding a diesel engine that would happily power a double decker bus, probably with turbo charger and intercooler. and an induction genset (to keep the overall shaft length down). The solution to the impulsive load, as it is for quite a few other pieces of  kit with nasty current demand profiles in green painted boxes,  is to request the genset makers to fit the extra large flywheel option and over sized crankshaft bearings on the prime mover relative to the same engine for road use.
  • 0
    To get a closer actual measurement in the field measure the fault current phase to phase with a decent loop tester and divide the reading by 0.87.

    Agreed. Noting L-L not L-N.


    L-L PFC would typically be √3 times the L-N result - as usually there would be similar impedances (presuming N is similar to L) but driven by √3 times the voltage (433 rather than 250, or 400 rather than 230).


    0.87 I think comes from √3 / 2 - so dividing the L-L PFC result by 0.87 is the same as multiplying it by 2/√3. If the L-L result is √3 times the L-N result then you get L-N result * √3 * 2/√3 - the √3s cancel out so you end up with just L-N result *2.


    I think the rule of thumb doesn't actually give much of a margin for error at all. (Indeed might under-estimate if the N has a higher impedance than L - say old cables with a reduced N).


       - Andy.

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