How much Time is Required to Cool Grain

Let’s work out an example of how long it will take to cool down a 3500 bushel bin of barley from 30 ⁰C to 20 ⁰C with air that is 15 C and flowing into the bin at 3000 CFM. We will make a wild assumption here and assume that the air will leave the grain at 30 C, making a 15 C increase. For this exercise we will use conservation of energy and we will ignore any drying or wetting that will result from latency heat.

First task is to find the energy that would be lost if the barley is cooled from 30 to 20 C. We must know the specific heat, Cp, of barley and I looked in a couple of places, but it was about 1.36 kJ/kg C. The weight of the barley is 3500 bu x 48 lbs/ bu x 1/2.204 kg/lb = 76,225 kg   times 1.36 = 103,666 kJ for every degree C and we are changing ten degrees so the energy to be removed from the barley will be 1,036,660 kJ.

3000 CFM is 50 cubic feet per second. Air is 0.0807 lbs per cubic foot.   4.035 lbs/s or 1.83 kg/sec

The specific heat for air is close to 1 kJ/kg C so for every degree C we could acquire 1.83 kJ/s. For 15 C we could remove 27.45 kJ/s. How many seconds to remove 1,036,660 kJ?   1,036,660/27.45 =37765 sec, 629 min, 10.5 hours.

This shows that it is possible to cool the grain over a one night period (12 hours, 9 PM to 9 AM).

Now, if we used a smaller fan and produced only 2000 CFM, we would expect the cooling time to increase accordingly to 16 hours. Even for a much larger bin, of 10,000 bushels, we could cool it down in 48 hours (four 12-hour nights). One could argue that this would be for air that is consistently 15 C and it assumed ideal energy transfer. On the other hand there will be cooling affects resulting from drying; but that’s for another blog.

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