Challenge

Project 2

1. Aim:-

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
Time Time Step Battery Current
00:00.4 0.1 -0.9632
00:00.5 0.2 -0.952
00:00.6 0.3 -0.9072
00:00.7 0.4 -0.9632
00:00.8 0.5 -1.0304
00:00.9 0.6 -0.9632
00:01.0 0.7 -1.0304
00:01.1 0.8 -1.008
00:01.2 0.9 -0.9856
00:01.3 1 -0.9632
00:01.4 1.1 -0.9184
00:01.5 1.2 -0.9296
00:01.6 1.3 -0.9296
00:01.7 1.4 -0.9296
00:01.8 1.5 -0.9184
00:01.9 1.6 -0.9408
00:02.0 1.7 -0.896
00:02.1 1.8 -0.9072
00:02.2 1.9 -0.9072
00:02.3 2 -0.9408
00:02.4 2.1 -0.9296
00:02.5 2.2 -0.9296
00:02.6 2.3 -0.9968
00:02.7 2.4 -0.9968
00:02.8 2.5 -0.9408
00:02.9 2.6 -0.9408
Battery resistance 2 milli ohm

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

2. Theory and calculation:-

1. The max heat generation of the battery
Heat Generated = Joule's heat + Entropic Heat

joules heat

from ohms law we know that

V=IR

when the current flowing through a resistance, there is a heat dissipated in the resistor this heat dissipation is called joules heating.

joules heating is also known as ohmic heating

power

P=V*I=I^2*R

heat

H=∫P.dt with the limits of time 0 to t

H=I^2*R*t

where,

P=power

V=voltage

I=current

R=resistance

H=heat

T=time

Entropic heat

The heat generation due to entropy change inside a battery occurs when electro chemical reaction are performed. the entropy heat is reversible heat resulting from change in open circuit voltage with respect to temperature at two electrodes.

So, first we need to find the Max. Joule's heat
Max. Current (I) = -0.896 A
Time = 1.7
Resistance (R) = 2 milli ohm
Max. Joule's Heat = I(max)^2Rt
= (-0.896)^2 * 2*10^-3 * 1.7
Max. Joule's Heat = 2.7295* 10^-3 J
Entropic heat = 2
Heat generation = Joule's heat + Entropic heat
= 2.7295*10^-3 + 2
Heat Generation = 2.0027 j
So, here the max. heat generation for the battery pack is 2.0027 j .
2. The SOC of the battery at 2 *104second of the battery operation

SOC = State of Charge is defines the how energy is usable for application

SOC = Q / Qmax

where,

Q = Available capacity

Qmax = Overall capacity

for the simulation model following blocks are used

Lithium-ion Battery :- Implements a generic battery model for most popular battery types. Temperature and aging (due to cycling) effects can be specified for Lithium-Ion battery type.

Controlled Current Source:- Converts the Simulink input signal into an equivalent current source. The generated current is driven by the input signal of the block.You can initialize your circuit with a specific AC or DC current. If you want to start the simulation in steady-state, the block input must be connected to a signal starting as a sinusoidal or DC waveform corresponding to the initial values
Signal Bulider :- It will run the module upto the 208s and the stop.
Scope:- It shows or display the result.
Powergui :- Set simulation type, simulation parameters, and preferences
3. Matlab Module:-
Simulink design module as shown below
4. Result:-
In the above result we can see that we are getting the result through the signal builder, In that result we can see that the vaule of current is changing with respect ot time, on the X axis we have taken the time value from the 0.1 to 2.6 and on the y axis we have taken the current value, in the garph we can see that the value of current is changing with respect to the time, it is increasing or decreaasing.
The lithium ion battery +ve is conneccted to the +ve of the controlled current source and -ve of the battery is connected to the -ve of the current sourece. Then the output of the current source is conneccted to thesignal builder and the output of the batter is connected to the scope block . Also power gui is used to run the simulation. At the end after run the simulation the result is shown in the scope block.
Conclusion:-
Maximum heat generation is obtained by calculatiton of the how much heat generated in the battery pack. The current is flowing through the circuit that time the resistance generates the heat which is in the form of heat dissipation. So here 2.0027J maximum heat generation occured.
SOC we are getting through the signal builder block which is connected to the battery . In that result we can see the value of current is increasing or decreasing with respect ot the time.
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