The Ohio State University Hyper Tech Research, Inc. Prospects to further improve Jc of Nb3Sn wires by refining grain size and improving Birr M2OrA-02 The Ohio State University X. Xu M. D. Sumption E. W. Collings Hyper Tech Research, Inc. X. Peng M. Tomsic This work was supported by the U.S. Department of Energy, under an SBIR project.

Motivation Outline: Internal oxidation of Nb3Sn wires. Prospects for further improvement of Nb3Sn strands: Birr ↑ or Fp,max ↑ => Jc ↑ Nb3Sn strands Birr, T Grain size & Fp-B peak at Present level 23-26 T (4.2 K), depending on HT temperature d~100-200 nm 0.2Birr Idealized level ~31 T (0 K) or 27-28 T (4.2 K) ? 15 nm-30 nm? 0.5Birr? Outline: Internal oxidation of Nb3Sn wires. A prediction of prospects of Nb3Sn wire Jc. How to improve Birr of Nb3Sn. A model predicting what determines Sn content of Nb3Sn. 15-30 nm 50-100 nm D. R. Dietderich and A. Godeke, Cryogenics 48, 331 (2008).

A brief review of the application of internal oxidation in Nb3Sn wires “Internal oxidation”: oxygen diffuses into an A-B alloy, and selectively oxidizes the solute B, forming B-O precipitates. Requirement: B is less noble than A. Control sample Internally oxidized Inter-granular ZrO2 particles Intra-granular ZrO2 particles Nb-1Zr-2O ZrO2 Cu-Sn, Nb6Sn5 Nb3Sn Nb-1Zr Internal oxidation of Nb3Sn subelements: I. Use Nb-Zr or Nb-Al. II. Oxide powder supplying O during HT. A tube type subelement: A PIT subelement: RRP: A pending patent filed jointly with Hyper Tech: PCT/US2015/016431.

Layer Jc and Fp-B curves of Internal oxidation strands Nb-1Zr SnO2 Sn core 625 °C x 800 h: HYPER TECH CONDUCTOR Average grain size: 36 nm NbO2 control: average 90 nm Birr, T Grain size, nm Fp-B peak NbO2 -625 °C 20.9 ~90 0.22Birr SnO2 - 650 °C 23 45 0.26Birr SnO2 -625 °C ~20 36 0.34Birr As d >~50 nm, reducing grain size only shifts the curve upward. As d < 50 nm, reducing grain size shifts the curve both to the up and to the right.

The influence of Internal oxidation on Nb3Sn layer growth rate Internal oxidation decreases layer growth rate at low Ts, but enhances it at high Ts: 5 μm 13 μm 2 μm 6.2 μm 750 C – no internal oxidation 750 C – with internal oxidation Possible reasons: Internal oxidation refines the Nb3Sn grain size, it should enhance the diffusion rate. Besides, it reduces the reaction rate. Nb3Sn/Nb interface Nb-1Zr -2O ZrO2 2.5 μm 8 μm 850 C – no internal oxidation 850 C – with internal oxidation 2 μm 10 μm Previous experiments suggested that the Nb3Sn layer growth shifts from reaction-rate limited to diffusion-rate limited as the HT T is increased. This explains why internal oxidation decreases the layer growth rate at low HT Ts but enhances it at high Ts. The layer growth rate is somewhat reduced at low reaction temperatures, so prolonged reaction time is needed to grow a thick Nb3Sn layer. 8 μm 24 μm HYPER TECH CONDUCTOR Internal oxidation, fully reacted

A prediction of the limit of the Jc of Nb3Sn wires I. Present state-of-the-art RRP strands II. The wire with SnO2 - 625 C / 800h III. Only push Birr up to 25 T IV. Only refine the grain size to 25 nm V. Both improve the Birr to 25 T and refine the grain size down to 25 nm d, nm 100 - 120 36 25 Fp-B peak 0.2Birr 0.34Birr 0.5Birr Fp,max, GN/m3 ~90 180 ~250 Birr, T 20 Increase in Birr: The above wire is based on binary Nb3Sn: Birr can be improved by Ta or Ti addition. The above wire is under-reacted: if it is fully reacted, the Birr should be improved. So, how to refine grain size from 36 to 25 nm? Use Nb-1.5Zr alloy to produce more ZrO2 particles. Use a lower reaction temperature (605 °C).

What determines Birr (or Bc2) of Nb3Sn? Bulk samples and thin films: Wires: All wires are 4 at.% Ta doped. Optimized Sn content, but binary. With proper additions, but sub-stoichiometric. Question: if their advantages are combined, i.e., if the Nb3Sn phase has proper additives, and has Sn content of 24-24.5 at.%, could its Bc2 be higher? Answer: yes, which can be seen from the fact that Birr of wires increase with HT temperatures.

What determines the Sn content of Nb3Sn? First, Sn contents of Nb3Sn wires increase with HT temperatures. Second, the Cu-Sn source can influence the Sn contents of Nb3Sn, too. Observation: the bronze-process, RRP, and PIT strands have similar Sn contents (~24 at.%) as Nb3Sn layer is thin, but they have very different Sn content gradients. So, what determines Sn content of Nb3Sn formed in a Cu-Sn/Nb3Sn/Nb diffusion couple? A model is developed for the composition of a non-stoichiometric compound formed by diffusion reactions.

The composition profile of a non-stoichiometric compound In analogy to the system of Nb3Sn, Let us consider such a system: A-B forms a non-stoichiometric compound AnB with a B content range of (Xl - Xu) (Xl < Xu <1). In a diffusion couple of M-B/AnB/A, suppose B is the diffusing species: B atoms diffuse across AnB layer, to the AnB/A interface, where B atoms react with A to form new AnB layers. The solubility of B in A is negligible. M does not dissolve in AnB lattice. No XB gradient in M-B alloy. A B Xu Xl Temperature AnB B at.% 1 A: Nb B: Sn M: Cu A AnB layer AnB/A interface (II) M-B alloy M-B/AnB interface (I) JB l x XB distance M-B A Xu Xl XM-B AnB XII XI Prior to heat treatment While AnB layer is growing I II Note that: 1, let us use a planar geometry here, and suppose x= 0 and l at the interfaces I and II, respectively; 2, we use XI (or μI) and XII (or μII) to represent the XBs (or μBs) of AnB at the interfaces I and II, respectively; 3, XM-B is not necessarily larger than XI: as long as µM-B > µI, M-B can supply B to AnB.

Calculation of the composition profile in AnB AnB layer M-B alloy JB I II l x Fick’s second law: ∂XB/∂t = D∙(∂2XB/∂x2). Mass conservation, dn/dt|I = JI∙AI, and dn/dt|II = JII∙AII. The mass transport rate dn/dt across an interface equals to k∙Aint∙exp(-Q/RT)∙[1-exp(-Δμ/RT)]. For interfaces I and II, Δμ values are μs-μI and μII-μl, respectively, where μs and μl are μB(M-B) and μB(A-Xl B), respectively. Reactant Product Q Δμ Besides, since the B atoms diffusing to the interface II are used form new AnB layers, we have: dl/dt = JII∙VmAnB/XII = -D/XII∙(∂XB/∂x)|x=l. From this equation, we can calculate the layer growth rate of AnB. Summary of the equations: ∂XB/∂t = D∙(∂2XB/∂x2) (1) 1-exp[-(μs-μI)/RT] = -(D/φI)∙(∂XB/∂x)|I (2) 1-exp[-(μII-μl)/RT]}= -(D/φII)∙(∂XB/∂x)|II (3) dl/dt = -(D/XII)∙(∂XB/∂x)|x=l (4) Let us denote φI and φII represent the interface reaction rates, with unit of m/s. Equations (1)-(4) are the governing equations for the said system. Solutions of equations (1)-(4) give both the composition profile XB(x, t) and layer growth rate l(t) of AnB. However, to solve the equations, we need first to know the μ(X) relation for Nb3Sn.