Quote of the Day

We didn't lose the war for that, but I don't know why we didn't.

— Admiral Leahy on Admiral Halsey's actions at the Battle of Leyte Gulf. As I have read more history of WW2, I have come to see that Admiral Halsey had issues – he was way too aggressive. I have also come to admire the thoughtful leadership of Spruance and Leahy.

## Introduction

I recently had an engineer ask me how to determine the ventilation requirements for a battery room that contains lead-acid batteries being charged. As I have discussed in previous posts (here), lead-acid batteries often release hydrogen gas while being charged. Because hydrogen gas is explosive for concentration levels from 4% to 94% (reference), care must be taken to ensure that the hydrogen gas levels in a battery room do not rise to these concentration levels. Safe operation is usually maintained by ensuring that the battery room is properly ventilated. Of course, ventilation systems can fail, which means that battery rooms must have hydrogen sensors to generate alarms in the event of a ventilation failure. I have included references to some well-known hydrogen explosions in Appendix B.

In this post, I will be looking at how these ventilation calculations are performed. My focus here is on standard wet-cell batteries, which simply release any gases generated during charging. Other battery types, like Absorbed Glass Mat (AGM), will attempt to recombine the H_{2} and O_{2} released during charging. The AGM battery's internal gas recombination will reduce the amount of H_{2} released by these batteries – example shown here. I will compare the output of my model to those generated by a number of web-based and textbook sources. The results are in good agreement.

I have included my Mathcad (source and PDF) and Excel version here. The Excel version includes a number of scenarios that compare my worksheet results to the output from various web-based tools. There is a small macro in the worksheet that allows me to choose the scenario desired from a pick list.

## Background

### Definitions

- Outgassing
- In the case of a battery, outgassing is the undesired release of H
_{2}and O_{2}gas during the charging process. - Float Voltage
- Float voltage is the voltage at which a battery is maintained after being fully charged to maintain that capacity by compensating for self-discharge of the battery. (Source)
- Charge Capacity
- The charge capacity of a battery is defined as the total charge available from a battery under a constant current load over a specific time interval – usually 20 hours, but discharge intervals of 4, 6, 8, and 10 hours are also used. The choice of time interval is driven by the application. For example, telecom applications are usually required to have a backup time guarantee of 8 hours with a battery that has lost 20% of its charge capacity due to aging. This means the battery must have an initial charge rating specified for 10 hours (i.e. 10 hours · [100% - 20%]=8 hours).
- The charge is measured in units of amp-hours (A-hr).
- C-Rate
- C-rate is the theoretical current that can be drawn in one hour from a battery of nominal capacity. For example, a 10 A-hr battery has a c-rate of 10 A. A battery's charge and discharge currents are often normalized with respect to c-rate. For example, we would refer to 1 A load from a 10 A-hr battery as a 0.1 c-rate load (=1 A/ 10 A).

### Battery Outgassing Basics

#### Why Do Batteries Outgass?

Charging a battery means redepositing lead on the negative terminal and lead oxide on the positive terminal. Because no chemical process is perfectly efficient, some of the charge current inevitably ends up electrolyzing the water in the electrolyte instead of charging the battery. The electrolysis process releases H_{2} and O_{2} molecules – an explosion hazard will exist if the H_{2} is allowed to accumulate. This electrolysis is inevitable because water electrolyzes at voltages greater than 1.227 V and the battery has a cell voltage greater than 1.75 V. The rate of outgassing increases dramatically with the cell voltage.

Many battery applications involve using batteries as a backup energy source. Ideally, these batteries are kept in a state of full charge until needed. However, all batteries have internal loss mechanisms that require a small charge current be continually applied to make up for these internal losses – we call this float charging. Feeding a constant charge current into a fully charged battery causes the battery to continuously generate H_{2} and O_{2}. If you drive the battery with too much current at too high a temperature, it can also cause the battery to thermally runaway.

#### Key Points to Remember

The following list summarizes the key points associated with battery outgassing:

- A battery will release H
_{2}when it is being charged. - When being overcharged, each cell will release H
_{2}at a rate proportional to the amount of excessive charge current. The rate of gas generation,*R*, is given by Equation 1._{G}

Eq. 1 where

*N*is the number of cells per battery._{CellsPerBattery}*N*is the number of batteries in the room._{Battery}*I*is the amount of current going into gas generation._{Overcharge}*T*is the reference temperature (77 °F) for the nominal gas generation rate of 7.607 mL/min·amp. (derivation)_{Ref}*T*is the battery temperature.

- The charge current used under float conditions is usually specified as a percentage of the c-rate of the battery. I commonly see float charge currents from 1% to 5% of the c-rate. The choice of charge rate depends on the self-discharge rate of the lead-acid battery chosen. This rate can vary widely based on the chemistry of the battery (e.g. lead-calcium vs lead-antimony).

### Ventilation Basics

Ventilation requirements are usually expressed in terms of the rate of air movement (e.g. cubic feet per minute or CFM) or the air exchange rate, which is the rate at which the entire volume of air in the room is replaced.

The required air flow rate, *F*, is dependent only the rate of *H _{2}* generation and the required dilution level (Equation 2).

Eq. 2 |

where

*k*is the maximum percentage of_{H2Limit}*H*gas allowed in the room._{2}

The rate of air exchange is simple to compute given the volume of the battery room (*V _{Room}*) and the flow rate (

*F*).

Eq. 3 |

## Analysis

### Setup

Figure 2 shows how I setup the calculations. It also includes some reference links that I used to test my routine. The gas generation rate function is the key utility function.

### Flow and Exchange Rate Functions

Figure 3 shows the important ventilation function: air flow rate (*F*) and rate of exchange (*R _{Exchange}*). These functions are related by the volume of the room (

*V*).

_{Room}### Worked Example

I worked the following example using the SBS battery room ventilation site and obtained the same result as my Mathcad and Excel routines (Figure 4). I actually worked many more examples, which are included in the attached Mathcad and Excel material.

## Conclusion

In this post, I provided Excel and Mathcad models for computing the ventilation requirements for a battery room. I included worked examples that showed that my routine provides the same results as some web-based tools – I even found an error in one online example.

I should mention that the Excel version provides a nice example of how to use Excel's scenario manager with data validation to provide a easy to use tool for engineers.

## Appendix A: Rate of Gas Generation Per Ampere of Current

Figure 5 shows how to derive the hydrogen gas generation constant.

## Appendix B: Examples of Hydrogen Gas Explosions

- USS Cochino, a US diesel-electric submarine that experienced a hydrogen gas explosion.
- LZ129 Hindenburg, famous airship explosion.
- USS Scorpion, a US nuclear submarine that some feel (e.g. Rear Admiral David Oliver) was lost due to a hydrogen explosion.

Hi

In your online article https://www.mathscinotes.com/2016/10/battery-room-ventilation-math/ , in Appendix A you show a sample calculation with result of 7.60224 mL/min.

How did you get from 6.96479 mL/min at 0 deg C to 7.60224 mL/min at 25 deg C using formula: 6.964791 x (1 + (T - Tref / Tref)) with a Tref of 0, as dividing by 0 is infinity ?