Archive for September, 2010

Gas Cooling: Part 4

Sunday, September 12th, 2010

Reading the Fire

Before returning to discussion of the science underlying gas cooling as a fire control technique, I wanted to share a video of an industrial fire in Maidencreek Township, Pennsylvania that provides an excellent illustration of smoke and air track indicators. Watch the first minute (1:00) of the video and answer the following questions:

  • Consider how you would read the smoke and air track indicators (particularly the level of the neutral plane and velocity) if this was a single family dwelling. How is air track indicators are different in a large building (with multiple ventilation openings) such as was the case in this incident?
  • What stage of development (incipient, growth, fully developed, or decay) and burning regime (fuel or ventilation controlled) is this fire in?
  • Watch the remainder of the video and examine the effectiveness of the master stream application? Are the streams effective? Why or why not? What could be done to increase the effectiveness of application?

For additional information on reading the fire, see the following posts:

Gas Laws

Paraphrasing Albert Einstein, British science writer Simon Singh stated that, “Science has nothing to do with common sense. Common sense is a set of prejudices” (Capps, 2010, p. 115). One of the challenges faced by firefighters engaging with the science of their craft is the common sense understanding of the fire environment and firefighting practices. This post continues examination of gas cooling as a fire control technique, by peeling off a few more layers and digging deeper into the underlying science related to the behavior of gases.

Readers who have worked through Gas Cooling Part 1, Part 2, and Part 3 have a reasonable idea how a small volume of water can reduce the temperature of the upper layer in a compartment and also reduce the volume of the upper layer (raising the level of the lower boundary of the layer). In addition, readers are likely to also understand the limitations of the simple explanation provided in prior posts.

In Water and Other Extinguishing Agents (Särdqvist,2002), Dr. Stefan Särdqvist provides a fairly detailed explanation of volume changes during smoke cooling and examines how the percentage of water vaporizing in the upper layer influences these changes. Understanding Stefan’s explanation requires a good understanding of the ideal gas law and a willingness to work through the math.

Gas Laws

The introduction to the gas laws and overview of Charles’s Law was provided in Gas Cooling: Part 3. This content has been repeated in this post, to save you from going back to the previous post.

While gases have different characteristics and properties, the behavior of gases can be described in general terms using the ideal gas law. This physical law describes the relationship between absolute temperature, volume, and pressure of a given amount of an ideal gas.

Figure 1. Temperature, Volume, Pressure & Amount

The concept of an ideal gas is based on the following assumptions:

  • Gases consist of molecules in random motion
  • The volume of the molecules is negligible in comparison to the total volume occupied by the gas
  • Intermolecular forces (i.e., attractive forces between molecules) are negligible
  • Pressure is the result of gas molecules colliding with the walls of its container

The ideal gas law is actually a synthesis of several other physical laws that each describes a single characteristic of the behavior of gases in a closed system (enclosed in some type of container).

Charles’s Law: In the 1780s, French scientist Jacques Charles studied the effect of temperature on a sample of gas at a constant pressure. Charles found that as the gas was heated, the volume increased. As the gas was cooled, the volume decreased. This finding gave rise to Charles’s Law which states that at a constant pressure the volume of a given amount (mass or number of molecules) of an ideal gas increases or decreases in direct proportion with its absolute (thermodynamic) temperature. The symbol  is used to express a proportional relationship (much the same as = is used to express equality), so this relationship can be expressed as:

Where:

V=Volume

T=Temperature

When two values (such as volume and temperature in Charles’s Law) are proportional, one is a consistent multiple of the other. For example If one value was consistently eight times the other, the values would also be proportional. In the case of Charles’s Law when the absolute temperature of a gas doubles, the pressure doubles. Figure 2 illustrates the relationship between absolute temperature in Kelvins (K) and volume in cubic millimeters (mm3).

Figure 2. Charles’s Law

This relationship can also be stated using the following equation:

Where

V=Volume

T=Temperature

Subscript of 1 refers to initial conditions

Subscript of 2 refers to final conditions

Gay-Lussac’s Law: When Jacques Charles discovered the relationship between temperature and volume, he also discovered a similar relationship between temperature and pressure. However, Charles never published this discovery. Charles’s work on temperature and pressure was recreated by French chemist Joseph-Louis Gay-Lussac. Gay Lussac’s Law states that if the volume of an ideal gas is held constant, the pressure of a given amount (mass or number of molecules) of an ideal gas increases or decreases proportionally with its absolute temperature. As with Charles’s Law, Gay-Lussac’s law can be expressed mathematically as:

Where

V=Volume

P=Pressure

Figure 3. Gay-Lussac’s Law

Boyle’s Law: in the 1660s, Irish physicist Robert Boyle studied the relationship of pressure and volume of gases. Boyle discovered that as pressure on a gas was increased, its volume decreased. Boyle’s Law states that if the temperature of an ideal gas is held constant, the pressure and volume of a given amount (mass or number of molecules) of an ideal gas are inversely proportional, as pressure increases, the volume occupied by the gas decreases. Boyle’s Law can be expressed mathematically as:

Where:

V=Volume

P=Pressure

Figure 4. Boyle’s Law

General Gas Law: The General Gas Law simply integrates Charles’s, Gay-Lussac’s, and Boyle’s Laws to state that the volume of an ideal gas is proportional to the amount (number of molecules) and absolute temperature and inversely proportional to pressure. The General Gas Law can be expressed mathematically as:

Where:

V=Volume

n=Mole (mol)

T=Temperature

P=Pressure

The General Gas Law defines the amount of gas in terms of the number of molecules, measured in moles (which has nothing to do with the animal having the same name).

Mole: While related to Avogadro’s Law, the term mole as a unit of measure was conceived by German chemist Wilhelm Ostwald in 1893. Unlike liters or grams, a mole is not a unit of volume or mass, but a counting unit. A mole is defined as the quantity of anything that has the same number of particles found in 12 grams of carbon-12. As atoms and molecules are extremely small, a mole is a large number of molecules. Specifically a mole contains 602,510,000,000,000,000,000,000 (more commonly written 6.0251 x 1023 in scientific notation) molecules of a substance. The number of moles of a substance is denoted by the letter n. In SI units, a kilogram mole (Kmol) is often used instead of the mole. A Kmol is 1000 mol or 6.0251 x 1026 molecules of a substance.

It may seem that using the mole to measure an amount of a substance makes this more complicated. After all, why not use a measure of volume such as liters or cubic meters or mass such as grams or kilograms? Chemical formula (such as H2O for water) describes the makeup of a chemical compound in terms of the numbers of atoms of each element comprising a single molecule of the substance.

While not a unit of mass, moles can be related to mass (just as you can determine the mass of a dozen eggs of a given size, by multiplying the mass of one of the eggs by 12).

Molar Mass: The molar mass of a compound is the mass of 1.0 moles of the substance in grams. Molar mass is determined by the sum of the standard atomic weights of the atoms which form the compound multiplied by the molar mass constant (Mu) of 1 g/mol. Figure 5 illustrates how the molar mass of water is calculated.

Figure 5. Molar Mass of Water

Molar mass can also be calculated for mixtures of substances. When dealing with mixtures, the molar mass of each constituent is calculated and applied proportionately on the basis of the percentage of that substance in the mixture. For example air is comprised of 78% Nitrogen, 21% Oxygen, and 1% of other gases such as Argon (Ar) and Carbon Dioxide (CO2). Nitrogen (N2) and Oxygen (O2) molecules are each comprised of two atoms (and are referred to as diatomic molecules). This means that the molar mass of Nitrogen and Oxygen molecules is twice the atomic mass.

Figure 6. Molar Mass of Air

Hopefully how the concepts of the mole and molar mass can be applied will become clear after examining the expansion of water when turned to steam and application of the gas laws to integrate steam expansion and changes in volume of the upper layer during gas cooling under a variety of circumstances.

Avogadro’s Law: In 1811, Italian physicist and mathematician Amedeo Avogadro published a theory regarding the relationship of the number of molecules in a gas if temperature, pressure, and volume are held constant. Avogadro’s Law states that samples of ideal gasses, at the same absolute temperature, pressure and volume, contain the same number of molecules regardless of their chemical nature and physical properties. More specifically, at a temperature of 273 K (0oC) and absolute pressure of 101300 Pa, 22.41 L (0.001 m3) of an ideal gas contains 6.0251 x 1023 molecules (1.0 mol)

Ideal Gas Law: This gas law integrates Avogadro’s law with the Combined Gas Law. If the number of molecules in a specific volume of an ideal gas at a consistent temperature and pressure (273 K and 101300 Pa) is always the same, then the proportional relationship between pressure, volume, temperature, and amount can be defined as having a constant value (Universal Gas Constant).

Where:

P=Pressure (Pa)

V= Volume (m3)

T=Temperature (K)

n=Moles

Ru=Universal Gas Constant (8.3145 J/mol K)

Universal Gas Constant (Ru): This physical constant identifies the internal kinetic energy per mole of an ideal gas for each Kelvin of temperature (J/mol K). As it is universal this constant is the same for all gases that demonstrate the properties of an ideal gas.

If the pressure, volume, and temperature of an ideal gas can be observed and Avogadro’s Law is accepted as being true (making the amount of gas also known), the value of the Universal Gas Constant can be determined empirically (based on observation) by solving the ideal gas law equation for Ru.

Where:

V=Volume

Ru=Universal Gas Constant

n=Moles

T=Temperature

P=Pressure

Figure 7 illustrates each of the gas laws and how they are integrated into the Ideal Gas Law.

Figure 7. Gas Laws

Application-Steam Expansion

As stated in Gas Cooling: Part 3, the 5th Edition of the Essentials of Firefighting (IFSTA, 2008) states that the volume of water expands 1700 times when it is converted to steam at 100o C (212o F). However, this information is presented as a fact to be memorized and no explanation is provided as to why this is the case or that if temperature is increased further, that the volume of steam will continue to expand. In the previous post, I asked the reader to accept this assumption with assurance that an explanation would follow. Application of the ideal gas law to expansion of steam provides an excellent opportunity to exercise your understanding of the gas laws and other scientific concepts presented in this post.

What we know:

  • Molecular Mass of Water: 18 g/mol
  • Boiling Point of Water at Atmospheric Pressure: 100o C (373.15 K)
  • Density of Water at 20o C (293.15 K): 1000000 g/m3
  • Atmospheric Pressure: 101325 Pa
  • Ideal Gas Constant (Ru): 8.3145 J/mol K

What we need to find out:

  1. What is the volume of 1 mole of steam
  2. What is the density (mass per unit volume) of steam at 100o C
  3. What is the ratio between the density of water and the density of steam at 100o C

The volume of 1 mole of pure steam can be calculated by solving the ideal gas equation for V.

As 1 mole of water (in the liquid or gaseous phase) contains the same number of molecules, it’s molar mass will be the same. 1 mole of water has a mass of 18 grams. Density is calculated by dividing mass by volume, so the density of steam at 100o C can be calculated as follows:

Dividing the density of water by the density of steam at 100o C determines the expansion ratio when a specific mass of water is vaporized into steam at this temperature.

This means that if a specific mass of water is vaporized into steam at 100o C, its volume will expand 1700 times. So the 5th Edition of the Essentials of Firefighting (IFSTA, 2008) is correct, but now you know why. However, what would happen if the steam continued to absorb energy from the upper layer and its temperature increased from 100o C to 300o C, the mass of the steam would remain the same, but what would happen to the volume? You can use the Ideal Gas Law to solve this question as well.

The Next Step

Just as the Ideal Gas Law can be used to determine the expiation ratio of steam, it can also be used to calculate contraction of the upper layer as it is cooled. The next post will examine how Dr. Stefan Särdqvist integrates these two calculations to determine changes in the volume of the upper layer under a variety of conditions.

New Book

Greg Gorbett and Jim Phar of Eastern Kentucky University (EKU) have written a textbook titled Fire Dynamics focused on meeting the Fire and Emergency Services Higher Education (FESHE) curriculum requirements for Fire Behavior and Combustion. I just received my copy and at first glance it appears to be an excellent work (as I would expect from these outstanding fire service educators). One useful feature of the text is a basic review of math, chemistry, and physics as it relates to the content of the course. I will be dong a more detailed review of the book in a subsequent post, but wanted to give readers of the CFBT-US Blog a heads up that it had been released.

Ed Hartin, MS, EFO, MIFireE, CFO

References

International Fire Service Training Association (IFSTA). (2008). Essentials of firefighting (5th ed). Stillwater, OK: Fire Protection Publications.

Särdqvist, S. (2002) Water and other extinguishing agents. Karlstad, Sweden: Räddnings Verket

Gas Cooling: Part 3

Sunday, September 5th, 2010

The first post in this series, Gas Cooling, began the process of providing a conceptual explanation of the fire control technique of gas cooling. As previously discussed, gas cooling reduces the hazards presented by the upper layer in a compartment fire by cooling hot gases and reducing the potential that they will ignite. Water is an effective fire control agent for this purpose because a tremendous amount of energy is required to raise its temperature and vaporize it at its boiling point.

Gas Cooling: Part 2 identified the amount of water that is theoretically necessary to cool the upper layer of a compartment containing 40 m3 (4 m wide  x 5 m long x 2 m deep) from 500o C (932o F) to 100o C (212o F). In addition, this post identified practical limitations with the efficiency of typical combination nozzles used and determined the duration of application necessary to cool the upper layer to 100o C (212o F) at different flow rates.

This raises the question, what would happen if you didn’t apply sufficient water to cool the upper layer to 100o C (212o F)?

What If?

Steam continues to absorb energy if its temperature is increased above 100o C (212o F). Some firefighters are under the impression that you cannot have steam at a temperature above 100o C (212o F) at normal atmospheric pressure. This is incorrect. Water (in liquid form) will not increase above 100o C (212o F) as this is it’s boiling point at normal atmospheric pressure, but steam acts as any other substance in the gaseous form and can increase in temperature beyond that which it changed phase from liquid to gas.

Figure 1. Properties of Water, Steam, & Smoke

Properties of Water, Steam, and Smoke

1 100 kg/M3 =1 kg/l

2 Not applicable as smoke and steam are in the gas phase

3 TCC is based on heating water from 20o C to 100o C and conversion to steam

4 Steam will continue to absorb energy until reaching temperature equilibrium

As illustrated in Figure 1, a kilogram of steam (slightly under 1.69 m3 at 100o C) will absorb 2.0 kJ of energy for each oC that the temperature of the steam is increased. The temperature of steam will continue to increase as long as the surrounding gases and/or surfaces that it is in contact with are of higher temperature. This process will continue until the steam, gases, and surfaces that the steam is in contact with reach equilibrium (i.e., the same temperature).

So even if insufficient water is applied to lower the temperature of the upper layer to 100o C (as described in Gas Cooling: Part 2 [LINK]), the combined effects of heating and vaporizing the water (the major cooling mechanism) and heating the steam produced to a temperature higher than 100o C (212o F), can have a significant cooling effect. This effect is often sufficient to extinguish flames in the upper layer and slow or reduce pyrolysis caused by heating of fuel packages due to radiative and conductive heat transfer from the flames and hot gases in the upper layer.

Gas Laws

When water as a liquid is vaporized to form steam, it expands and becomes less dense. Fire service texts such as the 5th Edition of the Essentials of Firefighting (IFSTA, 2008) commonly state that the volume of water expands 1700 times when it is converted to steam at 100o C (212o F). These texts state this as a fact to be memorized, but do not explain why this is the case or that if temperature is increased further, that the volume of steam will continue to expand. While having a number of different characteristics as illustrated in Figure 1, steam and smoke are both in the gas phase, they behave somewhat similarly. In chemistry and physics, the behavior of gases is described by a number of physical laws collectively described as the gas laws. Understanding the gas laws provides an explanation of why smoke and the steam produced during firefighting operations behave the way in which they do.

While gases have different characteristics and properties, behavior of gases can be described in general terms using the ideal gas law. This physical law describes the relationship between absolute temperature, volume, and pressure of a given amount of an ideal gas.

Figure 2. Temperature, Volume, Pressure & Amount

The concept of an ideal gas is based on the following assumptions:

  • Gases consist of molecules in random motion
  • The volume of the molecules is negligible in comparison to the total volume occupied by the gas
  • Intermolecular forces (i.e., attractive forces between molecules) are negligible
  • Pressure is the result of gas molecules colliding with the walls of its container

The ideal gas law is actually a synthesis of several other physical laws that each describes a single characteristic of the behavior of gases in a closed system (enclosed in some type of container). Of these gas laws, Charles’s Law provides the simplest explanation of the phenomena that occur during gas cooling.

Charles’s Law: In the 1780s, French scientist Jacques Charles studied the effect of temperature on a sample of gas at a constant pressure. Charles found that as the gas was heated, the volume increased. As the gas was cooled, the volume decreased. This finding gave rise to Charles’s Law which states that at a constant pressure the volume of a given amount (mass or number of molecules) of an ideal gas increases or decreases in direct proportion with its absolute (thermodynamic) temperature. The symbol  is used to express a proportional relationship (much the same as = is used to express equality), so this relationship can be expressed as:

Where:

V=Volume

T=Temperature

When two values (such as volume and temperature in Charles’s Law) are proportional, one is a consistent multiple of the other. For example If one value was consistently eight times the other, the values would also be proportional. In the case of Charles’s Law when the absolute temperature of a gas doubles, the pressure doubles. Figure 3 illustrates the relationship between absolute temperature in Kelvins (K) and volume in cubic millimeters (mm3).

Figure 3. Charles’s Law

This relationship can also be stated using the following equation:

Where

V=Volume

T=Temperature

Subscript of 1 refers to initial conditions

Subscript of 2 refers to final conditions

It is important to remember that absolute temperature is measured in Kelvins (K), not degrees Celsius or Fahrenheit, because the Kelvin scale places the zero point at absolute zero, so that doubling the temperature in K, is actually doubling the temperature. As illustrated in Figure 4, the same does not hold true when using the Celsius scale (the Fahrenheit scale presents the same problem).

Figure 4. Absolute Temperature

Application of Charles’s Law provides a simple approach to examining the question of why application of water into the upper layer does not necessarily result in an increase to upper layer volume (by adding steam) and increasing its thickness (with the bottom of the layer moving closer to the floor). This requires the assumption that while the higher temperature inside the fire compartment results in increased pressure, this increase is fairly small and does not have an appreciable outcome on volume changes during gas cooling.

As a first step in answering the question, consider what is known at this point (as illustrated in Figure 5):

  • The initial volume of the upper layer (Vu1) is 40 m3.
  • The initial temperature of the upper layer (Tu1) is 500o C (932o F)
  • The ending temperature of the upper layer (Tu2) is 100o C (212o F)

The answer we are in search of is the ending volume of the upper layer (Vu2), the volume of fire gases (Vfg) plus the volume of steam produced (Vst) during application of water for gas cooling.

Figure 5. Compartment Temperature and Volume

Expanding Steam

As discussed in Gas Cooling: Part 2 [LINK], 4.35 kg (4.35 l) of water must be vaporized in the upper layer in order to lower the temperature to 100o C (212o F). The volume of water in liters must be converted to cubic meters (the same units of measure used for the volume of the compartment and upper layer). A liter is 0.001 m3, so 4.35 l equals 0.00435 m3. For now, we will accept that conversion of water to steam results in a 1700:1 expansion ratio (a later post in this series will explain why). With an expansion ratio of 1700:1, 0.00435 m3 of water expands to 7.395 m3 of steam at 100o C (212o F) (see Figure 6)

Figure 6. Expansion of Steam at 100o F

Figure 7 illustrates the volume of steam produced when 4.35 l of water is vaporized in the upper layer of the example compartment relative to the initial volume of the upper layer.

Figure 7. Steam Expansion in a Compartment

Contracting Upper Layer

Why doesn’t the 7.397 m3 of steam that results from vaporization of the 4.35 liters of water applied for gas cooling simply increase the volume of the upper layer by 7.397 m3? Charles’s law provides the key. Charles’s Law indicates that as a gas is heated its volume will increase in direct proportion to the increase in its absolute temperature. However, the reverse is also true. The volume of a gas will decrease in direct proportion to the decrease in its absolute temperature.

Cooling the upper layer from 500o C (932o F) to 100o C (212o F) results in a 52% decrease in absolute temperature from 773 K to 373 K. The volume of the upper layer which was initially 40 m3 is reduced in direct proportion to the reduction in absolute temperature.

The volume of the upper layer (fire gases) after cooling from 500o C (932o F) to 100o C (212o F) can be calculated by solving for Vu2:

Reduction in temperature from 500o C (932o F) to 100o C (212o F) results in reduction of the volume of fire gases from 40m3 to 19.3 m3 as illustrated in Figure 8.

Figure 8. Contraction of the Upper Layer

Putting it All Together

If the water applied to cool the upper layer expands to form 7.395 m3 of steam and the final volume of the cooled upper layer is 19.3 m3, the total upper layer volume is 26.95 m3.

Figure 8. Total Upper Layer Volume

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Dividing the Total Upper Layer Volume (Vu2) of 26.95 m3 by the area of the compartment (20 m2) determines the depth of the upper layer as being 1.347 m. Therefore, cooling the upper layer from 500o C (932o F) to 100o (212o F) C will cause the bottom of the upper layer to rise 0.6525 m (2.1’).

The Short Answer

The following points summarize the last three posts dealing with gas cooling as a fire control technique:

  • The volume of water required to cool the upper layer is quite small due to its specific heat and latent heat of vaporization in its liquid form and the specific heat of steam.
  • The expansion ratio of steam at 100o C (212o F) is 1700:1, but as the volume of water used to cool the upper layer is small, the expanded volume is still relatively small (in comparison to the contraction of the upper layer).
  • In the process of reaching equilibrium, the temperature of the upper layer is reduced to a greater extent than the temperature of the water increases due to the cooling capacity of the water and the relatively low specific heat of fire gases and air.
  • The large temperature drop in the upper layer results in a proportional reduction in volume (which works out be greater than the increase in volume resulting from the expansion of steam from water vaporized in the hot gas layer for cooling).

Based on each of these factors, a small amount of water can cool the upper layer and reduce its volume, resulting in the lower boundary of the upper layer rising as its depth decreases.

A Few Little Wrinkles!

The preceding example may conflict with your personal experience. Many of us have been in a hot, smoke filled compartment and had steam and smoke bank down on top of us after application of water. Why might this be the case?

The outcome of the preceding example depends on all of the water being vaporized while traveling through the upper layer. In this case, energy to vaporize the water is transferred from the hot gases in the upper layer, cooling the layer and causing it to contract. If the water passes through the upper layer without vaporizing, the temperature of the upper layer is not reduced and it does not contract. Water vaporizing on contact with hot compartment linings results in the steam produced being added to the volume of the upper layer. This steam cools the upper layer to some degree, but far less than using the energy of the hot gases to vaporize the water as it passes through the upper layer (compare the specific heat of steam to the specific heat and latent heat of vaporization of water in Figure 1).

When applying water fog into the upper layer, some of the water vaporizes as it travels through the hot gases and some reaches the compartment linings. Determining changes in the volume of the upper layer under these conditions is a bit more complex and requires a deeper examination of the gas laws.

Continuing the Discussion

The next post in this series will examine the other gas laws that lead to the development to the Ideal Gas Law and how this law can be used to answer questions about changes in upper layer volume as a result of gas cooling under a variety of different conditions.

Spanish Translation of Effective and Efficient Fire Streams

Thanks to Firefighters Tomá Ricci and Martín Comesaña from San Martín de Los Andes, Argentina for translating the series of posts on Fire Stream Effectiveness and Efficiency into Spanish. They can be downloaded in PDF format:

Ed Hartin, MS, EFO, MIFireE, CFO

References

International Fire Service Training Association (IFSTA). (2008). Essentials of firefighting (5th ed). Stillwater, OK: Fire Protection Publications.