Project 4

AIM: To Calculate the capacity of the battery pack required and the number of series-parallel connections.For a vehicle of gross weight 300 kg, the motor ratings are 1200Wh, 65V. The range is 225km and the maximum speed is 60kmph.Give a stepwise discussion on how a battery pack is built.

Solution:-

Given Specification of Motor :-

Vehicle weight = 300 kg

Motor ratings = 1200 W

Voltage = 65 V

Range = 225 km

Top speed = 60 kmph

Calculation:-

Motor current = V * I ………(Formula)

1200 W = 65 V * I

1200 W / 65 V = I

18.46 = A

Current = 18.46 A

Current withdrawn by Motor is Current = 18.46 A

Consider Losses In Motor :- 5-10%

New Current = Current(A) * Losses

New Current = 18.46 * 1.08

New Current = 19.93 A

Now by considering the losses Actual Current withdrawn by Motor is Current = 19.93 A

Power = Voltage * Current ………(Formula)

Power = 65 V * 19.93 I

Power = 1295.45 W

Travel factor = Total range/ Max speed. ……(Formula)

Travel Factor = 225 / 60

Travel Factor = 3.75

Travel factor gives exact data about power required

New power = Power * Travel factor ……. (Formula)

New Power = 1295.45 * 3.75

New Power =4857.93 W

Based on Total range and Maximum speed calculate the Actual power required by Motor.

Consider Actual power required by Motor is 5 Kw

So Battery pack should be able to provide 5 Kw power.

Battery pack Efficiency = 90% = 0.9

Recalculate New Power because of the efficiency of the battery pack

Battery power = New power/battery pack efficiency

= 4857.93/0.9 = 5.39 kWh

Battery power = 5.39 Kwh

Motor Efficiency = 90% = 0.9

Recalculated capacity of the battery, Ah = Battery power (Wh)/voltage

= 5.39 (kWh)/65 V = 5397/65 = 83.03 Ah

capacity of the battery, Ah = 83.03 Ah

Assume this lithium ion datasheet.

From the above data sheet;

Cell voltage = 3.7 V

Cell capacity = 2.6 Ah

No of series connection = Total Voltage/Cell Voltage

= 65/3.7 = 17.56 (Approx 18 cells in series)

No of parallel connections = Total capacity/Cell capacity

= 18.46/2.6 = 7.1 (Approx 7 cells in parallel)

It can be expressed as 17S7P

Total No of cells = No of series Cells*No of parallel cells

= 17*7 = 119

Observations:-

• For calculating the capacity of the battery pack required and the number of series-parallel connections, firstly the following parameters are the motor ratings (1200 W, 65 V), the gross weight of the vehicle (300 kg), its expected total range (223 km) and the top speed (60 kmph) of the vehicle are already given. Also, a datasheet of lithium ion battery is also given.
• Firstly calculate the Motor current by formula = V * A, by this we get Current in A
• Then find out the new current by considering a losses of motor, Considering losses in motor is about 5-10%
  By this new current calculate the power.
• Calculate Travel factor of the vehicle, Travel factor gives exact data about power required
• For getting exact date again multiply the travel factor with power to get New exact power.
• Here the C-rate is considered to be 1, so the Charging/Discharging current in Amperes = C-rate*motor capacity, which is the same value 18.46 A as the C-rate is taken as unity.
• Then, the battery pack efficiency is considered as 90%, so the battery power is calculated in kWh
  Lastly, the motor efficiency is also considered as 90%, then the recalculated battery capacity is given as; Recalculated capacity of the battery, Ah = Battery power (Wh)/voltage.