Post on 03-Jan-2016
HD target
HD target overview
Characteristics of polarized HD targetPolarization MethodHD target is polarized by the static method using “brute force” at low temperature (10 mK) and high magnetic field (17 T). It takes about 2-3 months to polarize the target.
Advantage and disadvantageHD molecule does not contain heavy nuclei such as Carbon and Nitrogen. Good for experiments observing reactions with small cross sectionThe HD target needs thin aluminum wires (at most 20% in weight) to insure the cooling.
Polarization H : 90 % D : 60 %
Relaxation Time30 days at 200 mK and 1 T during the experiment.
Target Size25 mm in diameter 50 mm in thickness
HD target transportationRCNP Osaka university
From RCNP to SPring-8
BL33LEP SPring-8
K+
K-
In Beam Cryostat
Transfer Cryostat 1
Transfer Cryostat 2Experiment
HD Target
Dilution Refrigerator
StorageCryostat
Dilution refrigeratorLeiden Cryogenics
DRS-3000 (He3-He4)
Cooling power
3000μW at 120 mK
Lowest temperature
6 mK
Magnetic Field
17 T
Homogeneity of Magnetic Field
5×10-4 for 15 cm
Two Transfer Cryostats
Right : used at RCNPLeft : used at SPring-8
Al wire
2.5cm
IBC （ In Beam Cryostat ） for BL33LEP
NEXT
• Develop NMR system
• To advance the dilution refrigeration
• pure HD gas
• To advance the connector from target to cooling system
Temperature calculation of HD target
Temperature calculation of HD target
by M. M. Lowry.
M M Lowry, F Lincoln, L Miceli, T Saitoh, A Sandorfi, X Wei, C Whisnant. Progress on the LEGS polarized HD target. Int'l workshop on polarized beams and polarized targets, PST 2001, Vol Sept 2001, (2001)
Thermal conductivity of Aluminum
*
*
AT
LQ
（W/k-m）
Q = heat flow rate = 0.5 μWΔT = temperature difference =T2 -T1
L = the length of aluminum wire = 5 cmA = (section area of one aluminum wire) × (number of aluminum wire)
= × (Diameter of wire)2 × (Number of wire)4
L
Area
Q
T1
T2
Al
Aluminum wire temperature
We cut the aluminum wire to N partκ(figure) = κAl / TAl
κAl = κ(figure) ×TAl
ΔL = (the length of aluminum wire ) /N
T1
AlΔL
T0
TN
T2
AT
LQT
Alfigure **
*
)(
AT
LQTT
figure **
*
0)(01
AT
LQTT
figure **
*
1)(12
AT
LQTT
NfigureNN **
*
1)(1
...
T of wire at ring 17 mk
T of wire at top of HD 113.249 mk
T of wire at middle of HD 152.539 mk
2.5cm
T of wire at end of HD 183.604 mk
2.5cm
3cm
Q = Ortho Decay Heating = 0.5 μWWire material = Al 5056Number of wire = 3000Diameter of wire = 38.1 μmThermal conductivity of Al wire = 0.7 W/m/K^2
T of wire at ring 17 mk
T of wire at top of HD 17.359 mk
T of wire at middle of HD 17.653 mk
2.5cm
T of wire at end of HD 17.941 mk
2.5cm
3cm
Q = Ortho Decay Heating = 0.5 μWWire material = 99.999% AlNumber of wire = 2000Diameter of wire = 50.8 μmThermal conductivity of Al wire = 600 W/m/K^2
T of wire at middle of HD
HD temperature next to aluminum wire
Kapitza resistance = Thermal boundary resistance =Contact resistance
TAT
QR
surface
3kapitza
HD temperature next to aluminum wire
surfaceAlAlHD AT
RQTT
3
1
Q (A. Honig) = heat flow rate = 2.8×10-8 (W)Dwire (A. Honig) = diameter of wire = 50 μm
Nwire (A. Honig) = number of wire = 60
L wire (A. Honig) = length of wire = 5 cm
A surface (A. Honig) = (surface area of one aluminum wire) * Nwire
= π × (Diameter of wire) × (Length of wire) × (Number of wire) = π × (0.000050) × (0.05) × (60) (m^2)
= 0.000471 (m^2)
A. Honig, Q. Fan ,X Wei, A. M. Sandorfi, C. S. Whisnant, Nuclear Instruments and Methods in Physics Research Section A 356 39-46 (1995)
HD temperature next to aluminum wire
• The Kapitza resistance is 405422304 (W/ (m^2) (mk^3)). • The error of kapitza resistance is 32802524 (W/ (m^2) (mk^3))
• Q = Ortho Decay Heating = 0.5 μW• Wire material = 99.999% Al• Number of wire = 2000• Diameter of wire = 50.8 μm• Length of wire = 5 cm• TAl = 17.653 mk
• Asurface = π × (Diameter of wire) × (Length of wire) × (Number of wire) • =π × (0.0000508) × (0.05) × (2000) (m^2)• = 0.0159593 (m^2)
• THD =
• = (mk)
• = 19.9619 ±0.186813 (mk)
surfaceAlAl AT
RQT
3
1
3653.170159593.0
)32802524405422304(0000005.0653.17
T of wire at middle of HD
thermal conductivity of HD
TTTTcTccT 31271 )185.0(
Where α, β, γ, δ, σ are constants, which given by REF3, α=0.099, β=1.32×10-3, γ= 0.57, δ=0.0031, σ=3.34 ×10-3
Where “T” is the temperature of HD.
Where “c” is the (J=1) concentration.
J = rotational quantum number
o-H2 (J=1)
p-H2 (J=0)
time → ∞ , c→ 0
HD temperature farthest to aluminum wire
J. H. Constable and J. R. Gaines, Phys. Rev. B 8,3966 (1973)
Al ri
ro
HD
q
HD temperature farthest to aluminum wire
rLAr 2
dr
dTAq r
dr
dTLrq 2
this boundary conditions T = Ti at r = riT = To at r = ro
)/ln(
)(2
io
io
rr
TTLq
HD temperature farthest to aluminum wire
L
rrqTT
2
)/ln( 0101
L
rrqTT
2
)/ln( 1212
L
rrqTT NNNN 2
)/ln( 11
wiresofnumber
targetHD of areasection per wire areasection effective N
AA
wiresofnumber
targetHD of radiussection per wire radiussection effective
N
rr
...
Use computer to calculate temperature
Al 5056 99.999% Al
Numer of wire 3000 2000
L 5 cm 5 cm
rsection radius of HD target 1.25 cm 1.25 cm
r0 19.05 μm 25.4 μm
rN
(reffective section radius per wire)
228.218 μm 279.509 μm
T0 133.8 mk 19.84 mk
TN in figure 1 133.8 mk 19.94mk
TN in our calculation 133.8 mk 19.9757 mk
Calculation of various HD target
cm 5 target ofLength ; cm 1.25 target of Radius
target)of(Length ) targetof Radius(
) wireofLength () wireof Radius() wireofNumber ()
V
V( ratio Volume
2
2
HDAl
Al
T of wire at ring 17 mk
T of wire at top of HD 17.359 mk
T of wire at middle of HD 17.653 mk2.5cm
T of wire at end of HD 17.941 mk
2.5cm
3cm
Q = Ortho Decay Heating = 0.5 μWWire material = 99.999% AlNumber of wire = 2000Diameter of wire = 50.8 μmThermal conductivity of Al wire = 600 W/m/K^2Volume ratio ( Al /(HD+Al)) = 0.00825805
T of HD next to HD 19.9619±0.186813 mk
T of HD farthest of wire 20.0952 mk
Q (μW) 0.5
Wire material 99.999% Al
Length of Wire (cm) 5
T of wire at ring (mk) 17
Number of wire 500 2000
Diameter of wire (μm) 25 50 100 200 25 50 100 200
Volume ratio ( Al / HD+Al) 0.0005 0.002 0.008 0.032 0.002 0.008 0.032 0.128
T of wire at top of HD (mk) 22.197 18.437 17.37 17.093 18.437 17.37 17.093 17.023
T of wire at middle of HD (mk)
25.739 19.554 17.673 17.171 19.554 17.673 17.171 17.043
T of wire at end of HD (mk) 28.849 20.611 17.971 17.248 20.611 17.971 17.248 17.062
T of HD next to HD (mk) 31.7934±0.489861
26.4582±0.558612
22.3488±0.37831
8
19.72±0.2062
4
23.0061±0.279306
20.0109±0.18915
9
18.4455±0.10312
17.6947±0.05273
02
T of HD farthest of wire (mk) 32.0039 26.757 22.7311 20.1154 23.1222 20.1441 18.5668 17.7757
Number of wire 1000 4000
Diameter of wire (μm) 25 50 100 200 25 50 100 200
Volume ratio ( Al / HD+Al) 0.001 0.004 0.016 0.064 0.004 0.016 0.064 0.256
T of wire at top of HD (mk) 19.77 17.733 17.186 17.047 17.733 17.186 17.047 17.012
T of wire at middle of HD (mk)
21.812 18.322 17.34 17.086 18.322 17.34 17.086 17.021
T of wire at end of HD (mk) 23.678 18.892 17.492 17.124 18.892 17.492 17.124 17.031
T of HD next to HD (mk) 26.7863±0.402466
22.5183±0.339523
19.8152±0.20026
8
18.3796±0.1046
66
20.4202±0.169762
18.5776±0.10013
4
17.7328±0.05233
31
17.3481±0.0264674
T of HD farthest of wire (mk) 26.9467 22.7365 20.0508 18.5762 20.5001 18.6541 17.7929 17.3832
purify HD
purify HD
Distillator
topD2 solid, H2 gas, HD liquid
middleD2 solid, H2 gas, HD gas
bottomD2 solid, H2 gas, HD gas
HD～ 96％
D2～ 2％
H2～ 2％
H2～ 50％
HD～ 50％
2 week
H2
H2～ 4~6％
HD~95%
2 week
HD～ 99.999％
principle of concentration sensor
• use filament to ionize gas
ionize gas here
H2+
D2+
HD+
sensor
digital data
AMP(amplifier)
In amplifier, use different gain for each gasPC
concentration
sensor
pressure controller
pressuremonitorDistillator
top
middlebottom
throw gas out
concentration measurement
A
B
C
D
E
F
G
H I
1. open A, get gas for top of Distillator. J
2. close D, open B, then close B, now gas in C. 3. open D, then close D, now gas from C to E
4. control pressure by F, see J the pressure should be 9.0x10-7 mbar
6. close the software RGA
gas here
gas here
gas here
gas here
0. A ,B should closed, open D and G, measure background.
5. measure gas concentration by RGA (Residual gas Analyzer)
7. open G, then close G throw the gas away
gas outside
gas here
Physics objectives1 To investigate the ss content in the nucleon by the p -> p (n -> n) reaction To know the structure of the proton and neutron correctly is the
fundamental desire.
2 To determine the spin-parity of + particle Although I do not follow recent theoretical studies, to fix the initial
nucleon spin and photon polarization must be important.
3 To study the reaction mechanism of the hyperon photoproduction
Recently some interesting results measuring the double polarization
observables appeared. Advanced studies need the polarized nucleon target.