Project 2

                                                                                project 2

objective 

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

PROJECT DISCUSSION :

avg battery current : -20.84 Ah

battery resistance = 0.2 milli ohm

entropic heat = 2 times joules heat 

joules heat = (i^2) *R*t

t= 129 seconds

Substituting the given values, we get: joules heat = (-20.84)^2 * 0.2 * 129 = 11,298.9895.=1.1 kW

average heat generated = joules heat + entropic heat =1.1 kW+2.2 kW = 3.3Kw

maximum heated generated = ((0.8)^2)*(0.2)*(120) +2*(((0.8)^2)*(0.2)*(120))=15.36 w + 2*(15.36)= 46W

 

RESULTS 

CONCLUSION 

Using MATLAB/simulink and  the drive cycle from the attached excel sheet, we were able to find The max heat generation of the battery ad soc