## How Much Heat Does A Lead Acid Battery Generate?

Occasionally we are asked very interesting questions. Recently we were asked how much heat an industrial standby battery generates. It is fair to say, it depends on whom you ask. Different battery manufacturers have different answers to this question and the different method of calculation gives significantly different answers.

The heat emitted or generated is sometimes referred to as “heat loss”.

The author of this article makes no recommendations to the methods given below. The article is produced to show that there is conflict between the various methods used.

In general terms, the question is being asked to calculate ventilation requirements and this article explores different methods and demonstrates the variability of the results.

Heat is generated on recharge, float charge and discharge. The heat generated on charge is finite, i.e. once the battery is fully charged no more heat is generated but at this point the battery enters the float charge phase and as long as the battery is on charge, heat is being generated. Heat generated on discharge is also finite because once the battery is fully discharged no more heat will be generated. Therefore, we have three conditions to consider:

**1) heat on recharge.**

**2) heat on float charge.**

**3) heat on discharge.**

We all know that lead-acid batteries are heavy and have a large thermal mass. Because of this, during recharging, float charge and discharge, the heat generated within the cells will not dissipate to the surrounding atmosphere immediately and there is a difference of opinion on how quickly this will be. Part of the differing opinion is the result of different sizes and shapes of the cells or monoblocs making up the battery, and whether they are VRLA AGM, VRLA GEL or vented types.

In basic terms, heat is watts and watts can be calculated from V x I (volts x amperes) or we can use I2R (amperes x amperes x resistance). This principal these formulas may used to calculate the heat generated.

In this article the following battery system is used in the examples. The examples consider the following: -

**a) A battery supplying 300kW for 15m at 20°C to not less than 408V (1.70Vpc average).**

**b) The battery consists of 3 parallel strings, each comprising 40 x 12V monoblocs; i.e. 240 cells.**

**c) Float voltage 2.27Vpc = 545V.**

**d) The nominal capacity of each string is 110Ah i.e. 330Ah total battery capacity.**

**e) Each monobloc internal resistance is 3.8mOhms. This value is from the battery manufacturer’s information. Therefore the battery resistance is 3.8mΩ x 40 blocs / 3 strings = 50.7mΩ total resistance.**

**f) Fully charged float current 1mA per Ah = 330mA. The value of 1mA per Ah is the I float. (note below) value from BS EN 50272.**

**g) The recharge parameters are 10% current (33A) and 2.27Vpc (544.8V) constant voltage.**

(Note) - The fully charged float current may be obtained from the battery manufacturer. However, within BS EN 50272 (Safety Requirements for Secondary Batteries and Battery Installations) the typical value can be found in Table 1. The table gives the values for current when charging with IU or U chargers. While these values are used to calculate the gas emission on charge, they may also be used to estimate the current when fully charged. In practical terms, the values are worst-case scenarios with a safety margin built in.

For vented lead-acid batteries, VRLA lead acid batteries, and for NiCd batteries, the value is given as 1mA per Ah for float voltage conditions. We should consider the Ah as the nominal at the 10h rate for lead acid product and 5h rate for NiCd product.

# 1) Heat on Recharge.

First, we need to define “recharge” and in this context, we refer to the current / time required to return the capacity removed for the previous discharge. We are only considering the time to fully charged.

The amount of heat generated does not change appreciably even though the recharging parameters may be different. For example, the charger current i.e. 5% or 10% or 15% C10 amperes or using true float voltage (e.g. 2.27Vpc) or elevated voltage (e.g. 2.40Vpc), do not significantly change the heat generated or heat loss from the battery. However, the heat generated will be substantially different depending on the depth of previous discharge. For industrial standby batteries and in this article, we consider a constant voltage / limited current recharge characteristic; otherwise known as an IU or modified constant potential method such as 2.27vpc or 2.40Vpc, or similar, with a current limit.

It is worth noting at this stage that some battery manufacturers consider the heat generated on recharge may be calculated using the same method as if the battery was on float charge. This method is used in 1.1) below. This view is taken because any heat generated on recharge will not be released immediately because of the thermal mass of the battery.

The heat calculations are complicated when we take into consideration the specific heat characteristics of the battery and at least one battery manufacturer has produced results based on actual battery type and configuration. This does not help in establishing the heat generated for every battery configuration and we need something much simpler to use in an everyday situation. After all, we are looking at a typical value that may be used for room cooling purposes not a finite “laboratory evaluation”. In practical terms a good approximation is sufficiently accurate.

It follows that if the heat generated on recharge varies with the previous discharge then all other parameters are broadly irrelevant. We can then estimate the heat generated on recharge as a function of the previous discharge. To make the calculation a little more accurate we should estimate the time to fully charged based on the IU characteristics and previous depth of discharge. Most manufacturers have tables or even a software based method to determine the time to different states of charge including the time to fully charged. However, generally speaking it can be said that the time to fully charged will be many hours but the time to 80% charged will depend on the IU characteristic. During recharging, most of the heat will be generated as losses up to the battery reaching 80% charged which will be the “constant current” part of the recharge. During the constant current phase i.e. up to 80% charged, the heat may be estimated using the I2R principal. From 80% to 100%, the float current may be used to calculate the heat. Some battery manufacturers consider the current from 80% charged to 100% charged as twice the theoretical float current. In context with the actual heat, this may be considered as a reasonable method. This method is used in 1.2) below.

1.1) Considering the heat to be the same as if the battery is on float charge we have: -

** V x I = W**, or the alternatively method of

*I2R = W.*1.1.1) *V x I = watts.*

The only issue is deciding what voltage and what current to use.

For voltage, it is reasonable to consider the voltage as the actual float voltage across the battery terminals.

For current, it is reasonable to use the I float value as defined in BS EN 50272.

Calculate for 1 bloc: -

*2.27Vpc x 6 cells x 110mA = 1,498.2mW*

Therefore, for 40 x 3 blocs = *1,498.2 x 40 x 3 = 179,784mW = 179.784W.*

This heat will be for the recharge time of 76h. Therefore the heat may be expressed as ** 180W x 76h = 13,680Wh** but over

*76h = 180W.*1.1.2) *I2R = watts*

We can use the same current as above, i.e. I float and for voltage R we can use the resistance of the bloc i.e. 3.8 mΩ. Calculate for 1 bloc: -

*110mA x 110mA x 3.8mΩ. = 0.04598mW*

Therefore, for 40 x 3 blocs = *5.5176mW.*

This heat will be for the recharge time of 76h. Therefore the heat may be expressed as ** 5.5176mW x 76h = 0.42Wh** but over the 76h recharge time =

*5.5mW.*1.2) Heat to 80% Charged plus Heat from 80% to 100% charged

1.2.1) Heat to 80% Charged

Considering the battery system described above, we know that the recharge current will be 33A up to 80% charged and from 80% we will use the 2 x float current which is if we use the 2 x float current method, a current of 330 x 2 = 660mA. We need to establish the state of charge after the discharge. Assume the worst case of maximum current for 15m: -

The maximum current = *300kW x 1000 / 408V = 735A*

The capacity removed = *(735A x 15m) / 60 = 184Ah or 146Ah charged (330Ah – 184Ah).*

This 184Ah represents 56% discharged or 44% charged.

We know that the recharge current of 33A (11A per string) will flow until the battery is 80% charged.
The 80% charged condition is = *330Ah x 0.8 = 264Ah.*

Time from 146Ah in the battery at the end of the previous discharge to 264Ah in the battery = *118Ah / 33A = 3.6h.*

We can now estimate the heat from the commencement of recharging up to 80% charged as below.

Using I2R per bloc: -

*11A x 11A x 3.8mΩ = 495.8mW.*

Therefore for 40 x 3 blocs = *59,496mW*

This current will flow for 3.6h, which can be expressed as *214Wh.*

NOTE: The internal resistance of industrial batteries does not change significantly from 100% charged to 10% charged. Therefore, the I2R principal is valid.

1.2.2) Heat from 80% to 100% charged

We need to establish the time from 80% charged to fully charged and the battery manufacturer should supply this information. However, a reasonable assumption for the purposes of estimating the heat would be 72h. It is generally accepted that a fully discharged battery can be recharged using float current and between 5% and 15% recharge current within 72h. If we assume the full 72h, we are considering a worst-case scenario.

The heat per bloc can now be estimated to be: -

*110mA x 110mA x 3.8mΩ. = 0.04598mW*

Therefore, for 40 x 3 blocs = *5.5176mW.*

This heat will be for the recharge time of 72h. Therefore the heat may be expressed as ** 5.5176mW x 72h = 0.40Wh** and if we double this we get

*0.79Wh.*Adding 1.2.1) with 1.2.2) we get 214Wh + 0.79Wh = 215Wh. This is over the full recharge time which equates to *215Wh / 76h = 2.83W*

# 2) Heat on Float Charge.

Most battery manufacturers consider the heat on float charge as a simple volts x current. V x I = W, i.e. volts x current = watts. Alternatively, the I2R principal may be used.

For current, we can contact the battery manufacturer or we could refer to International Standards such as BS EN 50272.

We can now make a calculation. Below is the calculation for the same battery considered in above, i.e. a battery comprising 40 x 12V monoblocs of 330Ah. Two alternative calculations can be made. In 2.1) we use the V X I method and in 2.2) we use the I2R method.

2.1) Considering the V x I method: -

Considering for 1 bloc: *2.27Vpc x 6 cells x 1mA per Ah x 110Ah = 1.496W.*

Therefore, for the complete battery of 40 blocs and 3 strings: -

*1.496W x 40 x 3 = 180W.*

This heat will be generated for as long as the battery is on float charge.

2.2) Considering the I2R method: -

Consider for one bloc: *110mA x 110mA x 3.8mΩ = 0.04598mW*

Therefore, for 40 x 3 blocs = *5.5176mW or 0.005W.*

This heat will be generated for as long as the battery is on float charge.

# 3) Heat on Discharge.

Interestingly, many battery manufacturers do not quote a value for the heat generated on discharge because lead acid batteries are considered as endothermic. However, manufacturers generally accept that the internal components and external connections all have a resistance and will generate heat when a current is flowing.

Again, a simple mathematical calculation may be used and most battery manufacturers accept I2R as a reasonable approximation to the heat loss on discharge. We need to know the discharge current and internal resistance of the battery system.

Using the same 40 x 12V battery discharged at 300kW for 15m we first need to modify the 300kW to a current that can be used in the calculation. The “safe option” is to consider the end of discharge voltage and then calculate the maximum current. The end of discharge voltage has been given as 408V (see above). Therefore, the maximum current is *300kW x 1000 / 408V = 735A.*

The heat loss is calculated as: -

*735A x 735A x 50.7mΩ = 27.4kW.*

This may be expressed as Wh, i.e. *27.4kW x 0.25h = 6.85kWh*

Because the battery has a thermal mass it may be many hours before this heat is transmitted to the surrounding air. The battery in this article would weigh approximately 4800kg. Some manufacturers consider the heat being dissipated to the room will be spread over 10 x the discharge time. In this example, this will be 2.5h. This would calculate out to be *2.74kW for 10h.*

# 4) Battery Dimensions and Weight

It is worth looking at the total battery dimensions and weight to give an appreciation of the heat loss when compared with the physical parameters of the battery. If the heat were generated within 1m3 it would be considerable. However, if the heat were within a 10m3 volume the impact would be minimal. The following parameters are real for the battery of 3 x 40 x 110Ah x 12V blocs give this perspective.

While the dimensions and weights given below are real, we have to remember that the stand is the open type with a large free volume around the monoblocs. The total volume considering the open area within the cells and between rows and tiers is calculated as: -

*3.7 x 0.8 x 1.3 = 3. 8m3*

*Stand type: 2 row x 3 tier open steel type.*

*Length:**3.7m*

*Depth:**0.8m*

*Overal Height:**1.3m*

*Volume:**3.8m3*

*Weight:**4000kg*

# 5) Conclusions

It is difficult to substantiate the results of the heat when the battery is on recharge or float charge because batteries do not follow standard electrical characteristic and therefore the results must be questionable. We know that ohms law when applied to batteries does not work. This is largely because of the BACK EMF characteristics of batteries, which makes the V x I calculations questionable. Therefore, any mathematical results that rely on this principal must be suspect. Accordingly the V x I calculations must be suspect. To understand this more fully, we can calculate the theoretical float current using the I = V / R method. In our examples, we know that the applied float voltage is 2.27Vpc i.e. 13.62V for the 12V, 6 cell bloc and we know the resistance is 3.8mΩ. Applying ohms law, the float current should be I = V / R = 13.62V / 3.8mΩ = 3584A. Clearly, this is not correct.

If the V x I calculations are not reliable then we must also question the I2R results. What we do know, is that the current is real value and the internal resistance is also real. Therefore, the results must be more accurate, we hope!

# HEAT ON RECHARGE

The I2R results are more real because we know what the current is and we know the internal resistance of the product. The I2R results for recharging are very small and in all practical terms, the heat may be ignored. This is only 5.5mWh in the example.

# HEAT ON FLOAT CHARGE

Again, if the I2R results are more real and the V x I method is unreliable then the 0.005W heat on float charge may again be considered irrelevant.

# HEAT ON DISCHARGE

The only method that seems to be used for heat on discharge is I2R and, as expected, the heat on discharge is considerably higher than that on recharge or float charge. What we have to remember is that the heat will not be given up immediately and some estimation of the time this is given up has to be made. Without doubt this will be hours rather than minutes but this is a matter of opinion without consulting a heating engineer.

For recharging and float charging the heat is very small, particularly when we consider the mass of the battery. This is fortunate because although there are different methods in use, the results are insignificant when put into the context of removing heat from a battery room.

For the heat generated on discharge, the situation is different because most battery manufacturers accept the I2R method as the most accurate. In addition, we can more readily accept the results because there is no BACK EMF on discharge. In this example, the heat generated can be expressed as 27.4kWh but when considering the mass of the battery we must consider this heat to be given up over a longer time than the actual discharge period of 15m. Not all manufacturers consider a time of 10 x the discharge time, but it is clear that the heat will not be given up instantly.