Project 2
Using MATLAB/simulink and the drive cycle from the attached excel sheet, find-
- The max heat generation of the battery
- 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
AIM:
Using MATLAB/simulink and the drive cycle from the attached excel sheet, find-
- The max heat generation of the battery
- 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
1.
heat generated = joules heat + entropy heat
joules heat
p= v*i
i^2*r
heat= h= i^2*r*t
where
p= power
v= voltage
i= battery current
r= battery resistance
h= heat
t= time
for average heat generation in battery pack
average heat = ave joules heat+2*ave joules heat
ave i= -0.9507076923
time = 3.0
r= 2
ave joules heat= -0.9507076923^2*0.5*3.0
= -0.9038451162*2*3.0
= - 5.423 j
average heat generation = -5.423+2* -5.423
= -5.423+ -10.846
= -16.2 j
for maximum heat generation in the battery pack
max = max joules heat + 2*(max joule heat)
assuming that the entropic heat is twice as the Joules heat
max joule heat = max i ^2* r*tmax
i max = -1.0304
r= 0.5
tmax= 0.7
max joules heat = (-1.0304)^2*0.5*0.7
= (-1.06172416)*0.5*0.7
= -0.371603456 j
maximum heat generation in battery = max joules heat+2* max joules heat
= -0.371603456+2*-0.371603456
= -1.114810368 j
2.
