Author: ron.palmer@uregina.ca

Grain Dries Much More at Bottom than Top! Why?

We have all seen it, the bottom of the bin having a moisture content three or four percentage points below that of the top. Why?

To explain this phenomenon, we will use our setup with a 2200 bushel bin, a 5 HP aeration fan and we will look at two examples one with barley @ 20 ⁰C and MC 15%, and another with canola @ 20 ⁰C and MC 11%. In both cases the fan produced a pressure that supported a six inch column of water. This was measured with a home- made manometer which consisted of a plastic tube shaped into a U with coloured water in it. The air flow was 3000 CFM.

Even though the outside temperature is 20 ⁰C, the temperature of the air behind the fan is warmer because of heat given off by the motor and because of compression. Let’s first look at the heat from the motor. What air temperature rise can we expect from it?

I am guessing here, but let’s say that the 5 HP motor is 90% efficient, which means that 10% of the energy from the 5 HP motor will be going into heat and this is caused by wire resistance, bearing friction, and even air friction on the fan blades.   1 HP = 0.7475 kW so 5 HP = 3.737 kW and ten percent of that would be 0.3737 kW or 0.3737 kilojoules/sec will go into heating the air. The air is flowing at 3000 CFM or 50 ft3//sec and the weight of one cubic foot of air is 0.0807 lbs or 0.0366 kg, so 50 cubic feet would be 1.83 kg of air going by per second. The specific heat of air is close to 1 kJ/kg.K⁰ So the temperature rise of the air would be: ( 0.3737kJ/s / 1.83kg/s ) = 0.2 ⁰C. The heat from the motor would increase the temperature of the air from 20 to 20.2 ⁰C.

But the big increase in temperature is not from the motor but that of the compression. We know that the pressure behind the fan is enough to support a column of water six inches high. We know this because we measured it with our home-made manometer.   How much of temperature rise will we get from this pressure or compression? There is a thermodynamic formula that relates pressure to temperature: PV = nRT.   Pressure and Temperature are proportional. A typical pressure of 1 atmosphere will support a column of water 406.8 inches. And a typical temperature is 273 ⁰Kelvin. So an increase in pressure will produce a corresponding increase in temperature:   6”/4068” = x / 273 and this gives an x of 4 ⁰K or 4⁰ C. The air and grain at the bottom of the bin would be 24.2 ⁰C, and as the air flows to the top the compression would get less and less until at the top there would be none, and the temperature of the grain and air would be back to 20 ⁰C.

Does this increase in temperature affect the MC? Yes it does, and we will look at the barley at 15% MC and the outside air is 20 C and so is the barley at the top of the bin. 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 at the top of the bin. However at the bottom of the bin the temperature is 4.2 ⁰ C warmer at 24.2. We are assuming that no drying or wetting is taking place so the absolute humidity at the bottom will be 12 grams per cubic meter; the only thing that has changed is the temperature. Using the humidity calculator again by setting the RH to 100 and the temp to 24 gives a saturation humidity of 21.8 gr, and since our absolute humidity is 12, the RH must be 12/21.8 = 49.6%. By using the grain drying calculator with an air and grain temp of 24.2 entered, and by trial and error entering MC until the RHthres is close to 49.6. I found that if I entered 11.6% for the MC, I got 49.4% for RHthres – close enough. This means that the barley at the bottom of the bin will be in equilibrium with the air around it at a MC of 11.6 and at the same time we have barley at the top that is 15% MC in equilibrium with the cooler air. This is a spread in MC of 3.4%.   And yes we have seen this type of spread in our trial runs. The top to bottom spread in MC is quite commonly three or four percentage points different. And now we know why.

Does the same thing apply to oil seeds? It might even be worse, because the pressure might be higher with a grain that has more resistance; but let’s see what happens for a pressure of six inches. Let’s look at tough canola at 11% MC at the top of the bin and again at 20 ⁰C. The outside temperature is also 20, but after it gets heated and compressed by the fan it is now 24.2 C. Using the drying calculator, we plug in 11 for MC and 20 for both the grain and air temp, and it gives us 78.1% for RHthres. And then we use the humidity calculator to calculate the absolute humidity, 13.5 gr/m3. Since there is no drying taking place, the absolute humidity will be the same at the bottom of the bin: 13.5. The saturation humidity for 24.2 C is 21.8 and this then gives us an RH of 13.5/21.8 = 55.8%. Again with trial and error by plugging in different MC we find that a MC of 7.1% is the MC of the canola at which equilibrium is reached at 24.2 C and RH of 55.8%.   The top of the bin is at 11% MC and the bottom is at 7.1 % MC – a whopping 3.9% difference.

I have seen fact sheets and guidelines for drying that suggest this difference in MC from top to bottom is actually some sort of front, and that to get the drying front right through to the top, one must use a bigger fan with lots of air flow, lots of pressure.   But in understanding that an increased pressure will only result in more pressure, more compression; the spread in MC from top to bottom will only be worse. To mitigate this MC spread, I think there are a couple of things we can do. First, use smaller fans with less pressure, and secondly don’t run the fans continuously; give the moisture a chance to equalize. We also don’t want to have a pressure drop across the screen or perforated pipe. I would certainly be an advocate for open bottom pipes or louvers.

There is another complicating factor that aggravates this situation.   I assumed in this analysis that no drying was taking place. We were at that point in time when the drying was done, and equilibrium was achieved at the top and bottom. However when we first start the fan, the grain at the bottom was just as tough as the grain at the top and the bottom would dry first. And when it dries, it must give up energy and heat to vaporize the water. The drying will be another cooling agent, cooling the air as it goes to the top. This will keep the top cool, the bottom warm – the top wet, the bottom dry. It also means that water will be added to the air, and that the absolute humidity of the air at the top of the bin will not be the same as at the bottom; it will be higher. Now we have air at the top that is even colder and wetter than it would be if no drying was taking place at the bottom. It is no wonder that the top grain has no chance of drying until the bottom is finished drying and much over dried.

 

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.

Excel Column Definitions for Data Collected — for the record

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.  RHthres % 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. EMChum , 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 RHthres, 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: (Dchargehum – Airhum)/(EMChum – Airhum). The Dchargehum will be trying to get to the EMChum 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.

Spoilage Index Results from our Trials

The spoilage index is an accumulation of the reciprocal of Safe Days. Safe Days depends on the temperature, T , and moisture content, MC. Safe days is defined as:
Number of days until germination capacity is reduced to 95% (Fraser & Muir 1981)
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
Example, if safe days is 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.
We modified the spoilage index in order to accumulate every hour instead of every day:
Spoilage Index (SI) = Ʃ ( 1/ safe days)/24 x 100
What we got from our trial runs from 2007 to 2015:
File      SI     trial time (hours)
071      22.6   214
072      13.7   259
073      11.6   14.8
0809P  3.1      54
0809W 23.8    290
0810B 60.6     291
0810W 24.9    215
0909B   3.6       151
0909W 149.8  268
1010     250     674
1110     71.6     239
1216      85.3    263 bin had no temp cables, so used discharge T
1217      364.6   335 used discharge T, so SI is questionable
1218      45        417 used discharge T
1219      85.3     405 used discharge T
1309B    3494    1274 certainly had spoilage here, cause high MC
1316      216.9    1724
1317      371.4     1910
1318      72.6       1741
1319      38.7       951
1409      1284      2020
1410     350         1296
1416      315.7     1969
1417     86.5        1969
1418     51.2        1600
1419     109.7      1600
1509     426.7      1940  but using sample tube MC 14.5% gives SI of 72.7
1510     104         1862                                                    15.1                      110.2
1516     242         1941                                                     15.8                     145
1517     276         1941                                                     16.2                      196
1518     399         2069                                                     15.8                      267.5
1519     397         1941                                                     15.3                      113.2
As you can see many of our trials had an SI > 100 and this was mainly due to starting with a high or very high MC so the SI grew quickly in the first few days.

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.

When do we get condensation, dripping and crusting in the bin?

The grain drying calculator gives the precise conditions under which condensation will occur; and that is whenever RHthres is greater than 100. (See the blog on dripping).  In playing around with the calculator today, I can give you a rule of thumb for this without using the calculator.

For oilseeds like canola and flax, that is just dry 10% MC; if the grain is more than 5 deg C warmer than the outside air (roof), there will be condensation.  I tried the calculator with GrainTemp – AirTemp,  30 – 25,  20 -15, 10 -5.  So a difference of 5 C gave me an RHthres close to 100%.

For cereals like wheat and barley, that is just dry at 14.5% MC; if the grain is more than 7 deg C warmer than the outside air (roof  & walls) , you have conditions for condensation.  Tried 27 – 20,  17 – 10, 7  – 0 and they gave RHthres close to 100.

This is interesting, because we see that oilseeds are more sensitive to condensation, and we certainly don’t need much of a difference in temperature before condensation starts to form on the top layer of grain and the inside of the roof —  only 5 deg C  difference!!  For example, if you have wheat that is 28 C and the outside temp is 20 C, you could turn on the fan and you are in for a bit of condensation on the inside of your roof, and on the top layer of wheat.

For years and years we have had spoilage from condensation because we did not turn the fans on immediately to cool the grain with a temperature that was close to the actual temperature of the grain?

The Natural Air Grain Drying Project — a WGRF project

This project was started in 2007 at the Indian Head Experimental Farm as an IHARF (Indian Head Agricultural Research Foundation) project under the supervision of Guy Lafond.  Data was collected every year from two instrumented bins.  I was given the data in 2010 with a vague mandate to find out what was happening in the bins as the fans ran continuously.  Did the fans have to run continuously?  Were there times during the day in which more drying was taking place?

The experiment was set up brilliantly such that readings were taken every hour and most importantly that  RH and T sensors were placed to measure the discharge or exhaust air at the top of the bin as well as RH and T sensors measured the air entering the bin.  This lead to the ability to measure the water going into the bin and the water leaving the bin and thus we could precisely track the drying, hour by hour.  And the Diurnal Drying Cycle was determined.

In 2012 we got a grant from WGRF (Western Grains Research Foundation) and 4 more bins were instrumented to collect more data.  Another three year grant from WGRF was obtained in 2015 to discover more.  It was called:

New Insights into Natural Air Grain Drying  (2015 – 2018)

Objective: To develop a fan control strategy using natural air that results in the safe storage of grain, that is efficient and results in less fan running time, and that results in more uniform drying of grain.

Benefit: Reducing the risk of grain spoilage and preventing revenue loss with on farm storage.

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.