The battery efficiency is the ratio of the energy retrieved from the battery, to the energy provided to the battery, when coming back to the same SOC state.
Coulombic (or Faradic) efficiency
We define the coulombic efficiency as the ratio of the current provided to the current retrieved. This ratio is usually rather high, of the order of 97% for Lead Acid batteries.
For the Li-Ion technologies, we don't have explicit data. We have supposed a value of 96%, for approaching the global energetic efficiency of 92-95% reported by some few manufacturers. As the Resistance is very low, this contribution should represent a big part of the global inefficiency.
This parameter is due to the Electrochemical conversion current efficiency.
In lead-acid batteries we have a "gassing" phenomenon,due to the dissociation of the water molecules when the battery is overcharged. This consumes an additional current, which is accounted independently (not in the efficiency) in the PVsyst model.
We define also the ohmic inefficiency, as the loss due to the internal resistance during such a cycle (return to the original SOC).
This will depend on the currents involved during the cycle.
If we admit a constant charge/discharge current (Icharge = -Idisch) , and an identical resistance when charging and discharging, we have:
Voltage drop between Charge and Discharge DV = (Voc + Icharge * R) - (Voc - (-Idisch) * R) = 2 · R · I
=> energy loss = DV · I = 2 · R · I²
=> Ohmic efficiency = 1 - (2 · R · I² / (I · Voc)) = 1 - (2 · R · I / Voc)
Therefore, the battery efficiency depends on the instantaneous current: during the simulation it should be evaluated at each time step (integrated).
If the open circuit voltage is dependent on the temperature, there may also be an efficiency deficit when the temperature at discharging time is lower than the temperature when charging. This could be the case for an electric car in cold climates. In static solar systems, the battery bank is usually at a relatively stable temperature (indoor).
Finally we should have:
Battery efficiency = Coulombic efficiency * Ohmic efficiency * Temperature efficiency
Simulation: Efficiency loss
However this doesn't make sense to evaluate the efficiency for a given hour, because it is related to long-term processes.
During the simulation, the input and output energies are accumulated hour by hour. The final result will give a balance - over a significant time - of all these contributions. The global efficiency loss will be part of the final results (monthly or yearly).
NB: The ohmic voltage drops are easily visible, for example, on the V = f (SOC) curves (see especially in the linear region, not the gassing region).
These voltages differences are much less pronounced with Li-Ion batteries, which have a very low internal resistance.