Month: February 2016

In a previous blog, I explained how condensation could be a problem in the fall with warm grain and cold outside air.  The rough rule of thumb was that there was a potential for condensation on the roof if the grain was ten degrees warmer than the outside; but a more precise determination could be made with the grain drying calculator.  The recommendation was not to turn your fan on if the RH from the calculator was above 100%.

There is another time, in the spring that condensation could be a problem. When the grain is very cold and the outside air temp is much warmer and holding a fair amount of water. This air enters the bin through the fan entrance, the top or vents; and it hits the cold grain — it will cool past its dew point and condensation may well occur.  This will happen mostly on the poor old top layer, that may well have had water dumped on it from fall condensation, and it is the last layer in the bin to dry.  No wonder the crust always forms on the very most top layer.

To prevent condensation in the spring, the best thing to do would be to not let the warm air in.   The bin should be sealed.  The top lid closed tight, the fan covered and all vents covered.  Indeed, this is a time when vents will actually be the cause for grain damage.  Sealing the bin in mid winter, after the grain has been cooled to its lowest possible temperature, will also keep the grain as cold as possible for as long as possible.  Yes the grain will still warm up from conduction through the walls; but it will be a warming without adding water.  At the end of the summer, when the grain has warmed up and the nights again are getting cold, the bin can be opened up and the grain again cooled.  And this process could be done for years of very safe grain storage.

With this blog, we would like to know what it might cost to use supplemental electrical energy to remove one percent moisture content, MC, from one bushel of wheat.

Assumptions:

1. All the energy goes into evaporation.
2. wheat  60 lb/bu
3. Cost of electrical energy  is 0.12/kW hr.
4. Latent heat of evaporation,  energy to evaporate  2257 kJ/kg

One bushel is 60 lbs, so 1% MC would have a weight of  0.6 lbs or 0.2722 kg

2257/.2722 = 614.4 kJ to evaporate 1% MC from one bushel of wheat  and to put this into hours divide by 3600  =  0.17 kW hr.   and in terms of dollars,  0.17 x $ 0.12 = $ 0.02 to evaporate enough water out of a bushel of wheat to lower the MC by 1%.    That doesn’t sound like much but for 3500 bushels, it would cost $71.68, and to drop by 2% MC it would be $143.36.  It must also be remember that this is the ideal efficiency.  In the real world, we may have to double our cost again.

Could we use the free heat in the air to dry grain?  —  I think so; but it will take careful management and control of the aeration fans.

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.

For this exercise we will use one cubic meter of barley at 30 C. And we know that it takes energy to evaporate water; this is called latent heat of evaporation, which is 2257 kJ/kg. Let’s assume that all the energy in the barley in going from 30 C to 5 C is used in evaporating water from the barley. How much water (in %MC) can be removed?

1 bu = 1.2446 cubic feet       35.31 cubic feet = 1 cubic meter     2.204 lbs/kg

One cubic meter of barley weighs 35.31 ft3/1.2446 = 28.57 bu     x 48 lb/bu = 1361 lbs, 618 kg

One percent moisture, 1% MC would then be 6.18 kg.   If barley was 15% MC then 92.6 kg would be water.

The barley in going from 30 C to 5 C has how much energy to give. What is the specific heat? It can vary, but I found one source: 1.36 kJ/kgC

1.36 x 618 kg x 25 C = 21012 kJ and divide by how much energy is needed to evaporate

21012 kJ /2257 kJ/kg = 9.3 kg of water can be evaporated with the energy in the barley.

But 1%MC was 6.18 kg/%MC ,   9.3 kg/6.18 = 1.5% reduction in MC. If the barley was 15% and all the energy went into evaporating water in going from 30 to 5 C, then the MC would be reduced by 1.5% and be dried from 15% MC to 13.5%.

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 the case above, 1.5% resulted from a 25 C. A 1% reduction would require 16.66 deg reduction. This demonstrates that aeration (at least with our experimental trials) is very efficient at using the energy in 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.

 

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.

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 RH_thres 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.

 

 

 

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.