Project 2

                 PROJECT 2

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

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 0.5mOhm, delta time is 0.1, and entropic factor is 2 

HEAT GENERATED =JOULE HEAT + ENTROPY HEAT

from ohms law 

v=IR

when the current flowing through a resistance, there is a heat dissipated in the resistor this  heat dissipation is called joules heating.

joule heating is also known as ohmic heating

power

p=v*I=I^2R

Heat 

H=∫P.dt with the limits of time 0 to t

H=I^2*R*t

where,

P=power 

V=voltage

I=current

R=resistance

H=heat

T=time

The heat generated in the resistor is

directly proportional to the square of the current flowing through it

Directly proportional 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 0.5mOhm=0.0005Ohm

Maximum current=-0.896A

Max current time 1.7sec

Entropic factor=2

Total heat generation=jouleheat +Entropic heat

Max joule heat=I^2*Rt

EXCEL CALCULATION ATTACHED

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

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

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

SOC = Q / Qmax

where, 

Q = Available capacity

Qmax = Overall capacity

for the simulation model following blocks are used 

1.generic Battery

2.controlled current source

3.Bus selector

4.Signal builder block

5.scope

6. power gui

 HEt

resistance is chaging depend on current vaiation 

so we consider resistance 0.005oh as constant

heat genration calculation p

art of this system is shown below

the calculation of heat generation is I^Rt 

integration is used to determine the total amount of heat generated over a certain time period, given the rate of heat generation at each point in time.

RESULT 

TOTOL HEAT GENERATION IS 

0.001132 J

BATTERY SOC

RESULT

Estimate totol heat generation by excel calculation

estimate total heat generation using simulink