Project 2

AIM:

To find

a)The max heat generation of the battery

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

Solution:

a)

There are two heat source for battery heat generation. They are

• Joule heat
• Entropy Heat

Heat generated = joule heat + entropic heat

Joule heating is due to the heat dissipation in resistor when current pass through them.

Joule Heat, H = I^2*R*t

Where I = curreny

R = Resistance

H = Heat

t = time

Entropy Heat

Heat Generation due to entropy change inside a battery due to electro chemical reaction.

Given

battery resistance = 0.5 mOhm

delta time = 0.1

entropic factor = 2

entropic heat is twice as the Joules heat

Now, Maximum heat generation = Maximum joules heat + 2*(Maximum joules heat )

Maximum joules heat = Imax^2*R*t_total

Maximum joules heat = 1.034^2*0.5*10^-3*129

Maximum joules heat = 0.06896 Joule

Max heat generated = Max joules heat + 2*(Max joules heat)

Max heat generated = 0.06896 + 2*0.06896

Max heat generated = 0.20688 Joules

b)

Given

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

The figure below shows the simulink model for finding SOC

Signal Builder

Result

The figure above shows the SOC , Voltage and Current of the Battery.

We can see that the at 208 sec the SOC of the battery is around 98.985 %

Conclusion:

Calculated the Max heat generation of the battery and built a simulink model of battery for find the SOC at 208 sec and analysed the result