sPHENIX discussion of the superconducting solenoid

["Yip, Kin" <kinyip AT bnl.gov>]

Hi Carl,

Thank you very much for your circuit calculation !    

> 1 Hr and 45 Min.

I like this because it seems to be the same if I use Martin Berndt's 
methodology about the accelerated decay.  I attach his pdf file Babar P.S. 
Section1.pdf 
(BaBar P.S. Manual: Section 1) along with your resultant jpg file to the 
mailing list.

I generalize his equation on p.15 to  : i(t) = (I0 + V/R) exp ( -t / (L/R) ) 
- V/R

where I0 ~4600 A, V is the diode's forward voltage, R the cable resistance 
and L the Magnet's Inductor.

Setting i(t) to whatever target current (i_target, eg. 0), you can get the 
total time: .   t = - L/R * ln[ ( i_target + V/R)/(I0 + V/R) ]

I wrote a html/javascript page for it :   
https://www.c-ad.bnl.gov/kinyip/sPHENIX_Ramp_Down.html

So, if I set the R = 0.25 mOhm + 0.063*2 = 0.376 mOhm, Diode voltage ~1 V ( 
you et. al. told me ),

I get the time to reach 0 A is   6302.68 seconds or 1.7507 hours or 1 hour 
and 45 minutes !   This is barely bigger than the so-called
time constant L/R ~ 6276.6 seconds.


I'm not sure whether the current leads (etc.) really has  0.063 mOhm * 2.  If 
I ignore this and just use 0.25 mOhm, the total decay time
is ~2.0 hours (shorter than the L/R, time constant of 9440 seconds or 2.62 
hours).

Kin

-----Original Message-----
From: carls AT bnl.gov [mailto:carls AT bnl.gov] 
Sent: Wednesday, June 28, 2017 8:44 AM
To: Yip, Kin <kinyip AT bnl.gov>
Subject: Re: [Sphenix-magnet-l] how long ?

Hi Kin,

   With a total of 250 uOhms in the cables the decay time is about 1 Hr and 
45 Min. The lower resistance added about a half-hour to what I told you. At 
that time we didn't have an estimate of the cable length for the high-current 
test.
   The circuit is non-linear, if you use linear circuit techniques, aka. L/R 
you will get the wrong answer.
   Attached are pictures of the circuit I used and the result.

Carl


On 27 Jun 2017 at 17:12, Kin Yip wrote:

> 
> Hi,
> 
> Carl told me that the two current leads may have 63 micro-Ohm (each).  
> He read current/voltage somewhere and did the division to find the 
> resistance.  But he's not very sure about these values as the data 
> might be noisy and they seem high because 63 micro-Ohm at
> 4600 A will give ~1333 W.   1333 W next to the superconductor ??
> 
> Just for fun,
> if I include these 2*63 micro-Ohm, total resistance ~ 0.376 mOhm 
> (instead of 0.25 mOhm) => time-constant ~ 6276.6 seconds (1.74 hours)
> 
> => using Martin Berndt's model, it'll be only 3038.2 seconds (~0.84
> hours) to reach 1300 A .
> 
> Carl is trying to fix his circuit diagram in his software to find his 
> nos.
> 
> Kin
> 
> On 06/27/2017 04:00 PM, Kin Yip wrote:
>     Hi,
> 
>     I know you're probably busy with your test today ...
> 
>     Have you been able to solve electrical diagram model problem (when
>     Dave and I were in your office) ??
> 
>     Please let us know how long you've got for the current to decay.
> 
>     Our cable resistance (in the high field test) of 0.25 mOhm (not
>     BaBar's 1.25 mOhm) and L= 2.36 H (that in our Magnet simulation)
>     --- slightly smaller than BaBar's 2.58 H,
> 
>     the time-constant (tau) would be 9440 second (2.622... hours).
> 
>     If I use Martin Berndt methodology in 
>     "\\bnl.gov\c-adnas\babar\BaBar P.S. Manual\Babar P.S.
>     Section1.pdf" (p.15), and
> 
>     and assuming -1 V on the diode's forward voltage drop (instead of
>     -0.7 V),
> 
>     => i(t) = 8600*exp(-t/tau) - 4000 (Amperes)
> 
>     it'd take 4569.5 seconds or 1.27 hours to get to   ~1300 A  
>     (where we may do fast discharge).
> 
>     Kin
> 
> 
> 
> 
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>

Attachment: SlowDecay_250uOhmsCable.jpg
Description: SlowDecay_250uOhmsCable.jpg

Attachment: SlowDecay_250uOhmsCable_Circuit.jpg
Description: SlowDecay_250uOhmsCable_Circuit.jpg

Attachment: Babar P.S. Section1.pdf
Description: Babar P.S. Section1.pdf