## CFM calculations for 2017

The following are calculations for the CFM and CFM/bu. for the eight bins we had trials on for 2017. We measured the airflow into the fans in km/hr. so the first conversion is to get this into ft/min because we eventually want to get to cubic feet per minute.

kmph –> ft./min 1 kmph = 0.9113 ft/sec = 54.678 ft./min   H2O pressure

Bin 18 (3500 bu, Diam 23″) 11.2 km/hr x 54.678 = 612.4 ft/min x Area (pi r^2) 23/12/2 2.88sq ft = 1767 CFM /3500 0.50 CFM/bu          5.5″/7.4″

Bin 19 (3500 bu, Diam 23″) 11.2 km/hr x 54.678 = 612.4 ft/min x Area (pi r^2) 23/12/2 2.88sq ft = 1767 CFM /3500 0.50 CFM/bu          5.5″/7.4″

Bin 9 (2200 bu, Diam 15″) 30 km/hr x 54.678 = 1640 ft/min x Area (pi r^2) 15/12/2 1.22 sq ft = 2013 CFM /2000   1.0 CFM/bu                3.25″/5.0″

Bin 10 (2200 bu, Diam 15″) 30 km/hr x 54.678 = 1640 ft/min x Area (pi r^2) 15/12/2 1.22 sq ft = 2013 CFM /2000 1.0 CFM/bu              3.25″/5.0″

Bin 15 (10000 bu, Diam 15″) 70 km/hr x 54.678 = 3827 ft/min x Area (pi r^2) 15/12/2 1.22 sq ft = 4669 CFM /10000 0.466 CFM/bu  6″ (inside )

Bin 14 (5000 bu, Diam 15″) 85 km/hr x 54.678 = 4647 ft/min x Area (pi r^2) 15/12/2 1.22 sq ft = 5669 CFM /5000 1.13 CFM/bu                            5″

Bin 16 (3500 bu, Diam 16″) 18 km/hr x 54.678 = 984 ft/min x Area (pi r^2) 16/12/2 1.39 sq ft = 1373 CFM /3500 0.3925 CFM/bu                  5″/6″

Bin 16 (3500 bu, Diam 16″) 18 km/hr x 54.678 = 984 ft/min x Area (pi r^2) 16/12/2 1.39 sq ft = 1373 CFM /3500 0.3925 CFM/bu            4.8″/5.75″

The pressure is in inches of water ; the first number is outside the plenum and the number after the / is the inside of the plenum, so you can see the pressure drop across the perforated holes in the pipe.

## Duty Cycle of Absolute Humidity Controller

I took a look at the duty cycle (on time/total time) for the different bins:

Delage Bins Start End Percent Duty Cycle (on time/total time)

Bin 14 Aug 11 to Aug 21 73.6%
Bin 15 Aug 11  to Aug 21 64.8%

IHARF Bins
Bin 17 Aug 30 to Sep 6   94%
Bin 19 Aug 22 to Sep 6   95.6%

The IHARF bins were on almost constantly. And what is really strange is that the absolute humidity in the bin would increase, even when it was blasted with air that had a lower absolute humidity. I can’t explain it? Maybe it has to do with sensor placement in the bin? Were we really sensing or getting a true sample of the the exhaust air? Or perhaps we just had really good drying conditions for the period of time that we did the run? I would think that the Delage bins would be closer to the norm as far as duty cycle goes.

## Grain Drying Calculator Update

I have added soybeans and yellow peas to my grain drying calculator at planetcalc.com/4959/   This calculator gives the relative humidity, below which, drying will occur.  It only requires the temperature and moisture % of the grain, as well as the outside temperature and by using EMC equations it calculates the absolute moisture in the air, entering and leaving the bin.  It can be easily loaded onto an Iphone.

## Weyburn Presentation

The following is my latest presentation, done originally in power point.  The presentation typically takes one hour to deliver and is intended to be interactive with questions during and after it. It is somewhat dry without having commentary but can be instructive just the same.

## 2017 MFSA Forage Seed Conference: Grain Aeration — The Truth Revealed

On January 9 I will presenting at the MFSA Seed Conference at the Victoria Inn in Winnipeg.  I am scheduled for 11:30 and I will be presenting the fundamentals of grain aeration and even a little bit about supplemental heat.  If you happen to be in the area, drop by.

## Supplemental Heat: Act V Using the Grain Drying Calculator, A Balancing Act

I have blogged about the grain drying calculator, http://planetcalc.com/4959/

I explained how to use it without supplemental heat, but not how to use it when supplemental heat is used.  The short answer is that you use it in the same fashion.  You still enter the moisture content of the grain, the temperature of the grain, and the ambient outside temperature of the air (before it is heated). The resulting RHthres will indicate whether or not drying will occur.

One might conclude, that you will get the same answer for RHthres whether you are applying heat or not; and this is quite true — at first, but if heat is applied the temp of the grain will increase.  My Act III blog on Supplemental Heat indicated that it would take 12.3 hours to heat the grain 5 ºC with a 50,000 btu heater.  However I failed to mention the two implied assumptions: 1) that the grain body heated up uniformly and 2) the ambient outside temp remained the same.  Neither of these assumptions are true.

The grain will heat at the bottom of the bin first, and the so called “warming-front” will slowly work its way to the top.  If you have a cable with multiple temperature sensors, you will see this first hand.

The ambient air temperature also varies by 10-15 ºC.  The air going through the heater will get a boost in temperature, above the ambient, by maybe 20-30 ºC. So, let’s say the outside air temperature has a low of 5, and a high of 15; and the temperature rise through the heater is 30 C.  Let’s say we turn the heater on when it is cold outside, 5 C and so is the grain, 5 C.  After one hour, the bottom of the bin might increase to 10 C, while the top stays at 5.  After two hours the grain will be even warmer at the bottom, maybe 15, and as time goes on the bottom will get warmer and warmer with a wave of heat slowly creeping up the bin.  We have a string of temperature sensors, so we can actually watch this wave.  Eventually the top of the grain will also be getting warmer.

Now comes the tricky part, when should the heat be turned off in order to curtail condensation?  Remember we never want the top of the grain temperature to exceed the outside ambient temperature by more than 5 C.   Let’s say that we are approaching the highest temp of the day, 15.  Maybe the top of the grain is 10 C, and the bottom 30 C.  Everything is fine — no condensation — yet.  As the day cools off, to 5 C, the wave of heat will continue to heat the grain at the top, even if we turn the heat off. In a few hours the top of the grain may become 15 C and as we approach the low of the day, 5C — we will have conditions for condensation — the grain is 10 C higher than the ambient temp.

What would be ideal would be to apply heat such that the top layer of grain is always 5 ºC warmer than the outside air.  There are two difficulties with this. First there is a huge delay from the time we apply heat at the bottom until it reaches the top layer.  It would be hours, and it depends on so many things, such as air flow rate, heater size, bin size, etc. And to anticipate this delay is really difficult.  The other complicating factor is that the ambient temperature changes, and sure we can get a forecast of the temperature, it is not entirely accurate. And then to monitor and control this whole mess.

Let’s take on the heat wave delay problem.  The outside air goes up and down in temperature in a somewhat predictable fashion.  Somewhat of a sinusoid in shape, with the warmest part of the day taking place a couple of hours after noon, and the coldest part of the day a couple hours after midnight. Typically we see a difference in the high to low temperature of 10 – 15 ºC.   We ideally would like to apply supplemental heat such that the top layer of grain is always 5 to 6 degrees warmer than the outside temp.  The problem is there is a delay in applying the heat to the bottom of the bin to when it gets to the top.  What is this delay?  To determine the delay we could model the system, but this is a trickly onerous task — and there are so many variables.  The other way to find out what this delay is, would be to review the data of our runs, and see what it is.  I did just that, and I examined two trial runs, 9 10P and 09 09W.  Clearly the delay was close to 6 hours.  The flow was close to 1.2 CFM/bu and if we lowered the flow to 0.3 CFM/bu we would increase the delay accordingly to 24 hours.  The temperature of the top layer of grain will be synchronized with the daily cycle of temperature change of the outside air. We can use the rate of flow of air to adjust the delay and set it where we get a 24 hour delay.

Now the other problem, we want to add supplemental heat to keep the top layer of grain  5 to 6 ºC above ambient, we want a 5-6 ºC rise, throughout the whole day.  We would keep the fan running continuously, and the heater running continuously to provide a 5 C rise.  What size heater do we need for this?

In my earlier blog: “Big Heater for supplemental heat Example” I had a farmer with a 1.4 million btu/hr heater at 5,000 CFM. It produced a 30 C rise in temp.  Supposing we have a 3500 bushel bin, we would need 1000 CFM to get the synchronizing daily delay of 0.3 CFM/bu.   The rise we would get with a 50,000 btu furnace would be 50,000/1,400,000 x 5000/1000 x 30 = 5.35 ºC — Perfect.  A 50,000 btu furnace with 1000 CFM should do the job to give a 5 C rise with a 24 hour delay.  We might have to use some trial and error to fine tune this.

## Review: Aerating Stored Grain — Australian Guide

I was going through my files, and I came across an aeration guide that Guy Lafond had given me in 2011, along with his annotations.  It can be found at http://www.graintec.com.au/media/34545/Aerating%20stored%20grain%20-%20A%20Grains%20Industry%20Guide.pdf

It is called Aerating Stored Grain, Cooling or Drying for Quality Control, A grains Industry Guide by GRDC (Grains Research & Development Corporation) in Australia.

Guy passed away a couple of years ago, and in going through the report and reading his notes, it is almost as if he is talking to me.  I must have skimmed the report, but it is only now that I have slowly read the report and have appreciated the annotations.  Many of the findings in the report are in agreement with what we have found, and then there are guidelines that do not. I will now go through the report and perhaps in a somewhat disjointed way touch on the issues of agreement and disagreement.  You be the judge, and try to discern where the truth lies. Quotes from the booklet will be in italics, followed by my comment and Guy’s note.

This booklet explains the key differences and processes involved in aeration cooling and aeration drying.  We discovered right from the start of our analysis that cooling the grain also dries it. The two are not separate process, cooling is drying.  In fact this relationship can be approximately be quantified: cooling the grain by 15 ºC reduces the moisture content by one percent.

38 per cent is air space around each grain.  You may have noticed that we used 40% in the calculations done in my blogs.

Without circulation, the air surrounding the grain will reach a moisture (relative humidity) and temperature equilibrium within a few days.  In other words, you must leave your grain in a sealed bin for three days before you will get an accurate reading on your moisture cables.  Moisture cables use Equilibrium Moisture Content to calculate moisture; the air and the grain must be in equilibrium. The fans must be off for a long time to achieve equilibrium.  In other literature, I have seen times like 3 hours to reach equilibrium.  In actual fact, equilibrium is never reached, it is said to be asymptotic. It gets closer and closer but never reaches it.  That is why in my blogs I introduced the notion of a time constant, which is the time it takes to get about two thirds the way there.  Anyway the huge lesson here is that moisture cables can only give an accurate reading of Moisture Content if the fan is off for a considerable time.

The air in the head space, heats and cools each day creating ideal conditions for condensation to form, wetting the grain at the top of the stack.  The good news is that this guide does recognize a problem with condensation; the bad news is that they don’t give the specifics of the conditions that create condensation.  Using my calculator, http://planetcalc.com/4959/  we do have the specifics of when condensation occurs: it is whenever the RHthres is above 100%.  The more it is over 100, the more the condensation. It turns out that the rule of thumb is this: If the grain in the bin is more than 5º C warmer than the outside air, there are conditions for condensation.  The more the temp. difference, the more the condensation.

Grain aeration systems are generally designed to carry our either a drying or cooling function — not both.  This is what Guy had in his note on this: “Not so, drying occurs as grain is cooling”. Guy and I learned this by analyzing the data collected from bins with aeration since 2007.

Aeration cooling can be achieved with airflow rates of 2-3 litres per second per tonne delivered from fans driven by a 0.37 kilowatt (0.5 horsepower) electric motor.  I agree that low flows can achieve cooling within a matter of hours, and look at the size of the motor to do this , half a HP.  What I don’t agree with is that only cooling will take place with low flows; drying also takes place. Remember the rule of thumb that was developed from analyzing our data: Cooling the grain by 15 ºC removed 1% moisture content — regardless of the air flow rate.  In fact, I propose that more water will be removed with lower fan rates in terms of cooling.  We can make better use of the energy in the grain, to evaporate water if we do it a little slower, with lower air flow rates.  See my blog on air flows.

Aeration drying can be achieved with fans delivering 15-25L/s/t, typically powered by 7kW (10hp) electric motor. Again the message here is that only large air flows can dry grain; I disagree. Lower flows are actually more effective; not in terms of time, but in terms of actual moisture taken out.  Once we have taken the energy (heat) out of the grain, drying will no longer occur — with high or low flows. The trick is to use as much energy as we can for drying.  I think we can use this energy more efficiently to remove the water, if we go a little slower with lower flows.

Grain that is dry enough to meet specifications for sale (12.5% for wheat) can be cooled without drying.  I beg to differ, cooling is drying.  Guy’s comment was “really?”

After drying to the required moisture content, cool the grain to maintain quality. Guy’s comment to this: “Cooling and drying occur together”

Aeration cooling moves the air pockets around the grain, which evens out any hot or moist areas, creating a uniform stack.  Notice here they are talking about aeration cooling that use low air flows.  Therefore low air flows even things out, they do not create pockets or fronts of moisture or temperature.  I agree.

One way of measuring change in grain quality over time is seed germination. Exactly, this is what I have done by calculating the number of safe days, and the spoilage index.  It is sometimes uncanny how the same conclusion can be reached independently.

Stored grain deteriorates with time under any conditions, but poor storage conditions (high temperature and moisture) accelerate the deterioration process markedly.  I agree, you can’t stop spoilage, you can only slow it down. Grain in storage can never improve in quality, it can only get worse.

If aeration cooling is being used to hold moderately high-moisture grain temporarily until drying equipment is available, run fans continually while the ambient relative humidity is below 85 percent.  Guy notes: “Interesting Concept, Not sure if I agree.”  Again we have the assumption that low flows do not dry — but they do.  And where did the magic 85% come from?  What is it based on? I can show you tons of examples where drying takes place with RH above 85, and tons of examples where wetting takes place with the RH below 85.  We are missing some important parameters here like grain temp, and air temp.

However, do not operate the aeration fans on continuous mode for more than a few hours, if the ambient relative humidity is higher than 85 per cent.  Again, where does this magic 85% come from? And again, there are important parameters missing.  Guy notes: “Depends on Tº “.

After aeration fans have been running continuously to flush out the warm, humid air for 2-3 days, reduce run time to 9-12 hours per day for the next 3-5 days.  The difficulty is selecting the coolest air to run the fans and being on site to turn the fans on and off.  This is interesting and very close to my recommendation that the fan should be turned on, even as the bin is being loaded, and run until 9 the next day after which you only run the fan during the night, or the coldest hours.  Not during the day, with the hottest hours?  We learned that cooling is drying, and you can only cool if the air is colder than the grain.  The Aussies already knew this long before we did.

During this final phase they continually monitor ambient conditions and run fans on average during the coolest 100 hours per month.  Every time we pull the temperature of the grain down, we will be doing some drying, and by keeping it as cold as possible, the grain will be as safe as possible with the least spoilage. It is interesting that monitoring the ambient conditions is required, but exactly what are these ambient conditions in which one should run the fan??  I am suggesting that the fans be on if, and only if the air temp is less than the grain temp. and the RH < 80%.  We have found that if you follow this rule, that the fan duty cycle will be < 20% which is about 144 hours per month– amazingly close to 100 hours.  Again,  we have arrived at the same conclusion — independently.

Growers and bulk handlers need to  have an understanding of the effects of relative humidity and temperature when aerating grain.  I could not agree more, but the problem is, the report, is not giving us any understanding of the effects. Specifically, what temp and RH should I run the fan?  My calculator provides exactly what the temp and RH of the grain and ambient air should be to provide the conditions for drying AND for the conditions of condensation.

However, drying depends completely on the airflow through the grain.  I do not agree.  Grain will dry quite nicely, perhaps even more efficiently with low flows.

Operating in drying mode, aeration controllers select for air with low relative humidity.  Guy notes “Why only RH and not T as well ?”  They seem to keep missing the point that the amount of water in the air is determined not only by its RH but also, and maybe even more so, by its temperature.  They are missing the pyschrometric equation that relates RH and T to absolute humidity.

In rare situations aeration cooling fans can reduce grain moisture slightly, but they cannot reliably reduce grain moisture to a safe level.  I disagree.  The amount of drying depends on how efficiently we can use the energy in the grain to push the water out and evaporate it into the air.  Low air flows will do this as well as high air flows.  I would argue that low air flows actually do it better.

Much higher airflow rates are required for aeration drying in order to push a drying front through the grain bulk.  I disagree; in fact it is the larger airflow with the increased compression on the bottom of the bin that leads to over-drying the bottom and under-drying the top.  And there really is no drying front, it is more like a continuum, with the bottom being dried first and the top last.  And this is done more evenly if we have less compression on the bottom with a smaller fan and less flow.

Wheat at 16.5% MC at a temperature of 28 ºC was put into a silo with no aeration. Within hours the grain temperature reached 39 ºC and within two days it reached 46 ºC providing ideal conditions for mould growth and grain damage.  Guy notes: “I am not sure I agree with this.”

By monitoring the temperature and moisture content of the grain in storage, and reading the equilibrium tables for wheat or sorghum at the back of the booklet, a suitable relative humidity trigger point can be set.  Guy notes: “How to consider equilibrium moisture content?”  This suggests that I can take the temperature of the grain (let’s say 25) and the MC (let’s say 15%) and apply this to the EMC table to get a trigger RH of 82%.  What could be simpler?  Well, we missed a very major point:  The ambient air is NOT the same temperature as the wheat, therefore one can not consider the system to be in equilibrium and therefore the EMC table does not apply.  The trigger RH of 82 is for the air inside the bin, at equilibrium and at the same temperature as the wheat.   We can and do calculate the corresponding RH for the air outside with my calculator — but I would say that this important point is almost always overlooked or perhaps just not understood.

Firstly, warm air can transfer moisture from the grain more efficiently than cold air.  Guy notes: “Depends on RH?”  This has been a hang up with people for some time, and at first glance it seems obvious, of course warm air can dry better than cold air!!  But, wait a minute let’s examine this more carefully.  Drying occurs when the vapour pressure of the grain is larger than the vapour pressure of the air.  Vapour pressure depends on the moisture content and more importantly the temperature of the grain or the air.  When the grain temp is higher than the air temp, it most likely has a higher vapour pressure and drying occurs.  There are exceptions and one can conjure up an example where the air is just slightly cooler than the grain, at 100% RH and then wetting will occur.  But, as has been said in this report, with air,  RH < 85%, and a temp less than the grain, you are assured that you have drying conditions.    Sure warm air can hold more water than cold air, but the point that has been missed is that the warm air becomes the same temp as the grain as soon as it hits it.  If the grain is cold, the air will also become cold, and the air will not be able to hold the water it has, exceed 100% RH and dump the water onto the grain.  The warm air, that we thought was capable of carrying more water, and it could have if it stayed warm — but it doesn’t, it becomes almost the same temperature of the grain.  Sure eventually the grain will become the same temperature of the outside air, after hours and hours, but guess what by then the outside temperature has changed.  The outside ambient air temperature and the grain temperature are NEVER the same.  Yet people assume they are.  Big mistake.

## Big Heater for Supplemental Heat Example

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.

## AirFlow: How Much is Enough? Making a Case for Lower Rates.

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:

————————————————————————————————————————-

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.

## Supplemental Heat: Act IV Using a Gen Set for Heat

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.