MICE Note-331
Cold Mass Issues in the MICE Coupling Magnet
Michael A. Green
Lawrence Berkeley Laboratory
19 January 2011
I have identified two potential issues in the design of the attachments to the MICE
coupling magnet cold mass [1]. One of these attachment issues may extend into the cold
mass. Both issues came to light during the analysis of what happened to the spectrometer
solenoids. We must not make the same mistakes that were made on the spectrometer
solenoids. The two questions that result from the identified issues are:
1) Are the superconducting low temperature superconducting leads between the
coupling coil and the lower end of the high temperature superconducting (LTS)
leads robust enough not quench and burn out? In the spectrometer solenoid one
part of the LTS lead was not robust enough to prevent quenching and a burn out
of the LTS lead. (See MICE Note 324.) [2]
2) Will there be problems with the quench protection resistors and diodes when the
magnet quenches as a result of an HTS lead burning out or a disconnect of the
magnet from its power supply? The second question is very important because
the coupling coil has a large stored energy and inductance. As result, when
current flows through the diodes and the resistors, both can be over heated. We
observed resistor overheating in spectrometer magnet 2. This heating probably
happened when the HTS lead the LTS lead burned out during the tests of magnet
2A and 2B respectively. (See MICE Note 324.) [2]
The answer to the first question is simple and straightforward. This is the issue that
is primarily dealt with in MICE Note 324. If the resistance across the coil sub-divisions
is high enough, the whole magnet will turn normal through quench-back. Making the
resistance across a coil sub-division high enough is not simple.
Robust LTS leads from the Coupling Coil to the HTS Leads
There are three conditions that are necessary for making an LTS lead from the coil to
the bottom of the HTS lead robust enough for the lead not to quench and not to burn out,
if it does quench. Ideally these conditions are:
1) The minimum propagation zone length of the LTS lead inside of the coil must be
greater than the distance from the coil inner layer to the center of the outside
surface of the cold mass. For the coupling magnet this distance is about 300 mm.
2) The minimum propagation zone length of the LTS lead from the outside surface
of the cold mass to the bottom of the HTS leads must be greater than the LTS lead
length from the outside surface of the cold mass to the bottom of the HTS lead.
For the coupling magnet, this distance is about 450 mm.
-1-
3) The LTS lead between innermost layer of the coil to the bottom of the HTS lead
should be cryogenically stable when cooled by conduction.
Either having a long enough minimum propagation zone length or having cryogenic
stability is sufficient to prevent the LTS lead from quenching and burning out. Have both
conditions as the same time ensures that the lead will be stable even at the full operating
field of the coupling magnet. For the coupling magnet the peak induction in the LTS lead
conductor is about 6 T (at the inner corner of the coil) when the magnet operates at its
maximum design current of 210 A (p = 240 MeV/c in the flip mode).
MPZ Length and Minimum Quench Energy
The minimum propagation zone length equation is derived in MICE Note 324. The
minimum propagation zone length is a function of the conductor overall current density
JM, the matrix material resistivity ρM, and the copper to superconductor ratio r. The
minimum propagations zone length equation for a composite conductor takes the
following form (see Equation 5 in MICE Note 324);
0.5
LMPZ
€
 L (T 2 − T 2 )  r  3 
o
= O c
 ,
2 
 r + 1 
(J
ρ
)
 M M
-1-
where LMPZ is the minimum propagation zone length, L0 is the Lorenz number, Tc is the
conductor critical current, and TO is the operating temperature. JM = IO/Ac, where IO is the
operating current and Ac is the cross-section area of the conductor.
The quench energy QE can be estimated using the following expression (see
Equation 6 in MICE Note 324.);
QE = LMPZ Ac ΔH ,
€
-2-
where ΔH is the enthalpy change per unit volume to quench the conductor. ΔH is from
104 J m-3 to 3 x 104 J m-3 depending on the magnetic induction in the wire (the local TC).
In order to look at the minimum propagation zone length and quench energy, five
cases were calculated. They are:
Case 0: A single conductor that has a cross-section area of 1.5 mm2 as is used
within the coupling coil. At a coil current of 210 A, JM = 1.414 x 108 A m-2,
r = 4 and ρM = 2.2 x 10-10 Ω m (RRR = 70). For this case LMPZ = 13.2 mm.
The quench energy for this case is from 0.2 mJ to 0.6 mJ within the coil.
Case 1: A double conductor that has a cross-section area of 3.0 mm2 with no added
copper stabilizer. At a coil current of 210 A, JM = 7.07 x 107 A m-2, r = 4,
and ρM = 2.2 x 10-10 Ω m (copper RRR = 70). For this case LMPZ = 26.4 mm.
The quench energy for this case is from 0.8 mJ to 2.4 mJ. An LMPZ = 26.4
mm is probably adequate for the flexible leads that connect from the
spreader to the HTS lead bottoms.
Case 2: This case is the same as case 1 with 33 mm2 of added copper cladding. The
cross-section area is 36 mm2. The value of JM = 5.99 x 106 A m-2, r = 57,
and ρM = 2.2 x 10-10 Ω m (RRR = 70). For this case LMPZ = ~420 mm.
The quench energy for this case is about 420 mJ.
-2-
Case 3: This case is case 0 with about 16.5 mm2 of added copper cladding. This
case is shown in the coupling magnet design report of December 2008.
The cross-section area is 18 mm2. The value JM = 1.17 x 107 A m-2, r = 59,
and ρM = 2.2 x 10-10 Ω m (RRR = 70). For this case LMPZ = ~205 mm.
The quench energy for this case is about 110 mJ.
Case 4: This case is the same as case 1 with 35 mm2 of added copper cladding. The
cross-section area is 38 mm2. The value of JM = 5.52 x 106 A m-2, r = 62,
and ρM = 2.2 x 10-10 Ω m (RRR = 70). For this case LMPZ = ~450 mm.
The quench energy for this case is about 500 mJ.
The MPZ length of the lead conductor in the end plate is too short unless it is clad
with the 33 mm2 of extra copper. On the other hand the quench energy is quite high and
the conductor is well supported. The conductor in case 1 is unlikely to quench, but by
adding an extra 33 mm2 copper to the superconductor, as shown in the design report,
would increase the MPZ length to about 420 mm. This is more than adequate to achieve
an MPZ length of 300 mm as required.
The proposed lead from the top of the cold mass package through the spreader to
where the conductor connects to the HTS leads is not adequate from the standpoint of the
MPZ length. The added copper cross-section area should be increased from 16.5 mm2 to
about 35 mm2. In addition, there should be a second piece of the standard superconductor
so that the LTS conductor can be connected to the HTS leads on both sides of the 25 mm
long paddle at the bottom of the HTS-110 leads.
The jumper between the bottom of the HTS leads and the spreader can be made from
two pieces of the standard conductor soldered together or not soldered to together. If the
superconductor is not soldered to together a thin (0.5 mm thick) copper strip can be
soldered to each superconductor on the flat (1.6 mm wide) face. The copper strip can be
up to 10 mm wide. This will increase the MPZ length (up to 100 mm) of the jumper
while retaining the jumper flexibility in one direction. Having two pieces of
superconductor with added copper will reduce the resistance of the joint between the LTS
lead and the HTS lead. This will reduce the resistive heating in the joint.
Are the LTC Coil Leads Cryogenically Stable?
Cryogenic stability requires one to remove the heat generated in the conductor such
that its temperature is not above Tc after the disturbance that caused the quench has
passed. The heat generated in the conductor per unit surface area can be estimated using
the following expression (This is a variation of Equation 16 in MICE Note 324) [2];
Q
ρ r +1 2
= M
IO
A Ac PW r
€
-3-
where Q/A is the heat transfer per unit area; ρM is the resistivity of the matrix material
(copper), r is the copper to superconductor ratio; AC is the cross-section area of the
conductor (copper plus superconductor), PW is the heat transfer perimeter of the
conductor; and IO is the operating current of the lead.
The temperature drop ΔTi across the insulation can be estimated using the following
expression;
-3-
ΔTi =
€
Q
UA
-4-
The heat transfer coefficient (U factor) for a conduction cooled system U = ki/ti,
where ki is the insulation thermal conductivity and ti is the insulation thickness between
the LTS lead and the 4 K cooling sink. For an organic insulation ki = 0.1 W m-1 K-1. The
U factor will be 1000 W m-2 K-1 when the insulation thickness is 0.1 mm. With an
insulation thickness of 1 mm the value of U = 100 W m-2 K-1. If ΔTi < TC-TO, the
conductor is cryogenically stable. If one looks at the three cases one sees the following:
Case 0: A single conductor with ti = 0.14 mm, P = 5.1 mm, Q/A = 1617 W m-2, U =
~700 W m-2 K-1. The estimated ΔTi = ~2.3 K. The system is not stable.
Case 1a: A double conductor with ti = ~0.12 mm, P = 7.2 mm, Q/A = ~575 W m-2, U
= ~830 W m-2 K-1. The estimated ΔTi = ~0.7 K. The system may be stable.
Case 1b: A double conductor with ti = ~1 mm, P = 7.2 mm, Q/A = ~575 W m-2, U =
~100 W m-2 K-1. The estimated ΔTi = ~5.7 K. The system is not stable.
Case 2: Double conductor with 33 mm2 of added copper cladding with ti = 1.0 mm,
P = 33 mm (single surface), Q/A = ~8.3 W m-2, U = 100 W m-2 K—1. The
estimated ΔTi = ~0.09 K. The system is always stable.
Case 3: Single conductor with 16.5 mm2 copper cladding added with ti = ~0.3 mm,
P = 12 mm (double surface), Q/A = 45.7 W m-2, U = 333 W m-2 K—1. The
estimated ΔTi = ~0.14 K. The system is always stable.
Case 4: Double conductor with 35 mm2 of added copper cladding with ti = 0.3 mm,
P = 24 mm (double surface), Q/A = 10.9 W m-2, U = 333 W m-2 K—1. The
estimated ΔTi = ~0.03 K. The system is always stable.
In case 0, the conductor in the coil is not cryogenically stable. As the quench region
grows, the heat flow per unit area goes up because the quench volume grows faster that
the heat transfer perimeter.
Case 1 can be stable if the insulation thickness between the conductor and the heat
sink (the mandrel) is small. The insulation thickness around the conductor in the coil end
plates is about 1 mm. With an insulation thickness of 1 mm, the lead conductor is not
cryogenically stable.
In Case 2 (the actual leads described in the design report of 3 December 2008) [1]
the LTS leads have an added 33 mm2 of good copper soldered to the superconducting
wire. As a result, the MPZ length for this case is close to the MPZ length for case 4. The
heat transfer perimeter is also much larger than in case 1 (33 mm versus 7.2 mm) as a
result, the design will be cryogenically stable for insulation thicknesses up to ~0.5 mm.
Since the MPZ length is greater than the length of the copper strip, leads coming out of
the coil should be cryogenically stable, because of the added copper.
In case 3 (the actual leads described in the 3 December 2008 report) have an added
16.5 mm2 of copper added to a single conductor. This case is cryogenically stable even if
the insulation thickness in increased to 1 mm. The MPZ length (205 mm) is too short.
Case 4 is both cryogenically stable and it has a MPZ length that is greater than the
length from the cold mass outer surface to where the leads cross the space to the bottoms
of the HTS lead.
-4-
The flexible LTS leads that jump from the top of the spreader (case 2) to the bottom
of the HTS leads may be doubled standard conductor. Since the cooling is only at the
ends of these leads, the MPZ length must be longer that the distance along the conductor
from the spreader to the bottoms of the HTS leads. If the flexible LTS lead is designed to
be flexible in one direction, strips of copper can be added to increase the MPZ length.
Quench Propagation along the LTS Leads
The quench propagation velocity becomes an issue when the MPZ length is less than
the lead section length and when the lead is cryogenically stable. (MPZ length has no
bearing on a lead that is cryogenically stable, because the heat is conducted away from
the quench zone in the lateral (radial) direction. Heat flow at the ends of the normal
region only improves the cryogenic stability. The only case where the quench velocity
along the wire becomes important is in case 1 where there is insufficient copper to
lengthen the MPZ length to the proper value or make the lead cryogenically stable. An
expression for the adiabatic quench propagation velocity (with no transverse heat
transfer) along the wire is given as follows (see Equation 28 in MICE 324) [2];
v ≈ (5.7x10−14 )[1+ B]
€
0.62 1.65
M
J
,
-5-
where v is the quench velocity along the wire with no transverse heat transfer, B is the
magnetic induction at the wire; and JM is the current density in the conductor (matrix plus
superconductor). Transverse heat transfer slows down the quench propagation. If half of
the heat generated in the conductor can be transferred in the transverse direction the
quench velocity is reduced by a factor of 1.4.
From Equation 6 one can estimate the propagation velocity along the lead. In the
region at the end of the coupling coil, 4T < B < 6T. At a current of 210 A for the double
conductor in case 1, JM = 7.07 x 107 A m-2. The adiabatic quench propagation velocity
along the lead is from 1.39 to 1.72 m s-1. The quench passes along the 0.30 m long
section of conductor in ~0.2 s, at which point the magnet quenches. With the added
copper in case 2, the quench velocity along the wire is of the order of 0.02 to 0.03 m s-1.
The conductor will not burn out, even if the quench must propagate into the coil. It can
be concluded that the 300 mm long LTS leads into the coil are safe.
What Happens to the Quench Protection Resistors and
Diodes when an HTS Lead Fails?
During the first of spectrometer solenoid magnet 2 there was the failure of an HTS
lead due to overheating (the test of magnet 2A). The upper end of the HTS lead was at
93 K due to excess heat to the cooler first stages (not enough cooling for the actual heat
load, which was more than three times the estimated heat load). In the magnet 2A test
the lead quenched and the HTS lead burned out about 100 mm below high temperature
end of the lead. The HTS lead failure is discussed in MICE Notes 279 and 292 [3], [4].
During the second test of spectrometer magnet 2 there was a failure of an LTS lead
due to the lead quenching and overheating (the test of magnet 2B). During the test of
magnet 2B, there was plenty of cooling at the tops of the HTS leads. As a result the
copper plate that was connected thermally to the leads was much colder (<50 K). The
LTS lead that burned out was one of the leads to coil M2. This lead failed just inside of
-5-
the cold mass feed-through. The probable reason for the quenching of the LTS lead was
conductor motion in a section of conductor that was not well cooled. The normal region
was longer than the MPZ length for the lead conductor, so the quench propagated so that
the whole lead section became normal. After 4.3 seconds the LTS lead burned out
severing the magnet circuit. The LTS lead failure is discussed in MICE Note 324 [2].
The magnet 2 lead failures broke the magnet circuits causing the diodes to fire so
that current could pass through the shunt resistors. The full current of the magnet flowed
for a long period of time (>10 s). As a result, the resistors over heated. We don’t know if
the diodes overheated too. They appear to be OK, but they haven’t been tested to see if
the forward voltage and backward voltage have changed. This appears to be the most
likely reason that the six quench overheated. The overheating burned the G-10 insulation
and caused the ribbon resistors to warp and become discolored. When magnet 1 is
disassembled we will see if there was any overheating of the resistors in that magnet.
When the 0.02-ohm resistor carries current, it heats up rapidly until it reaches an
equilibrium temperature determined by radiation heat transfer. In magnet 2, the resistors
were in helium, but the dominant form of heat transfer was by radiation. The peak
temperature of the resistors could have been as high as 1300 K.
The diodes and resistors in the spectrometer solenoid didn’t overheat during a
normal quench. The resistors and diodes have enough mass to absorb heat energy put
into them while the magnet is quenching. The estimated temperature of the resistor after
a normal quench is about 200 K. The diodes, which are more massive, end up at a lower
temperature during a normal quench.
The voltage generated across the resistors and diodes will eventually cause the
spectrometer magnet to quench through quench-back. When the spectrometer solenoid
carries its design current of 275 A, the voltage drop across all of the resistors and diodes
is about 40 V. The heating in the resistors and diodes is about 10.9 kW at full current.
Heating of the cold mass from the heating of the resistors and diodes can eventually
cause the magnet to quench due to quench back. The quench-back is due to the heating
of the mandrel due induced circulating currents from the dB/dt within the coil and due to
direct heating of the cold mass from the resistors and diodes themselves. The magnet
eventually quenched after the lead burned out (in both cases), but the delay before the
magnet quench led to an overheating of the resistors used to protect the magnet.
The coupling coil will behave in the same way as the spectrometer solenoid (with all
five coils connected in series) did, except that the time constants associated with the
event are much longer due to the larger self-inductance (590 H versus 110 H). This
means that the resistors and diodes will carry currents longer before the magnet quenches
due to quench back from the inner mandrel and flanges. Earlier calculations suggest that
the coupling magnet powered at a current of 210 A will quench back in 1.8 seconds, if
the magnet is not sub-divided and a voltage of 590 V is put across the leads using a
resistance of about 2.8 ohms (see MICE Note 114) [5].
When the magnet is sub-divided, the 2.8 ohms needed to cause the magnet to quench
due to quench back must be put into the resistors across the sub-divisions. Since the
coupling magnet has eight sub-divisions, the resistance across each sub-division must be
about 0.35 ohms. Since these resistors absorb some of the stored energy from the magnet
during the quench process, the active elements of each resistor must have a mass of about
0.70 kg. The current design calls for resistors of 0.02 ohms with mass that is > 0.5 kg.
-6-
Quench-back from the Magnet Mandrel
The resistance needed to induce quench-back RQB can be estimated using the
following expression (see Equation 29 in MICE Note 324) [2] [5];
0.5
RQB
€
 1  0.5  N1 2πRc  ρ 2ΔH 2 
=  
 ,

 δ   N 2 io  tQB 
-6-
where RQB is the resistance needed to induce quench-back in a time tQB. δ is the coil to
mandrel coupling coefficient (N1 is the number of turns in the coil (N1 = 15272 for the
coupling coil); N2 is the number of turns in the secondary (N2 = 1); RC is the average
radius of the coil (RC = 0.801 m); io is the starting current (io = 210 A); ρ2 is the
secondary circuit resistivity; and ΔH2 is the enthalpy change per unit volume in the
secondary circuit material (mandrel). When a magnet is designed to quench back the
quench back time tQB is short (~1 s). The mandrel resistivity ρ2 = 1.42 x 10-8 Ω m. The
design enthalpy change ΔH2 for aluminum from 4 K to 10 K is ~11700 J m-3. With these
values the calculated value for the resistance is 5 ohms or 0.625 ohms per sub-division.
The expression in equation 6 can be turned around in order to estimate the time that
it would take to cause quench back a coil tQBC when there is an effective resistance Reff
across the coil. This expression is as follows (see Equation 30 in MICE Note 324) [2];
2
tQBC
€
 1  N 2πRC 
=   1
 ρ ΔH
 δ  N 2 io Reff  2 2
-7-
The effective resistance of the diodes, after they have fired and heated up to 300 K,
is about 0.0045 ohms. The resistance for the diodes and resistors at 300 K across the
coupling magnet is 8(0.03 + 0.005) = 0.28 ohms. With a resistance of 0.28 ohms, the
quench-back time tQB = ~300 seconds, which is a very long time. Since the volume of the
mandrel is 0.12 m3, the total energy put into the mandrel is 1405 J. All this will do is boil
away some helium in the tubes. This is not a good way of quenching the magnet.
Quench-back from the Resistors and Diodes attached to the Cold Mass
If one looks at the heat produced in the resistors over a period of 586 seconds, one
sees that there is much more heat dissipated in the eight resistors that cover 40 percent of
the cold mass assembly than is produced by circulating currents in the cold mass. The
peak i2R heating in the resistors and diodes is ~8.8 kW (using the 4.2 K resistance for the
resistors. As the resistors heat up, the resistance goes up. At 300 K the resistance of the
quench protection resistors increases to ~0.029 ohms. At 900 K, the resistor resistance
will go up another factor of two. If the heat produced in the resistors and diodes heat can
be directed into the cold mass assembly, the quench back time will be greatly reduced.
Before the magnet can be quenched by the resistors and diodes, one must boil away
the helium in the cooling pipes and the two helium tanks ant the top and bottom of the
magnets cooling circuit. The 40 liters of helium in the cooling system requires ~105 kJ
of energy to boil away at 4.2 K. Some additional energy may be needed before the relief
devices vent the helium to the atmosphere through the relief device.
-7-
Once the helium is boiled away, it takes only about 1.5 kJ to start the quench of the
magnet. Under this scenario, the magnet quench can start in as little as 12 s. (This
assumes that the heat flow from the resistors into the cold mass is 100 percent efficient in
getting from the heat source into the cold mass and that 105 kJ was all that was required
to boil away the helium.) In reality, it will take longer for the magnet to quench from the
heat supplied by the resistors and diodes. Once the quench of the magnet has started the
current in will decay to one percent of the starting current in about 16 seconds regardless
of the starting point for the magnet quench.
There are time constants associated with the resistor and diode heating process and
the process of transferring the heat from the diodes and resistors to the superconductor in
the coil. The key to making this scenario work is directing the heat energy produced by
the resistors and the diodes into the cold mass in an efficient way. There are two
approaches that can be used to get the heat into the cold mass. These approaches are:
1) The resistors and diodes can be surrounded by a copper box that is black on the
inside, so heat produced in the resistors and diodes is transferred to the box from
the resistors and diodes by radiation. The heat transferred to the copper box is
transferred by conduction to the cover plate and flanges of the magnet cold mass.
2) The resistors are insulated with insulation capable of withstanding temperatures of
150 C (423 K) and are sandwiched between copper plates. The heat in the copper
is transferred to the magnet cold mass cover plate and the flanges by conduction.
The first approach may be achievable without changing the basic design of the
resistor system. Because the resistors can get quite hot (1100 K) in this case, the
fiberglass epoxy (G-10) insulation above and below the resistors must be replaced by a
ceramic insulation that is black to infrared radiation. The quench resistors will expand
~1.4-percent as they get hot. The resistors must be spring loaded to keep them taut.
The second approach requires that copper resistor sandwich to conform to the
outside of the cold mass. A Kapton type of insulation around the resistor is suitable in
this case, if the heat transfer coefficient between the copper resistor sandwich and the
coupling magnet cold mass is high enough. The second approach would work much
better if the resistors were between the banding and the outer surface of the coil. Full
ground insulation would be required between the coil and the resistors.
The Time Constants associated with Radiation Heat Transfer
If the resistor gets hot enough to radiate the heat produced in the resistor, there is a
time constant needed to get to the operating temperature of the resistor. The equilibrium
temperature for the resistor is determined by radiant heat transfer, which can be
calculated using the following expression.
Q
= εσ [TR4 − TO4 ] ,
AR
€
-8-
where Q/AR is the heat transfer per unit of the resistor area, ε is the emissivity of the
resistor surface; σ is Stefan Boltzmann constant (σ = 5.67 x 10-8 W m-2 K-4), TR is the
resistor temperature, and TO is the sink temperature.
-8-
Since the sink temperature is low (<150 K) compared to the resistor temperature, the
sink temperature can almost be ignored. If one ignores the sink temperature and sets it to
zero and one assumes that the resistor will radiate from four surfaces (the top and bottom
and the two sides), one can estimate the equilibrium temperature of the resistor TR as it
radiates the heat being generated by the current. An expression for TR is as follows;
0.25
Q 1 
TR ≈ 

 AR εσ 
€
-9-
∫
TR
0
C(T)
dT =
ρ(T)
tc
2
∫ j(t) dt ,
-10-
0
where F* is the adiabatic heating function, C(T) is the volume specific heat as a function
of temperature, ρ(T) is the resistor material resistivity as a function of temperature, and
j(t) is the current density across the resistor cross-section as a function of time.
If one makes a three simplifying assumptions, one can come up with for an equation
for the time that it would take for resistor to come close to its equilibrium operating
temperature. The first assumption is the enthalpy change to go from 0 to 1000 K is about
four times the enthalpy change for going from 0 to 300 K (based on a constant specific
heat between 300 K and 1000 K). The second assumption is that the resistivity of the
resistor is the average mean resistivity ρm of the stainless steel between 0 and 1000 K.
The third assumption is that the current is the resistor does not change with time. Using
the assumption above one can use the following expression to estimate the time tC needed
to heat the resistor from 4 K to 1000 K;
tC ≈
€
,
where Io is the operating current of the magnet, R(TR) is the resistance of the resistor at TR
(roughly three times the 4.2 K resistance for stainless steel), LR is the resistor length, aR is
the resistor width and bR. When LR= 1.6 m, aR = 0.02 m, and bR = 0.002 m, the surface
area of the resistor AR = 0.0704 m2. When R = 0.06 ohms and ε = 0.6, the maximum
operating temperature for the resistor TR is about 1030 K. It this temperature, all of the
insulation that may be touched by the resistor must be a ceramic. Insulation such as G-10
would burn, as did the insulation that touched the resistors in the spectrometer solenoid.
There is a time constant that is associated with getting the resistor up to its nominal
operating temperature. This time constant can be calculated using the same adiabatic
equation that is used to calculate the hot-spot temperature for quench [5].
The
expression for F* takes the following form [6];
F * (TR ) =
€
0.25


io2 R(TR )
=

εσ (LR (2aR + 2bR )) 
4ΔH 4−300
.
ρ m j R2
-11-
For stainless steel, ΔH4-300 = 6.3x108 J m-3and, ρm = 9.8x10-8 Ω m. For the resistor
proposed at a current of 210 A, the value of JR = 5.25x106 A m-2. The time needed for the
resistor to reach 1000 K is about 90 seconds. During this time about 1.26 MJ of heat will
have been generated within the eight resistors. An additional 0.25 MJ will be generated
by the eight diodes. It is clear that the magnet will quench before the resistor reaches
1000 K, if the thermal connection between the magnet and the copper boxes around the
resistors is at all good.
-9-
The Time Constants Associated with Thermal Conduction to the Coil
The thermal time constant for the heat to get into the coil from the copper box
around the resistors to the coil is a function of the thermal diffusivity of the material
between where the heat is collected and the coil [7]. The thermal diffusivity α(T) of a
material as a function of temperature T can be calculated using the following expression;
α (T) =
€
k(T)
,
C(T)
-12-
where k(T) the thermal conductivity of the material as a function of T, and C(T) is the
specific heat per unit volume as a function of T. The temperature range for estimating
the time constant to quench the magnet is the temperature range from 4 to 10 K. In this
temperature range C is quite small. The three materials of interest are copper, aluminum,
and fiberglass epoxy. For copper at 7 K, k = ~800 W m-1 K-1, C = ~ 3560 J m-3 K-1, and α
= 0.22 m2 s-1. For 6061 aluminum at 7 K, k = ~12.2 W m-1 K-1, C = ~ 2320 J m-3 K-1, and
α = 0.0052 m2 s-1. For G-10 at 7 K, k = ~0.2 W m-1 K-1, C = ~ 1750 J m-3 K-1, and α =
0.00012 m2 s-1. The thermal diffusivity for these materials is two to four orders of
magnitude higher in the 4 to 10 K range than at room temperature.
The thermal diffusion distance λ(t) into the coil at a temperature T as a function of
time t can be estimated using the following expression [7];
0.5
λ(t) = (α (T)t ) .
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The thermal diffusion distance in the temperature range from 4 to 10 K is as follows for
the three primary materials in the cold mass package: For OFHC copper λ(1) = 0.47 m;
for 6061 aluminum, λ(1) = 0.072 m; and for G-10, λ(1) = 0.0073 m. This means that
heat will travel from the copper box to the cold mass in <1 s. The heat will be transferred
to coupling coil from the outer surface of the cold mass in ~1 s. The heat pulse will cross
all of the G-10 barriers between the copper and the coil in <1 s. Once the liquid helium
in the cold mass has been dispensed with the time needed to quench the magnet is <3 s.
The longest time constant in the system is the time needed to heat the resistor to a
temperature that will transport significant heat into the cold mass. This time is of the
order of 45 seconds. This time constant can be reduced by over a factor of two making
by reducing the cross-section a factor of 1.5 (increase the width to 25 mm and reduce the
thickness to 1 mm. This will increase the resistance of each resistor to 0.03 Ω. The peak
temperature of the resistor will increase from 1030 K to 1140 K. The next longest time
constant is the time needed to boil away the liquid helium. Once the helium has boiled
away, the quench will occur 2 or 3 seconds later. For the resistors that are in the design
report, the total time to quench is ~55 s. If the resistance is increased a factor of 1.5 with
the same heat transfer area, the time to quench will be reduced to ~35 s.
If the resistors are imbedded in insulators between two copper plates, the time
constant for heating up the resistors to a temperature where heat can be conducted into
the cold mass is eliminated. The primary time constant when the resistors are imbedded
is the time needed to boil away the liquid helium in the system. Depending on the
resistance this can be from 8 to 12 seconds. When 0.02 ohm resistors are used, the
quench time is ~15 s. When 0.03 ohm resistors are used, the quench is ~ 11 s. It would
be useful to do an FEA analysis of the problem to verify the quench times.
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Quench Protection Diodes after an HTS lead Fails
It is useful to look at the diodes after they fire and are subjected to current for a long
period of time. The protection diodes have a forward voltage of ~8 V at 4.2 K. The
effective resistance of the diode at 4.2 K at 210 A is ~0.038 Ω. The diode rapidly heats
up because its specific heat is low. At 77 K the forward voltage is ~2.3 V. The effective
resistance is ~0.011 Ω. The rate of heating slows down because the resistance is lower
and the specific heat is much higher. By the time the diode reaches 300 K, the forward
voltage is ~1 V. At 300 K, the effective resistance is ~0.0045 Ω.
At room temperature the specific heat of the silicon in the diode is 710 J kg-1 K-1.
The thermal conductivity of the silicon in the diode is 148 W m-1 K-1. The density of
silicon is 2330 kg m-3. One can make an estimate for the time needed to heat a diode
from 4 K to 300 K when there is current in the diode. The diode specific heat is high
when the forward voltage is low. From 77 K to 300 K the average heating is ~340 W.
The diode mass is ~75 g. The copper that is attached to the diode is ~30 g. The energy
change for diode (135 J/g) and its copper (74 J/g) from 77 K to 300 K is ~12300 J. The
diode will heat up adiabatically from 4 K to 300 K in ~40 s. To heat up another 100 K to
400 K, will take an additional 30 s, at a heating rate of ~210 W. Under either scenario
mentioned in the previous section, the coupling magnet should have quenched before the
diodes reach 400 K.
The diodes have a high thermal conductivity. If the frame for the diodes is designed
to take the heat away from the diodes and transport that heat to the cold mass, the heat
will be conducted to the cold mass before the diode temperature reaches 200 to 250 K.
The estimated temperature drop within the diodes that are in the frame is about 40 K
when the diodes are carrying 210 A at or near room temperature. The diode frame design
should be checked to see that the heat generated in the diodes can be conducted from the
interior of the silicon diodes to the cold mass.
The Copper leads that Connect the Resistors and Diodes to the magnet
The leads that connect from the superconducting leads coming out of the coil to the
diodes and resistors will be made from annealed copper. Equation 10 can be used to
determine the cross-section of these leads (assuming that the system adiabatic with no
radial cooling). The typical times that are needed to quench the magnet after an HTS
lead or LTS lead has burned out is up to 100 seconds (worst case). If the maximum
allowable temperature for the annealed copper leads is set to 300 K, the value of F* for
RRR = 100 copper is about 1.5 x 1017 A2 m-4 s. The maximum current density in these
copper leads is ~39 MA m-2. An RRR = 100 copper lead that carries 210 A must have a
minimum cross-section area of ~5.4 mm2 (number 9 wire). If the copper RRR = 10, the
value of F* = 1017 A2 m-4 s at 300 K. An insulated copper cable is used to make the
connection between the coil superconducting taps and the quench protection system has
an unknown RRR. To allow for this and other things that are not known, the copper area
of the cable should be increased to something of the order of 20 mm2 (number 4 cable)
for complete safety. An insulated number 4 annealed copper-cable can carry 60 to 80 A
continuously in air at 300 K depending on the insulation system.
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Concluding Comments
The LTS current leads within cold mass are adequate. The leads have an MPZ
length that is greater than the 300 mm needed to keep them from quenching. In addition
the LTS leads within the coil cold mass are cryogenically stable.
The design of the LTS leads from the surface of the cold mass to the flexible jumper
that goes from the spreader to the base of the HTS leads can be improved. The baseline
design has an MPZ length that is too short. The design should be modified so that there
are two superconductors in the HTS lead. The amount of copper should be increased to
make the MPZ length long enough (see case 4). Having two LTS superconducting wires
permits on to connect the LTS conductor to both sides of the paddle at the bottom of the
HTS leads. This improves the stability of the jumper by making its MPZ length longer
and it reduces the joint resistance between the LTS lead and the HTS lead. Increasing the
MPZ length of the jumper by adding a flexible copper strip is desireable.
The baseline quench protection resistor design is adequate for quench protection of
the coupling magnet during a normal quench. The design is not adequate for causing the
coupling magnet to quench in the event of an HTS lead failure. The resistor design in the
design report calls for cooling the resistors by radiation heat transfer. The resistors can
get very hot (up to ~1140 K) when an HTS lead fails and the resistors are cooled by
radiation. Any electrical insulation that is in contact with the resistors must be a
ceramic. If the quench protection resistors are radiation cooled, they must be spring
loaded to take up the 1.4 percent (~12 mm) of expansion that will occur when heated to
1140 K. The resistors must be in a copper box that is black to infrared radiation. The
copper box must collect the heat and transfer it to the surface of the cold mass. The
longest time constant for magnet quench using the radiation-cooled resistors is the time
that it takes for the resistors to get hot enough to start a magnet quench. Using the design
report resistors in copper box will result in a quench time of ~55 s.
A better approach would be to sandwich the resistors between two plates of copper
that would conduct the heat generated in the resistors directly to the cold mass. The
resistor temperature in this case would not exceed 400 K. The time to quench the magnet
would be much shorter ~15 s. The dominant time constant is the time needed to boil
away the helium that is in the cold mass. Making a resistor sandwich and connecting that
sandwich thermally to the cold mass is not without its difficulties. The electrical
insulation is one issue. The voltages across the resistors and to ground are low (~50 V),
but the insulation to ground must be high potting to 1100 V. The thermal connection of
the sandwich to the cold mass is another issue.
The diodes don’t have to be cooled during a normal magnet quench. Diode cooling
may be needed for a quench caused by a failure of the HTS leads. Fortunately the diodes
have a high thermal conductivity. The temperature drop in the diodes is reasonable as
long as the diode frame is well connected thermally to the cold mass.
The annealed copper leads that connect the resistors and diodes to the coil taps
should have a minimum area of 5.4 mm2, when the RRR = 100. Since one often doesn’t
know the RRR of the copper cable used to make the connection, the cross-section should
be increased to something that is of the order of 20 mm2. An insulated number 4 coppercable is more than adequate for connecting the coupling magnet quench voltage taps to
the quench diodes and resistors.
-12-
Acknowledgment
This work was also supported by the Office of Science of the US Department of
Energy under DOE contract DE-AC-02-05CH11231.
References
[1]
L. Wang, et al, “Coupling Solenoid Magnet Engineering Design Report,” Institute of
Cryogenics and Superconductive Technology, Harbin Institute of Technology, Dec. 2008
[2]
Michael A. Green, “What caused the Lead burn-out in Spectrometer Magnet 2B, MICE
Note 324. http://www.mice.iit.edu, November 2010.
[3]
S. P. Virostek, and M. A. Green, “The Results of Tests of the MICE Spectrometer
Solenoids,” to be published in IEEE Transactions on Applied Superconductivity 20, No. 3,
p 377, (2010), MICE Note 279, http://www.mice.iit.edu.
[4]
M. A. Green, “What Happened with Spectrometer Magnet 2B.” published as MICE Note
292, http://www.mice.iit.edu, LBNL-3927E, May 2010.
[5]
M. A. Green and H. Witte, “Quench Protection and Magnet Power Supply Requirements
for the MICE Focusing and Coupling Magnets,” MICE Note 114, http://www.mice.iit.edu,
LBNL-57580, June 2005
[6]
W. H. Cherry and J. I. Gittlemen, “Thermal and Electrodynamic Aspects of the
Superconductive Transition Process,” Solid State Electronics 1, p 287 (1960)
[7]
M. A. Green and S. Q. Yang, “The Effect of a Hydrogen Spill inside of a 300 K Vacuum
Vessel on MICE Absorber Hydrogen Safety,” MICE Note 100, http://www.mice.iit.edu,
(March 2004).
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DISCLAIMER
This document was prepared as an account of work sponsored by the United States
Government. While this document is believed to contain correct information,
neither the United States Government nor any agency thereof, nor The Regents of
the University of California, nor any of their employees, makes any warranty,
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The views and opinions of authors expressed herein do not necessarily state or
reflect those of the United States Government or any agency thereof, or The Regents
of the University of California.
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