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.

Objective:

• The maximum heat generation of battery.
• State of charge(SOC) of the battery at 2*109sec of the battery operation.

a)

Heat generation in a battery:

Batteries also generate heat during charge and discharge due to enthalpy changes, electrochemical polarization and resistive heating inside the cell [2, 3]. Temperature variation inside the batteries can lead to uneven temperature distribution which creates uneven charge/discharge behaviour within the pack.

Exothermic and Endothermic:

An exothermic process releases heat, causing the temperature of the immediate surroundings to rise. An endothermic process absorbs heat and cools the surroundings.

Enthalpy ,Entropy and Joules heat:

Enthalpy is the sum total of all the energies, whereas entropy is the measure of the change in enthalpy/temperature.

Joule heating, also known as resistive, resistance, or Ohmic heating, is the process by which the passage of an electric current through a conductor produces heat.

Heat generated = Joules Heat + Entropy Heat

Joule heat,

H = I^2*R*t

Where, p = Power

V = Voltage

I = Current

R = Resistance

H = Heat

t = Time

Given

Entropic factor = 2

Battery resistance = 0.5 mOhm = 0.0005 Ohm

Maximum current = -0.896 A

Maximum time = 2.0 min

Calculations:

Maximum joule heat = I^2 *R*t

= (-0.896) ^2*0.0005*2

= 0.000802816

Maximum heat generation in the battery pack = joule heat + Entropy heat

= 0.000802816 + 2

Maximum heat generation in the battery pack = 2.000802816

b)

Battery resistance

2 milli ohm

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

1. The signal buider is given as the input to the model.
2. Here the drive cycle stop time is 2*104 i.e 208sec
3. so the simulation run time is 208sec
4. Here the output of the signal buider is given to the current controlled source which is connected to the battery as a load
5. so here the battery is dicharged accroding to this siginal given by the data.
6. Here the power gui block is rquired for any simulation of the simulink model.
7. It mainly used to store the equivalent simulink circuit which represents the state space model.
8. The ouput of a battery is given to the scope through the bus selector the bus selector outputs the battery siginals to the scope