Question

Consider a cell having nominal voltage Vnom = 3.3V and total charge capacity Q = 25Ah. If the BMS is designed to keep the cell's state of charge between 5% and 95%, and if the cell's present state of charge is 70%, what is the approximate remaining total energy in the cell (in Wh)? Round your answer to the nearest Wh.

Answer 1

The remaining total energy in the cell with a nominal voltage of 3.3V, total charge capacity of 25Ah, and a current state of charge of 70%, is approximately 52Wh when rounded to the nearest whole number.

Step-by-step explanation:

To calculate the remaining total energy in the cell, we first determine the usable charge capacity based on the state of charge (SoC) limits imposed by the Battery Management System (BMS), which is between 5% and 95%. Given that the nominal voltage (Vnom) is 3.3V and the total charge capacity (Q) is 25Ah, we can calculate the available energy within these limits:

First, determine the usable capacity:
90% of 25Ah = 0.9 × 25Ah = 22.5Ah
Then, considering the cell's present state of charge is 70%, the remaining energy is:
70% of 22.5Ah = 0.7 × 22.5Ah = 15.75Ah
To find the energy in Watt-hours (Wh), multiply the remaining charge by the nominal voltage:
15.75Ah × 3.3V = 51.975Wh
Round this to the nearest Wh gives us approximately 52Wh of remaining energy.