Project 2
Here is a pragmatic approach:
Your cells feature an internal resitance.
Heat is the sum of losses inside the cells, so Pv(t) =Rcell*i(t)^ ².
Now it depends on the current profile and the thermal resistances of the setup, how the temperature in the cell
| Time | Time Step | Battery Current | Battery Resistance | P= R*I^2 | Average Heat | Maximum Heat |
| 00:00.4 | 0.1 | -0.963 | 0.5 | 0.463877 | 0.45260441 | 0.5308621 |
| 00:00.5 | 0.2 | -0.952 | 0.5 | 0.453152 | ||
| 00:00.6 | 0.3 | -0.907 | 0.5 | 0.411506 | ||
| 00:00.7 | 0.4 | -0.963 | 0.5 | 0.463877 | ||
| 00:00.8 | 0.5 | -1.03 | 0.5 | 0.530862 | ||
| 00:00.9 | 0.6 | -0.963 | 0.5 | 0.463877 | ||
| 00:01.0 | 0.7 | -1.03 | 0.5 | 0.530862 | ||
| 00:01.1 | 0.8 | -1.008 | 0.5 | 0.508032 | ||
| 00:01.2 | 0.9 | -0.986 | 0.5 | 0.485704 | ||
| 00:01.3 | 1 | -0.963 | 0.5 | 0.463877 | ||
| 00:01.4 | 1.1 | -0.918 | 0.5 | 0.421729 | ||
| 00:01.5 | 1.2 | -0.93 | 0.5 | 0.432078 | ||
| 00:01.6 | 1.3 | -0.93 | 0.5 | 0.432078 | ||
| 00:01.7 | 1.4 | -0.93 | 0.5 | 0.432078 | ||
| 00:01.8 | 1.5 | -0.918 | 0.5 | 0.421729 | ||
| 00:01.9 | 1.6 | -0.941 | 0.5 | 0.442552 | ||
| 00:02.0 | 1.7 | -0.896 | 0.5 | 0.401408 | ||
| 00:02.1 | 1.8 | -0.907 | 0.5 | 0.411506 | ||
| 00:02.2 | 1.9 | -0.907 | 0.5 | 0.411506 | ||
| 00:02.3 | 2 | -0.941 | 0.5 | 0.442552 | ||
| 00:02.4 | 2.1 | -0.93 | 0.5 | 0.432078 | ||
| 00:02.5 | 2.2 | -0.93 | 0.5 | 0.432078 | ||
| 00:02.6 | 2.3 | -0.997 | 0.5 | 0.496805 | ||
| 00:02.7 | 2.4 | -0.997 | 0.5 | 0.496805 | ||
| 00:02.8 | 2.5 | -0.941 | 0.5 | 0.442552 | ||
| 00:02.9 | 2.6 | -0.941 | 0.5 | 0.442552 |