a) Consider a lead acid battery with 100Ah capacity and a rated voltage of 24V.

What is the total capacity of energy in Watt-hours that can be stored in the battery?


b) Assume that the battery is completely empty. The battery is now charged at a C-rate of 0.50C. What is the charging current in Amps that is going into the battery?


c) How much time in minutes will it take for the battery to go from 50% SOC to 100% SOC, assuming a constant C-rate of 0.5C? You may assume a linear rate of charging.

d) If the battery is charged at an average constant voltage of 26V, and discharged at an average constant voltage of 23.4V, what is the voltaic efficiency of the battery in %? :


e) If the battery takes in a total charge of 100Ah, and discharges a total of 90Ah over one charge cycle, what is the coulombic efficiency of the battery in %? :


f) What is the roundtrip efficiency of storage in % for the battery? Assume the same voltaic efficiency and coulombic efficiency as you calculated above.

a)2400

b)50
c)60
d)90
e)90
f)81

a: 2400

b: 50
c: 60
d: 90
f:81

e: 90

i need step by step solution please

a) To calculate the total capacity of energy in Watt-hours that can be stored in the battery, you need to multiply the battery's capacity in Amp-hours (Ah) by its rated voltage (V).

Watt-hours (Wh) = Capacity (Ah) * Voltage (V)

In this case:
Capacity = 100 Ah
Voltage = 24 V

Watt-hours (Wh) = 100 Ah * 24 V = 2400 Wh

Therefore, the total energy capacity of the battery is 2400 Watt-hours.

b) The charging current in Amps can be calculated using the C-rate and the battery's capacity. The C-rate is the rate of charge or discharge relative to the battery capacity.

Charging Current (A) = C-rate * Capacity (Ah)

In this case:
C-rate = 0.50C
Capacity = 100 Ah

Charging Current (A) = 0.50C * 100 Ah = 50 A

Therefore, the charging current going into the battery is 50 Amps.

c) To calculate the time it takes for the battery to go from 50% State of Charge (SOC) to 100% SOC, you need to know the charging current and the battery capacity. Assuming a constant C-rate of 0.5C, you can set up the following equation:

Time (minutes) = (Desired SOC - Initial SOC) / (C-rate * Capacity) * 60

In this case:
Desired SOC = 100% = 1
Initial SOC = 50% = 0.5
C-rate = 0.5C
Capacity = 100 Ah

Time (minutes) = (1 - 0.5) / (0.5C * 100 Ah) * 60 = 30 minutes

Therefore, it will take 30 minutes for the battery to go from 50% SOC to 100% SOC.

d) Voltaic efficiency can be calculated by dividing the average discharge voltage by the average charge voltage and multiplying by 100 to get a percentage.

Voltaic Efficiency (%) = (Average Discharge Voltage / Average Charge Voltage) * 100

In this case:
Average Charge Voltage = 26V
Average Discharge Voltage = 23.4V

Voltaic Efficiency (%) = (23.4V / 26V) * 100 = 90%

Therefore, the voltaic efficiency of the battery is 90%.

e) Coulombic efficiency can be calculated by dividing the discharged charge by the charged charge and multiplying by 100 to get a percentage.

Coulombic Efficiency (%) = (Discharged Charge / Charged Charge) * 100

In this case:
Discharged Charge = 90 Ah
Charged Charge = 100 Ah

Coulombic Efficiency (%) = (90 Ah / 100 Ah) * 100 = 90%

Therefore, the coulombic efficiency of the battery is 90%.

f) Roundtrip efficiency of storage can be calculated by multiplying the voltaic efficiency and the coulombic efficiency and dividing by 100 to get a percentage.

Roundtrip Efficiency (%) = (Voltaic Efficiency * Coulombic Efficiency) / 100

In this case:
Voltaic Efficiency = 90%
Coulombic Efficiency = 90%

Roundtrip Efficiency (%) = (90% * 90%) / 100 = 81%

Therefore, the roundtrip efficiency of storage for the battery is 81%.