Project 2

Project Problem :

Using MATLAB/simulink and the drive cycle from the attached excel sheet, find-

1. The max heat generation of the battery
2. The SOC of the battery at 2 *104second of the battery operation

Consider the battery resistance is 0.5 mOhm, delta time is 0.1 and entropic factor is 2

Solution:

First of all we need to go through the time scaling which I'm going to use to solve this problem.

From above reference in our given data, If we assume that the time is in the seconds than the given time step is (1/10)th of second. So 10 time step represents 1 second.We have to calculate the heat generated according to this time taking into
consideration.

To understand completely what we have to calculate/evaluate let's first go through some basics of the heat generation of the Battery system in EV.

The total heat generated in Battery of EV/ES sytem can be bifurcated into 2 type.
1) Entropic Heat
2)Joule heat

Entropic Heat:

I wonder why this entropic heat comes into picture ? So the theory behind this entropic heat can be explained as follow..

In the laws of thermodynamics, Gibbs energy is the maximum possible nonexpansion work output done by a closed system under a process with constant temperature and pressure. In a closed electrochemical system, nonexpansion work output is electric energy,
so when the chemical energy is converted into electric energy under a reversible process, the electric energy equals the Gibbs energy, i.e., Δ G = – nFE . When the chemical energy is converted into electric energy under an irreversible process, the electric energy is less than the Gibbs energy, i.e., nFE < –Δ G . The residual Gibbs energy is converted into heat energy.The heat generated during
charge–discharge is separated into reversible heat ( Q rev ) and irreversible heat ( Q irrev ).The reversible heat ( Q rev ) is calculated by the change of entropy (Δ S ): Q rev = T Δ S = nFT ( E / T ).

This reversible heat is generated because of chraging the cell / Battery pack.The reversible losses inside the cell are a function of the current, the temperature, and the entropic heating coefficient, which itself is dependent on the cell chemistry.

The entropic heating coefficient is determined by exposing the cell to a range of temperatures at each SOC value of interest. The OCV is recorded at each combination of SOC and temperature, and (∂OCV/∂T) is calculated from the measurements. Since a ∆T of 20°C may result in a ∆OCV of 100μV or less.

Joule Heat :

It is nothing but ohmic losses that occuring at each battery cell terminal due to flow of current. Due to this flow of current it generates I^2*R*t losses for the time period for which it is flowing.This is due to the resistance offered by internal electrode connections of the battery pack. It is completely unavoidable except battery is in not running condition.Though some amount of internal current would flow in cell balancing process.

a) Lets calculate Maximum heat generated in Battery:

This maximum temperature of the battery is occuring in the charging state, because at this time both entropic heat & joule heat acted simultaneously. So We have to calculate both individually & then adding up both.

The data given in the problem as,

battery nternal resistance = 0.5 mili ohms

I(maximum) = 1.0304 Amps

Time (t) = 0.1 Seconds

So, joule heat would be I^2*R*t = I (maximum)^2*0.002*(0.1)

= (-1.0304)^2*0.002*0.1

= 0.00021234 watt-seconds

The entropic heat factor is given in the problem itself. So the Maximum heat would be added directly as,

0.00021234 + 2 = 2.00021234 watt-seconds or Joule.

b) SOC (State of Charge at 2*104 seconds):
Here time @ battery SOC we want to know is 2*104 = 208 second. Also we need to assume some parameters regarding battery.
Assuming battery initial SOC = 100%
Battery Nominal Voltage = 100 V
Battery capacity = 50 Ah

We will take Average current drawn back from battery to be 0.9507 Amperes and calculating SOC after 208 second.

0.9507 Amps * (208/60*60) hours divided by 50 Amp-Hours. = (0.0549/50)*100 = 0.1098 battery percent will be drained from 100 %

So the actual (SOC) after 208 Second the battery will have to be (100 % - 0.1098) = 99.8902 %

I have imported the excel file in the MATLAB & have made the script file for the same both problems.The script file for the same has been attached here underneath after running the script file one can verify the output with calculations.

For part a:

For Part b :

References:

Entropy and heat generation of lithium cells/batteries (iphy.ac.cn)

Battery Entropic Heating Coefficient Testing and Use in Cell-Level Loss Modeling for Extreme Fast Charging (sae.org)

Study on the Reversible and Irreversible Heat Generation of the Lithium-Ion Battery with LiFePO4 Cathode | Request PDF (researchgate.net)

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