Tutorial - Protecting Building Occupants from Outdoor Toxic Hazards and Chemical Attack

Martin W Liddament
March 2011

Algorithms

This program uses the VEETECH interactive algorithms:

http://phptoxic.veetech.org.uk/phptoxicgas.php

http://airqualityreservoir.veetech.org.uk/aqres.php

Introduction

Sometimes a space needs to be protected from outdoor chemical hazards. Such contaminants could be associated with transient problems, such as the pollutant emitted from vehicles in peak periods, or an accidental pollutant emission from nearby industrial or storage premises, or even pollution from a chemical attack.

There are two principle issues that need to be addressed i.e:

• For how long can a building provide occupant protection from an outdoor pollutant hazard?
• For how long will there be sufficient fresh air in the building for occupants?

Protection Time from Outdoor Pollutant

To answer the first question, a building intended to provide passive protection against pollutant ingress must be as airtight as possible. If the building itself cannot be made airtight then there must be an airtight safety zone within the building. In practice some air infiltration (and hence pollutant) ingress will inevitably occur and therefore it is necessary to know how long it will take for the safety threshold of any toxic pollutant to be reached.  A detailed analysis requires airtightness testing of the building and a comprehensive understanding of the prevailing weather conditions.

Fortunately high driving pressures for air infiltration (primarily wind in this case) will also rapidly disperse any transient outdoor pollution, therefore the most important analysis will probably be for fairly low wind speed conditions. Programs such as phpaida ( http://phpaida.veetech.org.uk/phpaida.php ) can give some pre-design guidance on infiltration rates while more complex multi-zone calculation techniques, such as CONTAM, can be used in a detailed analysis.

For a basic study, however assume the following:

• For an extremely airtight space, assume residual infiltration is 0.05 L/s for each 10m² of space. Such a level of airtightness would require specific design and construction and hence this infiltration level should only be applied in those instances where the airtightness level has been validated. In technical terms, this rate applies to an airtightness level corresponding to less than 0.1 air changes/hour (ac/h) at a test pressure of 50 Pa.
• For a very airtight space assume that the residual air infiltration into the building is equivalent to 0.5 Litre/second of outdoor air for each 10m² of floor area. This corresponds to an airtightness of about 0.5 ac/h at 50 Pa.
• For a building built to a standard airtightness specification (e.g. equivalent to 3 ac/h at 50 pa) assume an infiltration rate equivalent to 1.5 L/s for each 10 m² of floor area.
• For a building built to airtightness principles but not tested assume 4 L/s of infiltration/ 10 m² of floor area (approximately equivalent to 8 ac/h at 50 Pa. This is perhaps typical of a modern UK house built to Part L Building Regulations..
• For a building not built to airtightness standards but with weather stripping and obvious openings such as chimneys and vents sealed assume 8 L/s;
• For a leaky building with chimneys and vents in the open position assume 16 L/s of infiltration /10 m².

For this basic study, also assume a ceiling height of 3.00 m. hence the volume of each 10m ² of  floor area will be 30 m³.

Finally assume that the outdoor peak concentration of the pollutant is ‘normalised’ to the value 1. Thus, for example, if a toxic cloud has a pollutant concentration of 40 μg/m³ of air and the toxicity level is 4 μg/m³, the normalised value of this pollutant then becomes 1.0 and the toxicity level becomes 0.1.

In each of the above cases the standard dilution equation, such as that given in BS5925, can be applied to calculate the rate at which the indoor concentration of outdoor pollutant increases over time. In this exercise a worse case scenario is considered in which the outdoor pollutant level rises to 1.0 at time t = 0 and then remains at this level throughout the analysis. In reality the outdoor pollution level will, at some stage, fall back to zero. The design task, therefore, is to secure sufficient airtightness to ensure that the pollutant level indoors does not reach a critical level of 0.1 before the outdoor concentration has had time to return to zero. The interactive program http://phptoxic.veetech.org.uk/phptoxicgas.php performs this calculation and presents the approximate time taken to reach 0.1, 0.2, 0.3, 0.4, 0.5 0.6, 0.7 0.8 and 0.9 of the outdoor concentration.

Entering the above data in this algorithm (setting volume to 30 m3) yields the following results:

This shows that a very extreme level of airtightness is needed to provide protection that extends beyond an hour.

Fresh Air Time

Having identified how long it takes for the indoor pollution concentration to reach a toxic level, the next task is to determine the life support time of the air pocket in the space. Oxygen is consumed as part of the metabolic process and is converted into carbon dioxide. Under normal circumstances carbon dioxide is non-toxic but it becomes toxic before the point that oxygen depletion becomes a problem. Therefore the life support time is governed by the time it takes for metabolic carbon dioxide concentration to reach its toxic limit. These days, concentrations above 1000 ppm are associated with a loss of cognitive ability but, in practice, much higher concentrations can be tolerated for medium length periods. Health and safety 8-hour limits are typically set at 5000 ppm but a peak of 10,000 ppm should be acceptable under extreme conditions. Therefore the threshold for this analysis is set at 10,000 ppm. The rate at which the carbon dioxide concentration rises depends on

• Metabolic rate;
• The number of occupants in the space.

Thus negative factors are the number of people present and their level of activity. Clearly occupants must be as inactive as possible (i.e. in a sedentary state).

The algorithm http://airqualityreservoir.veetech.org.uk/aqres.php may be used to calculate the reservoir time. It presents the time taken to reach increments of 1000 ppm up to 10,000 ppm. It also gives the value at 72 hours. 

The data are given in the Table below for the above enclosure and for:

      1.      1 person/10m² (medium to light office occupant density);
2.      2 persons/10m² (heavy occupant density);
3.      3 persons/10m² Class room occupant density).

For the above examples only in the extreme airtightness case does CO₂ present more of a risk than the outdoor pollutant. However it is imperative that each case is considered for the specific toxicity of pollutant and actual level of airtightness.

Disclaimer

This tutorial presents is a pre-design and evaluation approach only. In critical cases much more detailed analysis will be required. Results are not guaranteed.

           
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