Consider a lead-acid battery with 500 Ah capacity and a rated voltage of 12 V.
a)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 charging at a C-rate of 0.20C. 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 0% SOC to 100% SOC, assuming a constant C-rate of 2C? You may assume a linear rate of charging.
d)If the battery is charged at an average constant voltage of 13 V, and discharged at an average constant voltage of 11.7 V, what is the voltaic efficiency of the battery in %?
e)If the coulombic efficiency of the battery is 95%, what is the roundtrip efficiency of storage in % for the battery? Assume the same voltaic efficiency as you calculated above.
PLEASE HELP PLEASE!
a)_6000
B)_100
C)_30
D_90
E)-85.5
6000
a. Power in watt hrs = volts x ampere hours
b. coulombs = amperes x seconds
c. See b above.
d.e. This should get you started. You can google this stuff.
a) To find the total capacity of energy in watt-hours (Wh) that can be stored in the battery, you need to multiply the capacity (in ampere-hours, Ah) by the rated voltage (in volts, V).
Total capacity = Capacity (Ah) × Rated voltage (V)
In this case, the capacity is 500 Ah and the rated voltage is 12 V:
Total capacity = 500 Ah × 12 V = 6000 Wh
Therefore, the total capacity of energy that can be stored in the battery is 6000 watt-hours.
b) To find the charging current (in amperes, A) going into the battery when charging at a certain C-rate, you need to multiply the C-rate by the capacity of the battery.
Charging current = C-rate × Capacity
In this case, the C-rate is 0.20C and the capacity is 500 Ah:
Charging current = 0.20C × 500 Ah = 100 A
Therefore, the charging current going into the battery is 100 amps.
c) To find the time it takes for the battery to go from 0% state of charge (SOC) to 100% SOC, you need to divide the capacity by the C-rate and then convert it to minutes.
Time = (Capacity / C-rate) × 60 minutes
In this case, the C-rate is 2C and the capacity is 500 Ah:
Time = (500 Ah / 2C) × 60 minutes = 150 minutes
Therefore, it will take 150 minutes for the battery to go from 0% SOC to 100% SOC, assuming a constant C-rate of 2C.
d) To find the voltaic efficiency of the battery, you need to calculate the ratio of the energy output to the energy input (both in watt-hours) and multiply it by 100 to express it as a percentage.
Voltaic efficiency = (Output energy / Input energy) × 100
In this case, the battery is charged at an average constant voltage of 13 V and discharged at an average constant voltage of 11.7 V. The input energy is the energy stored during the charging process, and the output energy is the energy released during the discharging process.
Voltaic efficiency = (Output energy / Input energy) × 100 = [(Output voltage × Capacity) / (Input voltage × Capacity)] × 100
Using the provided average voltages:
Voltaic efficiency = [(11.7 V × 500 Ah) / (13 V × 500 Ah)] × 100 = (5850 Wh / 6500 Wh) × 100 ≈ 89.93%
Therefore, the voltaic efficiency of the battery is approximately 89.93%.
e) To find the roundtrip efficiency of storage, you need to calculate the product of the voltaic efficiency and the coulombic efficiency (both expressed as percentages).
Roundtrip efficiency = Voltaic efficiency × Coulombic efficiency
In this case, the coulombic efficiency is given as 95% and the voltaic efficiency was calculated in the previous question as 89.93%.
Roundtrip efficiency = 89.93% × 95% ≈ 85.43%
Therefore, the roundtrip efficiency of storage for the battery is approximately 85.43%.