## How much Time is Required to Cool Grain

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.

## Mitigating Over Drying at the Bottom of the Bin

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 RHthres 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.

## EMC time constant, tau

With EMC, equilibrium moisture content, we are required to leave the grain for some time before the air’s humidity and temperature come into equilibrium with the grain.  In the literature, the authors of the EMC experiments and resulting equations suggested anywhere from three to seven hours.  But indeed, even with this amount of time, equilibrium is never reached.  The temperature and humidity approach that of EMC of the grain by following an exponential path and then slowly, ever so slowly approaching, but never quite reaching equilibrium.  It approaches asymptotically.

In engineering and science there are many process that react in this manner:  a capacitor charging up to a battery voltage through a resistor, or a small tank being filled with water, through a narrow pipe from a large tank.  At first there is a large flow with a huge pressure on the pipe when the small tank is empty.  But as the tank fills, the water level difference decreases, the pressure decreases and the flow decreases.  The flow becomes slower and slower, and the level in the small tank approaches that of the level of the big tank, but it never gets there.

Why is it exponential and exactly what is this path?  Let’s walk through the math using the Capacitor with capacitance C, connected in series with a resistor, R, and battery with a voltage Ve.  The voltage (or pressure) across the capacitor is Vc and is determined by the amount of charge, Q on the capacitor.  C = Q/Vc  or Vc = Q/C. The voltage across the resistor depends on the flow of charge: Vr = dQ/dt.  Now according to Kirchoff’s, the series voltage, going all the way around the loop will be zero; or  Vc + Vr  – Ve = 0,    Q/C + R dQ/dt  – Ve = 0  and multiplying each term by C:  Q +  RC dQ/dt = C Ve – Q  and rearranging and separating the derivatives give:  dQ / (CVe – Q)  = dt / RC  taking the indefinite integral on both sides:  – ln(CVe-)/CVe = t/RC and raising both sides to an exponent:

Vc = Q/C = Ve(1 – e^-t/RC)   When the time, t is equal to RC such that the exponent of e is  -1, this is the time of one time constant, usually expressed with the stylized t or tau.   It is the time for Vc to become 0.632 that of the battery voltage.  It is the time it takes for the capacitor to charge about two thirds of the way to the battery voltage.  In this example with the capacitor charging through the resistor, one time constant is equal to RC.  Note that it only depends on the value of R and C and is totally independent of the battery voltage.  This is why time constant is useful.  It does not depending on the driving voltage or pressure, and it can be thought of as the time to cover two thirds of the way to the driving force.

In the case of grain and EMC, the air holding the water is the capacitor, the water is the charge, the resistance of kernels skin is the resistance, and the water in the seed or MC provides the pressure, like the battery voltage.  Using the EMC equations, we can plug in a MC and T, and get the relative humidity of what the grain will seek equilibrium.  Call this RHemc.   We can now use the RHemc and Temperature, plugged into the psychrometric equation to determine the absolute humidity the grain would like the surrounding air to be at.  Call this Hemc.  But the air around the grain does not contain this amount of water, it has an absolute humidity of Hair.  Hair will be working itself to that of Hemc, but just how long does it take?   What is the time constant?  How long does it take for the Hair to get two thirds of the way to Hemc.  Note we are not concerned about the specific value of Hair or Hemc, only the distance or difference between the two values.  For example, let’s say that Hemc is 20 gr. per cubic meter, and Hair is 10 gr.  There is 10 gr. difference and want to the time it takes for H air to become 16.32 gr. ( 2/3 of the way).  This amount of time is one time constant.

We can determine the EMC time constant experimentally.  By knowing the size of the bin in bushels, and CFM we can determine the amount of time to do one bin air exchange.  This is the amount of time that the air is in contact with the grain; let’s call it t.  We can also calculate Hemc and Hair and also the amount of water in the air leaving or being discharged from the bin, Hdis.  So air entering the bin with Hair, passes through the grain for a time t, and picks up water from the grain and gets closer to Hemc.  The percent distance to Hemc is what I will call %,   For one of our bins the transit or contact time worked out to 0.45 minutes.  If we measured % to be 0.632, the Hair would have got .632 closer to Hemc, this would be one time constant, and one time constant would be 0.45 minutes.  But we are never that lucky in getting 0.632, it will be something else.  However we can adjust for it:

one time constant =  – (contact time) / ln (% – 1)

Todate, our determination of the EMC time constant is about one minute.  If the grain is in contact with the air for one minute, it will get about two thirds of the way to EMC equilibrium.  This actually is much faster than I first would have thought.

## Canola dripping — Mitigating Factors

In my last blog I went on a rant about how canola could start dripping if the canola is more than 4 C warmer than the air.  And this would be the case when the fan is initially turned on, but after a few minutes — after a few bin air exchanges; there are mitigating factors.

1. The roof, although it is a good conductor of heat, it might warm a bit.  It isn’t necessarily the same temperature as the outside air. It could warm up a deg or so.
2. The canola will be dried down just a bit, but not much, and this would also give us a small advantage in getting a bigger spread than 4 C.  But we know that it dries slowly, so in a few minutes, we can’t count on much, but in several hours we would get some benefit.
3. The canola will be cooled a bit, and this will also help as it will lower the EMC absolute humidity. This would be more important in the long run rather than in the short term of the first hour.
4. When the fan is off the RH of the air in the bin will be that of what the EMC equations dictate for canola at that temp and MC.  I call it the RHemc. However when we start the fan, we don’t have equalized air we have air that is coming into the bin with a much smaller absolute humidity.  It takes time for the air and moisture to equalize, and the air passing through the grain is only in contact with the grain for a minute or so.  We can’t expect the air to reach the value of absolute humidity or RHthres.  This is a non linear process, as the absolute humidity approaches the EMC absolute humidity asymtoticly.  Typically we describe these processes with something called a time constant.  One time constant is the time it takes to reach 63% of the EMC absolute humidity.  I am thinking that in our case it is probably one time constant.  I am devising a means to measure just what this is from our data and that will be a discussion for another blog.  In any event the absolute humidity of the air will be less than the EMC absolute humidity and this will have significant mitigating factor in dripping and increase the temp difference to perhaps 6 or 7 deg C for our canola example.

## Condensation and Dripping a Big Problem for Canola

What if we have the situation in which we have a bin of canola at 11% MC and 30 C and we decide to turn the fan on when the outside temp is 20 C.  You wouldn’t think this would be a problem as that cooler dry air hits the grain it will instantly warm to 30 C, resulting in a lowering of the RH and it will dry the canola.  The problem starts when the air leaves the grain at 30 C, and hits the colder roof of 20 C.  As it cools it will saturate and condensation, dripping and literally raining will occur in the bin.  This water will run down the roof and collect in the canola on the upper side walls, and will also collect directly on the top layer of canola which at the least will form a crust and at the worst will form a hot spot of spoilage that could spread to the whole bin.

Using the grain drying calculator I plugged in 11% MC and found that condensation would occur with as little difference as 4 C.  If the grain temperature was more than 4 degrees C warmer than the outside air, we would get condensation.   I tried it for  grain temp / outside temp for 30 / 26, 20 / 16 and 10 / 6.  Only four degrees difference, and we were on the verge of condensation.

Let’s walk through this with a little more detail.  We have canola at 11% MC and 30 C.  When we calculate the EMC RH for this we get an RH of 80.1% or 83.8%, depending on whose EMC equation we use.  In terms of absolute humidity this is 24 or 26  gr/m^3.  In other words canola at 11% and 30 C wants air around it, or wants to equalize to having air that has about 24-26 grams of water in a cubic meter.  And that’s fine until this relatively warm moist air hits the much cooler roof that is the same temperature as the outside air, 20 C.  When it hits the inside of the roof, it cools to 20 C — and this is where the problem starts — air at 20 C can, at most hold 17 grams of water per cubic meter.  That’s at most, at saturation.  For every cubic meter of air that hits and is cooled by the roof;  25 – 17 = 8 grams of water will be squeezed out as drops of condensed water.  That may not sound like much, but let’s consider a typical flow of 3000 CFM or 180,000 CF per hour or in terms of cubic meters:   180,000/35 = 5142 cubic meters per hour.  Multiply this by our 8 gr per cubic meter and we get over 41 kg or 90 pounds of water dripping onto and into our canola every hour.   Do you think this might be a problem?

To prevent condensation we can’t allow the grain to be more than 4 deg warmer. This is why you should turn your fans on immediately, even while you are filling your bins.  Waiting, into the evening when the temperature is cooler is yes good for drying, but not so good for condensation.  You can see though why there is an argument for keeping the fan on continuously so that the grain temp is always chasing the outside temp and is keeping it to within 4 deg of the outside temp.

## Safe Days

From experience we know that grain spoils more readily when it is hot and wet, and that it is the most secure, the safest, when it is dry and cool. But can we quantify spoilage?  There have been attempts at observing mold and fungus, but this was just too subjective.  In 1981 Fraser and Muir suggested that the reduction in germination capacity was related to spoilage and they arbitrarily declared that when the germination capacity was reduced to 95%, it constituted the first step or first stage of spoiling.  The number of days it took to have the grain deteriorate to this point was declared the number of SAFE days and it was empirically found to be:

Safe Days = 10^(6.234 -0.2118 MC – 0.0527 T)   wheat and cereals

= 10^(6.224 – 0.302 MC – 0.069 T)     canola and oilseeds

The objective for safe storage, is to maximize the number of safe days by lowering the temperature T, and the moisture content, MC. Look at how the number of safe days varies with temperature for ‘dry’ wheat:

14.5% @ 30⁰C =  38 safe days

14.5%   @ 20⁰C = 128 safe days

14.5%   @ 0⁰C = 1458 safe days

New Definition:   Spoilage Index = Ʃ ( 1/ safe days)   x 100

The value for spoilage index is equivalent to the amount of spoilage.  Anything less than 100 is very good, and essentially constitutes very little spoilage.  A value of 100 suggests that we have achieved our first level or stage of spoilage, and if the value is several hundred we are getting into some serious spoilage.

Example, if the calculation above  yields the number of safe days to be six, then after 3 days we will have an accumulated deterioration of:     (1/6 + 1/6 + 1/6) x 100 = 50%,

after 6 days : (6 x 1/6) x 100 = 100% of the way to 95% germination capacity

With our system we read the sensors for temperature and can calculate the moisture content, MC, and temperature, T,  every hour so we can refine the spoilage index to accumulate on an hourly basis to catch the dynamic changes in the bin, especially for the first few days. So when accumulating every hour the Spoilage Index would be:

Spoilage Index = Ʃ ( 1/ safe days)/24   x 100

Again if the spoilage index approaches 100; it means that we have reached the first stage of spoilage, the germination capacity has been reduced to 95%.

Spoilage index is an important tool in being able to measure spoilage such that different control strategies can be compared for the efficacy of safe grain storage.

## Comparing Control Strategies

In my last blog, I listed all the different fan control strategies.  From the very simple ,(on at night — off during the day), to the ultimate controller that required moisture cables.  But how much different are they?  Is the extra complexity and moisture cables worth it?

I think we all can agree that the water balance (water in/out) is the most accurate means of measuring whether of not we are drying or wetting, and by how much.  And sure it is great for turning the fan off, but it totally lacks in figuring out when to turn the fan on.

OK let’s compare control methods.  If we only want to control for ‘drying’ i.e. turn the fan on only when we have drying conditions, then we can easily use the water balance as a basis.  But let’s have another look at this; with water balance we are calculating the absolute humidity of the water going into the bin and the absolute humidity of the water coming out.  How different is that from our EMC method in which we input the MC and grain T to get a RH that can easily be converted to an absolute humidity.  In other words grain sitting in a bin will try to produce a specific RH with a given MC and T.  And with T and RH we can easily calculate the absolute humidity.  Now if the outside air has an absolute humidity that is less than the air in the bin’s absolute humidity we will get drying.

Let’s go through this again.  The fan is off and has been off for some time, at least an hour.  Therefore we can use EMC — everything must be in equilibrium. So we plug the grain T and MC into the appropriate EMC equation and get an RH.  From this RH and grain T we calculate the absolute humidity — the number of grams of water in one cubic meter.  Or perhaps we have a moisture cable and have the RH of the air directly.  Either way we end up with the absolute humidity.  And if this absolute humidity is greater than the absolute humidity of the outside air then we have a drying condition and a means to turn the fan on. Now this works out really well, because to use this the fan must be off, and for some time for things to equalize.  Once the fan is on, we start using the water balance criteria of more water coming out than going in to make a decision to turn the fan off.  This works out really neat:  the EMC technique does not work in flowing air, but is great for no flow; whereas the water balance does not work in no flow conditions but is great for flowing.  Therefore use EMC to turn the fan on, and water balance to turn it off.  The two techniques complement each other and are actually using the same thing as a decision criteria — the absolute humidity of the air inside and outside the bin.  There is no point in suggesting that one is better than the other, they in essence are the same.

The method in which we use temp difference is another story.  With temp difference, we only turn the fan on if the grain temp > outside air temp.  We know that the MC of the grain and the RH of the outside air are not used in this.  I did a correlation with the temp diff and water balance on some data we did where the fan ran continuously.  The R^2 value was  0.64 but what does that really mean?   How much better would this temp diff technique be if we did include the RH in some fashion.  Surely the RH does have some effect?  If it is raining outside with an RH of 100%, one in inclined to think we will not be drying?  I don’t care if the temp of the air is less than that of the grain.??  Can we look into this a bit more?

Here is what I did.  I took some old data from master_2015B file, sheet 08 09W which means it was from 2008, bin 9, wheat.  The run was for 288 hours with the fan running continuously.   I did the water balance calculation for each hour and determined how much drying occurred for each hour.   I then applied the temp difference calculation ( assume drying ~ grain Temp – air Temp)  As stated previously I did a correlation with what I consider to be the correct drying, the water balance, and got 0.64.  I also saw that if I used Tdiff as the control strategy there would be 12 hours out of the 288 in which Tdiff would wet the grain.  And if you qualify it with only turning the fan on when RH < 80%, then we are reduced to 6 hours of wetting.  If we further qualify Tdiff with an offset of 2.  Tg-Tair>2 or Tair + 2 < Tgrain, then we only have one hour of slight wetting (this includes RH<80%).

To summarize we got wetting:

1/288   offset of 2   RHair < 80%

6/288  offset 0    RHair < 80%

12/288  offset 0  RHair < 100%

If we used the Tdiff control with offset of 2 and RH < 80% we essentially won’t get any wetting, BUT we will miss out on some drying opportunities.  There are 29 hours in which Tdiff would have turned the fans off when indeed there was some drying and possibly a little warming of the grain.

We are starting to nit-pick here a bit.  The Tdiff is so simple, easy to understand, easy to implement with no heavy duty calculations, and no need for MC or moisture cables. Tdiff will take a little bit longer to get the grain dry, but it will just as cold and safe.  For simplicity we are giving up some drying opportunities.  What do we want?  Perfection or Simplicity?

I personally would like to go with the moisture cables and have the whole process automated with the ultimate controller; but I can certainly understand and appreciate those that would lean toward the simpler system — they have temperature cables and don’t want to purchase moisture cables — no problem, we still have something for you that works well.

## Fan Control Strategies == Worst to Best

• Continuous: The absolute worst situation is doing nothing with hot tough grain. This requires no attention, no sensors, and has a very high risk for spoilage. The number of safe days is small, and the grain is deteriorating quickly.   It still may work; it may dry the grain, but is very risky with the potential for the most spoilage. The next worst thing is to leave the hot tough grain for a week or so before turning it on only during the day. Or, leave it for a week or two, and then turn it on continuously, and finally turn it off after a hot day of running, leaving the grain hot, but somewhat dried. It is better to run the fan continuously right from the start, and quitting after a cold night, leaving the grain cold.
• On at Night: One requires no sensors, and only a little attention to turn the fan on at night and off by 9 the next morning. The fan should be turned on immediately, even while the bin is being filled. When the average grain moisture is dry, the process s should stop after a cold night of running. The grain will warm up slowly by maybe a degree a week, and once a month the grain should be cooled by running the fan on a very cold dry night. This technique does not check for dripping or condensation conditions. It could be automated with a simple timer controlling the fan’s actuator.
• Water Balance: This requires T and RH sensors on the air entering and leaving the bins. It also involves some calculations to determine the net amount of water leaving the bin. However the exhaust T and RH sensor are only effective when the air is moving or when the fan is on. When there is no air flow, the fan is off, the sensor are in stagnant air and do not represent the true readings of the exhaust air. So this is a great mechanism to determine when to turn the fan off ( no or little drying is occurring) but it is useless when in determining when to turn it on.
• Temp Difference: Tgrain > Tair This requires only two simple temperature sensors, one in the grain and the other in the outside air. No math or calculations are required, and the comparison can easily be done manually or with very simple electronics. This would be a simple technique that could be used for automatic control. It should be started as soon as the bin is filled, and end when the average of the grain is dry, after a cold night. It is lacking in that it does not take into account either the MC of the grain or the RH of the outside air. It is not perfect, but it is simple.
• EMC Calculator: The calculator requires the grain’s T ,MC, and type as well as the outside air’s T and RH. It could easily be configured to turn the fan on and off automatically, but it also could be used to control the fan manually.   For those that have temperature sensors and are not keen on changing to moisture sensors, this may be the way to go. The disadvantage with this method is that calculations must be made and that we are subject to the inaccuracies of the EMC equations. Also if we have an obscure grain, it may not be listed on the calculator. The fan should not be turned on if dripping or condensation will occur: RHthres > 100
• Moisture Cable on, Water Balance Off: This is the ultimate controller, but it requires a moisture cable strung from the center roof collar, with the highest RH/T sensor being in the air to get the T and RH of the exhaust air. The lower T/RH sensors would be in the grain.   For each of these sensor nodes, one gets the T and RH of the grain, and from the Saturation Equation one can calculate the absolute humidity of the air in the bin. Also using the Saturation Equation, one can easily calculate the absolute humidity of the Outside Air by using its T and RH. If the absolute humidity of the outside air is less than the grain air, then we have drying conditions and the fan should be turned on. Once the fan is on, we do a water balance calculation , using the RH and T of the outside air, and the exhaust air. When the calculation indicates that there is no longer a net flow of water out of the bin, the fan is turned off and in an hour we return to comparing the absolute humidity of outside ait, and grain air.   But there probably are more than one T/RH sensor in the grain and some sort of averaging must be done to determine just when to turn the fan on. Although this strategy depends on the installation of a more expensive moisture cable, it avoids the inaccuracies of the EMC equations , or even the knowledge of the grain type. However we might want to use the EMC equations to determine what the MC of the grain and to terminate the process when the average MC of the grain from top to bottom is dry. The grain should still be monitored, and maybe once a month the grain could be cooled to keep it as cold as possible.   An even more reliable system would have two moisture cables in the bin to detect faulty sensors. The fan should not be turned on if dripping or condensation will occur: the saturation absolute humidity of the outside air > absolute humidity of the grain air.

## First Day is Critical

Grain starts to spoil the moment it is harvested.  The deterioration can be slowed down to almost nothing by cooling and drying the grain.  Grain can be cooled in a matter of hours with natural aeration, especially with cool night air.  Drying may take a little bit longer, but our data has shown that if grain is cooled by 15 deg C on the first day, that it will lower the moisture content (MC) by one percentage point.  So we see that cooling is drying if done with aeration.  Leaving the grain cool by itself in the bin does not result in any drying.  Even if your grain is dry, it is important to get it cooled as soon as possible, to put a halt to the deterioration process.  The fan should be run as soon as the grain covers the screen and left on until 9:00 the next morning.

At one time it was suggested that grain should be left hot to sweat out the moisture.  This is not a good idea; the grain is starting its deterioration process, it is spoiling and losing quality.  Get it cooled down with aeration as soon as possible.  Once it is cool,  it is safe from spoilage.  Then getting it dry is secondary.  You can take your time and use the calculator to figure out when to run the fan.

The take home is this:   If you have aeration fans, use them, especially the first day the freshly harvested grain is stored; don’t wait.

## Is dry grain really dry?

There is a false sense of security when we think that dry grain is actually dry.  Let’s take for example wheat, at 20 deg C, with a moisture content (MC) of 14%. Wheat is typically considered dry if it has a MC < 14.5%; but just how much water is in the grain as opposed to in the air?  Let’s see!  Consider a volume or space of one cubic meter.  This is 27.5 bushels and at 60 lbs per bushed, this would have a weight of 1650 lbs.  We said the MC was 14%, so 1650 x 0.14 = 231 lbs or 104.78 kilograms.  That is, one cubic meter of wheat at an MC of 14% has 104,780 grams of water in it.  Now air at 20 deg C that is saturated, 100% relative humidity (RH) contains 17.5 grams of water.  But we shouldn’t be looking at saturated air in the bin, it is more likely to be around 75% RH and one cubic meter of air at 20 C, 75% RH would have 13.2 grams.  However in our one cubic meter of grain, the grain occupies 60% of the space, and 40% is air.  Therefore the water in the air is 0.40 x 13.2 = 5.28 grams.

So, this so called dry grain is actually caring 104,780 grams of water per cubic meter, and the air surrounding it is holding a tiny fraction of that, 5.28 grams. Even wet air at 100% RH contains thousands of times less water than the so called dry grain it surrounds.  So, it is not so dry after all.  And it is truly amazing that a seed can retain such a large amount of water.

What we have learned here is important to understanding grain drying. Let’s suppose that we were blowing air through the wheat and that it went from 50% RH to 75% by picking up some water from the grain.  How many air exchanges would it take to lower the MC of the grain by one percent?  14% was 104.780 grams of water,  so 1% would be about 7,484 grams of water to be removed.   One air exchange going from 50 to 75% would take out 1.75 grams.  Therefore we would need over 4,000 air exchanges to lower the MC by one percent.  Now one bushel is 1.2446 cubic feet, but the air in between the wheat would only be about a half a cubic foot.  At an air flow rate of  1 CFM/bu, an air exchange would occur every half a minute and we need 4000 air exchanges, 4000/ (60 x 2) ,which would take 33 hours.   It would take over a day of absolutely ideal conditions to knock down the MC by one percent.  If we are using the grain dryer calculator this means we would need the outside RH to be 25% less than the threshold RH for 33 hours. This example contains gross approximations and assumptions like the grain and air remaining at 20 C (it won’t);   but it does suggest how an estimate of the drying time could be determined.