Project 2

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=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.

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

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

SOC = State of Charge

SOC = Q / Qmax

where,

Q = Available capacity

Qmax = Overall capacity

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