To: e.mckigney@ic.ac.uk Hi Ed: Here is my NuFact02 talk for the WG1 session. It is in ASCII and I have it on plastic. I hope you can accept this format. Best Regards, Don Summers 1) Muon Acceleration with Fast Ramping Synchrotrons D.~J.~Summers University of Mississippi--Oxford NuFact02: Neutrino Factory Workshop 2002 Session WG1, Lecture Theater 2, 4 July 2002, 10:00 a.m. Department of Physics, Blackett Laboratory Imperial College--London, Prince Consort Road London SW7 2BW, UK 1--6 July 2002 2) ACCELERATE WITH A RING FROM 2 to 20 GeV/c Take a 1.7 Telsa field and a 70% packing fraction. 1) 95% iron lamination packing 2) field loss due to gradient dipoles 3) space for one kicker 4) 11 spaces for RF cavities 1.7 Tesla x 70% = 1.19 Tesla For a ring with p = 20 GeV/c, Radius = p/.3*B = 20/.3*1.19 = 56 meters Circumference = 2*pi*R = 352 meters A small ring. 3) 15 MV/m SUPERCONDUCTING RF CAVITIES Scenario #1 1/2 GeV of 200 MHz cavities in 11 spaces Accelerate from 2 to 20 GeV in 36 orbits Highest decay losses during the first few orbits 77% of the muons survive, a bit low Scenario #2 Add beam loading. A 75cm long, 200 MHz cavity stores a kilojoule (see J. Delayen et al. (JLAB), PAC2001, page 3380, accelconf.web.cern.ch/AccelConf/p01/PAPERS/RPPH067.PDF) To get 1/2 GeV, use 44 11.25 MeV cavities 44 kilojoules of stored energy add 18 GeV to 10**13 muons (18 x 10**9) (1.6 x 10**-19 Joules/eV) (10**13) = 29 kJ so 29/44 = 66% of the energy goes to the muons The stored energy goes as E**2 So 0.33 of the Joules remain at 20 GeV/c And SQRT(0.33) = 0.57 of the gradient or 8.5 MV/m remains Accelerate from 2 to 20 GeV in 48 orbits 73% of the muons survive, a bit lower Scenario #3 Add 1/2 GeV of less costly 400 MHz SCRF. Lowers losses in early orbits, while not increasing the end ramp rate at high B**2. 11 kilojoules of stored energy in the 400 MHz SCRF Accelerate from 2 to 20 GeV in 30 orbits Gains 1/3 GeV in orbit 30. 84% of the muons survive. 4) CHOICES FOR IRON LAMINATIONS See D. Summers, http://arXiv.org/pdf/physics/0109002 and H. Sasaki, KEK-91-261 for more details. Material Composition rho B_Max Hc Thickness (uOHM (Tesla) (Oer- (microns) -cm) steds) Grain Oriented 3Si 97Fe 47 2.0 .1 50 100 175 NKK Super E-Core 6Si 94Fe 82 1.8 50 100 Metglas 2605SA1 2C 3Si 14B 81Fe 135 1.6 .03 30 Choose 100 micron thick grain oriented silicon steel because it has the highest saturation, has a steep B-H curve, and I need to stamp laminations good to 10 microns. Metglas laminations pose an R&D project. Ramping at 1/3 GeV / orbit Field increases 1,7% in last orbit, a time of 1.2 microSEC 58 microSEC for a quarter cycle or f = 4300 Hertz Calculate the eddy current losses in the magnets. Power = (Volume) (2pi f B thickness)**2 / (24 * rho) = (300*.4*.4) (2pi 4300 1.2 10**-4)**2 / (24 * 47 x 10**-8) = (48) (10.5) / (1.13 x 10**-5) = 45 Megawatts But its only "on" for 15 half cycles per second!!!!!!!!!!!!! Duty factor = (15) (30 orbits) (352 meters) / (3 x 10**8 m/s) = 5.3 x 10**-4 Power = (45 Megawatts) (5.3 x 10**-4) = 24 Kilowatts <<<---------- Hysteresis losses are much less than the 15 Hz Fermilab Booster. Eddy currents losses go as f**2, but hysteresis losses go as f**1. Carbon steel is a lot lossier than grain oriented silicon steel. Lamination shape is changed within a single magnet to minimize end losses while still alternating gradients. 5) EFFECT OF SKIN DEPTH ON IRON LAMINATIONS References: Lorrain and Corson, 3rd edition, pages 537-542. K. L. Scott (Lucent Tech), "Variation of the Inductance of Coils due to the Magnetic Shielding Effect of Eddy Currents in the Cores," Proceedings of the Institute of Radio Engineers, 18 (1930) 1750-1764. First get the skin depth. Take mu = 4000 mu0 = 4pi x 10**-7 rho = 47 x 10**-8 f = 4300 SKIN = SQRT(2 rho / [2pi f mu mu0]) = SQRT(2 47 x 10**-8 / [2pi 4300 4000 4pi x 10**-7] = 83 microns Now get the inductance kept; following Scott R = THICK/SKIN L/L0 = (SKIN/THICK)*((SINH(R) + SIN(R)) / (COSH(R) + COS(R))) = 0.936 So one is off by 7% with 100 micron laminations With 50 micron laminations, L/L0 = 0.996 or very little loss; but they cost more per kilogram and packing fraction is lower. 100 microns may still be OK. As saturation occurs, mu goes down, and the skin depth goes up. And its only at the top end that complete penetration is needed. 6) TWO IRON REQUIREMENTS A) B Field Must Follow Grain Orientation B) Pole Faces Must Be Held Rigidly. ******************* * * ** * * 2 * * * **** **** * * * * * * * * * ******* * * *1 * * 3* * * ******* * * * * * * * * * **** **** * * * 4 * * * * ** ******************* Even Lamination (gradient not shown). 4 piece Grain Oriented Silicon Steel Lamination for an H-frame magnet. Flux travels to the right. The "C" on the left gives mechanical support. ******************* ** * * * * 2 * * * **** **** * * * * * * * * * ******* * * *3 * *1 * * * ******* * * * * * * * * * **** **** * * * 4 * * ** * * ******************* Odd Lamination (gradient not shown). 4 piece Grain Oriented Silicon Steel Lamination for an H-frame magnet. Flux travels to the left. The backward "C" on the right gives mechanical support. 7) MAGNET POWER SUPPLIES -- LC CIRCUITS See D. Summers, http://arXiv.org/pdf/physics/0109002 Calculate the energy stored in an 0.06m high by 0.08m wide gap of a 25 meter long magnet. Gradients are alternated by changing pole shapes within magnets to minimize the number of ends. Twelve magnets will be required. W = [Volume] B**2 / (2 u0) = [25 x ,06 x .1] 1.7**2 / (2 4pi x 10**-7) = 173,000 Joules NI = B h / u0 = 1.7 .06 / 4pi x 10**-7 = 81,000 Ampere-turns Let N=1 L = 2 W / I**2 = 5.3 x 10**-5 Henries C = 1/L(2pi f)**2 = 2.6 x 10**-5 Farads V = SQRT(2W/C) = 115,000 Volts So the VA requirement is 115kV x 81kA, which is met by 520 ABB Semiconductor 4500V 4000A GTO Thyristors costing $850 each. For 12 magnets the parts cost is $5.3 Million. 173,000 / 520 = 340 Joules in the capacitor attached to each thyristor. C = 2 W / V**2 = (2 340) / 4500**2 = 33 uF So buy two capacitors at $25 each from H & R 4500V 16uF GE 28F2009 4.5" x 2.81" x 10" http://www.herbach.com $320K for capacitors New capacitors will cost more. 8) CALCULATE LOSSES IN THE COPPER COILS See H. Sasaki, KEK-91-261 for more details. First calculate the skin depth in copper at 4300 Hz to set a maximum wire size. SKIN = SQRT(2 rho / [2pi f mu0]) = SQRT(2 1.7 x 10**-8 / [2pi 4300 4pi x 10**-7] = 1.0 mm So see how 30 gauge, width = 0.25mm copper wire does. Supplier: http://www.mwswire.com/litzmain.htm Take one turn of 0.05m square copper The length is 50 meters. R = 50 (1.8 x 10**-8) / .05**2 = 360 uOHMS P = I**2 R (INT cos**2) = 81000**2 360 x 10**-6 .5 = 1.2 MegaWatts Divide by 2000 for the duty cycle --> P = 600 Watts Next do the eddy current loss. Take an 0.1 Tesla fringe field, which is half the reason to use iron!!!! P = [Volume] (2pi f B w)**2 / (24 rho) = [50 .05**2] (2pi 4300 0.1 .00025)**2 / (24 1.8 x 10**-8) = [0.125] (0.46) (2,300,000) = 132,000 watts/magnet So multiply by 12 for 12 magnets and divide by the duty factor of 2000. P = 800 watts. The two choices for water cooling tubes are either 316L stainless steel with a resistivity of 74 uOHM-cm or Titanium 6Al-4V with a resistivity of 171 uOHM-cm In physics/0109002, even 316L loses less than the copper. 9) ACCELERATE WITH A RING FROM 20 to 180 GeV/c Take a 1.7 Telsa field and a 70% packing fraction. 1) 98% iron lamination packing 2) field loss due to gradient dipoles 3) space for one kicker 4) 11 spaces for RF cavities 1.7 Tesla x 70% = 1.19 Tesla For a ring with p = 180 GeV/c, Radius = p/.3*B = 180/.3*1.19 = 504 meters Circumference = 2*pi*R = 3167 meters The same size as the Fermilab main injector. Again accelerate in about 30 orbits. 10) ACCELERATE WITH A RING FROM 180 to 1600 GeV/c See D. Summers, http://arXiv.org/pdf/physics/0109002 Use interleaved fixed 2 meter long 8 Tesla superconducting dipoles and 3/6 meter long iron magnets ramping form -1.7 to +1.7 Tesla. Short lengths give small sagittas. 3m 2m 6m 2m 3m |_____|_____________|______|__________________|______|____________|____| |V SC | +-1.7T Grad-|8T SC | +-1.7T Grad- |8T SC |+-1.7T Grad-|H SC| |Quad | ient Dipole |Dipole| ient Dipole |Dipole|ient Dipole |Quad| |-----|-------------|------|------------------|------|------------|----| ^ | Gradient changes here The gradient dipole magnetic field starts at -1.7 Telsa and ends at +1.7 Tesla. At the start of an acceleration cycle, the gradient dipoles oppose the bending and focusing of the superconducting quadrapoles and dipoles. At the end of an acceleration cycle, the gradient dipoles bend and focus in the unison with the superconducting quadrapoles and dipoles. Take a 3.5 Telsa field and a 70% packing fraction. 3.5T x 0.7 = 2.45 Tesla For a ring with p = 1600 GeV/c, Radius = p/.3*B = 1600/.3*2.45 = 2200 meters Circumference = 2*pi*R = 13,800 meters The same size as the the FNAL or BNL sites. Accelerate in about 60 orbits. Note that doubling the B field, halves the decay losses per orbit. Now there is some time to refill an RF cavity during the acceleration cycle. The cost of 25 MV/m, 1300 MHz TESLA cavities is appealing, if they can be made to work in this application. 11) SUMMARY Scenarios to ramp from 2 to 20 GeV/c 20 to 180 GeV/c 180 to 1600 GeV/c Low Duty Cycle, Fast Ramping Magnets Seem to Warrant Further Study. What is new? 1) Exploit the low duty cycle!!!!!!!!!! 2) Exploit Grain Oriented Silicon Steel. High Saturation, Steep B-H curve. Use combination "C" / "H" laminations to make rigid magnets. 3) Use thin laminations to minimize eddy currents. 4) Switch gradients inside dipoles to minimize ends. 5) Use thin copper wire to minimize eddy currents. Sacrifice some I**2 R losses. Don't worry too much about packing fraction. 6) Accelerate faster in the first few orbits with cheaper, higher frequency superconducting RF when the gamma boost is low. 7) Accelerate muons in opposite directions in alternate cycles to save polarity reversing switches in the 2 to 20 and 20 to 180 GeV/c rings. 8) See if low cost, 25 MV/m 1300 MHz TESLA SRF can be used in the higher energy rings. 9) Interleaved fixed superconducting and fast ramping magnets for the 180 to 1600 GeV/c ring. Twice the energy for a given site size.