Project-2

Using MATLAB/Simulink and the drive cycle from the attached excel sheet, find-

1. The max heat generation of the battery
2. The SOC of the battery at 2 *104second of the battery operation

Consider the battery resistance is 0.5 mOhm, delta time is 0.1 and entropic factor is 2

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

SOL: -

1. The max heat generation of the battery

From ohm's Law, V = IR

When a current is flowing through a resistance, there is heat dissipated in the resistor. This heat dissipation is called joule heating. Joule heating is also known as ohmic heating.

Power =V*I =I²*R

Heat H=∫P.dt= I² Rt

Where P = Power

V = Voltage, I = Current, R = Resistance, H = Heat, t = Time

The heat produced in a resistor is

• directly proportional to the square of the current flowing through it.
• directly proportional to resistance.
• directly proportional to time for which current flows through it.

From the given drive cycle data Maximum current was recorded at 1.7 sec = -0.896A

Maximum Heat Generated in Joules = I^2*R*t = (-0.896) ^2*0.002*1.7= 0.00273 joules

Total Heat Generated = Joule heat+ Entropic heat

Entropic Heat = 2 joules

Entropy heat:

The heat generation due to entropy change inside a battery occurs when electrochemical reactions are performed. The entropy heat is reversible heat resulting from changes in open circuit voltage with respect to temperature at two electrodes.

Total Heat Generated =0.00273+2 =2.00273 joules

b.The SOC of the battery at 2 *104second of the battery operation

The blocks used in the Model

1. From Spreadsheet
2. Controlled Current Source
3. Battery
4. Bus Selector
5. Scope

The parameters of the Blocks as below

Drive Cycle Input

In the given drive cycle data, all the currents are negative which means the vehicle is in regenerative braking and the battery is in charging position, in parameters, we have given that the SOC of the battery is 90%, the resultant SOC of the battery would be increased in SOC.

From the above plot, we see that SOC of battery = 90.025%

The SOC of the battery at 2 *104s =208s

Developed the given data time with the 0.1 step time and the simulation time of 208

Consider the current SOC = 90%

After the simulation of 208 sec, the current SOC = 91.83%