Project 2

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

a. The max heat generation of the battery

b. 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

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

Heat Generation Calculation:

There are two heat sources for battery heat generation.

• Joule heat
• Entropy heat

Heat generated = Joule heat + Entropy heat

Joule heat:

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.

Joule heat,

H = I²Rt

Where, P = Power

V = Voltage

| = Current

R = Resistance

H = Heat

t = Time

The heat produced in a resistor is

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

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.

As entropic heat is twice the joule heat, the formula will be as shown below.

Maximum heat generated = maximum joule heat + maximum entropic heat

From the Given data

The maximum current in the given drive cycle is -0.896 at 2 minutes (120 seconds)

Battery Resistance = 2 milli ohm

Entopic factor = 2

Maximum Joule Heat = (-0.896)² x 0.005 x 120 = 0.4816 J

Maximum Heat generated = 0.4816 + 2(0.4816) = 1.445 J

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

The blocks used in the Model

Signal Builder

Constant Current Source

Battery

Bus Selector

Scope

Model

As given in the drive cycle data all the currents are negative that means the vehicle is in regenerative braking and the battery is in charging position

From the graph we can observe that the SOC has increased from 95 to 95.014