Doublevault latrines
The design of a doublevault latrine is similar to that of a pit latrine, i.e., the volume of each vault is calculated using the formula:
V = N × P × R
where 
V = the effective volume of the vault (m^{3}) 

N = the number of years the vault must last before becoming full 

P = the average number of users 

R = the estimated sludge accumulation rate for a single person (m^{3} per year). 
The difficulty with vault design is that very little information exists on the sludge accumulation rate in vaults where excreta are mixed with ash and other organic material, and there has been little research into the pathogen survival rate in such an environment.
Desludging period
Pit latrines are usually designed such that excreta are not handled for two years. Since the inside of a composting toilet is similar to that of a pit latrine, it is reasonable to assume that it should be designed using similar parameters. However, some researchers disagree with this, saying that the low moisture content of the compost produces very alkaline conditions that destroy the pathogens in a much shorter time. Times as low as four months have been suggested. In the absence of more accurate information, however, a twoyear retention time is recommended.
Sludge accumulation rates
The accumulation rate for the excreta component of the compost can be determined in the same way as for a doublepit latrine. In the absence of more accurate local information, figures 50% greater than those given in Table 5.3 are suggested.
Estimating the volume of ash and other organic material is more difficult. Experience in Viet Nam indicates that approximately twice the volume of faeces has to be added (Jayaseelan et al., 1987). Rybczynski (1981) suggested five times the volume of faeces, and Kalbermattan et al. (1980) recommended allowing 0.3 m^{3} per person per year for all wastes.
In the absence of information to the contrary, it is suggested that the total sludge buildup rate is calculated as three times the estimated faecal buildup rate.
Example 8.8
Design a doublevault composting toilet for a family of six who use paper for anal cleaning.
The effective volume of each vault (V) must be:
2 × 6 × (0.06 × 1.5 × 3) = 3.24 m^{3}
Vaults are usually sealed when they are threequarters full, therefore the actual volume of the vault must be:
_{}
If the vault has a plan area of 1.3 × 1.3 m, the depth will be:
_{}
Continuous composting toilets
Even fewer design data are available for continuous composting toilets than for doublevault latrines. Past designs have been empirical and little published information exists indicating the level of their success. It is suggested that, until more data are available, the size of the primary tank in the toilet should be based on the formulae and factors used for doublevault latrines. The second tank should be 1020% of the size of the first tank. The floors of both tanks should slope at an angle of 30° to the horizontal. No design data exist for calculating the size and number of aeration channels or the diameter and height of the ventilation pipe.
Example 8.9
Using the information given in Example 8.8, design an appropriate continuous composting toilet.
From Example 8.8 the volume of the primary tank should be 4.32 m^{3}.
The volume of the second tank will be:
4.3 × 0.15= 0.65 m^{3}
Assuming the first tank is 1.2 m wide and 2.2 m long then its depth will be 1.65 m.
The length of the second tank will be:
_{}
This is short and would make emptying very difficult; increase the length to 0.5 m.
Since the vault floor must slope at an angle of 30°, the depth of excavation at the outlet end will be greater than the depth at the inlet.
Assuming the floor of the second tank is horizontal the internal floor level will be at a depth of:
1.65 + 2.2 tan 30° = 2.9 m
Fig. 8.2 shows the final internal dimensions of the tank.
Fig. 8.2. Internal dimensions of the continuous composting toilet designed in example 8.9
WHO 91499