Project 2
AIM:-----The max heat generation of the batter
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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
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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%