2
$\begingroup$

everyone, I'm using nanohub's introduction to the material science of rechargeable batteries course to do some self-studying until I can go back into the lab. On one of the practice problems, it asks to calculate the energy density of a given battery system. For the first problem, it asks this for a lithium metal anode, LiCoO2 cathode system. We are asked to assume an electrode thickness of 100 micrometers and a porosity of 35%. I assumed no porosity contributions from other components such as binders. When I calculate the value I don't get something close to the answer given in the solution which is 453.6 J/kg.

I have tried calculating this and the other problems a few different ways with different units but I am not getting the correct answer. The formula I am using is

Energy density = (potential difference between anode and cathode) * cathode charge (capacity)* electrode thickness (this was given in the PS)*(1-porosity(this was also given)) below is an image of what the given information and potential answers looks like

enter image description here

$\endgroup$
0
$\begingroup$

Cathode and anode will have different specific charge capacities. These capacities can be made equal by tailoring the thickness of the electrodes. The question is asking you to make sure that the charge capacity ratio is equal to one, which means you will need to find what anode thickness you must have for the capacities match (they give you cathode thickness of 100 microns)

This is the equation to find the energy density of a battery: $$E=VC/m$$

$E$ is energy density in Wh/kg, $V$ is nominal voltage in volts, $C$ is capacity in Ah and $m$ is mass in kg. Why are you multiplying this equation with thickness and porosity?

You were given some electrode geometric parameters and you will have to find the others. Remember, you are looking for the electrodes masses so that you can compute energy density (hint: start by looking into densities of your materials).

Keep in mind that not all electrodes will be porous (for example for Li metal you can assume porosity is 0%). Also, you will have to report results in $J/kg$, so make sure you keep track of your units and convert if necessary.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.