We looked at the Battery Power Demand Solution but then thought it would be even better to work through an example.
Most applications of battery cells and packs give a power requirement. Sometimes that is just a single peak value and sometimes that is a power versus time history.
Here we see a speed versus time trace for a vehicle and the associated power demand from the battery.
This is great, but what we need is the current demand.
That depends on the battery cell and pack design, hence we need some information and ideas.
Let us start by looking at the battery.
We have the power demand as given by the above time history. The open circuit voltage of the pack is determined by the number of cells in series and the particular cell. The Internal resistance of the pack is determined by the number of cells in series and parallel, plus the resistance of busbars, joints, fuses etc.
The internal resistance (DCIR) of the cell will be available from the cell supplier and should be on the specification sheet.
We can write down a relationship for the above diagram:
Now that we have the voltage of the battery pack under load, VLpack we can calculate the current as that is simply:
Let us look at a simple pack design that could deliver 10kW of power. Maybe we could use a typical 21700 that we would see in the Lucid Air or Tesla Model 3: Capacity = 5 Ah, OCV = 3.6 V and DCIR = 0.025 Ohm. For the resistance of everything else, 0.01 Ohm is a good starting point.
A 48V pack would be interesting and might be a good starting point. With a cell nominal of 3.6V that would require 14 cells in series.
Using the Battery Calculations Workbook we can quickly look at the option of a 14s 1p pack. What we see is that VLpack is not solvable. We need to increase the number of cells in parallel.
In the workbook we can rapidly change the configuration until we get a solution.
Above we see that with 14s 7p we have a pack that with a 10kW demand has a terminal voltage of 31.1V and the current demand is 321A for the pack and hence nearly 46A for each cell and the cell voltage at 2.45V is below the normal 2.5V minimum.
Another useful parameter that we estimate here is the time for the cells to increase 10°C in temperature and at this 10kW power demand that is 11 seconds.
This post has been built based on the support and sponsorship from: Thermo Fisher Scientific, Eatron Technologies, About:Energy, Quarto Technical Services and Alvatek.
Next we increase the pack to 9p and this reduces the peak current to below 30A (at 50% SoC). With the reduced current the time to heat the cells 10°C as more than doubled.
Now we have a starting point from which we can iterate and look at an optimal solution.
This is a very first step and just gets you to a configuration that you can start to model in more detail and with more supporting data. It is a very simple estimate, however, this approach lets you iterate quickly through some initial ideas. A task that is difficult with a more complex model.