Lecture 2: Quenching. Plan. the most likely cause of death for a superconducting magnet. the quench process. decay times and temperature rise

Transcription

1 Lecture : Quenching Plan the quench prcess decay times and temperature rise the mst likely cause f death fr a supercnducting magnet prpagatin f the resistive zne resistance grwth and decay times quench prtectin schemes case study: LHC prtectin Martin Wilsn Lecture slide Supercnducting Magnets fr Acceleratrs JUAS Feb 005

2 Magnetic stred energy Magnetic energy density B E = at 5T E = 0 7 Jule.m -3 at 0T E = 4x0 7 Jule.m -3 µ LHC diple magnet (twin apertures) E = ½LI L = 0.H I =.5kA E = 7.8 x 0 6 Jules the magnet weighs 6 tnnes s the magnetic stred energy is equivalent t the kinetic energy f:- 6 tnnes travelling at 88km/hr Martin Wilsn Lecture slide Supercnducting Magnets fr Acceleratrs JUAS Feb 005

3 The quench prcess resistive regin starts smewhere in the winding at a pint - this is the prblem! it grws by thermal cnductin stred energy ½LI f the magnet is dissipated as heat greatest integrated heat dissipatin is at pint where the quench starts internal vltages much greater than terminal vltage ( = V cs current supply) maximum temperature may be calculated frm the current decay time via the U(θ) functin (adiabatic apprximatin) Martin Wilsn Lecture slide 3 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

4 The temperature rise functin U(θ) r the 'fuse blwing' calculatin (adiabatic apprximatin).6e+7 J ( T ) ρ( θ ) dt = γ C( θ ) dθ J(T) = verall current density, T = time, ρ(θ) = verall resistivity, γ = density, θ = temperature, C(θ) = specific heat, T Q = quench decay time. U( ) (A sm -4 ).E+7 8.0E+6 J ( T ) dt = J T Q θm θ = U ( θ γ C( θ ) dθ ρ( θ ) = U ( θ m ) m ) GSI 00 diple winding is 50% cpper, % NbTi, 6% Kaptn and 3% stainless steel 4.0E+6 0.0E+00 GSI 00 diple winding pure cpper temp (K) NB always use verall current density Martin Wilsn Lecture slide 4 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

5 Measured current decay after a quench (with calculated vltages) current (A) current V lwer cil IR = L di/dt V tp cil cil vltage (V) time (s) -80 Diple GSI00 measured at Brkhaven Natinal Labratry Martin Wilsn Lecture slide 5 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

6 Calculating the temperature rise frm the current decay curve J dt (measured) U(θ) (calculated) 6.E+6 6.E+6 integral (J dt) 4.E+6 4.E+6 U(θ) (A sm -4).E+6.E+6 0.E time (s) temp (K) 0.E+00 Martin Wilsn Lecture slide 6 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

7 Calculated temperature temperature (K) time (s) calculate the U(θ) functin frm knwn materials prperties measure the current decay prfile calculate the maximum temperature rise at the pint where quench starts we nw knw if the temperature rise is acceptable - but nly after it has happened! need t calculate current decay curve befre quenching Martin Wilsn Lecture slide 7 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

8 Sensitivity f temperature rise 8.E+6 U( ) (A sm -4 ) 6.E+6 4.E+6 increase decay time by / 3 duble the temperature rise.e+6 0.E temp (K) Martin Wilsn Lecture slide 8 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

9 * Martin Wilsn Lecture slide 9 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

10 Quench prpagatin velcity resistive zne starts at a pint and spreads utwards the frce driving it frward is the heat generatin in the nrmal zne, tgether with heat cnductin alng the wire write the heat cnductin equatins with resistive pwer generatin G per unit vlume in left hand regin and G = 0 in right hand regin. θ θ k A γ C A hp( θ θ0 ) + GA = 0 x x t where P = cled perimeter, A = area f crss sectin, k = thermal cnductivity, h = heat transfer cefficient, C = specific heat, γ = density assume x s mves t the right at velcity v and take a new crdinate ε = x-x s = x-vt d θ + dε vγ C k dθ dε h P k A ( θ θ 0 ) + G k = 0 Martin Wilsn Lecture slide 0 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

11 d θ vγ C + dε k Quench prpagatin velcity dθ hp ( θ θ0) + dε k A = 0 we apprximate t a sharp transitin between supercnductivity and resistance at θ s, althugh in cpper cmpsites it is a prgessive current transfer G k when h = 0, the slutin fr θ which gives a cntinuus jin between left and right sides gives the adiabatic prpagatin velcity v ad J ρk = γ C θ s θ J Lθ s = 0 γ C θ θ0 s the resistive zne als prpagates sideways thrugh the inter-turn insulatin; calculatin is similar, but with a lwer thermal cnductivity and (pssibly) a higher specific heat - thus the velcity rati α is: recap Wiedemann Franz Law ρ(θ).k(θ) = L θ because C varies s strngly with temperature, it is better c C (, ) t calculate an average C frm the enthalpy change av θ θc = ( θ θ ) α = v v trans lng = H ( θ ) H ( θ ) γ C γ C lng c trans k k trans lng Martin Wilsn Lecture slide Supercnducting Magnets fr Acceleratrs JUAS Feb 005

12 Resistance grwth and current decay Apprximate thery based n sme simplifying assumptins: a) current remains cnstant at it starting value until all the inductive stred energy energy f the magnet has been dissipated, then it falls t zer b) temperature rises given by the parablic U(θ) apprximatin, define U at θ c) resistivity increases linearly with temperature after time T the resistive zne has grwn t an ellipse f semi axis x = vt and ellipticity α resistance f the zne substitute R = X 4π x α ρ( θ ) dx A J dt = J T d = U ( θ ) U ( θ / θ ) α.v x X / v θ U J ρ ( θ ) = ρ ρ = ρ θ = U U 4 τ where τ is the elapsed time since nrmality: at the centre τ = T, at the edge t = 0 and in between τ = T - x / v Martin Wilsn Lecture slide Supercnducting Magnets fr Acceleratrs JUAS Feb 005

13 R Resistance grwth and current decay = ( T x ) 4 4 π x α ρ J 4 v 4π ρ J v dx = α A U 30 A U vt 0 3 T 5 recap τ is the elapsed time since nrmality: where: v = lngitudinal velcity, α = rati lngitudinal/transverse velcity, ρ = resistivity at θ, U = U functin at θ, J = current density at start characteristic time T Q f current decay is btained by setting energy dissipated in the nrmal regin equal t the initial stred energy E (assuming I = I is cnstant thrughut) T Q I R( T ) dt = 0 E T Q π ρα J 0 v T J A dt = 30A U E thus T Q = 45U E 3 J 0 π ρ α v 6 characteristic decay time f the quench Health Warning: this rugh mdel ignres what happens when nrmal zne reaches cil bundaries (extends τ) fr a gd answer use the cmputer! Martin Wilsn Lecture slide 3 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

14 Resistance grwth and current decay 3 maximum temperature (at centre f nrmal zne) frm J TQ = U ( θ ) with the parablic apprximatin θ = m J 4 0 T U Q θ substitute back t get the current decay curve I( t) = I e T 6 T 6 Q = I e t 6 I / I 0.5 where t = T/T Q perhaps the riginal apprximatin (a) wasn't s bad after all! T / TQ Health Warning: this rugh mdel ignres what happens when nrmal zne reaches cil bundaries (extends τ) fr a gd answer use the cmputer! Martin Wilsn Lecture slide 4 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

15 Cmputatin f resistance grwth and current decay start resistive zne in time δt zne grws v.dt lngitudinally and α.v.dt transversely * vδt αvdt temperature f zne grws by δθ = J ρ(θ )δτ / γ C(θ ) resistivity f zne is ρ(θ ) calculate resistance and hence current decay δi = R / L.δt in time δt add zne n: v.δt lngitudinal and α.v.δt transverse vδt αvdt temperature f each zne grws by δθ = J.dt / γc(θ ) δθ = J.dt / γc(θ ) δθ n = J.dt / γc(θ n ) resistivity f each zne is ρ(θ ) ρ(θ n ) ρ(θ n ) calculate resistance R = r. + r +. r n.. and hence current decay δi = (I R /L).δt when I 0 stp Martin Wilsn Lecture slide 5 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

16 ** * Martin Wilsn Lecture slide 6 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

17 * Martin Wilsn Lecture slide 7 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

18 Cmputer simulatin f quench (diple GSI00) ple blck nd blck mid blck current (A) measured ple blck nd blck mid blck time (s) Martin Wilsn Lecture slide 8 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

19 Cmputer simulatin f quench temperature rise ple blck nd blck mid blck temperature (K) frm measured ple blck nd blck mid blck time (s) Martin Wilsn Lecture slide 9 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

20 Methds f quench prtectin: ) external dump resistr detect the quench electrnically pen an external circuit breaker frce the current t decay with a time cnstant I t I e τ = where τ = L R p calculate θ max frm J τ = U ( θ m ) Nte: circuit breaker must be able t pen at full current against a vltage V = I.R p (expensive) Martin Wilsn Lecture slide 0 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

21 Methds f quench prtectin: ) quench back heater detect the quench electrnically pwer a heater in gd thermal cntact with the winding this quenches ther regins f the magnet, effectively frcing the nrmal zne t grw mre rapidly higher resistance shrter decay time lwer temperature rise at the ht spt Nte: usually pulse the heater by a capacitr, the high vltages invlved raise a cnflict between:- - gd themal cntact - gd electrical insulatin methd mst cmmnly used in acceleratr magnets Martin Wilsn Lecture slide Supercnducting Magnets fr Acceleratrs JUAS Feb 005

22 Methds f quench prtectin: 3) quench detectin recap current decay T 6 T I = I e Q = I 6 e t 6 internal vltage V = L di dt = 5 6 T 3LI 6 T 3 t T LI 5 T Q T Q e Q = T Q t e 6 V I nt much happens in the early stages -small di /dt small V but imprtant t act sn if we are t reduce T Q significantly I/I & VTQ/3LI 0.5 s must detect small vltage supercnducting magnets have large inductance large vltages during charging detectr must reject V = LdI/ dt and pick up V = IR 0 0 t detectr must als withstand high vltage - as must the insulatin Martin Wilsn Lecture slide Supercnducting Magnets fr Acceleratrs JUAS Feb 005

23 Methds f quench prtectin: 3) quench detectin i) Mutual inductance D ii) Balanced ptentimeter adjust fr balance when nt quenched unbalance f resistive zne seen as vltage acrss detectr D if yu wrry abut symmetrical quenches cnnect a secnd detectr at a different pint detectr subtracts vltages t give V = di L dt + IR Q M di dt D adjust detectr t effectively make L = M M can be a trid linking the current supply bus, but must be linear - n irn! Martin Wilsn Lecture slide 3 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

24 Methds f quench prtectin: 4) Subdivisin I resistr chain acrss magnet - cld in crystat current frm rest f magnet can by-pass the resistive sectin effective inductance f the quenched sectin is reduced reduced decay time reduced temperature rise current in rest f magnet increased by mutual inductance effects quench initiatin in ther regins ften use cld dides t avid shunting magnet when charging it dides nly cnduct (frwards) when vltage rises t quench levels cnnect dides 'back t back' s they can cnduct (abve threshld) in either directin Martin Wilsn Lecture slide 4 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

25 Case study: LHC diple prtectin It's difficult! - the main challenges are: ) Series cnnectin f many magnets In each ctant, 54 diples are cnnected in series. If ne magnet quenches, the cmbined inductance f the thers will try t maintain the current. Result is that the stred energy f all 54 magnets will be fed int the magnet which has quenched vaprizatin f that magnet!. Slutin : put cld dides acrss the terminals f each magnet. In nrmal peratin, the dides d nt cnduct - s that the magnets all track accurately. At quench, the dides f the quenched magnet cnduct s that the ctant current by-passes that magnet. Slutin : pen a circuit breaker nt a dump resistr (several tnnes) s that the current in the ctant is reduced t zer in ~ 00 secs. ) High current density, high stred energy and lng length As a result f these factrs, the individual magnets are nt self prtecting. If they were t quench alne r with the by-pass dide, they wuld still burn ut. Slutin 3: Quench heaters n tp and bttm halves f every magnet. Martin Wilsn Lecture slide 5 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

26 LHC quench-back heaters stainless steel fil 5mm x 5 µm glued t uter surface f winding insulated by Kaptn pulsed by capacitr x 3.3 mf at 400 V = 500 J quench delay - at rated current = 30msec - at 60% f rated current = 50msec cpper plated 'stripes' t reduce resistance Martin Wilsn Lecture slide 6 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

27 LHC pwer supply circuit fr ne ctant circuit breaker dides allw the ctant current t by-pass the magnet which has quenched circuit breaker reduces t ctant current t zer with a time cnstant f 00 sec initial vltage acrss breaker = 000V stred energy f the ctant =.33GJ Martin Wilsn Lecture slide 7 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

28 Dides t by-pass the main ring current Installing the cld dide package n the end f an LHC diple Martin Wilsn Lecture slide 8 Supercnducting Magnets fr Acceleratrs JUAS Feb 005

29 Quenching: cncluding remarks magnets stre large amunts f energy - during a quench this energy gets dumped in the winding intense heating (J ~ fuse blwing) pssible death f magnet temperature rise and internal vltage can be calculated frm the current decay time simple analytic mdel gives rugh decay time but ignres bundaries cmputer mdelling f the quench prcess gives a gd estimate f decay time but must decide where the quench starts if temperature rise is t much, must use a prtectin scheme active quench prtectin schemes use quench heaters r an external circuit breaker - need a quench detectin circuit which must reject LdI/dt and be 00% reliable passive quench prtectin schemes are less effective because V grws s slwly - but are 00% reliable prtectin f acceleratr magnets is made mre difficult by series cnnectin - all the ther magnets feed their energy int the ne that quenches fr acceleratr magnets use by-pass dides and quench heaters remember the quench when designing the electrical insulatin always d the quench calculatins befre testing the magnet Martin Wilsn Lecture slide 9 Supercnducting Magnets fr Acceleratrs JUAS Feb 005