The BC ** Compost Facility Requirements Guideline** offers the following scenario (http://www.env.gov.bc.ca/epd/mun-waste/regs/omrr/pdf/compost.pdf):

“A windrow composting operation with 3 m deep and 7.6 m wide piles and 1.5 m aisles averages about 195 kg of degradable organic matter per square meter. This provides about 2840 MJoule of evaporation energy per square meter, which has the potential to evaporate 1125 mm of water per square meter. Twenty five millimeters (25 mm) of rain on the active composting area at this facility produces about 37,000 litres per hectare of polluted runoff”

In a previous blog, I suggest that this was invalid, and promised to provide detailed calculations to demonstrate this? The model that predicted this does not appear to be available on the web, so let us see if we can reconstruct it based on theory.

- What is the average depth of compost on this site? If we assume a 1 hectare site with 10 windrows as described above, with 3 m along each side and 6 m along each end, we have 10 windrows that are 92 m long. Based on the above dimensions and using a windrow shape factor of 0.66, we get an average of 1.37 m or 4.5 ft.
- How much degradable organic matter can we expect in this 1.37 m depth? If we assume a bulk density of 700 kg per cubic meter, and a moisture content of 65%, we have a total of 336 kg of dry material per square meter. If we assume that 90% of that is organic matter (10% ash), of which 50% is biodegradable, we come up with a total of 134 kg of degradable organic material.
- How do we calculate the 2840 MJoule of evaporation energy per square meter? If we assume that all of the biodegradable carbon is a carbohydrate such as cellulose, we obtain the following equation: C
_{6}H_{12}O_{6}+ 6O_{2}= 6CO_{2}+ 6 H_{2}O, (carbohydrate 342.34 g/mole). The heat of reaction of glucose (the components of cellulose) is -2822 KJ/mole (http://faculty.ncc.edu/LinkClick.aspx?fileticket=FJO362tolqA%3D&tabid=1939 ), or 15.7 MJ/kg. Multiplying 15.7 MJ/kg of biodegradable solids by 195 kg of biodegradable solids per square equals 3062 MJ/square meter, which is close to the 2840 estimated in this scenario. - How much water can 2840 MJ of energy per square meter evaporate (this is assuming that all the available energy will be used for evaporation). The heat of vaporization of water at 25 C is 2260 kJ/kg (http://en.wikipedia.org/wiki/Enthalpy_of_vaporization). This means that the estimated 195 kg of biodegradable organic matter has enough energy to vaporize 1265 L of water which corresponds to 1265 mm of rain. Because the energy of vaporization is temperature dependent, the actual amount of water vaporized at 5 or 10 C will be slightly less than the 1265 L estimated, so 1125 mm of rain is a reasonable estimate.
- How many days does it need to rain 25 mm to vaporize the 1265 L of water per square meter? If we divide the 1265 mm of rain by 25 mm per day, we get 50 days or 7 weeks of composting required to degrade the 50 % of biodegradable carbon. This would assume a fast composting process, which is not likely to be achieved with a passively aerated windrow process of the size described in this scenario.
- How can we ensure that all precipitation gets to the sites of energy release in the composting mass so that all of the water can be vaporized? In the case of the windrows in this example, the precipitation is going to fall on the windrow itself as well as on the impermeable surface between the windrows. There may also be some runoff down the side of the windrow due to the shape of the windrow (Compost Facility Requirements Guideline). This suggests that up to 50% of the precipitation will not come in contact with energy being released from decomposing organic material. The one solution is to orient the windrows across the slope, so that the precipitation can pool between the windrows, and gradually become absorbed into the windrow as the heat energy vaporizes the water in the windrow. While this sounds like a great idea, the reality is that the Compost Facility Requirements Guideline suggests that the windrows be oriented up and down the slope to allow leachate and runoff between the piles rather than through them. We can safely assume that a minimum of 50% of the precipitation will run off the site before it has a chance to evaporate after coming in contact with energy released by decomposing carbon.

- How are we going to ensure that all of the heat energy is going to be used to vaporize moisture and will not be lost by other means? The Cornell Compost Engineering website (http://compost.css.cornell.edu/physics.html) suggests that heat in a compost pile can be lost via conduction, convection and radiation.
- Evaporation of water depends on saturation vapor pressure which is affected by temperature and airflow above the composting material. Energy, temperature, humidity and wind are described as the four universal factors affecting evaporation
*.*In a compost facility, the energy is primarily the heat energy. - The effect of temperature on evaporation potential can best be described in the following diagram, where the moisture content of air is primarily affected by temperature. This has the greatest influence on how much moisture will actually be removed from the composting process.

In summary, the energy created by the composting process is not all going to be used for moisture removal because not all of precipitation comes in contact with the energy created by the composting process, and a significant amount of the energy is lost as conductive heat loss. The process of evaporation is much more complex than simply accounting for the energy in the compost. It must also consider temperature, humidity and air movement through and around the composting material.

Full details can be found here: Evaluation of Evaporation as Suggested by Compost Facility Requirements Guideline