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Justin, from the Qu’Appelle area, phoned me yesterday and asked how he should be using his big heater to dry his very wet wheat. Here is his situation: He is renting a big industrial diesel heater, 1.4 million btu/hr. and 5,000 CFM. It burns 10 gallons of diesel every hour.  He is going to connect the heater to 4 bins, each holding 4500 bushels of wheat @ 20% MC.  How long will it take to heat his grain by 1 C ?

We will consider only one bin of 4500 bushels x 60 lbs/bu  divide by 2.204 lbs/kg to get 122,505 kg of wheat.  The specific heat of wheat is about 1.36 kJ/kg C .  To raise the temp of the wheat by 1 C, we need 1.36 kJ x 122,505 kg = 166,606 kJ.

Our heater puts out 1,400,000 btu/hr. but we only get 1/4 of this, 350,000 btu/hr and there is 1.055 kJ/btu  or 369,250 kJ/hr. put into one bin and since we need 166,606 kJ to raise the temp of the grain by 1 deg C; it will take 166,606/369,250 = 0.45 hours or 27 mins to raise the temp by one degree.  Let’s call it a half hour or 30 mins.   I told Justin that he should not heat the grain by more than 6 C to prevent problems with condensation.  He would be able to do this in 3 hours.

Justin also wanted to know how hot the air would be coming out of the heater; or the temperature rise.  We have to know the specific heat of air, 1kJ/kg and the weight of the air which is 0.0807 lbs per cubic foot.   The heater has an air flow of 5000 CFM, so in one hour, 300,000 cubic feet of air will pass through it which weighs 300,000 x 0.0807 = 24210 lbs or 10,984 kg.  To raise the temp of this air by 1 C requires 10,984 kJ/hr  but we have 369,250 kJ/hr.    So 369,250 / 10,984 = 33 º C. If the air entering the heater was 0 ºC, it would come out around 30 ºC  That’s not too bad, you won’t be scorching the wheat.

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 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.

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.

In my last blog it was established that the grain should not be heated by more than 5 C and that a 50,000 btu furnace could do this in 12 hours. What if we used the heat from a small gas driven generator set?  The electric power generated could be used to power the aeration fan,  and this set up would be ideal for remote bins where power from the grid is not available.  OK let’s consider this.  But how big should the gen set be?  What will it cost for the gen set, and the fuel to dry Jim S’s 3500 bushels of wheat that is now 17%?

Jim is using a 5 HP fan.  1 HP is  .7457 kW, so the gen set must be able of producing at least  5 x .7475 = 3.7 kW.  At Princess Auto, I found a 7500 Watt, Westinghouse, gas Generator that had a 6.6 gallon (25 liter) fuel tank that would last 11 hours at  half load.  It cost about $1000.  We would be burning 25/11 or 2.27 liters per hour.

The specific energy of gasoline is 46.4 MJ/kg and 1 liter of gas weighs 1.64 lbs/2.2 = 0.743 kg.  One liter of gas will produce 46.4 x .743 = 34.47 MJ or 34,470 kJ.  But we burn 2.2 liters per hour, 2.2 x 34,470 = 75,845 KJ expended as heat and electricity in one hour.  We said we need 3.7 kW to power the fan, but a Watt is in terms of per second and we want it as per hour, so multiply by 3600.  3.7 x 3600 = 13,320 kJ per hour.  The energy produced as heat will be the total energy produced by the burning gas, minus the electrical energy:  75,845 kJ – 13,320 kJ = 62,525 kJ per hour.

In the previous blog, we saw that our 50,000 btu furnace put out 52,750 kJ/hr and we saw that this would be the energy necessary to raise the temperature of the wheat by 5 C.  That is at 100% efficiency.  However we will be losing some of the heat; not all the heat energy will be going into raising the temp of the grain.  Our gen set produces a little more heat than the furnace, and doing 11 hour runs would be enough.  A 5 C rise in temp, is like pulling the temperature of the grain down by 5C.  Using our rule of thumb that 15 C decrease reduces the MC by 1%, then 5C would reduce it by 0.3 percentage points.  To go from 17 to 14.4 would take 2.6/.3 = 8.6 eleven hour runs.  Let’s say 9 runs. Each run uses 25 liters, and let’s say a liter costs us $1.  The cost for the fuel would be 9 x 25 = $225.  One must bear in mind that this is for both the electricity to drive the fan but also for the supplemental heat.

One way to make this better would be to convert the gas gen set to natural gas.  With an inexpensive conversion kit, this can easily be done.  The cost of gas is about 2.5 times that of natural gas, so we could easily get our price of fuel down to $100.

Another idea, would be to use a car, truck or tractor that runs on gas, and simply park it right up against the fan so that the radiator is right next to the input of the fan.  Shroud the whole thing with a tarp, so that the fan sucks in all the heat from the idling vehicle.  If it is burning about 2 liters per hour, it will be giving off about the same amount of heat as our gen set, and will raise the temperature of the wheat by 5 C.  We would avoid the cost of the gen set, but we would be paying over $200 in gas to get the grain dried.

In all of this estimating, it was assumed that the daily temp was constant.  But we all know that the temp changes by 5, 10 or even 15 deg over a day.  Can we use this variation in temp to save some money.  Also on any given day we will have better and worse drying conditions.  The humidity can be high or it can be low. Maybe we could do some careful cherry-picking to only do our runs when we will maximize our drying and reduce our overall cost?  Should we be cycling heating and cooling?  Just more to consider for future blogs.

Flax                12                           4.5                        0                          100.4            4.5
Flax               12                            24                         20                         102.6             4

So, I think this is a serious problem with those attempting to use supplemental heat with their aeration fans:  heating the grain by more than 5C can result in serious condensation problems at the top of your bin.

How much water will be condensing out as rain.  Let’s say that Jim’s wheat is heated to 15 C and the outside air is 5 C; the wheat is 10 degrees higher in temperature than the outside air, so we suspect there will be condensation; but how much. Using the grain calculator (EMC equations) we discover that wheat @ 17% moisture at 15 C will produce an RH of 79.1% and an absolute humidity of 10.2 grams of water per cubic meter.  When this air hits the cooler roof that is the same temp as the outside air, 5 C, it will cool to 5 C.  The most water that air can hold at 5 C is 6.86 grams of water per cubic meter.  The air when it was warmer at 15 C was holding 10.2 grams.  The difference is the amount of water that will condense as liquid water on the roof, or fall onto the top layer of wheat as rain. For every cubic meter of air that is blown into the bin, 10.2 – 6.86 = 3.34 grams of water will rain down on the wheat.  In one hour, it was determined earlier, that 5100 cubic meters of air flows through the bin.  In one hour 5100 x 3.34 = 17,034 grams, or 17.034 kg, or 37.54 lbs of water is deposited onto the wheat.  Can you imagine filling up the better part of a 5 gallon pail with water and throwing it onto your wheat — every hour??  And why do we get spoilage and a crust on the top??

I got a call a couple of days ago from Jim S from Wainwright, Alberta.  He said this night drying thing wasn’t working for him, and that he was considering buying a natural gas furnace to add some supplemental heat to the process. Would that work?  I told him I thought so, but I would have to work out the numbers to see; but I’ll save that for another blog. First let’s have a look at why this night drying isn’t working out.

Jim has tough wheat, 17%, that was harvested cold and put in a 3500 bushel hopper bin. Attempts at night drying brought the temperature down, but because of the nasty wet weather they have been having, the moisture has remained close to 17.  Jim thought that water may have been added.  I told Jim that once you got the temperature of the grain down, you have taken the energy out of it that could have been used for drying.  So it is quite possible that now that the grain is cold, that drying will cease.

Here is what I did to get a handle on what was happening in Jim’s case.  I went to the website for BINcast and got the hourly temperature and relative humidity for Wainwright from Sunday Oct 14 til Thurs Oct 18.  The temperature was bouncing around the freezing point, from +4 to -6 C.  The relative humidity was high throughout this time, 90% –  95%.  We didn’t know the temperature of the wheat, but we knew it was cold, so I assumed it was 4 C.   I modeled the temperature of grain, so that it would chase the outside temp in a similar fashion as I saw how the grain temperature changed in our trials at Indian Head. At the end of this simulation, the grain temp was close to -1.5 C.

I now had all the information I needed to calculate the amount of hourly drying. To make a long story short, I used the principle of absolute humidity, and EMC equations to calculate the amount of drying for each hour. For the first few hours we did take out a bit of moisture until the temp of the grain came down, and then we started adding small amounts of water. We did get some drying when the temp went down to -5, but this was short lived.  In the end we did indeed add water, 7.52 kg.  This actually isn’t that much, it would raise the MC from 17% to 17.001% however we can definitely say that night drying did not work in these conditions.   Maybe we should consider supplemental heat?  Maybe Jim should buy his furnace.  Using supplemental heat brings forth a whole host of other questions.  What fuel should I use? How much will this cost?  At what time of day do we apply the heat?  How hot should we get the grain?  Should we have a cooling cycle, or should we just apply the heat continuously. How long should the cooling cycle be?

In my previous blogs, I showed that a 15 deg C cooling of the grain resulted in a 1 percent decrease in moisture.  So let’s say we want to raise the temp of the wheat by 15 C. The specific heat of wheat varies, but it is about 1.36 kJ/kg C. I will spare you the details but to heat 3500 bushels, 15 C, requires 2 GJoules.

I checked with SaskEnergy and used the internet to check out the cost of the different fuels for 1 GJ.

Natural Gas   $6.48 /GJ

Gasoline          $16 /GJ

Fuel Oil (Diesel)  $21.65 /GJ

Propane            $14.66 /GJ

Electricity           $32.81/ GJ

So, we can see there is no decision, if you have natural gas, it is by far the cheapest supplemental fuel.  Also 1 GJ , at least, is the required energy to remove 0.5% MC.  Maybe we can get Jim’s wheat dry for $40 — if we do everything just right? As our benchmark,  and challenge, we will use the same nasty weather conditions above, with the supplemental heat and see what happens?

Stay tuned for the next blog on playing with supplemental heat.

When should the fan be turned out?  There is another calculator — other than mine that tells you when the fan should be operated.  But it is based on assumptions that are not realizable — judge for yourself.

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On Thu, Sep 22, 2016 at 3:34 PM, Ron Palmer <Ron.Palmer@uregina.ca> wrote:

Sarah:
I have not seen BINcast as a webpage, but have followed the theory of operation.  It does not work; because they do not take into consideration the grain temperature. Here is the reasoning behind it, and at first glance, it appears to be a rational theory:
If one takes the outside air temp and relative humidity, and inserts these into an EMC equation for — let’s say barley — you will get an Equilibrium Moisture Content. Let’s assume this is 16%.  If I blow this air at my barley, it will eventually reach a moisture level of 16%.  And that is true, if this air is blown at the grain long enough, eventually the barley and air would reach equilibrium.  Reaching equilibrium means that the barley must be the same temperature as the air.  One can not assume that the barley is the same temp as the outside air, in fact it never is.  It takes hours and hours of blowing air (probably a thousand air exchanges) to change the barley’s temperature.  But the outside air is not a constant it is changing hour by hour, and the barley is always playing catch-up.  The outside air and the barley are NEVER in equilibrium.  Sorry but this is not a good calculator.  I can come up with example after example, where this calculator will give the wrong answer.
I use EMC equations, but for temperature I use the temperature of the grain, (because the air becomes the temp of the grain) and I use the EMC equations differently.  You will notice I ask for the moisture content of the grain; using the moisture content and the temperature of the grain, I get the equilibrium relative humidity. That is where the grain wants to put the air’s RH.
If you are really interested, I can give you concrete examples of how the BinCast is totally different than my calculator, and it makes the huge false assumption that the air and the grain are at the same temperature.
Ron Palmer

We have all seen those little diagrams of curved pointed lines that indicate convection currents in a grain bin.  How in the winter the air flows up through the middle and down the sides, and then in the summer where it flows down the center of the bin and up along the warm side walls.  We can also get convection currents right through the bin with air flowing through the fan, vents, and center opening.  But just how fast is the air moving with convection currents? Let’s see.

First of all, the only reason we get convection currents is because of the air being at two different temperatures.  Air is heavier if it is colder.  A column of cold air has more weight than a column of warm air, and consequently will push down with a pressure equal to the difference in weight.

According to http://www.engineeringtoolbox.com/air-density-specific-weight-d_600.html , the weight of air is:

ºC         ºF         lbs/ft^3

10        50          7.786

21.1     70          7.492

32.2      90         7.219

If the outside air and the grain are the same temp, the air in the bin and air outside will have the same weight and there will be no pressure difference and therefore no convection current through the bin.   But let’s say that the grain is at room temperature of 21.1ºC , 70ºF, and the outside air is at 10º C.  That is quite a difference in temperature and also a difference in weight, 7.786 – 7.492 = 0.294 pounds per cubic feet.  But the grain in the bin is 20 feet high, so 20 x 0.294 = 5.88.  A column of air, 20 feet tall,  inside the bin weighs 5.88 lbs more than a corresponding 20 foot column of air outside. But 60% of this column is occupied by grain, and only 40% is air; so this column of air is only 2.35 lbs heavier.  But that is the pressure exerted over one square foot, psi or pounds per square inch is a more common unit of measure for pressure. 2.35/144 = .016 psi  This isn’t very much pressure, when you think of our car tires have 30 psi.   Another common pressure unit is inches of water.  How many inches of water would this pressure support? One psi would support a column of water 27.68 inches.  0.016 psi is equivalent to 0.452 “H20.

To determine the flow, one must have some appreciation for the grain resistance. From our past trials we saw that 3000 CFM required a pressure that would support 6 inches of water. Now if we said that pressure and flow were in a linear relationship, (it is not –it is closer to a squared),  the convection current would be about one thirteenth or 226 CFM.  This is a reasonable flow.  The bin we had was about 200 sq feet. so a flow of 226 CFM would be a vertical speed of 1 ft per second or about 1 kmph; but then again one must remember that this is only for a very significant temperature spread; as the temperature of the grain and the outside temperature become less, so too would there be a reduction in the convection current.

Here is a question I got from Quebec:

“Mr. Palmer, Last fall we discussed a lot about grain ventilation. I remember how you where generous on this subject. I want to share with you what we do. This year we will do other observations. The producer where we will make observations will get a 4 noodles cable with Stormax. (20’ x 20’ silo). However, control will be executed by water balance at least for the stop time. Re-start will be done manually or by timer followed automatic running. I will try to share you these informations including stormax data if possible. Do you think this approach is good ? Do you have suggestions ?

The producer is supposed to fill it with 2-3 crops (first : rye…temporary storage, second : Oats…temporary storage, third : soya). In the litterature I did not find EMC equation for rye. Is it similar to wheat ? Do you think Opi systems have it ?”

I gave the following reply:

Good to hear from you. I will try to answer your questions. I am not sure what a 4 noodle cable is, I am assuming it is one with 4 senors, that is 4 sensor temp and 4 senors for relative humidity. I am assuming that you are doing a water balance to stop by calculating the absolute humidity of the discharge air and comparing it to the absolute humidity of the inlet air. And you only stop the fan if the absolute humidity of the inlet air is greater than the absolute humidity of the discharge air? Great. To turn the fan on we can’t use the discharge air, but we do want to know what the absolute humidity of the air in the silo is? There are two ways to do this. 1. this is the best way and it can be done if you have a temp and RH sensor in the core of your grain or crop. Use the T and RH as you did before to determine the absolute humidity. 2. If you don’t have the RH sensor in the body of grain, and only the temp of the grain, you can use EMC equations. By plugging in the moisture content and temp of your grain, you can determine the RH to which it will find equilibrium. Use this RH and T to determine the absolute humidity. If this absolute humidity is greater than the absolute humidity of the outside air, then you have a drying condition and the fan should be turned on. You can use the calculator that I made to do this for you. Just plug in the temp and moisture content of your grain, and the outside temp, and the calculator gives you an RH thrreshold, if the outside RH is below this then you will be drying, anything above this, and it will be wetting. Actually the only thing I am doing with the calculator is the math for the EMC equations. And no I don’t know the EMC equation for rye, but with a little digging I am sure I could find it. I think it would be OK to use wheat — it would be close.
My calculator can be found at planetcalc.com Grain Drying Calculator

Grain Drying all comes down to something pretty simple. If the outside air has an absolute humidity less than the absolute humidity of the air inside your bin or silo, then you have a drying condition. The only thing you need to calculate the absolute humidity is the temperature and the relative humidity. When the fan is on, the relative humidity is read at the point of discharge and when the fan is off you get the relative humidity by reading an RH sensor directly in the body of grain, or if you don’t have this RH sensor then you must resort to using an EMC equation (or my calculator) to determine the RH.

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Still thank you Mr Palmer, Yes it is supposed to be an Opi Cable with 4 points reading of T, RH and EMC. When you talk about a T&RH sensor in the grain, what position do you suggest : at the top, at the surface or in the grain and if so how deep ?

The sensor in the grain, the RH & T sensor, should represent as much as possible the entire grain bulk. We know that the bottom has more variation than the top. So I am thinking in the center of the bin or silo, about two thirds to three quarters the way up. You don’t want it too high because shrinkage may lower the grain and leave the sensor exposed to the air, which can be greatly influenced by the open discharge port. We also considered averaging several sensors to get an average T and RH, but one can get into trouble really quickly because the pyschrometric equations to calculate absolute humidity are non-linear, so I think using just a single sensor might be better.
Also the fan must be off for some time to use this sensor, as one must wait for the air in the grain to ‘equalize’ with the grain. I am thinking at least a half an hour or maybe an hour..
Ron Palmer

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Integris of Opi Systems has the capability to start and stop automatically the ventilator by comparison of grain EMC with air intake. To stop ventilator, how this decision is good considering air is not equalized with the grain. To be correct and to take the good decision, ventilator should it be off for a while and then make the calculation after 0,5-1 h ? Do you think so ? Another explanation : Would they correct the reading, considering it is not in a stabilized situation ?

I also ask myself if the decision to stop or to start is depending of one, a combination or the average of all sensors in the bin.

I hope all these questions are not bothering you. Your informations are very usefull to understand.

EMC is equilibrium moisture content, notice the word equilibrium! You must wait some time for equilibrium to occur. Our data shows that you can not correctly do an EMC while the fan is running. You get the wrong value.  The correct way to control the fan is with the absolute humidity, and if you have RH sensors, there is no need to even do an EMC.

Excel Column Definitions for Data Collected at Indian Head 2007 to 2015 from Reliable System.  This is for the record and is only useful to those having the data.

Column         Definition

A. Date & time stamp of sensor readings on the hour Nov 03/15 17:27:23

B. lowest temperature sensor close to perforated tube, ⁰C

C. Tm, mid temperature sensor in the grain half way up, ⁰C

D.  Tt, highest temperature sensor in the grain ⁰C

E.  T, temperature of discharge air just inside roof opening, ⁰C

F.  RH, relative humidity of the discharge air %

G. Pressure in inches of water or CFM, fans running if pres > 0.05

H. Temperature of outside air, ⁰C

I.   RHair, relative humidity of outside air, %

J.  Not usually used but in bigger bins pressure is sometimes here

K. Net H₂0 leaving the bin per cubic meter, grams per cubic meter

L.  =IF(G1 > 0.05,750 * K1 * 60/35310,0) if the fan is running, multiply K by the number of cubic feet that are leaving the bin. In this case CFM is 750 and there are 35.31 cubic feet in a cubic meter

M.  MC , moisture content of the grain,%. The initial moisture content is manually put in Row 1 and then we subtract from that using the amounts from L. We need to know the number of bushels and weight of the grain so that the ratio can be calculated.

N.  RH_thres % Knowing the MC and Temp of the grain, we can plug these into an EMC equation and get the relative humidity that this grain wants to be at.

O. EMC_hum , EMC absolute humidity, grams of water per cubic meter that the grain wants the air surrounding it to contain this much water, and we calculate this from the psychrometric equation or the saturation equation factored by RH_thres, just calculated in column N.

P.  This is the ratio that shows how much water was added to the air as it passed through the grain and how close it got to the EMC value: (Dcharge_hum – Air_hum)/(EMC_hum – Air_hum). The Dcharge_hum will be trying to get to the EMC_hum but will never get there, so this ratio will less than one.

Q. Ƭ, Time Constant for how long it takes the outside air to acquire moisture from the air and be 63% of the way to EMC We know how close it got from P. and we can calculate the time, t, that is was in contact with the grain by knowing the number of bushels and the CFM. The smaller bins hold 2200 bushels, and one bushel is 1.2446 ft³. So the bin is 2738 ft³. But the grain takes about half this space so there is 1369 ft³ of air. At 3000 CFM we would get an air exchange every 1369/3000 = 0.45 min. Or, the air is in contact the grain for 0.45 min. The time constant, Ƭ = -t / ln( P – 1) where t is 0.45 min and P is ratio calculated in column P. This time constant is important in understanding the amount of time that is needed to reach equilibrium.

R. Ratio of t/Ƭ. The bigger this number, the closer we are to equilibrium. It is a measure of the efficiency to which we are using the energy in the grain to dry. The bigger this number, the more energy we are using from the grain to dry.

S. Safe days. See Spoilage Index Report on how safe days and spoilage index is calculated.

T.  Spoilage Index.