The objective of this project is to analyze and design an optimum battery pack for a two-wheeler EV which can
also be used to retrofit an ICE two-wheeler to EV.
This battery pack should meet the below requirements & assumptions:
1. Able to provide a drive range of 80km per charge.
2. Able to drive a traction motor which can provide a top speed of 90kmph and pickup of 0-60kmph in 10 sec.
3. Battery pack rated voltage – 48V.
4. Cell Chemistry – Lithium-Ion.
1. MATLAB – Simulink
2. QSS tool box
3. MS Office – Excel
Vehicle selection and vehicle parameter collection
Here we are designing a battery pack to retrofit the ICE two-wheeler Suzuki Access 125 to EV.
Drive Cycle selection
We are considering FTP-75 drive cycle (Federal Test Procedure) as it meets our top speed (90kmph) and
acceleration (0-60kmph in 10 sec) requirements.
Top speed – 25mps = 90kmph
Top Acceleration – 1.7m/s2 or 0-60kmph in 10 seconds
Total distance – 17.77km
Calculate the average energy consumption per km
We have used the qausi-static approach to calculate the average energy consumption of our proposed EV. Using the
QSS tool box library in MATLAB we designed below model.
Note: Here we have not considered any transmission block as we are considering to use BLDC hub motor.
The P_EM output from the Motor block shows the motor power consumption as below
Below are the findings from the QSS simulation:
1. The maximum motor power consumption = 4.5kw
2. The total energy consumption for the drive cycle = 398.2watts
3. The average energy consumption per km = 21.96watts
4. Top speed = 90kmph
5. Top Acceleration = 1.7m/s2 or 0-60kmph in 10 seconds
6. Battery capacity = 2.62kwh/100km
From the graph it’s clear that the maximum power consumed by the motor for FTP-75 drive cycle is 4.5kw. Hence
in our application we will use a BLDC hub motor of 2kw continuous power and 5kw peak power.
Battery calculations using MS excel
Requirements & assumptions of our battery pack:
1. Able to drive the BLDC motor (2kw continues power & 5kw peak power) through the FTP-75 drive cycle.
2. Should have a range of 80km per charge for energy consumption of 22wh/km
3. Battery pack rated voltage = 48v
4. Battery depth of discharge to be considered as 85%
Battery total energy = (Energy consumption per km * Total range of vehicle) / Battery depth of discharge
= (22wh/km * 80km) / 0.85
Battery total capacity = Battery total energy / Battery rated voltage
= 21000kwh / 48v
= 43 Ah
Cell level calculations
Pack level calculations
From the analysis we found out that Kokam pouch cells are the best suited for our application. As it has very low
internal resistance and also the BP peak power capability and BP capacity can seamlessly satisfy our 5kw motor
Considering the selected kokam pouch cells our pack configuration will be 3P14S
A bus bar (also spelled busbar, buss bar, or busbar), is a strip or bar of copper, brass, or aluminium that conducts
electricity within a switchboard, distribution board, substation, battery bank, or other electrical apparatus. Its main
purpose is to conduct electricity, not to function as a structural member.
Busbars are typically either flat strips or hollow tubes as these shapes allow heat to dissipate more efficiently due to
their high surface area to cross-sectional area ratio. A hollow section has higher stiffness than a solid rod of equivalent
current-carrying capacity, which allows a greater span between busbar supports in outdoor switchyards.
A busbar may either be supported on insulators, or else insulation may completely surround it. Busbars are protected
from accidental contact either by a metal earthed enclosure or by elevation out of normal reach. Power neutral busbars
may also be insulated.
Bus bar Calculation
Ratings for designing bus bar for battery pack:
1. The battery pack current at 2C rate = 93.6A.
2. The material used is Copper (which has a current density of 1.6A/mm2).
3. The thickness of the copper is 10mm as per standard size
For 1 mm2 area, the current density is 1.6 A.
Therefore for 93.6 A what will be the area?
93.6 / 1.6 = 58.5 mm2
So to find the width of the bus bar with the standard thickness of copper material ie 10mm.
Therefore Area = (width) * (thickness)
58.5 = ? * 10mm
Width = Area / thickness
= 58.5 / 10
= 5.85mm ~ 6mm
Therefore for 93.6 A = The copper bus bar size is 6mm * 10 mm
As per the standard size of copper bus bar size available in market, we will select 12 * 10 mm
Voltage Drop in Bus bar
Current = 93.6A
Width = 12 mm.
Thickness = 10 mm.
Length = 150mm.
Therefore, the Cross-sectional area = 12mm * 10mm
Bus bar Resistance = (Copper resistivity * Length of busbar) / (cross sectional area of busbar)
= ((1.68*10-8) * (150*10-3)) / (12*10*10-6)
= 0.000021 Ω
Voltage drop = 0.000021 * 93.6
= 0.0019V ~ 0.002V
Hence, we conclude that the voltage drop across the busbar is negligible.
Battery pack structural design
• Our Cell configuration is 3P14S.
• Let’s split the pack into 3 modules each of 1P14S configuration and place it as two modules in bottom line
and one module on top.
1. Battery pack length = (thickness of cell + space for swelling) * 14
= (0.0077 + 0.003) * 14 = 0.15m
2. Battery pack breadth = (cell width + spacer) * 2
= (0.082 + 0.002) * 2 = 0.17m
3. Battery pack height = (height of cell + spacer) * 2
= (0.272 + 0.002) * 2 = 0.55m
Volume of the battery pack = (Pack thickness * Pack width * Pack height)
= (0.15 * 0.17 * 0.55) = 0.014 m3 = 14 Liters
In Suzuki Access 125 we have under seat space of 21.8 L, thus our battery pack can be placed under the seat space
Base plate Calculation
• Area of base plate, A = 150 * 170mm
• q(load) = Total force/Total area = (battery pack mass * 9.8) / 25500
= (14*9.8) / 25500
= 5.4*10-3 N/mm2
= 5.4*103 pa
• Density = 7.85g/cc
• E, young’s modulus = 210Gpa = 210 * 109 pa
• Material Index = 21.8 * 10-3
• Cost = 34 Rs/kg
Plate thickness calculation
According to Roark’s Formula for stress
In our case,
a / b = 170 / 150 = 1.2
alpha = 0.0616
beta = 0.3762
Maximum deflection, Ymax = (-alpha * q * b4) / (E * t3)
Therefore, Plate thickness, t = 3√ ((-alpha * q * b4) / (E * Ymax))
Let’s assume max deflection Ymax as 1mm in downward direction (-ve)
:. Plate thickness, t = 3√ ((-0.0616 * 5.4 * 103 * 0.154 ) / (210 * 109) * (-1 *10-3))
Calculating max stress generated
Maximum stress on plate, σmax = (beta * q * b2) / t
2 = (0.3762 * 4,827 * 103 * 0.152)/ (1*10-6)
= 40.85 * 106 pa
= 40.85 Mpa
Allowable bending strength of mild steel = 350 Mpa
Max deflection occurring in my 1mm base plate by BP is 0.89mm
Heat load determination
Consider discharging of cell at 2C rate from SOC 100% to 20%
Cell nominal capacity = 15.6Ah
Cell nominal voltage = 3.6V
Cell internal resistance = 2.50mΩ = 0.0025Ω
Heat generated by cell at 2C discharge rate, Qcell = I2 * R
= (15.6 * 2)2 * 0.0025
Heat generated by battery pack at 2C discharge rate, Qpack = Qcell * no of cells
= 2.43 * 42
= 102.21 Watts
Energy lost by cell as heat
Time for 100% to 20% SOC at 2C discharge rate, t = (3600 sec * 0.5 * 0.80) = 1440 sec
Energy loss in cell, Ecell = (t * Qcell) = (2.43 watts * 1440 sec)
Energy loss in pack, Epack = (no of cells * Ecell) = (3.5 * 42)
= 147 KJ
Comparison with total pack energy
Pack energy from 100% to 20% SOC = (n * V * I * t)
= 42 * 3.6 * 15.6 * 2 * 1440
= 6793.11 KJ
Percentage of energy lost as heat = 147KJ / 6793.11KJ = 2.2%
Heat transfer coefficient calculation
Assume pack temperature is at 50°C when discharging at 2C rate, and ambient temperature as 25°C
Convective area for our battery pack, A = base area + top area + side areas
Heat generated from our pack at 2C discharge rate, Qpack = 102.21W
Pack Temperature, Tpack = 50°C
Ambient Temperature, Tamb = 25°C
ΔT = Tpack – Tamb = 25°C
Thermal resistance, Rth = ΔT/Q = 25 / 102.21
= 0.245 K/W
Convective Rth = 1 / (convective heat transfer coefficient * A)
Convective heat transfer co-efficient, h = 1/ (Rth * A) = 1 / (0.245 * 0.35)
= 11.7 W/m²K
Our convective heat transfer coefficient 11.7 W/m2K falls under the range for free convection process, hence
we use natural air convection methods to manage our battery pack temperature.