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.

Answer:-

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Li-ion cell(ANR26650M1-B):

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EV design - Battery calculation:

Battery energy = 18 kwh

Nominal voltage capacity of single cell = 3.3V

Current Capacity of single cell = 2.5Ah

Assumptions:

Voltage (to run the vehicle) = 400V

Energy = 18 kWh

(i) Number of cells (N) = energy / (nominal voltage capacity)

N=18000/(3.3*2.5)

N=2182

We need a minimum of 2182 cells to get 18 kWh energy, now based on that, we need to decide configuration-series and parallel In series configuration capacity remains the same and voltage will be the sum of individual cell voltage. In a parallel configuration, the voltage remains the same and capacity accumulates.

And we assumed 400 Volts are need to run the vehicle.

The nominal voltage of one cell is 3.3 V

Number of series cells required to produce 400 V (Ns) = voltage required / nominal voltage of one cell

Ns = 400/3.3

Ns = 121.21

Ns = 122

122 cells are connected in series will generate 400 voltage and the capacity of the whole module will be 2.5 Ah

Now we need to find the number of cells which will be connected in a parallel configuration

Np (cell in parallel) = total cell/Ns

Np = 2182/122

Np = 17.8

Np = 18

Hence we take 18 cells for each parallel row, although it is exceeding the required capacity, that is negligible.

Cell architecture 122 Series and 18 Parallel = 122S18P

(ii) Battery capacity:

We know parallel configured cell multiplies individual cell capacity so.

Capacity parallel cell row * individual cell capacity

Capacity 18 * 2.5

Capacity- 45 Ah

(iii) Current supplied by the battery

We have already calculated battery capacity which is 45 Ah means it is going to supply 45 A current per hour. which can be considered as C-rate.

(iv) Range:

Let's powertrain energy consumption = 100 Wh/km at 70% powertrain efficiency

Range = Capacity / energy consumed per kilometre

Range=18000/100

Range = 180 km

So,our vehicle will be able to cover around 180 km with current parameters.

(v) Internal Resistance of Battery:

The entire resistance encountered by a current as if it flows through a battery from the negative terminal to the positive terminal is known as internal resistance of battery.

(vi) Battery Dimensions:

ANR26650 cell has 26mm diameter and 65mm height. On the basis of that we need to decide battery size of battery casing.

Length of casing (L) = number of cells at series * [cell diameter (26) + margin at both side of cell (4)] + clearance at side (diameter of edged cell +margin)

L=122*(30+60)

L = 3720mm

Breadth of casing (B) = number of cells at parallel * [cell diameter (26) + margin at both side of cell (4)] + clearance at side (diameter of edged cell + margin)

B=18*(30+60)

B = 600mm

Height of casing (H) = 65 * 35 (Height of cell + margin for wiring Cell holders and metal strips)

H = 100mm

(viii) Weight of battery pack

Cell weight-76g

Number of the cells - 2182

Battery pack weight (W) = cell weight * number of cells + weight of accessories

W=76 * 2182 + 4.5

W =165.8k + 4.5

W =170.3 kg

Mechanical design of battery pack:-