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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…
Vijay S
updated on 02 Feb 2023
Using MATLAB/simulink and the drive cycle from the attached excel sheet, find-
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-
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 flowing 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 electro chemical reaction are performed. the entropy heat is reversible heat resulting from change in open circuit voltage with respect to temperature at two electrodes.
batterry resistance = 0.5m ohm = 0.0005ohm.
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
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
You can initialize your circuit with a specific AC or DC current. If you want to start the simulation in steady-state, the block input must be connected to a signal starting as a sinusoidal or DC waveform corresponding to the initial values.
SIGNAL BUILDER:
ouptuts:
CONTROLLLED CURRENT SOURCE:
BLOCK PARAMETERS:
BATTERY:
PARAMETERS:
BUS SELECTER:
RESULT:
SIMULINK MODEL:
RESULTS:
OUTUT:
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