In GitLab by @mkovari on Dec 12, 2019, 13:26
After further discussions, I have a new set of proposed equations. (See #714 for background.)
Cables from energy store to power supply. This is high voltage, so the resistive loss in this cable = 0.
(A) Storage. Energy stored in storage unit = $E_s$. Power loss depends on rate of energy storage or delivery:
$$Loss_s = k_s \left|\frac{dE_s}{dt}\right|$$
During the flat-top the power requirement is small, and we could assume this comes directly from the grid. During the pre-charge, ramp-up, and ramp-down phases we will need the energy storage system. In these phases, between time points $t_n$ and $t_{n+1}$, we have,
$$\frac{dE_s}{dt} = - \frac{E_{PF}(t_{n+1})-E_{PF}(t_n)}{t_{n+1}-t_n},$$
where $E_{PF}$ is the energy stored in the poloidal field. The energy loss between time points $t_n$ and $t_{n+1}$ is:
$$ELoss_s = k_s {\left|E_{PF}(t_{n+1})-E_{PF}(t_n)\right|}.$$
The following loss terms are summed over all the PF coils.
(B) Power supply, where $P_i$ = power delivered by by the power supply for coil $i$
$$Loss_{ps} = k_{ps}\sum_{i}|P_i|$$
To determine $P_i$ we need to know the instantaneous voltage and current in each coil, $V_i$ and $I_i$
$$P_i = V_i I_i$$
The current $I_i$ is already calculated, but the voltage $V_i$ needs to be derived using the mutual inductance matrix $M$ for all the coils and the plasma current,
$$V_i = \sum_{j} M_{ij} \frac{dI_j}{dt}$$
PROCESS calculates the current at each time point. For each phase of the plasma pulse between time points $t_n$ and $t_{n+1}$ we can estimate the rate of change of current in each coil,
$$\frac{dI_j}{dt} = \frac{I_j(t_{n+1})-I_j(t_n)}{t_{n+1}-t_n},$$
and the mean current
$$I_i = \frac{I_i(t_{n+1})+I_i(t_n)}{2},$$
Therefore:
$$V_i = \frac{1}{t_{n+1}-t_n}\sum_{j} M_{ij} \left(I_j(t_{n+1})-I_j(t_n)\right)$$
and
$$P_i = \frac{1}{2}\frac{I_i(t_{n+1})+I_i(t_n)}{t_{n+1}-t_n}\sum_{j} M_{ij} \left(I_j(t_{n+1})-I_j(t_n)\right)$$
The energy loss time points $t_n$ and $t_{n+1}$ in the power supply is then
$$ELoss_{ps} = \frac{k_{ps}}{2}\left| \left[I_i(t_{n+1})+I_i(t_n)\right] \sum_{j} M_{ij} \left[I_j(t_{n+1})-I_j(t_n)\right] \right|$$
(C) Coil bus-bar, where bus-bar resistance in coil $i$ = $R_i$:
$$Loss_{bus} = \sum_{i} I_i^2 R_i$$
Therefore the energy loss in the bus-bars is
$$ELoss_{bus} = \left[t_{n+1} - t_n\right] \sum_{i} \left(\frac{I_i(t_{n+1})+I_i(t_n)}{2}\right)^2 R_i$$
(D) Main substation. Loss depends on energy delivered. This could be dealt with by adding a fixed 'tax' on all the losses listed above:
$$ELoss_m = k_m (ELoss_s+ELoss_{ps}+ELoss_{bus})$$
Since Stevie Wray suggests that the substation is 99% efficient ($k_m=0.01$), I will ignore this contribution.
The total energy loss through over all phases $n$ is
$$EnergyLoss = \sum_{n} \left(ELoss_s+ELoss_{ps}+ELoss_{bus}\right)$$
Stevie Wray has proposed the following values for the coefficents:
$k_s$ = energy storage % power loss (on each trip) = 10% (i.e. 90% efficient)
$k_{ps}$ = magnet power supply % loss (% of output) = 10% (i.e. 90% efficient)
While these quantities can be calculated for each phase of the ideal plasma pulse, in practice the demand on the PF coil system will be dominated by the vertical stabilisation system, whose properties are not known. Consequently I am not sure it is worth implementing this model.
Any comments? @ajpearcey @stuartmuldrew @jmorris-uk ?
