Plasma flow

In deriving the Grad-Shafranov equation, we have assumed that the plasma flow is zero. Next, we examine whether this assumption is well justified for the EAST plasmas. Note that $\nabla P \approx P_0 / a$, where $P_0$ is the thermal pressure at the magnetic axis and $a$ is the minor radius of plasma. For typical EAST plasmas (EAST#38300@3.9s), $P_0 \approx 2 \times 10^4 \ensuremath{\operatorname{Pa}}$, $a = 0.45 m$. Then $\nabla P \approx P_0 / a = 4 \times 10^4 N / m^3$. One the other hand, the centrifugal force due to toroidal rotation can be estimated as

$$F_{\ensuremath{\operatorname{rot}}} = \rho_m U^2 / R \approx 4 \t... ...10^{- 27} \ensuremath{\operatorname{kg}} U^2 / 1.85 = 3.6 \times 10^{- 8} U^2,$$

where $U$ is the rotation velocity. To make the centrifugal force be comparable with $\nabla P$, i.e.,

$$3.6 \times 10^{- 8} U^2 = 4 \times 10^4,$$

the rotation velocity needs to be of the order $U \approx 1.0 \times 10^6 m / s$, which corresponds to a rotation frequency $f = U / (2 \pi R) = 86 \ensuremath{\operatorname{KHz}}$ which is larger by one order than the rotation frequency observed in EAST experiments (about 10 $\ensuremath{\operatorname{KHz}}$). This indicates the rotation has little influence on the static equilibrium. In other words, the EAST tokamak equilibrium can be well described by the static equilibrium.

Note in passing that the complete momentum equation is given by

$$\rho_m \left[ \frac{\partial \mathbf{U}}{\partial t} +\mathbf{U} ... ...ht] = \rho_q \mathbf{E}+\mathbf{J} \times \mathbf{B}- \nabla \cdot \mathbbm{P},$$ (85)

where $\rho_q \mathbf{E}$ term can be usually neglected due to either $\rho_q \approx 0$ or $\mathbf{E} \approx 0$, $\mathbbm{P}$ is a pressure tensor, which is different from the scalar pressure considered in this note. The equilibrium with pressure tensor can be important for future burning plasma, where the pressure contributed by fast ions from the neutral beam injection can be a tensor.

[Mass density of EAST#38300@3.9s, $\rho_m = n_D m_D = 4 \times 10^{19} m^{- 3} \times 2 \times 1.6726 \times 10^{- 27} k g = 1.3 \times 10^{- 7} k g / m^3$.

$$\nabla p \approx p_0 / a = 2 \times 10^4 \ensuremath{\operatorname{Pa}} / 0.45 m = 4 \times 10^4 N / m^3$$ (86)

]