(5)30% (i) Li and graphite (C) are the anode materials for Li ion batteries, whe
ID: 1075370 • Letter: #
Question
(5)30% (i) Li and graphite (C) are the anode materials for Li ion batteries, when they are . used for anode, please calculate their theoretical capacities, and their advantages and disadvantages for using as anodes; (ii) if the capacity of a full Li battery is determined by the amount of LiCo02 cathode, the weight of LiCo02 cathode is 391.2g, (a) what is actual capacity of this battery (considering the % of delithiation ); (b)what charge current (A) should be used when this battery wants to be charged in 30 min and 2h, respectively, (c) assume this battery has a constant OCV of 3.5V and internal resistance of 0.2 ohm, if you want to charge this battery in Sh, and 20 min, using constant-current mode, respectively, determine the applied voltage needed.Explanation / Answer
As in this question firstly we have to calculate the theoritical capacities for the anode. Theoritical capacities can be calculated by Faraday's law
Qtheoritical = nF / Wc = 26800 X 1 / 6 X 12 = 372 mAh/g
Adavantages of using graphite as anode material is basically these are rechargeable batteries. Provide different levels of graphitization so they have high capacity. They provide long cycle life and high rate capablity.
Disadvantage is associated with extreme expansion of graphite matrix leading to deterioration of the graphite as a result drastically decreased charge storage capablity.
Actual capacity of battery = nF/ 3600 Mw
Weight of LiCoO2 = 391.2 g
In g/mol = 97.87 g/mol
Actual capacity of battery = 1 X 96485/ 3600 X 97.87 = 352332 = 273 mAh/g
q = it
Q = Battery capacity
I = Charge Current
t = Time
237 = I X 150
I = 237 /150 = 1.58 mA
According to Ohm's law
I= V/R
I = 3.5 /0.2
I = 17.5 A
We want to charge this for 5h 20 min so the current applied is 17.5 A.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.