Using Calculus to Optimize Ventilation for Sustainable Food Storage

Vegetable, root, grain and seed storage is an important process of sustainable agriculture. For agriculture to be sustainable there has to be successful mechanisms of efficient growth but also of storage. In a very real sense we can think of agriculture as being a generation and storage of food energy and for this to be a process which can “break-even” in terms of energy invested on energy returned.

So in order to avoid a waste of energy we must have a way to store the food generated by photosynthesis using the minimum amount of energy possible. This means we must design storage tanks with optimization in mind but also to tap into the energy transfer channels in the environment in order to create stable temperatures and environments inside storage vats and tanks, inside basements, vaults etc.

Initial Optimization - Most Efficient Air Siphon Through Tank:

Let us then consider the problem of storing vegetables or roots in a tank where we want to have a cool dry atmosphere sustained inside.

We can, first and most obvious of all, have an inlet faced in the opposite direction the wind is blowing most prominently (with appropriate shielding such as an air-permeable netting to prevent windblown debris, spores and animals getting insdie). 

The wind then drives cool, dry air into the inlet tubing that we feed into the storage tank (fig-1)

fig-1: cool,dry air inlet faced in the direction of wind

The cool, dry outside air fed into the tank has a greater density, ρ, than the moist, warmer air inside the tank, that we want removed to prevent rotting by generation of mold and most other rotting agents.

In other words:

So the cool, dry air fed into the tank will be heavier and sink relative to the warm moist air which will rise. Since we want to move any hot, moist air from the tank, i.e, by keeping the food store well ventilated, we can assist the process by positioning the warm air exhaust higher than the cool air intake (fig-2):

Fig-2: air circulation driven by difference in inlet and outlet heights.

The two pipes then create an effective siphoning of cold outside air into the insulated cellar, allowing the temperature to remain cool and remove moisture as it is circulated around the food store. 

Moreover, the siphon also creates the necessary ventilation to remove ethylene gases that fruit and vegetables can produce as they ripen. Venting ethylene gases slows down the ripening process and so the fruit and vegetables can be stored longer this way.

For the siphon to function optimally, space the interior ends of the cool air intake and warm air exhaust pipes as far apart as possible.

However, we must also consider  the nature of heat capacity of the volume of gas (i.e. volumetric heat capacity). The volume heat capacity is the average randomized kinetic energy of constituent particles of matter (e.g. atoms or molecules) of the gas relative to the centre of volume of the system. 

The siphoning system is itself the transfer of heat energy,  ∆Q , across the volume from the temperature difference, ∆T, between intake to the exhaust by passive means, i.e. with no net energy added to the system. 

Volume, rotational symmetry, and vibrational symmetry of atoms represent the degrees of freedom of motion which classically contribute to the heat capacity of gases.  

For this case we shall simply consider volume, V. The heat capacity, C_heat, is therefore:

Hence an increase in the volume of the transfer will mean that there will be less heat capacity of the gas across the intake to exhaust, i.e. the heat capacity is inversely proportional to volume. This means that although the density differences between the cool dry air intake and the hot moist air exhaust will create an effect siphon if they are kept far apart, an increase in the transnational stress will conversely result in that the effect of siphoning will decreasing with the distance above a critical point.

Therefore we must optimize the distance based on the geometry. This can be can be optimized with the following optimization by volume coordinate transformation formula:

Our goal is to find the volume optimization constant λ:

We then look at the dimensions of the storage tank (fig-3), which for our case is 

(X=2.5m,Y=1.3m,Z=0.9m)

Fig-3: food storage tank dimensions

The Total Volume of the tank is therefore:

Hence, by the volume optimization formula:

Or:

Therefore, if we have the outlet exhaust pipe at the origin, i.e. at x=0, y=0, z=0 (i.e. at the top right hand corner of the tank) the optimized coordinates are for the  position of the inlet pipe:

This can be displayed on the following diagram (fig-4):

Fig(4) - Optimized pipe locations for optimum air siphon.

So given 2 pipes of equal size, this is the optimum position to have then which has them at the farthest distance to create the best siphoning effect but not too far as to diminish the relative heat capacity which creates the potential energy needed to move the air inside the tank volume.

Further Optimization - Most Efficient Air Transfer Through Pipes

To optimize the air transfer efficiency for a volume of changing air, i.e. air flow, we must relate the rate of change of the height of a block of air over time to the rate of change in volume over time (the flow rate)

First we describe volume in a deconstructed fashion as being:

                                                                Volume=(Area)(height)

Hence, the rate of change of volume:

By calculus:

By separating the variables we have:

Where k is a fixed scalar constant related to the height (i.e. the scalar height of a column of air through the pipe)

Which we can integrate:

Solving this:

So for any time, t, a “block” of air of a height, h, is transferred out of the tank as of the above formula.

At the primary minima (i.e. when a block of warm, damp air has been completely vented from the tank) we have the above formula at:

so that:

If we then remember that we have fixed the scalar constant k to be a fixed height of an air column flowing through the pipe, then the value kt is a scalar value of the speed of air flow.

Therefore the speed at which the air flows is therefore directly proportional to twice the relative height of the air block.

What does this mean?

It means that if we want to increase the speed of air flow through the pipe we must make the height at which the air flows into the pipe twice as high. 

We can do this by making the inlet pipe twice as small in diameter relative to the outlet pipe (or making the outlet pipe twice as wide in diameter as the outlet pipe (fig-5) which will increase the relative height of the air sent in from the input by 2 times which will then result in the linear increase in the velocity of air flow, as we have determined.

Fig(5)- Inlet-Outlet Pipe Size Difference for optimum air flow at speed kt.

It is hoped that the following optimization calculations can yield a good example of how to design a well-ventilated food storage system for use in sustainable agriculture.