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 0.5mOhm, delta time is 0.1 and entropic factor is 2
Aim:
Using MATLAB and Simulink tool and drive cycle find the maximum heat generated by the battery and soc of the battery.
Given:
<!-- [if !supportLists]-->· <!--[endif]-->Battery resistance 0.5mohm / 0.0005ohm.
<!-- [if !supportLists]-->· <!--[endif]-->Delta time 0.1
<!-- [if !supportLists]-->· <!--[endif]-->Entropic factor 2
<!-- [if !supportLists]-->· <!--[endif]-->Maximum time tmax 129 sec.
<!-- [if !supportLists]-->· <!--[endif]-->Maximum current Imax -0.896
|
Time in second |
Battery Current |
|
|
4 |
-0.9632 |
|
|
5 |
-0.952 |
|
|
6 |
-0.9072 |
|
|
7 |
-0.9632 |
|
|
8 |
-1.0304 |
|
|
9 |
-0.9632 |
|
|
60 |
-1.0304 |
|
|
61 |
-1.008 |
|
|
62 |
-0.9856 |
|
|
63 |
-0.9632 |
|
|
64 |
-0.9184 |
|
|
65 |
-0.9296 |
|
|
66 |
-0.9296 |
|
|
67 |
-0.9296 |
|
|
68 |
-0.9184 |
|
|
69 |
-0.9408 |
|
|
120 |
-0.896 |
|
|
121 |
-0.9072 |
|
|
122 |
-0.9072 |
|
|
123 |
-0.9408 |
|
|
124 |
-0.9296 |
|
|
125 |
-0.9296 |
|
|
126 |
-0.9968 |
|
|
127 |
-0.9968 |
|
|
128 |
-0.9408 |
|
|
129 |
-0.9408 |
|
|
Maximum |
129 |
-0.896 |
Heat Generation = joule heat + entropic heat
Joule heating is the physical effect by which the pass of current through an electrical conductor/resistance produces thermal energy. This thermal energy is called Joule heating. Joule heating is also known as ohmic heating.
From ohms law
V = IR
The entropic heating coefficient is the rate of change of the cell's open-circuit voltage (OCV) with respect to temperature; it is a function of state-of-charge (SOC) and temperature and is often expressed in mV/K.
Power:
P= VI
Then
P= I^2 * R
Heat,
H = ʃP.dt
H = I^2*R*t
- The max heat generation of the battery
H = I^2 x R x t
H = Maximum juoles Heat.
I = Battery current
R = Battery resistance
t = time.
= (-0.896) ^2 x 0.0005 x 129
= 0.802816 x 0.0005 x 129
= 0.051781632 joules
Heat Generation =maximum joule heat + entropic heat
Entropic factor =2
= 0.051781632 + 2
=2.051781632 J
- The max heat generation of the battery 2.051781632 J
- The SOC of the battery at 2 *104second of the battery operation
SOC: State of charge, energy available in the battery.
SOC= (total charge input (Available capacity) / Total battery capacity) x 100
Simulink Blocks:
Powergui, signal editor, controlled current source, generic battery, bus selector and scop.
Simulink model:
Results:
We are find the Maximum heat generation by the battery is 2.051781632 J and soc is increasing with time because of as per driving cycle current in negative mode (-) then battery is on charging mode.