Home Based Recycling Business
For the best results in a composting process, it is important to have appropriate mixing of sludge cake with bulking agents and recycled materials. For the process to operate in good condition, it needs to have the optimum mass balance, moisture, temperature, pH, nutrients, and air. Figure 7.11 shows a mass balance diagram that can be used for all three composting systems: windrow, aerated static pile, and in-vessel. In the diagram:
New Bulking Agent
New Bulking Agent
A' = weight of ready compost in one day A = weight of recycled compost in one day B = weight of dewatered sludge in one day
D = weight of a mixture of dewatered sludge, bulking agent, recycled bulking agent, and recycled compost in one day
F = weight of bulking agent in one day
C^1 , C2 , C3 , C4 = concentration (%) of dry solids in dewatered sludge, recycled compost, composting mixture, and bulking agents, respectively
Or1, Or2, Or3, Or4 = concentration (%) of organics in dewatered sludge, recycled compost, composting mixuture, and bulking agents, respectively
The quantity of mixture being composted is described by the equation
The quantity of mixture without bulking agents is
The mass of dry solids being composted without bulking agents is
Incorporating equation (7.2) yields
That is,
Therefore,
The mass of recycled compost divided by the mass of dewatered sludge is
where R1 is the coefficient of compost recycling. Therefore,
From equation (7.5),
Dividing equation (7.11) by BC 1 gives
AC 2
where R2 is the coefficient of compost recycling on a dry weight basis. By incorporating equation (7.7), we obtain
Equations (7.10) and (7.15) can be used to determine the quantity of recycled compost.
The quantity of added bulking agent can be determined by the equation
Adding the mass of bulking agents and the mass of recycling compost to dewatered sludge, we have f1 = A + B (7.17)
Part of the bulking material has been lost by organic decomposition in the composting process:
From equations (7.10) and (7.15), it can be seen that one of the most important factors is the dry solids concentration in dewatered sludge. Reducing the concentration of dry solids in dewatered sludge increases the coefficient R1. Increasing the concentration of dry solids in recycled compost decreases the quantity of recycled compost to be used (coefficient R2 decreased). The concentrations of dry solids in the mixture of dewatered sludge and bulking agents exert a big influence on the duration of composting process and the quantity of air required for aeration. The concentration of dry solids in the mixture has to be not less than 35% and not more than 50%. Seventy percent moisture in the composting mixture will increase the quantity of air required by several times.
Decomposition of organics by microorganisms releases carbon dioxide and water, resulting in decreased mass and volume. Assuming that Or3 is fully decomposed as carbon dioxide and water, from the mass balance diagram,
Using equation (7.2) yields
Therefore,
Dividing numerator and denominator by B and using equation (7.9) gives us r-. _ OrC 1 + Or2C2R1 (n
Or by using the coefficient of recycling in dry solids, we obtain
The organics contained in the composting mixture depend on the initial organics contained in sludge. If the quantity of dry solids in sludge is decreased, the quantity of recycled compost has to be increased, but that will decrease the concentration of organics in the composting mixture. Decomposition of organic material in composting process decreases organics by 25 to 30%. The heat generated by decomposition of organics is approximately 20 to 21 MJ/kg. Because raw sludge contains more organics than digested sludge, composting of raw sludge is more effective than composting digested sludge. One of the effective ways to increase organics in composting process is to use bulking agents. Fresh sawdust and wood chips contain up to 98% organics, and when recycled from ready compost, up to 80% organics. However, stored sawdust and wood chips including recycled wood chips can raise temperature in a composting pile much faster than fresh sawdust and wood chips because they contain more living microorganisms. That is one of the reasons for recycling wood chips from the compost. Also, this recycling of bulking agent puts the carbon-to-nitrogen ratio in the composting mixture within the optimum range of 25 : 1 to 35 : 1.
The quantity of air in the composting process in the absence of a bulking agent, per the mass balance, can be determined indirectly from the quantity of water evaporated in one day by the equation
H2O = ( B - C1 B)-( A'- C2 A') (7.24) The mass balance for the inorganic substance is
(1 - Or )C1B = (1 - Or2 )C2 A' (7.25) or by using equation (7.24),
The airflow rate has to be monitored and controlled based on the temperature, oxygen, or carbon dioxide levels. The amount of air needed for the composting process is approximately 15 to 20 m3/h for each metric ton of organic material. More air is required as the temperature rises to keep it below 70°C. If the temperature reaches 70°C, the quantity of microorganisms decreases and the process of composting stops. The carbon dioxide level should not exceed 8%, and oxygen should not be less than 5 to 15% in the pile. The stoichiometric demand of oxygen can be determined by the equation
Ammonia (NH3) generated during the decomposition of organics is volatilized. One gram of organics needs 2 g of oxygen for decomposition. However, during the first few days of the process, as the temperature increases, the demand for oxygen can be higher. This can increase the demand for oxygen by 3 to 6 g/g of organics. When moisture in the pile increases due to the production of water in the composting reaction, the demand for air also increases, but increasing the air can cause a decrease in temperature, and that in turn can decrease the decomposition rate.
During the decomposition of organics, heat will be lost through the surface of the pile such that where
Ql = heat lost
K = coefficient of heat exchange Tc = temperature of composting mass Ta = ambient temperature outside the pile
Several scientists have developed equations (Ettlich and Lewis, 1977; Sicora et al., 1981; Wilson, 1983; Epstein, 1997) for heat loss and heat exchange.
Design Example 7.1 A wastewater treatment plant composts combined primary and waste activated sludge dewatered on belt filter press. Given the following data, calculate the quantity of the bulking material, wood chips (F), and the dimensions of an aerated static pile composting system.
wet weight of dewater sludge (B) = 100 tons/day wet wight of recycled compost (A) = 20 tons/day concentration of dry solids in dewatered sludge (C1) = 25% concentration of dry solids in recycled compost (C2) = 50% concentration of dry solids in composting mixture (C3) = 40% mass of dry solid in wood chips = 0.25 ton/m3 or 25%
It is also assumed that the process variables have the following values:
organics in sludge cake: 75%
organics in sludge cake decomposed : 45%
solids in wood chips : 70%
organics in wood chips : 90%
organics in wood chips decomposed : 10%
organics in recycled compost : 75%
organics in compost decomposed : 10%
From equation (7.2), the weight of the composting mixture without a bulking agent
From equation (7.16), the quantity of wood chips r (120X0.40) _ (20X0.50) + (100)(0.25)
0.25
= 208 m3/day ratio of dewatered sludge to recycled compost = B: F
quantity of new wood chips = 52 _ 20 = 32 tons/day
The ratio
(100)(0.25)(0.7)(0.45) + (20)(0.5)(0.75)(0.1) + (52)(0.7)(0.9)(0.1) = 8.0
Since W is less than 10, no additional recycling of compost or addition of external amendment is required.
The following bulk densities are used in this example:
dewatered sludge cake (25% solids): 890 kg/ m wood chips : 250 kg/m3
screened (recycled) compost: 520kg/m3
total volume = 358.9m3/day
For a 2.5-m-high and 20-m-long pile, the width of the pile extended each day is
The amount of wood chips to construct the 0.3-m-thick pad for the pile is
Unscreened compost required to cover the pile with a thickness of 0.2 m is
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