HEAT DISSIPATION ON DISCHARGE AND CHARGE

Transcription

1 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. 1

2 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: where Therefore Q cal = 3600 (s) * C Ah * ( U V ) 4.18 Q cal is the heat dissipated in Calories C Ah is the discharge capacity in Ah U V is the average discharge voltage in Volt Q cal = 860 * C Ah * ( U V ) 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: 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. 2

3 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 Cell Voltage (V) over-charge charge 1.44 V EXOTHERMIIC 1.4 ENDOTHERMIC 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 Current (A) over-charge 0.12 charge 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. 3

4 Thus, the Quantity of Heat generated in calories/cell = Q cal = -860 * I c * (1.44 -U c ) where, and I c = the charge current in amperes U c = 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 * I f * (1.44 *R v ) where, and I f = floating current R v = 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: 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 4

5 Table of Floating Currents Floating current as a function of floating voltage and type of cell Floating Voltage Floating currents in ma/ah SBH SBM SBL SPH 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 Q cal = 860 * C Ah * ( U V ) = 860 * (60 * 2) * ( ) 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 = 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 5

6 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. ( ) / 2 = 8.8 amperes Thus during the charging, the heat generated is : Q cal = -860 * I c * (1.44 -U c ) = -860 * 8.8 * ( ) = k cal and the temperature increase is therefore : = / (8.4 * 0.35) = 0.3 C During the floating at 1.45 volts the heat generated is : Q cal = -860 * I c * (1.44 -U c ) = -860 * 0.29 * ( ) = k cal and the theoretical temperature increase is therefore : = / (8.4 * 0.35) = C / hour. 6