Project 2

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.

Solution:

For this the maximum current at any given point is taken

The maximum current in the drive cycle is -0.896 at 2 minutes(120 seconds)

Therefore

maximum heat generated=Imax^2*RBat*Time

=(-0.896)(-0.896)*2 mill ohm*120

= 192.676 milli Joule

This is the joule heat , It is given that the entropic heat is twice that of joule heat

Therefore total heat generated is Joule heat+Entropic heat

=192.676 milli+2(192.676 milli)

=74.248 milli Joules.

Average heat is calculated by taking the average value of current and time

Average value of time is= 1929/26

= 74.19 seconds

Average value of current is= -24.7184/26

= -0.9507

Therefore Joule heat generated= IsquarsquarR*t

= -0.9507*-0.9507*2 milli* 74.19 sec

= 134.11 milli Joule

Total heat generated= Joule heat + Entropy heat

= 134.11milli+2(134.11milli)

= 0.402 milli Joule.

2.The SOC of the battery at 2 *104second of the battery operation