Nb3Sn Dipole for LHC Upgrade
Case Study 4
Presented by: M. Brown
Group 4A:
E. Tulu, R. Betemps, M. Sugano,
A. Sublet, P. Zhang, M. Brown
14.3 m Nb-Ti
LHC Upgrade
5.5 m Nb3Sn
3m
Collim.
5.5 m Nb3Sn
t
Initial Parameters
Given Values
aperture
Bop
Bss
Operational %
κw-c
κc-i
Chosen Values
coil width, w
strand diameter
number of strands
VCu/noCu
pitch angle
cable width, wcable
cable thickness, t
insulation thickness
filling factor, κ
Cu/SC Ratio
value
60 mm
11 T
13.75 T
0.8
0.85
0.85
value
30 mm
0.85 mm
35
1.2
15 deg
14.7 mm
1.5 mm
0.15 mm
0.328
1.2
Source
given
given
calculated
given
Todesco, slide 17
Todesco, slide 17
wcable
Source
Flukiger II, slide 23
Flukiger I, slide 22
instructor
estimate
calculated
estimate
calculated
w
-
a
+
r
-
+
Coil Width Calculation
Central Coil Field graph with
following parameters:
• b = 22 T
• γc0 = 6.63(10-7) Tm/A
• α = 0.045
• s = 3.90(109) A/m2/T
• r = 0.03 m
• κ = 0.33
Bpeak
B
Field in Coil
Central Field
20
15 13.75 T
10
5
r = 30 mm
30 mm
0
0
SS field is not used in operation!!!

4b
 1  1

2  s c

SS Central Field vs. Coil Width, Nb3Sn 1.9 K
Central field @ SS (T)
Dipole Fields
Bss 
s c 
10
20
30
Coil width (mm)
40
50
Calculating SS and OP Conditions
5000
𝐵𝑜𝑝 = 11 𝑇
𝐵𝑜𝑝 = 0.8 𝐵𝑠𝑠
𝐵𝑠𝑠 = 13.75 𝑇
4500
Critical current density Jc (A/mm2)
4000
𝐵𝑝𝑒𝑎𝑘
Nb3Sn, 1.9 K
Load line Nb3Sn
Nb3Sn
Nb
3Sn BBpeak_op
peak_op
≈ 1.05
𝐵𝑝𝑒𝑎𝑘_𝑠𝑠 = 1.05 𝐵𝑠𝑠 = 14.44 𝑇
𝐵𝑝𝑒𝑎𝑘_𝑜𝑝 = 1.05 𝐵𝑜𝑝 = 11.55 𝑇
3500
3000
𝐵
2500
Jsc_ss = 2049 A/mm2
2000
Jsc_op = 1639 A/mm2
1500
1000
Bpeak_op = 11.55 T
500
Bpeak_ss = 14.44 T
0
0
5
10
15
Field (T)
20
25
Calculating SS and OP Conditions
5000
NbTi@
1.91.9
KK
NbTi
4500
Nb3Sn,
Nb
1.9KK
3Sn @1.9
Critical current density Jc (A/mm2)
4000
LoadLine
line Nb
Nb3Sn
Load
3Sn
3500
Nb3Sn
Nb
3Sn BBpeak_op
peak_op
3000
2500
Jsc_ss = 1561 A/mm2
2000
Jsc_op = 1248 A/mm2
NbTiBBpeak_op
NbTi
peak_op
1500
1000
Bpeak_op = 7.86 T
500
Bpeak_ss = 9.83 T
0
0
5
10
15
Field (T)
20
25
Calculation and Comparison
Nb3Sn Values
NbTi Values
Comparison
Bpeak_ss = 14.44 T
Bpeak_ss = 9.83 T
ΔBpeak_ss = 4.61 T
Bpeak_op = 11.55 T
Bpeak_op = 7.86 T
ΔBpeak_op = 3.69 T
Jsc_ss = 2049 A/mm2
Jsc_ss = 1561 A/mm2
ΔJsc_ss = 488 A/mm2
Jsc_op = 1639 A/mm2
Jsc_op = 1248 A/mm2
ΔJsc_op = 391 A/mm2
Jo_ss = Jsc_ss (κ) = 673 A/mm2
Jo_ss = Jsc_ss (κ) = 512 A/mm2
ΔJo_ss = 161 A/mm2
Jo_op = Jsc_op (κ) = 538 A/mm2
Jo_op = Jsc_op (κ) = 409 A/mm2
ΔJo_op = 129 A/mm2
Iss = Jsc_ss (Asc) = 18,467 A
Iss = Jsc_ss (Asc) = 14,064 A
ΔIss = 4403 A/mm2
Iop = Jsc_op (Asc) = 14,773 A
Iop = Jsc_op (Asc) = 11,244 A
ΔIop = 3529 A/mm2
Nb3Sn out-performs NbTi by generating higher fields!!!
Comparison: Quench Margins
Nb3Sn Margins
Load line crosses
Bpeak_op at 6.9 K
ΔT = -5 K
ΔB1.9 -> 6.9 = 2.89 T
ΔJ1.9 -> 6.9 = 410 A/mm2
NbTi Margins
Load line crosses
Bpeak_op at 4.1 K
ΔT = -2.2 K
ΔB1.9 -> 4.1 = 1.97 T
ΔJ1.9 -> 4.1 = 313 A/mm2
Critical current density Jc (A/mm2)
Magnet quenches
when Bpeak_op > Bpeak_ss
Critical Surface Curve
5000
4500
4000
3500
3000
2500
2000
1500
1000
500
0
NbTi 1.9 K
Nb3Sn, 1.9 K
Nb3Sn 6.9 K
Load line Nb3Sn
Nb3Sn Bpeak_op
NbTi 4.1 K
NbTi Bpeak_op
ΔJ
ΔJ
ΔB
0
5
ΔB
10
15
20
25
Field (T)
Important: More margin before quench with Nb3Sn!!!
Possible Coil Layout
50.0
45.0
B3 
B5 
 0 jR
2
ref

 0 jR
4
ref

sin 3a 3  sin 3a 2  sin 3a 1
3
sin 5a 3  sin 5a 2  sin 5a 1
5
1 
1



r
r

w


 1
1
 3 
(r  w) 3
r
40.0
35.0
30.0
25.0



20.0
15.0
10.0
5.0
0.0
sin( 3a 3 )  sin( 3a 2 )  sin( 3a1 )  0

sin( 5a 3 )  sin( 5a 2 )  sin( 5a1 )  0
There are a family of solutions… We choose:
α1 = 48˚
α2 = 60˚
α3 = 72˚
72˚
60˚
48˚
Equations given in presentation by Dr. Todesco
a3 a
2
a1
Force Calculation – Sector Coil
Fx  

3  2  3 3
3 a2 3 4  3 3 
2
a

ln
a

a

a
a

2
1
1
2 1
2  36
12 a1
36
6

2 0 J 02
Fy  

20 J 02

  _ mid  plane_ av
3  1 3 1 a1 3 1 3 
 a2  ln a1  a1 
2 12
4 a2
12 
 1
2 0 J 02 3  5 3 1  a1 2  3 1
2



 a  ln  a  a a 
 4  36 2 6  a2 3  1 4 2 1  a2  a1
Values
μo = 1.26(10-6)
Jo = 5.38(106) A/m2
α1 = 0.03 m
α2 = 0.06 m
r = 0.045
Equations given in presentation by Dr. Toral
α2
r
α1
Fx = 2.49(106) N/m
Fy = -2.22(106) N/m
σθ = -20 MPa
Iron Yoke/Collar/Shrinking Cylinder
Coil width: 30 mm (given previously)
Collar : generally 2 - 3 cm.
Function is to apply compression stress to fix the coil.
Iron Yoke :
𝑟𝐵𝑜𝑝
𝑡𝑦𝑜𝑘𝑒 ∼ 𝐵
𝑠𝑎𝑡𝐹𝑒
=
30 𝑚𝑚 11 𝑇
2𝑇
= 165 𝑚𝑚
LHC uses a 50% margin for yoke width:
165 mm * 1.5 = 250 mm
Shrinking Cylinder:
Iss @ 90% = 605.2 A/mm2
Fx = 3.16 MN/m
σ ≈ 200 MPa
δ = Fx/σ = 15.78 mm ≈ 16 mm
Coil Collar Iron Shrinking
Yoke Cylinder
Width
 High temperature superconductor: YBCO VS Bi2212
Bi2212
YBCO
Anisotropy of Jc
Small
Large
Mechanical strength
Low
Axially high,
Transversely low
Rutherford cable
Possible
Impossible
Magnetization current
Large
Too large
Quench protection, cost, length are problems for both conductors
 Superconducting coil design: block vs cos
cos: Radial stress  supported by external structure
Azimuthal stress  compression
Stress at the pole  to be kept compression by pre-stress
block: Coil support technique is critical issue
 Support structure: collar-based vs shell-based
collar-based: Support for Lorentz force and high accuracy coil positioning
shell-based: Large compressive pre-stress can be applied by thermal contraction of shell
 Assembly procedure: high-stress vs low stress
For high field magnet, high pre-stress is necessary to keep stress at pole compressive
Too much pre-compression can cause degradation at midplane