Reservoirs and Distribution System

Layout means the arrangement of mains, sub-mains and branches in a distribution system. The four common types are dead end, grid iron, circular (ring) and radial. The best layout depends on the road pattern, topography and size of the town, so different layouts are recommended for different places.

1. Dead end (tree) system

 R ====+====+====+====   Branches end at dead ends
       |    |    |

• Merits: simple design; low cost (less pipe and valves); suited to irregular towns; discharge in each pipe is known.
• Demerits: stagnation at dead ends; whole area beyond a break loses water; poor pressure at ends; limited fire flow.

2. Grid iron (reticulation) system

 R ==+====+====+==
     |    |    |
   --+----+----+--   interconnected grid
     |    |    |

• Merits: water circulates, no dead ends; supply continues during repair; uniform pressure; good fire flow.
• Demerits: difficult design (Hardy-Cross); high cost (more pipes and valves); needs skilled maintenance.

3. Circular (ring) system

      +------------+
      |  Main ring |
 R ===+--+----+----+  inner sub-mains
      |  |    |    |
      +------------+

• A closed ring main runs round the town, with sub-mains inside it. Suited to well planned towns with a circular or square pattern.
• Merits: each point can be fed from two directions, so supply continues during repair; good pressure and fire flow.
• Demerits: more pipe length and valves; costly; design difficult.

4. Radial system

        \  |  /
         \ | /
      ----( R )----   water flows outwards
         / | \
        /  |  \

• The service reservoir is at the centre and pipes radiate outwards along the roads. Zones are fed from separate reservoirs.
• Merits: quick supply and high pressure; the design is simple; small pipe sizes.
• Demerits: many reservoirs are needed; more cost; works only for planned radial towns.

Why different layouts?

A town may be planned and regular (grid or ring), haphazardly grown (dead end), hilly with a linear pattern, or large with several zones (radial). Cost, reliability, pressure and ease of maintenance have to be balanced against each other, so no single layout suits every town.

In the continuous system water is available in the pipes for all 24 hours of the day, while in the intermittent system water is supplied only for fixed hours (for example 2–4 hours in the morning and evening) and the pipes are empty or shut for the rest of the day.

Merits of the continuous system

1. Safe, because positive pressure always prevents contamination.
2. Water is available for fire fighting at any time.
3. Consumers get water at any hour, without household storage.
4. Smaller pipe sizes are enough.

Demerits of the continuous system

1. Wastage and leakage can be heavy, so a larger source and metering are needed.
2. Needs a larger quantity of water and a higher cost of treatment.
3. Burst pipes cause large losses before they are detected.

Merits of the intermittent system

1. Less water is wasted, so a smaller source is enough.
2. Needs less treated water, so it suits places where the source is scarce.

Demerits of the intermittent system

1. Contamination risk during empty periods; poor water quality.
2. Large pipes, valves and fittings are needed to deliver the daily demand in a few hours.
3. Consumers must store water; no water at hydrants during the off period.
4. Frequent wetting and drying of pipes causes corrosion and damage; air in the pipes gives false meter readings.

The four layouts differ in suitability, and each has strong and weak points. Nepalese towns usually grow in an unplanned way, so the traditional dead end layout is often improved by looping and zoning.

Types, suitability, pros and cons

Key improvements over the traditional (dead end) layout

1. Loop the ends: connect neighbouring dead ends to form loops, so water circulates and flow comes from two directions.
2. Scour/flushing valves at every dead end, and regular flushing.
3. Sectional (isolating) valves so that a repair cuts off only a small area.
4. Zoning (district metered areas): divide the town into pressure zones with separate reservoirs or pressure-reducing valves, so that pressure is even and leakage is easy to measure.
5. Ring main around the town centre, fed from the reservoir, with branches off it.
6. Replace small, old pipes with larger HDPE/DI pipes where the pressure is low and add booster pumps if needed.
7. House connection metering and leakage control to cut waste.
8. Hydrants placed at regular spacing with adequate flow for fire fighting.
9. Use of hydraulic modelling and GIS maps for design and operation.

A water supply system consists of intake, treatment plant, pumps, pipes, reservoirs and valves, all of which deteriorate with time and use. Maintenance keeps it working properly.

Why maintenance is necessary

1. To give a continuous supply of safe, wholesome water at proper pressure.
2. To prevent breakdown, and to increase the life of the structures and machines.
3. To reduce water loss through leakage and save treatment and pumping cost.
4. To protect public health from contamination (cross connection, broken pipes).
5. It is cheaper to do regular maintenance than to repair major failures.

Types of maintenance

1. Regular (routine and preventive) maintenance - planned, done at fixed intervals.

• Cleaning of intake screens and catchment, removal of silt from sedimentation tanks, cleaning/backwashing of filters (sand scraping for slow sand filters).
• Checking and refilling chemical dosing (alum, chlorine) and residual chlorine tests.
• Greasing and oiling of pumps and valves; operating (exercising) sluice valves to keep them from jamming.
• Flushing of mains and dead ends; cleaning of reservoirs once or twice a year.
• Painting of exposed steel pipes and tanks; checking meters; leak detection survey.
• Keeping records and maintaining the catchment area fence.

2. Emergency (breakdown/corrective) maintenance - unplanned, done after a failure.

• Repair of burst or leaking pipes and replacing joints.
• Repair of pumps and motors after breakdown; arrangements for standby generator.
• Repair of the intake or the spring chamber after a flood or landslide.
• Disinfection after contamination, and supply of water by tankers.

An emergency team with spare pipes, fittings, tools and transport should always be kept ready, and the community should be trained and fees collected to pay for repairs (water user committees in Nepal).

Distribution system design is the process of finding the layout, pipe sizes and appurtenances that supply the required quantity of water at the required pressure to all consumers at least cost.

Factors to consider

1. Population now and the design year; growth rate and settlement pattern.
2. Per capita demand, peak factors, fire demand and institutional/commercial needs.
3. Topography and ground levels; position of source and the reservoir.
4. Road pattern and the town plan.
5. Available head (gravity or pumping) and required residual pressure.
6. Cost, availability of pipe materials, and ease of operation and maintenance.
7. Future expansion and water quality (to avoid stagnation).

Procedure

1. Collect data: map with contours, road plan, population, source yield and level.
2. Forecast the population for the design period and find the demand: average daily, maximum day (1.5–2 × average) and peak hour (about 3 × average, or 2–3 × maximum day).
3. Select the system: gravity, pumping or dual; continuous or intermittent.
4. Choose the layout (dead end, grid iron, ring or radial) as per the town.
5. Fix the location and capacity of the service reservoir (balancing + fire + emergency storage).
6. Divide the area into zones and find the discharge in each pipe for the peak demand.
7. Size the pipes with Hazen-Williams (or Darcy-Weisbach); for looped networks use Hardy-Cross or software.
8. Check velocity and pressure everywhere; revise the sizes until the criteria are satisfied.
9. Provide valves, hydrants, air valves, scour valves, break pressure tanks and house connections.
10. Prepare drawings, estimate the cost, and compare the alternatives.

Design criteria

• Velocity: 0.6–3.0 m/s (minimum 0.6 to prevent silting; maximum limited to avoid water hammer and erosion).
• Residual head: at least 10–15 m at the tap (about 7–10 m for rural stand posts); maximum static pressure about 60–70 m, otherwise use BPT or pressure-reducing valves.
• Minimum pipe diameter: 100 mm for fire hydrant lines in towns; 15–25 mm for rural branches.
• Hydrants every 100–150 m; sectional valves at every junction.
• Pipe laid with at least 0.75–1 m cover.

Given data

• Daily requirement = 0.25 MLD = 250 m³/day
• Pumping at 0.032 ML/h = 32 m³/h from 8:00 to 16:00 (8 h), so pumped volume = 256 m³
• Demand: 7–8 h = 30% (75 m³), 8–17 h = 35% (87.5 m³), 17–18.5 h = 30% (75 m³), 18.5–7 h = 5% (12.5 m³)

Hourly demand rates

• 18.5–7 h (12.5 h): 12.5/12.5 = 1.0 m³/h
• 7–8 h: 75 m³/h
• 8–17 h (9 h): 87.5/9 = 9.72 m³/h
• 17–18.5 h (1.5 h): 75/1.5 = 50 m³/h

Mass-curve table (analytical method)

The cumulative at 24 h is +6.0 m³, which is the small excess (256 − 250) of pumping over demand.

Capacity

• Maximum cumulative deficit = 82.00 m³ (at 8:00)
• Maximum cumulative surplus = 96.22 m³ (at 16:00)
• Balancing capacity = 82.00 + 96.22 = 178.22 m³

Answer: Required balancing reservoir capacity ≈ 178 m³ (0.178 ML), i.e. about 71% of the daily demand. The tank must hold 82 m³ at midnight so that the 7–8 h peak can be met before pumping starts.

Layout preferred

For a haphazardly growing city of Nepal, the dead end (tree) system is the most suitable. Roads are narrow, irregular and unplanned, the ground is hilly, and funds are limited. A tree system follows the existing roads and lanes, and can be extended as the town grows, without a regular pattern. It is cheap and simple to design.

   Reservoir
      ||
   ===++===========+====== main
      ||   |       |   |
      ||   +-- b1  |   +-- b3   (branches end
      ||           +-- b2         in dead ends)

Advantages

1. Simple design and low cost (shorter pipe length, fewer valves).
2. Fits irregular roads and hilly land; follows existing lanes.
3. Easy to extend branch by branch as new houses come up.
4. Discharge in each pipe is easily calculated.

Disadvantages

1. Stagnant water at dead ends, with taste, odour and bacterial growth.
2. A break in a pipe cuts the supply to all consumers beyond it.
3. Low pressure at the ends, and limited fire fighting flow.
4. Extension reduces pressure in the existing branches.

Improvement with minimum works

1. Provide scour valves (blow-offs) at the dead ends and flush weekly.
2. Join the nearby dead ends with small connecting pipes, which forms loops and creates circulation (a low-cost partial grid).
3. Install sectional valves at the junctions, so that a repair isolates only a small area.
4. Divide the town into zones fed from separate tanks, or use break pressure tanks on steep land.
5. Use pressure-reducing or flow control valves, and fix leaks quickly.

Service (distribution) reservoirs are storage structures built near the distribution area, to store treated water and supply it to the consumers at the required pressure during demand variations. A clear water reservoir is the tank at the treatment plant that stores the filtered, disinfected water.

Purposes

1. Balancing storage: supply is at a uniform rate, but demand varies hourly; the reservoir stores the surplus and supplies the deficit.
2. Emergency storage: supply continues during breakdown of pumps, power failure or repair of the main.
3. Fire reserve: keeps water for fire fighting.
4. Maintains pressure in the system (elevated or hill-top tanks give pressure by gravity).
5. Lets the treatment plant and pumps work at a steady rate, which saves cost.

Types

• Underground or ground-level (on a hill), RCC, masonry or steel
• Elevated tanks (towers) for flat towns

Construction

• Site: as high as possible and near the centre of the demand area.
• Walls and floor: RCC (M20 or richer) or brick/stone masonry with cement plaster; watertight, with floor sloping to the washout.
• Roof: RCC slab or dome, covered with earth, to prevent contamination and keep the water cool.
• Compartments: divided into two cells so one can be cleaned while the other works.
• Fittings: inlet, outlet (above the floor), overflow, washout (scour) pipe, ventilation pipes, manhole with cover, ladder, level indicator and water-level gauge.

        vent        manhole
   ___ ___|___________|____ roof
  |                          |-- overflow
  | inlet-> water level ~~~~ |
  |                          |
  |__________________________|-- outlet
            washout   floor sloped

Clear water reservoir (CWR)

It is built at the treatment works after filtration, usually partly underground, with baffles. It stores the filtered water, gives chlorine contact time (30 minutes or more), supplies the pumps for back washing and for pumping to the town, and balances the difference between plant output and pumping rate. Its construction is the same as above (RCC, roof, baffles, manholes, vents, inlet/outlet, overflow, washout).

A service connection is the pipe that carries water from the street distribution main to the tap or meter inside a private building.

 Street                 Boundary         House
 =====Main=====+=========|==M==|=====>  tap
   (CI/DI)  ferrule  stop cock meter
                  \ goose
                   neck

Components and their use

The service line is laid with a small slope, with all joints made leak-tight; the meter is kept in a protective box.

Given data

• Design-year demand = 20,000 L/day = 20 m³/day
• Safe yield of river = 0.25 lps = 0.25 × 86,400 = 21,600 L/day = 21.6 m³/day, i.e. 0.9 m³/h continuous inflow
• Consumption: 5–7 h 25%, 7–12 h 35%, 12–17 h 20%, 17–19 h 20%, 19–5 h 0%

Check of source and need of storage

The source yield (21.6 m³/day) exceeds the demand (20 m³/day), so the source is adequate for the day. But the demand rate in 5–7 h is 5/2 = 2.5 m³/h and in 7–12 h it is 1.4 m³/h, both more than the inflow of 0.9 m³/h. A balancing tank is therefore necessary.

Mass-curve table (analytical method)

Inflow: 0.9 m³/h for 24 h. Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 20.00 m³; total inflow = 21.60 m³.

Capacity

• Maximum cumulative surplus = 1.60 m³
• Maximum cumulative deficit = 7.40 m³
• Balancing capacity = max. surplus + max. deficit = 1.60 + 7.40 = 9.00 m³
• Because the inflow in a day (21.60 m³) is more than the demand, the extra water simply overflows once the tank is full. The minimum storage that still meets every demand is the largest draw-down, 7.40 m³. The textbook value 9.00 m³ is on the safe side.

Water level in the tank

The tank is empty at 19:00, when the cumulative deficit is largest (−7.40 m³), and full (9.00 m³) at 05:00. So storage at any time = cumulative value + 7.40 m³.

• At 10:00: cumulative = −3.20 + (3/5)(−2.50) = −4.70, storage = 2.70 m³
• At 18:00 (6 PM): cumulative = −5.20 + (1/2)(−2.20) = −6.30, storage = 1.10 m³

For a tank of plan area $A$ (m²): level at 10:00 $= 2.70/A$ m and level at 6 PM $= 1.10/A$ m. For example, with a 3 m × 3 m tank ($A = 9$ m²), these are 0.30 m and 0.12 m above the floor.

Answer: Balancing tank capacity ≈ 9 m³ (practical 10 m³). Water depth = 2.7/A m at 10:00 and 1.1/A m at 6 PM.

Residual pressure at each tap = 1.1 kg/cm² = 11 m of water. Static water level at A = RL 985 m, so the HGL at A is 985 m. The figure gives only pipes AB and BC; for pipe AD the data of the companion problem is assumed (L = 160 m, Q = 0.5 lps, tap RL 940 m).

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

Design

Pipe AB

• Discharge $Q = 0.350$ lps, length $L = 120$ m
• HGL at A = 985.00 m; ground RL at B = 960 m; residual head required = 11 m
• Available head $H = 985.00 - 960 - 11 = 14.00$ m
• Gradient $S = H/L = 14.00/120 = 0.11667$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 21.3$ mm, adopt commercial size 25 mm
• Actual $h_f = 6.383$ m, HGL at B = 978.62 m, residual head = 18.62 m (≥ 11 m, OK)
• Velocity $v = Q/A = 0.71$ m/s

Pipe BC

• Discharge $Q = 0.150$ lps, length $L = 150$ m
• HGL at B = 978.62 m; ground RL at C = 920 m; residual head required = 11 m
• Available head $H = 978.62 - 920 - 11 = 47.62$ m
• Gradient $S = H/L = 47.62/150 = 0.31745$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 12.6$ mm, adopt commercial size 15 mm
• Actual $h_f = 19.990$ m, HGL at C = 958.63 m, residual head = 38.63 m (≥ 11 m, OK)
• Velocity $v = Q/A = 0.85$ m/s

Pipe AD

• Discharge $Q = 0.500$ lps, length $L = 160$ m
• HGL at A = 985.00 m; ground RL at D = 940 m; residual head required = 11 m
• Available head $H = 985.00 - 940 - 11 = 34.00$ m
• Gradient $S = H/L = 34.00/160 = 0.21250$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 21.5$ mm, adopt commercial size 25 mm
• Actual $h_f = 16.474$ m, HGL at D = 968.53 m, residual head = 28.53 m (≥ 11 m, OK)
• Velocity $v = Q/A = 1.02$ m/s

Summary

Maximum static head: B = 985 − 960 = 25 m, D = 45 m and C = 985 − 920 = 65 m. Pipe BC must therefore be of a class that can bear at least 65 m (about 0.65 MPa), e.g. GI medium or HDPE PN 10, while AB can be a lower class.

Answer: AB = 25 mm, BC = 15 mm, AD = 25 mm (GI), all giving residual head above 11 m and velocity below 1.1 m/s.

Given data

• Population 1,300,000 × 110 lpcd = 143,000 m³/day
• Pumping at a uniform rate from 5 AM to 6 PM (13 h): rate = 143,000/13 = 11,000 m³/h
• Fire demand neglected

Mass-curve table (analytical method)

Inflow: 11,000 m³/h during 5–18 h. Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 143000.00 m³; total inflow = 143000.00 m³.

Capacity

• Maximum cumulative surplus = 21450.00 m³
• Maximum cumulative deficit = 23650.00 m³
• Balancing capacity = max. surplus + max. deficit = 21450.00 + 23650.00 = 45100.00 m³

Answer: Balancing reservoir capacity = 45,100 m³ (31.5% of the daily demand).

Given data

• Population 1,200,000 × 45 lpcd = 54,000 m³/day
• Pumping in two shifts, 5–10 h and 17–20 h (8 h in all), at a rate 54,000/8 = 6,750 m³/h
• Fire storage = 1% of the total daily demand

Mass-curve table (analytical method)

Inflow: 6,750 m³/h during 5–10 h and 17–20 h. Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 54000.00 m³; total inflow = 54000.00 m³.

Capacity

• Maximum cumulative surplus = 8910.00 m³
• Maximum cumulative deficit = 9450.00 m³
• Balancing capacity = max. surplus + max. deficit = 8910.00 + 9450.00 = 18360.00 m³
• Fire storage = 1% of daily demand = 0.01 × 54000 = 540.00 m³
• Total capacity = 18360.00 + 540.00 = 18900.00 m³

Answer: Balancing capacity 18,360 m³ + fire storage 540 m³ = 18,900 m³ total.

The grid iron (reticulation) system is a layout in which the mains, sub-mains and branches are interconnected in a rectangular grid, so that every pipe is fed from at least two directions and there are no dead ends.

  Main (from reservoir)
   ||====+======+======+=====
   ||    |      |      |
   ||====+======+======+=====   sub-mains
   ||    |      |      |
   ||====+======+======+=====   branches
   Valves at every junction

Description

• The main runs along the principal road, sub-mains branch off it at right angles, and branches connect the sub-mains. Pipes meet at junctions to form loops.
• Suitable for well planned, regularly laid out towns with a rectangular road pattern.
• Design is done by loop methods such as Hardy-Cross because flow can reach a point by more than one path.

Advantages

1. Water reaches any point from two or more sides, so there is no stagnation and quality is good.
2. If one pipe is closed for repair, the rest of the area still gets water (only a short section is cut off by sectional valves).
3. Pressure is more uniform and losses are lower, because the flow divides between several paths.
4. Good discharge for firefighting from several directions.
5. Easy to extend the network.

Disadvantages

1. Design is difficult and needs trial-and-error methods (Hardy-Cross) or software.
2. More pipe length and many valves are needed, so the cost is high.
3. Many cross-connections, so it is hard to find the leaks and the burst section.
4. Not suited to irregular, haphazardly grown towns.
5. Needs skilled laying and maintenance.

The dead end (tree) system is a layout in which one main runs from the reservoir through the town and sub-mains branch from it like a tree; each branch ends at a dead end and has no connection with other branches.

         Sub-mains / branches
              |   |   |
  Reservoir ==============+=== Main
              |   |   |   |
           branches end (dead end)

Merits

1. Simple to design: the discharge in each pipe is known, so sizes are fixed without trial and error.
2. Fewer valves and less pipe length are needed, so the cost is low.
3. Easy to lay and maintain; the discharge and head loss are easily found.
4. Suited to towns that have grown without a planned pattern and to hilly or linear settlements, where it follows the roads.
5. Individual pipe diameters are economical, as the pipe size decreases away from the main.

Demerits

1. Water in the dead ends is stagnant, which gives bad taste, bacterial growth and sediment deposits. Scour/blow-off valves are needed at the ends and must be flushed regularly.
2. If a pipe breaks or is shut for repair, all consumers beyond that point lose supply.
3. Pressure at the far ends is low, and it varies with the demand on other branches.
4. Discharge for fire fighting is available from one direction only, so it is limited.
5. Extension of the system reduces the pressure in existing branches.

General considerations in planning a distribution system

1. Demand: population and demand for the design period, with peak factors and fire demand.
2. Source and topography: position of the source and reservoir, and ground levels, to decide between gravity, pumping or a combination.
3. Layout: choose dead end, grid iron, ring or radial layout to suit the roads and the town plan.
4. Pressure: keep the residual head at least 10–15 m at the tap (7–10 m for rural stand posts) and the maximum pressure about 60–70 m, using zoning or BPT if more.
5. Velocity: 0.6–3 m/s in the pipes.
6. Reservoir: provide balancing, emergency and fire storage at a suitable high point close to the demand.
7. Appurtenances: sectional valves, air valves, scour valves at low points and dead ends, and hydrants.
8. Quality protection: avoid dead ends, cross connections and low spots where water stagnates.
9. Materials and cost: pipe material, class and cost; ease of operation and maintenance; future expansion.

When intermittent supply is recommended

1. When the source yield is limited and cannot provide for the continuous 24 h supply (for example in dry-season springs).
2. When the pumping is costly or the power supply is available only for a few hours.
3. When the head available in the system is small and only part of the area can be given water at a time (zone-by-zone rotation).
4. When the treatment capacity is small, or in the early years when the network is incomplete.
5. In small or rural towns where consumers can store water without difficulty.

Drawbacks of the intermittent system

1. Pipes are empty during the off period, so polluted water can be sucked in through leaks, and quality is poor.
2. Larger pipes, valves and meters are required, because the whole day's demand is delivered in a few hours.
3. Consumers must store water, which wastes space and can become contaminated.
4. No water for fire fighting during the off period.
5. Frequent pressure change and wet–dry cycles cause corrosion and pipe damage; air in the pipes distorts meter readings.
6. Supply is unfair to people at the far ends or on high ground.

Water is delivered from the service reservoir to the consumers by one of three methods, depending on the position of the source and the ground levels.

1. Gravity system

The source or the service reservoir is at a level high enough above the town, so that water flows by gravity at the required pressure.

 Source/Reservoir
   \
     \  gravity flow
       \_______ town

• Merits: most economical and reliable, since there is no pumping and no power cost; simple to operate; very low maintenance; pressure is steady.
• Demerits: possible only where a high source exists; pressure at low ground may be too high so BPT or PRV are needed; large initial pipe cost for long transmission lines.

2. Pumping system (direct pumping)

Treated water is pumped directly into the mains without any storage; the pumps run according to the demand.

• Merits: high pressure can be achieved easily, useful for fire fighting; the reservoir cost is saved; suited for flat towns.
• Demerits: pumps must run all the time with varying load, so the cost is high and efficiency is low; any breakdown or power failure stops the supply; standby pumps are needed; pressure fluctuates with the demand.

3. Combined (dual) system - pumping with storage

Pumps lift water at a uniform rate into an elevated service reservoir, which supplies the town by gravity and also receives the excess flow in the low-demand hours.

 Source -> Pump -> [Elevated reservoir] -> distribution
                        |  (surplus stored in the off-peak hours)

• Merits: the pumps work at a constant, efficient rate; the supply continues for a time even if the pump fails; peak demands and fire are met from storage; the pump size and the running cost are less. This is the most common system.
• Demerits: needs the cost of the reservoir and also of the pumps and power; more complex operation and maintenance.

Choice

The gravity system is preferred where the topography allows; the combined system is the most practical in flat towns; direct pumping is used only for small schemes or as a temporary arrangement.

Given data

• Population 800, demand 45 lpcd: daily demand = 800 × 45 = 36,000 L = 36 m³/day
• Spring yield = 0.5 lps = 0.5 × 3600 = 1.8 m³/h = 43.2 m³/day
• Demand pattern (% of the daily demand) for six 4-hour periods (the table gives % although the heading says million litres)

Check of source and need of storage

The yield 43.2 m³/day is more than the demand of 36 m³/day, so the source is enough for the whole day. But in 4–8 h the demand is 30% × 36 = 10.8 m³ in 4 h = 2.7 m³/h, and in 8–12 h it is 2.25 m³/h; both exceed the inflow of 1.8 m³/h. Therefore a reservoir is necessary.

Mass-curve table (analytical method)

Inflow: 1.8 m³/h (spring yield). Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 36.00 m³; total inflow = 43.20 m³.

Capacity

• Maximum cumulative surplus = 7.20 m³
• Maximum cumulative deficit = 0.00 m³
• Balancing capacity = max. surplus + max. deficit = 7.20 + 0.00 = 7.20 m³
• Because the inflow in a day (43.20 m³) is more than the demand, the extra water simply overflows once the tank is full. The minimum storage that still meets every demand is the largest draw-down, 5.40 m³. The textbook value 7.20 m³ is on the safe side.

Answer: Reservoir is necessary; balancing capacity ≈ 7.2 m³ (adopt 8 m³). The cumulative mass curve rises to +7.2 m³ because the spring over-supplies in the night hours; if the overflow is wasted the smallest storage that meets every demand is 5.4 m³, so 7.2 m³ is on the safe side.

Given data

• Daily demand = 80,000 L = 80 m³/day
• Spring safe yield = 0.8 lps = 0.8 × 86,400 = 69,120 L/day = 69.12 m³/day (2.88 m³/h)
• Demand: 5–7 h 20%, 7–12 h 30%, 12–17 h 20%, 17–19 h 25%, 19–5 h 5%

Check of source and need of storage

The yield (69.12 m³/day) is less than the demand (80 m³/day), a shortage of 10.88 m³/day (13.6%). A storage tank cannot remove this shortage; another source or a reduced demand is needed. For balancing the hourly variation, the inflow is taken equal to the demand (80/24 = 3.33 m³/h). The maximum hourly demand is 8 m³/h (5–7 h) and 5.33 m³/h (17–19 h), more than the inflow, so a balancing tank is necessary.

Mass-curve table (analytical method)

Inflow: 3.33 m³/h (daily demand spread over 24 h). Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 80.00 m³; total inflow = 80.00 m³.

Capacity

• Maximum cumulative surplus = 0.00 m³
• Maximum cumulative deficit = 29.33 m³
• Balancing capacity = max. surplus + max. deficit = 0.00 + 29.33 = 29.33 m³

Answer: Balancing tank is necessary; capacity ≈ 29.3 m³ (adopt 30 m³), provided the source shortage of 10.9 m³/day is made up.

Given data

• Daily demand = 20,000 L = 20 m³/day
• Spring yield = 0.25 lps = 21,600 L/day = 21.6 m³/day = 0.9 m³/h
• Demand: 5–7 h 25%, 7–12 h 35%, 12–17 h 20%, 17–19 h 20%, 19–5 h 0%

Check of source and need of storage

The yield (21.6 m³/day) exceeds the demand (20 m³/day), so the source is adequate. The demand rate in 5–7 h is 2.5 m³/h and in 7–12 h it is 1.4 m³/h, which are greater than the inflow rate 0.9 m³/h. The balancing tank is therefore necessary.

Mass-curve table (analytical method)

Inflow: 0.9 m³/h (spring yield). Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 20.00 m³; total inflow = 21.60 m³.

Capacity

• Maximum cumulative surplus = 1.60 m³
• Maximum cumulative deficit = 7.40 m³
• Balancing capacity = max. surplus + max. deficit = 1.60 + 7.40 = 9.00 m³
• Because the inflow in a day (21.60 m³) is more than the demand, the extra water simply overflows once the tank is full. The minimum storage that still meets every demand is the largest draw-down, 7.40 m³. The textbook value 9.00 m³ is on the safe side.

Answer: Balancing tank is necessary because peak demand exceeds the continuous inflow; capacity ≈ 9 m³ (adopt 10 m³).

Given data

• Daily demand = 22,000 L = 22 m³/day
• River safe yield = 0.25 lps = 21.6 m³/day = 0.9 m³/h
• Demand: 5–7 h 25%, 7–12 h 35%, 12–17 h 20%, 17–19 h 20%, 19–5 h 0%

Check of source and need of storage

The printed question gives no explicit question; the capacity of the balancing tank is found. The safe yield (21.6 m³/day) is slightly less than the demand (22 m³/day), a deficit of 0.4 m³/day (1.8%), which a tank cannot remove (it needs a small extra supply or a demand control). The hourly demand rates (2.75 and 1.54 m³/h) exceed the inflow of 0.9 m³/h, so the tank is necessary.

Mass-curve table (analytical method)

Inflow: 0.9 m³/h (safe yield). Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 22.00 m³; total inflow = 21.60 m³.

Capacity

• Maximum cumulative surplus = 0.00 m³
• Maximum cumulative deficit = 9.40 m³
• Balancing capacity = max. surplus + max. deficit = 0.00 + 9.40 = 9.40 m³
• Because the inflow in a day (21.60 m³) is more than the demand, the extra water simply overflows once the tank is full. The minimum storage that still meets every demand is the largest draw-down, 11.00 m³. The textbook value 9.40 m³ is on the safe side.

Answer: Balancing tank capacity ≈ 9.4 m³ (adopt 10 m³). If the inflow were 22/24 = 0.917 m³/h (shortage made up), the capacity would be 9.17 m³, nearly the same.

Given data

• Population 3,000,000 × 110 lpcd = 330,000 m³/day
• Pumping 10 h/day (5–10 h and 14–19 h), at a constant rate of 330,000/10 = 33,000 m³/h
• Supply from the tank to the city is continuous, with the % consumption given

Mass-curve table (analytical method)

Inflow: 33,000 m³/h during 5–10 h and 14–19 h only (pumped). Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 330000.00 m³; total inflow = 330000.00 m³.

Capacity

• Maximum cumulative surplus = 82500.00 m³
• Maximum cumulative deficit = 0.00 m³
• Balancing capacity = max. surplus + max. deficit = 82500.00 + 0.00 = 82500.00 m³

Answer: Balancing reservoir capacity = 82,500 m³ (25% of the daily demand).

Given data

• Population 150,000 × 110 lpcd = 16,500 m³/day
• Pumping 6–9 h and 18–21 h (6 h), at a rate 16,500/6 = 2,750 m³/h
• Fire demand neglected

Mass-curve table (analytical method)

Inflow: 2,750 m³/h during 6–9 h and 18–21 h. Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 16500.00 m³; total inflow = 16500.00 m³.

Capacity

• Maximum cumulative surplus = 3300.00 m³
• Maximum cumulative deficit = 3300.00 m³
• Balancing capacity = max. surplus + max. deficit = 3300.00 + 3300.00 = 6600.00 m³

Answer: Service reservoir capacity = 6,600 m³ (40% of the daily demand).

Given data

• Daily demand = 10,000 m³/day
• River water enters the reservoir by gravity at a uniform rate: 10,000/24 = 416.67 m³/h
• Demand: 5–7 h 25%, 7–12 h 30%, 12–17 h 15%, 17–19 h 20%, 19–5 h 10%

Mass-curve table (analytical method)

Inflow: 416.67 m³/h for 24 h. Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 10000.00 m³; total inflow = 10000.00 m³.

Capacity

• Maximum cumulative surplus = 0.00 m³
• Maximum cumulative deficit = 3166.67 m³
• Balancing capacity = max. surplus + max. deficit = 0.00 + 3166.67 = 3166.67 m³

Water level in the reservoir

The reservoir is empty at 19:00 (largest deficit −3,166.67 m³) and full (3,166.67 m³) at 05:00. Storage at any time = cumulative value + 3,166.67 m³.

The plan area $A$ of the reservoir is not given, so the level is storage ÷ $A$. For example, if the tank has 4 m water depth for 3,167 m³, $A = 3167/4 = 792$ m², giving levels of 2.95 m, 0.74 m, 0.74 m and 2.00 m.

Answer: Service reservoir capacity ≈ 3,167 m³ (31.7% of the daily demand).

Given data

• Population 100,000 × 110 lpcd = 11,000 m³/day
• River water flows in by gravity at a uniform rate: 11,000/24 = 458.33 m³/h
• Tank diameter = 30 m, area = π × 30²/4 = 706.86 m²
• Demand: 5–7 h 25%, 7–12 h 30%, 12–17 h 15%, 17–19 h 20%, 19–5 h 10%

Mass-curve table (analytical method)

Inflow: 458.33 m³/h for 24 h. Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 11000.00 m³; total inflow = 11000.00 m³.

Capacity

• Maximum cumulative surplus = 0.00 m³
• Maximum cumulative deficit = 3483.33 m³
• Balancing capacity = max. surplus + max. deficit = 0.00 + 3483.33 = 3483.33 m³

Depth and water level

• Water depth for full capacity $= 3483.33/706.86 = 4.93$ m (adopt about 5 m plus 0.5 m free board)
• The reservoir is empty at 19:00 and full at 05:00. At 12:00 the cumulative value is −2,841.67 m³ and storage = −2,841.67 + 3,483.33 = 641.67 m³
• Water level at 12 h $= 641.67/706.86 = 0.91$ m above the floor

Answer: Capacity ≈ 3,483 m³; water level in the tank at 12 h ≈ 0.91 m.

Given data

• Daily demand = 20,000 L = 20 m³/day
• Intermittent supply in 7–10 h and 17–20 h (6 h in all): inflow rate = 20/6 = 3.33 m³/h
• Demand: 5–7 h 20%, 7–12 h 35%, 12–17 h 10%, 17–20 h 25%, 20–5 h 10%

Mass-curve table (analytical method)

Inflow: 3.33 m³/h during 7–10 h and 17–20 h only. Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 20.00 m³; total inflow = 20.00 m³.

Capacity

• Maximum cumulative surplus = 2.00 m³
• Maximum cumulative deficit = 4.00 m³
• Balancing capacity = max. surplus + max. deficit = 2.00 + 4.00 = 6.00 m³

Answer: Balancing reservoir capacity = 6 m³ (30% of the daily demand).

Given data

• Population 500 × 45 lpcd = 22,500 L/day = 22.5 m³/day
• Spring yield = 0.28 lps = 24,192 L/day = 24.19 m³/day = 1.008 m³/h
• Demand: 5–7 h 15%, 7–12 h 45%, 12–17 h 10%, 17–19 h 20%, 19–5 h 10%

Check of source and need of storage

The yield (24.19 m³/day) is more than the demand (22.5 m³/day), so the source is adequate. The demand rate in 7–12 h is 0.45 × 22.5/5 = 2.03 m³/h, and in 5–7 h it is 1.69 m³/h, both more than the inflow rate of 1.008 m³/h. So a balancing reservoir is necessary.

Mass-curve table (analytical method)

Inflow: 1.008 m³/h (spring yield). Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 22.50 m³; total inflow = 24.19 m³.

Capacity

• Maximum cumulative surplus = 1.69 m³
• Maximum cumulative deficit = 6.44 m³
• Balancing capacity = max. surplus + max. deficit = 1.69 + 6.44 = 8.14 m³
• Because the inflow in a day (24.19 m³) is more than the demand, the extra water simply overflows once the tank is full. The minimum storage that still meets every demand is the largest draw-down, 6.44 m³. The textbook value 8.14 m³ is on the safe side.

Answer: Balancing reservoir is necessary; capacity ≈ 8.1 m³ (adopt 9 m³ or 10 m³). It is needed because the peak demand in the morning is about twice the continuous spring flow.

Given data

• Demand in million litres for six 4-hour periods: 0.48, 0.67, 1.33, 1.00, 0.82, 0.54
• Two digits are unclear in the question paper; they are read as 0.67 and 1.33. Total demand = 4.84 ML
• Inflow is uniform over 24 h: 4.84/24 = 0.2017 ML/h = 0.8067 ML per 4 h

Mass-curve table (analytical method)

Inflow: 0.807 ML per 4-hour period. Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 4.84 ML; total inflow = 4.84 ML.

Capacity

• Maximum cumulative surplus = 0.46 ML
• Maximum cumulative deficit = 0.27 ML
• Balancing capacity = max. surplus + max. deficit = 0.46 + 0.27 = 0.73 ML

Answer: Storage required ≈ 0.73 ML (730 m³). If the unclear digits are different, repeat the same table with the correct values: capacity = maximum surplus + maximum deficit of the cumulative column.

Given data

• Population 600 × 65 lpcd = 39,000 L/day = 39 m³/day
• Spring yield = 0.5 lps = 43,200 L/day = 43.2 m³/day = 1.8 m³/h
• Demand: 5–7 h 30%, 7–12 h 30%, 12–17 h 15%, 17–19 h 20%, 19–5 h 5%

Check of source and need of storage

The yield (43.2 m³/day) exceeds the demand (39 m³/day). But the demand in 5–7 h is 0.30 × 39/2 = 5.85 m³/h and in 7–12 h it is 2.34 m³/h, both more than the inflow of 1.8 m³/h. So the balancing reservoir is necessary.

Mass-curve table (analytical method)

Inflow: 1.8 m³/h (spring yield). Time is counted from the start of the first demand period; a positive cumulative value means surplus in the tank.

Total demand = 39.00 m³; total inflow = 43.20 m³.

Capacity

• Maximum cumulative surplus = 4.20 m³
• Maximum cumulative deficit = 11.85 m³
• Balancing capacity = max. surplus + max. deficit = 4.20 + 11.85 = 16.05 m³
• Because the inflow in a day (43.20 m³) is more than the demand, the extra water simply overflows once the tank is full. The minimum storage that still meets every demand is the largest draw-down, 11.85 m³. The textbook value 16.05 m³ is on the safe side.

Answer: Balancing reservoir is necessary; capacity ≈ 16.1 m³ (adopt 16–17 m³).

Data: C = 100, peak (maximum) demand = 3 × average, 75 lpcd, minimum residual head 5.0 m. The tank at A stands 6 m high, so the HGL at A = 2160 + 6 = 2166 m. The population of each section is taken as drawn at its end point (B for AB, C for BC), so pipe AB carries the demand of 850 + 480 = 1330 people and BC carries 480 people.

• $Q_{AB} = 1330 \times 75 \times 3/86400 = 3.464$ lps
• $Q_{BC} = 480 \times 75 \times 3/86400 = 1.250$ lps

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

Pipe AB

• Discharge $Q = 3.464$ lps, length $L = 400$ m
• HGL at A = 2166.00 m; ground RL at B = 2125 m; residual head required = 5 m
• Available head $H = 2166.00 - 2125 - 5 = 36.00$ m
• Gradient $S = H/L = 36.00/400 = 0.09000$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 53.7$ mm, adopt commercial size 65 mm
• Actual $h_f = 14.143$ m, HGL at B = 2151.86 m, residual head = 26.86 m (≥ 5 m, OK)
• Velocity $v = Q/A = 1.04$ m/s

Pipe BC

• Discharge $Q = 1.250$ lps, length $L = 500$ m
• HGL at B = 2151.86 m; ground RL at C = 2115 m; residual head required = 5 m
• Available head $H = 2151.86 - 2115 - 5 = 31.86$ m
• Gradient $S = H/L = 31.86/500 = 0.06371$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 39.1$ mm, adopt commercial size 40 mm
• Actual $h_f = 28.483$ m, HGL at C = 2123.37 m, residual head = 8.37 m (≥ 5 m, OK)
• Velocity $v = Q/A = 0.99$ m/s

Summary

Velocities are between 0.6 m/s and 3 m/s, so the sizes are acceptable.

Answer: AB = 65 mm and BC = 40 mm; residual heads 26.9 m at B and 8.4 m at C (both more than 5 m); velocities 1.04 m/s and 0.99 m/s.

Residual pressure at each tap = 1 kg/cm² = 10 m. Source A has RL 1050 m, so HGL at A = 1050 m. Flows: AB = 0.35 lps, BC = 0.20 lps.

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

Pipe AB

• Discharge $Q = 0.350$ lps, length $L = 320$ m
• HGL at A = 1050.00 m; ground RL at B = 970 m; residual head required = 10 m
• Available head $H = 1050.00 - 970 - 10 = 70.00$ m
• Gradient $S = H/L = 70.00/320 = 0.21875$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 18.7$ mm, adopt commercial size 20 mm
• Actual $h_f = 50.456$ m, HGL at B = 999.54 m, residual head = 29.54 m (≥ 10 m, OK)
• Velocity $v = Q/A = 1.11$ m/s

Pipe BC

• Discharge $Q = 0.200$ lps, length $L = 250$ m
• HGL at B = 999.54 m; ground RL at C = 910 m; residual head required = 10 m
• Available head $H = 999.54 - 910 - 10 = 79.54$ m
• Gradient $S = H/L = 79.54/250 = 0.31818$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 14.0$ mm, adopt commercial size 15 mm
• Actual $h_f = 56.761$ m, HGL at C = 942.78 m, residual head = 32.78 m (≥ 10 m, OK)
• Velocity $v = Q/A = 1.13$ m/s

Summary

Pipe class

The pipe class must resist the maximum static (zero-flow) head:

• Section AB: static head at B = 1050 − 970 = 80 m
• Section BC: static head at C = 1050 − 910 = 140 m (14 kg/cm² = 1.4 MPa)

So pipe BC must be of a high class that can withstand at least 140 m (for example GI heavy or HDPE PN 16 / PN 20), while AB needs only a class for 80 m (GI medium or PN 10). The high-class pipe is costly.

Reducing the high class

Provide a break pressure tank (BPT) at B (RL 970 m). The tank breaks the pressure to zero (atmospheric) and starts a new hydraulic line from RL 970 m, so the static head in BC becomes 970 − 910 = 60 m only, and the whole scheme can use a pipe class for 80 m (PN 10 / GI medium).

Redesign of BC with BPT at B: available head $H = 970 - 910 - 10 = 50$ m, $S = 0.200$, $D_{calc} = 15.4$ mm, adopt 20 mm, $h_f = 13.98$ m, residual head = 46.02 m, $v = 0.64$ m/s. A pressure-reducing valve at B could also be used.

Answer: AB = 20 mm, BC = 15 mm (without BPT). Highest class needed is for 140 m static head in BC; with a BPT at B the BC static head falls to 60 m and a lower class (for 80 m) is enough, with BC = 20 mm.

Hazen-Williams constant C = 120; the residual pressure at B, C and D must be at least 12 m. HGL at A = RL of A = 500 m.

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 120.

Pipe AB

• Discharge $Q = 0.250$ lps, length $L = 700$ m
• HGL at A = 500.00 m; ground RL at B = 480 m; residual head required = 12 m
• Available head $H = 500.00 - 480 - 12 = 8.00$ m
• Gradient $S = H/L = 8.00/700 = 0.01143$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 28.1$ mm, adopt commercial size 32 mm
• Actual $h_f = 4.281$ m, HGL at B = 495.72 m, residual head = 15.72 m (≥ 12 m, OK)
• Velocity $v = Q/A = 0.31$ m/s

Pipe BC

• Discharge $Q = 0.150$ lps, length $L = 500$ m
• HGL at B = 495.72 m; ground RL at C = 465 m; residual head required = 12 m
• Available head $H = 495.72 - 465 - 12 = 18.72$ m
• Gradient $S = H/L = 18.72/500 = 0.03744$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 18.2$ mm, adopt commercial size 20 mm
• Actual $h_f = 11.711$ m, HGL at C = 484.01 m, residual head = 19.01 m (≥ 12 m, OK)
• Velocity $v = Q/A = 0.48$ m/s

Pipe CD

• Discharge $Q = 0.100$ lps, length $L = 600$ m
• HGL at C = 484.01 m; ground RL at D = 440 m; residual head required = 12 m
• Available head $H = 484.01 - 440 - 12 = 32.01$ m
• Gradient $S = H/L = 32.01/600 = 0.05335$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 14.5$ mm, adopt commercial size 15 mm
• Actual $h_f = 26.922$ m, HGL at D = 457.09 m, residual head = 17.09 m (≥ 12 m, OK)
• Velocity $v = Q/A = 0.57$ m/s

Summary

The pipe sizes are small because the flows are only a few tenths of a litre per second, and the velocities (0.3–0.6 m/s) are low; they are acceptable for a rural system with the limited head available. Maximum static head is at D: 500 − 440 = 60 m, so a pipe class for 60 m (about 0.6 MPa) is needed.

Answer: AB = 32 mm, BC = 20 mm, CD = 15 mm.

Residual pressure = 1 kg/cm² = 10 m at all taps; C = 110. HGL at source P = 950 m. The flows given for each pipe are taken as the discharge carried by that pipe (PR = 0.7 lps, RS = 0.3 lps, PQ = 0.25 lps).

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 110.

Pipe PR

• Discharge $Q = 0.700$ lps, length $L = 150$ m
• HGL at P = 950.00 m; ground RL at R = 860 m; residual head required = 10 m
• Available head $H = 950.00 - 860 - 10 = 80.00$ m
• Gradient $S = H/L = 80.00/150 = 0.53333$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 19.5$ mm, adopt commercial size 20 mm
• Actual $h_f = 71.565$ m, HGL at R = 878.43 m, residual head = 18.43 m (≥ 10 m, OK)
• Velocity $v = Q/A = 2.23$ m/s

Pipe RS

• Discharge $Q = 0.300$ lps, length $L = 160$ m
• HGL at R = 878.43 m; ground RL at S = 845 m; residual head required = 10 m
• Available head $H = 878.43 - 845 - 10 = 23.43$ m
• Gradient $S = H/L = 23.43/160 = 0.14647$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 18.5$ mm, adopt commercial size 20 mm
• Actual $h_f = 15.894$ m, HGL at S = 862.54 m, residual head = 17.54 m (≥ 10 m, OK)
• Velocity $v = Q/A = 0.95$ m/s

Pipe PQ

• Discharge $Q = 0.250$ lps, length $L = 180$ m
• HGL at P = 950.00 m; ground RL at Q = 930 m; residual head required = 10 m
• Available head $H = 950.00 - 930 - 10 = 10.00$ m
• Gradient $S = H/L = 10.00/180 = 0.05556$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 21.0$ mm, adopt commercial size 25 mm
• Actual $h_f = 4.303$ m, HGL at Q = 945.70 m, residual head = 15.70 m (≥ 10 m, OK)
• Velocity $v = Q/A = 0.51$ m/s

Summary

Static heads: Q = 950 − 930 = 20 m, R = 90 m, S = 105 m. The pipe PR and RS must be of a class for at least 105 m; where high pressure is a problem a break pressure tank near R can be provided.

Answer: PQ = 25 mm, PR = 20 mm, RS = 20 mm. (Velocity in PR is 2.23 m/s, still within the 3 m/s limit.)

Per capita demand 160 lpcd, peak factor 3, C = 100, minimum residual head 10 m. HGL at A = 1000 m. The population along each section is taken as supplied at its end node: AB carries the demand of 3600 + 4400 = 8000 people, and BC carries 4400 people.

• $Q_{AB} = 8000 \times 160 \times 3/86400 = 44.44$ lps
• $Q_{BC} = 4400 \times 160 \times 3/86400 = 24.44$ lps

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

Pipe AB

• Discharge $Q = 44.444$ lps, length $L = 200$ m
• HGL at A = 1000.00 m; ground RL at B = 950 m; residual head required = 10 m
• Available head $H = 1000.00 - 950 - 10 = 40.00$ m
• Gradient $S = H/L = 40.00/200 = 0.20000$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 120.2$ mm, adopt commercial size 150 mm (raised from the head-based size because velocity would exceed 3 m/s)
• Actual $h_f = 13.595$ m, HGL at B = 986.40 m, residual head = 36.40 m (≥ 10 m, OK)
• Velocity $v = Q/A = 2.52$ m/s

Pipe BC

• Discharge $Q = 24.444$ lps, length $L = 500$ m
• HGL at B = 986.40 m; ground RL at C = 900 m; residual head required = 10 m
• Available head $H = 986.40 - 900 - 10 = 76.40$ m
• Gradient $S = H/L = 76.40/500 = 0.15281$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 101.2$ mm, adopt commercial size 125 mm
• Actual $h_f = 27.296$ m, HGL at C = 959.11 m, residual head = 59.11 m (≥ 10 m, OK)
• Velocity $v = Q/A = 1.99$ m/s

Summary

Answer: AB = 150 mm and BC = 125 mm. For AB the head-based size is 125 mm, but its velocity would be 3.6 m/s, which is too high, so 150 mm is adopted (v = 2.5 m/s). Residual heads are above 10 m at B and C.

C = 100; minimum residual pressure at B and C = 10 m. HGL at A = 950 m. The discharge in BC is not given in the figure; it is assumed that half of the 1.3 lps is used at B, so BC carries 0.65 lps (if the actual discharge is known, substitute it in the same steps).

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

Pipe AB

• Discharge $Q = 1.300$ lps, length $L = 2750$ m
• HGL at A = 950.00 m; ground RL at B = 920 m; residual head required = 10 m
• Available head $H = 950.00 - 920 - 10 = 20.00$ m
• Gradient $S = H/L = 20.00/2750 = 0.00727$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 62.0$ mm, adopt commercial size 65 mm
• Actual $h_f = 15.836$ m, HGL at B = 934.16 m, residual head = 14.16 m (≥ 10 m, OK)
• Velocity $v = Q/A = 0.39$ m/s

Pipe BC

• Discharge $Q = 0.650$ lps, length $L = 1700$ m
• HGL at B = 934.16 m; ground RL at C = 894 m; residual head required = 10 m
• Available head $H = 934.16 - 894 - 10 = 30.16$ m
• Gradient $S = H/L = 30.16/1700 = 0.01774$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 39.6$ mm, adopt commercial size 40 mm
• Actual $h_f = 28.847$ m, HGL at C = 905.32 m, residual head = 11.32 m (≥ 10 m, OK)
• Velocity $v = Q/A = 0.52$ m/s

Summary

Answer: AB = 65 mm and BC = 40 mm (GI), with residual heads of 14.2 m at B and 11.3 m at C.

Per capita demand 200 lpcd, peak factor 3, C = 100, residual head ≥ 15 m. HGL at R = 890 m. The population along each section is treated as drawn at its end node (A for RA, B for AB): pipe RA carries 3600 + 4400 = 8000 people and AB carries 4400 people.

• $Q_{RA} = 8000 \times 200 \times 3/86400 = 55.56$ lps
• $Q_{AB} = 4400 \times 200 \times 3/86400 = 30.56$ lps

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

Pipe RA

• Discharge $Q = 55.556$ lps, length $L = 200$ m
• HGL at R = 890.00 m; ground RL at A = 840 m; residual head required = 15 m
• Available head $H = 890.00 - 840 - 15 = 35.00$ m
• Gradient $S = H/L = 35.00/200 = 0.17500$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 134.5$ mm, adopt commercial size 200 mm (raised from the head-based size because velocity would exceed 3 m/s)
• Actual $h_f = 5.063$ m, HGL at A = 884.94 m, residual head = 44.94 m (≥ 15 m, OK)
• Velocity $v = Q/A = 1.77$ m/s

Pipe AB

• Discharge $Q = 30.556$ lps, length $L = 500$ m
• HGL at A = 884.94 m; ground RL at B = 790 m; residual head required = 15 m
• Available head $H = 884.94 - 790 - 15 = 79.94$ m
• Gradient $S = H/L = 79.94/500 = 0.15987$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 109.1$ mm, adopt commercial size 125 mm
• Actual $h_f = 41.264$ m, HGL at B = 843.67 m, residual head = 53.67 m (≥ 15 m, OK)
• Velocity $v = Q/A = 2.49$ m/s

Summary

The head-based diameter of RA is 134.5 mm, but 150 mm would give a velocity of 3.14 m/s, which is above the usual limit of 3 m/s; 200 mm is therefore adopted (v = 1.77 m/s). All velocities are between 0.6 and 3 m/s and all residual heads are above 15 m.

Answer: RA = 200 mm, AB = 125 mm.

Residual pressure = 1 kg/cm² = 10 m; C = 100. HGL at A = 985 m. Discharges as given: AB = 0.25 lps, BC = 0.10 lps, AD = 0.50 lps.

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

Pipe AB

• Discharge $Q = 0.250$ lps, length $L = 120$ m
• HGL at A = 985.00 m; ground RL at B = 960 m; residual head required = 10 m
• Available head $H = 985.00 - 960 - 10 = 15.00$ m
• Gradient $S = H/L = 15.00/120 = 0.12500$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 18.5$ mm, adopt commercial size 20 mm
• Actual $h_f = 10.146$ m, HGL at B = 974.85 m, residual head = 14.85 m (≥ 10 m, OK)
• Velocity $v = Q/A = 0.80$ m/s

Pipe BC

• Discharge $Q = 0.100$ lps, length $L = 150$ m
• HGL at B = 974.85 m; ground RL at C = 920 m; residual head required = 10 m
• Available head $H = 974.85 - 920 - 10 = 44.85$ m
• Gradient $S = H/L = 44.85/150 = 0.29902$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 10.9$ mm, adopt commercial size 15 mm
• Actual $h_f = 9.434$ m, HGL at C = 965.42 m, residual head = 45.42 m (≥ 10 m, OK)
• Velocity $v = Q/A = 0.57$ m/s

Pipe AD

• Discharge $Q = 0.500$ lps, length $L = 160$ m
• HGL at A = 985.00 m; ground RL at D = 940 m; residual head required = 10 m
• Available head $H = 985.00 - 940 - 10 = 35.00$ m
• Gradient $S = H/L = 35.00/160 = 0.21875$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 21.4$ mm, adopt commercial size 25 mm
• Actual $h_f = 16.474$ m, HGL at D = 968.53 m, residual head = 28.53 m (≥ 10 m, OK)
• Velocity $v = Q/A = 1.02$ m/s

Summary

Static head is highest at C (985 − 920 = 65 m), so the pipes BC (and AB feeding it) must be of a class suitable for at least 65 m.

Answer: AB = 20 mm, BC = 15 mm, AD = 25 mm.

C = 100, minimum residual head = 10 m. HGL at A = RL of A = 210 m. A to B: 700 m, 44 lps, RL of B = 154 m. B to C: 550 m, 18 lps, RL of C = 126 m.

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

Pipe AB

• Discharge $Q = 44.000$ lps, length $L = 700$ m
• HGL at A = 210.00 m; ground RL at B = 154 m; residual head required = 10 m
• Available head $H = 210.00 - 154 - 10 = 46.00$ m
• Gradient $S = H/L = 46.00/700 = 0.06571$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 150.5$ mm, adopt commercial size 200 mm
• Actual $h_f = 11.506$ m, HGL at B = 198.49 m, residual head = 44.49 m (≥ 10 m, OK)
• Velocity $v = Q/A = 1.40$ m/s

Pipe BC

• Discharge $Q = 18.000$ lps, length $L = 550$ m
• HGL at B = 198.49 m; ground RL at C = 126 m; residual head required = 10 m
• Available head $H = 198.49 - 126 - 10 = 62.49$ m
• Gradient $S = H/L = 62.49/550 = 0.11363$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 95.7$ mm, adopt commercial size 100 mm
• Actual $h_f = 50.500$ m, HGL at C = 147.99 m, residual head = 21.99 m (≥ 10 m, OK)
• Velocity $v = Q/A = 2.29$ m/s

Summary

The head-based size of AB is 150.5 mm, which is slightly above 150 mm, so 200 mm is adopted (other option: 150 mm in part length and 200 mm in the rest). Velocity in BC is 2.29 m/s (< 3 m/s).

Answer: AB = 200 mm and BC = 100 mm.

Source A has RL 180 m; minimum pressure 1.5 kg/cm² = 15 m; C = 100. A to B: 44 lps, 700 m, RL of B = 154 m. B to C: 18 lps, 550 m, RL of C = 126 m.

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

Pipe AB

• Discharge $Q = 44.000$ lps, length $L = 700$ m
• HGL at A = 180.00 m; ground RL at B = 154 m; residual head required = 15 m
• Available head $H = 180.00 - 154 - 15 = 11.00$ m
• Gradient $S = H/L = 11.00/700 = 0.01571$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 201.9$ mm, adopt commercial size 250 mm
• Actual $h_f = 3.881$ m, HGL at B = 176.12 m, residual head = 22.12 m (≥ 15 m, OK)
• Velocity $v = Q/A = 0.90$ m/s

Pipe BC

• Discharge $Q = 18.000$ lps, length $L = 550$ m
• HGL at B = 176.12 m; ground RL at C = 126 m; residual head required = 15 m
• Available head $H = 176.12 - 126 - 15 = 35.12$ m
• Gradient $S = H/L = 35.12/550 = 0.06385$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 107.7$ mm, adopt commercial size 125 mm
• Actual $h_f = 17.035$ m, HGL at C = 159.08 m, residual head = 33.08 m (≥ 15 m, OK)
• Velocity $v = Q/A = 1.47$ m/s

Summary

The small available head (11 m) for the 700 m pipe AB needs a large pipe. A 250 mm pipe gives a velocity of 0.9 m/s, which is acceptable.

Answer: AB = 250 mm and BC = 125 mm.

Given data

• Triangle A–B–C at the same RL. Inflow 33 lps at A; outflow 25 lps at B and 8 lps at C. $L_{AB} = 200$ m, $L_{BC} = 200$ m, $L_{AC} = 270$ m, $C = 110$
• Pressure at A = 15 kg/cm² = 150 m of water; minimum pressure required at B and C = 15 m
• Head that can be lost between A and B (or C) $= 150 - 15 = 135$ m

Step 1: Choose diameters

The head available (135 m) is much more than what is needed, so the velocity (about 1–2.5 m/s) governs the choice. The first estimate of flows: AB = 25 lps (to B), AC = 8 lps (to C), and BC just a connection. Adopt (commercial sizes): AB = 125 mm, AC = 80 mm, BC = 80 mm.

Step 2: Flow distribution by Hardy-Cross

Loop A→B→C→A. Resistance $r = 10.67L/(C^{1.852}D^{4.87})$ and $h = rQ^{1.852}$, $\Delta Q = -\sum h/(1.852\sum |h/Q|)$. Trial flows: AB = 25, BC = 0, AC = 8 lps (A to B, B to C and A to C).

Final flows: AB = 26.05 lps, BC = 1.15 lps (B to C), AC = 6.95 lps.

Step 3: Check heads and velocities

• Head loss in AB $= 10.30$ m; head loss A to C via AC $= 10.57$ m and in BC $= 0.28$ m; the loop closes because AB + BC $\approx$ AC
• Pressure head at B $= 150 - 10.30 = 139.7$ m; at C $= 150 - 10.57 = 139.4$ m; both are far more than 15 m
• Velocities: AB $= 2.12$ m/s, AC $= 1.38$ m/s, BC $= 0.23$ m/s (all below 3 m/s)

Answer: AB = 125 mm, AC = 80 mm, BC = 80 mm. The pressure at B and C (about 140 m) is high, so a pressure-reducing valve or high-class pipe must be used.