Project 2

Aim of the Project:-

Using MATLAB/simulink and the drive cycle from the attached Excel sheet, find the following

1. The max heat generation of the battery
2. The SOC of the battery at 2 *104 seconds of the battery operation. The battery resistance is 2mΩ and we need to consider the 0.5mΩ,`delta time is 0.1 entopic 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

Calculation:-

we need to calculate a few things to solve this problem statement.

1) Heat generated = Joules heat + enotopy heat.

in the above equation, we need to calculate the Joules heat for which we need to go to basic law i.e., Ohm law

V=I*R

as we all know when the current flows through the resistance it generates heat, this heat is known as joule heating which is also known as ohmic heating.

another infamous law is also there which is known as the power equation i.e,

P=`I*R`

which can also be written as

P=`I^2 * R`

Similarly, the heat equation is also calculated as

H = `int_0^1P.dt` where the t is from 0 to 1

accumulating the power into the heat equation we will get the new equation

`H = I^2 *R*t`

from the above equation, the annotation is known as

• P = Power
• I = Current
• V = Voltage
• t = time
• H = Heat
• R = Resistance

Explanation:-

1. from the Ohm law
   1. Directly proportion to the square of the current flowing through it
   2. Directly proportional to the resistance
   3. Directly proportional to the time for which the current flows through it.
2. Entropy Heat:- The production of heat resulting from changes in entropy within a battery occurs during electrochemical reactions. Entropy heat, in this context, refers to the reversible heat generated due to alterations in the open circuit voltage concerning temperature at the two electrodes.

Solution:-

1) the maximum heat generated by the battery

Maximum Heat Generated = Maximum joule heat + 2(maximum joule heat)

`I_(max) = -0.896A`

`t_(max) = 1.7 sec`

`R_(st) =0.0005`Ω

Maximum joule heat = `I_(max)^2 * R *t_(max)`

=`(-0.896)^2 * 0.0005 * 1.7`

Maximum joule heat = 0.000682393 (Ans)

2) The SOC of the battery at 2*104 seconds of the battery operation

SOC = State of Charge is defined the how energy is usable for application

SOC = Q / Qmax

where,

Q = Available capacity

Qmax = Overall capacity

Observation:-

Model

Battery Config

Signal input for current control

Signal Output

Conclusion:-

1)The maximum heat generated in which the current flows in and out i.e., charge and discharge due to internal resistance and chemical reaction during charging and discharging of the battery generates the heat of 2.0006823936 Jouls.

2)The above graph gives the results of SOC voltage and current. Here we operated the soc of the battery at 2*104 and we can also see how the soc of the battery changes it mainly decreases from 100 to 99 and voltage also decreases