Project 2
Using MATLAB/Simulink and the drive cycle from the attached Excel sheet, find-
- The max heat generation of the battery
- 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
Ans-
heat generated = joules heat + entropy heat
joules heat
from ohms law, we know that
V=IR
when the current flows through a resistance, there is a heat dissipated inn the resistor this heat dissipation is called joules heating.
joules heating is also known as ohmic heating
power
P=V*I=I^2*R
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
entropy heat
the heat generation due to entropy change inside a battery occurs when electrochemical reactions are performed. the entropy heat is reversible heat resulting from change in open circuit voltage concerning temperature at two electrodes.
batterry resistance = 0.5m ohm = 0.0005ohm.
- a) Maximum heat generated = maximum joule heat + 2(maximum joule heat)
Imax = -0.896 A
tmax = 1.7 sec
Rint = 0.0005 ohm
maximum joule heat = (Imax)^2 * R * tmax
maximum joule heat = (-0.896)^2 * 0.0005 * 1.7 = 0.000682393J
Maximum heat generated = maximum joule heat + 2(maximum joule heat)
Maximum heat generated = 0.000682393J+2=2.000682393J
-
2. The SOC of the battery at 2 *104second of the battery operation:
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.

