Project 1 Mechanical design of battery pack
Battery pack capacity: 18 kWh
Cell: ANR26650M1-B
Prepare a detailed battery pack drawing along with its enclosure. State your
assumptions.
Solution:
Cell: ANR26650M1-B Specifications:-
| Manufacturer: | LithiumWerks |
| Mfg PN: | 300732-006 |
| Chemistry: | Lithium Iron Phosphate (LiFePO4) |
| Voltage: | 3.3 Volts |
| Capacity @ 25°C: | 2.56 Ah |
| Energy @ 25°C: | 8.25 Wh |
| Energy @ 25°C: | 8.25 Wh |
| Specific Power: | >4000W/kg | (@ 25°C, 2 sec pulse) |
| Impedance: | >6mΩ | (1kHz AC) Typical) |
| Cycle Life: | > 4000 Cycles | (@ 1C/1C, 100% DOD) |
| Dimensions | |
| Height (max): | 65.15 (+/- 0.5 mm) |
| Diameter (max): | 25.96 (+/- 0.5mm) |
| Mass: | 39.5 (+/- 1.0 g) |
| Warranty: | DOA |
Given that-
Battery pack capacity: 18 kWh
Cell dimensions-
Cell calculation:
1) Volume of the cell: For the calculation of the volume of the cell we use the formula of Volume of cylinder. so the formula is:πr^2h From Datasheet of Cell, we have 'r ' and 'h ' values
r=0.013 m and h = 0.065m
volume of the cell is= 3.14*(0.013)^2*0.065 = 0.0000344929 m^3
Basically, the value of the volume of the cell is used to find out the total volume of the battery pack.
Cell Energy= Cell Capacity (Ah) * Cell Voltage (V) = 2.5*3.3=8.25Wh
Cell Energy is 8.25Wh.
The value of a single cell is used to calculate the total energy of the battery pack.
Single Cell Energy Density-
Volumetric energy= Single Cell Energy / Volume of each cell
= 8.25/0.0000344929
= 2,39,179.65 Wh/m^3
Single cell eneregy density is= 2,39,179.65 Wh/m^3.
Battery Pack Assumptions:
There are 50 cell are in series and 44 cells are in parallel, and as above maintioned we are going to use Cell: ANR26650Ml-B this cells.
Battery pack calculation:
Now we calculated parallel and series combinations so we can now calculate total no of cells required
Total no of cells: Series cells * parallel cells = 50*44 = 2200.
Total no of cells required are 2200.
So from series and parallel combination, we can calculate the voltage of battery pack and the current of the battery pack.
Voltage = no of cell in series* nominal voltage = 50*3.3= 165 V.
The total value of voltage is 165 V.
Current=No of parallel cells * cell capacity = 44*2.5 =110Ah. The total value of current is 110Ah So now we can calculate the how much amount of energy this combination have by product of voltage and current, and check if it will satisfied or not our requirement.
Energy= voltage*current
The total value of Energy is 18150WAh.
So our calculation are satisfied requirement what we need. now we have total no of cells and volume of single cell so we can calculate the total volume required for battery pack
Total volume= no of cells * Volume of single cell = 2200*0.0000344929 = 0.07588438 m^3.
We also have single cell weight from that we can calculate the total weight of battery pack without other mechanical and electrical connections.
Weight = No of cells * weight of single cell = 2200*0.076 =167.2 kg Weight of battery pack is 167.2kg
Energy density =volume battery*pack energy = 0.07588438*18150 = 1377.301497 KWh/m^3.
Assembly of the full battery pack with 50s/44P. The arrangement of the pack is as follow:
Then, 2112 cells will have 50 S/ 44P.
The detail technical specification of 2112 battery pack is as follow:
- Package: 50S / 44P
- Capacity: 110 Ah
- Voltage:165V
- Weight: 167.2kg (Only Cells)
- No of Cells: 2100
- Power: 18150Wh or 18.15kWh
Design of cell holder (bottom and top)-
Design of PACK COVER-
RESULTS-
| NO | Calculated Parameters | Value Get |
| 1 | Cell Volume | 0.0000344929 m^3 |
| 2 | Cell Energy | 8.25 Wh |
| 3 | Cell Density | 2,39,179.65 Wh/m^3. |
| 4 | Pack assumptions | 50 s 44 p |
| 5 | Total no of cells | 2200 |
| 6 | Voltage | 165 V |
| 7 | Capacity | 110 Ah |
| 8 | Energy | 18150WAh |
| 9 | Total volume | 0.07588 m^2 |
| 10 | Total Density | 1377.3 kWh/m^2 |
| 11 | Weight | 167.2 kg |