Project 2
AIM:-----The max heat generation of the batter
.......................
1-The max heat generation of the battery
The heat generated = Joules Heat + Entropy Heat
- the conversion of electrical energy into heat energy by the resistance in a circuit is known as joules heating
- entropy is a measure of the heat transfer of energy into a system.
Consider the battery resistance is 0.5 mOhm, delta time is 0.1 and entropic factor is 2
battery resistance = 0.5mohm
Battery resistance = 0.5 mOhm = 0.0005 Ohm
delta time = 0.1
entropic factor = 2
I= -0.896
calculation:
Maximum jouls heat generated = I^2*R*t
=(-0.896)^2*0.0005*1.7
= 0.0006823936
Max heat generation = 0.0006823936 + 2 ------------------>2.0006823936 Jouls
................................
b)The SOC of the battery at 2 *104second of the battery operation
|
ime |
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 |
result of the scope -------
conclusion
the soc won't discharge because the current values fed to the drive cycle are negative and in the negative current case is considered to be a regenerative case
so soc will be 100%