Current density versus maximum magnetic field curves:

LCC: Who wrote this? ZD to CK? LB? When? Title says TF Coil Conference Minutes…

  1. Are the current density curves for superconductors the actual data that we can use in our studies?

A  Yes, these are experimental curves.

  1. In the HTS worksheet, the superconductor (SC) current density is given in two separate rows. Values in the row 6 are half of the values in the row 5. The row 6 is used for all the calculations. Why?

A  The row 6 includes a safety factor of 0.5.

  1. In the HTS worksheet, the current density of copper is defined as half of the current density of the superconductor. Why?

A  In the high temperature superconductor, the copper is not used for quench protection, only for the protection of the material. To insure the proper protection, the decision was made to use a thickness of copper which is twice as much as the thickness of the superconductor. For example, if the superconductor is 100 mm thick, the copper would be 200 mm thick.

  1. In the worksheets featuring low temperature superconductors, the current density of copper is defined as We have no idea why this is. Should we use this procedure in the systems code?

A  This has to do with how much copper we need to use in order to minimize the heating of material. A magnet typically stores 50-100 GJ of energy. This energy is removed via heat dissipation, which occurs through the copper and needs to be quick enough so that the temperature of the superconductor does not increase by more than 200K. The heat dissipation is proportional to , where is the current density and is the dissipation time. A value of s, chosen in the spreadsheet can be considered sufficiently low for our purposes.

Structural support by sheath and coil casing:

  1. In all the worksheets, the fractional thickness of sheath is obtained by dividing the by the stress in sheath . Is this sufficient to insure that the TF coil will have all the structural support it needs? Please explain.

A  The sheath is not meant to give the primary structural support to the entire TF coil, only to the winding pack. The thickness of the sheath is only about 100 mm, and it is made of NiW or a similar material. The stress in the sheath is partially due to pulling and partially due to compression.

2.  What do we need to do in order to provide a sufficient structural support?

A  For the time being, we can use the equations given in Leslie’s presentation from the June 14 ARIES meeting, page 18:

Here t1 is the thickness of the inboard side, t2 is the thickness of the outboard side, a is the thickness of the top, sm is the allowable stress and the material is the stainless steel 316. The coil casing is to be assumed as square in shape. The side thickness of the casing is much thinner than those defined by t1, t2 and a and can be assumed to be a couple of centimeters.

Regarding the external support, we have two options here: either use a wedged coil or a bucking cylinder. In case we use a bucking cylinder, we can apply the equations above but the bucking cylinder needs to be calculated separately. In case we use a wedged coil, the equations above will need to be modified.

Regarding the shape of the coil, Princeton-D profile would not be a good idea since it is much taller than we need. Instead, we should arbitrarily choose a coil shape that is tightest to the plasma and allows the power core elements to be placed within the coil. Any smooth looking curvature will do.

We can also calculate the ripple, based on Leslie’s presentation at the June 14 (YEAR?) ARIES meeting. A guideline is that the ripple should not exceed 0.1-0.2%.

Fig 1: horizontal cross section of the TF coil in the (r-z plane). In the vertical cross section that corresponds to the plasma major radius, t1 and t2 should be replaced by a, as given in the equations above.

Value of the current in the TF coil

  1. In fields A30 and A31, a value of current of 40 kA was assumed. Is this a reasonable value to use for Tokamak and why? Since the number of Ampere-turns follows from the plasma physics, it seems that one can technically achieve the current of 40 kA by selecting an appropriate number of turns. For the design that uses 16 coils, this would determine the number of turns per coil. The only question is why 40 kA?

A  The choice of 40kA for the current has to do with the ability to dissipate the energy of the magnet out while providing the quench protection (?). It seems like quenching impairs the ability of the superconductor to conduct electricity. In order to prevent this, one has to put lots of copper into the winding pack in order to minimize heating. The idea for using 40 kA came from the experience with designing the Stelarator, which has 36 electrical circles and dumps 50 GJ of energy into an external resistor. The goal was to discharge the magnet as quickly as possible. A value of 20 kV was also mentioned in the conversation. With this voltage, the total electric power would be 20kV*40kA = 0.8 GW. A role of leads during the cooling was mentioned (few kW of cooling due to the leads), which I didn’t understand. It seems that given the voltage, magnet energy and circuit configuration, the choice of 40 kA was necessary in order to prevent the damage to the magnet in the Stelarator. We need to decide which value would be the best for our own studies.