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HEAT DISSIPATION ON DISCHARGE AND CHARGE
During charging and discharging a battery a certain quantity of heat is generated. In
general, this is relatively small and does not cause a significant increase in the
battery temperature. However, in some cases there may be a significant short term
increase. The actual thermodynamics of this heat generation can be quite complex
and the following two sections give a simple method of calculation for the heat
generated and also the temperature rise which would occur if there was no heat
loss. In practice, the actual temperature rise will be less than calculated, or even
zero, as the normal heat losses due to conduction, convection and radiation will
easily dissipate the small level of heat generated when the battery is in a steady
state condition. Thus the only significant temperature rise occurs during discharge or
if the battery has severe over-charge.
1 Discharge
The main data to take into account when dealing with heat dissipation in a Ni-Cd cell
is the potential of zero heat dissipation.
This is a thermodynamic value (V°) which is dependent on the electrochemical Ni-Cd
couple and has a value equal to 1.44 Volt.
During a discharge, the dissipated heat in a cell is directly related to the difference
between the V° value and the discharge voltage. See figure below.
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The instantaneous heat produced in the cell during the discharge is related to the
voltage difference mentioned above, the discharge current and the duration.
Over the complete discharge, an average value of the discharge voltage has to be
taken into account, and the following formula can be used:
Qcal = 3600 (s) * C Ah * (1.44 - UV )
4.18
where
Qcal is the heat dissipated in Calories
CAh is the discharge capacity in Ah
UV is the average discharge voltage in Volt
Therefore
Qcal = 860 * CAh * ( 1.44 - UV )
As Uv < V° = 1.44 Volt, the electrochemical reaction in the Ni-Cd cell is exothermic
during all the discharge .
The theoretical temperature elevation inside the cell taking no account of external
cooling can be derived by using the following formula:
∆ °K = Qcal / ( m * Cp )
where
Q is the heat dissipated on discharge in Calories
m is the mass of the cell in g
Cp is the specific heat of the cell in cal.g-1.°K-1
For nickel cadmium flooded industrial cells a typical average value is taken for Cp of
0.35 cal.g-1.°K-1.
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2 Charge
When a battery is charged, the energy produced by the charger is stored in the
battery. During the first part of the charge, up to the gassing step, there is virtually
no heat dissipation as, during this phase, the electrochemical charging process is
endothermic. Thus any small heating effect due to resistance loss is masked by the
cooling effect of the reaction.
1.55
Cell Voltage (V)
over-charge
1.5
charge
EXOTHERMIIC
1.44 V
1.45
ENDOTHERMIC
1.4
1.35
1.3
0
20
40
60
80
100
120
Capacity %C5(Ah)
After the gassing stage, the charging efficiency of the battery decreases and falls to
zero when the fully charged state is reached. Part of the excess energy which is not
being used to charge the battery, called overcharge, is used to decompose water to
form gas and the rest is converted to heat.
0.24
Current (A)
0.2
over-charge
0.16
charge
0.12
0.08
0.04
0
0
20
40
60
80
100
120
Capacity %C5(Ah)
The voltage at which the charge changes from an endothermic to an exothermic
reaction is 1.44 volts per cell (the « zero heat » voltage). Thus, to estimate the
dissipated heat, the difference between the cell voltage and 1.44 volts per cell is
used.
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Thus, the Quantity of Heat generated in calories/cell =
Qcal = -860 * Ic * (1.44 -Uc)
where,
and
Ic = the charge current in amperes
Uc = charge voltage
In the case of charging voltages which are below 1.44 volts per cell are concerned,
the heat dissipation is not necessarily zero. This is because at low levels of voltage
and current there is a certain level of gas recombination and this gives a heating
effect.
In these cases, a certain proportion of the zero heat voltage (1.44 volts) must be
used to calculate the heat generated.
In this case the quantity of heat generated in calories/cell =
860 * If * (1.44 *Rv)
where,
and
If = floating current
Rv = recombination value = typically 0.6 for sintered/PBE
= typically 0.3 for pocket plate
and, again, as in the discharge calculation, the theoretical temperature elevation
inside the cell taking no account of external cooling can be derived by using the
following formula:
∆ °K = Qcal / ( m * Cp )
where
Q is the heat dissipated on discharge in Calories
m is the mass of the cell in g
Cp is the specific heat of the cell in cal.g-1.°K-1
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Table of Floating Currents
Floating current as a function of floating voltage and type of cell
1.4
Floating Voltage
1.41
1.42
1.45
1.6
Floating currents in mA/Ah
SBH
1.00
1.15
1.32
2.20
20
SBM
0.85
0.98
1.13
1.80
15
SBL
0.65
0.75
0.86
1.30
12
SPH
0.40
0.50
0.65
0.70
1.5
Example
A battery consists of 40 cells type SBM161
Charge 15 hours at 62 volts (1.55 vpc) with a current limit of 16 amperes, followed
by floating at 58 volts (1.45 vpc).
The discharge is 60 amperes for 2 hours.
Calculation for Discharge
Qcal = 860 * CAh * ( 1.44 - UV )
= 860 * (60 * 2) * (1.44 - 1.20)
the average discharge voltage 1.20 is obtained from the discharge curve in the SBM
brochure for about 0.75C .
= 25 kcal
and the increase of temperature during this discharge is
∆ °K = Qcal / ( m * Cp )
= 25 kcal / (8.4 kg * 0.35) = 8.5°C (weight of SBM161 = 8.4 kg)
Calculation for Charge and Floating
Float current for SBM161 from table
= 1.8 * 161 / 1000 amperes
= 0.29 amperes
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Charging 15 hours at 16 amperes and 1.55 voltage limit
The first approximately 10 hours will be endothermic and will not generate heat.
For the final 5 hours, the voltage limit will be reached and the current will fall rapidly
to the floating current of 1.55 voltage limit.
Charge current = 16 amperes
Floating current for 1.55 volts (from table estimate 10 mA/Ah), so floating current =
10 * 161 / 1000 = 1.6 amperes
So, for the 5 hours we will assume the average between the charge current and the
floating current i.e. (16 + 1.6) / 2 = 8.8 amperes
Thus during the charging, the heat generated is :
Qcal = -860 * Ic * (1.44 -Uc)
= -860 * 8.8 * (1.44 - 1.55) = 0.832 k cal
and the temperature increase is therefore :
∆ °K = Qcal / ( m * Cp )
= 0.832 / (8.4 * 0.35) = 0.3°C
During the floating at 1.45 volts the heat generated is :
Qcal = -860 * Ic * (1.44 -Uc)
= -860 * 0.29 * (1.44 - 1.45) = 0.0025 k cal
and the theoretical temperature increase is therefore :
∆ °K = Qcal / ( m * Cp )
= 0.0025 / (8.4 * 0.35) = 0.00085 °C / hour.
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