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3D Fire Suppression Calculations  

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3D Water Fog Attack - To achieve effective results the 'fog-cone' and application angles are as important as the practical aspects of nozzle 'pulsing'. For example, a 60-degree fog-cone applied at a 45-degree angle to the floor into an average room (say 50 m³) will contain about 16 m³ of water droplets. A one second spurt from a 100 LPM flow hose-line will place approximately 1.6 litres of water into the cone.

For the purposes of this explanation let us suggest a single 'unit' of air heated at 538^oC weighs 0.45kg and occupies a volume of one cubic metre. This single 'unit' of air is capable of evaporating 0.1kg (0.1 litre) of water, which as steam (generated at this, a typical fire temperature in a compartment bordering on flashover) will occupy 0.37 m³.

It should be noted that a 60-degree fog-cone, when applied, would occupy the space of 16 'units' of air at 538^oC. This means that 1.6kg (16 x 0.1kg), or 1.6 litres of water can be evaporated - ie; the exact amount that is discharged into the cone during a single one second burst. This amount is evaporated in the gases before it reaches the walls and ceiling, maximising the cooling effect in the overhead. It may be seen that too much water will pass through the gases to evaporate into undesirable amounts of steam as it reaches the hot surfaces within the compartment.

Now, by resorting to Charles Law calculations we are able to observe how the gases have been effectively cooled, causing them to contract. Each 'unit' of air within the cone has now been cooled to about 100^oC and occupies a volume of only 0.45 m³. This causes a reduction of total air volume (within the confines of the cone's space) from 16 m³ to 7.2 m³. However, to this we must add the 5.92 m³ of water vapour (16 x 0.37) as generated at 538^oC within the gases. The dramatic effect has created a negative pressure within the compartment by reducing overall volume from 50 m³ to 47.1 m³ with a single burst of fog! Any air inflow that may have taken place at the nozzle will be minimal (around 0.9 m³) and the negative pressure is maintained.

Paul Grimwood

Fog Attack 1991

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Indirect Water-fog Attack - Giselsson and Rosander present a calculation to explain the action of indirect firefighting attack (the application of water to hot surfaces to create a steam rich atmosphere, displacing oxygen, and controlling a fire), this has been taken up by Grimwood, with a few corrections in his book 'Fog Attack'. The explanation needs some embellishment to aid understanding due to a lack of rigour in the original (for example a statement such as 90° =380kW is nonsensical). In addition some steps in the calculation and associated values are missing. This is an attempt to rewrite the indirect fog attack example calculation clearly.

 A REVISED CALCULATION

Consider a room with a 40m² floor area, 2.5m high filled with burning gases. Application of water is intended to create an atmosphere of 10% water vapour at 180°C (supply water at 10°C).

Volume of steam at 180°C = 10 m³ (10% of 100m³)

Using the ideal gas laws to correct this volume to a temperature of 100° C

V100 = V180(100+273)/(180+273) = 0.823 V180 = 8.23 m³

This is 8230 litres of steam at 100°C

A litre of water will vaporise to 1700 litres of steam at 100°C. To create the 10% steam atmosphere

8230/1700 = 4.84 litres of water must be vaporised.

To heat 4.84 litres of water from 10°C to steam at 180°C energy must be provided to:

raise the water temperature from 10° to 100°C

provided latent heat of vaporisation

raise steam temperature from 100°C to 180°C

Generally

E = m (Cp(water)Dqw+L+cp(steam)Dqs)

where m         Mass of water (kg)

Cp(water)          Specific heat capacity of water (J/kg/K)

Dqw                 Temperature rise of the water (K)

L                     Latent heat of water (J/kg)

Cp (steam)        Specific heat capacity of steam (J/kg/K)

Dqs         Temperature rise of the steam (K)

NB. the mass of 1 litre of water is 1kg

Evaluating gives

E = 4.84 (4180*90 + 2260000 + 2020*80) = 13.541 MJ

Giselsson and Rosander assume that in the first instance all this heat is held in the first 1mm of the wall. The available energy in this slab of wall may be found from:

Ewall = pwallAdcpwallDqw Joules

Where pwall Density of the wall material

A         Area of wall/ceiling

d          Depth

Cp(wall) Specific heat capacity of the wall material

Dqw     Temperature change of the wall

Assuming an initial wall temperature of 500°C and final temperature of 180°C, density of 1000 kg/m³ specific heat capacity of 1000 J/kg/K and the depth of 1mm then the area required to provided the required amount of heat is:

A=      (          Ewall          ) =(                      13.5 x106          )
 (pwallcpwall dDqw )    (1000.0x1000.0x0.001x (500.0-180.0))

= 42.2 m²

Therefore 4.9 litre of water should be applied to 42.0²of wall to achieve the required concentration of steam, an application of 0.11 litre/m² as calculated by Giselsson and Rosander and reproduced by Grimwood.

A transient model for heat losses from the walls could significantly improve this analysis as the reheating time and hence the time between applications and the duration of subsequent applications of the spray could be estimated.

Several fire suppression/control actions have occurred, firstly as stated by Giselsson and Rosander the oxygen concentration in the room is reduced inhibiting combustion reactions. In addition the compartment temperature will have been reduced decreasing thermal feedback to the fuel surface and the heat losses to the boundary increased. These thermal factors may be sufficient for the fire to jump to a lower stable equilibrium (a reverse of the flashover mechanism).

Giselsson and Rosander continue to warn of the effects of over drenching (causing the wall temperature to fall below 100°C) and observing that fuel rich atmospheres will require less water since they will be oxygen depleted already and leaner mixtures will require more. It is then stated that the opening should be kept as small as possible during the fire fighting procedure, presumably to reduce incoming oxygen. The reignition hazard is emphasised.

Richard Chitty UK

A Survey of Backdraught

ODPM

Reconnaissance and reading the fire

The first step of the operations is a reconnaissance to search for the origin of the fire (which may not not be obvious for an indoor fire, especially when there are no witnesses), and spot the specific risks and the possible casualties. Any fire occurring outside may not require reconnaissance; on the other hand, a fire in a cellar or an underground car park with only a few centimeters of visibility may require a long reconnaissance to spot the seat of the fire.

The "reading" of the fire is the analysis by the firefighters of the forewarnings of a thermal accident (flashover, backdraft, smoke explosion), which is performed during the reconnaissance and the fire suppression maneuvers. The main signs are:

• hot zones, which can be detected with a gloved hand, especially by touching a door before opening it;
• the presence of soot on the windows, which usually means that combustion is incomplete and thus there is a lack of air
• smoke goes in and out from the door frame, as if the fire breathes, which usually means a lack of air to support combustion;
• spraying water on the ceiling with a short pulse of a diffused spray (e.g. cone with an opening angle of 60°) to test the heat of the smoke;
• when the temperature is moderate, the water falls down in drops with a sound of rain;
• when the temperature is high, it vaporises with a hiss.

Use of water

Often, the main way to extinguish a fire is to spray with water. The water has two roles:

• in contact with the fire, it vaporizes, and this vapour displaces the oxygen (the volume of water vapour is 1,700 times greater than liquid water); the fire has no combustive agent anymore;
• the vaporization of water absorbs the heat; it cools the smoke, air, walls, objects, etc. and prevents an extension of the fire.

The extinction is thus a combination of "asphyxia" and cooling. The flame itself is suppressed by asphyxia, but the cooling is the most important element to master a fire in a closed area.

Closed volume fire

The use of constant flow water-fog can have unfortunate and dramatic consequences: the water pushes air in front of it, so the fire is supplied with extra oxygen before the water reaches it. This activation of the fire, and the mixing of the gases produced by the water flow, can create a flashover.

The most important issue is not the flames, but control of the fire, i.e. the cooling of the smoke that can spread and start distant fires, and that endanger the life of people, including firefighters. The volume must be cooled before the seat is treated. This strategy, originally of Swedish origin (Mats Rosander & Krister Giselsson), was further adapted by London Fire Officer Paul Grimwood following a decade of operational use in London's busy west-end district between 1984-94 (www.firetactics.com) and termed three-dimensional attack, or 3D attack.

Use of a diffused spray was first proposed by Chief Lloyd Layman of Parkersburg, West Virginia Fire Department, at the Fire Department Instructor's Conference (FDIC) in 1950 held in Memphis, Tennessee, U.S.A.

Using Grimwood's modified '3D attack strategy' the ceiling is first sprayed with short pulses of a diffused spray:

• it cools the smoke, thus the smoke is less likely to start a fire when it moves away;
• the pressure of the gas drops when it cools (law of ideal gases), thus it also reduces the mobility of the smoke and avoids a "backfire" of water vapour;
• it creates an inert "water vapour sky" which prevents roll-over (rolls of flames on the ceiling created by the burning of hot gases).

Only short pulses of water must be sprayed, otherwise the spraying modifies the equilibrium, and the gases mix instead of remaining stratified: the hot gases (initially at the ceiling) move around the room and the temperature rises at the ground, which is dangerous for firefighters. An alternative is to cool all the atmosphere by spraying the whole atmosphere as if drawing letters in the air ("pencilling").

Calculation of the amount of water required to suppress a fire in a closed volume (Compartment)

From Wikipedia - In the case of a closed volume, it is easy to compute the amount of water needed. The oxygen (O₂) in air (21%) is necessary for combustion. Whatever the amount of fuel available (wood, paper, cloth), combustion will stop when the air becomes "thin", i.e. when it contains less than 15% oxygen. If additional air cannot enter, we can calculate:

• The amount of water required to make the atmosphere inert, i.e. to prevent the pyrolysis gases to burn; this is the "volume computation";
• The amount of water required to cool the smoke, the atmosphere; this is the "thermal computation".

These computations are only valid when considering a diffused spray which penetrates the entire volume; this is not possible in the case of a high ceiling: the spray is short and does not reach the upper layers of air. Consequently the computations are not valid for large volumes such as barns or warehouses: a warehouse of 1,000 m² (1,200 square yards) and 10 m high (33 ft) represents 10,000 m³. In practice, such large volumes are unlikely to be airtight anyway.

Volume computation

Fire needs air; if water vapour pushes all the air away, the fuel can no longer burn. But the replacement of all the air by water vapour is harmful for firefighters and other people still in the building: the water vapour can carry much more heat than air at the same temperature (one can be burnt by water vapour at 100 °C (212 °F) above a boiling saucepan, whereas it is possible to put an arm in an oven—without touching the metal!—at 270 °C (520 °F) without damage). This amount of water is thus an upper limit which should not actually be reached.

The optimal, and minimum, amount of water to use is the amount required to dilute the air to 15% oxygen: below this concentration, the fire cannot burn.

The amount used should be between the optimal value and the upper limit. Any additional water would just run on the floor and cause water damage without contributing to fire suppression.

Let us call:

• V_r the volume of the room,
• V_v the volume of vapour required,
• V_w the volume of liquid water to create the V_v volume of vapour,

then for an air at 500 °C (773 K, 932 °F, best case concerning the volume, probable case at the beginning of the operation), we have¹

and for a temperature of 100 °C (373 K, 212 °F, worst cas concerning the volume, probable case when the fire is suppressed and the temperature is lowered):²

For the maximum volume, we have:

V_v = V_r

considering a temperature of 100 °C. To compute the optimal volume (dilution of oxygen from 21 to 15%), we have³

for a temperature of 500 °C. The table below show some results, for rooms with a height of 2.70 m (8 ft 10 in).

Note that the formulas give the results in cubic meters; which are are multiplied by 1,000 to convert to liters.

Of course, a room is never really closed, gases can go in (fresh air) and out (hot gases and water vapour) so the computations will not be exact.

Notes

Note 1: indeed, the mass of one mole of water is 18 g, a liter (0.001 m³) represents one kilogram i.e. 55.6 moles, and at 500 °C (773 K), 55.6 moles of an ideal gas at atmospheric pressure represents a volume of 3.57 m³.

Note 2: same as above with a temperature of 100 °C (373 K), one liter of liquid water produces 1.723 m³ of vapour

Note 3: we consider that only V_r - V_v of the original room atmosphere remains (V_v has been replaced by water vapour). This atmosphere contains less than 21% of oxygen (some was used by the fire), so the remaining amount of oxygen represents less than 0.21·(V_r-V_v). The concentration of oxygen is thus less than 0.21·(V_r-V_v)/V_r; we need it to be 0.15 (15%)

Thermal computation

In the case of a fire in a closed volume, the first concern is to lower the temperature. In the worst case, we can consider that it is necessary to absorb all the heat produced by the fire (in practice, only a part of the heat must be absorbed to extinguish the fire). The heat is transferred to the smoke, walls, ceiling, floor; part of it is carried away with the smoke by ventilation, or through poorly insulated walls. The most critical point is to absorb the heat of the smoke inside the room, and to lower the temperature, although not down to the normal ambient temperature of 20°C (68°F). The computation made with this hypothesis is thus the calculation of a maximum, the amount that is really required is smaller.

If the room is totally airtight, the fire will stop spontaneously when the concentration of oxygen drops below 15%. The volume of oxygen used for this is 0.06·V_l.⁴

A cubic meter of oxygen combined with a fuel typically produces 4,800 kcal, i.e. 20 MJ.⁵ The rise in temperature from 20 to 100 °C (68 to 212 °F) and the vaporization of one liter of water absorbs 539,000 kcal (2,260 MJ).

The volume of water V_w' that is required to absorb the heat is thus:⁶

Note that the formula gives the result in cubic meters; it is multiplied by 1,000 to convert to liters.

Notes

Note 4: the concentration of oxygen dropped from 21% to 15%, the volume of oxygen involved represents 21-15 = 6% of the volume of the room

Note 5: for example, the combustion of 1 m³ of methane requires 2 m³ of pure O₂ and generates 35.6 MJ ; 1 m³ of O₂ thus contributes to the creation of 17.8 MJ (4,250 kcal);

Note 6: V_w'·2260 = 0.06·V_r·20 in megajoules, thus V_w' = 5.31·10^-4·V_r ;
V_w'·539000 = 0.06·V_r·4800 in kilocalories, thus V_w' = 5.34·10^-4·V_r ;
the difference of 0.6% between the values is due to the approximations, and is negligible

Conclusion

Let us compare the calculated values:

We can see that both computations give closely similar values. This means that the amount of water required to cool the smoke is sufficient to make the atmosphere inert, and thus to suppress the fire.

Taken from http://en.wikipedia.org/wiki/Fire_fighting#Closed_volume_fire

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