Project 2
Question 1:
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
Solution
Description of project:
Heat generated in the battery pack can be calculated in maximum heat generated . For considering the values of current and time. we should calculate the maximum values of both current and time period from the excel sheet given. The battery pack heat generated can be obtained from the addition of joules heat and entropy heat. The joule heat is called as the heat which is generated by the battery pack due to I2R losses. The current flows through the battery pack whether the battery pack is in charging case and discharging condition. The entropy heat is the change of open circuit volatge with respective to the temperature of the battery pack. So the joule's heat and entropy heat summation can be identified as maximum heat generation of the battery pack.
fig(a) Identification of Battery pack heat generation using colour effects
The SoC of the battery pack can be obtained by bus selector which enables the user to use the factors of the battery which are SoC, voltage, current of the battery pack. So by that procedure we can determine the SOC, voltage, current of the battery pack.
Fig(b) SoC of the battery pack
Specifications of the project:
The Specifications of the project are considered from the excel sheet which is attached below. The excel sheet consists of data of the battery pack in which the time step and battery current are identified respectively. From the excel data, Specifications are noted.
| 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 |
Maximum current = -1.0304 A
Maximum time = 0.7sec
Resistance of battery pack = 2mΩ = 0.002 Ω
Entropy heat = 2
Required Formulations for project:
Maximum Joule heat = H = I*I Rt ----------->(1)
Maximum heat generated = Maximum Joule heat generated + Entropy heat ----------->(2)
Joule heat calculations:
Joule heat = H = I*I Rt -------->(1)
where
H is the Joule heat
I is the current with respect to max, avg
R is the resistance of battery pack
t is the time period
Maximum heat generation of battery pack calculations:
Now let's calculate the Maximum Joule heat generated using equation (1)
Maximum Joule heat generated= H = Maximum current^2 x resistance of battery pack x Maximum time period
Maximum Joule heat generated= H = (-1.0304)^2 x 0.002 Ω x 0.7sec
Maximum Joule heat generated= H = 0.001485 joules
Now let's calculate the Maximum heat generated using equation (2)
Maximum heat generated = Maximum Joule heat generated + Entropy heat factor
Maximum heat generated = 0.001485 + 2 = 2.00148
Maximum heat generated from the battery pack = 2.00148 joules
Simulating the project:
fig(a) Simulink model for Maximum heat generation of the battery pack of Question 1
fig(b) Simulink model for SoC of the battery for Question 2
fig(c) Battery specifications of lithuim ion chemistry type
fig(d) Signal builder response from the excel sheet data
fig(e) SoC response of the battery pack
fig(f) Current response of the battery pack
fig(g) Voltage response of the battery pack
Observations of the project:
* The project has pratical calculation and here I also simulated the practical calculation, so both values are matching with each other as shown above.
* The signal builder has been developed by the excel data which is given in the question. The data from the excel sheet is given as input to the signal builder to get the response as shown above. When the signal builder obtains the response then the battery would discharges particularly.
* The Maximum heat generated of the battery pack is identified as 2.00148 joules, this is the calculation in a practical manner.
* The Maximum heat generated of the battery pack is identified as 2.001 joules, this is the calculation in simulation. Both the values are attached above for consideration.
* As the question 2 were asked to obtain the SoC of the battery pack, the SoC is identified as the 99.5% approx is the state of charge of battery pack.
* The battery pack's Current response is identified as the 1.03A as highest current drawn and 0.94A is at a constant drawn current of pack
* The battery pack's voltage response is identified as the 56V as highest voltage attained and it got steep to 55 approx volatge of pack
Conclusion of the project:
As the project suggested to Use MATLAB/simulink and there is a drive cycle attached in a excel sheet. The maximum heat generation of the battery and The SOC of the battery operation is suggested to obtained. Thus the maximum heat generated and SoC of the battery, voltage and current of the battery pack are also identifed. The simulation time is also set to 208 seconds as suggested. Hnece the output are matched with each other, pratical calculation, simulation respectively matched.
The Maximum heat generated from the battery pack is 2.00148 joules. The battery pack used is of 48V battery pack and of 10Ah capacity pack. The SOC at the initial stage is set to 100%.