# Presentation: 684488

## Handy Dandy Guide to Gas Calculations

**Dave Quigley, Ph.D.**
**B&W Y-12**
**March 11, 2009**

## Principles and Assumptions

- Principles do not change
- Assumptions may change with event being modeled

## Principle 1

- Gases do not mix instantly

## Issues with Principle 1

- Gases don’t mix homogeneously unless well agitated for an extended period
-  Results in
- Concentrations vary from point to point
- Light gases tend to rise
- Heavy gases tend to sink
- Homogeneous gas mixtures do not tend to separate according to density
- Because of this principle, be conservative!

## Principle 2

- See Principle 1

## Assumptions

- Know your assumptions
- Some are universal
- Some are dependent upon situation being analyzed
- State all assumptions prior to analysis
- Ensure all assumptions err on the conservative side

## Keep Analysis Simple

- Gases not mixing well leads to uncertainty
- Fine details become meaningless when uncertainty is present
- Goal should be to obtain an analysis that is reasonably accurate and errs on the side of conservatism

## Example 1: Effect of Asphyxiant Release

- Question:
-    If a standard size nitrogen cylinder containing 225 ft3 were to leak into a room (20’ x 12’ 9’), would it result in an oxygen deficient atmosphere?

## Reality

- HVAC adds fresh air to the room at a rate of 2 – 6 air exchanges per hour depending upon the system
- Gas release will typically take up to a minute to release and will be very loud
- Release rate will slow as tank pressure decreases
- Gas released will become colder as it is released and tend to sink
- Non-homogeneous mixing will occur

## Assumptions

- Instant release of gas into room
- No air exchanges in room from HVAC
- Homogeneous mixing of gas in room
- Gas leaving the room due to overpressurization is homogeneous mixture
- Gases obey the Ideal Gas Law
- Air is 78% nitrogen and 22% oxygen

## Calculations

- Volume of room  equals 2160 ft3
- Volume of cylinder equals 225 ft3
- Using Dalton’s Law of Partial Pressures: Quantity of oxygen present is
-             2160 x 22% = 475 ft3
-   Quantity of nitrogen initially present is
-          2160 x 78% = 1685 ft3
- Quantity of nitrogen present after tank release is
-          1685 ft3 + 225 ft3 = 1910 ft3

## Calculations (cont.)

- Total volume of gas after tank release =
- 475 ft3 oxygen + 1910 ft3 nitrogen = 2385 ft3
- % oxygen present after tank release =
-  475 ft3/2385 ft3 = 19.9% oxygen
- 19.9% is above the 19.5% minimum oxygen level required therefore the room is not oxygen deficient

## Example 2: Toxic Gas Mixed with an Inert Gas

- Question: Is a toxic gas mixed diluted with  a nontoxic gas considered to be toxic?
- LC50M = _____1____
-                 (C/LC)
- LC50M = lethal concentration of the mixture
- C = Concentration of toxin in % vol
- LC = the LC50 of toxin based on 1 hour exposure (rat model preferred) in ppm
- If LC50M < 2000 ppm, then the mixture is toxic
- From Appendix E, Hazard Categories, 2003 International Fire Code

## Example

- 15% chlorine in 85% Nitrogen
- LC50 Chlorine (1 hour – rat) = 293 ppm
- LC50M = 1/(0.15/293 ppm) = 1,953 ppm
- Since 1,953 ppm < 2,000 ppm
- this mixture is toxic

## What if I Have Multiple Toxics in My Mixture

- Use the same equation in an expanded form
- LC50M = _________1__________________  _
-                 (C1/LC1) + (C2/LC2) + (C3/LC3)....
- Ci = Concentration of component (i) in % vol
- LCi = the LC50 of component (i) based on 1 hour exposure (rat model preferred) in ppm

## Data for Exposures for Other than 1 Hour

- Use the following normalization factors

| Time (hrs) | Multiply by |
| --- | --- |
| 0.5 | 0.7 |
| 1 | 1 |
| 1.5 | 1.2 |
| 2 | 1.4 |
| 3 | 1.7 |
| 4 | 2.0 |
| 5 | 2.2 |
| 6 | 2.4 |
| 7 | 2.6 |
| 8 | 2.8 |

## Time to Change Gears

- Question:
- How do I know if a gas mixture is flammable or nonflammable?
- Use
-    ISO 10156, Gases and Gas Mixtures – Determination of Fire Potential and Oxidizing Ability for the Selection of Cylinder Valve Outlets

## Formula to Use

- If Ai is the molar fraction of the ith flammable gas, Bi is the molar fraction of the ith inert gas, Fi is the ith flammable gas, Ii is the ith inert gas, and Ki is the coefficient of nitrogen equivalency for the ith inert gas, then
- (ΣAiFi + ΣKiBiN2)(1/ ΣAi + ΣKiBi)
- Becomes the statement expressing the sum of the component gas fractions expressed in nitrogen equivalents

## Formula to Use (cont.)

- Then
- Ai’ = Ai/(ΣAi + ΣKiBi)
- The maximum heat content becomes
- Σ Ai’/Tc and if Σ Ai’/Tc x 100 < 1 then the
-  mixture is nonflammable

## HUH?

## Let’s simplify the situation by simply looking at some examples

## Step 1 – Coefficient of Equivalency (Ki)

- A numerical value that compares an inert gas to nitrogen

| Gas | N2 | CO2 | He | Ar | Ne | Kr | Xe | SO2 | SF6 | CF4 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
| Ki | 1 | 1.5 | 0.5 | 0.5 | 0.5 | 0.5 | 0.5 | 1.5 | 1.5 | 1.5 |

## Step 2 – Determine TC for Flammable Gas

- Value is maximum content of flammable gas which when mixed with nitrogen is not flammable in air (list is not complete – see ISO 10156 for complete list)

| Gas | TC | Gas | TC |
| --- | --- | --- | --- |
| Hydrogen | 5.7 | Ethylene | 6 |
| CO | 20 | Propane | 6 |
| Methane | 14.3 | Propenes | 6.5 |
| Ethane | 7.6 | Acetylene | 4 |
| Butanes | 5.7 | n-Hexane | 3.5 |

## Step 3 – Determine Constituents in Gas Mixture and Modify

- Assume a mixture of 7% H2 in CO2
- Coefficient of equivalency of CO2 is 1.5
- Therefore 93% CO2 is equivalent to 93 x 1.5 or 139.5% nitrogen
- Since 7% H2 + 139.5% N2 equivalents >100%, this must be normalized to 100%

## Step 4 – Normalize to 100% Nitrogen and Compare with TC

- 7% H2 + 139.5% N2 = 146.5%
- 100/146.5 = 0.682
- 7% H2 (0.682) + 139.5% N2 (0.682) = 100%
- 4.78% H2 + 95.22% N2 = 100%
- T**C**** **for H2 = 5.7
- 4.78/5.7 < 1 therefore mixture is nonflammable

## Adventures with Mixtures

- Assume a mixture of 2% H2 + 8% methane in 65% He and 25% Ar
- Ki (He) = 0.5  Ki (Ar) = 0.5
- 2%H2 + 8%CH4 + (0.5)(25%)Ar + (0.5)(65%)He
- 2%H2 + 8%CH4 + 12.5%N2 + 32.5%N2 = 55%

## Adventures with Mixtures (cont.)

- 100/55 = 1.82
- (2%H2 + 8%CH4 + 12.5%N2 + 32.5%N2)(1.82)
- 3.6%H2 + 14.5%CH4 + 22.8%N2 + 59.1%N2
- Since TC (H2) = 5.7 and TC (CH4) = 14.3
- 3.6/5.7 (H2) = 0.63  14.5/14.3 (CH4) =1.01

## Adventures with Mixtures (cont.)

- Results are
- 0.63 + 1.01 = 1.64
- Since 1.64 > 1, the mixture is flammable
- Note:  These calculations are conservative.  When calculated values are near “1” or when this determination may result in significant costs, actual testing may be required .

## More Complex Calculations

- Is a gas more or less oxidizing then air –
- Use ISO 10156
- Similar in complexity to the flammability gas mixture calculations above
- Gas line break
- Many more factors such as size of the room, rate of air exchanges in the room, volume flow of gas from pipeline, etc that must be modeled
- Can become very complex

## Questions?

## Disclaimer

-     This work of authorship and those incorporated herein were prepared by Contractor as accounts of work sponsored by an agency of the United States Government. Neither the United States Government nor any agency thereof, nor Contractor, nor any of their employees, makes any warranty, express or implied, or assumes any legal liability or responsibility for the accuracy, completeness, use made, or usefulness of any information, apparatus, product, or process disclosed, or represents that its use would not infringe privately owned rights. Reference herein to any specific commercial product, process, or service by trade name, trademark, manufacturer, or otherwise, does not necessarily constitute or imply its endorsement, recommendation, or favoring by the United States Government or any agency or Contractor thereof. The views and opinions of authors expressed herein do not necessarily state or reflect those of the United States Government or any agency or Contractor thereof.