Water pollution

Water pollution is the contamination of water bodies (rivers, lakes, groundwater, oceans) by substances that degrade its quality and make it unfit for its intended use, harming humans and aquatic life.

Sources

• Point sources: discharge from a specific location: sewage outfalls, industrial effluents (tannery, textile, distillery, paper), hospital waste.
• Non-point sources: diffuse runoff from agricultural land (fertilizer, pesticides), urban storm water, mining areas and open dumping.

Types of pollutants and their effects

Water-borne diseases are the main health burden in developing countries; the aquatic ecosystem suffers from oxygen depletion, loss of species and eutrophication.

Water quality is judged by comparing measured parameters with standards for the intended use. Nepal's National Drinking Water Quality Standards (NDWQS, 2062 BS / 2005) give the maximum concentration allowed for drinking water; the values below are typical limits (as per NDWQS).

The bacteriological test is the most important since pathogens cause immediate disease; coliforms are used as indicator organisms because they are easy to detect and live in the human gut.

(a) BOD and COD

(b) BOD5 and ultimate BOD

Dilution factor $P = \dfrac{\text{volume of sample}}{\text{volume of mixture}} = \dfrac{10}{300} = 0.0333$.

$$\text{BOD}_5 = \frac{D_1 - D_2}{P} = \frac{8.6 - 4.1}{0.0333} = 135\ \text{mg/L}$$

First-order law: $\text{BOD}_t = L_0\,(1 - e^{-kt})$, so

$$L_0 = \frac{\text{BOD}_5}{1 - e^{-0.23\times5}} = \frac{135}{1 - e^{-1.15}} = \frac{135}{1 - 0.3166} = 197.6\ \text{mg/L}$$

(c) COD

$$\text{COD} = \frac{(A - B)\times N \times 8000}{V_{sample}}$$

where $A$ = blank (24.0 mL), $B$ = sample titre (14.0 mL), 8000 is the equivalent weight of oxygen in mg/eq.

$$\text{COD} = \frac{(24.0 - 14.0)\times0.1\times8000}{20} = 400\ \text{mg/L}$$ $$\frac{\text{BOD}_5}{\text{COD}} = \frac{135}{400} = 0.34$$

A ratio of about 0.34 indicates that a considerable part of the organic load is not easily biodegradable (domestic sewage is about 0.4-0.8).

Answer: BOD$_5$ = 135 mg/L; ultimate BOD $L_0$ = 198 mg/L; COD = 400 mg/L; BOD$_5$/COD = 0.34.

When organic waste enters a river, bacteria oxidise it and use up dissolved oxygen (DO) from the water (deoxygenation), while the stream takes oxygen from the atmosphere (reaeration). The plot of DO against time or distance downstream of the discharge point is the oxygen sag curve. It is a "sag" because DO first falls to a minimum and then recovers.

 DO
 DOsat |------\                      _______ recovery
       |       \                  _--
       |        \              _-
       |         \          _-
 DOmin |...........\______--  <- critical point (tc)
       |  clean | degrad. | active decomp. | recovery | clean
       +---------------------------------------> distance/time

Processes

• Deoxygenation rate $= k_1 L_t$ where $L_t = L_0 e^{-k_1 t}$ is remaining BOD (first order, $k_1$ = deoxygenation constant, $d^{-1}$).
• Reaeration rate $= k_2 D$ where $D = DO_{sat} - DO$ is the oxygen deficit (mg/L) and $k_2$ the reaeration constant.
• Net change of deficit:

$$\frac{dD}{dt} = k_1 L_0 e^{-k_1 t} - k_2 D$$

Integrating gives the Streeter-Phelps equation:

$$D_t = \frac{k_1 L_0}{k_2 - k_1}\left(e^{-k_1 t} - e^{-k_2 t}\right) + D_0 e^{-k_2 t}$$

The critical time (at $dD/dt = 0$) and deficit are:

$$t_c = \frac{1}{k_2 - k_1}\ln\left[\frac{k_2}{k_1}\left(1 - \frac{D_0 (k_2 - k_1)}{k_1 L_0}\right)\right], \qquad D_c = \frac{k_1}{k_2}L_0 e^{-k_1 t_c}$$

Zones of a polluted river

1. Clean zone (upstream), 2. Zone of degradation (turbid water, DO falling), 3. Zone of active decomposition (DO lowest, may be zero, foul smell, anaerobic), 4. Zone of recovery (DO rising, fish return), 5. Zone of clean water.

Factors affecting the critical point

• Waste load ($L_0$) and initial deficit ($D_0$): more waste, lower DO.
• $k_1$ and $k_2$: reaeration is higher in fast, shallow, turbulent streams, so the sag is smaller.
• Temperature (lower solubility and faster decay at high temperature), stream velocity and depth, and DO of the river water upstream.

Given: $L_0 = 20$ mg/L, $D_0 = 2$ mg/L, $DO_{sat} = 9.1$ mg/L, $k_1 = 0.2\ d^{-1}$, $k_2 = 0.4\ d^{-1}$, $u = 0.3$ m/s.

(a) Critical time

$$t_c = \frac{1}{k_2 - k_1}\ln\left[\frac{k_2}{k_1}\left(1 - \frac{D_0(k_2 - k_1)}{k_1 L_0}\right)\right]$$ $$\begin{aligned} \frac{D_0(k_2 - k_1)}{k_1 L_0} &= \frac{2\times0.2}{0.2\times20} = 0.1 \\ t_c &= \frac{1}{0.2}\ln\left[2\,(1 - 0.1)\right] = 5\ln(1.8) = 5\times0.5878 = 2.94\ \text{days} \end{aligned}$$

(b) Critical deficit and minimum DO

$$D_c = \frac{k_1}{k_2}L_0\,e^{-k_1 t_c} = \frac{0.2}{0.4}\times20\times e^{-0.2\times2.94} = 10\times e^{-0.5878} = 10\times0.5556 = 5.56\ \text{mg/L}$$ $$DO_{min} = DO_{sat} - D_c = 9.1 - 5.56 = 3.54\ \text{mg/L}$$

(c) Distance

Velocity $= 0.3\times86\,400 = 25\,920$ m/day.

$$x_c = 25\,920 \times 2.94 = 76\,200\ \text{m} \approx 76\ \text{km}$$

(d) Deficit after 1 day

$$\begin{aligned} D_1 &= \frac{0.2\times20}{0.4 - 0.2}\left(e^{-0.2} - e^{-0.4}\right) + 2\,e^{-0.4} \\ &= 20\,(0.8187 - 0.6703) + 2\times0.6703 = 2.968 + 1.341 = 4.31\ \text{mg/L} \end{aligned}$$

DO after 1 day $= 9.1 - 4.31 = 4.79$ mg/L.

Answer: $t_c$ = 2.94 days; $D_c$ = 5.56 mg/L; minimum DO = 3.54 mg/L; at about 76 km downstream; deficit after 1 day = 4.31 mg/L. The minimum DO is below the 4-5 mg/L usually needed for fish, so further treatment would be needed.

Municipal wastewater (sewage) is treated to remove solids, organic matter, nutrients and pathogens before discharge or reuse.

Raw sewage
   |
 [Screen] -> [Grit chamber]          PRELIMINARY
   |
 [Primary sedimentation tank] -> sludge ---+   PRIMARY
   |                                       |
 [Aeration tank] <--return sludge--+       |   SECONDARY
   |                              |        |
 [Secondary clarifier] -> excess sludge ->[Digester]->drying bed
   |
 [Disinfection (chlorine/UV)]               TERTIARY
   |
 Treated effluent -> river / reuse

Preliminary treatment

• Screens (bar racks) remove rags, plastics and large floating matter.
• Grit chamber (velocity about 0.3 m/s) settles sand and gravel; skimming removes oil and grease. It protects pumps and pipes.

Primary treatment

• Primary sedimentation (detention 1.5-2.5 h) settles suspended solids by gravity. It removes about 60% of SS and 30-35% of BOD. The settled sludge goes to the digester.

Secondary (biological) treatment

Removes dissolved and colloidal organic matter using microorganisms: activated sludge process, trickling filter, oxidation ponds, aerated lagoons. It removes 85-95% of BOD.

Activated sludge process: effluent from the primary tank is mixed with a recycled "activated" sludge rich in bacteria in the aeration tank, to which air is supplied by diffusers or surface aerators for 4-8 h. Bacteria oxidise the organic matter and form flocs. The mixture goes to a secondary clarifier where the flocs settle; part of the sludge (return activated sludge, RAS) goes back to maintain the biomass (MLSS about 2000-3000 mg/L), and the rest (waste sludge) is treated. Key design parameters: F/M ratio, sludge age and detention time.

Tertiary (advanced) treatment

• Removes nutrients (nitrogen and phosphorus), remaining solids (sand filtration) and pathogens by disinfection (chlorination, UV, ozone). Used when effluent standards are strict or the water is reused.

Sludge treatment

Thickening, anaerobic digestion (producing biogas), drying beds and then land application or disposal.