Project 1 Mechanical design of battery pack
AIM: Prepare a detailed battery pack drawing of [Battery pack capacity: 18 kWh, Cell: ANR26650M1-B] along with its enclosure. State your assumptions.
Details of Cell: ANR26650M1-B:
ANR26650M1-B specification:
Nominal Rating:
- voltage :3.3 volts
- capacity:2.5Ah
- energy: 8.25Wh
- specific power: 2600W/kg
- impedence(1KHz AC typical) 6 mΩ
- cell type: 26650 lithium ion power cylindrical cell.
Discharging:
- max.continuous discharage: 50A
- max pulse discharge current > 50% SOC (10sec): 120A
- minumum voltage:2 volts
- temperature: -30 to 55 degree celcious
charging:
- recommended std.charge: 2.5A
- max charge rate:10A
- max pulse charge current<50% SOC (10sec) :25A
- float voltage: 3.45 volts
mechanical parameters:
- diameter: 25.96 +/- 0.5 mm
- length: 65.15+/- 0.5 mm
- mass : 76g
calculation as per assumptions: 18 KWh battery pack.
calculation for designing an 18 KWh battery pack basically 18 KWh is energy require to achive the vehicle it proper range.and its depend on the voltage(V) and amp-hr capacity(Ah)
1) single cell energy(Wh) [the value of a single cell energy is used to calculate the total energy of the battery pack]:
single cell energy(Wh)= cell capacity(Ah)* cell voltage(V)
= 2.5Ah *3.3V
= 8.25 Wh
2) single cell volume calculation:
single cell volume = pi(r^2)*h
= 3.14(0.013^2)*0.065
= 0.03451 m^3
3) single cell energy density:
* volumetric energy(Wh/m^3) = single cell energy / volume of each cell
= 8.25Wh / 0.03451 m^3
= 239.0611 Wh/m^3
* gravimetric energy(Wh/kg) = single cell energy / single cell mass
= 8.25Wh / 0.076 kg
=108.5526 Wh/Kg
calculation of high voltage battery pack
for design and calculation of battery pack here,i assume a vehicle to achive 200km in 18 KWh battery pack also i assume 145.2 nominal voltage for series connection.
so we assume that,
vehicle range = 200 Km
battery pack capacity = 18.1210KWh
now,
4) we will calculate per Km average energy requirement = 18120.96 (Wh)/200 (Km)
= 90.60 Wh/Km
therefore,
battery pack total energy = average energy * km
= 90.60Wh * 200km
= 18120.96 Wh
let's assume nominal voltage for seres connection= 145.2 V basically this depends upon the motor power,we can decide the value according to the motor.
5) number of cell connected in series = nominal voltage of seres connection/nominal voltage of a single cell
= 145.2V/ 3.3V
= 44 cell
44 cell will be connected in a series connection to provide a proper voltage.
6) energy for battery pack(Wh) = number of cell connected in series * single cell energy
= 44* 8.25Wh
= 363 Wh
7)number of cell connected in parallel = battery pack total energy(Wh)/energy for battery pack(Wh)
= 18120.96 Wh/363 Wh
= 50 cell
50 cell will be connected in a parallel to provide a proper capacity power.
8) battery pack total energy(Wh) = no. of cell connected in parallel * energy for battery pack
= 50* 363 Wh
= 18.150 KWh
9) battery pack capacity(Ah) = no. of cell connected in parallel * single cell capacity(Ah)
= 50*2.5 Ah
= 125 Ah
the value of battery pack capacity is used for the total power required by the vehicle.
10) total number of cell for battery pack = no. of cell connected in parallel * number of cell connected in series
= 50*44
=2200 cells
2200 cells will be used to make a complete battery pack.
11) mass of battery pack = total number of cell for battery pack * mass of single cell
= 2200 * 0.076 kg
= 167.2 kg
167.2 kg its the total weight of cell connected in parallel and series
12) total volume of the battery pack = total number of cell for battery pack * volume of single cell
= 2200 * 0.0345 m^3
= 75.9 m^3
75.9 m^3 its the volume occupied by total cell used in battey pack.
13) cell peak current(A) = C rate(max pulse charge current)* single cell capacity
= 20* 2.5
= 50A
14) battery pack peak current(A) = cell peak current(A) * no. of cell connected in parallel
=50*50
= 2500A
15) battery pack peak power(W) = battery pack peak current(A) * voltage of series connection
= 2500 A * 145.2
= 363 KW
16) cell continuous current(A) = C rate(max continuous charge current)* single cell capacity
= 10* 2.5
= 25A
17) battery pack continuous current(A) = cell continuous current(A) * no. of cell connected in parallel
= 25*50
= 1250 A
18) battery pack continuous power(W) = battery pack continuous current(A) * voltage of series connection
= 1250 A * 145.2
= 181.5 KW
Details of Cell: ANR26650M1-B:
Battery holder:
Electric connection:
Battery pack with BMS[battery management system]:
