Use of Simple Analytic Expression in Tokamak Design Studies John Sheffield,

July 29, 2010, ISSE, University of Tennessee, KnoxvilleInspiration Needed

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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 PF (MW)• Plasma profiles n = n0(1-r2/a2)n and T = T0(1-r2/a2)T

• Fusion fuel fraction nD/ne, where nD = nT (1020 m-3) and the impurities ∑nZZ.• fG the fraction of the Greenwald limit used. Greenwald limit nG = I/(a2) (1020 m-3)

• T = Te = Ti (10keV)—these are volume averaged

= a/R, , d, q95, the gap between the plasma and inner leg of the TF coil, and

• fa = aw/a the scrape-off layer thickness (assumed to be constant around the plasma).

• Bm (T) the maximum field on the coil and Rm (m) is TF coil inner leg outer radius

• I = 2.5(BR)2f1/q95 (MA) BR = Bm Rm

• where f1 = {1 + 2(1 + 22 – 1.23)}{1.17 – 0.65}/{1 - 2}2

Fusion Power

PF = 0.247 f(n,T) f12.0 (q95

)-2 f22 (f3f4)-2 N

2 4 (BR)4/R (MW)

f(n,T) = {(1 + n)2.(1 + 2n + 3T)2/ (1 + 2n + 2T)3} x{(1 + n +T). (1 + 2n + 2T)}2/{(1 + n).(1 + 2n + 3T)}2

The beta limit is given by <> = NI/aB (%)

<neT> + 2<nDT> + ∑<nZTz> = neT.f3.f4

where f3 = (1 + n)(1+ T)/ {(1 + n +T) and f4 = 1 + 2 f2 + <nZTz>/<neT>

R = Rm/[1 – (1 +fa)- /R] (m)

Confinement and Power to Sustain Plasma

ITER H-mode scaling is

E = 0.1444 HH I0.93 B0.15 n0.41 M0.19 R1.97 0.58 X0.78P-0.69

= 4.737Ra2<nT>/P (s)

Note that in the ITER-FEAT studies X/≈ 1.1, and with M = 2.5

P = [1.023 HH-1fG

-0.41N I-0.34(BR)0.85 1.24 0.22]3.23 (MW)

For Q =20, Pa = P/5 and 0.2PF/P = 0.8

This kind of level is needed for Qeng ~ 1 in a Pilot Plant.

e.g., PF =600 MW, P = 150 MW, and Pa = 30 MW

Balancing Alpha Power Against Power Requirement

With the formulae for PF, P, I, and fG < 1, and T = 1 (10 keV)

0.2PF/P = 4.43 x 10-3 (Plasma)x(Shaping)x(Impurities)x(Size)

Plasma Parameters HH3.23N

0.094f(n,T)/(q953.1f3

3.32f43.32)

Shaping f13.1 3.5 0.29

Impurity Depletion (nD/ne)2

Size (BR)3.68/R

where f1 = {1 + 2(1 + 22 – 1.23)}{1.17 – 0.65}/{1 - 2}2

Example Case with R/a =3.5, = 2.2, = 0.8, q95 = 3.7Bm =10 T, = 0.8, R = 4.23 m, a = 1.21 m, HH = 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 - IBS/I – ICD/I) hrsIBS/I ≈ 0.24 N

0.79 , ICD/I ≈ PauxT/fG,

PF = 64 N2 MW,

Pelg = 0.3(1.1PF+ Paux) MWe

Preq = 50 + 0.1PF + Paux/0.4 (see ITER)

P = 0.75 N0.094 MW (see ARIES-AT)

fG = 0.234 N 0.9, nG = 2.65 (1020 m-3)

N PF

(MW)

P

(MW)

P

(MW)

Paux

(MW)

P/R

(MW/m)

P/Aw

(MW.m-2)

Pn/Aw

(MW.m-2)

2.5 400 80 98 18 23 0.33 1.1

3.0 580 115 138 23 33 0.46 1.6

3.5 790 157 186 29 44 0.62 2.1

4.0 1030 205 240 35 57 0.80 2.8

4.5 1300 260 301 41 71 1.00 3.5

5.0 1600 321 367 46 87 1.22 4.3

N fG IBS/I ICD/I F

(hrs)

Peg

(MWe)

Qeng

2.5 0.59 0.49 0.06 0.4 140 1.0

3.0 0.70 0.57 0.07 0.5 200 1.2

3.5 0.82 0.65 0.07 0.7 270 1.3

4.0 0.90 0.72 0.08 1.0 350 1.45

4.5 0.90 0.79 0.11 2.0 440 1.6

5.0 0.90 0.86 0.13 S.S. 540 1.7

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.

ConsiderationsWhat 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 withRm = 2.0 m and Bm = 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

nD/ne = 0.35, T = 1.0, N = 3.5, q95 = 3.5,

f1 = 10.6, f3 = 1.17, f4 = 1.72, fa = 1.1I =12.3 MA, IR = 50.2 MA.m,

fG = 0.025 NBR/T f3 f4 = 0.78

To get Q= 20 requires HH = 1.08 and PF = 716 MW, Pa = 40 MWWith thermal-electric conversion efficiency 0.3, blanket gain of 10%,

auxiliary power efficiency of 0.4, and 100 MWe for everything else, Qeng 1.24.

Fine, so what if we have difficulty providing the shaping? aPF/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 aPF/a.

In the mid plane on the inside, aPF/a ≈ (1.1a + + TF)/a

On the outside, aPF 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 aPF/a and bPF/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, fG = 0.84

To get Q= 20 requires HH = 1.0

PF = 956 MW, Pa = 44 MW, Qeng 1.57

aPF/a ≈ 2.18

Parameter R/a = 3.5 R/a =3.25 R/a = 3.25 R/a =3.0

R (m) 4.08 4.23 4.23 4.42

q95 3.5 3.5 3.8 4.0

I (MA) 12.3 14.5 13.4 15.2

IR (MA.m) 50.3 61.3 56.5 67.2

fG 0.78 0.84 0.84 0.91

HH 1.08 1.00 1.08 1.03

PF (MW) 716 956 845 1019

Pa (MW) 40 44 39 51

Qeng 1.24 1.57 1.47 1.55

aPF/a 2.30 2.18 2.18 2.05

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.