Wastewater Disposal

Necessity (importance) of wastewater management

Wastewater produced by a community contains pathogens, organic matter, nutrients and toxic substances. If it is left on the street or let into rivers without treatment, it:

• spreads waterborne diseases (cholera, typhoid, dysentery, hepatitis, diarrhoea),
• pollutes surface water and groundwater that people use for drinking, bathing and irrigation,
• kills fish and aquatic life by depleting oxygen, and causes eutrophication,
• gives bad smell, mosquitoes and flies, and spoils the look of the city,
• blocks the drains and causes flooding, and
• reduces land value and tourism.

So scientific management (collection, treatment and safe disposal) is necessary for public health, protection of the environment and economic development, and is required by law (Environment Protection Act 2053 and Rules 2054).

Objectives of sewage (wastewater) disposal

1. To protect public health by preventing contact of people with pathogens and by controlling disease carriers.
2. To prevent pollution of water sources: streams, lakes and groundwater, so that the quality of the water supply remains safe.
3. To protect aquatic life and the ecological balance of the receiving water (by maintaining dissolved oxygen).
4. To avoid nuisance: bad smell, flies, mosquitoes, and unsightly conditions.
5. To meet the legal standards for effluent (Nepal's generic standards).
6. To recover resources: reuse of treated water for irrigation and industry, biogas from sludge, and sludge as manure.
7. To protect the soil and crops from clogging and contamination.
8. To dispose of the wastewater in the cheapest and safest way, with the least treatment that meets the above aims.

Assumptions

Streeter-Phelps equation with rate constants to base 10; the BOD values given are taken as the ultimate (first-stage) BODs, so $L_0$ is the BOD of the mixture; $K_1=0.1$ day⁻¹; the reaeration constant is found from the O'Connor-Dobbins formula (base $e$, 20°C):

$$K_{2(e)}=\frac{3.9\,v^{0.5}}{H^{1.5}}\ \text{day}^{-1}\quad(v\text{ in m/s},\ H\text{ in m}),\qquad K_2(\text{base }10)=\frac{K_{2(e)}}{2.303}$$

Step 1: Mixing at the outfall

$$\begin{aligned} Q_{mix}&=1.9+0.6=2.5\ \text{m}^3/\text{s}\\ DO_{mix}&=\frac{1.9(9.17)+0.6(1.8)}{2.5}=7.401\ \text{mg/L}\\ D_0&=9.17-7.401=1.769\ \text{mg/L}\\ L_0&=\frac{1.9(2.5)+0.6(210)}{2.5}=52.30\ \text{mg/L} \end{aligned}$$

Step 2: Reaeration constant and time of flow

$$\begin{aligned} K_{2(e)}&=\frac{3.9\times0.15^{0.5}}{1.4^{1.5}}=0.912\ \text{day}^{-1}\Rightarrow K_2=0.396\ \text{day}^{-1}\ (\text{base }10)\\ t&=\frac{x}{v}=\frac{50000\ \text{m}}{0.15\ \text{m/s}}=333333\ \text{s}=3.858\ \text{days} \end{aligned}$$

Step 3: Deficit at the section

$$D_t=\frac{K_1L_0}{K_2-K_1}\left(10^{-K_1t}-10^{-K_2t}\right)+D_010^{-K_2t}$$ $$\begin{aligned} D_t&=\frac{0.1\times52.30}{0.396-0.1}\left(10^{-0.3858}-10^{-1.5278}\right)+1.769\times10^{-1.5278}\\ &=6.80\ \text{mg/L} \end{aligned}$$

Answer: DO deficit 50 km downstream = 6.80 mg/L (DO = 2.37 mg/L).

Note: if the given BODs were 5-day values, $L_0$ would be $76.5$ mg/L, and the deficit would rise further (to about 9.9 mg/L, which is more than the saturation value, so the stream would turn septic).

Assumptions

Streeter-Phelps equation with rate constants to base 10; the BOD values given are taken as the ultimate (first-stage) BODs, so $L_0$ is the BOD of the mixture; $K_1=0.1$ day⁻¹; the reaeration constant is found from the O'Connor-Dobbins formula (base $e$, 20°C):

$$K_{2(e)}=\frac{3.9\,v^{0.5}}{H^{1.5}}\ \text{day}^{-1}\quad(v\text{ in m/s},\ H\text{ in m}),\qquad K_2(\text{base }10)=\frac{K_{2(e)}}{2.303}$$

Step 1: Mixing at the outfall

$$\begin{aligned} Q_{mix}&=2+0.5=2.5\ \text{m}^3/\text{s}\\ DO_{mix}&=\frac{2(9.17)+0.5(2)}{2.5}=7.736\ \text{mg/L}\\ D_0&=9.17-7.736=1.434\ \text{mg/L}\\ L_0&=\frac{2(3)+0.5(200)}{2.5}=42.40\ \text{mg/L} \end{aligned}$$

Step 2: Reaeration constant and time of flow

$$\begin{aligned} K_{2(e)}&=\frac{3.9\times0.2^{0.5}}{1.2^{1.5}}=1.327\ \text{day}^{-1}\Rightarrow K_2=0.576\ \text{day}^{-1}\ (\text{base }10)\\ t&=\frac{x}{v}=\frac{60000\ \text{m}}{0.2\ \text{m/s}}=300000\ \text{s}=3.472\ \text{days} \end{aligned}$$

Step 3: Deficit at the section

$$D_t=\frac{K_1L_0}{K_2-K_1}\left(10^{-K_1t}-10^{-K_2t}\right)+D_010^{-K_2t}$$ $$\begin{aligned} D_t&=\frac{0.1\times42.40}{0.576-0.1}\left(10^{-0.3472}-10^{-2.0008}\right)+1.434\times10^{-2.0008}\\ &=3.93\ \text{mg/L} \end{aligned}$$

Answer: DO deficit 60 km downstream = 3.93 mg/L (DO = 5.24 mg/L).

Note: if the given BODs were 5-day values, $L_0$ would be $62.0$ mg/L, and the deficit would rise further (to about 5.7 mg/L).

The critical point is where the oxygen deficit is maximum (Streeter-Phelps oxygen sag). The river DO is 80% of saturation, so $DO_r = 0.8\times 9.17 = 7.336$ mg/l. Rates are base 10: $K=0.1$/day, $R=0.5$/day.

Step 1: Conditions just after mixing

$$\begin{aligned} Q_{mix} &= Q_s+Q_r = 50 + 500 = 550\ l/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{50\times 400 + 500\times 4}{550} = 40.00\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{50\times 0 + 500\times 7.336}{550} = 6.67\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 6.67 = 2.50\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and self-purification constant

$$\begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{40.00}{1-10^{-5\times 0.1}} = 58.50\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.5}{0.1} = 5.00 \end{aligned}$$

Step 3: Critical time and critical deficit

$$\begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(5.00-1)}\log_{10}\left[5.00\left\{1-(5.00-1)\frac{2.50}{58.50}\right\}\right] = 1.544\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{58.50}{5.00}\times 10^{-0.1\times 1.544} = 8.20\ \text{mg/l} \end{aligned}$$

Answer: critical time $t_c \approx 1.54$ days (about 74.1 km downstream at 48 km/day), with critical deficit $D_c = 8.20$ mg/l, so the lowest DO is $9.17-8.20=0.97$ mg/l.

River DO $=0.7\times 9.17=6.419$ mg/l. Rates are base 10: $K_1=0.1$/day, $K_2=0.5$/day, so $f=5$. Flows are in l/s.

Step 1: Conditions just after mixing

$$\begin{aligned} Q_{mix} &= Q_s+Q_r = 100 + 1000 = 1100\ l/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{100\times 450 + 1000\times 4}{1100} = 44.55\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{100\times 0 + 1000\times 6.419}{1100} = 5.84\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 5.84 = 3.33\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and self-purification constant

$$\begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{44.55}{1-10^{-5\times 0.1}} = 65.15\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.5}{0.1} = 5.00 \end{aligned}$$

Step 3: Critical time and critical deficit

$$\begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(5.00-1)}\log_{10}\left[5.00\left\{1-(5.00-1)\frac{3.33}{65.15}\right\}\right] = 1.499\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{65.15}{5.00}\times 10^{-0.1\times 1.499} = 9.23\ \text{mg/l} \end{aligned}$$

Answer: critical DO deficit $D_c=9.23$ mg/l (at $t_c=1.499$ days, about 89.9 km downstream at 60 km/day). The lowest DO is $9.17-9.23=-0.06$ mg/l,

Note: the computed deficit is larger than the saturation DO (9.17 mg/l). In reality DO cannot fall below zero, so the stream would go anaerobic near the critical point and the effluent needs treatment before discharge. The formula is a numerical result of the data given.

Sewage (after suitable treatment) is finally disposed of by two natural methods.

 Sewage disposal
   |-- Dilution  (into a river, lake or sea)
   '-- Land treatment (irrigation / sewage farming, broad irrigation, filtration)

1. Disposal by dilution

Sewage is discharged into a river, lake or sea, where the large volume of water dilutes it and natural self-purification (with the oxygen dissolved in the water) oxidises the organic matter.

Essential conditions: a river with enough flow and DO; no downstream water supply intakes near the outfall; good mixing; no unsightly floating matter; dilution factor according to the table (as per common practice, Garg):

Advantages: cheap, simple, uses natural resources, no land needed. Disadvantages: pollutes the river if flow is small; odour and floating matter; harmful to downstream users; not feasible in dry season of small streams; sea disposal may affect the coast.

2. Disposal on land (land treatment)

Sewage is applied on the land (sewage farming/irrigation). Organic matter is oxidised by soil bacteria, the plants use nutrients, and the water is partly evaporated and partly percolates.

Conditions: sufficient land at a distance of several hundred metres to 1 km, porous soil (sandy loam), low groundwater table, suitable climate, and a skilled operation. Sewage must first be pre-treated (sedimentation) to avoid clogging ("sewage sickness" occurs when the soil pores clog).

Advantages: recovers manure, nutrients and water for crops, an income to the municipality, good removal of BOD and bacteria, no river pollution. Disadvantages: needs large land area (the rate is 1 ha for 250-500 persons or about 50 m³/ha/day); smell and mosquitoes; risk to health of workers and consumers of vegetables; contamination of groundwater; not useful in the rainy season or on frozen ground; sewage sickness.

Methods of land application: broad irrigation (sewage farming), surface/spray irrigation and intermittent sand filtration.

Self-purification of a river

Self-purification is the natural process by which a polluted stream recovers its original quality by physical, chemical and biological actions, without human help. When sewage enters a river, the oxygen in the water is used for decomposition, but atmospheric oxygen and photosynthesis restore it as the stream flows downstream.

Process

1. Dilution by river water reduces concentration of pollutants.
2. Sedimentation of the suspended matter.
3. Oxidation of organic matter by aerobic bacteria (deoxygenation): $\text{organic matter}+O_2\to CO_2+H_2O+\dots$
4. Reaeration (reoxygenation) from the atmosphere through the surface, aided by turbulence, and oxygen from aquatic plants (photosynthesis).
5. Sunlight kills bacteria; reduction by anaerobic decomposition of the deposited sludge; and predation by protozoa.

Oxygen sag curve and zones of pollution

The DO is used by bacteria (deoxygenation) at a rate $K_1L_t$ and replaced from air (reaeration) at a rate $K_2D_t$. The net result is the oxygen sag curve:

$$D_t=\frac{K_1L_0}{K_2-K_1}\left(10^{-K_1t}-10^{-K_2t}\right)+D_010^{-K_2t}$$

 DO
 sat |----.                          ______ recovery
     |     '.                      .'
     |       '.                  .'
     |         '.    Dc        .'
 min |           '.___v_____.-'
     |  Degradation | Active |Recovery| Clean
     |    zone      |decomp. | zone   | water
     +---------------------------------------> distance/time
   outfall          critical point

Zones

1. Zone of degradation (just downstream of the outfall): water turns turbid and grey, DO falls to about 40% of saturation, aerobic bacteria and fungi grow, and fish may disappear.
2. Zone of active decomposition: DO falls to a minimum (zero in a heavily polluted stream, then anaerobic with black colour, bubbles of gas and bad smell). The critical point (maximum deficit $D_c$) is at the lowest DO.
3. Zone of recovery: reaeration exceeds the deoxygenation; DO rises, water clears, fish return, and nitrates form.
4. Zone of clear water: the river is restored to its original state with DO near saturation, and with the BOD satisfied.

The critical time is $t_c=\dfrac{1}{K_2-K_1}\log_{10}\left[\dfrac{K_2}{K_1}\left(1-\dfrac{(f-1)D_0}{L_0}\right)\right]$, where $f=K_2/K_1$.

Factors affecting self-purification

Dilution (river flow), temperature, sunlight, velocity and turbulence, nature and amount of pollutant, DO of the river, rate of reaeration, sedimentation, and the presence of algae (explained under factors).

Self-purification is the natural recovery of a polluted stream. Its rate depends on the following factors.

1. Dilution: the greater the flow and volume of the river compared to the sewage, the more the pollutants are diluted and the more oxygen is available. A high dilution factor (above 500) means a fast recovery.
2. Temperature: high temperature increases the rate of bacterial decomposition ($K_T=K_{20}\,1.047^{T-20}$) and so the self-purification is quicker, but oxygen solubility falls, which reduces the DO. In cold weather the process is slow.
3. Sunlight: kills bacteria by UV rays and allows algae to produce oxygen by photosynthesis. Turbid water lowers this effect.
4. Velocity and turbulence of the stream: turbulence mixes the sewage and increases reaeration by exposing the water to the air. Fast, shallow, rocky streams recover faster than slow, deep ones.
5. Rate of reaeration: depends on the surface area, depth, wind and temperature; the constant $K_2$ is high in shallow, fast streams.
6. Dissolved oxygen in the river water: a higher initial DO gives more capacity for oxidising the pollutants without going anaerobic.
7. Sedimentation: the settleable solids settle in a quiet stream, which clarifies the water (but the deposited sludge then decomposes and may use oxygen).
8. Nature and quantity of the pollutant: a strong waste with high BOD, toxic industrial wastes, oil, and non-biodegradable matter reduce the rate. The more biodegradable the waste, the faster is the purification.
9. Presence of bacteria, algae, protozoa and plants: microorganisms decompose waste; algae give oxygen by photosynthesis; protozoa feed on bacteria.
10. Chemical actions: oxidation, reduction, coagulation by dissolved chemicals, and pH.
11. Time of travel / length of stream: self-purification needs time; a long stream with no new pollution allows full recovery.

Assumptions

The river upstream is saturated with oxygen (DO = 8.38 ppm) and has no BOD; the sewage has zero DO. The constants are in base 10 and refer to 25°C. Let $Q_r$ be the river flow required (m³/day).

Step 1: Sewage and its ultimate BOD

$$Q_s=1500\times200\ \text{L/d}=300\ \text{m}^3/\text{day}$$ $$L_s=\frac{BOD_5}{1-10^{-K_1\times5}}=\frac{350}{1-10^{-0.15\times5}}=\frac{350}{1-0.1778}=425.7\ \text{ppm}$$

Step 2: Allowable critical deficit

Minimum DO to be kept = 4 ppm, so the maximum (critical) deficit is

$$D_c=8.38-4=4.38\ \text{ppm}$$

Step 3: Mixture (let $x=\dfrac{Q_s}{Q_s+Q_r}$)

$$L_0=x\,L_s=425.7\,x,\qquad D_0=x\,(8.38-0)=8.38\,x$$

Self-purification factor $f=\dfrac{K_2}{K_1}=\dfrac{0.27}{0.15}=1.8$.

Critical time:

$$t_c=\frac{1}{K_2-K_1}\log_{10}\left[\frac{K_2}{K_1}\left(1-\frac{(f-1)D_0}{L_0}\right)\right]$$

Since $D_0/L_0=8.38/425.7=0.01969$ is independent of $x$,

$$t_c=\frac{1}{0.12}\log_{10}\left[1.8\left(1-0.8\times0.01969\right)\right]=2.07\ \text{days}$$

Critical deficit: $D_c=\dfrac{L_0}{f}10^{-K_1t_c}$

$$4.38=\frac{425.7\,x}{1.8}\times10^{-0.15\times2.07}=115.71\,x\ \Rightarrow\ x=0.03785$$

Step 4: River flow

$$Q_r=Q_s\left(\frac{1}{x}-1\right)=300\left(\frac{1}{0.03785}-1\right)=7625\ \text{m}^3/\text{day}$$

Answer: Minimum river flow = about 7625 m³/day (= 0.0883 m³/s or 88.3 l/s) for the 1500 people, i.e. a dilution of about 25.4 times the sewage flow.

The oxygen sag (Streeter-Phelps) analysis is applied to the mixture of river and wastewater. Constants are in base 10.

Step 1: Conditions after mixing

$$\begin{aligned} Q_{mix}&=0.3+5.5=5.8\ \text{m}^3/\text{s}\\ DO_{mix}&=\frac{0.3(0)+5.5(8.5)}{5.8}=8.06\ \text{mg/L}\quad(\text{putrefied sewage: DO}\approx0)\\ D_0&=9.1-8.06=1.04\ \text{mg/L}\\ BOD_5^{mix}&=\frac{0.3(450)+5.5(0)}{5.8}=23.28\ \text{mg/L}\\ L_0&=\frac{BOD_5}{1-10^{-0.1\times5}}=\frac{23.28}{0.6838}=34.04\ \text{mg/L} \end{aligned}$$

Step 2: Critical deficit without treatment

$$f=\frac{K_2}{K_1}=\frac{0.4}{0.1}=4$$ $$t_c=\frac{1}{K_2-K_1}\log_{10}\left[f\left(1-\frac{(f-1)D_0}{L_0}\right)\right]=\frac{1}{0.3}\log_{10}\left[4\left(1-\frac{3\times1.04}{34.04}\right)\right]=1.87\ \text{days}$$ $$D_c=\frac{L_0}{f}10^{-K_1t_c}=\frac{34.04}{4}\times10^{-0.1\times1.87}=5.54\ \text{mg/L}$$

Since $D_c=5.54$ mg/L is greater than the allowable deficit of 4.5 mg/L, treatment is necessary.

Step 3: Required degree of treatment

Find the maximum $L_0$ for which $D_c=4.5$ mg/L (with the same $D_0$, by trial using the same equations): $L_0=27.45$ mg/L. Then

$$BOD_5^{mix}=27.45\times0.6838=18.77\ \text{mg/L}$$

The effluent BOD₅ that can be allowed is:

$$BOD_5^{eff}=18.77\times\frac{5.8}{0.3}=362.8\ \text{mg/L}$$ $$\text{Efficiency of treatment}=\frac{450-362.8}{450}\times100=19.4\%$$

Answer: Treatment is necessary; the plant must remove about 19.4% of the BOD (effluent BOD₅ not more than about 362.8 mg/L).

The BOD is taken as the 5-day BOD at 20°C. Rates are base $e$ and are corrected to the mixture temperature with $K_T=K_{20}(1.047)^{T-20}$ and $R_T=R_{20}(1.016)^{T-20}$.

Step 1: Mixture properties

$$\begin{aligned} Q &= 0.18+0.46 = 0.64\ \text{m}^3/\text{s}\\ T_{mix} &= \frac{0.18\times 25+0.46\times 22}{0.64} = 22.84^\circ\text{C}\\ DO_{mix} &= \frac{0.18\times 1.5+0.46\times 8.3}{0.64} = 6.39\ \text{mg/l}\\ BOD_5^{mix} &= \frac{0.18\times 32+0.46\times 2}{0.64} = 10.44\ \text{mg/l}\\ DO_{sat}(22.84^\circ\text{C}) &= 8.855\ \text{mg/l (interpolated between 22 and 23}^\circ\text{C in the table)}\\ D_0 &= 8.855-6.39 = 2.47\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and corrected rate constants

$$\begin{aligned} L_0 &= \frac{BOD_5}{1-e^{-5k'}} = \frac{10.44}{1-e^{-5\times 0.2}} = 16.51\ \text{mg/l}\\ k'_T &= 0.2(1.047)^{2.84} = 0.2279\ \text{/day}\\ r'_T &= 0.4(1.016)^{2.84} = 0.4185\ \text{/day} \end{aligned}$$

Step 3: Critical time, deficit and location

$$\begin{aligned} t_c &= \frac{1}{r'-k'}\ln\left[\frac{r'}{k'}\left\{1-\frac{D_0(r'-k')}{k'L_0}\right\}\right] = 2.488\ \text{days}\\ D_c &= \frac{k'L_0}{r'}e^{-k't_c} = 5.10\ \text{mg/l}\\ x_c &= v\,t_c = 0.16\times 86400\times 2.488/1000 = 34.40\ \text{km} \end{aligned}$$

Answer: critical oxygen deficit $D_c = 5.10$ mg/l, occurring about $34.40$ km downstream of point A (after $2.488$ days). The minimum DO is $8.855-5.10 = 3.75$ mg/l.

The BOD is the 5-day BOD at 20°C. Rates are base $e$, corrected to the mixture temperature by $K_T=K_{20}(1.047)^{T-20}$ and $R_T=R_{20}(1.016)^{T-20}$. The saturation DO is read from the table at the mixture temperature.

Step 1: Mixture properties

$$\begin{aligned} Q &= 0.35+0.60 = 0.95\ \text{m}^3/\text{s}\\ T_{mix} &= \frac{0.35\times 29+0.60\times 22}{0.95} = 24.58^\circ\text{C}\\ DO_{mix} &= \frac{0.35\times 2+0.60\times 7}{0.95} = 5.16\ \text{mg/l}\\ BOD_5^{mix} &= \frac{0.35\times 155+0.60\times 2}{0.95} = 58.37\ \text{mg/l}\\ DO_{sat} &= 8.440\ \text{mg/l (interpolated between 24 and 26}^\circ\text{C)}\\ D_0 &= 8.440-5.16 = 3.28\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and corrected constants

$$\begin{aligned} L_0 &= \frac{58.37}{1-e^{-5\times 0.2}} = 92.34\ \text{mg/l}\\ k'_T &= 0.2(1.047)^{4.58} = 0.2468\ \text{/day}, \qquad r'_T = 0.35(1.016)^{4.58} = 0.3764\ \text{/day} \end{aligned}$$

Step 3: Critical time, deficit and location

$$\begin{aligned} t_c &= \frac{1}{r'-k'}\ln\left[\frac{r'}{k'}\left\{1-\frac{D_0(r'-k')}{k'L_0}\right\}\right] = 3.111\ \text{days}\\ D_c &= \frac{k'L_0}{r'}e^{-k't_c} = 28.09\ \text{mg/l}\\ x_c &= 0.25\times 86400\times 3.111/1000 = 67.21\ \text{km} \end{aligned}$$

Answer: critical oxygen deficit $D_c = 28.09$ mg/l at about $67.21$ km downstream of point A ($t_c=3.111$ days). Minimum DO $=8.440-28.09=-19.65$ mg/l.

Stream DO is 90% of saturation at 18°C: $DO_r = 0.9\times 9.54 = 8.586$ mg/l. The mixture temperature is $(1.5\times 30+7.5\times 18)/9 = 20^\circ$C, so the given $K=0.1$ and $R=0.5$ (base 10) and $DO_{sat}=9.17$ mg/l apply without correction.

Step 1: Conditions just after mixing

$$\begin{aligned} Q_{mix} &= Q_s+Q_r = 1.5 + 7.5 = 9\ m^3/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{1.5\times 280 + 7.5\times 1.2}{9} = 47.67\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{1.5\times 0.2 + 7.5\times 8.586}{9} = 7.19\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 7.19 = 1.98\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and self-purification constant

$$\begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{47.67}{1-10^{-5\times 0.1}} = 69.71\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.5}{0.1} = 5.00 \end{aligned}$$

Step 3: Critical time and critical deficit

$$\begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(5.00-1)}\log_{10}\left[5.00\left\{1-(5.00-1)\frac{1.98}{69.71}\right\}\right] = 1.616\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{69.71}{5.00}\times 10^{-0.1\times 1.616} = 9.61\ \text{mg/l} \end{aligned}$$

(a) Critical DO deficit

$D_c = 9.61$ mg/l (minimum DO $=9.17-9.61=-0.44$ mg/l).

(b) Location of the critical deficit

$$x_c = v\,t_c = 0.8\times 86400\times 1.616/1000 = 111.72\ \text{km downstream}$$

(c) BOD5 of a sample taken at the critical point

BOD remaining at time $t_c$ is $L_t = L_0\,10^{-Kt_c} = 69.71\times 10^{-0.1\times 1.616} = 48.05$ mg/l (ultimate). The 5-day BOD of that sample is

$$BOD_5 = L_t\left(1-10^{-5K}\right) = 48.05\times 0.6838 = 32.85\ \text{mg/l}$$

(d) Sag curve

Deficit at time $t$: $D_t=\dfrac{K L_0}{R-K}\left(10^{-Kt}-10^{-Rt}\right)+D_0\,10^{-Rt}$, and $DO_t = 9.17-D_t$.

 7.43 |*                                   * * *  *
 6.45 |                              * * *         
 5.47 |                         * *                
 4.48 |                     * *                    
 3.50 |                   *                        
 2.52 |  *            * *                          
 1.54 |            *                               
 0.56 |    *   * *                                 
-0.42 |      *                                     
      +--------------------------------------------
       0                                10  t, days

The DO falls from 7.19 mg/l to a minimum of -0.44 mg/l at $t_c=1.616$ days (the critical point), then recovers towards 9.17 mg/l as reaeration exceeds deoxygenation. Negative DO values only mean that the oxygen is exhausted (the stream turns anaerobic in that stretch); this effluent needs treatment.

Answer: $D_c=9.61$ mg/l; $x_c=111.72$ km; BOD5 at critical point $=32.85$ mg/l.

Treatment is necessary if the critical deficit of the mixture exceeds the permissible value, 4.77 mg/l (the limit given). Rates are base 10: $K=0.1$, $R=0.5$, so $f=R/K=5$.

Step 1: Deficit with no treatment

$$\begin{aligned} BOD_5^{mix} &= \frac{1.72\times 225+7.24\times 1.2}{8.96} = 44.16\ \text{mg/l}\\ L_0 &= \frac{44.16}{1-10^{-0.5}} = 64.59\ \text{mg/l}\\ t_c &= \frac{1}{0.1(5-1)}\log_{10}\left[5\left\{1-4\times\frac{2.16}{64.59}\right\}\right] = 1.591\ \text{days}\\ D_c &= \frac{64.59}{5}\,10^{-0.1\times 1.591} = 8.95\ \text{mg/l} \end{aligned}$$

Since $D_c = 8.95 > 4.77$ mg/l, treatment is necessary.

Step 2: Maximum permissible BOD of the mixture

The ultimate BOD $L_m$ of the mixture that just gives $D_c = 4.77$ mg/l (with $D_0=2.16$) is found from $D_c=\dfrac{L_m}{f}10^{-Kt_c}$ and the $t_c$ equation by trial. Solving gives $L_m = 33.06$ mg/l (check: $t_c=1.419$ days, $D_c=4.77$ mg/l).

Step 3: Permissible BOD of the sewage

River $L_r = 1.2/0.6838 = 1.75$ mg/l. From $Q_sL_s + Q_rL_r = L_m(Q_s+Q_r)$:

$$L_s = \frac{33.06\times 8.96 - 7.24\times 1.75}{1.72} = 164.85\ \text{mg/l}\quad (BOD_5 = 112.72\ \text{mg/l})$$

Raw sewage $L_s = 225/0.6838 = 329.06$ mg/l.

$$\text{BOD reduction} = \frac{329.06-164.85}{329.06}\times 100 = 49.9\%$$

Answer: treatment is necessary; BOD of the sewage must be reduced by about 49.9% (to about 112.72 mg/l BOD5).

Rates are base 10: $K=0.1$/day, $R=0.3$/day. The velocity is 2 km/h $=48$ km/day.

Step 1: Conditions just after mixing

$$\begin{aligned} Q_{mix} &= Q_s+Q_r = 2.0 + 20.0 = 22\ m^3/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{2.0\times 350 + 20.0\times 3.0}{22} = 34.55\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{2.0\times 0.8 + 20.0\times 7.140}{22} = 6.56\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 6.56 = 2.61\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and self-purification constant

$$\begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{34.55}{1-10^{-5\times 0.1}} = 50.52\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.3}{0.1} = 3.00 \end{aligned}$$

Step 3: Critical time and critical deficit

$$\begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(3.00-1)}\log_{10}\left[3.00\left\{1-(3.00-1)\frac{2.61}{50.52}\right\}\right] = 2.149\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{50.52}{3.00}\times 10^{-0.1\times 2.149} = 10.27\ \text{mg/l} \end{aligned}$$

Location of the critical deficit

$$x_c = v\,t_c = 48\ \text{km/day}\times 2.149\ \text{days} = 103.2\ \text{km}$$

Answer: the critical DO deficit ($D_c=10.27$ mg/l) occurs about $103.2$ km downstream of the industry, after $2.149$ days.

Flows are in m³/s ($Q_s=1.2$, $Q_r=5.0$). River DO $=0.9\times 9.17=8.253$ mg/l; sewage DO $=0$; the temperature is 20°C so $DO_{sat}=9.17$ mg/l.

Step 1: Mixing and permissible deficit

$$\begin{aligned} DO_{mix} &= \frac{1.2\times 0 + 5.0\times 8.253}{6.200} = 6.66\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17-6.66 = 2.51\ \text{mg/l}\\ D_{c(allowed)} &= DO_{sat}-DO_{min} = 9.17-4.0 = 5.17\ \text{mg/l} \end{aligned}$$

Step 2: Permissible BOD of the mixture

With $K=0.1$ and $R=0.3$ (base 10), $f=R/K=3.00$. The critical deficit is

$$t_c=\frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_m}\right\}\right],\qquad D_c = \frac{L_m}{f}10^{-Kt_c}$$

and $D_c$ must not exceed 5.17 mg/l. Solving these two equations for $L_m$ by trial (with $D_0=2.51$ fixed) gives $L_m=23.87$ mg/l, with $t_c=1.872$ days (check: $D_c=5.17$ mg/l).

Step 3: Permissible BOD of the effluent

$$\begin{aligned} L_{raw} &= \frac{BOD_5}{1-10^{-5K}} = \frac{160}{0.6838} = 234.00\ \text{mg/l}\\ L_r &= 2.92\ \text{mg/l (river)}\\ L_s &= \frac{L_m(Q_s+Q_r)-Q_rL_r}{Q_s} = \frac{23.87\times 6.200 - 5.0\times 2.92}{1.2} = 111.12\ \text{mg/l} \end{aligned}$$

So the treated effluent can have $BOD_5 \approx 75.98$ mg/l.

$$\text{Degree of treatment} = \frac{234.00-111.12}{234.00}\times 100 = 52.5\%$$

Answer: the sewage BOD5 must be reduced from 160 mg/l to about 75.98 mg/l, i.e. a treatment efficiency of about 52.5%.

Convert flows to m³/s: $Q_s = 2\times 10^6/(86400\times 1000) = 0.0231$ m³/s and $Q_r = 5.0$ m³/s. The river is at about 20°C, so I assume the standard constants $K=0.1$/day and $R=0.4$/day (base 10), $f=4$. Saturation DO after mixing is 9.2 mg/l, and the minimum DO to be kept is 4 mg/l.

Step 1: Mixing and permissible deficit

$$\begin{aligned} DO_{mix} &= \frac{0.0231\times 0 + 5.0\times 7.4}{5.023} = 7.37\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.2-7.37 = 1.83\ \text{mg/l}\\ D_{c(allowed)} &= DO_{sat}-DO_{min} = 9.2-4.0 = 5.20\ \text{mg/l} \end{aligned}$$

Step 2: Permissible BOD of the mixture

With $K=0.1$ and $R=0.4$ (base 10), $f=R/K=4.00$. The critical deficit is

$$t_c=\frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_m}\right\}\right],\qquad D_c = \frac{L_m}{f}10^{-Kt_c}$$

and $D_c$ must not exceed 5.20 mg/l. Solving these two equations for $L_m$ by trial (with $D_0=1.83$ fixed) gives $L_m=30.93$ mg/l, with $t_c=1.723$ days (check: $D_c=5.20$ mg/l).

Step 3: Permissible BOD of the effluent

$$\begin{aligned} L_{raw} &= \frac{BOD_5}{1-10^{-5K}} = \frac{5000}{0.6838} = 7312.38\ \text{mg/l}\\ L_r &= 0.00\ \text{mg/l (river)}\\ L_s &= \frac{L_m(Q_s+Q_r)-Q_rL_r}{Q_s} = \frac{30.93\times 5.023 - 5.0\times 0.00}{0.0231} = 6712.02\ \text{mg/l} \end{aligned}$$

So the treated effluent can have $BOD_5 \approx 4589.49$ mg/l.

$$\text{Degree of treatment} = \frac{7312.38-6712.02}{7312.38}\times 100 = 8.2\%$$

Answer: the effluent BOD5 must be brought from 5000 mg/l down to about 4589.49 mg/l, i.e. about 8.2% treatment (BOD removal) is required. The required degree is small only because the river dilution is very large (5 m³/s against 0.023 m³/s of waste); if the river flow were lower, much higher treatment would be needed.

Data and assumptions

• Sewage flow $=50000\times 140 = 7000\ \text{m}^3/\text{day} = 0.0810\ \text{m}^3/\text{s}$; river lean flow $Q_r=0.13$ m³/s.
• Raw sewage BOD: $\dfrac{0.07\times 50000\times 10^6\ \text{mg/day}}{7000\times 10^3\ \text{l/day}} = 500.0$ mg/l (taken as 5-day BOD).
• Assumed: upstream river is saturated ($DO_r=9.17$ mg/l, 20°C) and the treated sewage carries zero DO (conservative). $K_1=0.1$, $K_2=0.5$ (base 10), $f=5$. The deficit must not exceed $9.17-5=4.17$ mg/l.

Step 1: Mixing

$$\begin{aligned} DO_{mix} &= \frac{0.13\times 9.17+0.0810\times 0}{0.2110} = 5.65\ \text{mg/l}\\ D_0 &= 9.17-5.65 = 3.52\ \text{mg/l} \end{aligned}$$

Step 2: Permissible BOD of the mixture

Using $t_c=\dfrac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\dfrac{D_0}{L_m}\right\}\right]$ and $D_c=\dfrac{L_m}{f}10^{-Kt_c}=4.17$ mg/l, trial gives $L_m=25.51$ mg/l ($t_c=0.875$ days).

Step 3: Permissible BOD of the effluent

$$\begin{aligned} L_{r} &= \frac{3}{0.6838} = 4.39\ \text{mg/l}\\ L_s &= \frac{L_m(Q_s+Q_r)-Q_rL_r}{Q_s} = \frac{25.51\times 0.2110 - 0.13\times 4.39}{0.0810} = 59.39\ \text{mg/l}\\ L_{raw} &= \frac{500.0}{0.6838} = 731.24\ \text{mg/l} \end{aligned}$$ $$\text{Percentage purification} = \frac{731.24-59.39}{731.24}\times 100 = 91.9\%$$

Answer: about 91.9% BOD removal is required (effluent BOD5 about 40.61 mg/l).

River DO $=0.8\times 9.17=7.336$ mg/l. Both temperatures are 20°C, so no correction is needed. $K_1=0.1$, $K_2=0.4$ (base 10), $f=4$. Flows are in l/s.

Step 1: Conditions just after mixing

$$\begin{aligned} Q_{mix} &= Q_s+Q_r = 300 + 1300 = 1600\ l/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{300\times 400 + 1300\times 4}{1600} = 78.25\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{300\times 0 + 1300\times 7.336}{1600} = 5.96\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 5.96 = 3.21\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and self-purification constant

$$\begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{78.25}{1-10^{-5\times 0.1}} = 114.44\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.4}{0.1} = 4.00 \end{aligned}$$

Step 3: Critical time and critical deficit

$$\begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(4.00-1)}\log_{10}\left[4.00\left\{1-(4.00-1)\frac{3.21}{114.44}\right\}\right] = 1.880\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{114.44}{4.00}\times 10^{-0.1\times 1.880} = 18.56\ \text{mg/l} \end{aligned}$$

Answer: critical DO deficit $D_c = 18.56$ mg/l, reached after $t_c=1.880$ days (about 90.2 km downstream at 48 km/day). Minimum DO $=9.17-18.56=-9.39$ mg/l.

Note: the computed deficit is larger than the saturation DO (9.17 mg/l). In reality DO cannot fall below zero, so the stream would go anaerobic near the critical point and the effluent needs treatment before discharge. The formula is a numerical result of the data given.