Project 2
Ans1:-
|
Time |
Time Step |
Time In Sec |
Battery Current |
|
00:00.4 |
0.1 |
4 |
-0.9632 |
|
00:00.5 |
0.2 |
5 |
-0.952 |
|
00:00.6 |
0.3 |
6 |
-0.9072 |
|
00:00.7 |
0.4 |
7 |
-0.9632 |
|
00:00.8 |
0.5 |
8 |
-1.0304 |
|
00:00.9 |
0.6 |
9 |
-0.9632 |
|
00:01.0 |
0.7 |
60 |
-1.0304 |
|
00:01.1 |
0.8 |
61 |
-1.008 |
|
00:01.2 |
0.9 |
62 |
-0.9856 |
|
00:01.3 |
1 |
63 |
-0.9632 |
|
00:01.4 |
1.1 |
64 |
-0.9184 |
|
00:01.5 |
1.2 |
65 |
-0.9296 |
|
00:01.6 |
1.3 |
66 |
-0.9296 |
|
00:01.7 |
1.4 |
67 |
-0.9296 |
|
00:01.8 |
1.5 |
68 |
-0.9184 |
|
00:01.9 |
1.6 |
69 |
-0.9408 |
|
00:02.0 |
1.7 |
120 |
-0.896 |
|
00:02.1 |
1.8 |
121 |
-0.9072 |
|
00:02.2 |
1.9 |
122 |
-0.9072 |
|
00:02.3 |
2 |
123 |
-0.9408 |
|
00:02.4 |
2.1 |
124 |
-0.9296 |
|
00:02.5 |
2.2 |
125 |
-0.9296 |
|
00:02.6 |
2.3 |
126 |
-0.9968 |
|
00:02.7 |
2.4 |
127 |
-0.9968 |
|
00:02.8 |
2.5 |
128 |
-0.9408 |
|
00:02.9 |
2.6 |
129 |
-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: a)
P = VI= IR x I = I2R
Heat = H = Joule Heat = I2Rt
R:-Resistance = 2miliohm = 0.002 ohm
I:- Current t:-time in sec V:- Voltage C:- Current
Maximum Joule heat = I2Rt = (-0.896)2 x 0.002 ohm x 120 = 0.1926
Maximum heat generated= Maximum joule heat + Entropic heat
Maximum heat generated = 0.1926+2 = 2.1926
b)
Time given = 2 x104 = 208Sec
Battery Resistance = 0.5 mOhm = 0.0005 Ohm
A +ve signal is provided from Current Signal Controller to Battery +ve so it indicates the charging. So SOC (%) must be increasing. But in the above table, the signals are provided & all –ve current So logically it discharging the SOC (%) will be decreased.
Simulink Model:-
Result:-
Signal Provided:-
