Beam Operation Time [days]
0 10 20 30 40 50 60 70 80 90
Peak Temperature [K]
4.5 5 5.5 6
6.5 1 mm
2 mm 3 mm
Beam Operation Time [days]
0 10 20 30 40 50 60 70 80 90
Peak Temperature [K]
4.5 5 5.5 6 6.5
7 15 sheets 20 sheets
24 sheets 25 sheets
Figure 3.26: Effect of the thickness of innermost layer aluminum strip (left) and the aluminum strip filled along the azimuth.
the aluminum strips must be filled between the coil layers along the azimuth as much as possible to reduce the peak temperature.
Summary As for a novel method, the effect of aluminum strips is studied to search for the best way to align the thermal path for removing the nuclear heating from the coils, and the optimized alignment is proposed as follows:
1). Inserting the aluminum strips between each layer of coils, in which the last layer may be eliminated since the impact is relatively inefficiently.
2). Extracting the aluminum strips from the both ends and connecting to the cooling pipe.
3). Increasing the thickness of aluminum strip at inner layer to 3 mm.
4). Increasing the number of aluminum strips along the azimuth.
Chapter 3. Thermal Stability during the Beam Operation
conductors with a short duration owing to the required time interval between the bunch and bunch is only 1.17µsec, consequently, a local quench could be triggered by this fast energy deposition before the heat is removed. In order to ensure the thermal stability in a short duration, a simple calculation is preformed with the following procedures as follows:
- Regarding the aluminum-stabilized conductor or Rutherford cable as adiabatic, the characteristic time of the thermal diffusion is calculated.
- During a time duration of thermal diffusion, total bunches hit the production target are counted from the time structure of primary beam.
- The time structure of the energy deposition in the coil is estimated by the PHITS simulation.
- With the estimated total bunches and energy deposition in a time duration, the maximum temperature is found from the material property of specific heat.
Thermal Diffusion
Figure 3.27 gives the thermal diffusivity of aluminum, copper, NbTi and polyimide (Kapton) tape from 4.5 K to room temperature. The thermal diffusivityα is calculated
from the material properties as
α= k
%C (3.44)
where k is the thermal conductivity, % is the mass density, and C is the specific heat, respectively.
Temperature [K]
10 102
/sec]2 Thermal Diffusivity [m
−7
10
−6
10
−5
10
−4
10
−3
10
−2
10
−1
10 Copper
Aluminum NbTi Kapton
Figure 3.27: The thermal diffusivity for aluminum, copper, NbTi and Kapton tape at 5.4 Tesla.
RRR of aluminum and copper are 400 and 50, respectively.
From the view of heat transfer, the copper and aluminum has relatively higher thermal diffusivity at 4.5 K with respect to NbTi, which indicates that the heat is difficult to be removed in a short duration for NbTi. By solving the heat transfer equation at unsteady-state condition for an one-dimensional slab with a thickness of 2d, the temperature is obtained as a function of exponential time term from the exact solution [49]. The characteristic time of thermal diffusion τt can be written as
τt = 1 α(2d
π )2 (3.45)
where d is the half of thickness. In the PCS, the diameter of the strand contained with NbTi filaments and copper matrix is 1.15 mm. Assumed each filament has the diameter of 100 µm, τt is calculated to be 25 µsec at 4.5 K in a filament, which is longer than the characteristic time in copper (∼0.7 µsec calculated with a diameter of 1.15 mm) and aluminum stabilizer (∼11µsec calculated with a thickness of 4.73 mm). The characteristic time for insulation is 50 msec due to its low thermal diffusivity. A single bunch of beam contains the number of 5.1×108 protons with the time width of 100 nsec. Thereby, 4.4×104 bunches will hit the production during the maximum time duration of 51 msec.
Result
In order to investigate the worst case, the evolution of energy deposition in the innermost layer of CS1 coil where has the peak is calculated as plotted in Fig. 3.28. A lot of secondary particles deposit their energy in coil before 100µsec, and the integrated energy converges after 200µsec to be 1.1×10−6 mJ/cm3 per bunch. With 4.4×104 bunches in the thermal diffusion time, the energy deposited in coil is obtained to be 4.8×10−2 mJ/cm3 at maximum.
As shown in Fig. 3.28, the enthalpy of Rutherford cable and aluminum-stabilized conductor, which is integrated from the bath temperature of 4.5 K, is obtained from the specific heat. With the estimated total energy of 4.8×10−2 mJ/cm3, the temperature rise can be found to be lower than 0.1 K. Thereby, the quench will not be triggered during a time duration of thermal diffusion.
3.4.2 Irradiation Influence on Thermal Shield
In PCS, the superconducting coils are embedded in a cylindrical thermal shield to reduce the heat inleak from the thermal radiation. Considering the secondary particles can also deposit their energy in the thermal shield, which affects the thermal stability of magnet during the beam operation if the thermal shield is heated up to a high temperature.
Thus a thermal analysis is preformed to check the temperature rise in thermal shield when it is heated by both of the thermal radiation and nuclear heating.
Simulation Model
Since the thermal shield is not complex like magnet, the model is constructed in Matlab, and solved with a partial differential equation (PDE) toolbox. To solve the
Chapter 3. Thermal Stability during the Beam Operation
0 200 400 600 800 1000
Time [µsec]
10-16 10-15 10-14 10-13 10-12 10-11 10-10 10-9 10-8 10-7 10-6
Energy Deposition [mJ/cm3/bunch]
0.0 0.2 0.4 0.6 0.8 1.0 1.2
Integral Enery Deposition [mJ/cm3/bunch]
1e 6
4 5 6 7 8 9
Temperature [K]
0 5 10 15 20 25 30 35 40
Enthalpy [mJ/cm3]
Conductor Rutherford Cable
Figure 3.28: The time structure of energy deposition in one bunch hit (left) and enthalpy of Rutherford cable and aluminum-stabilized conductor (right). Left: red line is the integrated energy deposition and black line is the energy deposition with the time bin of 5 µsec. Right:
blue line is the enthalpy of conductor averaged from aluminum, copper and NbTi with a ratio of 7.3:0.9:1, and green line is the enthalpy of Rutherford cable averaged from copper and NbTi with a ratio of 0.9:1.
PDE, by considering an axis symmetrical model, the heat transfer equation at steady condition is given by5
− 1 r
∂
∂r(kr∂T
∂r)− ∂
∂z(k∂T
∂z) = Q. (3.46)
In this model, the radiation is considered a heat flux interacted at boundary, and the thermal shield is cooled down by the gas helium at 60 K with the cooling pipe. The Neumann boundary in PDE is implemented as
∇T =
hc(Tf −T), outer surfaces hr, cooling pipe 0, other boundaries
(3.47)
where hc is the heat transfer coefficient of the gas helium in cooling pipe, Tf is the temperature of the gas helium, and hr is the heat inleak in the unit of W/m2, respectively.
The thermal shield is cooled with the gas helium at 60 K. To determine the heat transfer coefficient for gas helium, the heat transfer coefficient is calculated from a dimensionless number, Nusselt number N u for an internal flow in a pipe:
hc=kfN u
D (3.48)
where D is the diameter of cooling pipe, and kf is the thermal conductivity of flow.
Assumed the gas helium flows in the cooling pipe with a mass flow rate ˙m, the Reynolds number Reis
Re=GD
µ (3.49)
5Since the model is an axis symmetry, the heat transfer along the azimuth is not allowed, thereby, the term of∂T /∂θ is equal to zero.
where µis the viscosity and G= ˙m/A is the mass flow rate per unit cross-sectional area A in cooling pipe. Since the Reynolds number is calculated to be higher than 3500 with the parameters given as follows, the gas helium flow is determined to be a turbulent flow.
As for the turbulent flow, the Gnielinski correlation of the heat transfer coefficient is expressed as [71]
8N u
ReP r1/3f = (1− 1000Re )P r2/3
1 + 1.27(f8)1/2(P r2/3−1) (3.50) where P r is the Prandtl number, f is the friction factor. The Prandtl number is a parameter is the ratio of momentum diffusivity to the thermal diffusivity, and given by
P r=µcp
kf (3.51)
where cp is the specific heat at constant pressure. As for a turbulent flow (Re > 3500) in a pipe, the friction factor is given by [72]
1
f1/2 = 0.782ln[ 1
0.135(e/D) + (6.5/Re)] (3.52)
where e is the roughness of the tube surface. The cooling pipe has a diameter of 25 mm and a thickness of 3 mm, and the roughness is assumed to be 0.0075 mm for brass [73].
Assumed the mass flow rate is 10 g/sec conservatively, the heat transfer coefficient can be determined to 867 W/m2/K with the properties of gaseous helium at 60 K and 0.7 MPa6 as follows:
Thermal conductivity,kf: 47.4 mW·m−1·K−1, Viscosity, µ: 7.219µPa·sec,
Specific heat at constant pressure,cp: 192 kJ·kg−1·K−1.
To reduce the thermal radiation from the room temperature, the thermal shield is covered with the multilayer insulation (MLI). The heat inleak hr through the MLI is determined to be 3 W/m2 (from room temperature to 77 K) conservatively in accordance with the data given in Ref. [74], [75]. The thermal shield is made of aluminum with a thickness of 10 mm and length of 4.3 m. The cooling pipe is settled at the place with a distance of 10 cm from the edge. Realized the degradation will be occurred in thermal conductivity, the thermal conductivity is determined by the RRR of 5 and the magnetic field of 5.4 Tesla. The heat generation is set to be 108 W/m3 uniformly.
Result
Figure 3.29 shows the temperature distribution in the thermal shield. The maximum temperature is about 69 K with both of thermal radiation and nuclear heating, in contrast, the maximum temperature becomes to 62 K with only the nuclear heating. Additionally,
6The helium properties are received from Dr. H. Yamaguchi and Prof. T. Okamura in a private conversation.
Chapter 3. Thermal Stability during the Beam Operation
Figure 3.29: Calculated the temperature distribution in thermal shield by assuming the thermal radiation and nuclear heating are 3 W/m2 and 108 W/m3, respectively.
excluding the degradation of RRR, the peak temperature is reduced to 61 K (RRR = 15).
Thereby, compared with the radiation, the thermal radiation is considered as the domain source for the temperature rise in thermal shield. Since the simulation is preformed with the conservative parameters as mentioned, the temperature is increased by 9 K at maximum, which is acceptable for thermal design.