1. Work with TSC Yong Guo

2. Introduction Non-inductive current for NSTX TSC model for EAST Simulation for EAST experiment Voltage second consumption for different maximum plasma current Methods of saving voltage second Future work

3. Introduction The Tokamak Simulation Code (TSC) is a numerical mode of an tokamak plasma and the associated control systems. The code simulates the time evolution of a free boundary plasma by solving the MHD equations on a rectangular computational grid. The MHD equations are coupled to the external circuits through the boundary conditions The TSC model evolves the surface-averaged thermodynamic variables relative to magnetic coordinate surface.

4. Non-inductive current TSC model for NSTX Reason: limitation of voltage-second Aim: improve the fraction of non-inductive current Non-inductive current: NB driven current, bootstrap current

5. Introduction to simulation: The simulation begins at 0.0sec, end at 5.0sec The initial plasma current is 25KA Plasma reach flattop at 0.2sec The flattop last 4.8sec The transition from L-mode to H-mode is at 0.08sec NB is turned on at 0.12sec, and the maximum amplitude is 6.15MW

6. . * The non-inductive current is more than the total plasma current.

7. Conclusion: compared to reference improve plasma confinement: increases central temperature and stored energy reduces the volt-second needed for plasma operation It improve bootstrap current and NB driven current It reach 100% non-inductive operation when anomalous transport coefficients equals to 0.8. lower total plasma current: reduce the resistive volt-second consumption increase the total volt-second reduce bootstrap current and NB driven current increase the non-inductive current fraction and bootstrap current fraction Increase plasma density: increase bootstrap current and reduce NB driven current the non-inductive current fraction is close more resistive volt-second consumption and less inductive consumption the total volt-second consumption is close

8. TSC model for EAST PF coils: 14 PF coils PF7 and PF9 connect together PF8 and PF10 connect together Vacuum vessel: divide into 4cm*4cm cross-section coils along vacuum vessel Limiter point: inner limiter point (1.3634, 0.0) at the mid-plane outer limiter point (2.3700, 0.0) at the mid-plane Active coils: two active coils inside the vacuum vessel for VDE control Computational region: 1.100m < R < 2.940m, 47 vertices -1.400m < Z < 1.400m, 71 vertices

9. PF 7 and PF9 are in series, they share one power supply same as PF 8 and PF 10

10. Simulation for EAST experiment Shot 7721 diagnostic data: surface voltage, PF current, plasma current, line average density EFIT: major radius, minor radius, elongation, Lack of diagnostic data central temperature, temperature profile, density profile, beta_p, internal inductance, …… Plasma evolution: Ohmic discharge initial plasma current is 109.4KA, maximum current is 304.0KA ramp-up: 0.00sec - 0.95sec limiter configuration: 0.95sec - 2.03sec transition to divertor configuration: 2.03sec - 2.89sec divertor configuration: 2.89sec - 3.03sec transition to limiter configuration: 3.03sec - 4.46sec ramp-down: 4.46sec - 5.23sec

11. Ip (A)Surface voltage (V)Line average density (m^-2) Group1 current (KA) Group2 current (KA) Group3 current (KA)

12. Group4 current (KA) Group5 current (KA)Group6 current (KA) Major radius (m) Minor radius (m) Elongation

13. q(0) Beta_torInternal inductance Internal energy Te(0) Te(0)/Te_av Volt-second Vsec-abs

14. Summary: Central safety factor is larger than 1 in the evolution. And in the EFIT calculation at chosen time slice, q0 is set as 1, and get a wrong li. vertical magnetic field, which mainly provided by group 6, is a function of beta_p+li/2. When I try to fit plasma position provided by EFIT, the simulation results of group 6 current disagree with Exp. The line average density trajectory is not flat between 2sec and 4.5sec. It cause the internal energy and beta_t have the same change The central electron temperature is about 900EV at flattop, and te(0)/Te_av, which signs electron temperature profile is about 2.75. this shot start at -2.0328v*sec, end at 1.8513v*sec. Resistive consumption is large, about 77% at 4.5sec

15. Voltage second consumption for different Ip_max Choose five time slices of shot 7721 [0.0sec, 0.95sec, 2.03sec, 2.89sec, 3.03sec] Design new equilibrium use the plasma shape, internal inductance, poloidal beta at the chosen time slice, design three sets of plasma equilibrium for 300KA, 500KA and 600KA, respectively Follow the course of shot 7721 ramp up – limiter configuration – transition – divertor configuration Note flux linkage: -3.5 V-sec – 4.8 V-sec (due to the PF coil) same ramp up rate keep similar central temperature and temperature profile flatten the central density trajectory at flattop

16. groupMaximum value 300KA 500KA 600KA 1 3920 KA-2.8373E3 KA-2.9925E3 KA-3.3413E3 KA 2 3920 KA-3.6261E3 KA-3.7562E3 KA-3.5469E3 KA 3 3920 KA-3.5953E3 KA-2.4462E3 KA-1.7664E3 KA 4 6944 KA 1.6247E2 KA 1.1638E3 KA 1.4409E3 KA 5 1680 KA-1.1976E3 KA-1.6689E3 KA-1.5441E3 KA 6 896 KA 1.1209E2 KA 1.8827E2 KA-1.0680E0 KA (1) For 300KA, group 2 current is close to 95% of maximum value (2)For 500KA and 600KA, group 2 and group 5 current are close to 95% of their maximum value So, I choose 4.8V-sec as the maximum flux linkage for these simulation The group coil current at 4.8V-sec for different I_max

17. 300KA500KA600KA First stage (ramp up) Time1.0s2.0s2.5s Consumed V-sec 1.64.15.275 Second stage (limiter configuration) Time1.0s Consumed V-sec 0.771.131.625 Third stage (transition) Time1.0s Consumed V-sec 0.7641.0851.2132 Fourth stage (divertor configuration) Time7.6s2.0s0.06s Consumed V-sec 5.1661.9850.1750 (1) In the ramp up part, the demand of voltage-second raises quickly as Ip_max increases (2) In the other three parts, the demand of voltage-second raises slowly as Ip_max increases (3)The flattop time reduces as Ip_max increases (4)Ip_max>600KA, there is no enough voltage-second to form divertor configuration as shot 7721 evolution The volt-second consumption at the different stage for different I_max

18. Ramp up: 300KA (1.0sec)500KA (2.0sec)600KA (2.5sec) Resistive 0.6582.1313.033 Inductive 0.3280.6560.705 External 0.6161.3131.537 Total 1.64.15.275 1. In ramp-up stage, plasma consumes 19.28% of total volt-second for 300KA, 49.4% for 500KA, and 63.55% for 600KA 2. With the increasing of Ip, the inductive component and external component increase slowly, and resistive component increases quickly 3. For 600KA case, resistive component take a large part of total flux needed, about 57.5%. For 300KA case, resistive component only take 39.25%

19. increase ramp up rate (taking 600KA as example): Ramp up time2.0s1.5s1.25s1.0s Voltage second 4.96234.20813.88653.5708 resistive2.4921.7461.4281.116 inductive0.6710.5490.6170.489 external1.7991.9131.8411.966 Note: (1)As ramp up time reduces, the total demand of voltage-second reduces. (2)The demand of resistive component reduces obviously. It takes 50.22% of total volt-second for ramp-up time=2.0sec, and only 31.25% of total volt-second for ramp-up time=1.0sec. (3)The inductive component change small, due to it is used to establish magnetic configuration to hold plasma (4)It save 1.3915v-sec when ramp up time reduces from 2s to 1s. It could make divertor configuration about 0.4771sec longer Methods of saving voltage second

20. Auxiliary heating and current driven : Methods of saving voltage second 600KA, 2.5sec ramp-up ICRF for heating, and LH for heating and current driven ICRF is launched at 0.5sec for heating, and LH is launched at 0.6sec when the plasma temperature is higher ICRF power profile is prescribed peaked profile For LH, the parallel index of refraction is limited by the plasma density profile and temperature profile. The density profile and temperature profile are not optimized to make driven current locate at somewhere. Here, I use the smallest n_para I find which can drive current. Due to the heating of ICRF and LH, the plasma internal energy will increase. Due to the LH driven current and bootstrap current, the plasma current profile will change. It is difficult to design plasma at the chosen time slices for evolution calculation. I use former evolution result as the input for next evolution, make it convergence.

21. 1234567 ICRF2MW 1MW 0.5MW LH2MW0MW2MW1MW0MW1MW0.5MW N_para5.2~6.06.7~7.8 q0 (t=2.5sec) 2.300.981.481.020.930.990.97 LH current8.63e+4~7.63e+43.98e+4~2.66e+41.37e+4 BS current1.57e+58.05e+41.12e+57.01e+45.97e+46.05e+45.19e+4 Resistive0.62190.69040.74150.98491.1171.2851.564 Inductive0.59010.83760.61550.76610.8570.7680.800 External1.60701.6151.6221.6321.6031.6311.634 Total2.8193.1432.9793.3833.57743.6843.988 Li(GA)0.811.0190.810.961.0610.940.99 Beta_p0.8250.470.610.440.330.370.32 Auxiliary heating and current driven (taking 600KA as example):

22. Comparison case 1 and case 2 Same ICRF power and different LH power LH driven current located at rou=0.4, make current profile off- axis Off-axis current profile make li small, reduce the inductive volt- second consumption Resistive volt-second consumption is close for two case Off-axis current profile make q(0) large (2.3 for case 1, and 0.98 for case 2) Some LH power deposit at rou=0.4 for case 1. It help to create more bootstrap current (the same results for the comparison of case 3,4,5 and comparison of case 6,7) T Case 1 j (rou) AUX heating j (rou) AUX heating T Case 2

23. Comparison case 1 and case 3 Same ICRF power and different LH power Less beta_p and a little more resistive consumption for case 3 Smaller temperature for case 3 Less BS current created for case 3 LH driven current is smaller and location is outer for case 3 Li is same for two case, and inductive consumption is close ( same results for comparison of case 4,6 ) j (rou) AUX heating TT Case 1Case 3

24. comparing case 2 and case 4 Same total power of LH and ICRF LH driven current located at rou=0.6 for case 4which turn on 1MW LH A little more q0 for case 4 Less Li and inductive volt-second consumption for case 4 Less temperature and more resistive volt-second consumption for case 4 Less BS current for case 4 More non-inductive current for case 4 due to the contribution of LH current (3.98e4 A LH + 7.01e4A BS) ( same results for comparison of case 5,7 ) j (rou) AUX heating T j (rou) AUX heating Case 2Case 4

25. Future work In the this month Ptransp TSC work with Transp together More practices with TSC such as: VDE study …

26. Thank you