A collector for use in removing metal ions from a plasma in a vacuum chamber includes a collector plate that is mounted inside the chamber and formed with an internal cooling channel. An injector introduces a dissociated salt into the chamber with a first throughput value, and it introduces a plasma including metal ions into the chamber with a lower second throughput value. A pump is used to pump a liquid coolant through the cooling channel to maintain the collector plate at a temperature that forms a portion of the salt as a protective layer on the collector plate, and causes the salt to thereafter deposit on the layer in a molten condition at a faster rate than evaporation therefrom to trap metal ions therein. The trapped metal ions are then removed with the molten salt from the chamber.

The present invention pertains generally to methods and devices for collecting ions from a multi-species plasma. More specifically, the present invention pertains to methods and devices for collecting ions from a multi-species plasma after the ions of the plasma have been separated according to the respective masses of the constituent elements. The present invention is particularly, but not exclusively, useful for collecting the relatively low mass particles in a multi-species plasma.
A plasma mass filter that is designed to separate low mass particles from high mass particles is disclosed in U.S. Pat. No. 6,096,220 which issued to Ohkawa for an invention entitled xe2x80x9cPlasma Mass Filterxe2x80x9d and which is assigned to the same assignee as the present invention. U.S. Pat. No. 6,096,220 is incorporated herein by reference. In overview, a plasma mass filter includes a substantially cylindrical shaped wall which surrounds a hollow chamber. A magnetic field is generated in the chamber which is generally axially oriented and an electric field is generated within the chamber which is oriented substantially perpendicular to the longitudinal axis of the chamber. Importantly, for operation of a plasma filter, the electric field has a positive potential on the axis relative to the wall which is usually at a zero potential. When a multi-species plasma is injected into the chamber, the plasma interacts with the crossed electric and magnetic fields, resulting in the rotation of the plasma about the chamber axis. In response to the crossed electric and magnetic fields, each ionized or charged particle in the multi-species plasma will travel on a predictable trajectory about the axis as it transits the chamber. The particular trajectory is dependent on the mass to charge ratio of the orbiting particle. With this in mind, the plasma mass filter is designed so that high mass particles will travel on unconfined orbits that allow the high mass particles to strike and be collected on the wall of the chamber. On the other hand, the low mass particles will have orbits that are smaller than the chamber radius, and hence are confined inside the chamber so as not to strike the chamber walls. Thus, the orbiting low mass particles eventually exit the chamber at one end of the cylinder where they need to be collected.
When salt vapors, such as sodium hydroxide vapors (NaOH), are introduced into the vacuum chamber, along with the multi-species plasma discussed above, a mechanism for collecting the low mass particles (light ions) in the plasma can be provided. To appreciate how this can be accomplished, certain characteristics of sodium hydroxide need to be considered. Specifically, these characteristics need to be considered in a context where ions derived from a salt, such as sodium hydroxide (NaOH) come into contact with a solid wall or plate. For this consideration, reference here is first made to FIG. 1.
With reference to FIG. 1, it is known that when plasma ions derived from a salt come into contact with a solid wall 10 they will be neutralized and reform into a salt. If the wall 10 is cold enough (e.g.  less than 100xc2x0 C.), the salt will deposit faster than it evaporates and a layer of the salt will grow. Initially, the salt will deposit as a solid, forming a protective layer on the wall. However, as this layer grows, its surface temperature will increase, since salts are good insulators and there is a heat load on the wall from the plasma. Eventually, however, the surface will melt to create a molten layer 12. Deposition of additional salt will increase the thickness of the molten layer 12. In a steady state, molten salt will flow away at the same rate as the plasma deposits it. In addition, other ion species from the plasma will be neutralized and collected. Under these conditions, if the throughput of ions from other species is small enough, compared to the throughput of ions from the salt, then the other species will be incorporated into the molten layer. Thus, the other species will either be dissolved in the molten salt or will form micro-crystals that will be carried along by the molten salt. This provides a mechanism of draining both the molten salt and the other species off of the collector.
It happens that the draining of the molten salt off of the collector depends on the rate at which the molten salt is deposited. If the deposition rate is small, then gravity is the primary force driving the flow of the molten salt. In this case, the molten salt can be drained off the bottom of a flat collector plate. On the other hand, if the deposition rate is high, then the primary force driving the flow of the molten salt is the xe2x80x9cplasma windxe2x80x9d force that arises because the molten layer absorbs momentum from the plasma. In that case, the molten salt can be drained off the side of a vertical collector plate. Or, the collector can be formed as a series of concentric cones and the molten layer can be drained off the back of the cones.
For the conditions mentioned above, the molten salt collector works best on salts that have a low vapor pressure in the molten state such as sodium hydroxide (NaOH). Accordingly, sodium hydroxide is used only as an example herein.
As mentioned above, if the deposition rate of the sodium hydroxide is relatively small, gravity is the primary force driving the molten layer. For example, if the plasma deposits sodium hydroxide onto a flat collector plate that is tilted at an angle xcex1 with respect to vertical, as shown in FIG. 1 in a steady state, the gravitational force is balanced by the viscous drag:       0    =                  ρ                          xe2x80x83                        g                          xe2x80x83                        cos                          xe2x80x83                        α            +              η                                                    ∂              2                                    u                                ∂                          z              2                                            ,
where xcfx81 is the mass density, xcex7 is the viscosity, and u is the velocity of the molten NaOH along the collector plate. Inertial terms in the fluid equations have been neglected. Solving this equation with the boundary conditions u(z=0)=0 and ∂u/∂z(z=d)=0 gives:             u                    (        z        )              =                            ρ                                xe2x80x83                              g                                xe2x80x83                              cos                                xe2x80x83                              α                η                                    d          2                                  (                                    z              d                        -                                          z                2                                            2                                                  d                  2                                                              )                      ,
where d is the local thickness of the molten layer 12.
The average axial velocity is then:       u    _    =                    1        d                                    ∫          0          d                                  u                                ⅆ            z                                =                            ρ                                xe2x80x83                              g                                xe2x80x83                                          d            2                                          xe2x80x83                              cos                                xe2x80x83                              α                          3                    η                    .
The thickness of the molten layer, d, is determined by conservation of mass:                     ∂                  ∂          y                                  (                  ρ                                xe2x80x83                                          u            _                                          xe2x80x83                              d                )              =          m                    xe2x80x83                  Γcosα        ,
where y is the distance along the plane of the collector, xcex93 is the ion flux, and m is the average ion mass. Solving this equation gives:                                           d            =                                                            (                                                                                3                                                η                                                                          xe2x80x83                                                                        m                                                                          xe2x80x83                                                                        Γ                                                                                ρ                          2                                                                        g                                                                                y                                    )                                                  1                  /                  3                                            ≈                              0.053                                                                      (                                          y                                              R                        w                                                              )                                                        1                    /                    3                                                                                xe2x80x83                                                mm                                                                    xe2x80x83                                                        u          =                                                                      (                                                                                g                                                                          xe2x80x83                                                                                m                          2                                                                                Γ                          2                                                                                3                                                ρη                                                                                y                      2                                                        )                                                  1                  /                  3                                                          cos                                            xe2x80x83                                          α                        ≈                          0.42                            cos                                            xe2x80x83                                                                            α                                                        (                                                                  y                        2                                                                    R                        w                        2                                                              )                                                                    1                  /                  3                                                                          xe2x80x83                                          cm                                            /                                          s                                            ,
where it is assumed that d=0 and hence {overscore (u)}=0 at y=0. Also, the following values for the parameters have been used:
Note that in this example the thickness d is not explicitly dependent on the tilt angle xcex1. This is because the reduction in the driving force is cancelled by the spreading out of the ion flux. Assuming that y less than 8RW, we find that d less than 0.1 mm. The flow velocity is less than 1 cm/s, indicating that it takes around a minute for a deposited atom to leave the collector plate. In this case, a trough can be placed below the collector plate to catch the molten sodium hydroxide.
As the plasma imparts a heat load on the molten NaOH layer, the surface temperature of the NaOH is given by:
Ts=Tm+qd/xcexa,
where Tm=322xc2x0 C. is the melting temperature of sodium, q≈0.5 MW/m2 is the heat load from the plasma, and xcexa≈1 W/m/K is the thermal conductivity of the molten NaOH. For d=0.11 mm the surface temperature is TS=372xc2x0 C., still well below the vaporization temperature of NaOH. Note that the viscosity decreases with temperature, but d is only weakly dependent on viscosity.
Unlike the conditions just described, if the deposition rate of sodium hydroxide is high, then the xe2x80x9cplasma windxe2x80x9d forces become the primary forces driving the molten layer. In this condition, the current density in the molten layer, j, crossed with the magnetic field, B, results in a Lorentz force on the molten layer that is small compared to the plasma wind since the molten NaOH is more resistive than the plasma. With this in mind, the plasma wind force arises because the molten layer 12 absorbs momentum from the flowing plasma. If the collector plate is vertical, then the rotation of the plasma drives the molten layer in the azimuthal direction. Averaged over the depth of the molten layer, the plasma wind force density is:             f      p        =                            m                                xe2x80x83                              Γω                d                          (                                            -              y                                                  xe2x80x83                                                  x              ^                                +                      x                                      xe2x80x83                                                  y              ^                                      )              ,
where xcfx89 is the rotation frequency of the plasma and the (x=0, y=0) refers to the axis of the machine. In this case it is assumed the rotation frequency and ion flux are uniform. Force balance in the x-y plane is then given by:                                                         ρ                                            xe2x80x83                                                          u                ·                                  ∇                                      u                    x                                                                        =                                                            -                                                            m                                                                    xe2x80x83                                                                  Γω                                        d                                                                  y                            +                              η                                                                                ∂                      2                                                                                u                      x                                                                            ∂                                          z                      2                                                                                xe2x80x83                                                                    ρ                                      xe2x80x83                                                  u              ·                              ∇                                  u                  y                                                              =                                                    +                                                      m                                                              xe2x80x83                                                            Γω                                    d                                                          x                        +                          η                                                                                ∂                    2                                                                          u                    y                                                                    ∂                                      z                    2                                                                        -                          ρ                                            xe2x80x83                                          g                                            ,
where gravity points in the xe2x88x92y direction. Assuming that d is uniform, we can remove gravity from the equations by defining       x    xe2x80x2    =      x    -                            ρ                                xe2x80x83                              g                                xe2x80x83                              d                          m                                xe2x80x83                              Γω                    .
The equations then become:                                           ρ                                      xe2x80x83                                                  u              ·                              ∇                                  u                                      x                    xe2x80x2                                                                                =                                                    -                                                      m                                                              xe2x80x83                                                            Γω                                    d                                                          y                        +                          η                                                                                ∂                    2                                                                          u                                          x                      xe2x80x2                                                                                ∂                                      z                    2                                                                                ρ                                      xe2x80x83                                                  u              ·                              ∇                                  u                  y                                                              =                                                    +                                                      m                                                              xe2x80x83                                                            Γω                                    d                                                                          x                xe2x80x2                                      +                          η                                                                                ∂                    2                                                                          u                    y                                                                    ∂                                      z                    2                                                                                .
To solve these equations, three things must be done. These are: first, convert to polar coordinates with the origin defined as (xxe2x80x2=0, y=0); second, neglect all inertial terms except centrifugal force; and third, change the form of the plasma wind force density to reflect the fact that it does not act volumetrically on the molten layer. Rather, it acts only on the surface. The equations are now:                                                                         -                ρ                                                                            u                  θ                  2                                r                                      =                          η                                                                                ∂                    2                                                                          u                    r                                                                    ∂                                      z                    2                                                                                xe2x80x83                                                        0          =                                                                      m                                                        xe2x80x83                                                      Γω                                a                                                                            exp                                                        (                                          z                      /                      a                                        )                                                                                exp                                                              (                                              d                        /                        a                                            )                                                        -                  1                                                          r                        +                          η                                                                                ∂                    2                                                                          u                    θ                                                                    ∂                                      z                    2                                                                                .
Here it is assumed that the plasma wind force increases exponentially with z towards the surface of the molten layer. Solving for uxcex8 in the limit where d greater than  greater than xcex1 gives:             u      θ                  (      z      )        =                    m                          xe2x80x83                        Γω                          xe2x80x83                        r            η                  z      .
The azimuthal velocity is just proportional to z. It is now possible to solve for the radial flow that is driven by the centrifugal force:       u    r    =                    ρ                          xe2x80x83                                  d          3                            3                η                                (                              m                                      xe2x80x83                                    Γω                    η                )            2            r              xe2x80x83                          z                          (                      1            -                                          z                3                                            4                                                  d                  3                                                              )                    .
The z-averaged velocities are:                                                                         u                _                            θ                        =                                                            m                                                        xe2x80x83                                                      Γω                                                        xe2x80x83                                                      d                                                  2                                    η                                                          r                                                      xe2x80x83                                                                                u              _                        r                    =                                                    3                                ρ                                                  xe2x80x83                                                                  d                  4                                                            20                                η                                                                                (                                                      m                                                              xe2x80x83                                                            Γω                                    η                                )                            2                                    r                                .
As before, the thickness of the layer is found using mass conservation:                     1        r                          ∂                  ∂          r                                  (                  ρ                                xe2x80x83                                                        u              _                        r                              d                )              =          s                    xe2x80x83                  m                    xe2x80x83                  Γ        ,
where s is the sticking coefficient of the incoming ions. Ions which do not stick are assumed to still impart their full momentum to the melt.
Solving the equation with typical numbers, and using a sticking coefficient of s=1 gives:                                           d            =                                                            (                                                            10                      3                                                                                s                                                                          xe2x80x83                                                                                η                          3                                                                                ρ                          2                                                                        m                                                                          xe2x80x83                                                                        Γ                                                                          xe2x80x83                                                                                ω                          2                                                                                )                                                  1                  /                  5                                            ≈                              0.15                                                  xe2x80x83                                                mm                                                                    xe2x80x83                                                                                u                _                            θ                        =                                                                                (                                                                  5                        48                                                                                s                                                                                xe2x80x83                                                                                m                            4                                                                                Γ                            4                                                                                ω                            3                                                                                R                            w                            5                                                                                ρ                            2                                                                                η                            2                                                                                )                                                        1                    /                    5                                                                                r                                      R                    w                                                              ≈                              25.1                                                  r                                      R                    w                                                                                xe2x80x83                                                cm                                                  /                                                s                                                                    xe2x80x83                                                                                u              _                        r                    =                                                                      (                                                            3                      320                                                                                s                          4                                                                                m                          6                                                                                Γ                          6                                                                                ω                          2                                                                                R                          w                          5                                                                                ρ                          3                                                                                η                          3                                                                                )                                                  1                  /                  5                                                                          r                                  R                  w                                                      ≈                          0.061                                            r                                  R                  w                                                                          xe2x80x83                                          cm                                            /                                          s                                            .
Above, it is assumed that xcfx89=34200 rad/s. Note that the azimuthal velocity is much faster than the gravity driven flow. From the values of uxcex8 and ur it can be seen that the molten NaOH flows in a tight spiral; the streamlines circle the origin over 100 times between r=0 and r=Rw. Gravity acts to shift this spiral in the +x direction a distance:       Δ              xe2x80x83            x    =                    ρ                          xe2x80x83                        g                          xe2x80x83                        d                    m                          xe2x80x83                        Γ                          xe2x80x83                        ω              =                            (                                    10              3                                                                  s                                                  xe2x80x83                                                                  g                  5                                                                  ρ                  3                                                                  η                  3                                                                              m                  6                                                                  Γ                  6                                                                  ω                  7                                                              )                          1          /          5                    ≈              12                          xe2x80x83                                  cm          .
At (x, y)=(xcex94x, 0), the plasma wind force balances the gravitational force, so the velocity is zero at this point. Because of this shift, the molten sodium hydroxide can be drained off of the collector at the point where the edge of the collector intersects the x-axis.
For the specific condition wherein the collector is conical section 14, then the plasma wind force along the magnetic field has a component along the surface of the cone that will drive the molten NaOH backwards and out radially (see FIG. 2). Ignoring gravity and the centrifugal force, the force balance equations are:                               0          =                                                                      m                                                        xe2x80x83                                                      Γ                                                        xe2x80x83                                                                          v                    b                                                  a                                                                            exp                                                        (                                          z                      /                      a                                        )                                                                                exp                                                              (                                              d                        /                        a                                            )                                                        -                  1                                                          sin                                            xe2x80x83                                          α                        +                          η                                                                                ∂                    2                                                                          u                    y                                                                    ∂                                      z                    2                                                                                0            =                                                                                m                                                              xe2x80x83                                                            Γ                                                              xe2x80x83                                                            ω                                    a                                                                                exp                                                              (                                              z                        /                        a                                            )                                                                                exp                                                                    (                                                  d                          /                          a                                                )                                                              -                    1                                                                  r                            +                              η                                                                                ∂                      2                                                                                u                      θ                                                                            ∂                                          z                      2                                                                                xe2x80x83                                ,
where ŷ=cos xcex1{circumflex over (r)}+sin xcex1{circumflex over (b)} is now the radially outward direction along the surface of the cone, {circumflex over (b)} is the direction of the magnetic field, xcex1 is the angle of the cone with respect to vertical (xcex1=0 is a vertical plate), z is the direction perpendicular to the surface of the cone, and xcexdb≈104 m/s is the plasma velocity parallel to the magnetic field. Solving the equations, letting axe2x86x920, and z-averaging gives:                                                                         u                _                            θ                        =                                                            m                                                        xe2x80x83                                                      Γ                                                        xe2x80x83                                                      ω                                                        xe2x80x83                                                      r                                                  2                                    η                                                          d                                                      xe2x80x83                                                                                u              _                        y                    =                                                    m                                                  xe2x80x83                                                Γ                                                  xe2x80x83                                                                  v                  b                                                            2                                η                                                  d                                      xe2x80x83                                    sin                                      xe2x80x83                                    α                                .
Again, the thickness d is found by using conservation of mass.
2xcfx80rdxcfx81{overscore (u)}y=mxcex93xcfx80(r2xe2x88x92r02),
where r0 is the radius of the leading edge of the cone. Solving the equations with the boundary condition d(r0)=0 gives:                                           d            =                                                            η                                                        (                                                                  r                        2                                            -                                              r                        0                        2                                                              )                                                                    ρ                                                        xe2x80x83                                                                          v                    b                                                      r                                                        xe2x80x83                                                      sin                                                        xe2x80x83                                                      α                                                                                xe2x80x83                                                                                u                _                            θ                        =                                                            m                                                        xe2x80x83                                                      Γ                                                        xe2x80x83                                                      ω                                2                                                                                r                                                              (                                                                        r                          2                                                -                                                  r                          0                          2                                                                    )                                                                            ηρ                                                              xe2x80x83                                                                                v                      b                                                            sin                                                              xe2x80x83                                                            α                                                                                xe2x80x83                                                                    u            y                    =                                                    m                                                  xe2x80x83                                                Γ                            2                                                                                v                      b                                                                                (                                                                        r                          2                                                -                                                  r                          0                          2                                                                    )                                                                          sin                                                        xe2x80x83                                                      α                                                  ηρ                                                        xe2x80x83                                                      r                                                                          .
The maximum value of the thickness and flow velocity is reached where |rxe2x88x92r0| is maximum. Defining D=max|rxe2x88x92r0, and assuming D greater than  greater than r0, the maximum values are given by:             max                    xe2x80x83                  d        =                                        2                        η                                      xe2x80x83                                    D                                              ρ                                            xe2x80x83                                                          v                b                                      |                          sin                                            xe2x80x83                                          α                        |                              ≈              6.8        xc3x97                  10                      -            3                                            xe2x80x83                        mm                        max                    xe2x80x83                                    u          _                θ              =                                        m                                      xe2x80x83                                    Γ                                      xe2x80x83                                    ω                    2                                                            2                            D                                            xe2x80x83                                                          R                w                2                                                                    ηρ                                                  xe2x80x83                                                                  v                  b                                            |                              sin                                                  xe2x80x83                                                α                            |                                          ≈              1.12                          xe2x80x83                        cm                          /                        s                        max                    xe2x80x83                                    u          _                y              =                                        m                                      xe2x80x83                                    Γ                    2                                                                            2                                                  xe2x80x83                                                                  v                  b                                                D                            |                              sin                                                  xe2x80x83                                                α                            |                        ηρ                              ≈              0.54                          xe2x80x83                        cm                          /                        s
The +/xe2x88x92 sign refers to the sign of the angle xcex1. In the limit that r0≈Rw greater than  greater than D, L=3D, sin xcex1={square root over (9/10)}, and D=0.1 m, the maximum values of d and u are independent of the sign of xcex1 and have the values shown. Note that the velocity at which the NaOH leaves the collector, uy, is much larger than in the flat plate cases discussed above. This is because the plasma wind directly contributes to the flow of the waste off of the collector. The sodium hydroxide can be drained off of the collector at the back edge 16 of the conical section 14.
In light of the above it is an object of the present invention to provide a molten collector which can be incorporated into the structure of a plasma mass filter for the collection and removal of low mass particles from a multi-species plasma. It is another object of the present invention to provide a molten collector that allows for the efficient and uninterrupted removal of low mass particles from the collector during a continual operation of the filter. Still another object of the present invention is to provide a molten collector which uses the xe2x80x9cplasma windxe2x80x9d force to direct the removal of low mass particles in a molten salt from a plasma mass filter. Yet another object of the present invention is to provide a molten collector which is easy to use, relatively simple to manufacture, and comparatively cost effective.
A device for collecting metal ions from a plasma and then removing them from a vacuum chamber includes at least one collector plate that is positioned inside the chamber. Preferably, there are a plurality of such collector plates and each collector plate is formed as a generally truncated conical section. In particular, all of the conical sections are concentrically mounted in the vacuum chamber relative to a defined axis of the chamber, and they are placed in a substantially coplanar arrangement. Also, the conical sections are angled relative to the axis with a separation space between them. In this arrangement, the conical sections overlap each other so as to effectively screen a cross section of the chamber. Importantly, each conical section is formed with an internal cooling channel.
The device of the present invention also includes a fluid pump that is connected in fluid communication with the cooling channels of the conical sections. For the purposes of the present invention, the pump is used for pumping a liquid coolant, such as water, through each of the cooling channels. The purpose here is to maintain each of the conical sections at a temperature of approximately 100xc2x0 C. during the operation of the vacuum chamber.
An injector, mounted on the chamber, is used to introduce a salt, which may or may not be vaporized, and a multi-species plasma into the chamber. Specifically, the salt is preferably sodium hydroxide, and the multi-species plasma will be generated with oxides from a group which may include, but is not limited to, aluminum oxide, silicon oxide, calcium oxide, iron oxide, chromium oxide and uranium oxide. Insofar as the salt is concerned, if sodium hydroxide (NaOH) is used, it is expected that the salt will be fully dissociated in sodium (Na), oxygen (O), and hydrogen (H) atoms. Further, the sodium atoms, and possibly the oxygen and hydrogen atoms, will be ionized along with the multi-species plasma.
In the operation of the present invention, the salt (e.g. sodium hydroxide) is introduced into the chamber by the injector so as to have a predetermined throughput. Further, as envisioned by the present invention the plasma will be introduced into the chamber as a multi-species plasma that includes both relatively light metal ions of low mass/charge ratios (M1) and relatively heavy metal ions of high mass/charge ratios (M2). Importantly, the multi-species plasma is introduced into the chamber by the injector so that the metal ions in the plasma will have a predetermined throughput that is less than the throughput value of the salt.
Initially, due to the temperature differential between the plasma and the cooled collector plate inside the vacuum chamber, as the salt collides with the collector plates (conical sections) they will reform to create a solid deposit on the collector plates. With a continued build-up of solid salt on the collector plates, however, and the consequent thermal insulation that results, subsequent deposits of salt on the now-insulated collector plates will not completely solidify. Instead, this salt (sodium hydroxide) will assume a molten state. Importantly, the molten state is maintained at a temperature that causes the salt to deposit on the collector plates at a higher rate than it will evaporate from the collector plates. For some conditions, this may require the use of heaters.
As indicated above, along with the salt (sodium hydroxide), both light ions (M1) and heavy ions (M2) of the plasma will be inside the chamber. For the purposes of the present invention, it is envisioned that the light ions and heavy ions will be separated from each other while they are in the vacuum chamber, such as by the use of plasma mass filter techniques disclosed by Ohkawa in U.S. Pat. No. 6,096,220. Specifically, in accordance with these techniques, the light metal ions and the heavy metal ions are directed onto different trajectories inside the chamber. More specifically, the heavy ions are directed into contact with the chamber wall, where they are subsequently collected. On the other hand, the light metal ions are directed onto trajectories that cause them to transit through the chamber. Accordingly, the collector plates (conical sections) of the device of the present invention are positioned so that the light metal ions will collide with the collector plates after passing through the chamber. In these collisions, the light metal ions become embedded in the molten sodium hydroxide that has become deposited on the collector plates. Then, the molten sodium hydroxide, along with the light metal ions that are trapped therein, are caused to be removed from the vacuum chamber under the influence of either the plasma wind in the chamber, or gravitational forces. Further, by properly aligning the angle of the conical sections (collector plates) with respect to the plasma wind, the plasma wind force can be used to channel the molten material to a desired location.

Molten salt collector for plasma separations


An electrical energy transmission device has phase conductors, which carry alternating current and have transfer impedance, and sheathed conductors, which are inductively coupled to the phase conductors. A first end and a second end of each sheathed conductor form a sheath circuit with a reactance. An electronic assembly for changing the impedance of the sheath circuit is provided. The electrical energy transmission device can be used flexibly given different demands placed on the energy transmission. The electronic assembly is also configured to increase the transfer impedance.

1. Field of the Invention:
The invention relates to an electrical power transmission device having phase conductors which carry alternating current and have a transmission impedance, and having casing conductors which are inductively coupled to the phase conductors, with a first end and a second end of each casing conductor together with a reactance forming a casing circuit, and with an electronic assembly being provided in order to vary the impedance of the casing circuit.
Cables or gas-insulated lines (GIL) are used for power distribution purposes, in order to transmit large amounts of energy between different grid systems. GILs comprise phase conductors which extend in tubular and gas-insulated casing conductors, and, because of their low transmission impedance, offer the capability to transmit power levels of several thousand megawatts with low losses. However, despite this low transmission impedance, the distribution of high-power currents between grid systems may be adversely affected, for example, when such GILs are integrated.
DE 199 37 661 A1 discloses a circuit arrangement of the generic type for a three-phase transmission system in which a casing conductor is provided for each phase conductor. A capacitive reactance is connected in series with the casing conductor and is used to compensate for the series impedance of the phase conductor. Compensation such as this reduces or minimizes the wattless component in the transmission system.
Furthermore, DE 199 37 661 A1 proposes that a plurality of capacitors be connected or disconnected as a function of the load on the transmission system. This circuit arrangement has the disadvantage that the impedance can only be reduced for power-factor correction.


Biomaterial medical devices or products, e.g., absorbable sutures or joint prostheses components, are gamma irradiation sterilized in the substantial absence of oxygen (e.g., vacuum treatment to 5.times.10.sup.-6 torr) while being maintained at a temperature about that of liquid nitrogen, with improved strength loss resistance compared to gamma irradiation sterilization under ambient conditions.

Synthetic absorbable sutures composed of biodegradable biomaterials including polyglycolic acid (e.g., sold under the tradename Dexon), copolymer of glycolide and lactide (e.g., sold under the tradename Vicryl), poly-p-dioxanone (sold under the tradename PDSII), copolymer of glycolide and trimethylene carbonate (sold under the tradename Maxon) and copolymer of glycolide and epsilon-caprolactone (sold under the tradename Monocryl) are currently sterilized by gas (ethylene oxide) sterilization because of the known adverse effect of gamma irradiation sterilization on the mechanical properties and hydrolytic degradation rate of these biomaterials. Gas sterilization is time consuming and costly. The toxicity of residual amounts of ethylene oxide in medical devices and products has been a concern and degassing is a long and tedious, costly process. The medical industry has expressed a desire to replace ethylene oxide sterilization for absorbable biomaterials, e.g., absorbable sutures, with gamma irradiation sterilization if the latter method would not significantly increase the strength loss during use of the biomaterials. However, all reported data from conventional gamma irradiation of synthetic polymeric absorbable biomaterials indicates that gamma irradiation sterilization of synthetic polymeric absorbable biomaterials would be unacceptable.
There is a similar problem for non-absorbable synthetic polymeric biomaterials, e.g., in the case of acetabular or tibia components of joint protheses made of ultra high molecular weight polyethylene, which are disadvantageously weakened by conventional gamma irradiation to the extent that gas sterilization has been required.

Gamma irradiation sterilizing of biomaterial medical devices or products, with improved degradation and mechanical properties


An extractor and a ventilation control window are coupled to the housing of a sheet delivery conveyor to extract unwanted heat, moisture, volatile vapors and obnoxious odors from the conveyor housing to eliminate the need for a separate venting system above the sheet delivery stacker. The suction airflow is varied by adjusting the speed of a vacuum source or motor driven fans, or by adjusting a ventilation window. A sheet control ventilation window is covered by a slidable, transparent panel which permits the operator to observe the orientation of the freshly printed sheets as the suction airflow is adjusted to precision. The sheet control window is also covered by a slidable screened panel which prevents introduction of objects into the press. Volatile vapors, moisture laden air and the like are also extracted from laterally opposite sides of the sheet delivery path, thus helping to control air turbulence at the delivery sheet stacker.

It has been traditional in the art of sheet-fed printing presses to provide systems for supporting freshly printed sheets when transferring the sheets from one printing unit to another or when handling the sheets as they are transferred by a press delivery system from the last printing unit to a sheet delivery stacker. A sheet transfer system comprises a support roller or cylinder disposed between one or more printing units in the press and which functions to receive a freshly printed sheet from one impression cylinder and transfer the sheet to the next printing unit for additional printing. The press delivery conveyor system usually includes chain driven gripper bars which receive the freshly printed sheets from the last impression cylinder of the press and deliver the sheets to the press delivery stacker.
Because the inks used with offset printing presses typically remain wet and tacky for some time, marking and smearing of the freshly printed ink is a concern in all sheet transfer and delivery systems. When transferring a sheet between printing units, marking or smearing of the printed side of the sheet is often caused by a fluttering motion of the sheet as it transfers through a reverse curvilinear path from the impression cylinder to the next transfer cylinder.
Turbulent air movement is caused by a delivery venting system which extracts moisture, volatile vapors and odors released from the freshly printed and/or coated sheets. Such delivery venting systems typically include a hood that is mounted above the delivery sheet stacker through which air is drawn up from the vicinity of the press delivery stacker. The resulting turbulent air flow in the delivery area of the press often causes fluttering motion of the sheets as they are released over the sheet delivery stacker. Moreover, after the grippers release, free fall of the sheet is retarded by the updraft of the delivery venting hood so that the trailing edge portion of a sheet floats momentarily, then contacts the next gripper bar assembly, thus resulting in a sheet jam-up in the delivery stacker.

Delivery conveyor with control window ventilation and extraction system