Project 1

Question 1.

Solution. Given the parameters for the battery pack:

• Power requirement: 150 kW
• Voltage requirement: 120 V
• Cell chemistry: NMC (Nickel Manganese Cobalt), 3500 mAh (3.5 Ah), 3.6 V nominal

Step 1: Calculate the Number of Cells in Series (Voltage)

Each individual NMC cell has a nominal voltage of 3.6 V. To achieve a total of 120 V, the number of cells in series (S) can be calculated as:

$$S = \frac{120 \, \text{V}}{3.6 \, \text{V/cell}} = 33.33 \approx 34 \, \text{cells in series}$$S=3.6V/cell120V​=33.33≈34cells in series

Thus, we will use 34 cells in series to achieve the required voltage.

Step 2: Calculate the Number of Cells in Parallel (Capacity)

The total energy needed is 150 kW, and we are aiming for a 120 V battery pack. The required total energy in Watt-hours (Wh) is:

$$\text{Total Power} = 150 \, \text{kW} \times 1 \, \text{hour} = 150,000 \, \text{Wh}$$Total Power=150kW×1hour=150,000Wh

The energy provided by one cell (assuming nominal voltage of 3.6 V and capacity of 3.5 Ah) is:

$$\text{Energy per cell} = 3.6 \, \text{V} \times 3.5 \, \text{Ah} = 12.6 \, \text{Wh}$$Energy per cell=3.6V×3.5Ah=12.6Wh

The total number of cells required to meet the total energy demand is:

$$\frac{150,000 \, \text{Wh}}{12.6 \, \text{Wh/cell}} \approx 11,905 \, \text{cells}$$12.6Wh/cell150,000Wh​≈11,905cells

Since we have 34 cells in series, the number of cells in parallel (P) would be:

$$P = \frac{11,905}{34} \approx 350 \, \text{cells in parallel}$$P=3411,905​≈350cells in parallel

So, for the battery pack, we will have 34 cells in series and 350 cells in parallel.

Step 3: Pack Configuration

To summarize, the battery pack configuration consists of:

• 34 cells in series (S) for voltage
• 350 cells in parallel (P) for capacity

This results in:

$$\text{Total number of cells} = 34 \times 350 = 11,900 \, \text{cells}$$Total number of cells=34×350=11,900cells