Dresner, Stability of superconductors.2002

Figure 8.6. A conceptual arrangement of a double-jacketed cable-in-conduit conductor. The inner jacket is perforated by small holes that bleed helium from region A to region B. (Redrawn from an original appearing in Dresner (1993, “Reducing”) with permission of Butterworth-Heinemann, Oxford, England.)

The mass balance is

(8.13.1)

which becomes in the limit dz → 0,

This is the same as Eq. (7.2.1) except for the appearance of the sink term –m on the right-hand side.

The momentum balance over the control volume is

The fifth term on the right-hand side is based on the view that the helium leaving space A carries its momentum with it. This seems plausible because as the helium enters the perforations its forward motion is arrested by collision with the walls and its momentum thereby annihilated. The last term on the right-hand side is the frictional retarding force exerted by the cable and the walls of the main channel. As dz → 0, Eq. (8.13.3) becomes

where, as in Sections 7.2ff., F=2fv2/D. If we multiply Eq. (8.13.2) by v and subtract it from Eq. (8.13.4.), we obtain

which is the same as Eq. (7.2.2). (Cf. Section 7.2 for the definition of the total (hydrodynamic) derivative, d( ) /dt.)

where e = u + v2/2 is the sum of the specific internal energy u and the specific kinetic energy v2/2. The fifth term on the right-hand side is the loss of energy accompanying the helium percolating through the walls. The idea here is that the kinetic energy of the helium departing space A is converted to internal energy by collision with the walls of the perforations and then that energy and the original internal energy are swept out of space A. The sixth term is the pressure work done by the helium remaining in space A in pushing the departing fluid through the perforations. The form of this term is based on the tacit assumptions that the helium percolating into space B does not significantly raise its pressure and that the pressure in space A is much larger than that in space B. As explained in Section 7.2, the frictional force F does not appear in the energy balance. In the limit as dz → 0 and after subtraction of e times Eq. (8.13.2) , Eq. (8.13.6) becomes

Except for the additional term –p(m/r) on the right-hand side, Eq. (8.13.7) is the same as Eq. (7.2.3).

Now we follow the procedure that led to the entropy Eq. (7.2.7), namely, multiplying the momentum Eq. (8.13.5) by v and subtracting from the energy Eq. (8.13.7); we obtain thereby

Now we use the mass balance Eq. (8.13.2) to eliminate ∂v/∂z, obtaining precisely Eq. (7.2.6), namely,

It follows then, as before, by using the second law of thermodynamics, that

Now if we eliminate r from the mass balance Eq. (8.13.2) using the thermodynamic identity (7.2.8), we obtain, with the help of Eq. (8.13.10)

which is the same as Eq. (7.2.9) except for the extra term –mc2 on the right-hand side.

8.14.PERFORATED JACKETS: REDUCTION OF THE QUENCH PRESSURE

Equations (8.13.5) and (8.3.11), like their counterparts Eqs. (7.2.2) and (7.2.9), are partial differential equations that serve to determine p and v. If we set the inertial term, r(dv/dt), to zero in Eq. (8.13.5), as we did at the beginning of this chapter in its counterpart Eq. (7.2.2), these two equations become

which are identical with Eqs. (8.1.1) and (8.1.2) except for the last term on the right-hand side of Eq. (8.14.2).

If we now follow the steps in Section 8.2 that led there to Eq. (8.2.4), we find instead

(8.14.3)

To go any further, we need to relate the sink strength m to the diameter D´ and the spacing of the perforations.

The perforations are short channels of length equal to the thickness of the inner conduit. The helium is presumed to traverse these channels in turbulent flow. If we let s be the porosity of the inner conduit (the ratio of perforation area to total

where f´ is the Fanning friction factor in the perforations, and D1 = 4AHe /PA, where PA is theperimeter of space A. Eq. (8.14.4) is based on the tacitassumptions, already noted in the last section in connection with Eq. (8.13.6), that the heliumpercolating into space B does not significantly raise its pressure and that the pressure in space A is much larger than that in space B. If we substitute Eq. (8.14.4) into Eq. (8.14.3), we can rewrite the latter as

Comparing Eq. (8.14.5) with Eq. (8.2.4) we see that the maximum value of po, pm, is reduced by the factor

compared with the result we would calculate from Eq. (8.2.4) for an unperforated inner conduit.

As a numerical example, let us again consider the Westinghouse conductor

=60 m, D = 0.4 mm, f = 0.02, AHe = 88 mm2, P A = 83 mm). Then

6.33x 105. If the correction factor (8.14.6) is chosen equal to, say, 0.25, then g must

have the value 1.58 x 10–6. The jacket thickness is 1.75 mm. Let us take D´ = 0.1 mm for the purpose of illustration and assume f´ = 0.005 (since the perforations are likely to be fairly smooth inside). Then the porosity s = 1.65 x 10–7. The required spacing d of the holes is given by

Then d = 0.573 m. Thus 100-µm holes spaced every 57 cm in the Westinghouse conductor should reduce the maximum quench pressure, when an entire hydraulic path goes normal, by a factor of 4.

8.15.PERFORATED JACKETS: EFFECTON THE STABILITY MARGIN

The next question is whether the perforations adversely affect the heating-in- duced flow that is responsible for the high upper stability margin. If we treat Eqs. (8.13.5) and (8.13.11) as we treated Eqs. (7.2.2) and (7.2.9) in Section 7.3, they become

We can preserve the entire work of Sections 7.3 and 7.4 by replacing q by the effective value

For the Westinghouse conductor (b = 0.043 K–1, r = 156 kg/m3, cp = 2690 J kg–1 K–1, q = 84 kW/kgHe), the quantity brq/cp = 210 kg m–3 s–1. According to Eq.

(8.14.4), when g = 1.58 x 10–6, m = 2.08 kg m–3 s–1 for p = 0.2 MPa, which is a typical pressure rise accompanying heating-induced flow. According to Eq. (8.15.3), the effective value of q to use in the determination of the heating-induced

174 CHAPTER 8

flow is less by only 1% than the actual value of q. This means that the perforations in the numerical illustration of the last section have virtually no effect on heatinginduced flow, or by extension, on the upper stability margin.

If the porosity of the inner jacket is greatly increased over what has been used in the foregoing illustration, it is likely that the perforations will affect the heatinginduced flow and the stability margin. Just what this effect will be is difficult to say. Imagine, for example, that the cable were disposed as a cylindrical annulus surrounding a large, inner, helium-filled plenum and held in place by a coiled spring. Then there would be almost no hydraulic resistance between the interstitial helium in the cable (region A) and the central plenum (region B). There would likely be strong heating-induced flow transverse to the conductor axis. Luongo et al. (1994, “Helium”) have suggested that such transverse flow might blow the helium out of the cable interstices and into the central plenum very quickly, and there is experimental evidence supporting their contention (Luongo et al., 1994, “Quench”). One presumes the effect on stability of such helium depletion of the cable would be deleterious.

Notes to Chapter 8

1It is shown in Appendix A that y ~ x–2 ~ t 2/dz-2 when x > > 1, i.e., when either z > > 1 or t < < 1 or both. But then v ~ t (a+2)/dz–2 = t z–2, which approaches 0 as t approaches 0.

2The condition for this generalization is that the instantaneous velocity V= dZ/dt should change little

during the time it takes sound waves to cross the length_ Dz of the column of moving fluid, i.e., that IdV/dtl(Dz/c) << V. Since V = nXtn –1, this becomes t >> ln-1l(Dz/c).

3The expulsion velocity has been calculated for early times, where it is small compared with the piston velocity by means of an asymptotic solution to Eq. (8.6.4) valid far from the piston. For a more detailed discussion, see Section 8.11.

4For the sake of completeness, let it be noted that the initial heat pulse was taken to be 10 m long and of 1 ms duration. The exponent m was chosen as 4/15, and the critical current density of pure NbTi at 4.2 K and 8 T was taken to be 1.34 kA/mm2 (from data of Larbalestier, 1986).

Although it may be something of a misnomer, I shall call the transport of heat in turbulent He-II superfluid heat conduction. Likewise, I shall call the coefficient K the superfluid thermal conductivity. When stagnant, subcooled He-II at atmospheric pressure (point P2 in Fig. 1.7) is used as the cryogen in a superconducting magnet, cooling takes place by superfluid heat conduction. Because Eq. (9.1.1) is different from Fourier’s classical law, we cannot use the wealth of known solutions to ordinary heat conduction problems to help us analyze the stability of superconducting magnets. Instead, we must start again at the beginning.

The ordinary diffusion equation for heat is derived by substituting Fourier’s expression for the heat flux q into the heat balance (continuity) equation

where S is the volumetric heat capacity of He-II. If instead we substitute Eq. (9.1.1) for q into Eq. (9.1.2), we find

I shall refer to Eq. (9.1.3) as the superfluid diffusion equation. It is this equation that governs the temperature distribution in turbulent He-II. Because it is strongly nonlinear (owing to the cube root), we cannot solve it with any of the classical methods designed for linear equations, such as separation of variables, expansion in series, or Laplace transforms. It can be treated by the method of similarity solutions, however, and this method will afford us the solutions we shall need.1

9.2.SUPERCONDUCTOR STABILITY: THE METHOD OF SEYFERT ET AL.

Let us consider the following generic problem of superconductor stability: a superconductor is cooled by contact with a closed channel of length L filled with He-II (cf. Fig. 9.1). The thermodynamic state is that denoted by point P2 in Fig. 1.7. Suppose the superconductor is driven normal by a sudden, uniform heat pulse density DH after which it produces a steady Joule heat flux qJ. If DH is small enough, the He-II cools the superconductor well enough to overcome the Joule power and the superconductor recovers. If DH is too large, the He-II cannot cool the superconductor well enough to overcome the Joule power and the superconductor quenches. We want the value of DH that separates these two alternatives, i.e., we want the stability margin.

It will prove convenient in what follows to calculate the related quantity E = (DH)V/A, where V is the volume of superconductor and A its area of contact with He-II. The quantity E is thus the bifurcation energy per unit wetted surface. In order to determine E, we need to calculate the rate of transfer of heat from the superconductor to the helium channel. In its full generality, this is a two-region problem, one

Figure 9.1. A closed, He-II-filled channel of length L in contact at one end with a superconductor (Dresner, 1987). (Redrawn from an original appearing in Dresner (1987) with permission of the IEEE; ©IEEE 1987.)

region being the superconductor, the other the helium. But owing to the peculiar properties of He-II, the problem can be reduced to a one-region problem, as Seyfert, Lafferranderie, and Claudet (1982) have shown.

If the temperature of the superconductor is high enough, the helium contacting it has a temperature higher than the He-I/He-II transition temperature. Thus a layer of liquid He-I (and possibly a vapor layer) forms adjacent to the superconductor, separating it from the He-II filling the channel. The downstream boundary of this He-I layer has the temperature of the line LL´ in Fig. 1.7 (approximately Tl, the lambda temperature of point L). In most practical situations, the heat flux down the channel (and thus across the He-I layer) is a few tens of kW/m2. The temperature difference between the superconductor and Tl is a few Kelvins. Since the thermal conductivity of He-I is ~10–2 W m–1 K–1, the thickness of the He-I layer is ~10–6 m. Thus the He-I/He-II interface is always very close to the surface of the superconductor. Furthermore, owing to the extreme thinness of the layer, it has an extremely small heat capacity (~0.4 J m–2 K–1) and so its width can respond to changes of the heat flux in times of the order of tens of microseconds. Looking back from the He-II-filled tube toward the superconductor, one therefore sees a surface whose temperature is clamped at Tl, and whose locations is, for practical purposes, that of the superconductor surface. This crucially important observation was first made by Seyfert, Lafferranderie, and Claudet (1982), who summarized the situation succinctly as follows: “At the onset of burnout [transition from He-II to He-I at the surface of the heat bath], formation of the thermal barrier starts. The He-II near the heated surface experiences a phase transition. A He-II/He-I interface appears which has its temperature locked at Tl. . . . We assumed that this barrier has a negligible thickness and that it only affected heat transport in He-II by the condition of a constant temperature, i.e., T = Tl, at the hot end of the channels in our test section.”

Accordingly, the temperature distribution in the channel is given by the solution of the one-region problem comprising Eq. (9.1.3), in one dimension,

9.3. SIMILARITY SOLUTION IN A LONG CHANNEL

The problem posed in Eqs. (9.2.1) and (9.2.2) has no simple solution as written; but when L → ∞ and S and K are treated as constants, it has a similarity solution (Dresner, 1984, “Transient”), namely,

The reader can verify by substitution that Eqs. (9.3.1) and (9.3.2) do indeed satisfy the partial differential Eq. (9.2.1) and the boundary and initial conditions (9.2.2). The manner in which Eq. (9.3.1) is obtained in the first place is explained in Appendix A.

According toEq. (9.3.1), the instantaneous heat flux into the superfluid channel q(0,t) is given by

(9.3.3)

This flux represents the cooling capacity of an infinite channel filled with superfluid helium.

Seyfert et al. (1982) have shown how this result can be used to calculate the stability margin E based on the balance of areas shown in Fig. 9.2. In this figure, the ordinate is the heat flux from the conductor into the helium and the abscissa is the time elapsed since the beginning of the heat pulse. The stepped curve depicts the power production in the superconductor. The initial heat pulse E, which has a duration t1, is the first part of the stepped curve. After the time t1 elapses, the superconductor is assumed to be normal and to be producing a steady Joule heat flux qJ (post-heating).

The smooth curve labeled “similarity solution” is the heat flux q(0,t) given by Eq. (9.3.3). At time t2, it crosses the level qJ of the post-heating flux. If the helium has not withdrawn all of the heat produced or deposited in the superconductor by the time t2, the superconductor will not have cooled enough to recover the super-

Figure 9.2. A sketch depicting the balance-of-areas argument of Seyfert et al. (1982). (Redrawn from an original appearing in Seyfert et al. (1982) with permission of Plenum Publishing Corp., New York.)

conducting state. Then the heat flux qJ will persist beyond t2, where it exceeds q(0,t), and the conductor will quench. It is clear from this argument that the largest value of E that still permits recovery is attained when all the heat produced or deposited up to time t2 just equals the heat withdrawn by the superfluid helium. This means that areas A and B in Fig. 9.2 must be equal, i.e., that

0

where t2 = (k/qj )4. If in Eq. (9.3.4) t2 >> t1, we find E = qJt2/3. Combining these last two results with Eq. (9.3.3), we obtain

This result applies to long (strictly speaking, infinite) channels. If the channel is short enough, the temperature profile in it may be taken as uniform and the maximum amount of heat that can be withdrawn from the superconductor without causing a phase transition in the helium is