Grain Aeration

1.  Grain Drying 101    Drying Conditions if the absolute humidity of the air inside the bin is greater than the absolute humidity of the outside air.  Explain what absolute humidity is:  amount of water in one cubic meter, explain how it is different from relative humidity,  and that absolute humidity is calculated from the saturation curve and pro rated with the relative humidity.  Diurnal cycle and the night drying and the yardlight rule.
2. Motor/Generator Set to make a portable system.  Especially good for remote sites that have no power.  Make use of the wasted heat from the motor.
3. Plenum design, show how we waste energy (pressure) in going across the screen.  Better to use the electricity with a heater in front of a small fan. The pipe or plenum should be open on the bottom for the least resistance in getting the air into the grain.
4. Time Constant Tau.  The time constant is the amount of time required for the outside air absolute humidity to get to 60% of the distance of the difference. A CFM/bu of 0.35 would be about 3 time contants and get us to over 90% of the way there.  This blog would show that 0.35 CFM/bu is kind of the sweet spot for using as much of the grains energy as possible, while still drying in a relatively short time.
5. The argument for small fans:  less capital cost, less power infrastructure required, more uniform drying, use more of the inherent grains heat to dry, more efficient  — the downside is that it is a bit slower — but what is the hurry? How slow can you go.  The ideal fan size. CFM/bu
6. More on condensation, and how to use the grain drying calculator to detect conditions for condensation.
7. More specifics on how to use the grain calculator.
8. Something for everybody, from the simplest strategy (yard light rule)  to the most advanced (absolute humidity) and the pros and cons of each. Depending on what you have.
9. Long term storage, use whatever fan strategy until the end of January, in which you will get the grain frozen and cold as possible, and from January til the end of July keep the bin as sealed as possible.
10. Roof vents don’t do much, and can actually result in condensation on the top layer of grain in the spring when the grain is cold and it is swiped by warm moist air circulating through the vents.
11. Vertical in grain tubes.
12. top dries first, compression causes heat
13. safe days
14. comparison of control strategies, continuous, nite, temp differential, absolute humidity.
15. Flow from natural convection
16. Energy in the grain to dry
17. Grain Resistance
18. Motor HP as we drop the flow in half, top down drying, HP 5 to 1
19. First 24 Hours is critical

In a previous blog, I explained how condensation could be a problem in the fall with warm grain and cold outside air.  The rough rule of thumb was that there was a potential for condensation on the roof if the grain was ten degrees warmer than the outside; but a more precise determination could be made with the grain drying calculator.  The recommendation was not to turn your fan on if the RH from the calculator was above 100%.

There is another time, in the spring that condensation could be a problem. When the grain is very cold and the outside air temp is much warmer and holding a fair amount of water. This air enters the bin through the fan entrance, the top or vents; and it hits the cold grain — it will cool past its dew point and condensation may well occur.  This will happen mostly on the poor old top layer, that may well have had water dumped on it from fall condensation, and it is the last layer in the bin to dry.  No wonder the crust always forms on the very most top layer.

To prevent condensation in the spring, the best thing to do would be to not let the warm air in.   The bin should be sealed.  The top lid closed tight, the fan covered and all vents covered.  Indeed, this is a time when vents will actually be the cause for grain damage.  Sealing the bin in mid winter, after the grain has been cooled to its lowest possible temperature, will also keep the grain as cold as possible for as long as possible.  Yes the grain will still warm up from conduction through the walls; but it will be a warming without adding water.  At the end of the summer, when the grain has warmed up and the nights again are getting cold, the bin can be opened up and the grain again cooled.  And this process could be done for years of very safe grain storage.

With this blog, we would like to know what it might cost to use supplemental electrical energy to remove one percent moisture content, MC, from one bushel of wheat.

Assumptions:

1. All the energy goes into evaporation.
2. wheat  60 lb/bu
3. Cost of electrical energy  is 0.12/kW hr.
4. Latent heat of evaporation,  energy to evaporate  2257 kJ/kg

One bushel is 60 lbs, so 1% MC would have a weight of  0.6 lbs or 0.2722 kg

2257/.2722 = 614.4 kJ to evaporate 1% MC from one bushel of wheat  and to put this into hours divide by 3600  =  0.17 kW hr.   and in terms of dollars,  0.17 x $ 0.12 = $ 0.02 to evaporate enough water out of a bushel of wheat to lower the MC by 1%.    That doesn’t sound like much but for 3500 bushels, it would cost $71.68, and to drop by 2% MC it would be $143.36.  It must also be remember that this is the ideal efficiency.  In the real world, we may have to double our cost again.

Could we use the free heat in the air to dry grain?  —  I think so; but it will take careful management and control of the aeration fans.

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 ft³ = 1 cubic meter     2.204 lbs/kg

One cubic meter of wheat has a weight of : 35.31 ft³/1.2446 = 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.

For this exercise we will use one cubic meter of barley at 30 C. And we know that it takes energy to evaporate water; this is called latent heat of evaporation, which is 2257 kJ/kg. Let’s assume that all the energy in the barley in going from 30 C to 5 C is used in evaporating water from the barley. How much water (in %MC) can be removed?

1 bu = 1.2446 cubic feet       35.31 cubic feet = 1 cubic meter     2.204 lbs/kg

One cubic meter of barley weighs 35.31 ft3/1.2446 = 28.57 bu     x 48 lb/bu = 1361 lbs, 618 kg

One percent moisture, 1% MC would then be 6.18 kg.   If barley was 15% MC then 92.6 kg would be water.

The barley in going from 30 C to 5 C has how much energy to give. What is the specific heat? It can vary, but I found one source: 1.36 kJ/kgC

1.36 x 618 kg x 25 C = 21012 kJ and divide by how much energy is needed to evaporate

21012 kJ /2257 kJ/kg = 9.3 kg of water can be evaporated with the energy in the barley.

But 1%MC was 6.18 kg/%MC ,   9.3 kg/6.18 = 1.5% reduction in MC. If the barley was 15% and all the energy went into evaporating water in going from 30 to 5 C, then the MC would be reduced by 1.5% and be dried from 15% MC to 13.5%.

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 the case above, 1.5% resulted from a 25 C. A 1% reduction would require 16.66 deg reduction. This demonstrates that aeration (at least with our experimental trials) is very efficient at using the energy in 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.

Let’s work out an example of how long it will take to cool down a 3500 bushel bin of barley from 30 ⁰C to 20 ⁰C with air that is 15 C and flowing into the bin at 3000 CFM. We will make a wild assumption here and assume that the air will leave the grain at 30 C, making a 15 C increase. For this exercise we will use conservation of energy and we will ignore any drying or wetting that will result from latency heat.

First task is to find the energy that would be lost if the barley is cooled from 30 to 20 C. We must know the specific heat, Cp, of barley and I looked in a couple of places, but it was about 1.36 kJ/kg C. The weight of the barley is 3500 bu x 48 lbs/ bu x 1/2.204 kg/lb = 76,225 kg   times 1.36 = 103,666 kJ for every degree C and we are changing ten degrees so the energy to be removed from the barley will be 1,036,660 kJ.

3000 CFM is 50 cubic feet per second. Air is 0.0807 lbs per cubic foot.   4.035 lbs/s or 1.83 kg/sec

The specific heat for air is close to 1 kJ/kg C so for every degree C we could acquire 1.83 kJ/s. For 15 C we could remove 27.45 kJ/s. How many seconds to remove 1,036,660 kJ?   1,036,660/27.45 =37765 sec, 629 min, 10.5 hours.

This shows that it is possible to cool the grain over a one night period (12 hours, 9 PM to 9 AM).

Now, if we used a smaller fan and produced only 2000 CFM, we would expect the cooling time to increase accordingly to 16 hours. Even for a much larger bin, of 10,000 bushels, we could cool it down in 48 hours (four 12-hour nights). One could argue that this would be for air that is consistently 15 C and it assumed ideal energy transfer. On the other hand there will be cooling affects resulting from drying; but that’s for another blog.

In my last blog, I showed how over-drying occurred from the increased temperature at the bottom of the bin caused by compression.  Maybe we cannot eliminate it; but can we at least reduce it.

In our previous example we used a 2200 bushel-bin of barley at 20⁰ C and MC of 15%. A 5 HP aeration fan produced 3000 CFM with 6” H₂0 pressure. This resulted in the bottom being 4.2⁰ C warmer and dried to 11.6% MC all while the top remained at 15%.

Browsing around the internet, looking at fan curves, I found out that the air resistance is not linear like electrical resistance. Static pressure is proportional to the square of CFM (flow).   And required horsepower, HP, is proportional to the cube of CFM. We can find these proportionality constants:

Pres = a CFM ²   6” = a 3 ² (3 is for 3,000 or CFM is in thousands)     a = 6/9 = 0.666

HP = b CFM ³     5 = b 3 ³   b = 5/27

Let’s see what happens if we reduce the CFM from 3 thousand to 2 thousand.

Pres = 0.66 (2 ² ) = 0.66 x 4 = 8/3 = 2.66 “ H₂0,    less than half of what it was, 6.

And the required HP = 5/27 ( 2 3 ) = 40/27 = 1.48 or call it 1.5 HP

Isn’t that incredible, we only decreased the flow by two thirds, and in return we get a pressure that is less than half, and a required horsepower that is much less.

This reduced pressure will also decrease the temperature rise from compression.

2.66/406.8 = x/293, where x must be 1.9; that is we get a 1.9 ⁰C rise from compression. Let’s say we also get an increase in temperature from motor inefficiency, for a combined total of 2⁰ C or an absolute temperature of 22 ⁰C.

So we had at the top of the bin barley at 20 C and 15% MC. Now we will use the grain drying calculator and plug 15 in for MC and 20 for both the outside air and grain temperature, this gives an RH_thres of 68.6% and now we can use the relative humidity to absolute humidity calculator to see that the absolute humidity is 12 grams per cubic meter. We have assumed that the air at the top of the bin has reached equilibrium with the grain; the relative humidity of the air is 68.6% and the absolute humidity is 12 grams per cubic meter. We are at equilibrium – no drying or wetting is taking place, so the air at the bottom has the same absolute humidity, or 12 grams per cubic meter. At the bottom of the bin, the temperature is 22 C, and the saturation humidity is 19.5 gr, giving an RH of 12/19.5 = 61.5%. Now using the grain drying calculator, find the MC for barley by trial and error with a temp of 22 and RH 61.5% and I get a MC of 13.7%. This is a spread of 1.3%.   This is better, remember with a higher pressure of 6, we got the bottom to be 11.6%, and a spread of 3.4%.

The conclusion to this story is that a relatively small decrease in flow, will greatly reduce the pressure, which has a profound decrease in over drying the bottom. Reducing the flow will also increase the transit or contact air-to-grain time, giving the grain more time to pass water into the air.