Handy Dandy Guide to Gas Calculations by donovantatehe


									Handy Dandy Guide to Gas

       Dave Quigley, Ph.D.
           B&W Y-12
         March 11, 2009
Principles and Assumptions
• Principles do not change
• Assumptions may change with event being
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
• Know your assumptions
 – Some are universal
 – Some are dependent upon situation being
 – 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
• 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?
• 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
• Gas released will become colder as it is
  released and tend to sink
• Non-homogeneous mixing will occur
• 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
• Gases obey the Ideal Gas Law
• Air is 78% nitrogen and 22% oxygen
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___

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
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
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
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
How do I know if a gas mixture is flammable
 or nonflammable?

 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
Formula to Use (cont.)

            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
  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
• 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%

TC 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
 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,
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 apparatus, product, or process disclosed, or represents that its
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 necessarily constitute or imply its endorsement,
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 authors expressed herein do not necessarily state or reflect
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