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Chapter 9 · 3 hours

Disposal of Sewage from Isolated Buildings

IOE past exam questions

Past questions and answers

21 questions set from this chapter, 3 of them more than once. Most repeated first.

  • Asked 2 times
  • 2076 Bhadra · 8 marks
  • 2070 Bhadra · 8 marks

What are the design considerations adopted for the design of septic tank and soak pit in Nepal (state the procedure for designing). Briefly describe them in neat sketch showing the zones and their purposes.

Answer

A septic tank is a watertight, single-storey sedimentation and digestion tank that gives primary treatment to the sewage of an isolated building. Its effluent is still polluted, so it is disposed of in a soak pit (or dispersion trench) where the soil gives further treatment.

Zones of a septic tank and their purposes

 Inlet                       Outlet
  |   (T-pipe)  | partition   |(T-pipe)
  v             |  wall       v
 ___|____________|_____________|___  <- cover slabs
 |  ^ scum   |  ^ opening  | scum  |  
 |-----------|-------------|-------|  <- water level
 |  Sewage zone (settling)  | zone  |
 |..........................|.......|
 |  Sludge digestion zone   |       |
 |  Digested sludge storage |       |
 |___________________________|______|
   Chamber 1 (2/3 L)      Chamber 2 (1/3 L)
  • Scum zone (top): grease, oil and light solids float here; the dipped T-pipes keep scum from leaving.
  • Sewage (settling) zone: sewage is held about 24 h so that suspended solids settle; designed for the daily flow.
  • Sludge digestion zone: settled solids are digested anaerobically by bacteria, producing gas and reducing volume.
  • Digested sludge storage zone: stabilised sludge collects between desludging periods (usually 2-3 years).

Design considerations (Nepal practice)

  1. Flow: sewage = 80% of water supply (or 80-100 L per person per day); users from the building occupancy.
  2. Detention time: 24 h (minimum 12-24 h) for the sewage zone.
  3. Sludge: digestion zone 0.0425 m³/person; digested sludge storage about 0.0283 m³/person/year (0.085 m³/person for 3 years).
  4. Shape and size: liquid depth 1.0-2.0 m, plus 0.3 m free board; width at least 0.75 m; length at least 1.5 m; L:BL:B = 2:1 to 4:1; usually 2 chambers, first = 2/3 length.
  5. Inlet/outlet: T-pipes dipping 0.3 m (up to 0.45 m) below the water level, with vent pipe and manhole cover for inspection and desludging.
  6. Structure: watertight masonry or RCC, 1:3 cement plaster, 75 mm thick floor sloping towards inlet; keep at least 6 m from water sources and 1.5 m from buildings.
  7. Effluent disposal: soak pit, dispersion trench or evapotranspiration mound, depending on soil.

Procedure for design

  1. Find the number of users and sewage flow QQ.
  2. Volume of sewage zone =Q×= Q \times detention time.
  3. Add digestion and sludge storage volumes (per-person rates ×\times users).
  4. Fix depth, then L×B=V/dL \times B = V/d with L:BL:B = 3:1; add free board.
  5. Divide into two chambers; fix T-pipe, vent and manhole.

Soak pit design

  • Area A=Q/infiltration rateA = Q/\text{infiltration rate} (L/m²/d, from percolation test).
  • Circular pit of diameter DD and effective depth hh: πDh=A\pi D h = A, usually DD = 1.5-3 m, hh = 2-3 m.
  • Pit is honeycomb-lined and filled with brick bats/gravel; bottom at least 1 m above ground-water; at least 15 m from a well. For larger areas use several pits or dispersion trenches.
  • Asked 2 times
  • 2071 Magh · 4 marks
  • 2068 Bhadra (old course) · 8 marks

Design a double pit VIP latrine for a family of 15 users. Assume the necessary data suitably.

Answer

A double pit VIP (ventilated improved pit) latrine has two pits side by side, used one at a time. When one pit is full it is closed and the second is used; the first pit's contents decompose for a year or more and are then safely emptied and reused as manure.

Design data (assumed)

  • Users: 15 (family/joint-use).
  • Sludge accumulation rate (wet pit, water used for anal cleansing): 0.04 m³/person/year.
  • Each pit is designed to last 2 years, so while one is in use the other rests for 2 years.
  • Free space above sludge: 0.5 m.

Pit size (each pit)

V=15×0.04×2=1.2 m3V = 15 \times 0.04 \times 2 = 1.2\ \text{m}^3

Adopt pit plan 1.2 m ×\times 1.0 m = 1.2 m², so effective depth = 1.2/1.2 = 1.0 m, plus 0.5 m free space = 1.5 m total depth.

Provisions

  • Two pits of 1.2 m ×\times 1.0 m ×\times 1.5 m, with a clear gap of 0.5 m or more between them; lined with brick/stone in cement mortar with open joints for the lower part, and a watertight upper 0.3 m collar.
  • Slab: RCC slab with a squat hole (or pedestal pan) over the working pit; the other hole is sealed with a concrete cover.
  • Y-shaped diversion chute/channel in the substructure to divert excreta to the pit in use.
  • Vent pipe: 150 mm dia (not less than 110 mm) PVC/GI/bamboo pipe, extending at least 0.5 m above the roof, fitted with fly screen (mesh) at the top; one vent per pit. Opening of the pipe on the sunny side.
  • Superstructure: about 1.0 m ×\times 1.2 m in plan, 2 m high, with a door facing away from the vent and a dark interior so that flies move to the light of the vent; roof sloping to the back.
  • Located at least 15 m from water sources and downhill of wells.
  vent pipe (fly screen)
        |   superstructure
   _____|______
  | [door]  O  |   O = squat hole
  |____________|__ slab
  | Pit A | Pit B |   1.2 x 1.0 x 1.5 m each
  | (use) | (rest)|
  • Asked 2 times
  • 2071 Bhadra · 8 marks
  • 2070 Magh · 8 marks

With a neat sketch describe the working and design procedure of ventilated improved pit latrine.

Answer

A ventilated improved pit (VIP) latrine is a pit latrine with a vent pipe that removes odours and traps flies, so it is practically free of smell and insect nuisance.

Working

  1. Excreta fall into the pit through the squat hole in the slab; the pit digests it anaerobically.
  2. Wind blowing across the top of the vent pipe creates suction, and warm air rising in the pipe (heated by the sun) draws air from the pit. Fresh air enters through the squat hole, passes down into the pit and up the vent, carrying smell out above the roof.
  3. The superstructure is kept dark inside. Flies are attracted to the light at the top of the vent pipe, but a fly screen (mesh) at its top stops them leaving. Flies trying to enter the pit are drawn to the light, get trapped and die.
  4. Liquid seeps into the soil; solids decompose and reduce in volume.
          wind -->
      fly screen  _
         |       | |  vent pipe, 0.5 m
      ___|_______|_|  above roof
     |  dark superstructure |
     |  O squat hole        |
 ____|______________________|__ slab
 |                           |
 |   pit (lined, open joints) |
 |___________________________|

Design procedure

  1. Find the number of users nn and the design period TT (years) between emptying (single pit: 5-10 years; double pit: 1-2 years each).
  2. Sludge accumulation rate rr: 0.04 m³/person/year (wet) or 0.06 m³/person/year (dry cleansing material).
  3. Pit volume V=n×r×TV = n \times r \times T, plus 0.5 m free space above sludge level.
  4. Choose plan (circular 1.0-1.5 m dia or rectangular) so that depth is within 1.5-3 m, and at least 1.5 m above ground-water level.
  5. Pit lining: brick or stone with open joints, with the top 0.3 m cement-mortar (watertight), slab on a raised collar to keep out surface water.
  6. Vent pipe: internal diameter 150 mm (min 110 mm), 0.5 m above the highest point of the roof, fitted with fly screen.
  7. Superstructure with a door opening away from the sun (so the inside stays dark), 1.0 m ×\times 1.2 m in plan, roof, ventilation and walls of local material.

Example: for 5 users, 3 years, wet pit: V=5×0.04×3=0.6V = 5 \times 0.04 \times 3 = 0.6 m³, add 0.5 m free space and use a 1 m dia circular pit, depth =0.6/0.785+0.5=1.3= 0.6/0.785 + 0.5 = 1.3 m (adopt 1.5 m).

  • 2079 Chaitra · 8 marks

Design a septic tank and soak pit for 15 users of a house at a place where infiltration capacity of soil is 100 liters per square meter per day. The peak sewage flow rate 110 lpcd and the detention time in septic tank is 1 day and sludge is cleaned at an interval of 3 years. The ground water table lies 10 m below the ground level.

Similar questions: Septic tank and soak pit, 10 users, 85 l/m²/d (2072 Asoj)

Answer

Given: 15 users, 110 L/c/d, detention 1 day, desludging every 3 years, infiltration 100 L/m²/d, ground-water 10 m below ground (so a deep pit is safe).

Septic tank

  • Sewage flow: Q=15×110 L/c/d=1650 L/d=1.650 m3/dQ = 15 \times 110\ \text{L/c/d} = 1650\ \text{L/d} = 1.650\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=1.650 m3V_1 = 1.650\ \text{m}^3
  • Sludge digestion zone: 0.0425×15=0.638 m30.0425 \times 15 = 0.638\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×15=1.275 m3(0.085/3) \times 3 \times 15 = 1.275\ \text{m}^3
V=1.650+0.638+1.275=3.56 m3V = 1.650 + 0.638 + 1.275 = 3.56\ \text{m}^3

Adopt liquid depth d=1.5d = 1.5 m and L:B≈3:1L:B \approx 3:1.

Plan area=3.56/1.5=2.38 m2B=2.38/3≈0.9 m,L=2.38/0.9≈2.7 m\begin{aligned} \text{Plan area} &= 3.56/1.5 = 2.38\ \text{m}^2 \\ B &= \sqrt{2.38/3} \approx 0.9\ \text{m}, \quad L = 2.38/0.9 \approx 2.7\ \text{m} \end{aligned}

Check: 2.7×0.9×1.5=3.65 m3≥3.56 m32.7 \times 0.9 \times 1.5 = 3.65\ \text{m}^3 \ge 3.56\ \text{m}^3 (safe).

Septic tank: internal size 2.7 m×0.9 m×1.8 m2.7\ \text{m} \times 0.9\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 1.801.80 m long (2/3), second 0.900.90 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

Soak pit

  • Required infiltration area: A=Q/rate=1650/100=16.50 m2A = Q/\text{rate} = 1650/100 = 16.50\ \text{m}^2 (only the side wall below the inlet is counted; the bottom is ignored as it clogs)
  • Adopt effective depth h=3h = 3 m
πDh=16.50⇒D=16.50π×3=1.75≈1.8 m\pi D h = 16.50 \Rightarrow D = \frac{16.50}{\pi \times 3} = 1.75 \approx 1.8\ \text{m}

Check: π×1.8×3=16.96 m2≥16.50 m2\pi \times 1.8 \times 3 = 16.96\ \text{m}^2 \ge 16.50\ \text{m}^2.

Soak pit: one circular pit, 1.8 m internal diameter and 3 m deep, lined with dry brick/stone masonry with open joints (honeycomb), filled with brick bats/gravel, and covered with a slab.

The ground-water table is 10 m deep, far below the pit bottom (3 m), so the 1 m clearance is satisfied.

  • 2076 Baisakh · 8 marks

Design a septic tank and soak pit to dispose the sewage generated from a household of 80 persons. The sewage is generated at the rate of 120 lit/person/day. Assume that septic tank is cleaned once in 2 years and infiltration rate of soil is 80 lit/m²/day.

Similar questions: Septic tank and soak pit, 8 persons (2069 Bhadra)

Answer

Given: 80 persons, 120 L/c/d, desludging every 2 years, infiltration 80 L/m²/d. Assume 24 h detention and soak pit effective depth 3 m (ground-water table deep).

Septic tank

  • Sewage flow: Q=80×120 L/c/d=9600 L/d=9.600 m3/dQ = 80 \times 120\ \text{L/c/d} = 9600\ \text{L/d} = 9.600\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=9.600 m3V_1 = 9.600\ \text{m}^3
  • Sludge digestion zone: 0.0425×80=3.400 m30.0425 \times 80 = 3.400\ \text{m}^3
  • Digested sludge storage for 2 years: (0.085/3)×2×80=4.533 m3(0.085/3) \times 2 \times 80 = 4.533\ \text{m}^3
V=9.600+3.400+4.533=17.53 m3V = 9.600 + 3.400 + 4.533 = 17.53\ \text{m}^3

Adopt liquid depth d=2.0d = 2.0 m and L:B≈3:1L:B \approx 3:1.

Plan area=17.53/2.0=8.77 m2B=8.77/3≈1.8 m,L=8.77/1.8≈4.9 m\begin{aligned} \text{Plan area} &= 17.53/2.0 = 8.77\ \text{m}^2 \\ B &= \sqrt{8.77/3} \approx 1.8\ \text{m}, \quad L = 8.77/1.8 \approx 4.9\ \text{m} \end{aligned}

Check: 4.9×1.8×2.0=17.64 m3≥17.53 m34.9 \times 1.8 \times 2.0 = 17.64\ \text{m}^3 \ge 17.53\ \text{m}^3 (safe).

Septic tank: internal size 4.9 m×1.8 m×2.3 m4.9\ \text{m} \times 1.8\ \text{m} \times 2.3\ \text{m} (including 0.3 m free board), in two chambers: first chamber 3.303.30 m long (2/3), second 1.601.60 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

Soak pit

  • Required infiltration area: A=Q/rate=9600/80=120.00 m2A = Q/\text{rate} = 9600/80 = 120.00\ \text{m}^2 (only the side wall below the inlet is counted; the bottom is ignored as it clogs)
  • Adopt effective depth h=3h = 3 m

A single pit would be too large, so adopt pits of D=3.0D = 3.0 m (area of each = π×3.0×3=28.27 m2\pi \times 3.0 \times 3 = 28.27\ \text{m}^2).

N=120.0028.27=4.24⇒5 pitsN = \frac{120.00}{28.27} = 4.24 \Rightarrow 5\ \text{pits}

Soak pits: 5 circular pits, each 3.0 m diameter and 3 m deep, honeycomb-lined and filled with brick bats/gravel, spaced at least 3 m (about 2 times the diameter) apart and connected through a distribution box. Soak-pit bottoms must stay at least 1 m above the highest ground-water level.

  • 2072 Asoj · 8 marks

Design a septic tank and soak pit for 10 number of users of a house at a place where infiltration capacity of soil is 85 liters per square meter per day. The peak sewage flow rate 80 lpcd and the septic tank is cleaned at an interval of 2 years. The ground water table lies 6 m below the ground level.

Similar questions: Septic tank and soak pit, 15 users (2079 Chaitra)

Answer

Given: 10 users, 80 L/c/d (peak), desludging every 2 years, infiltration 85 L/m²/d, ground-water 6 m deep. Detention 24 h assumed; soak pit depth 3 m keeps 3 m clearance above ground-water.

Septic tank

  • Sewage flow: Q=10×80 L/c/d=800 L/d=0.800 m3/dQ = 10 \times 80\ \text{L/c/d} = 800\ \text{L/d} = 0.800\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=0.800 m3V_1 = 0.800\ \text{m}^3
  • Sludge digestion zone: 0.0425×10=0.425 m30.0425 \times 10 = 0.425\ \text{m}^3
  • Digested sludge storage for 2 years: (0.085/3)×2×10=0.567 m3(0.085/3) \times 2 \times 10 = 0.567\ \text{m}^3
V=0.800+0.425+0.567=1.79 m3V = 0.800 + 0.425 + 0.567 = 1.79\ \text{m}^3

Adopt liquid depth d=1.5d = 1.5 m and L:B≈3:1L:B \approx 3:1.

Plan area=1.79/1.5=1.19 m2B=1.19/3≈0.7 m,L=1.19/0.7≈1.8 m\begin{aligned} \text{Plan area} &= 1.79/1.5 = 1.19\ \text{m}^2 \\ B &= \sqrt{1.19/3} \approx 0.7\ \text{m}, \quad L = 1.19/0.7 \approx 1.8\ \text{m} \end{aligned}

Check: 1.8×0.7×1.5=1.89 m3≥1.79 m31.8 \times 0.7 \times 1.5 = 1.89\ \text{m}^3 \ge 1.79\ \text{m}^3 (safe).

Septic tank: internal size 1.8 m×0.7 m×1.8 m1.8\ \text{m} \times 0.7\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 1.201.20 m long (2/3), second 0.600.60 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

Soak pit

  • Required infiltration area: A=Q/rate=800/85=9.41 m2A = Q/\text{rate} = 800/85 = 9.41\ \text{m}^2 (only the side wall below the inlet is counted; the bottom is ignored as it clogs)
  • Adopt effective depth h=3h = 3 m
πDh=9.41⇒D=9.41π×3=1.00≈1.0 m\pi D h = 9.41 \Rightarrow D = \frac{9.41}{\pi \times 3} = 1.00 \approx 1.0\ \text{m}

Check: π×1.0×3=9.42 m2≥9.41 m2\pi \times 1.0 \times 3 = 9.42\ \text{m}^2 \ge 9.41\ \text{m}^2.

Soak pit: one circular pit, 1.0 m internal diameter and 3 m deep, lined with dry brick/stone masonry with open joints (honeycomb), filled with brick bats/gravel, and covered with a slab.

  • 2069 Bhadra · 8 marks

Design a septic tank and soak pit to dispose the sewage generated from a household of 8 persons. The sewage is generated at the rate of 100 liters/person/day. Assume that septic tank is cleaned once in 3 years and infiltration rate of soil is 50 liters/m²/day.

Similar questions: Septic tank and soak pit, household of 80 (2076 Baisakh)

Answer

Given: 8 persons, 100 L/c/d, desludging every 3 years, infiltration 50 L/m²/d. Detention 24 h assumed; soak pit depth 3 m.

Septic tank

  • Sewage flow: Q=8×100 L/c/d=800 L/d=0.800 m3/dQ = 8 \times 100\ \text{L/c/d} = 800\ \text{L/d} = 0.800\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=0.800 m3V_1 = 0.800\ \text{m}^3
  • Sludge digestion zone: 0.0425×8=0.340 m30.0425 \times 8 = 0.340\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×8=0.680 m3(0.085/3) \times 3 \times 8 = 0.680\ \text{m}^3
V=0.800+0.340+0.680=1.82 m3V = 0.800 + 0.340 + 0.680 = 1.82\ \text{m}^3

Adopt liquid depth d=1.5d = 1.5 m and L:B≈3:1L:B \approx 3:1.

Plan area=1.82/1.5=1.21 m2B=1.21/3≈0.7 m,L=1.21/0.7≈1.8 m\begin{aligned} \text{Plan area} &= 1.82/1.5 = 1.21\ \text{m}^2 \\ B &= \sqrt{1.21/3} \approx 0.7\ \text{m}, \quad L = 1.21/0.7 \approx 1.8\ \text{m} \end{aligned}

Check: 1.8×0.7×1.5=1.89 m3≥1.82 m31.8 \times 0.7 \times 1.5 = 1.89\ \text{m}^3 \ge 1.82\ \text{m}^3 (safe).

Septic tank: internal size 1.8 m×0.7 m×1.8 m1.8\ \text{m} \times 0.7\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 1.201.20 m long (2/3), second 0.600.60 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

Soak pit

  • Required infiltration area: A=Q/rate=800/50=16.00 m2A = Q/\text{rate} = 800/50 = 16.00\ \text{m}^2 (only the side wall below the inlet is counted; the bottom is ignored as it clogs)
  • Adopt effective depth h=3h = 3 m
πDh=16.00⇒D=16.00π×3=1.70≈1.7 m\pi D h = 16.00 \Rightarrow D = \frac{16.00}{\pi \times 3} = 1.70 \approx 1.7\ \text{m}

Check: π×1.7×3=16.02 m2≥16.00 m2\pi \times 1.7 \times 3 = 16.02\ \text{m}^2 \ge 16.00\ \text{m}^2.

Soak pit: one circular pit, 1.7 m internal diameter and 3 m deep, lined with dry brick/stone masonry with open joints (honeycomb), filled with brick bats/gravel, and covered with a slab.

  • 2081 Chaitra · 3+5 marks

Describe about VIP latrine. Design a septic tank for 15 users. The rate of sewage flow is 80 lpcd. Assume sludge is cleaned from septic tank once in a three year.

Answer

VIP latrine

A VIP (ventilated improved pit) latrine is a pit latrine in which a vertical vent pipe (150 mm dia, 0.5 m above the roof, with fly screen at top) removes smell and traps flies.

  • Wind blowing across the pipe top and warm air rising in it draw air from the pit through the squat hole, carrying odour out above the roof.
  • The superstructure is dark, so flies are attracted to the light at the top of the pipe, are blocked by the screen and die.
  • The pit (lined with open-jointed brick/stone) is 1.5-3 m deep, designed at 0.04 m³/person/year; it has a slab with squat hole and a raised collar against surface water.
  • Double-pit versions allow one pit to rest and decompose while the other is used.
        vent (screen)
          |  ___________
          | | dark room |
   _______|_|_O_________|__ slab
   |        pit            |

Septic tank for 15 users

Given: 80 L/c/d, desludging every 3 years; detention 24 h assumed.

  • Sewage flow: Q=15×80 L/c/d=1200 L/d=1.200 m3/dQ = 15 \times 80\ \text{L/c/d} = 1200\ \text{L/d} = 1.200\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=1.200 m3V_1 = 1.200\ \text{m}^3
  • Sludge digestion zone: 0.0425×15=0.638 m30.0425 \times 15 = 0.638\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×15=1.275 m3(0.085/3) \times 3 \times 15 = 1.275\ \text{m}^3
V=1.200+0.638+1.275=3.11 m3V = 1.200 + 0.638 + 1.275 = 3.11\ \text{m}^3

Adopt liquid depth d=1.5d = 1.5 m and L:B≈3:1L:B \approx 3:1.

Plan area=3.11/1.5=2.07 m2B=2.07/3≈0.9 m,L=2.07/0.9≈2.4 m\begin{aligned} \text{Plan area} &= 3.11/1.5 = 2.07\ \text{m}^2 \\ B &= \sqrt{2.07/3} \approx 0.9\ \text{m}, \quad L = 2.07/0.9 \approx 2.4\ \text{m} \end{aligned}

Check: 2.4×0.9×1.5=3.24 m3≥3.11 m32.4 \times 0.9 \times 1.5 = 3.24\ \text{m}^3 \ge 3.11\ \text{m}^3 (safe).

Septic tank: internal size 2.4 m×0.9 m×1.8 m2.4\ \text{m} \times 0.9\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 1.601.60 m long (2/3), second 0.800.80 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

  • 2080 Chaitra · 5+3 marks

Assuming 80% contribution of water supply for the sewage, design septic tank and dispersion trenches for a house having 12 persons with an average water supply rate of 70 lpcd. The desludging period of septic tank is 3 years. The infiltration capacity of soil is 30 l/m²/d.

Answer

Given: 12 persons, water supply 70 L/c/d, sewage = 80% of supply, desludging every 3 years, infiltration 30 L/m²/d.

Septic tank

  • Sewage flow: Q=12×70×0.8 L/c/d=672 L/d=0.672 m3/dQ = 12 \times 70\times0.8\ \text{L/c/d} = 672\ \text{L/d} = 0.672\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=0.672 m3V_1 = 0.672\ \text{m}^3
  • Sludge digestion zone: 0.0425×12=0.510 m30.0425 \times 12 = 0.510\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×12=1.020 m3(0.085/3) \times 3 \times 12 = 1.020\ \text{m}^3
V=0.672+0.510+1.020=2.20 m3V = 0.672 + 0.510 + 1.020 = 2.20\ \text{m}^3

Adopt liquid depth d=1.5d = 1.5 m and L:B≈3:1L:B \approx 3:1.

Plan area=2.20/1.5=1.47 m2B=1.47/3≈0.7 m,L=1.47/0.7≈2.1 m\begin{aligned} \text{Plan area} &= 2.20/1.5 = 1.47\ \text{m}^2 \\ B &= \sqrt{1.47/3} \approx 0.7\ \text{m}, \quad L = 1.47/0.7 \approx 2.1\ \text{m} \end{aligned}

Check: 2.1×0.7×1.5=2.21 m3≥2.20 m32.1 \times 0.7 \times 1.5 = 2.21\ \text{m}^3 \ge 2.20\ \text{m}^3 (safe).

Septic tank: internal size 2.1 m×0.7 m×1.8 m2.1\ \text{m} \times 0.7\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 1.401.40 m long (2/3), second 0.700.70 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

Dispersion trenches

  • Infiltration area: A=672/30=22.40 m2A = 672/30 = 22.40\ \text{m}^2 (trench bottom area)
  • Trench width w=0.6w = 0.6 m, so total length =22.40/0.6=37.3= 22.40/0.6 = 37.3 m
  • Maximum length per trench about 30 m, so adopt 2 trench(es), each 19.0 m long (22.8 m² provided), 0.6 m wide and 0.6-1.0 m deep, with a 1 in 200 to 1 in 300 grade, 2 m centre-to-centre spacing.

Each trench has a 0.3 m deep layer of gravel (20-40 mm) with a 100 mm perforated (or open-jointed) distribution pipe laid in it, covered with 50 mm coarse sand/gravel, a filter layer such as geotextile or straw, and then soil. Effluent is fed from a distribution box.

  • 2078 Chaitra · 8 marks

Describe in brief about the purpose and design criteria of pit privy. Also, with a neat sketch, discuss the suitable disposal of septic tank effluent for the rocks area with high ground water table.

Answer

Pit privy: purpose

A pit privy is a simple on-site latrine that holds human excreta in a pit, keeping it away from people, flies and water so that it decomposes safely and prevents faecal-oral disease. Used in rural and low-income areas without sewers or piped water.

Design criteria

  • Location at least 15 m from a well/spring and 6 m from a house; pit bottom at least 1.5 m above ground-water level.
  • Pit size: 0.06 m³/person/year (dry), 0.04 for wet pits, designed for 5-10 years; usually 1.0-1.2 m wide and 2-3 m deep, plus 0.5 m free space.
  • Pit lined with brick/stone open joints (or bamboo/wood), top 0.3 m watertight and raised above ground against surface runoff.
  • Slab of RCC/wood, 1.2 m × 1.2 m, with squat hole 0.2-0.25 m and cover lid.
  • Superstructure about 1.0 × 1.2 × 2.0 m with door and roof; when the pit fills to 0.5 m below the slab, it is covered with soil and a new pit is dug.

Disposal of septic tank effluent for rocky area with high water table

Soak pits and trenches cannot be used, because the rock prevents absorption and the water table is high. The suitable method is an evapotranspiration (ET) mound / sand mound:

  1. Effluent from the septic tank goes to a distribution box and perforated pipes (75-100 mm) laid in a gravel bed inside a mound of clean coarse sand, built on the rock or ground above the high water table.
  2. The mound is 0.5-1.0 m high with side slopes 1:3 and is covered with 0.3 m topsoil and planted with grass or high-transpiring plants (banana, bamboo).
  3. Effluent moves upward and through the sand by capillary action; most of it evaporates or is taken up by plants, and the rest drains slowly.
  4. Surface water is diverted away and the mound is kept at least 1 m above the water table.
   plants / grass cover
   ___________________________
  /    topsoil + straw layer   \
 /-------------------------------\
 / gravel + perforated pipe <- from \
/        coarse sand bed       dist.box
================================ rock / natural ground
   ^ at least 1 m  GWT (water table)

Alternatively, a sand filter or constructed wetland can treat the effluent before release into a drain or stream.

  • 2073 Bhadra · 4 marks

With a neat sketch, describe the suitable septic tank effluent disposal method for the area of high ground water table and for the rocky area.

Answer

Septic tank effluent still has high BOD and pathogens, so it must be disposed of in a way that the soil or plants complete the treatment. Soak pits and trenches need deep, permeable soil, so for high water table or rocky ground the following are used.

  • Dispersion trench with raised (mound) bed / evapotranspiration mound: for high ground-water table, effluent is applied to an above-ground sand mound or shallow trenches (0.3-0.5 m deep) so that the bottom stays at least 1 m above the water table. Effluent is distributed through perforated pipes in a gravel bed inside a built-up sand mound, planted with grass or high-transpiring plants; part evaporates and the rest percolates.
  • Constructed wetland / reed bed: a lined gravel bed planted with reeds treats the effluent biologically; outflow is then used for irrigation or released to a drain.
  • Rocky area: soil is too shallow to absorb effluent, so a soak pit cannot be dug. Effluent is led to a sealed sand-filter bed built above the rock (sand filter then discharge to a nearby stream/drain after treatment and disinfection), or an evapotranspiration mound built on the rock surface with imported sand and soil. Where there is a bare rock face, the effluent can be pumped or piped to a collection pit for reuse in gardening.
   plants / grass cover
   ___________________________
  /    topsoil + straw layer   \
 /-------------------------------\
 / gravel + perforated pipe <- from \
/        coarse sand bed       dist.box
================================ rock / natural ground
   ^ at least 1 m  GWT (water table)
  • 2077 Chaitra · 8 marks

If you are asked to design septic tank and soak pit for a hostel with 400 users, calculate sizes. For septic tank, consider sludge digestion rate of 0.0425 m³/person and volume required for storage of digested sludge 0.085 m³/person for 3 years cleaning period.

Answer

Assumptions (flow and soil data not given): sewage 100 L/person/day, detention 24 h, infiltration 50 L/m²/d (sandy loam), digestion 0.0425 m³/person and storage 0.085 m³/person for 3 years as given. Soak pit effective depth 3 m, pit diameter 3 m.

Septic tank

  • Sewage flow: Q=400×100 L/c/d=40000 L/d=40.000 m3/dQ = 400 \times 100\ \text{L/c/d} = 40000\ \text{L/d} = 40.000\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=40.000 m3V_1 = 40.000\ \text{m}^3
  • Sludge digestion zone: 0.0425×400=17.000 m30.0425 \times 400 = 17.000\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×400=34.000 m3(0.085/3) \times 3 \times 400 = 34.000\ \text{m}^3
V=40.000+17.000+34.000=91.00 m3V = 40.000 + 17.000 + 34.000 = 91.00\ \text{m}^3

Adopt liquid depth d=2.0d = 2.0 m and L:B≈3:1L:B \approx 3:1.

Plan area=91.00/2.0=45.50 m2B=45.50/3≈3.9 m,L=45.50/3.9≈11.7 m\begin{aligned} \text{Plan area} &= 91.00/2.0 = 45.50\ \text{m}^2 \\ B &= \sqrt{45.50/3} \approx 3.9\ \text{m}, \quad L = 45.50/3.9 \approx 11.7\ \text{m} \end{aligned}

Check: 11.7×3.9×2.0=91.26 m3≥91.00 m311.7 \times 3.9 \times 2.0 = 91.26\ \text{m}^3 \ge 91.00\ \text{m}^3 (safe).

Septic tank: internal size 11.7 m×3.9 m×2.3 m11.7\ \text{m} \times 3.9\ \text{m} \times 2.3\ \text{m} (including 0.3 m free board), in two chambers: first chamber 7.807.80 m long (2/3), second 3.903.90 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

Soak pit

  • Required infiltration area: A=Q/rate=40000/50=800.00 m2A = Q/\text{rate} = 40000/50 = 800.00\ \text{m}^2 (only the side wall below the inlet is counted; the bottom is ignored as it clogs)
  • Adopt effective depth h=3h = 3 m

A single pit would be too large, so adopt pits of D=3.0D = 3.0 m (area of each = π×3.0×3=28.27 m2\pi \times 3.0 \times 3 = 28.27\ \text{m}^2).

N=800.0028.27=28.29⇒29 pitsN = \frac{800.00}{28.27} = 28.29 \Rightarrow 29\ \text{pits}

Soak pits: 29 circular pits, each 3.0 m diameter and 3 m deep, honeycomb-lined and filled with brick bats/gravel, spaced at least 3 m (about 2 times the diameter) apart and connected through a distribution box. Soak-pit bottoms must stay at least 1 m above the highest ground-water level.

Such a large number of soak pits is impractical on a small site; in practice a hostel of this size uses several septic tanks in parallel or a series of dispersion trenches / a small package treatment plant followed by disposal.

  • 2075 Bhadra · 8 marks

A household having 22 persons produces 135 liter/person/day of sewage. Design a septic tank and drain field to dispose the sewage in a soil having infiltration rate of 35 liters/m²/day. Assume that the septic tank is cleaned once in three years.

Answer

Given: 22 persons, 135 L/c/d, infiltration 35 L/m²/d, desludging every 3 years. Detention 24 h assumed.

Septic tank

  • Sewage flow: Q=22×135 L/c/d=2970 L/d=2.970 m3/dQ = 22 \times 135\ \text{L/c/d} = 2970\ \text{L/d} = 2.970\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=2.970 m3V_1 = 2.970\ \text{m}^3
  • Sludge digestion zone: 0.0425×22=0.935 m30.0425 \times 22 = 0.935\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×22=1.870 m3(0.085/3) \times 3 \times 22 = 1.870\ \text{m}^3
V=2.970+0.935+1.870=5.78 m3V = 2.970 + 0.935 + 1.870 = 5.78\ \text{m}^3

Adopt liquid depth d=1.5d = 1.5 m and L:B≈3:1L:B \approx 3:1.

Plan area=5.78/1.5=3.85 m2B=3.85/3≈1.2 m,L=3.85/1.2≈3.3 m\begin{aligned} \text{Plan area} &= 5.78/1.5 = 3.85\ \text{m}^2 \\ B &= \sqrt{3.85/3} \approx 1.2\ \text{m}, \quad L = 3.85/1.2 \approx 3.3\ \text{m} \end{aligned}

Check: 3.3×1.2×1.5=5.94 m3≥5.78 m33.3 \times 1.2 \times 1.5 = 5.94\ \text{m}^3 \ge 5.78\ \text{m}^3 (safe).

Septic tank: internal size 3.3 m×1.2 m×1.8 m3.3\ \text{m} \times 1.2\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 2.202.20 m long (2/3), second 1.101.10 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

Drain field (dispersion trenches)

  • Infiltration area: A=2970/35=84.86 m2A = 2970/35 = 84.86\ \text{m}^2 (trench bottom area)
  • Trench width w=0.6w = 0.6 m, so total length =84.86/0.6=141.4= 84.86/0.6 = 141.4 m
  • Maximum length per trench about 30 m, so adopt 5 trench(es), each 28.5 m long (85.5 m² provided), 0.6 m wide and 0.6-1.0 m deep, with a 1 in 200 to 1 in 300 grade, 2 m centre-to-centre spacing.

Each trench has a 0.3 m deep layer of gravel (20-40 mm) with a 100 mm perforated (or open-jointed) distribution pipe laid in it, covered with 50 mm coarse sand/gravel, a filter layer such as geotextile or straw, and then soil. Effluent is fed from a distribution box.

  • 2074 Bhadra · 8 marks

What would be the internal dimension of a septic tank and numbers of soak pits for an isolated hotel situated at mid-southern zone of Nepal having average 80 numbers of average users? Rate of sewage discharge is 210 lpcd. Cleaning period of septic tank is 3 years. Assume other necessary data if required.

Answer

Assumptions: mid-southern zone (Terai/inner Terai) with sandy loam soil, infiltration rate 50 L/m²/d, ground-water at least 5 m below the surface; detention 24 h, soak pit depth 3 m and diameter 3 m.

Septic tank

  • Sewage flow: Q=80×210 L/c/d=16800 L/d=16.800 m3/dQ = 80 \times 210\ \text{L/c/d} = 16800\ \text{L/d} = 16.800\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=16.800 m3V_1 = 16.800\ \text{m}^3
  • Sludge digestion zone: 0.0425×80=3.400 m30.0425 \times 80 = 3.400\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×80=6.800 m3(0.085/3) \times 3 \times 80 = 6.800\ \text{m}^3
V=16.800+3.400+6.800=27.00 m3V = 16.800 + 3.400 + 6.800 = 27.00\ \text{m}^3

Adopt liquid depth d=2.0d = 2.0 m and L:B≈3:1L:B \approx 3:1.

Plan area=27.00/2.0=13.50 m2B=13.50/3≈2.2 m,L=13.50/2.2≈6.2 m\begin{aligned} \text{Plan area} &= 27.00/2.0 = 13.50\ \text{m}^2 \\ B &= \sqrt{13.50/3} \approx 2.2\ \text{m}, \quad L = 13.50/2.2 \approx 6.2\ \text{m} \end{aligned}

Check: 6.2×2.2×2.0=27.28 m3≥27.00 m36.2 \times 2.2 \times 2.0 = 27.28\ \text{m}^3 \ge 27.00\ \text{m}^3 (safe).

Septic tank: internal size 6.2 m×2.2 m×2.3 m6.2\ \text{m} \times 2.2\ \text{m} \times 2.3\ \text{m} (including 0.3 m free board), in two chambers: first chamber 4.154.15 m long (2/3), second 2.052.05 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

Soak pits

  • Required infiltration area: A=Q/rate=16800/50=336.00 m2A = Q/\text{rate} = 16800/50 = 336.00\ \text{m}^2 (only the side wall below the inlet is counted; the bottom is ignored as it clogs)
  • Adopt effective depth h=3h = 3 m

A single pit would be too large, so adopt pits of D=3.0D = 3.0 m (area of each = π×3.0×3=28.27 m2\pi \times 3.0 \times 3 = 28.27\ \text{m}^2).

N=336.0028.27=11.88⇒12 pitsN = \frac{336.00}{28.27} = 11.88 \Rightarrow 12\ \text{pits}

Soak pits: 12 circular pits, each 3.0 m diameter and 3 m deep, honeycomb-lined and filled with brick bats/gravel, spaced at least 3 m (about 2 times the diameter) apart and connected through a distribution box. Soak-pit bottoms must stay at least 1 m above the highest ground-water level.

The large number of pits can be replaced by dispersion trenches or by two tanks in parallel if space is limited.

  • 2072 Magh · 8 marks

Design the septic tank and dispersion trenches in Nepalese perspectives for 20 users.

Answer

Assumptions (Nepal practice): water supply 100 L/c/d, sewage 80% of this = 80 L/c/d; detention 24 h; desludging every 3 years; infiltration rate 50 L/m²/d (sandy loam).

Septic tank

  • Sewage flow: Q=20×100×0.8 L/c/d=1600 L/d=1.600 m3/dQ = 20 \times 100\times0.8\ \text{L/c/d} = 1600\ \text{L/d} = 1.600\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=1.600 m3V_1 = 1.600\ \text{m}^3
  • Sludge digestion zone: 0.0425×20=0.850 m30.0425 \times 20 = 0.850\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×20=1.700 m3(0.085/3) \times 3 \times 20 = 1.700\ \text{m}^3
V=1.600+0.850+1.700=4.15 m3V = 1.600 + 0.850 + 1.700 = 4.15\ \text{m}^3

Adopt liquid depth d=1.5d = 1.5 m and L:B≈3:1L:B \approx 3:1.

Plan area=4.15/1.5=2.77 m2B=2.77/3≈1.0 m,L=2.77/1.0≈2.8 m\begin{aligned} \text{Plan area} &= 4.15/1.5 = 2.77\ \text{m}^2 \\ B &= \sqrt{2.77/3} \approx 1.0\ \text{m}, \quad L = 2.77/1.0 \approx 2.8\ \text{m} \end{aligned}

Check: 2.8×1.0×1.5=4.20 m3≥4.15 m32.8 \times 1.0 \times 1.5 = 4.20\ \text{m}^3 \ge 4.15\ \text{m}^3 (safe).

Septic tank: internal size 2.8 m×1.0 m×1.8 m2.8\ \text{m} \times 1.0\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 1.901.90 m long (2/3), second 0.900.90 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

Dispersion trenches

  • Infiltration area: A=1600/50=32.00 m2A = 1600/50 = 32.00\ \text{m}^2 (trench bottom area)
  • Trench width w=0.6w = 0.6 m, so total length =32.00/0.6=53.3= 32.00/0.6 = 53.3 m
  • Maximum length per trench about 30 m, so adopt 2 trench(es), each 27.0 m long (32.4 m² provided), 0.6 m wide and 0.6-1.0 m deep, with a 1 in 200 to 1 in 300 grade, 2 m centre-to-centre spacing.

Each trench has a 0.3 m deep layer of gravel (20-40 mm) with a 100 mm perforated (or open-jointed) distribution pipe laid in it, covered with 50 mm coarse sand/gravel, a filter layer such as geotextile or straw, and then soil. Effluent is fed from a distribution box.

  • 2068 Magh (old course) · 10 marks

Design a septic tank and numbers of soak pit for 15 numbers of average users. Rate of sewage discharge is 45 lpcd. Cleaning period of septic tank is 4 years interval. Assume other necessary data if required. Location of construction is in Terai area.

Answer

Assumptions: Terai has a high ground-water table (about 3 m) and flat land, so soak pits are kept shallow (effective depth 2.0 m) and infiltration rate for sandy soil is taken as 50 L/m²/d; detention 24 h; pit diameter at most 3 m.

Septic tank

  • Sewage flow: Q=15×45 L/c/d=675 L/d=0.675 m3/dQ = 15 \times 45\ \text{L/c/d} = 675\ \text{L/d} = 0.675\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=0.675 m3V_1 = 0.675\ \text{m}^3
  • Sludge digestion zone: 0.0425×15=0.638 m30.0425 \times 15 = 0.638\ \text{m}^3
  • Digested sludge storage for 4 years: (0.085/3)×4×15=1.700 m3(0.085/3) \times 4 \times 15 = 1.700\ \text{m}^3
V=0.675+0.638+1.700=3.01 m3V = 0.675 + 0.638 + 1.700 = 3.01\ \text{m}^3

Adopt liquid depth d=1.5d = 1.5 m and L:B≈3:1L:B \approx 3:1.

Plan area=3.01/1.5=2.01 m2B=2.01/3≈0.9 m,L=2.01/0.9≈2.3 m\begin{aligned} \text{Plan area} &= 3.01/1.5 = 2.01\ \text{m}^2 \\ B &= \sqrt{2.01/3} \approx 0.9\ \text{m}, \quad L = 2.01/0.9 \approx 2.3\ \text{m} \end{aligned}

Check: 2.3×0.9×1.5=3.11 m3≥3.01 m32.3 \times 0.9 \times 1.5 = 3.11\ \text{m}^3 \ge 3.01\ \text{m}^3 (safe).

Septic tank: internal size 2.3 m×0.9 m×1.8 m2.3\ \text{m} \times 0.9\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 1.551.55 m long (2/3), second 0.750.75 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

Soak pits

  • Required infiltration area: A=Q/rate=675/50=13.50 m2A = Q/\text{rate} = 675/50 = 13.50\ \text{m}^2 (only the side wall below the inlet is counted; the bottom is ignored as it clogs)
  • Adopt effective depth h=2.0h = 2.0 m
πDh=13.50⇒D=13.50π×2.0=2.15≈2.2 m\pi D h = 13.50 \Rightarrow D = \frac{13.50}{\pi \times 2.0} = 2.15 \approx 2.2\ \text{m}

Check: π×2.2×2.0=13.82 m2≥13.50 m2\pi \times 2.2 \times 2.0 = 13.82\ \text{m}^2 \ge 13.50\ \text{m}^2.

Soak pit: one circular pit, 2.2 m internal diameter and 2.0 m deep, lined with dry brick/stone masonry with open joints (honeycomb), filled with brick bats/gravel, and covered with a slab.

Because of the high water table, keep the pit bottom at least 1 m above the highest ground-water level; if this cannot be done use dispersion trenches or an evapotranspiration mound instead.

  • 2073 Magh · 4 marks

Design a septic tank for a household having average users of 15. Assume suitable data suitably.

Answer

Assumptions: water supply 100 L/c/d, sewage 80% = 80 L/c/d, detention 24 h, desludging every 3 years, digestion 0.0425 m³/person, storage 0.085 m³/person for 3 years.

  • Sewage flow: Q=15×100×0.8 L/c/d=1200 L/d=1.200 m3/dQ = 15 \times 100\times0.8\ \text{L/c/d} = 1200\ \text{L/d} = 1.200\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=1.200 m3V_1 = 1.200\ \text{m}^3
  • Sludge digestion zone: 0.0425×15=0.638 m30.0425 \times 15 = 0.638\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×15=1.275 m3(0.085/3) \times 3 \times 15 = 1.275\ \text{m}^3
V=1.200+0.638+1.275=3.11 m3V = 1.200 + 0.638 + 1.275 = 3.11\ \text{m}^3

Adopt liquid depth d=1.5d = 1.5 m and L:B≈3:1L:B \approx 3:1.

Plan area=3.11/1.5=2.07 m2B=2.07/3≈0.9 m,L=2.07/0.9≈2.4 m\begin{aligned} \text{Plan area} &= 3.11/1.5 = 2.07\ \text{m}^2 \\ B &= \sqrt{2.07/3} \approx 0.9\ \text{m}, \quad L = 2.07/0.9 \approx 2.4\ \text{m} \end{aligned}

Check: 2.4×0.9×1.5=3.24 m3≥3.11 m32.4 \times 0.9 \times 1.5 = 3.24\ \text{m}^3 \ge 3.11\ \text{m}^3 (safe).

Septic tank: internal size 2.4 m×0.9 m×1.8 m2.4\ \text{m} \times 0.9\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 1.601.60 m long (2/3), second 0.800.80 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

  • 2075 Baisakh · 8 marks

Design VIP latrine and septic tank for a family of 10 users. The detention time for septic tank is 24 hr. Sludge is cleaned in every three years.

Answer

1. VIP latrine for 10 users

Assumptions: wet pit, sludge accumulation 0.04 m³/person/year; double pit, each pit lasts 2 years; free space 0.5 m.

V=10×0.04×2=0.8 m3 per pitV = 10 \times 0.04 \times 2 = 0.8\ \text{m}^3 \text{ per pit}

Adopt pit 1.0 m×1.0 m1.0\ \text{m} \times 1.0\ \text{m} plan: effective depth 0.80.8 m + 0.5 m free space = 1.3 m (adopt 1.5 m deep). Provide two such pits, a 150 mm vent pipe 0.5 m above roof with fly screen, RCC squat slab and a 1.0×1.21.0 \times 1.2 m superstructure with a dark interior. Pits are lined with open-jointed brick/stone and the pits are at least 0.5 m apart.

2. Septic tank for 10 users

Assumed sewage rate 80 L/person/day (80% of 100 lpcd water supply), detention 24 h, desludging every 3 years.

  • Sewage flow: Q=10×80 L/c/d=800 L/d=0.800 m3/dQ = 10 \times 80\ \text{L/c/d} = 800\ \text{L/d} = 0.800\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=0.800 m3V_1 = 0.800\ \text{m}^3
  • Sludge digestion zone: 0.0425×10=0.425 m30.0425 \times 10 = 0.425\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×10=0.850 m3(0.085/3) \times 3 \times 10 = 0.850\ \text{m}^3
V=0.800+0.425+0.850=2.08 m3V = 0.800 + 0.425 + 0.850 = 2.08\ \text{m}^3

Adopt liquid depth d=1.5d = 1.5 m and L:B≈3:1L:B \approx 3:1.

Plan area=2.08/1.5=1.38 m2B=1.38/3≈0.7 m,L=1.38/0.7≈2.0 m\begin{aligned} \text{Plan area} &= 2.08/1.5 = 1.38\ \text{m}^2 \\ B &= \sqrt{1.38/3} \approx 0.7\ \text{m}, \quad L = 1.38/0.7 \approx 2.0\ \text{m} \end{aligned}

Check: 2.0×0.7×1.5=2.10 m3≥2.08 m32.0 \times 0.7 \times 1.5 = 2.10\ \text{m}^3 \ge 2.08\ \text{m}^3 (safe).

Septic tank: internal size 2.0 m×0.7 m×1.8 m2.0\ \text{m} \times 0.7\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 1.351.35 m long (2/3), second 0.650.65 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

  • 2073 Bhadra · 4 marks

Design a VIP latrine for a household with 10 numbers of people. Assume necessary data suitably.

Answer

Assumptions: wet pit (water used for cleaning), sludge accumulation 0.04 m³/person/year; double-pit system with each pit used for 2 years; free space above sludge 0.5 m; users = 10.

Pit size (each of two pits)

V=n×r×T=10×0.04×2=0.8 m3V = n \times r \times T = 10 \times 0.04 \times 2 = 0.8\ \text{m}^3

Adopt a pit 1.0 m ×\times 1.0 m in plan: depth for sludge =0.8/1.0=0.8= 0.8/1.0 = 0.8 m. Add free space 0.5 m: depth =1.3= 1.3 m, adopt 1.5 m.

Other provisions

  • Two pits of 1.0 m ×\times 1.0 m ×\times 1.5 m, 0.5 m apart, lined with open-jointed brick/stone masonry, top 0.3 m watertight.
  • RCC slab with squat hole and foot rests; hole over the second pit sealed with a lid.
  • Vent pipe: 150 mm dia (min. 110 mm) PVC pipe, 0.5 m above the roof, fly screen at top.
  • Superstructure 1.0 m ×\times 1.2 m ×\times 2.0 m, dark inside with door facing away from the sun.
  • At least 15 m from water source; pit bottom at least 1.5 m above ground-water table.
   vent pipe (150 mm, screen)
       |   ____________
       |  | superstruct |
  _____|__|____O________|__ slab
  | Pit 1 |    | Pit 2  |   1.0 x 1.0 x 1.5 m
  |  use  | 0.5 |  rest |
  • 2073 Magh · 4 marks

What is the purpose of pit privy? Describe the construction of pit privy with a neat sketch.

Answer

Purpose

A pit privy (pit latrine) is the simplest on-site sanitation facility, used where there is no sewerage or water supply for flushing. Its purpose is to collect and store human excreta in a pit so that it is isolated from people, flies and water sources, while the excreta decomposes in the pit and the liquid soaks into the soil. It prevents open defecation and the spread of faecal-oral diseases (diarrhoea, cholera, typhoid, worms).

Construction

  1. Pit: dug 1.0-1.2 m diameter or square and 2-3 m deep (size from 0.06 m³/person/year); lined with brick/stone with open joints, bamboo or wood to prevent collapse; the top 0.3 m is lined watertight and raised above ground to keep out surface water.
  2. Slab: RCC, wood or stone slab (about 1.2 m ×\times 1.2 m) over the pit with a squat hole of 0.2-0.25 m diameter with cover lid.
  3. Superstructure: about 1.0 m ×\times 1.2 m in plan, 2 m high, walls of brick, bamboo, corrugated sheet and a light roof, with a door.
  4. Location: at least 15 m from a well/spring, 6 m from a house, and downstream of water sources; pit bottom at least 1.5 m above ground-water.
  5. When the pit fills to 0.5 m from the slab, it is covered with soil and a new pit is dug.
    ______________
   | superstructure|
   |___ O (hole)___|__ slab, 1.2 x 1.2
 ##|               |##  <- raised collar
   |  pit 2-3 m    |
   | (lined,       |
   |  open joints) |
   |_______________|
  • 2071 Magh · 4 marks

Describe the purpose and construction of an evapo-transpiration mound.

Answer

Purpose

An evapo-transpiration (ET) mound is an above-ground sand-fill bed for disposing septic tank effluent where soil is shallow, has a high ground-water table, is rocky, or has a very slow infiltration rate, so that a soak pit or trench cannot work. The effluent is spread through the mound and is lost mainly by evaporation from the surface and transpiration by plants growing on it, with the rest percolating into the soil below.

Construction

  1. The site is ploughed, and a fill of clean coarse sand (and gravel) is built to form a mound about 0.5-1.0 m high with side slopes of 1:3 or flatter. The sand bed is placed on the loosened natural ground.
  2. Septic tank effluent is fed through a distribution box into perforated distribution pipes (75-100 mm dia) laid in a gravel bed (15-20 cm) inside the sand.
  3. The gravel is covered with a layer of geotextile/straw, then 0.3 m of topsoil, and the surface is grassed or planted with high-transpiring plants (banana, bamboo, canna etc.).
  4. The area is sized from effluent flow and local evaporation: plan area ≈Q/(infiltration + evapotranspiration rate)\approx Q/(\text{infiltration + evapotranspiration rate}).
  5. Surface runoff is kept off with a diversion drain and the mound is fenced.
     plants / grass
    __________________
   /  topsoil 0.3 m   \
  /--------------------\
 /  gravel + distr.pipe \
/  coarse sand fill      \
=========================== natural ground
   (poor or rocky soil)

Questions from Old Question Collection (CE 656) (IOE BCE exam papers (CE 656) from 2068 to 2081, 21 papers). Answers are written for this site; check them against your class notes.

Chapter titles and hours from the IOE syllabus ↗