Supplemental Heat: Be Careful

In my last blog on night drying does not work, I explained how Jim S. from Wainwright was considering adding a natural gas furnace to his aeration fan to provide supplemental heat.  He was getting no where with night drying, in fact the weather conditions were so bad that water was being added.  Off the top of my head, I thought adding a furnace would be a good idea, but I would run some numbers to check it out.

Here is the situation: 3500 bushels of tough wheat at 17% Moisture Content. Let’s say his grain is near the freezing point, 0 C, and his fan can produce 3000 CFM or not quite the 1 CFM/bu recommendation.  The immediate weather forecast is for cool temperatures and high humidity.

3500 bushels weighs  60 lbs/bu   2.204 lbs/kg     3500 x 60/2.204 = 95,281 kg

If we want to remove 1% MC we would need to remove 953 kg of water.

To evaporate 1 kg of water it requires 2257 k joules of energy, so to remove 953 kg of water,  2257 x 953     =  2,150,921 or about 2 GJ of energy.  Checking with SaskEnergy, a GJ of natural gas would cost about $6.50.  So, at least $30 of natural gas is required to lower the MC by 1% and $75 to bring Jim’s wheat down from 17 to 14.5%. This of course assumes that all of the energy we add will go into drying.

How big should the furnace be?  When shopping for furnaces, you will see the spec as to the heat they give off is usually in BTUs or British Thermal Unit.  And 1 btu = 1.055 kJ  What they really mean in the spec is btu/hr; the per hour is implied.  I found a ceiling furnace at Princess Auto with an output of 50,000 btu/hr.   or 52.75 MJ/hr.  We calculated above that we need 2,000 MJ of energy to get 1% moisture out.  To get this energy into the wheat, it would take 2000/52.75 = 38 hours of running the furnace.  Perhaps 100,000 btu –> 19 hours or even a 200,000 btu would take 9.5 hours.

OK so we have the furnace sized, and we have an idea of how much this is going to cost in natural gas; what’s next?  How much should we heat the grain?  Should we cycle the heating and cooling?  If we are going to cycle — how long should the heat/cooling cycle be?   I am going to assume here that the wheat is at 0 C and the outside air is also at 0 C, 76.2% RH.  I ran some numbers with the grain drying calculator with the grain heated to various temps.

Grain-Temp  RH-bin-air abs-hum-bin-air RHthres-outside-air  abshum dif  5100m

20                     79.9             23 gr H2O           302%                              19.3               98.43

15                     79.1%          10.2                     217.6%                           6.5           33.15 kg

10                    78.2%           7.4                       154.2%                           3.7           18.87 kg

5                      77.2               5.3                        108.4                              1.6            5.92 kg

0                     76.2                3.7                         76.2%                                0               0 kg

Looking at the first line, we heat the grain from 0 to 20 C.  Because the wheat is 17% MC at 20 C, EMC equations tell us that it wants the air to be at an RH of 79.9%, which gives an absolute humidity of 23 gr per cubic meter. The air entering the bin had an absolute humidity of 3.7, so for every cubic meter of air flowing through the bin 23 – 3.7 = 19.3 gr.  In one hour 5100 cubic meters of air flow through the bin so 19.3 x 5100 = 98.430 kg of water are removed in one hour. To remove one percent MC we need to remove 953 kg of water, and this would take about 10 hours.

This all looks really good, but we have one huge problem — CONDENSATION !  We see that the RHthres for the outside air is 302%.  Anything over 100, and we will get condensation on the roof and walls of the bin.  Here is what is happening: the air goes through the grain and becomes 20 C and RH 79.9. It is carrying 23 grams of water per cubic meter.  This air hits the cold roof and is cooled to 0 C.  But air can’t hold much water at 0 C, only 5 gr/m^3. The remaining 18 grams condenses and comes raining and running back into the wheat.  If we want to avoid condensation we should only have the grain about 5 degrees warmer than the outside air.  If we only raise the temp of the grain by 5 C, it will take a long time to remove 1% MC —  953/5.92 = 161 hours or about a week.  It would take over 2 weeks to get Jim’s wheat dry.  Trying to go faster with higher grain temps would be a dangerous exercise in creating condensation.   Maybe we should reconsider this supplemental heat strategy and go with cherry picking good drying conditions.  The wheat is safe, being as cold as it is, so what’s the hurry?

If you want to use supplemental heat, you better be careful.

I would tell Jim to use the smaller furnace, of 50,000 btu, which would give a temp rise of 8 C.  The fan should only be run when the temp of the day is the highest, maybe at 6 hours a day. The heat should then be shut off and the fan left running into the coldest part of the night.  I would do some cherry picking of good conditions using the drying calculator, and certainly not running the fan when Condensation conditions exist.

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