Dresner, Stability of superconductors.2002

190	CHAPTER 10

Figure 10.3. A sketch showing the temperature profiles of the fully superconducting state FS and the two partly normal states PN1 and PN2.

10.4.THE HEAT BALANCE EQUATION FOR VAPOR-COOLED LEADS

The thermal behavior of any vapor-cooled lead is governed by the heat balance Eq. (4.4.1). The vapor cooling term qP/A has a special form which we now calculate. Figure 10.4 shows the control volume between z and z + dz of the helium channel. The mass flow of helium up the channel is m [kg/s]. A heat balance over this control volume takes the form

Figure 10.4. The control volume between stations z and z + dz of the helium channel.

Miscellaneous Problems	191
qP = mCp(∂T/∂z)	(10.4.1)

where Cp is the specific heat of the helium and P is the perimeter of the helium channel. Eq. (10.4.1) is based on the assumption of thermal equilibrium between the helium vapor and the lead. In the actual construction of leads, the helium channel is in fact many fine channels in parallel having a very large ratio of exchange surface to volume. Thus the assumption of thermal equilibrium is quite a reasonable one.

The mass flow m is related to the heat leak [kA (∂T/∂z)]z=0 into the liquid helium

by

[kA (∂T/∂z)]z=0 = mCL

(10.4.2)

where CL[J/kg] is the latent heat of helium. In steady state, Eq. (4.4.1) then becomes d/dz[k(dT/dz)] + QP/A – [k(dT/dz)]z=0(Cp/CL)(dT/dz) = 0 (10.4.3)

which must be solved subject to the boundary conditions T(0) = Tb and T(L) = Tc.

10.5. COPPER LEADS

When the lead is made of copper, QP/A =ρJ2. Again, we introduce the auxiliary variable s = k(dT/dz) and find as before

s(ds/dT) + LoJ2T– sb(Cp/CL)s = 0

(10.5.1)

where sb = s(Tb) = [k(dT/dz)]z=0. Now we introduce special units in which Lo J 2 = Cp/CL = 1. The dimensions of these quantities are, respectively, P2L–4 Q–2 , and Q–1 , where P is power and Q is temperature. Then Eq. (10.5.1) becomes

s(ds/dT) + T– sbs = 0

(10.5.2)

The direction field of Eq. (10.5.2) is plotted in Fig. 10.5. The smallest value of sb, which is the heat leak into the helium, occurs for the integral curve C for

which s = 0 when T = Tc.1

Since Eq. (10.5.2) is homogeneous, its variables may be separated if we introduce u = s/T as a new dependent variable. Then we find

dT/T= –[u/(u 2 – s u + 1)] du	(10.5.3)
b

which gives upon integration from Tb to Tc

(10.5.4)

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Figure 10.5. The direction field of Eq. (10.5.2).

Eq. (10.5.4) is a transcendental equation for the determination of sb given Tc and Tb. The integral in Eq. (10.5.4) is tabulated and equals

ln(sb2 /T b2 – sb2 /Tb + 1)1/2
+sb(4–sb2 )–1/2arctan[sb(4–sb2)1/2/(2Tb–sb2 )]	(10.5.5)

where the arctan must be given its value in the interval (0,p). According to the tables of Arp and McCarty (1989), Cp = 5.2 J g–1 K–1 for helium at atmospheric pressure over most of the range from saturation (4.2 K) to room temperature. At the same pressure CL = 20.7 J g–1, so that the special unit of temperature is CL/Cp = 3.98 K. Thus Tb = 4.2 K = 1.055 in special units. If Tc = 77 K, then it follows from a little trial and error using Eqs. (10.5.4) and (10.5.5) that sb = 1.68 in special units. Now sb is the heat leak into the helium and has the units PL–2. The special unit of heat leak is then Lo1/2J(CL/Cp).Then, in ordinary units,

S	/J= 1.68L1	/2	(C	L	/C ) = 1.05 W/kA	(10.5.6)
b	o				p

When Tc = 300 K, sb = 1.79 is special units, yielding a heat leak of 1.12 W/kA. Thus when we use copper leads, little is to be gained by an intermediate heat intercept at nitrogen temperature, as was pointed out long ago by Williams (1963). These conclusions hold not just for copper leads, but for leads made of any metal which obeys the Wiedemann–Franz law, Eq. (2.9.1). The above figures are very close to the more precise results of Lock (1969), who took into account the weak temperature dependence of the Lorenz ratio kr/T for copper.

We can calculate the length L of the optimized copper lead from the formula

(10.5.7)

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193

using the values of s = uT we obtain by integrating Eq. (10.5.3). This calculation is tedious, but if we use an idea of Williams (1963) and assume k is constant, Eq. (10.5.7) can be written with the help of Eq. (10.5.3) as

(10.5.8)

The integral in Eq. (10.5.8) is likewise tabulated and equals

2(4–s2 )–1/2arctan[s	b	(4–s	2 )1/2/(2T		–s2 )]	(10.5.9)
b			b	b	b

where again the arctan must be given its value in the interval (0,p). When sb = 1.79 (Tc = 300K), L/k = 4.87 in special units. In ordinary units this last equality becomes LL1/2o J/k = 4.87, which in turn gives the ratio L/A for the optimized lead once the total current has been chosen.

10.6. SUPERCONDUCTING LEADS

When the lead is superconducting, Q = 0 in Eq. (10.4.3) and accordingly the second term (the one proportional to T is absent in Eqs. (10.5.1) and (10.5.2). The solution of Eq. (10.5.2) is then

s = (T– Tb + 1)sb

(10.6.1)

(special units), which can be written as

sb dz = k dT/(T– Tb + 1)

(10.6.2)

Thus,

(10.6.3)

Again using Williams’s approximation of constant k, we carry out this integral to obtain

sbL = k 1n(Tc – Tb + 1)	(10.6.4)
in special units, or
sbL =k(CL/Cp)1n[(Cp/CL)(Tc – Tb + 1)]	(10.6.5)

194	CHAPTER 10

in ordinary units. The heat leak sb thus varies inversely as the lead length L and so can be made as small as desired by making L long enough.

10.7. PARTLY NORMAL STATES

As noted earlier, if L exceeds some critical length, partly normal states become possible. Then in Eq. (10.4.3),

QP/A = 0, T < Tcr		(10.7.1a)
= rJ2, T >T	cr	(10.7.1b)

where Tcr is the critical temperature of the superconductor. For convenience, let us take Tcr = Tc, the saturation temperature of nitrogen. It will prove more convenient in what follows to work directly with Eq. (10.4.3) so we do not introduce the variable s. Again we assume that k is constant and take r = LoT/k in order to fulfill the Wiedemann–Franz law. Finally, we add k = 1 (dimensions PL–1 Q–1 ) to the definition of our special units. In special units, Eq. (10.4.3) then becomes

d2T/dz2 – sb(dT/dz) = 0,	0 < z < w	(10.7.2a)
d2T/dz2 + T – sb(dT/dz) = 0,	w < z < L	(10.7.2b)

where w is the length of the lower, superconducting segment of the lead. Eqs. (10.7.2a) and (10.7.2.b) must be solved subject to the boundary conditions (in special units)

T(0) = Tb; T(L) = Tc; T(w) = Tc; (dT/dz)z=0 = sb

(10.7.3)

plus the condition of continuity of dT/dz at z = w. Now Eq. (10.7.2a) can easily be solved to give

T = exp(sbz) + Tb – 1

(10.7.4)

which satisfies the first and fourth of the conditions (10.7.3). Then using the third of these conditions, we find

w = ln(Tc – Tb + 1)/sb	(10.7.5)
so that
(dT/dz)z=w=sb(Tc – Tb + 1)	(10.7.6)
When w < z < L, the solution of Eq. (10.7.2b) is given by

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T = exp(µz )[Acos(vz) + Bsin(vz)]	(10.7.7)

where k+ = m + iv and k_ = m – iv are the conjugate roots of the characteristic equation

k2 – sbκ + 1 = 0	(10.7.8)
The second and third boundary conditions (10.7.3) then give
Tc = exp( µw)[Acos(vw) + Bsin(vw)]	(10.7.9a)
Tc = exp( µL)[Acos(vL ) + Bsin(vL)]	(10.7.9b)
while the condition of continuity of dT/dz at z = w becomes
Sb(Tc – Tb+ 1) = mTc + vexp( µw)[Bcos(vw) – Asin(vw)]	(10.7.9c)

Now we proceed as follows: (1) Choose sb and calculate m and v from sb = 2m and m2 + v2 = 1. (2) Find w fromEq. (10.7.5). (3) SolveEqs. (10.7.9a) and(10.7.9c) for A and B. (4) Solve Eq. (10.7.9b) for L. These calculations are tedious but straightforward; the results are shown in Fig. 10.6 and discussed in Section 10.8. Similar calculations have been made by Matrone et al. (1989) and Hull (1989).

10.8. PARTLY NORMAL STATES: RESULTS AND DISCUSSION

When L exceeds some critical length, partly normal states become possible. The results of illustrative calculations described in Section 10.7 are shown in Fig. 10.6. The abscissa is the dimensionless lead length b = LL1o/2J/k and the ordinate is the dimensionless product of heat leak and lead length a = (sbL/k)(Cp/cL). These calculations refer to a case for which Tcr = Tc = 78 K, T b = 4.2 K, and CL/Cp = 3.98 K, k is independent of temperature, and r = LoT/k in the normal state. The numbers alongside the points on the curve are values of the dimensionless peak temperature Tmax; multiplication by C L /Cp converts them to Kelvins.

When the dimensionless lead length b is < 4.989, the only steady state is the superconducting state. Now

sb /J = L1o/2 (CL/Cp)(a/b)

(10.8.1)

so that for the longest cryostable superconducting vapor-cooled lead (b = 4.989), sb /J = 0.371 W/kA, which is only slightly better than an optimized vapor-cooled lead. If we use the difference in peak temperature of the lower, unstable partly normal state PN1 and the fully superconducting state FS as a measure of the stability of the latter, we see that increasing the length L of the lead, while reducing the heat leak (10.8.1), also sharply reduces the stability of the superconducting state.

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Figure 10.6. Aplot of the dimensionless producta of heat leak and lead length versus the dimensionless lead length b for the states FS, PN1, and PN2. (Redrawn from an original appearing in Dresner (1991, “Superconductor”) with permission of Butterworth-Heinemann, Oxford, England.)

Furthermore, as the lead length increases, the peak temperature of the upper, stable partly normal state PN2 increases sharply. This means that if a long lead should transfer from the state FS to the state PN2 owing to some perturbation, it may overheat and be destroyed.

Three factors mitigate the gloomy picture just presented. First, superconducting leads are often made of rods of high-temperature superconductors and are therefore relatively rigid. Second, they are located in a region where the magnetic field is weak so that they experience only a small Lorentz force. Third, owing to the rapid increase of specific heat with temperature, thermal runaway from state FS through state PN1 to state PN2 may be slow enough that we can intercede when a voltage first develops across the lead.

Notes to Chapter 10

1The proof of this statement depends on the following theorem: If s1 and s2 are two solutions of Eq. (10.5.2) and sb1 > sb2, then s1 > s2 for all T. It is sufficient to prove it only when s1 and s2 are infinitesimally close. We subtract Eq. (10.5.2) written for s2 from Eq. (10.5.2) written for s1 and set h = s1 – s2; then we find dh/dT + P h = hb, where P = (ds/dT – sb) /s . This first-orderlinear differential equation for h has the solution

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197

Since exponentials are always positive, the algebraic sign of h and that of hb are always the same, as was to be proved.

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Appendix A

The Method of Similarity Solutions

A.1. PARTIAL DIFFERENTIAL EQUATIONS INVARIANT TO

ONE-PARAMETER FAMILIES OF ONE PARAMETER

STRETCHING GROUPS

Roughly speaking, similarity solutions are special solutions of partial differential equations that can be obtained by solving a related ordinary differential equation. This makes them calculable with a great deal less work than other solutions of the partial differential equation, and that is the chief reason that they interest us. However, not every solution of the partial differential equation is a similarity solution. In fact, similarity solutions are a small subset of the totality of all solutions. So not every problem described by the partial differential equation leads to a similarity solution. But some do, and we profit from the method described here when we succeed in identifying a problem of technological interest that is described by a similarity solution.

The partial differential equations that we have treated with the similarity

method in this book are
(1)	ct = (cn)zz	(Chapter 3: critique of the critical-state model)
(2)	cct = czz	(Chapter 8: hydrodynamic phenomena)	(A.1.1)
(3)	ct = (c1z/3 )z	(Chapter 9: cooling with superfluid helium)

Here we have used c for the dependent variable and z and t for the independent variables for the sake of a uniform notation.

The common feature of these equations that allows us to find similarity solutions to them is their invariance to a family of groups of stretching transformations of the form

199