Have you ever wondered how to compute the dust equivalence ratio of a solid combustible fuel?
This post demonstrates the math and stoichiometry behind computing combustible dust equivalence ratio.
It begins by developing a mathematical relationship for equivalence ratio based on fuel concentration, air density, and the Fuel-To-Air ratio at stoichiometric conditions.
The second part of this post focuses on determining the stoichiometric Fuel-To-Air ratio for different dusts. Five cases are given for different combustible materials:
- Case #1: Coal Dust (40% CH4, 60% Ash) [Download Worksheet]
- Case #2: Coal Dust (40% CH4, 60% Carbon) [Download Worksheet]
- Case #3: Organic Dust (CH2.01O0.8) [Download Worksheet]
- Case #4: Aluminum (Al) [Download Worksheet]
- Case #5 Iron (Fe) [Download Worksheet]
Before jumping into the stoichiometric calculations, background on computing the dust equivalence ratio is required.
Combustible Dust Equivalence Ratio
The math required to compute reactive dust equivalence ratio is relatively straight forward.
The equivalence ratio is defined as the Fuel-To-Air-Ratio at the specified concentration, divided by the stoichiometric Fuel-To-Air-Ratio.
\(\mathit{\Phi} = \frac{\left( F/A\right)}{\left( F/A\right)_{\text{St}}}\)
The Fuel-To-Air-Ratio can be redefined as the ratio of the fuel concentration, \(\mathit{\sigma}_{\text{F}}\), divided by the density of air, \(\rho_{\text{a}}\).
\(\left(F/A\right) = \left( \mathit{\sigma}_{\text{F}}/\rho_{\text{a}}\right)\)
This allows the following relation for combustible dust equivalence ratio to be developed:
\(\mathit{\Phi} = \frac{\left( F/A\right)}{\left( F/A\right)_{\text{St}}} = \frac{\left( \mathit{\sigma}_{\text{F}}/\rho_{\text{a}}\right)}{\left( F/A\right)_{\text{St}}} = \frac{\mathit{\sigma}_{\text{F}}}{\rho_{\text{a}}\left( F/A\right)_{\text{St}}}\)
This relation has three parameters: the fuel concentration in the dust (\(\mathit{\sigma}_{\text{F}}\), kg/m3), the density of air (\(\rho_{\text{a}}\), kg/m3), and the stoichiometric Fuel-To-Air-Ratio (\(\left( F/A\right)_{\text{St}}\)).
The dust concentration is generally taken as the mass of dust divided by the system volume and must be specified in kg/m3.
The density of air can be calculated from the ideal gas law:
\(\rho_{\text{a}} = \frac{P}{R_{\text{spec}}T}\)
where \(P\) is pressure (Pa), \(T\) is temperature (K), and \(R_{\text{spec}}\) is the specific gas constant for air (universal gas constant divided by molecular weight of air, 287.058 J/kg-K).
At ambient pressure and temperature the density of air can be computed as
\(\rho_{\text{a}} = \frac{\text{101325} Pa}{287.058 \text{J/kg-K} \cdot 293 \text{K}} = 1.20 \text{kg/m}^{3}\)
The most difficult part of computing the combustible dust equivalence ratio is determining the stoichiometric Fuel-To-Air-Ratio, (\(\left( F/A\right)_{\text{St}}\)). The follow sections show this calculation for five different test cases.
Case #1: Coal Dust (40% CH4, 60% Ash)
In this case, the dust is assumed to contain only reactive volatiles and inert ash. The concentration of fuel is then the fraction of volatiles multiplied by the overall dust concentration.
\(\mathit{\sigma}_{\text{F}} =\psi_{\text{v}}\mathit{\sigma}_{\text{p}} \)
where \(\mathit{\sigma}_{\text{p}}\) is the overall dust concentration (kg/m3) and \(\psi_{\text{v}}\) is the fraction of volatiles.
Now that the fuel concentration is defined, the next step is to determine Fuel-To-Air ratio under stoichiometric conditions.
Table 1 shows the balanced chemical equation and resulting mass of each specie. It is also useful to perform some checks that mass and mols/mass fractions balance out and these are done in the attached worksheet.
| CH4 | O2 | N2 | \(\rightarrow\) | CO2 | H2O | N2 | |
|---|---|---|---|---|---|---|---|
| M (mol) | 1.0 | 2.0 | 7.523 | 1.0 | 2.0 | 7.523 | |
| Mw (g/mol) | 16.043 | 31.999 | 28.013 | 44.010 | 18.015 | 28.031 | |
| Mass (g) | 16.043 | 63.998 | 210.762 | 44.010 | 36.031 | 210.762 |
From the balanced chemical equation, the stoichiometric Fuel-To-Air ratio is calculated as the mass of methane divided by the combined mass of oxygen and nitrogen:
\(\left( F/A\right)_{\text{St}} = \frac{16.043}{63.998 + 210.762} = 0.0583\)
The final relation for combustible dust equivalence ratio can now be determined. Using the specified volatile fraction of 0.4 the equation can be written as
\(\mathit{\Phi} = \frac{0.4\cdot\mathit{\sigma}_{\text{p}}}{0.0583\cdot\left(1.2\text{kg/m}^{3}\right)} = 5.717\mathit{\sigma}_{\text{p}}\)
again, where \(\mathit{\sigma}_{\text{p}}\) is specified in kg/m3. This relation is shown in the plot at the top of this post along with the other materials explained in the following sections.
The stoichiometric dust concentration can be determined by setting the equivalence ratio to 1.0, and gives a value of 175 g/m3.
Case #2: Coal Dust (40% CH4, 60% Carbon)
Often both gas phase volatile reaction and solid phase surface reaction are considered for coal particles.
Assuming a coal particle that is 40% volatile and 60% carbon, the balanced chemical equation given in Table 2 is developed.
| CH4 | C | O2 | N2 | \(\rightarrow\) | CO2 | H2O | N2 | |
|---|---|---|---|---|---|---|---|---|
| M (mol) | 1.000 | 2.004 | 4.004 | 15.061 | 3.004 | 2.000 | 15.061 | |
| Mw (g/mol) | 16.043 | 12.011 | 31.999 | 28.013 | 44.010 | 18.015 | 28.031 | |
| Mass (g) | 16.043 | 24.064 | 128.108 | 421.896 | 132.186 | 36.031 | 421.896 |
Now the numerator in the stoichiometric fuel-to-air ratio includes both fuel components
\(\left( F/A\right)_{\text{St}} = \frac{16.043 + 24.064}{128.108 + 421.896} = 0.0729\)
and the combustible dust equivalence ratio is computed as
\(\mathit{\Phi} = \frac{\mathit{\sigma}_{\text{p}}}{0.0729\cdot\left(1.2\text{kg/m}^{3}\right)} = 11.431\mathit{\sigma}_{\text{p}}\)
again where \(\mathit{\sigma}_{\text{p}}\) is specified in kg/m3. Note that the entire particle now contains fuel instead of just a fraction as shown in Case #1.
Since the entire particle can contribute to combustion the equivalence ratio is substantially larger. The stoichiometric dust concentration under these conditions is 87 g/m3.
Case #3: Organic Dust (CwHxOy)
The same approach can be used for any combustible dust as long as the volatile products and combustion reaction are known.
A general equation for an organic combustible dust is given on Page 60 in the text “Dust Explosion Dynamics” by Dr. Russell Ogle [bibcite key=Ogle2017], along with a detailed description of how to compute dust equivalence ratio.
\(\text{C}_{w}\text{H}_{x}\text{O}_{y} + \left(w + \frac{x}{4} – \frac{y}{2}\right) \left(\text{O}_{2} + 3.76\text{N}_{2}\right) \rightarrow w\text{CO}_{2} + \frac{x}{2}\text{H}_{2}\text{O} + 3.76\left(w + \frac{x}{4} – \frac{y}{2}\right)\text{N}_{2}\)
The worksheet for this example allows \(w\), \(x\), and \(y\) to be modified. Example 3.3 of Ogle, 2017 [bibcite key=Ogle2017] uses the chemical species C1H2.01O0.8, proposed by Medina et al., 2013 [bibcite key=Medina2013]. Table 3 gives the balanced chemical equation for this material.
| C1H2.01O0.8 | O2 | N2 | \(\rightarrow\) | CO2 | H2O | N2 | |
|---|---|---|---|---|---|---|---|
| M (mol) | 1.000 | 1.102 | 4.147 | 1.000 | 1.005 | 4.147 | |
| Mw (g/mol) | 26.836 | 31.999 | 28.013 | 44.010 | 18.015 | 28.031 | |
| Mass (g) | 26.836 | 35.279 | 116.182 | 44.010 | 18.105 | 116.182 |
Similarly to coal dust the stoichiometric Fuel-To-Air ratio is determined from the mass of fuel and air in the table
\(\left( F/A\right)_{\text{St}} = \frac{26.836}{35.279 + 116.182} = 0.177\)
resulting in the following relation for equivalence ratio
\(\mathit{\Phi} = \frac{\mathit{\sigma}_{\text{p}}}{0.177\cdot\left(1.2\text{kg/m}^{3}\right)} = 4.703\mathit{\sigma}_{\text{p}}\)
From this the stoichiometric concentration of C1H2.01O0.8 can be calculated as 212 g/m3. This is within a few percent of the value calculated in the textbook of Ogle, 2017 [bibcite key=Ogle2017].
Case #4: Aluminum Dust (Al)
The equivalence ratio for metal dusts is calculated using the same approach as organic dusts. Assuming the aluminum reacts with oxygen to create aluminum oxide the balanced equation given in Table 4 is developed.
| Al | O2 | N2 | \(\rightarrow\) | Al2O3 | N2 | |
|---|---|---|---|---|---|---|
| M (mol) | 4.0 | 3.0 | 11.285 | 2.0 | 11.285 | |
| Mw (g/mol) | 26.981 | 31.999 | 28.013 | 101.96 | 28.031 | |
| Mass (g) | 107.924 | 95.996 | 316.142 | 203.92 | 316.142 |
The stoichiometric Fuel-To-Air ratio is the mass of aluminum divided by the mass of air
\(\left( F/A\right)_{\text{St}} = \frac{107.924}{95.996 + 316.142} = 0.261\)
which results in the following relation for dust equivalence ratio.
\(\mathit{\Phi} = \frac{\mathit{\sigma}_{\text{p}}}{0.261\cdot\left(1.2\text{kg/m}^{3}\right)} = 3.182\mathit{\sigma}_{\text{p}}\)
Using this relation the stoichiometric concentration of aluminum is calculated as 314 g/m3.
Case #5: Iron Dust (Fe)
The last case given in this post is for iron dust and shows that the equivalence ratio decreases with a decrease in metalic dust molecular weight.
Assuming that iron oxide is the combustion product, the balanced chemical equation in Table 5 is developed.
| Fe | O2 | N2 | \(\rightarrow\) | FE\e2O3 | N2 | |
|---|---|---|---|---|---|---|
| M (mol) | 4.0 | 3.0 | 11.285 | 2.0 | 11.285 | |
| Mw (g/mol) | 55.845 | 31.999 | 28.013 | 159.690 | 28.031 | |
| Mass (g) | 223.380 | 95.996 | 316.142 | 319.380 | 316.142 |
The stoichiometric Fuel-To-Air ratio is calculated from the table
\(\left( F/A\right)_{\text{St}} = \frac{223.380}{95.996 + 316.142} = 0.542\)
giving the following relation for dust equivalence ratio.
\(\mathit{\Phi} = \frac{\mathit{\sigma}_{\text{p}}}{0.542\cdot\left(1.2\text{kg/m}^{3}\right)} = 1.537\mathit{\sigma}_{\text{p}}\)
Due to the lower molecular weight, the stoichiometric concentration of iron is quite a bit higher than aluminum at 650 g/m3.
Conclusion
This post demonstrated how to compute combustible dust equivalence ratio using a simple mathematical relationship and stoichiometry.
The results demonstrate that the equivalence ratio is linearly related to dust concentration and the slope depends on the density of air (pressure and temperature) and stoichiometric Fuel-To-Air ratio.
If you have any questions or would like to help others understand combustible dust fundamentals please share and comment using the buttons/fields at the bottom of the page.
Thanks again and I look forward to sharing more soon!
Related Textbooks
Below is a list of related textbooks mentioned in this post. Note that the images and links are affiliate links and I will receive a small commission if you make a purchase after clicking.
Dust Explosion Dynamics
Russell A. Ogle
