Project 2

QUESTION 1:

AIM:

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

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

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

ANSWER:

1. The max heat generation of the battery

There are 2 heat sources for the battery heat generation

1. Joules heat
2. Entropy heat

Heat generated = Joules Heat + Entropy Heat

Joules Heat :

froms ohms law we know that

V = IR

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

Power,

P = V*I = I*I*R = I^2*R

Heat,

H = ∫P.dt

H = I^2*R*t

Joule heat,

H = I^2*R*t

where. p = Power

V = Voltage

I = Current

R = Resistance

H = Heat

t = Time

The heat Generated in the Reisitor is-

• Directly praportion to the square of the current flowing through it
• Directly praportion to the resistance
• Directly praportion to the time for which the current flows through it.

Entropy Heat:

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

Given data:

Battery resistance = 0.5 mOhm = 0.0005 Ohm

Maximun current = -0.896 A

Max current time = 120 sec

Entropic factor = 2

power loss =0.4mA

Total heat generation = Jouleheat + Entropy heat

The maximum heat generation in the battery pack

Max heat generation = Max Joule Heat + Entropy factor

Max Joule Heat = I^2 *R*t

Max Joule Heat = (-0.8962)⋅(0.0005)⋅(1)

Max Joule Heat = 48.168 J"

Max heat generation = 48.168+ 2 = 50 mJ

The Max heat generation is 50 mJ

QUESTION 2:

The SoC refers to the State of Charge of the battery, which is the capacity percent left in the battery which can be used. It a very improtent thing which teles about the battery status which helps in using the battery in a proper manner.

The SoC can be measued in different ways. Some of them are listed below

• Voltage Translation Method (VT)
• Coulamb counting method (CC)
• Combination of VT and CC

The VT and CC has its own advantages ad disadvantages so to get accurate value we have to use the Combination of VT and CC.

Voltage Translation Method (VT):

It is measued by Open Circut Voltge of the battery and compared it with the Data sheet of the battery and SoC is estimated. It cannot used while the battery is in operation.

Coulamb counting method (CC):

It is calculated by the formula

SoC = total charge input / Max capacity of the battery

Total charge input = charge current * time (H)

Or

Z(t) = Z(0) - 1/Cn ∫ Ni*i(t) dt with thw limits of 0 to t

where, Cn = Battery Nominal Capacity

Z(t) = State of charge

i(t) = Instantansous cell current

t = time

Ni = coulomb efficiency

Z(0) = initial state of charge

+ for discharging

- for charging

Due to Faid in the battery capacity over the time it shows error

Combination of VT and CC:

To avoid the calculation of SoC both the Voltage Translation Method (VT) and Coulamb counting method (CC) are used combinedly.

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

blocks used :

1) Battery :

• Here the battery block can be found in the director of electrical/specialized power systems/electrical drives/extra sources
• Here we used default values of the Li-ion battery.

2) Signal builder:

3) Controlled voltage source:

4) Bus selector

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

Output results:

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