Project 1 Mechanical design of battery pack

Battery energy: 18Kwh

Voltage to run the vehicle: 400V

1. Energy = 18Kwh
2. Number of cells = Energy/(Nominal voltage * Capacity)

N = 18000/(3.3 * 2.5) = 2182

We need minimum 2182 cells to get 18Kwh energy. Based on thislet us decide series & parallel configuration.

Since 400V is required to run the vehicle & each cell is having 3.3 as a nominal voltage, Number of series cells required = 400/3.3 = 121.21 = 122 cells.

As no of cells required to form a complete battery pack = 2182 cells Thus the no of parallel cells needed = 2182 / 122 = 17.8 = 18 cells.

Ns = 122 & Np = 18.

3. Battery Capacity = Parallel cells in a row * Individual cell capacity

= 18 * 2.5 = 45Ah

4. Current supplied by the battery =

We have calculated the battery capacity to be 45Ah. Hence battery can supply 45A current for one hour which is considered as the C rating of the battery.

5. Range =

Let’s powertrain energy consumption = 100Wh/Km at 70% of powertrain efficiency

Range = Capacity / Energy consumed per Km

= 18000 / 100 = 180Km

6. Internal resistance of the battery =

Entire resistance encountered by a current when it flows from Negative terminal to the positive terminal of the battery is it’s internal resistance.

7. Weight of the battery pack =

Cell weight = 76g

No of cells in a pack = 2182

Other accessories weight = 4.5 Kg

Overall weight of the pack = 2182 * 76 / 1000 + 4.5 = 170.3Kg