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

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

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

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 with the limits of time 0 to t

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,battery resistance = 2 milli ohm.

maximum current = -0.896

maximum current time = 2.0 sec

Entropic factor = 2

Total heat generation = joules heat + entropic heat

max joule heat = (Imax)^2*r*t

= (-0.896)^2*2*2

= 0.802816*4=3.2111264 mJ or 0.003211264 J

Entropic heat = 2*max joules heat = 2*3.2111264 = 6.422 mJ or 0.006422528 J

The maximum heat generation in the battery pack = Max joules heat+entropy heat

=9.6331 mJ OR 0.009633264 J

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

ANSWER: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

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

Simulink Model:Calculation of battery SOC:

BLOCKS USED:Battery,controlled current source,signal builder,bus selector,powergui ,scope.

Scope result:

Since the current values are negative,charging takes place.Here we can see that for the default values of battery ,it has charged from 0% to 5% due to small value of charging current.

Conclusion:The maximum heat generation of the battery is 9.6331 mJ OR 0.009633264 J.

And the battery SOC : charged from 0% to 5%.