Project 2

                                                                                                                                   PROJECT 2

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

1.Finding maximum hear generation of the battery

heat generated = joules heat + entropy heat

P=power ,V=voltage,I=current,R=resistance,t=time

from ohms law we know that V=IR

• when the current flowing through a electric conductor it dissipates heat which is called joules heat

P=V*I=I^2*R

Joules heat =∫P.dt time limit 0 to t

joules heat =I^2*R*t

• The entropic heat  of a battery is responsible for heat generation or consumption during operation and its prior measurement is mandatory for developing a thermal model,the heat generation due to entropy change inside a battery occurs when electro chemical reaction are performed

maximum heat generated = maximum joule heat + entropic heat

 we have the equation

maximum heat generated = maximum joule heat + entropic heat

        maximum joule heat=Imax^2*R*t 

                             Imax = -0.896 A

                                  R  = 0.5 milli Ohm

                                   t  = 2min=120sec

                 entropic factor = 2

therefore

     maximum joule heat  = (-0.896)^2*0.5*10^-3 *120

                                    =  48.16 milli joules

  maximum heat generated =48.16 milli joules + 2 

                                        =50.16 milliJoules   

2.FINDING SOC OF THE BATTERY AT 2*104 second of the battery operation

Block used: Signal builder, controlled current source , battery , bus selector , scope, power gui

Design

  we can calculate soc using coulumb counting method 

From battery measurements in simulink we will get the SOC 

Designing of block parameters in simulink

• Given excel sheet data is loaded to signal builder block
• Resistance value set to 1 Ohm
• In battery block initial state of charge set to 0%

remaining block parameters are set to default values.

Simulink Model :

MODEL EXPLANATION:

Here we loaded given excel sheet data to signal builder and output of signal builder is connected to a controlled current source which provide current according to the input value ,The positive terminal of controlled current source is connected to negative terminal of the battery through a resistor of given value and negative terminal of the controlled current source is connected to positive terminal of the battery inorder to charge the battery ,the battery measurements SOC , Current  and Voltage are given to Scope through a bus selector

• Run the simulation for 208 seconds

RESULT:

• From output plot we got SOC% is 100% when it charged for 208 sec
• Voltage is 6.6576 so battery is fully charged.

CONCLUSION: We found the maximum heat generated of the battery =50.16milliJoules and SOC % after 208 sec is 100% while charging.