I don’t know how many times that I have seen recommendations for airflow that go something like this: For drying, NAD, the airflow rate should be 1 CFM/bu and for cooling 0.1 CFM/bu. But where did this recommendation come from? Is it based on science? or experience? or data analysis?

Last year IHARF subcontracted PAMI to test different airflows as to the effect it has on drying. Wheat that was originally 17% MC was put under aeration with different airflows. The wheat dried to:

0.1 CFM/bu 16.5% MC

0.5 CFM/bu 14.5% MC

1.0 CFM/bu 14.5% MC

At first glance one would say that the prevailing recommendations of 1 CFM/bu for drying is correct. It did indeed dry the grain; but then again with half this flow, 0.5 CFM/bu. it dried just as much! And even the 0.1, which is only suppose to cool the grain, did more than just cool the grain, it also dried the grain by a half a point. It was not suppose to dry the grain at all; such a low flow was only suppose to cool the grain, yet one tenth of the flow did one fifth of the drying as the 1 CFM/bu. If 1 CFM/bu dries 2.5 points, then you would think that a tenth of the flow , 0.1, would do one tenth of the drying and reduce it by 0.25 points; but it actually took out 0.5 points. Looking at it this way, we see that the lower flow is actually more efficient at removing water. In terms of time, sure the higher flow dries more; but in terms of effort or energy expended, the lower flow is much better. To say that the lower flow does not dry, is simply not true, and I would argue that a slower removal of water is not only more efficient, it is also more effective, and here is why.

Grain sitting in a bin has inherent energy in it and it is expressed as heat, called its **specific heat**. If we heat the grain up; it takes energy, and the amount of energy required is so many joules per kilogram of grain per degree C. For grain this is about 1.36 KJ/kg C.

When grain is cooled with aeration fans, there are two avenues in which the grain can give up its energy: by conduction, or by putting it into the evaporation of water. It takes energy to evaporate water; this is called the latent heat of evaporation. To evaporate one kilogram of water requires 2257 K joules.

Ideally, to get the most drying done, we would like to use all of the energy in the grain for evaporation; but with higher air flows, the water in the grain does not have time to seep through the outer kernel of the grain and thus much of the energy of grain is taken away with conduction.

In my very first blog “**Wheat Energy Dries Itself**“, I show that in our trials we were using most of the energy in the wheat to dry — but not all of it. To use more of its energy for drying, we could and should use lower flows. I will repeat my original blog:

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We all know that it takes energy to dry grain and grain does contain energy. This energy is given up when the grain cools, but what if all the energy in the grain was used, as it cools for drying?

To answer this question we will use one cubic meter of wheat that will be cooled with aeration from 30⁰ to 5⁰ C. It has a MC of 15%. If all the energy in cooling could be used to dry the wheat, what could we get the MC down to – 14%? 13%? Let’s do the math and find out.

We know that it takes energy to evaporate water, the latent heat of evaporation; it is 2257 kJ/kg. Let’s assume that all the energy in the wheat, in going from 30⁰ C to 5⁰ C. will be used in evaporating water from the wheat. How much water (in %MC) can be removed?

1 bu = 1.2446 cubic feet 35.31 ft^{3} = 1 cubic meter 2.204 lbs/kg

One cubic meter of wheat has a weight of : 35.31 ft^{3}/1.2446 = 28.57 bu. x 60 lb/bu. = 1714 lbs, 777 kg

One percent moisture, 1% MC would then be 7.77 kg. If wheat was 15% MC, then 116 kg would be water. (Remember we are dealing with one cubic meter or 28.57 bushels)

The wheat in going from 30⁰ C to 5⁰ C has how much energy to give. What is the specific heat? It can vary, and I found that it increases as the MC increases but for a MC of 15% it is about 2 kJ/kgC

The wheat has got energy:

2 x 777 kg x 25 C = 38,850 kJ and divide by how much energy is needed to evaporate

38850 /2257 = 17.2 kg of water can be evaporated with the energy from the wheat.

But 1% MC is 7.77 kg , 17.2 kg/7.77 = 2.21% reduction in MC. If the wheat was 15% and all the energy went into evaporating water in going from 30 to 5⁰ C, then the MC would be reduced by 2.2% and be dried from 15% MC to 12.8%.

Our data showed just about the same thing. Cooling the grain with aeration, we found that reducing the temperature by 15 C would remove 1% moisture. In some trials with a high MC, we found this to be: 10C/%. In this case, for wheat, the amount of cooling to decrease the MC by one is: 11⁰/%MC. This demonstrates that aeration (at least with our experimental trials) is efficient in using the energy of the grain. I thought we might be losing more of the energy to conduction, but it appears that almost all the energy in the grain goes into drying. I thought that by increasing the contact time, (decreasing CFM) that we could get more efficient, but it appears that we are already doing a pretty good job in using almost all the energy in the grain for drying. It also tells me that we can’t expect much more than 2% decrease in MC in going from 30⁰C to 5⁰C. If we must lower the MC by more than two percent, we might have to use supplemental heat, or carefully allow the outside air to warm the grain (Hopefully without adding water) and then cooling it to remove more water.

The energy in the grain is precious, and we want to use it carefully for drying and not foolishly throw it away with conduction. So the original recommendation of: higher flows for drying, and lower flows for cooling is almost backwards. The higher flows result in more energy loss through conduction, and it is the lower flows that use more of the energy in the wheat for drying. Yes, it might take longer in calendar days to dry — or will it? Initially the higher flow will do more drying, but once the energy comes out of the grain, the higher flows won’t do any more drying than the lower air flows. In fact it might be one of those ‘turtle and the hare’ events’ — the slow and steady wins the day.

In looking through the literature and grain drying articles, I was hard pressed to find the basis for the 1 CFM/bu. I did find something in the Brooker book, “Drying and Storage of Grains and Oilseeds”. On page 169 he claims that the recommended rates — “the minimum airflow rate necessary to dry the grain mass before spoilage takes place.” I would argue, that even with low air flows, grain can be cooled to a safe condition in a matter of hours. Brooker is implying that the only way to make grain safe is by making it dry. Actually cooling is a better and faster means of getting grain to a safe condition. And with lower air flows, we may very well be more efficient at using the energy in the grain to dry.

What would the ideal air flow be? In considering the PAMI trials, and the IHARF trials, I would say the ideal air flow should be somewhere around 0.3 to 0.4 CFM/bu.

Reducing the flow, also has some other serious advantages: the power required is reduced considerably and the ‘over drying the bottom’ problem is mitigated. Let’s first look at the power.

Fan Laws from Brooker’s book:

- Air flow (CFM) is proportional to fan speed
- Static pressure is proportional to speed squared
- Power is proportional to speed cubed
- Power is proportional to air density
- Static pressure is proportional to air density

So using these rules, if a 5 HP fan produces 1 CFM/bu, then how much horse power is required for 0.3 CFM/bu? 0.4 CFM/bu? 0.5 CFM/bu?

Reducing the flow, has a drastic effect on reducing the required horse power. It is not linear! We see that power is proportional to the cube of the speed, and since the speed is proportional to the flow (CFM), the power is proportional to the cube of CFM. First find the proportionality factor, k.

5 hp = k ( 1 CFM )³ therefore k is 5.

For 0.3 CFM/bu, the required HP = 5 ( 0.3 )³ = 0.135 HP

for 0.4 CFM/bu, the required HP = 5 (0.4)³ = 0.32 HP

for 0.5 CFM/bu, the required HP = 5 (0.5) = 0.625

So we see that the lower the air flow reduces the required horse power dramatically. Where we have a 5 HP fan now, we could get by with a half or one hp!!

We also see that the static pressure is reduced according to the power reduction. In our trials, with a 5 hp fan, at 1 CFM/bu, we had a static pressure behind the fan of about 5 inches of water. A reduced flow to 0.4, would result in a pressure of 0.32 inches of water. Why is this important? Because the compressed air, behind your fan, at the bottom of your bin results in an increased air temperature according to the thermodynamic formula, PV=nRT. We found this to be an increase of a couple of degrees C, and this air can and will hold more water, resulting in the bottom over-drying. This much reduced compression, will mitigate the over-drying bottom phenomenon.

So, to conclude, I think the best aeration air flow rates should be about 0.4 CFM/bu in terms of the most efficient drying, the least hp, and in mitigating bottom over-drying.