6.2.1: Mixing ANFO and the Oxygen Balance

6.2.1: Mixing ANFO and the Oxygen Balance

The proportion of oxidizers and fuels in the mix is critical to the performance of the explosive, and we say that the explosive must be oxygen balanced.

Oxygen Balance

It is important to achieve an oxygen balance within the explosive. This means there is exactly enough oxygen present to completely oxidize the contained fuel, but none left over to react with the contained nitrogen. We can calculate the proportion of ammonium nitrate and fuel oil to achieve an oxygen-balanced reaction. The decomposition of an oxygen-balanced ANFO is:

We can find the atomic weights for each of the elements from the periodic table.:

N = 14.006

H = 1.0078

N = 14.006

C = 12.009

And then, we can calculate the molecular weight of the ammonium nitrate and the fuel oil.

${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$: molecular weight = 80.05 g/mol

${\text{CH}}_2$: molecular weight = 14.03 g/mol

The ideal mixture, i.e., the oxygen-balanced mixture, consists of 3 moles of ammonium nitrate and one mole of fuel oil.

Next, we should find the weight ratio of ammonium nitrate to fuel oil.  

For 3 moles of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ :  3*80.05 = 240.15 g

For 1 mole of ${\text{CH}}_2$:  1*14.03 = 14.03 g

The molecular weight of ANFO is, therefore: 240.15 + 14.03 = 254.18 g

The percent by weight is:

${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ :  (240.15 / 254.18) *100 = 94.5% by weight

${\text{CH}}_2$ :  (14.03 / 254.18) * 100 = 5.5% by weight

So, now we know that when we mix the ammonium nitrate with the fuel oil, we need to do it in this proportion if we want an oxygen balanced reaction.

We can calculate the density of ANFO mixed to this ratio. Why we would we want to do that? If we know the weight per unit volume of the correct mix, then we can tell the blasters who make the mixture in the field to weigh one cup of the mixture to ensure the weight matches the specification weight. Note, I used “cup” as the volume measurement where one cup equals eight fluid ounces. You can use any unit, as long as you are consistent.

Let’s look at an example.

We will assume that the industrial-grade ammonium nitrate and fuel oil have the following densities: 53 lb/ft³ for the ammonium nitrate and 8.0 lb/gal for the fuel oil. How much ammonium nitrate will be required for each gallon of fuel oil to make an oxygen-balanced batch of ANFO? We will need to mix 2.6 ft³ of ammonium nitrate with that gallon of fuel oil. You should verify this result.

Continuing with this example: we know that the blasters will have a scale on the mixing truck and they will have a “cup”, which they will use to do a “cup density” check. In other words, they will fill the cup with their batch of mixed ANFO and weigh it. They will compare that weight to the weight that they have been given as the specified weight. If the blaster is using an 8-ounce container for the cup density check, what weight should be displayed on the scale?

You don’t have sufficient information to answer this question. What is the calculation that you need to do to answer the question, and what are you missing?

Although you know the weight of each ingredient in the mix, you don’t know the density of the mix. The manufacturer of the ammonium nitrate would be able to give you that information, or you could do a simple experiment in the lab to determine the density of that mix. For our purposes here, let’s assume that the density of this mix is 6.68 lb/gal. Therefore, the number that you need to give to your blaster is 0.42 lb.

The mixing equipment on the truck requires regular calibration, and even if calibrated, sometimes it can malfunction. Thus, it is recommended that blasters perform a cup-density check frequently, e.g., every 5 or so holes in a surface mine application.

We’ve looked at the ideal, i.e., oxygen-balanced mixture. Let’s look at two other cases in which we have too little or too much ammonium nitrate for the amount of fuel oil that has been added. We’ll choose a 92: 8 ratio for the first case and a 96.6: 3.4 ratio for the second case I’ve picked these ratios so that we have an integer number of moles in our formula.

Thus, for the first case with a 92:8 ratio, we have:

And for the second case, with a 96.6 : 3.4 ratio, we have:

Let’s look at the right-hand side of the equations. In the case of ideal mix, the products of the reaction, in addition to the energy that is not shown, are water vapor, carbon dioxide and nitrogen. In the first case of too little oxidizer, i.e., the ammonium nitrate, we produce carbon monoxide, a deadly gas. In the second case of too much oxidizer, we produce NO. The gas, NO, changes to NO₂ when exposed to the atmosphere. NO₂ is a toxic, but unlike CO, which is colorless, NO₂ produces a bright yellow-orange cloud.

We are not going to calculate the volume of toxic gas produced, but be aware that it is significant. In addition to the production of toxic gas, the energy release is reduced.

The oxygen balance can be lost for varied reasons. The most common include:

• poor quality control at the batch mixing truck;
• incorrect formulation, for example by adding a sensitizer like aluminum, but not accounting for it in the addition of the fuel oil;
• water in the hole;
• lack of confinement;
• improper stemming;
• improper timing of adjacent holes, allowing the explosive to become unconfined by the time it detonates;
• open-air blasting;
• explosives at the collar of the hole (increases fume production, not fragmentation).