Question

Tesla Roadster utilizing the A123 2650 cells. in place of the commercial 18650 lithium ion batteries currently in use. The battery pack consists of 5000 A123 2650 cells with 100 cells in series by 50 cells in parallel with series and parallel power connections and mechanical containment as well as thermal management and electronic controls subsystems:

Open Circuit Voltage = 3.315 V

Resistance = 0.02 ohm = 20 mohm

Nominal Discharge Voltage = 3.30 V

Rated Discharge Capacity = 2.3 Ah

Weight = 80 g

Diameter = 26 mm

Length = 65 mm

1. What is the nominal discharge voltage of the battery pack?

2. What is the rated capacity of the battery pack?

3. What is the total battery pack discharge energy?

4. What is the open circuit voltage of the battery pack?

5. What is the total resistance of the battery pack?

6. What is the battery pack peak discharge power?

7. What is the energy efficiency discharging at this peak discharge rate?

8. What is the I²R heat generation discharging at this peak discharge rate?

9. If the battery cell voltage must not exceed 4.2 V, what is the maximum regen power the pack is capable of accepting?

Answer 1

1.Nominal discharge voltage of the battery pack = (Number of cells in series)*(Nominal voltage of each cell)

= 100*3.30= 330 Volts

2.Rated capacity of the battery pack= (Number of cells in parallel)*(discharge capacity of each parallel path)

= 50*2.3 Ah = 115 Ah

3.Total discharge energy= discharge power * Total voltage

= 115 * (100*3.315) Joules

= 38.1225 KJ

4.Open circuit voltage of the battery pack = (no of cells in series )*(open circuit voltage of each cell)

= 100*3.315 = 331.5 Volts

5.Total Resistance of the battery pack = (100*0.02)/50

= 2/50 = 0.04 Ohm

6.Battery pack peak discharge power = power supllied to such a load of a value equal to the equivalent resistance of the battery when voltage of battery is peak discharge voltage

= (330*330)/(4*0.04)

=680625 W = 680.62 KW

7.At this peak discharge rate, power discharged is maximum. The energy supplied at this condition is divided equally between the battery internal resistance and the external connected load. So the efficiency will be exactly 50%.

8.The Ohmic discharges are both internal and external and equal to each other.

so, total ohmic loss= 2*680.62 W = 1361.24 KW