Attachment G-1: Pit Latrine Diagram. Fig E.1a: Pit Latrine. Fig E.1b: Plan View of Twin Pits

Transcription

1 Attachment G-1: Pit Latrine Diagram Fig E.1a: Pit Latrine Fig E.1b: Plan View of Twin Pits

2 Fig E.1c: Section of a water-sealed pan Fig E.1d: 3D view of Overflow Pipe

3 Fig E.1e: 2D view of Overflow Pipe

4 Attachment G-2: Composting Latrine Diagram Fig E.2a: Compost Toilet Fig E.2b: Dimensions of vault

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6 Attachment G-3: Pros and Cons of Pit Latrine and Composting Latrine Pit Latrine Advantages 1. Convenient pour flush, no odor 2. Hygienic during emptying 3. Emptying done every 3 years 4. Minimal maintenance Disadvantages 1. Needs water 2. Emptying takes effort 3. Soil must be permeable Composting Latrine Advantages 1. Soil need not be permeable 2. Vault is above the ground, emptying is very easy 3. Waste product can be used as fertilizer Disadvantages 1. More costly to build the vault above the ground ($ vs $ 140 for pit latrine) 2. Empty more often than pit latrine (once a year) 3. Takes up a lot of space 4. High maintenance add sawdust, woodchip, ashes, etc 5. Problems with odor and insects

7 Attachment G-4: Exact Pit Calculation for Pit Latrine Pit Volume of discharge = 0.05 m 3 / person/ year 1 Size of one household = 9 persons Number of years before emptying = 3 years Volume for 9 persons for 3 years=(0.05 m 3 / person/ year)*(9 persons)*(3 years)=1.35 m 3 Pit area = (1.25 m) * (1.25 m) = 1.56 m 2 Depth = Volume / Area = (1.35 m 3) / (1.56 m 2 ) = 0.87 m The depth of the pit should be designed 0.20 m deeper from the surface to prevent wastes from coming too near to the surface after the designated time. Thus, Pit Depth = 0.87 m m = 1.07 m = 1.1 m In summary, the pit dimension is 1.25 m long * 1.25 m wide * 1.1 m deep Pit Cover The pit cover should extend 12 cm from each side of the pit. Thus: Length = length of pit + 2 * (0.12 m) = 1.25 m m = 1.49 m 2 Width = width of pit + 2 * (0.12 m) = 1.25 m m = 1.49 m 2 Thickness of pit cover = 6 cm = 0.06 m Thus, volume of pit cover = (1.49 m) * (1.49 m) * (0.06 m) = m 3 Cover Base The cover base should extend 10 cm from each side of pit. Thus: Area of cover base = (1.25 m m) * (0.10 m) * 4 = 0.54 m 2 Thickness of cover base = 2 cm = 0.02 m Thus, volume of cover base = (0.54 m 2 )* (0.02 m) = m 3 Slab Area of slab = (1.0 m) * (1.2 m) = 1.2 m 2 Thickness = 2 cm = 0.02 m Thus, volume of slab = (1.2 m) * (0.02 m) = m 2 1 Taken from Lifewater International website: 0.04 m 3 / person/ year. We add 0.01 m 3 factor of safety (March 2006)

8 Shelter Area of shelter base = (1.0 m) * (1.2 m) = 1.2 m 2 Height = 1.8 m Area of shelter wall = 2 * (1.0 m) * (1.8 m) + 2 * (1.2 m) * (1.8 m) = 7.92 m 2 The area of the roof extends 0.12 m from each side of the shelter Area of roof = (Length of shelter m) * (Width of shelter m) = (1.2 m m) * (1.0 m m) = 1.48 m 2

9 Attachment G-5: Calculation for Composting Latrine Vault Volume of discharge = 0.06 m 3 / person/ year 1 Size of one household = 9 persons Number of years before emptying = 1 year Volume for 9 persons for 2 year = (0.06 m 3 / person/ year)*(9 persons)*(1 year) = 0.54 m 3 Since for composting toilet, the two vaults are side by side, thus we do calculations for total volumes of both vaults instead of individual vault. Volume for 2 vaults = (0.54 m 3) * 2 = 1.08 m 3 Vault area = (1.10 m) * (1.60 m) = 1.76 m 2 Height = Volume / Area = (1.08 m 3 )/ (1.76 m 2 ) = 0.61 m For composting latrine, we allow an additional 0.10 m height of vault to prevent wastes from coming to close to the surface after the designated time. Thus, vault height = 0.61 m m = 0.71 m = 0.70 m In summary, the vault dimension is (1.10 m long) * (1.60 m wide) * (0.70 m high) Thickness of vault wall is 75 mm and thickness of divider between the two vaults is 100 mm Total area of vault wall = m * (2 * (2*0.80 m m m) + 2 * (1.10 m m) = 7.25 m 2 Total volume of vault wall = (7.25 m 2 ) * (0.70 m) = 5.08 m 3 1 Taken from Lifewater International website: 0.06 m 3 / person/ year. (March 2006)

10 Slab Slab length = length of vault + 2*0.075 m m = 1.60 m m m = 1.85 m Slab width = width of vault + 2*0.075 m = 1.10 m m = 1.25 m Thickness of slab = 10 cm = 0.10 m Thus, volume of slab = 1.85 m * 1.25 m * 0.10 m = m 3

11 Attachment G-6: Comparison between different materials Materials Unit Cost Material Unit Unit Cost Reference Cement 25 kg 5.3 FUCOHSO latest information (February 2006) Wood m FUCOHSO latest information (February 2006) Gravel m 3 40 Data collected from previous visit (August 2005) Sand - - FUCOHSO latest information (February 2006) Brick unit 0.04 FUCOHSO latest information (February 2006) Chicken wires m 2 2 Data collected from previous visit (August 2005) Zinc Sheet m 2 5 FUCOHSO latest information (February 2006) 8 inch PVC m 3 FUCOHSO latest information (February 2006) Toilet Seat unit 20 FUCOHSO latest information (February 2006) Hammer unit 2 FUCOHSO latest information (February 2006) Saw unit 5 FUCOHSO latest information (February 2006) Labour day 7 Data collected from previous visit (August 2005) Lock 2 Data collected from previous visit (August 2005) Hinges 2 Data collected from previous visit (August 2005) Nails 1 Data collected from previous visit (August 2005) Wire netting 2 Data collected from previous visit (August 2005) Table G.6a: Materials unit cost Different materials are used to build vault, slab and shelter Vault can be made out of concrete, brick or ferro-cement Slab can be made out of concrete, ferro-cement, or wood Shelter can be made out of concrete, brick, ferro-cement or wood We use the construction of the shelter to compare the cost of different materials Area of shelter base = 1.0 m * 1.2 m = 1.2 m 2 Height = 1.8 m Area of shelter wall = 2 * 1.0 m * 1.8 m + 2 * 1.2 m * 1.8 m = 7.92 m 2 Area of shelter wall minus door = 7.92 m m * 1.8 m = 6.12 m 2 Using concrete, the thickness of the wall can be 10 cm = 0.10 m Thus, volume of concrete used = 6.12 m 2 * 0.10 m = m 3 Concrete is a mixture of 1 part cement to 3 parts sand to 4 parts gravel Total volume of concrete used = m 3 Volume of wet cement = 1/ ( ) * m 3 = m 3 Volume of dry cement = m 3 * 1.65 = m 3 Weight of m 3 = m 3 * 50 kg / m 3 = 196 kg Cost of cement = 196 kg/ 25 kg * $ 5.3 = $ 42

12 Volume of gravel = 4/ 8 * m 3 = m 3 Cost of gravel = m 3 * $ 40 = $12.24 Total cost of concrete = $ 42 + $ = $ Using brick: Dimension of brick: 40 cm * 20 cm Area of one brick = 0.04 m * 0.02 m = m 2 No of brick needed = 6.12 m 2 / m 2 = 7650 Total cost of brick = 7650 * $ 0.04 = $ 306 Using ferro-cement, the thickness of the wall can be 5.0 cm = 0.05 m Thus, volume of ferro-cement used = 6.12 m 2 * 0.05 m = m 3 Ferro-cement is a mixture of 1 part cement to 3 parts sand Total volume of ferro-cement used = m 3 Volume of wet cement = 1/ ( ) * m 3 = m 3 Volume of dry cement = m 3 * 1.65 = m 3 Weight of m 3 = m 3 * 50 kg / m 3 = 98 kg Cost = 98 kg/ 25 kg * $ 5.3 = $ 20.8 Cost of chicken wire = 6.12 m 2 * $ 2/ m 2 = $ Total cost of ferro-cement = $ $ = $ Using wood: Cost of wood = 6.12 m 2 * $ 4.5 = $ Material Comparison Table: Material Cost Lifetime/ years Strength Concrete $54 40 good Brick $ good Ferro-cement $33 20 good Wood $27 5 poor

13 Attachment G-7: Family size of each household No No of people No of families No of families vs Family size Family size Series2 Total number of families (not including isolated house) = 29 Total number of people = 141 Average family size = 4.87

14 Attachment G-8: Materials Cost Material Unit Unit Cost/ $ Cement 25 kg 5.3 Wood m Gravel m 3 40 Sand - - Chicken wires m 2 2 Zinc Sheet m 2 5 Refer to Table G.6a for thorough unit cost and cost reference Cost for Pit Latrine The cost listed below is based on using ferro-cement for pit cover, cover base and slab and wood for shelter. Ferro-cement uses 1 part cement and 3 part sand. Wet volume is the final volume of cement and dry volume is the volume of ferrocement when it is not yet mixed with water. Dry volume is 1.65 * Wet volume m 3 of dry cement weigh 50 kg Total volume of ferro-cement used (pit cover, cover base, slab) = m m m 3 = m 3 Volume of wet cement = 1/ (1 + 3) * m 3 = m 3 Volume of dry cement = m 3 * 1.65 = m 3 Weight of m 3 = m 3 * 50 kg / m 3 (1) = 107 kg Material Unit Unit Cost Quantity Cost/ $ Cement 25 kg kg 23 Wood m Gravel m Sand Chicken wires m Zinc Sheet m inch PVC m Toilet Seat unit 20 Labour day Lock 2 2 Hinges 2 2 Nails 1 1 Wire netting 2 2 Total Cost 140 Cost of entire project = $ 140 * 29 = $ Taken from Lifewater website: (March 2006)

15 Cost for Composting Latrine The cost listed below uses concrete for vault and slab and wood for shelter. Concrete uses 1 part cement, 3 parts sand and 4 parts gravel. Wet volume is the final volume of cement and dry volume is the volume of cement when it is not yet mixed with water. Dry volume is 1.65 * Wet volume m 3 of cement weigh 50 kg Total volume of concrete used = 5.08 m m = 5.31 m 3 Volume of wet cement = 1/ ( ) * 5.31 m 3 = m 3 Volume of dry cement = m 3 * 1.65 = 1.10 m 3 Weight of m 3 = 1.10 m 3 * 50 kg / m 3 (1) = 1700 kg Volume of gravel used = 4/ 8 * 5.31 m 3 = 2.67 m 3 Material Unit Unit Cost Quantity Cost/ $ Cement 25 kg kg 360 Wood m Gravel m Sand Chicken wires m Zinc Sheet m inch PVC m Toilet Seat unit Labour day Lock Hinge Nails Wire netting Total Cost Cost of entire project = $ * 29 = $ Taken from Lifewater website: (March 2006)

16 Attachment G-9: Details of Technological Innovation A. Pit Alarm System The diagram above shows a mechanical pit alarm system. The detachable pipe allows detection of solid waste a few months before the pit gets filled up. User will find water being flushed down slower than usual. And this is the indicator that the pit is almost full and it is not long till he/ she has to change the pit used.

17 The second pit alarm system is more electronical. It relies on a voltage source, which can be solar powered battery. Once the solid reaches a certain height, there will be a layer of water on top of it. Water is a conductor of electricity, and it will close the circuit, lighting the light bulb, which will alert the user that it is time to empty the pit. Another alternative is to use chemical properties of the solid waste to generate small current, which will light up the light bulb. Research will be done on the area of electrochemical cell and chemistry content of the excreta to look into this possibility. B. Cost Saving Shelter

18 Generally a shelter is designed with ventilation on top. This is an extension to that idea. By making the ventilation bigger, while at the same time making sure that the angle of the netting is such that nobody can look to the level of the toilet seat, privacy and cost saving can be achieved at the same time. This is assuming that the cost of netting is much cheaper than the material wall. The design will be most effective if brick is used for shelter wall since a reduction of a few more bricks mean a lot of cost saving. C. Pit Emptying Tool

19 The pit emptying tool works with leg muscles, since it is stronger than arm muscle. The main two components are the wheel and the sliding mechanism. The function of the sliding mechanism is to lift the whole ploughing tool up when it hits the ground, thus allowing continuous motion. The earth will be collected at one site. Another design for the tool is shown below: A bucket is added to collect the earth.

20 Attachment G-10: Specific Soil Tests The two soil tests that need to be carried out in our next trip are the test for groundwater level and percolation test, which is a test for the permeability of the soil. Test for Groundwater Level The depth of the pit is to be 1 m above the groundwater level. Our design pit depth is 1.1 m. Thus the groundwater level should be at least 2.1 m deep. Procedure: 1. Dig a hole 2.5 m deep 2. Wait for two hours for the groundwater to enter the hole 3. Check for the depth of the water level inside the hole. This can be done by disturbing the water with a stick, or a stone tied on a string and note the length of the stick or string inside the hole 4. If the water level is more than 2.1 m deep, the site is suitable for pit construction 5. The test should be carried out during the wettest season of the year Percolation Test 1. Two percolation tests must be conducted at the proposed site. Two test holes are dug to the depth of the pit (1.1 m). Generally, the results of the two tests will be about the same. If they differ, use the slower of the two percolation rates to design the system. 2. Dig or bore a hole about 300mm in diameter, or 300mm square, to the proper depth. Do not use the same hole used for locating groundwater. Make the walls of the hole vertical. Scrape the walls to remove any patches of compacted soil. Place about 50mm of clean gravel in the bottom of the hole. 3. Fill the hole with water and let it soak overnight. This will allow ample time for soil swelling and saturation, and provide more accurate test results. 4. Place a board or piece of lumber across the center of the hole and anchor it firmly in place, perhaps by placing a rock on each end. The board must not be moved until the test is complete. Mark a point near the center of the board to be used as a guide for the remainder of the test. 5. Most or all of the water poured in the day before will have drained away. Pour in enough water so that the depth is 200mm. 6. Place a pointed slat or similar measuring stick next to the reference mark on the board and slide it down until it just touches the water surface. Ripples on the water can be observed when the slat touches. Note the exact time and draw a horizontal line on the slat, using the edge of the board for a guide, as shown in Figure Repeat step 6 at l0-minute intervals. If the water level drops rapidly, repeat at oneminute intervals. Do not allow the water to drop lower than 100mm. If it does, pour in more water to the 200mm depth and continue the test.

21 8. Note the spacing between the pencil marks on the slat. When at least three spaces become approximately equal, as shown in Figure 3, the test is completed. This may take as little as one-half hour or as long as several hours. 9. Using the measuring tape or ruler, measure the space between the equal pencil markings and compute how long it took the water level to drop 25mm. This step is necessary because percolation rates are described in terms of "minutes per 25mm." This can be approximated closely with the ruler and a series of equally spaced markings on the slat, as shown in Figure 3, or it can be calculated. 10. If the percolation rate for 25mm isbetween 10 and 60 minutes, the soil is acceptable.

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23 Attachment G-11: Site layout