sPHENIX discussion of the superconducting solenoid

[Kin Yip <kinyip AT bnl.gov>]

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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