Quantity of Wastewater

Sanitary sewage is the wastewater from the sanitary fixtures and activities of homes, offices, institutions and (to some extent) industries. It consists of human excreta, bath and kitchen water and washing water.

Potential sources

1. Residential areas: toilets, bathrooms, kitchens, washing; the largest part.
2. Commercial buildings: offices, hotels, restaurants, shops, markets.
3. Institutions and public places: schools, colleges, hospitals, bus parks, temples, public toilets.
4. Industries: sanitary wastes of workers and (after pretreatment) process waste.
5. Infiltration: groundwater entering through defective joints and cracks.
6. Other: illegal roof or yard drain connections and leakage from the water supply.

Factors affecting the quantity

• Population and its growth: more people, more sewage. The sewer is designed for the future population at the end of the design period.
• Rate of water supply: sewage is about 75-80% of the water supplied; the rest is lost in lawn watering, evaporation, leakage and consumption.
• Type of area: residential, commercial or industrial areas produce different flows; commercial and industrial areas have high per-hectare flows.
• Standard of living and habits: in rich households, the use of flush toilets, washing machines and showers increases the flow.
• Private water supplies: water from wells entering the sewer adds to the sewage that is not recorded by the public supply.
• Groundwater infiltration: depends on the water table, sewer material, quality of joints and age; it can be 5,000-15,000 litres per day per km of sewer per cm diameter.
• Unauthorised connections: storm water or other waters connected to the sewer.
• Climate and time variation: hourly, daily and seasonal peaks; summer use is higher than winter use.
• Water meters and tariff: metering reduces wastage, and so reduces sewage.

The quantity of storm water from a locality is estimated by the Rational method, which is the standard method for small and medium urban catchments (up to a few hundred hectares).

$$Q = \frac{C\,i\,A}{360}$$

where $Q$ = peak runoff (m³/s), $C$ = coefficient of runoff, $i$ = rainfall intensity (mm/hr) for a duration equal to the time of concentration, and $A$ = catchment area (ha).

Steps

1. Find the catchment area $A$ from the map, and divide it into surfaces (roofs, roads, lawns, open land).
2. Find the weighted coefficient of runoff: $C=\dfrac{\sum C_jA_j}{\sum A_j}$. Typical values: roofs 0.75-0.95, asphalt road 0.70-0.95, gravel 0.25-0.40, lawns 0.05-0.20, parks 0.10-0.25. For future development use values for the fully developed area.
3. Find the time of concentration: $t_c=t_e+t_f$, where $t_e$ is the inlet (entry) time (usually 5-20 min) and $t_f=L/v$ is the flow time in the drain.
4. Find the intensity $i$ for duration $t_c$ and a chosen return period (2-10 years for urban drains) from the intensity-duration-frequency (IDF) curve or formula of the local Department of Hydrology and Meteorology. A common empirical form is $i=\dfrac{a}{t+b}$.
5. Compute $Q=CiA/360$, and design each drain section for its own $t_c$ and area, adding the area of each branch downstream.

Example

A residential-commercial locality of 40 ha: 30% roads and pavements ($C=0.85$), 20% roofs ($C=0.90$), 20% yards ($C=0.30$), 30% gardens ($C=0.15$). Entry time = 10 min, flow time in drain = 10 min. Using $i=760/(t+10)$ mm/hr (assumed).

$$\begin{aligned} C &= 0.3(0.85)+0.2(0.90)+0.2(0.30)+0.3(0.15) = 0.540 \\ t_c &= 10+10 = 20\ \text{min} \\ i &= \frac{760}{20+10} = 25.33\ \text{mm/hr} \\ Q &= \frac{0.540 \times 25.33 \times 40}{360} = 1.520\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Q = 1.520 m³/s; the storm drain is designed for this discharge.

Time of concentration

The time of concentration ($t_c$) is the time taken by rain falling on the most remote point of the catchment to reach the point of the drain or sewer under design.

$$t_c = t_e + t_f$$

• $t_e$ = time of entry (inlet time): time for water to flow overland and through gutters to the first inlet (5-20 min).
• $t_f$ = time of flow: travel time in the sewer, $t_f=L/v$.

Why it is important

1. The peak discharge occurs when the whole catchment contributes, i.e. when the rain has lasted at least $t_c$. So the design storm duration is taken equal to $t_c$.
2. The rainfall intensity $i$ depends on duration: short storms are more intense. A wrong $t_c$ gives a wrong $i$, and so a wrong $Q = CiA/360$.
3. If $t_c$ is too short, $i$ is too high and the sewer is oversized and costly; if too long, the sewer is undersized and floods.
4. It is needed to fix the size of every downstream section, since $t_c$ increases down the sewer.
5. It decides how far a storm drain of a given size can serve.

Time-area graph

It is a plot of the contributing area against time, used to find how the runoff builds up.

Area
  A |                    ________ whole area
    |                 .'
    |              .'
    |           .'
    |        .'
    |     .'
    +----+-----+-----+-----> time
    0   t1    t2    tc

• The catchment is divided by lines of equal travel time (isochrones) to the outlet, 0, t1, t2 ... tc.
• The area between two successive isochrones is measured and the cumulative area is plotted against time.
• The graph rises with time and becomes constant at $t_c$, when the entire area A contributes.
• For a rain of duration less than $t_c$, only the area read from the graph at that time contributes, and runoff is smaller.
• For a uniform rainfall, the runoff hydrograph is obtained as $Q = C\,i\,A_t$, where $A_t$ is the area read from the graph at time $t$. Its shape shows that the peak occurs at $t_c$ (rational method).

For a catchment which is nearly rectangular, the graph is a straight line; for fan-shaped ones it is curved.

Dry weather flow (DWF) is the flow in a sewer in the absence of rain, made of domestic and industrial sewage plus groundwater infiltration. Wet weather flow (WWF) is the total flow in a sewer in rain, DWF plus the runoff entering from the surface.

Why both are important in sewer design

1. Maximum capacity: in a combined or partially separate system the sewer must carry the peak WWF, $Q = \text{peak DWF} + \dfrac{CiA}{360}$. If only DWF were used, the sewer would be too small and overflow, flooding streets.
2. Self-cleansing velocity: a sewer sized for WWF is almost empty in dry weather. The minimum DWF (about one-third of average) is used to check that velocity is at least 0.6-0.75 m/s, so that solids do not settle.
3. Choice of sewer system: the large ratio of WWF to DWF (often 20 to 100 in monsoon areas) is the reason a separate system is chosen.
4. Sizing of treatment works: the treatment plant is designed for about 3 x DWF, and the excess is released through storm overflows. Both flows are needed to design overflows, storm tanks and regulators.
5. Pumping stations: pumps are selected for DWF range; storm pumps for WWF.
6. Infiltration and illegal connections: the difference between DWF and expected flow shows leakage, and extra flow in rain shows illegal roof connections.
7. Economy and safety: a design based on both gives a sewer which is neither costly nor unsafe, and prevents pollution from overflows.

Dry weather flow (DWF)

DWF is the flow in sewer when there is no storm runoff. It consists of domestic sewage, industrial wastewater and groundwater infiltration. It is the average flow during a dry period, usually after seven days without rain.

Wet weather flow (WWF)

WWF is the flow in a sewer during and after rain. It is the sum of DWF and the storm water that enters the sewer:

$$\text{WWF} = \text{DWF} + \text{storm runoff}$$

Estimation of WWF

1. Find the DWF from the design population, the water supply rate (sewage = 75-80% of supply) and a peak factor, plus infiltration.
2. Find the storm runoff by the rational method:

$$Q_s = \frac{C\,i\,A}{360}\ \ (\text{m}^3/\text{s})$$

with $C$ the weighted runoff coefficient, $i$ the rainfall intensity (mm/hr) for duration $t_c = t_e + t_f$, and $A$ the area in hectares. 3. Add them: $Q_{WWF} = Q_{DWF}+Q_s$.

For a combined system the whole of it is taken. In a partially separate system only the runoff from roofs and courtyards (say one-third to one-half of the area) is used. A rough rule is: sewer is designed for 6 x DWF (or the DWF + runoff of 3 to 6 mm/day), but the rational method is preferred.

Definitions

• Dry weather flow (DWF): the flow of sewage in a sewer in dry weather, consisting of domestic and industrial sewage and groundwater infiltration, without any storm water.
• Wet weather flow (WWF): the flow in a sewer during rain: DWF plus storm water runoff.

Factors affecting storm water (runoff)

1. Rainfall intensity and duration: heavier and longer storms give more runoff. Peak runoff happens when the duration equals the time of concentration.
2. Area of the catchment: runoff is directly proportional to the area drained.
3. Imperviousness (type of surface): roofs, roads and pavements give 70-95% runoff, but lawns and forests only 5-20%. Urbanisation increases runoff.
4. Slope and shape of catchment: steep and short catchments give quick, high peaks; flat and long catchments give delayed, flat peaks.
5. Time of concentration: depends on size, slope and surface roughness; a shorter time gives higher intensity.
6. Initial wetness of soil and infiltration: wet or saturated soil gives more runoff.
7. Depression storage, evaporation and interception by vegetation reduce runoff.
8. Distribution of rainfall: a storm that is uneven over the area or moves along the drainage direction changes the peak.
9. Condition and capacity of drains: blocked drains, back water and ponding delay and reduce the flow.

Difference between DWF and WWF

Sources of sanitary sewage

1. Residential: water closets, bathrooms, kitchens and laundry. This is the main source.
2. Commercial: offices, hotels, restaurants, markets, shops.
3. Institutional: schools, hospitals, colleges, hostels, public buildings.
4. Industrial: sanitary wastes of employees, and process wastes where allowed.
5. Public places: public toilets, bus parks, parks.
6. Groundwater infiltration through leaking joints and manholes, and unauthorised connections such as roof drains.

Sources of sanitary sewage

• Residential buildings (toilets, baths, kitchens, washing), the largest source.
• Commercial and institutional buildings: offices, shops, hotels, restaurants, schools, hospitals.
• Industries: sanitary waste of workers and process waste.
• Public places: public toilets, markets, bus parks.
• Infiltration of groundwater and unauthorised connections.

Impact of time and city development

With time

• Daily: the flow is minimum at night (about 1/3 of average), has a first peak in the morning (7-10 am) and a second peak in the evening; the maximum rate may be 2 to 3 times the average.
• Seasonal: flow is larger in summer (more water use) and in the rainy season (higher infiltration and illegal storm connections).
• Long term: as the population grows over the design period, flow increases, so sewers must be designed for the future population.

With development activities

• Better living standard, piped water, flush toilets, washing machines and hotels increase water use and sewage per head from, say, 60 lpcd to 135 lpcd or more.
• Growth of commercial and industrial areas, hospitals, hotels and institutions add heavy and sometimes harmful flows, and raise the pollution load.
• Urbanisation increases impervious area, so more storm water reaches sewers in a combined system.
• Densification (high-rise buildings) increases flow per hectare.
• New connections and expansion of sewer network raise infiltration if construction quality is poor.

Factors affecting discharge

1. Population and its density, and the future growth over the design period.
2. Rate of water supply (sewage is 75-80% of supply).
3. Type of area (residential, commercial, industrial) and standard of living.
4. Private and public water supplies, and water metering.
5. Groundwater infiltration, depending on water table and condition of joints.
6. Unauthorised connections (roof, yard drains).
7. Climate, season and hour of the day.

Calculation of sanitary sewage discharge

1. Design population $P$ at the end of the design period (20-30 years), by arithmetic, geometric or incremental increase method.
2. Average sewage flow:

$$Q_{avg} = \frac{P \times q \times 0.8}{86400}\ \ \text{l/s}$$

where $q$ = water supply (lpcd) and 0.8 = fraction becoming sewage. 3. Peak (maximum) flow: $Q_{max}=\text{PF}\times Q_{avg}$. The peak factor is 2 to 3 (about 3 for small, 2 for big towns; or by Harman's formula $PF = 1+\dfrac{14}{4+\sqrt{P/1000}}$). 4. Minimum flow is about $\tfrac13 Q_{avg}$ (used to check the self-cleansing velocity). 5. Add infiltration (e.g. 5,000-15,000 l/day/km per cm diameter of sewer, or a percentage) and industrial flows if present.

Design discharge = peak sewage flow + infiltration (+ storm runoff $CiA/360$ for combined systems).

The quantity of sewage is estimated from the future population, the water supply per capita, and the factors that raise or reduce the flow. For a city in Nepal (e.g. Pokhara, Biratnagar) the method is as follows.

Step 1: Design population

Collect census data (CBS, Nepal) and project the population for the design period (about 20-25 years after completion) using the arithmetic, geometric or incremental increase method:

• Arithmetic: $P_n = P_0 + n\bar{x}$
• Geometric: $P_n = P_0(1+r/100)^n$

Add floating population (tourists, commuters, students) if large.

Step 2: Water supply rate

Use the design water demand of the city (Nepal Water Supply Corporation/NWSC design guidelines): 45 lpcd for rural and 85-135 lpcd for urban areas, with additions for institutional, commercial and industrial demand and leakage.

Step 3: Average sewage flow

About 80% of supply returns as sewage (the rest is lost in gardening, evaporation, leakage and consumption):

$$Q_{avg}=\frac{0.8\,P\,q}{86400}\ \ \text{l/s}$$

Step 4: Peak and minimum flow

• Peak factor 2 to 3 (higher for small towns); $Q_{max}=PF\cdot Q_{avg}$ for sewer design.
• Minimum flow about $\tfrac13 Q_{avg}$ for velocity check at start of the design period.

Step 5: Additions

• Infiltration depending on groundwater table and pipe quality (about 5-10% of DWF in Nepali cities, or 5,000-10,000 l/km/day/cm diameter).
• Industrial wastewater from large industries (from their own records).
• Storm water with $Q_s=\dfrac{CiA}{360}$ if the system is combined or partially separate. The intensity for the Nepal city is taken from the DHM intensity-duration-frequency curves, with $t_c=t_e+t_f$.

Step 6: Design discharge

$$Q_{design} = Q_{max} + Q_{infiltration} + Q_{industrial}\ (+Q_{storm})$$

The sewers are designed for $Q_{design}$ at the end of the design period and checked for minimum flow at the beginning (velocity at least 0.6 m/s).

Example

For $P=60{,}000$, $q=135$ lpcd, PF = 3: $Q_{avg}=0.8\times60000\times135/86400=75$ l/s and $Q_{max}=225$ l/s.

Determination of storm water quantity

For a densely built urban area the rational method is used:

$$Q=\frac{C\,i\,A}{360}\ \ \text{(m}^3/\text{s)}$$

1. Measure the catchment area $A$ (ha) of each drain, from a contour map and drainage layout.
2. Find the weighted runoff coefficient $C$ from the land use (roofs, paved roads, courtyards, open land). For a dense urban area $C$ is high, about 0.7-0.9.
3. Find the time of concentration $t_c=t_e+t_f$ (entry time 5-10 min in dense areas plus flow time in the drain).
4. Read the intensity $i$ for $t_c$ and the design return period (5-10 years for main drains) from the IDF curve of the nearest rain gauge station.
5. Compute $Q$ for each section and design the drains using Manning's formula.

For large or complex areas, hydrograph methods or computer models (unit hydrograph, SWMM) are used.

Limitations in a highly populated urban area

1. Assumption of uniform rainfall over the whole catchment and uniform intensity during $t_c$ is not true for large areas.
2. Runoff coefficient is uncertain. It changes with soil moisture, storm intensity and quick urban growth. A fixed $C$ may under- or overestimate.
3. Time of concentration is hard to find in a complex network with many inlets, varied slopes and encroached drains.
4. Lack of rainfall data: short-duration records are often few or missing for Nepali towns, so IDF curves are doubtful.
5. The method gives only the peak, not volume or hydrograph; it cannot account for storage in ponds or detention.
6. Backwater and blockage: the method ignores blocked inlets and drains, solid waste in drains, river backwater and surcharging, so actual flow differs.
7. Valid only for small catchments (generally below 400-500 ha).
8. Changing land use: future urbanisation, filling of ponds and drains, climate change increasing rainfall, are difficult to predict.
9. Interaction with the sewage flow and illegal connections are not included.

To reduce errors a factor of safety is used, local data are collected, and drains are checked after construction.

In a combined system, the wet weather discharge is the dry weather sewage flow plus the storm runoff.

Given

Population = 500, water supply = 160 lpcd, area $A$ = 1.5 ha, rainfall intensity $i$ = 50 mm/hr. Assume that 80% of water supplied becomes sewage.

Step 1: Weighted coefficient of runoff

$$C=\frac{0.4(0.8)+0.6(0.2)}{1.0}=0.44$$

Step 2: Storm runoff (Rational method)

$$Q_s=\frac{C\,i\,A}{360}=\frac{0.44\times 50\times 1.5}{360}=0.0917\ \text{m}^3/\text{s}$$

Step 3: Sanitary sewage (dry weather flow)

$$Q_d=\frac{500\times160\times0.8}{86400\times1000}=0.00074\ \text{m}^3/\text{s}$$

Step 4: Wet weather discharge

$$Q=Q_s+Q_d=0.0917+0.00074=0.0924\ \text{m}^3/\text{s}$$

Answer: Wet weather discharge is about 0.0924 m³/s (92.4 l/s); storm water is 99% of it, so the sewage part is almost negligible.

The runoff from the area is found by the Rational formula $Q = CiA/360$, with $A$ in hectares, $i$ in mm/hr and $Q$ in m³/s.

Weighted runoff coefficient

$$C=\frac{0.40(0.8)+0.60(0.2)}{1.00}=0.32+0.12=0.44$$

Runoff

$$Q=\frac{C\,i\,A}{360}=\frac{0.44\times50\times20}{360}=1.222\ \text{m}^3/\text{s}$$

Answer: Runoff = 1.222 m³/s (about 1222 l/s).

The design discharge of a combined sewer is the peak sanitary sewage plus the storm water runoff.

Assumptions

• Permeability 60% means the area is 40% impervious, so $C=1-0.60=0.40$.
• Rainfall intensity for the given storm duration of 15 min is taken from $i=\dfrac{760}{t+10}$ mm/hr (assumed empirical formula, $t$ in min).
• Time of concentration $t_c=t_e+t_f=5+15=20$ min (the storm of 15 min is used for intensity as given).

Step 1: Sanitary sewage

Average sewage = 80% of 100 lpcd = 80 lpcd. Peak = 3 x average.

$$Q_{sewage}=\frac{75000\times100\times0.8\times3}{86400\times1000}=0.2083\ \text{m}^3/\text{s}$$

Step 2: Storm water

Area $A=8\ \text{km}^2=800$ ha.

$$\begin{aligned} i&=\frac{760}{15+10}=30.40\ \text{mm/hr}\\ Q_{storm}&=\frac{C\,i\,A}{360}=\frac{0.40\times30.40\times800}{360}=27.022\ \text{m}^3/\text{s} \end{aligned}$$

Step 3: Design discharge

$$Q=Q_{sewage}+Q_{storm}=0.2083+27.022=27.231\ \text{m}^3/\text{s}$$

Answer: Design discharge = 27.231 m³/s (storm water 27.022 m³/s plus sewage 0.2083 m³/s). If the examination supplies its own intensity formula, replace $i$ and the same steps apply.

Design discharge of a combined sewer = peak dry weather sewage flow + storm runoff.

Step 1: Peak sewage flow

Average flow = 45 lpcd (given as the sewage flow), peak factor 2.7.

$$Q_{sewage}=\frac{30000\times45\times2.7}{86400\times1000}=0.0422\ \text{m}^3/\text{s}$$

Step 2: Time of concentration and intensity

$$t_c=t_e+t_f=25+5=30\ \text{min}$$

No rainfall formula is given, so I use the common empirical form $i=\dfrac{760}{t+10}$ mm/hr (assumed):

$$i=\frac{760}{30+10}=19.0\ \text{mm/hr}$$

Step 3: Storm water (Rational formula)

$$Q_{storm}=\frac{C\,i\,A}{360}=\frac{0.60\times19.0\times60}{360}=1.900\ \text{m}^3/\text{s}$$

Step 4: Design discharge

$$Q=0.0422+1.900=1.942\ \text{m}^3/\text{s}$$

Answer: Design discharge = 1.942 m³/s (the sewage is only about 2% of it).

Storm water quantity by the Rational method: $Q=\dfrac{CiA}{360}$.

Step 1: Weighted runoff coefficient

$$C=\frac{\sum C_jA_j}{\sum A_j}=0.5025$$

Step 2: Time of concentration and intensity

$$t_c=t_e+t_f=20+10=30\ \text{min}$$

No intensity formula is given, so the common formula $i=\dfrac{760}{t+10}$ mm/hr is assumed:

$$i=\frac{760}{30+10}=19.0\ \text{mm/hr}$$

Step 3: Discharge

$$Q=\frac{0.5025\times19.0\times45}{360}=1.193\ \text{m}^3/\text{s}$$

Answer: Storm water quantity = 1.193 m³/s (1193 l/s).

Assumptions

• 80% of water supplied becomes sewage; peak factor = 3; infiltration neglected.
• No rainfall formula is given, so $i=\dfrac{760}{t+10}$ mm/hr is assumed for the intensity at $t_c$.

(a) Separate system (sewage only)

$$Q_{sew}=\frac{45000\times150\times0.8\times3}{86400\times1000}=0.1875\ \text{m}^3/\text{s}=187.5\ \text{l/s}$$

Design the sewer running full ($S=1/500$, $n=0.013$). From Manning's formula for a full circular pipe, $Q=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$:

$$D=\left(\frac{Q\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.1875\times0.013}{0.3117\times0.04472}\right)^{3/8}=0.520\ \text{m}$$

Adopt $D=0.525$ m (standard size). Full velocity $v=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=0.89$ m/s, which is above 0.6 m/s.

(b) Combined system

$$\begin{aligned} i&=\frac{760}{30+10}=19.0\ \text{mm/hr}\\ Q_{storm}&=\frac{0.45\times19.0\times20}{360}=0.475\ \text{m}^3/\text{s}\\ Q_{comb}&=Q_{sew}+Q_{storm}=0.1875+0.475=0.662\ \text{m}^3/\text{s} \end{aligned}$$ $$D=\left(\frac{0.662\times0.013}{0.3117\times0.04472}\right)^{3/8}=0.835\ \text{m}$$

Adopt $D=0.90$ m. Full velocity $=1.27$ m/s (below the 3 m/s maximum).

Answer: Separate system: $Q=0.1875$ m³/s, sewer about $D=0.525$ m. Combined system: $Q=0.662$ m³/s, sewer about $D=0.90$ m, both at slope 1 in 500.