Septic Tank Design
Assignment
In
Advanced Water and Wastewater Wastewater Engineering By Raji, M.O. (05/30GB090) Department of Civil Engineering University of Ilorin, Nigeria
CVE 691 Course Lecturer: Dr. A.A. Aremu
March, 26th 2013
Calculation 1. Design a septic tank for the following data; Number of people = 100 Sewage/Capacity/day =120 litres De-sludging period =1 year What would be the size of its soak well if the efluent from septic tank is to be discharged in it, assume percolation rate of the soak well is to be 1250 l/m 3/d? Solution Quantity of the sewage produced per day =12,000litres/day The capacity of the tank = quantity of the sewage produced during the detention period
= 12000× 24⁄24 = 12,000 Assuming: The rate of sludge deposit = 30litres/capita/year De-sludging period = 1 year
= 30× 100 ×1 = 3,000 Total required capacity of the tank = 12,000+3,000 = 15,000 = 15 = Assuming the depth of the tank as 1.5m, the cross -sectional area of the tank = . The quantity of sludge deposited
10 Using an L: B as 4:1, we have
4 = 10 = √ 2.5 = 1.5 = 4×1.5 = 6 Adding 0.3m for freeboard, overall depth 1.5+0.3 = 1.8 Hence, tank size= .× . ×. Design of soak well Assuming percolation rate =
1250// 2
Sewage outlow =
12,000/
= = 9. 6 .× = 2.47 = 2.5 Therefore, the diameter of soak well required, = Therefore, the diameter of soak well = . Volume of iltering media required for the soak well
Question 2. Estimate the size of a septic tank (Length to Width ratio 2.25, liquid depth 2m with 300mm free board), De -sludging interval in years and the total trench area (m 2) of the percolation ield for a small colony of 300 people. Assume water supply of 100 litres per capital per day. Waste water low at 80% of water consumption, sludge production of 0.04m 3/capital/year and retention time of 3 days at start up. De sludging is done when the tank is one-third full of sludge. A percolation test indicated an allowable hydraulic loading of 100l/m2/day. Solution 1. Septic tank Given
= 2.25
= 2 = 0.3 = 300 = 100// = 80% = 100// = 0.04 // = 3 = 100 ×300 = 30,000/ Sewage produced= 80% 30000 = 24,000/ Sewage retained= 24000 ×3 = 72,000 = 72 If de-sludging is done when the tank is illed up to 1⁄3 of its capacity. Capacity= + = 72+ ⁄3 ∴ 2⁄3 = 72 Water supplied to the colony
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= 108 Assume , = 2 = 108⁄2= 56 = 2.25 ∴ = 2.25 ×2.25 = 56 = 4.9 = 2.25 ×4.9 = 11.10 Overall depth = 2.0 + 0.3( ) = 2.3 Hence, dimension of the tank = .×. ×. 2. Soak well
= ⁄3 = 108⁄3= 36 = 0.04 ⁄. ×300 = 12/ Sewage outlow= 72 36 of sludge will therefore be produced in 1⁄12 × 36 = 3 Hence, desludging interval= Sludge volume removed in desludging
Hydraulic loading of percolation trench
= 100//
= 72 Outlowing sewage per 1 day = 24 = 24,000/ Outlowing sewage per 3 days
Trench area required
, / = = //
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