Use of Simple Analytic Expression in Tokamak Design Studies John Sheffield, July 29, 2010, ISSE, University of Tennessee, Knoxville Inspiration Needed

Use of Simple Analytic Expression in Tokamak Design Studies In designing future tokamaks it is necessary to use a portfolio of computer codes, including what Pat Diamond describes as the codes that ate Manhattan Nevertheless, considerable insight can be obtained analytically--using studies such as ITER and ARIES-AT to benchmark the results. Of course, this, like other systems codes, only produces a sketch until a fully engineered design, supported by R&D, is completed.

Important Parameters Fusion power P F (MW) Plasma profiles n = n 0 (1-r 2 /a 2 )  n and T = T 0 (1-r 2 /a 2 )  T Fusion fuel fraction n D /n e, where n D = n T (10 20 m-3 ) and the impurities ∑n Z Z. f G the fraction of the Greenwald limit used. Greenwald limit n G = I/(  a 2 )(10 20 m -3 ) T = T e = T i (10keV)— these are volume averaged  = a/R, , d, q 95,  the gap between the plasma and inner leg of the TF coil, and f a = a w /a the scrape-off layer thickness (assumed to be constant around the plasma). B m (T) the maximum field on the coil and R m (m) is TF coil inner leg outer radius I = 2.5(BR)  2 f 1 /q 95 (MA) BR = B m R m where f 1 = {1 +  2 (1 + 2  2 – 1.2  3 )}{1.17 – 0.65  }/{1 -  2 } 2

Fusion Power P F = f(  n,  T ) f (q 95 )-2 f 2 2 (f 3 f 4 ) -2  N 2  4  (BR) 4/ R (MW) f(  n,  T ) = {(1 +  n ) 2.(1 + 2  n + 3  T ) 2 / (1 + 2  n + 2  T ) 3 } x{(1 +  n +  T ). (1 + 2  n + 2  T )} 2 /{(1 +  n ).(1 + 2  n + 3  T )} 2 The beta limit is given by =  N I/aB (%) ∑ = n e T.f 3.f 4 where f 3 = (1 +  n )(1+  T )/ {(1 +  n +  T ) and f 4 = f 2 +  / R = R m /[1 – (1 +f a )  -  /R] (m)

Confinement and Power to Sustain Plasma ITER H-mode scaling is  E = H H I 0.93 B 0.15 n 0.41 M 0.19 R 1.97  0.58  X 0.78 P = 4.737Ra 2  /P (s) Note that in the ITER-FEAT studies  X /  ≈ 1.1, and with M = 2.5 P = [1.023 H H -1 f G  N I -0.34( BR) 0.85  1.24  0.22 ] 3.23 (MW) For Q =20, P a = P/5 and 0.2P F /P = 0.8 This kind of level is needed for Qeng ~ 1 in a Pilot Plant. e.g., P F =600 MW, P = 150 MW, and P a = 30 MW

Balancing Alpha Power Against Power Requirement With the formulae for P F, P, I, and f G < 1, and T = 1 ( 10 keV ) 0.2P F /P = 4.43 x 10-3 (Plasma)x(Shaping)x(Impurities)x(Size) Plasma Parameters H H 3.23  N f(  n,  T )/(q f f ) Shaping f  3.5  0.29 Impurity Depletion (n D /n e ) 2 Size (BR) 3.68 /R where f 1 = {1 +  2 (1 + 2  2 – 1.2  3 )}{1.17 – 0.65  }/{1 -  2 }2

Example Case with R/a =3.5,  = 2.2,  = 0.8, q 95 = 3.7 B m =10 T,  = 0.8, R = 4.23 m, a = 1.21 m, H H = 1.1  n = 0.4,  T = 1.0, I = 12.2 MA, V.s = 144 Volume averaged temperature ≥ 1 (10 keV)  F ≈ 0.2(T/1) 1.5 /(1 - I BS /I – I CD /I) hrs I BS /I ≈ 0.24  N 0.79, I CD /I ≈ P aux T/f G, P F = 64  N 2 MW, P elg = 0.3(1.1P F + P aux ) MW e Preq = P F + P aux /0.4 (see ITER) P = 0.75  N MW (see ARIES-AT) f G =  N  0.9, n G = 2.65 (10 20 m-3)

NN P F (MW) P  (MW) P (MW) P aux (MW) P/R (MW/m) P/A w (MW.m -2 ) P n /A w (MW.m -2 )

NN fGfG I BS /II CD /I  F (hrs) P eg (MW e ) Q eng S.S

First Conclusions Simple analytic formulae make it easy to understand how a Pilot Plant might have staged operation. In the example, initial operation was under ITER-like conditions. Ultimate operation was under ARIES-AT-like conditions.

Since I have time left, as they said on Monty Python’s Flying Circus, “And now for something totally different.”

Issue of Replacing a TF Coil I am not convinced that it will be possible to replace an SC toroidal coil on a reactor on any reasonable timescale i.e., < 1 year. Maybe it would be a good idea to view the TF coil set like a reactor pressure vessel. It can be replaced but only every 20 years or so.

Considerations What part of an SC TF coil set would fail? What is the data base? Unless there is a fundamental design problem as at CERN, do we expect a coil to fail after the shakedown phase? An SC coil does not suffer corrosion problems like a water-cooled copper coil. Pulsed field and radiation damage could be a problem. Argues for steady state operation and hefty shielding. Current feeds could be main area for concern, but could be accessible.

Neutron Shielding If the TF coils are considered to be “permanent” fixtures, then a fixed shield and VV could be installed around the inner legs e.g., 0.50 m thick. No gaps, no streaming, and something to attach the remaining blanket/shield to. I suggest that it’s worth considering.

Another Example Case with R m = 2.0 m and B m = 10 T  n = 0.4 and  T = 1.0 R/a = 3.5,  = 0.80 m, R = 4.08 m, a = 1.17 m,  = 2.2,  = 0.8 n D /n e = 0.35, T = 1.0,  N = 3.5, q 95 = 3.5, f 1 = 10.6, f 3 = 1.17, f 4 = 1.72, f a = 1.1 I =12.3 MA, IR = 50.2 MA.m, f G =   N BR  /T f 3 f 4 = 0.78 To get Q= 20 requires H H = 1.08 and P F = 716 MW, P a = 40 MW With thermal-electric conversion efficiency 0.3, blanket gain of 10%, auxiliary power efficiency of 0.4, and 100 MW e for everything else, Q eng  Fine, so what if we have difficulty providing the shaping? a PF /a ≈ 2.30

Issue of Shaping When the PF coils are outside the TF coils, there can be the issue of providing the strong shaping characteristic of Advanced Tokamak scenarios. An important factor is the ratio a PF /a. In the mid plane on the inside, a PF /a ≈ (1.1a +  +  TF )/a On the outside, a PF is determined by the outer blanket/shield/VV and the constraint set by TF ripple. Once the inner build of solenoid and TF coil inner leg +  +  TF are given, the only way to decrease a PF /a and b PF /b is to increase a and b. In practice, the ratio in the outer mid-plane doesn’t change much at constant TF ripple at the plasma edge. Nevertheless there is a gain from decreasing R/a. This is also beneficial because the lower R/a naturally allows higher k.

Example with R/a = 3.25  = 0.80 m, R = 4.23 m, a = 1.30 m,  = 2.2,  = 0.8 I =14.5 MA, IR = 61.3 MA.m, f G = 0.84 To get Q= 20 requires H H = 1.0 P F = 956 MW, P a = 44 MW, Q eng  1.57 a PF /a ≈ 2.18

ParameterR/a = 3.5R/a =3.25 R/a =3.0 R (m) q I (MA) IR (MA.m) fGfG H P F (MW) P a (MW) Qeng a PF /a

Conclusions Simple analytic formulae allow one to understand rapidly the effect of small changes. The example used looked at what happens if a particular design presents difficulties for the PF coils in providing the required shaping.