Characteristics and Examination of Wastewater

Importance of BOD value

BOD tells the amount of biodegradable organic matter in wastewater, i.e. its strength and polluting power. It is needed to

• design the treatment units and assess their performance (percentage BOD removal),
• decide the degree of treatment before disposal into a river, from the oxygen demand on the stream,
• check effluent against standards, and
• know the oxygen the sample would take from receiving water (cause of fish kill and odours). A high BOD shows heavily polluted wastewater (raw sewage 200-400 mg/L; treated effluent should be below 30-50 mg/L).

Principle

A diluted sample is incubated for 5 days at 20°C in the dark in a standard bottle, and the oxygen consumed by the microorganisms in oxidising the organic matter is measured by the dissolved oxygen (DO) before and after incubation.

Apparatus and reagents

BOD bottles (300 mL, glass-stoppered), incubator at 20 ± 1°C, pipettes, DO meter or Winkler titration set. Dilution water is prepared from distilled water, with 1 mL each per litre of phosphate buffer, $MgSO_4$, $CaCl_2$ and $FeCl_3$ solutions (nutrients for bacteria), and aerated to saturate with oxygen. Seed (settled domestic sewage) is added if the sample has no bacteria.

Procedure

1. Select dilutions. Sewage is too strong to hold the oxygen, so it is diluted in such a way that, after 5 days, at least 2 mg/L of DO is used up and at least 1 mg/L of DO remains. Take 1-5% for strong wastes (raw sewage).
2. Prepare the diluted samples: pour a measured volume of sample (e.g. 5 mL) into each of the BOD bottles and fill to the neck with dilution water without trapping air bubbles. Make a blank with only dilution water.
3. Initial DO ($D_1$): measure the DO of one set of bottles immediately (Winkler azide method or probe).
4. Incubate the second set, with stoppers sealed by a water seal, for 5 days at 20°C in the dark (to stop algae making oxygen).
5. Final DO ($D_2$): after 5 days measure the DO of the incubated bottles.
6. Calculate

$$BOD_5=\frac{D_1-D_2}{P}\ \text{mg/L},\qquad P=\frac{\text{volume of sample}}{\text{volume of diluted mixture}}$$

For example, 5 mL of sample in a 300 mL bottle gives $P=5/300=0.0167$. With $D_1=9$ and $D_2=5$ mg/L, $BOD_5=4/0.0167=240$ mg/L. For seeded samples, a seed correction is applied, and the blank is used to check the dilution water (its depletion should be below 0.2 mg/L).

The ultimate BOD can be found by $BOD_t = L(1-10^{-kt})$ if $k$ is known.

Necessity of examination of wastewater

1. To know the characteristics and strength (BOD, COD, solids, pH, nitrogen, bacteria) of the wastewater.
2. To design the treatment plant and choose the process.
3. To control and monitor the operation and efficiency of the units.
4. To decide the degree of treatment necessary before disposal into streams, or reuse in irrigation.
5. To check compliance with effluent standards and laws.
6. To study pollution of the receiving water and the effect on the public health and the environment.
7. To trace infiltration and industrial wastes in sewers, and to detect toxic wastes.

Sampling of wastewater

A sample must truly represent the wastewater, because the flow and strength change hourly.

Types of samples

• Grab (catch) sample: a single sample collected at one place and one time; used for pH, DO, temperature, chlorine, and where the flow is uniform.
• Composite sample: many small samples are collected at equal intervals (hourly) over 24 hours and mixed in proportion to the flow at that time. It gives the average quality and is used for BOD, solids, etc.
• Integrated sample (several points at one time) is used for wide channels.

Procedure and precautions

1. Select the sampling point: at an inlet works, a place with good mixing (flume, drop, manhole) and not at a deposit or a dead pool; sample at mid-depth and mid-stream.
2. Use clean polyethylene or glass bottles, rinse them 2-3 times with the wastewater, and fill completely (for BOD and DO, no air gap).
3. Take the sample quantity of 1-2 litres (more for composite).
4. Label with place, date, time, temperature and name of collector.
5. Preserve the sample at about 4°C (in ice), and analyse BOD within 6 hours (24 hours at the latest); add acid or other preservative for some tests as per Standard Methods.
6. Note site data (flow, weather, colour, odour) at the time.

The scanned value after incubation is read as 5.0 mg/L.

Dilution

$$P=\frac{\text{volume of sample}}{\text{volume of mixture}}=\frac{2.5}{250}=0.01$$

5-day BOD of raw sewage

$$BOD_5=\frac{D_1-D_2}{P}=\frac{8.0-5.0}{0.01}=300\ \text{mg/L}$$

Mass of BOD in 5 million litres

$$\text{Mass}=300\ \text{mg/L}\times5\times10^{6}\ \text{L}=1500000000\ \text{mg}=1500\ \text{kg}$$

Answer: BOD₅ = 300 mg/L; BOD in 5 ML = 1500 kg.

The scanned value after incubation, "51.0", is not possible (it is more than the initial DO of 7.8 mg/L), so it is read as 5.1 mg/L.

Dilution

$$P=\frac{2.5}{250}=0.01$$

5-day BOD of raw sewage

$$BOD_5=\frac{D_1-D_2}{P}=\frac{7.8-5.1}{0.01}=270\ \text{mg/L}$$

Mass of BOD in 3.2 million litres

$$\text{Mass}=270\ \text{mg/L}\times3.2\times10^{6}\ \text{L}=864000000\ \text{mg}=864\ \text{kg}$$

Answer: BOD₅ = 270 mg/L; BOD in 3.2 ML = 864 kg.

Practical approach to determine solids

1. Total solids (TS): a measured volume (100 mL) of well-mixed sample is evaporated in a pre-weighed dish on a water bath and dried at 103-105°C to constant weight. TS = (final − empty weight) / volume.

2. Suspended and dissolved solids: another sample is filtered through a weighed glass fibre filter. The dried residue on the filter is the suspended solids (SS); $TDS = TS - SS$.

3. Volatile and fixed solids: the dried residue is ignited in a muffle furnace at 550°C. The weight lost is the volatile (organic) solids; the ash remaining is the fixed (inorganic) solids.

4. Settleable solids: 1 litre of mixed sample is poured into an Imhoff cone and left undisturbed for 1 hour (after 45 min, the sides are gently stirred). The volume of sludge settled in the tip is read in mL/L. (The gravimetric way: settleable solids = SS of the raw sample − SS of the clear supernatant left after 1 h of settling.)

5. Non-settleable solids: the SS of the supernatant after 1 hour of settling, which are mainly colloids and fine particles; i.e.

$$\text{Non-settleable solids} = \text{Total suspended solids} - \text{Settleable solids}$$

Derivation of first-stage BOD expression

The first stage (carbonaceous) BOD is the oxygen used to oxidise organic carbon. It is a first order reaction: the rate of oxidation at any time is proportional to the organic matter still remaining (not yet oxidised).

Let $L_t$ be the oxygen equivalent of organic matter remaining at time $t$ and $K$ the rate constant (day⁻¹).

$$\frac{dL_t}{dt}=-K\,L_t$$

Separate variables and integrate between $t=0$ ($L_t=L$, the ultimate first-stage BOD) and time $t$:

$$\int_{L}^{L_t}\frac{dL_t}{L_t}=-K\int_0^t dt\ \Rightarrow\ \ln\frac{L_t}{L}=-Kt$$ $$L_t=L\,e^{-Kt}=L\cdot10^{-k t}\quad(k=K/2.303)$$

BOD exerted up to time $t$ is the amount used up:

$$y_t = L - L_t = L\left(1-e^{-Kt}\right)=L\left(1-10^{-kt}\right)$$

For example, with $k=0.1$/day, $BOD_5=0.68L$. The rate constant depends on temperature as $K_T=K_{20}(1.047)^{T-20}$.

BOD (Biochemical Oxygen Demand) is the amount of oxygen required by bacteria to oxidise (stabilise) the biodegradable organic matter in a water or wastewater sample under aerobic conditions at a specified temperature and time (standard: 5 days at 20°C). It is given in mg/L.

COD (Chemical Oxygen Demand) is the amount of oxygen needed to oxidise all organic matter (biodegradable and non-biodegradable) by a strong chemical oxidant, potassium dichromate ($K_2Cr_2O_7$) in boiling acid medium, expressed in mg/L. The test takes about 2-3 hours.

Significance of BOD

1. It measures the strength (organic pollution load) of sewage; raw domestic sewage has BOD₅ of 200-400 mg/L.
2. It is used to design treatment plants (trickling filter, activated sludge, ponds) and to find their efficiency.
3. It is used to find the oxygen demand on a receiving stream, and so the degree of treatment needed before discharge (self-purification calculations).
4. It is the standard of effluent regulation (for example, Nepal's generic effluent standard allows BOD of about 50 mg/L for discharge to inland surface water).
5. It measures only the biodegradable organic matter and gives an idea of how much will be decomposed in nature.

Significance of COD

1. It is a rapid test (hours against 5 days) and can be used for plant control.
2. It includes also non-biodegradable matter, so for industrial wastes (toxic to bacteria) where BOD cannot be found, COD is the measure.
3. COD is always greater than BOD. The ratio BOD/COD shows biodegradability: above about 0.5 the waste is easily treated biologically; below 0.3 it is difficult and contains toxic or refractory matter.
4. It is used with BOD to estimate the non-biodegradable part: $COD-BOD_u$.

Numerical (BOD at 30°C)

Given: $y_3$ at 25°C = 250 mg/L, $K_{20}=0.1$ day⁻¹ (base 10). Temperature coefficient $\theta=1.047$. The ultimate BOD $L$ does not change with temperature; only the rate constant changes.

$$\begin{aligned} K_{25}&=K_{20}(1.047)^{25-20}=0.1\times1.047^{5}=0.1258\ \text{day}^{-1}\\ L&=\frac{y_3}{1-10^{-K_{25}t}}=\frac{250}{1-10^{-0.1258\times3}}=430.5\ \text{mg/L}\\ K_{30}&=0.1\times1.047^{10}=0.1583\ \text{day}^{-1}\\ y_5(30^\circ\text{C})&=L\left(1-10^{-K_{30}\times5}\right)=430.5\left(1-10^{-0.7915}\right)=360.9\ \text{mg/L} \end{aligned}$$

Answer: 5-day BOD at 30°C = 360.9 mg/L (ultimate BOD = 430.5 mg/L).

BOD (Biochemical Oxygen Demand) is the amount of oxygen required by bacteria to oxidise (stabilise) the biodegradable organic matter in a water or wastewater sample under aerobic conditions at a specified temperature and time (standard: 5 days at 20°C). It is given in mg/L.

COD (Chemical Oxygen Demand) is the amount of oxygen needed to oxidise all organic matter (biodegradable and non-biodegradable) by a strong chemical oxidant, potassium dichromate ($K_2Cr_2O_7$) in boiling acid medium, expressed in mg/L. The test takes about 2-3 hours.

Significance of BOD

1. It measures the strength (organic pollution load) of sewage; raw domestic sewage has BOD₅ of 200-400 mg/L.
2. It is used to design treatment plants (trickling filter, activated sludge, ponds) and to find their efficiency.
3. It is used to find the oxygen demand on a receiving stream, and so the degree of treatment needed before discharge (self-purification calculations).
4. It is the standard of effluent regulation (for example, Nepal's generic effluent standard allows BOD of about 50 mg/L for discharge to inland surface water).
5. It measures only the biodegradable organic matter and gives an idea of how much will be decomposed in nature.

Significance of COD

1. It is a rapid test (hours against 5 days) and can be used for plant control.
2. It includes also non-biodegradable matter, so for industrial wastes (toxic to bacteria) where BOD cannot be found, COD is the measure.
3. COD is always greater than BOD. The ratio BOD/COD shows biodegradability: above about 0.5 the waste is easily treated biologically; below 0.3 it is difficult and contains toxic or refractory matter.
4. It is used with BOD to estimate the non-biodegradable part: $COD-BOD_u$.

Numerical

Given: $y_5=480$ mg/L, $K=0.25$ day⁻¹ (base $e$). First-stage BOD: $y_t=L\left(1-e^{-Kt}\right)$.

$$L=\frac{y_5}{1-e^{-Kt}}=\frac{480}{1-e^{-0.25\times5}}=\frac{480}{1-0.2865}=672.7\ \text{mg/L}$$

BOD remaining (unsatisfied) after 20 days: $L_{20}=Le^{-Kt}=672.7\times e^{-0.25\times20}=672.7\times0.00674=4.53$ mg/L.

$$\text{Proportion un-oxidised}=\frac{L_{20}}{L}=e^{-5}=0.00674\ (\approx0.67\%)$$

Answer: Ultimate BOD = 672.7 mg/L; proportion remaining after 20 days = 0.00674 (0.67%), i.e. about 4.53 mg/L.

BOD (Biochemical Oxygen Demand) is the amount of oxygen required by bacteria to oxidise (stabilise) the biodegradable organic matter in a water or wastewater sample under aerobic conditions at a specified temperature and time (standard: 5 days at 20°C). It is given in mg/L.

COD (Chemical Oxygen Demand) is the amount of oxygen needed to oxidise all organic matter (biodegradable and non-biodegradable) by a strong chemical oxidant, potassium dichromate ($K_2Cr_2O_7$) in boiling acid medium, expressed in mg/L. The test takes about 2-3 hours.

Significance of BOD

1. It measures the strength (organic pollution load) of sewage; raw domestic sewage has BOD₅ of 200-400 mg/L.
2. It is used to design treatment plants (trickling filter, activated sludge, ponds) and to find their efficiency.
3. It is used to find the oxygen demand on a receiving stream, and so the degree of treatment needed before discharge (self-purification calculations).
4. It is the standard of effluent regulation (for example, Nepal's generic effluent standard allows BOD of about 50 mg/L for discharge to inland surface water).
5. It measures only the biodegradable organic matter and gives an idea of how much will be decomposed in nature.

Significance of COD

1. It is a rapid test (hours against 5 days) and can be used for plant control.
2. It includes also non-biodegradable matter, so for industrial wastes (toxic to bacteria) where BOD cannot be found, COD is the measure.
3. COD is always greater than BOD. The ratio BOD/COD shows biodegradability: above about 0.5 the waste is easily treated biologically; below 0.3 it is difficult and contains toxic or refractory matter.
4. It is used with BOD to estimate the non-biodegradable part: $COD-BOD_u$.

Difference in brief

BOD (Biochemical Oxygen Demand) is the amount of oxygen required by bacteria to oxidise (stabilise) the biodegradable organic matter in a water or wastewater sample under aerobic conditions at a specified temperature and time (standard: 5 days at 20°C). It is given in mg/L.

COD (Chemical Oxygen Demand) is the amount of oxygen needed to oxidise all organic matter (biodegradable and non-biodegradable) by a strong chemical oxidant, potassium dichromate ($K_2Cr_2O_7$) in boiling acid medium, expressed in mg/L. The test takes about 2-3 hours.

Significance of BOD

1. It measures the strength (organic pollution load) of sewage; raw domestic sewage has BOD₅ of 200-400 mg/L.
2. It is used to design treatment plants (trickling filter, activated sludge, ponds) and to find their efficiency.
3. It is used to find the oxygen demand on a receiving stream, and so the degree of treatment needed before discharge (self-purification calculations).
4. It is the standard of effluent regulation (for example, Nepal's generic effluent standard allows BOD of about 50 mg/L for discharge to inland surface water).
5. It measures only the biodegradable organic matter and gives an idea of how much will be decomposed in nature.

Significance of COD

1. It is a rapid test (hours against 5 days) and can be used for plant control.
2. It includes also non-biodegradable matter, so for industrial wastes (toxic to bacteria) where BOD cannot be found, COD is the measure.
3. COD is always greater than BOD. The ratio BOD/COD shows biodegradability: above about 0.5 the waste is easily treated biologically; below 0.3 it is difficult and contains toxic or refractory matter.
4. It is used with BOD to estimate the non-biodegradable part: $COD-BOD_u$.

Derivation of BOD equation

The oxidation of organic matter by bacteria follows a first-order reaction: the rate of reaction at any time is proportional to the oxygen equivalent of organic matter remaining.

Let $L$ = ultimate first-stage BOD (total organic matter at $t=0$), $L_t$ = BOD remaining at time $t$, $y_t$ = BOD exerted (satisfied) in time $t$, and $K$ = rate constant (day⁻¹).

$$\frac{dL_t}{dt}=-K L_t$$ $$\int_{L}^{L_t}\frac{dL_t}{L_t}=-K\int_{0}^{t}dt\ \Rightarrow\ \ln\frac{L_t}{L}=-Kt\ \Rightarrow\ L_t=L e^{-Kt}$$

In base 10 form, $L_t=L\,10^{-kt}$, where $k=K/2.303$.

BOD remaining after time $t$ is therefore related to the ultimate BOD by

$$\boxed{L_t = L\,e^{-Kt}=L\,10^{-kt}}$$

and the BOD exerted is

$$y_t=L-L_t=L\left(1-e^{-Kt}\right)=L\left(1-10^{-kt}\right)$$

For $k=0.1$ per day at 20°C, $y_5=0.68L$ (about two-thirds of the ultimate BOD is satisfied in 5 days). The curve of $y_t$ against $t$ rises quickly at first, and approaches $L$ slowly.

First-stage BOD (carbonaceous)

In the first stage, bacteria oxidise the carbonaceous organic matter into CO₂ and water. It takes about 20 days at 20°C to be complete (about 70% in 5 days, 95% in 20 days). The first stage is called carbonaceous BOD (CBOD).

Second-stage BOD (nitrogenous)

After about 8-10 days, nitrifying bacteria (Nitrosomonas, Nitrobacter) oxidise ammonia to nitrite and then to nitrate, using more oxygen:

$$NH_4^+\xrightarrow{\text{Nitrosomonas}}NO_2^-\xrightarrow{\text{Nitrobacter}}NO_3^-$$

This is the nitrogenous BOD (NBOD). Because the nitrifiers grow slowly, it does not appear in the standard 5 day test at 20°C.

 BOD exerted
   |                  ______ second stage
   |            _____/ (nitrification)
   |      _____/___ 
   |   __/ first stage ends (L)
   |  /
   | /
   +----+----+----+----+----> days
   0    5   10   15   20

The first-stage curve is $y_t=L(1-10^{-kt})$. The second-stage curve starts at about the 8th-10th day and rises above it.

Ultimate BOD calculation

The rate constant is not given, so the standard value $K_{20}=0.1$ day⁻¹ (base 10) is assumed.

$$L=\frac{y_5}{1-10^{-kt}}=\frac{400}{1-10^{-0.1\times5}}=\frac{400}{1-0.3162}=585.0\ \text{mg/L}$$

Answer: Ultimate first-stage BOD = 585.0 mg/L (with $K=0.1$/day base 10; with $K=0.23$/day base $e$ the result would be the same).

Given: $y_1$ at 20°C = 300 mg/L, $k_{20}=0.1$ day⁻¹ (base 10). Assume $\theta=1.047$ for the temperature correction. The ultimate BOD $L$ is independent of temperature.

Step 1: Ultimate BOD

$$L=\frac{y_1}{1-10^{-k_{20}\,t}}=\frac{300}{1-10^{-0.1\times1}}=\frac{300}{1-0.7943}=1458.6\ \text{mg/L}$$

Step 2: Rate constant at 30°C

$$k_{30}=k_{20}(1.047)^{30-20}=0.1\times1.047^{10}=0.1583\ \text{day}^{-1}$$

Step 3: 5-day BOD at 30°C

$$y_5=L\left(1-10^{-k_{30}\times5}\right)=1458.6\left(1-10^{-0.7915}\right)=1222.9\ \text{mg/L}$$

Answer: 5-day BOD at 30°C = 1222.9 mg/L.

Given: $y_1$ at 20°C = 100 mg/L and $y_5$ at 25°C = 210 mg/L. Assumptions: first-order (base 10) reaction, ultimate BOD $L$ independent of temperature, and $k_T=k_{20}(1.047)^{T-20}$.

Step 1: Equations

$$\begin{aligned} 100&=L\left(1-10^{-k_{20}\times1}\right)\\ 210&=L\left(1-10^{-k_{25}\times5}\right),\qquad k_{25}=k_{20}(1.047)^{5}=1.2582\,k_{20} \end{aligned}$$

Dividing the two equations gives one unknown, $k_{20}$:

$$\frac{1-10^{-1.2582\,k_{20}\times5}}{1-10^{-k_{20}}}=2.1$$

Solving by trial (trial values of $k_{20}$ are tried until the ratio is 2.1; e.g. $k_{20}=0.2$ gives 2.560, $k_{20}=0.3$ gives 1.979):

$$k_{20}=0.273\ \text{day}^{-1},\qquad k_{25}=0.344\ \text{day}^{-1}$$

Step 2: Ultimate BOD

$$L=\frac{100}{1-10^{-0.273}}=214.1\ \text{mg/L}$$

(Check at 25°C: $y_5=214.1(1-10^{-0.344\times5})=210.0$ mg/L, as given.)

Step 3: BOD$_4$ at 22°C

$$\begin{aligned} k_{22}&=0.273\times1.047^{2}=0.2997\ \text{day}^{-1}\\ y_4&=L\left(1-10^{-k_{22}\times4}\right)=214.1\left(1-10^{-1.1986}\right)=200.5\ \text{mg/L} \end{aligned}$$

Answer: Ultimate BOD = 214.1 mg/L; rate constant $k_{20}$ = 0.273 day⁻¹ (base 10) ($k_{25}$ = 0.344); BOD₄ at 22°C = 200.5 mg/L.

Given at 25°C: $y_5=250$ mg/L, $y_{11}=325$ mg/L. Assume a first-order reaction in base 10, $y_t=L(1-10^{-kt})$, and $\theta=1.047$.

Step 1: Rate constant at 25°C

Divide the two equations to remove $L$:

$$\frac{y_{11}}{y_5}=\frac{1-10^{-11k}}{1-10^{-5k}}=\frac{325}{250}=1.3$$

Solving by trial: $k_{25}=0.1$ gives 1.346, $k_{25}=0.2$ gives 1.104; refining gives

$$k_{25}=0.1118\ \text{day}^{-1}\ (\text{base }10)$$

Step 2: Ultimate BOD

$$L=\frac{250}{1-10^{-0.1118\times5}}=345.4\ \text{mg/L}$$

Step 3: At 30°C

$$k_{30}=k_{25}(1.047)^{5}=0.1118\times1.2582=0.1406\ \text{day}^{-1}$$

Organic matter remaining after 6 days (the ultimate BOD is the same at any temperature):

$$L_6=L\cdot10^{-k_{30}\times6}=345.4\times10^{-0.8439}=49.5\ \text{mg/L}$$

Answer: $k_{25}$ = 0.1118 day⁻¹; organic matter remaining after 6 days at 30°C = 49.5 mg/L (BOD exerted = 295.9 mg/L).

Effect of temperature on the rate of decomposition

Decomposition of organic matter is done by bacteria, and their activity increases as temperature increases (up to about 35-40°C). The rate constant follows the van't Hoff-Arrhenius relation:

$$K_T = K_{20}\,\theta^{\,(T-20)}$$

where $\theta=1.047$ is the temperature coefficient (for the BOD reaction between 20 and 30°C; some books use 1.056 for 20-30°C). With $K_{20}=0.1$ per day:

• Higher temperature: the reaction is faster, the BOD is satisfied in a shorter time, and also less oxygen dissolves in water. The ultimate BOD $L$ remains the same.
• Lower temperature: bacterial activity slows, the BOD exerted in 5 days is less, and the sewage decomposes more slowly.

 BOD exerted
   |        ______________ 30 C
   |     __/  _____________ 20 C
   |   _/  __/  ___________ 10 C
   |  / __/ __/
   +--------------------------> time

All curves reach the same ultimate BOD $L$, but at different times. For this reason the standard BOD test is done at 20°C.

Numerical

Given $y_1=100$ mg/L at 20°C, $K_{20}=0.1$ day⁻¹ (base 10).

$$\begin{aligned} L&=\frac{y_1}{1-10^{-K t}}=\frac{100}{1-10^{-0.1}}=\frac{100}{1-0.7943}=486.2\ \text{mg/L}\\ y_5&=L\left(1-10^{-0.1\times5}\right)=486.2\left(1-0.3162\right)=332.5\ \text{mg/L} \end{aligned}$$

Answer: Five-day BOD at 20°C = 332.5 mg/L (ultimate BOD = 486.2 mg/L).

Step 1: BOD measured in the test (7 days at 25°C)

Dilution: 5 mL of sample in a 300 mL bottle, so the decimal fraction is

$$P=\frac{5}{300}=0.0167$$ $$BOD_7^{25}=\frac{D_1-D_2}{P}=\frac{9.2-5.3}{0.0167}=234\ \text{mg/L}$$

Step 2: Ultimate BOD

Here $K=0.1$ day⁻¹ (base 10) is taken as the rate constant at 20°C; the temperature coefficient is $\theta=1.047$. At 25°C:

$$K_{25}=0.1\times1.047^{5}=0.1258\ \text{day}^{-1}$$ $$L=\frac{BOD_7}{1-10^{-K_{25}t}}=\frac{234}{1-10^{-0.8807}}=269.5\ \text{mg/L}$$

Step 3: 4-day BOD at 30°C

$$\begin{aligned} K_{30}&=0.1\times1.047^{10}=0.1583\ \text{day}^{-1}\\ BOD_4^{30}&=L\left(1-10^{-K_{30}\times4}\right)=269.5\left(1-10^{-0.6332}\right)=206.8\ \text{mg/L} \end{aligned}$$

Answer: 4-day BOD at 30°C = 206.8 mg/L (ultimate BOD = 269.5 mg/L).

Given: $y_1$ at 22°C = 310 mg/L, $K_{20}=0.12$ day⁻¹ (base 10). Temperature coefficient $\theta=1.047$. The ultimate BOD $L$ is independent of temperature.

Step 1: Rate constant at 22°C and ultimate BOD

$$\begin{aligned} K_{22}&=0.12\times1.047^{2}=0.1315\ \text{day}^{-1}\\ L&=\frac{y_1}{1-10^{-K_{22}\times1}}=\frac{310}{1-10^{-0.1315}}=1186.3\ \text{mg/L} \end{aligned}$$

Step 2: Rate constant at 30°C

$$K_{30}=0.12\times1.047^{10}=0.1900\ \text{day}^{-1}$$

Step 3: 5-day BOD at 30°C

$$y_5=L\left(1-10^{-K_{30}\times5}\right)=1186.3\left(1-10^{-0.9498}\right)=1053.1\ \text{mg/L}$$

Answer: BOD₅ at 30°C = 1053.1 mg/L.

Given: $y_1$ at 23°C = 105 mg/L, $K_{20}=0.1$ day⁻¹ (base 10), $\theta=1.047$. The ultimate BOD does not change with temperature.

Step 1: Ultimate BOD

$$\begin{aligned} K_{23}&=0.1\times1.047^{3}=0.1148\ \text{day}^{-1}\\ L&=\frac{105}{1-10^{-K_{23}\times1}}=\frac{105}{1-10^{-0.1148}}=452.1\ \text{mg/L} \end{aligned}$$

Step 2: 5-day BOD at 30°C

$$\begin{aligned} K_{30}&=0.1\times1.047^{10}=0.1583\ \text{day}^{-1}\\ y_5&=L\left(1-10^{-K_{30}\times5}\right)=452.1\left(1-10^{-0.7915}\right)=379.0\ \text{mg/L} \end{aligned}$$

Answer: Five-day BOD at 30°C = 379.0 mg/L.

The data are read as: the 1-day BOD at 30°C is 170 mg/L, $K_{20}=0.12$ day⁻¹ (base 10), and the 5-day BOD at 20°C is required. $\theta=1.047$.

Step 1: Rate constant at 30°C

$$K_{30}=K_{20}(1.047)^{30-20}=0.12\times1.047^{10}=0.1900\ \text{day}^{-1}$$

Step 2: Ultimate BOD

$$L=\frac{y_1}{1-10^{-K_{30}t}}=\frac{170}{1-10^{-0.1900\times1}}=\frac{170}{1-0.6457}=479.9\ \text{mg/L}$$

Step 3: 5-day BOD at 20°C

$$y_5=L\left(1-10^{-K_{20}\times5}\right)=479.9\left(1-10^{-0.12\times5}\right)=479.9\left(1-0.2512\right)=359.3\ \text{mg/L}$$

Answer: The 5-day BOD at 20°C = 359.3 mg/L (ultimate BOD = 479.9 mg/L).

Given: $y_5=200$ mg/L at 37°C and rate of deoxygenation $k=0.17$ day⁻¹ at that temperature (taken as base 10, the usual form $y_t=L(1-10^{-kt})$, so no temperature correction is needed).

Ultimate BOD

$$L=\frac{y_5}{1-10^{-kt}}=\frac{200}{1-10^{-0.17\times5}}=\frac{200}{1-0.1413}=232.9\ \text{mg/L}$$

BOD remaining after 5 days

$$L_5=L-y_5=232.9-200=32.9\ \text{mg/L}\quad\left(=L\,10^{-kt}\right)$$

Answer: Ultimate BOD = 232.9 mg/L; BOD remaining after 5 days = 32.9 mg/L. (If the constant is taken as base $e$, $L=200/(1-e^{-0.85})=349.3$ mg/L.)

Check of the data

For a first-order reaction, $y_t=L(1-10^{-kt})$, the ratio $y_5/y_1$ cannot exceed 5 (it is 5 only when $k\to0$). Here

$$\frac{y_5}{y_1}=\frac{293.55}{56.15}=5.228>5$$

so the two readings cannot both belong to a single first-order curve; one of the printed values is doubtful. Therefore $K_{20}$ cannot be found from the pair, and I use the more reliable 5-day value together with the standard rate constant $K_{20}=0.1$ day⁻¹ (base 10), $\theta=1.047$.

Step 1: Rate constant at 25°C

$$K_{25}=K_{20}(1.047)^{25-20}=0.1\times1.047^{5}=0.1258\ \text{day}^{-1}$$

Step 2: Ultimate first-stage BOD

$$L=\frac{y_5}{1-10^{-K_{20}\times5}}=\frac{293.55}{1-10^{-0.5}}=\frac{293.55}{0.6838}=429.3\ \text{mg/L}$$

The ultimate BOD does not depend on temperature, so the ultimate BOD at 30°C is the same, 429.3 mg/L (only the rate constant changes with temperature; $K_{30}=0.1\times1.047^{10}=0.1583$ day⁻¹, so the 5-day BOD at 30°C would be 359.9 mg/L).

Answer: $K_{25}$ = 0.1258 day⁻¹ (base 10); ultimate BOD ≈ 429.3 mg/L.

Initial DO of the diluted sample

The dilution water (blank) has DO = 5 mg/L, the raw sample has DO = 0.5 mg/L, and the dilution is 5% ($P=0.05$ sample, 0.95 dilution water). Initial DO of the mixture:

$$D_1=5(0.95)+0.5(0.05)=4.775\ \text{mg/L}$$

The DO after 5 days at 20°C is $D_2=1$ mg/L.

5-day BOD

$$BOD_5=\frac{D_1-D_2}{P}=\frac{4.775-1}{0.05}=75.5\ \text{mg/L}$$

Ultimate BOD

The rate constant is not given, so the usual $K_{20}=0.1$ day⁻¹ (base 10) is assumed:

$$L=\frac{BOD_5}{1-10^{-0.1\times5}}=\frac{75.5}{0.6838}=110.4\ \text{mg/L}$$

Answer: BOD₅ = 75.5 mg/L; ultimate BOD ≈ 110.4 mg/L.

Sampling of wastewater

A sample must truly represent the wastewater, because the flow and strength change hourly.

Types of samples

• Grab (catch) sample: a single sample collected at one place and one time; used for pH, DO, temperature, chlorine, and where the flow is uniform.
• Composite sample: many small samples are collected at equal intervals (hourly) over 24 hours and mixed in proportion to the flow at that time. It gives the average quality and is used for BOD, solids, etc.
• Integrated sample (several points at one time) is used for wide channels.

Procedure and precautions

1. Select the sampling point: at an inlet works, a place with good mixing (flume, drop, manhole) and not at a deposit or a dead pool; sample at mid-depth and mid-stream.
2. Use clean polyethylene or glass bottles, rinse them 2-3 times with the wastewater, and fill completely (for BOD and DO, no air gap).
3. Take the sample quantity of 1-2 litres (more for composite).
4. Label with place, date, time, temperature and name of collector.
5. Preserve the sample at about 4°C (in ice), and analyse BOD within 6 hours (24 hours at the latest); add acid or other preservative for some tests as per Standard Methods.
6. Note site data (flow, weather, colour, odour) at the time.

BOD determination in the laboratory (dilution method)

A diluted sample is incubated for 5 days at 20°C in the dark in a standard bottle, and the oxygen consumed by the microorganisms in oxidising the organic matter is measured by the dissolved oxygen (DO) before and after incubation.

Apparatus and reagents

BOD bottles (300 mL, glass-stoppered), incubator at 20 ± 1°C, pipettes, DO meter or Winkler titration set. Dilution water is prepared from distilled water, with 1 mL each per litre of phosphate buffer, $MgSO_4$, $CaCl_2$ and $FeCl_3$ solutions (nutrients for bacteria), and aerated to saturate with oxygen. Seed (settled domestic sewage) is added if the sample has no bacteria.

Procedure

1. Select dilutions. Sewage is too strong to hold the oxygen, so it is diluted in such a way that, after 5 days, at least 2 mg/L of DO is used up and at least 1 mg/L of DO remains. Take 1-5% for strong wastes (raw sewage).
2. Prepare the diluted samples: pour a measured volume of sample (e.g. 5 mL) into each of the BOD bottles and fill to the neck with dilution water without trapping air bubbles. Make a blank with only dilution water.
3. Initial DO ($D_1$): measure the DO of one set of bottles immediately (Winkler azide method or probe).
4. Incubate the second set, with stoppers sealed by a water seal, for 5 days at 20°C in the dark (to stop algae making oxygen).
5. Final DO ($D_2$): after 5 days measure the DO of the incubated bottles.
6. Calculate

$$BOD_5=\frac{D_1-D_2}{P}\ \text{mg/L},\qquad P=\frac{\text{volume of sample}}{\text{volume of diluted mixture}}$$

For example, 5 mL of sample in a 300 mL bottle gives $P=5/300=0.0167$. With $D_1=9$ and $D_2=5$ mg/L, $BOD_5=4/0.0167=240$ mg/L. For seeded samples, a seed correction is applied, and the blank is used to check the dilution water (its depletion should be below 0.2 mg/L).

The ultimate BOD can be found by $BOD_t = L(1-10^{-kt})$ if $k$ is known.

Why examination of wastewater is necessary

1. To know the nature and strength (organic load, solids, nutrients, bacteria) of the wastewater.
2. To design the treatment plant and select the treatment process.
3. To control the operation and check the efficiency of each unit.
4. To decide the degree of treatment needed before disposal to a river or the land, or for reuse.
5. To check the standards (effluent limits) and to prevent water pollution and diseases.
6. To study industrial wastes, infiltration and illegal connections.

Total solids (TS)

A well-mixed sample of 100 mL is poured into a clean, dried, pre-weighed evaporating dish (weight $W_1$). It is evaporated on a steam bath and dried in an oven at 103-105°C for 1 hour, cooled in a desiccator and weighed ($W_2$).

$$TS\ (\text{mg/L})=\frac{(W_2-W_1)\times10^6}{\text{volume of sample (mL)}}$$

Total fixed and volatile solids

The same dish with the residue is ignited in a muffle furnace at 550 ± 50°C for 15-20 min, cooled and weighed ($W_3$). Organic matter burns away.

$$\text{Volatile solids}=\frac{(W_2-W_3)\times10^6}{V},\qquad \text{Fixed solids}=\frac{(W_3-W_1)\times10^6}{V}$$

so $TS = \text{fixed} + \text{volatile}$.

Dissolved and suspended solids (for completeness)

A filtered sample (glass fibre filter) is dried at 103°C to give suspended solids, and total dissolved solids = TS − suspended solids. Fixed and volatile solids are found in the same way for each fraction.

Typical values for raw domestic sewage: total solids about 700-1000 mg/L, of which about 50% are volatile.

All the solids are determined by weighing the residue after evaporation, filtration, settling or ignition.

Total solids (TS)

A well-mixed sample of 100 mL is poured into a clean, dried, pre-weighed evaporating dish (weight $W_1$). It is evaporated on a steam bath and dried in an oven at 103-105°C for 1 hour, cooled in a desiccator and weighed ($W_2$).

$$TS\ (\text{mg/L})=\frac{(W_2-W_1)\times10^6}{\text{volume of sample (mL)}}$$

Total fixed and volatile solids

The same dish with the residue is ignited in a muffle furnace at 550 ± 50°C for 15-20 min, cooled and weighed ($W_3$). Organic matter burns away.

$$\text{Volatile solids}=\frac{(W_2-W_3)\times10^6}{V},\qquad \text{Fixed solids}=\frac{(W_3-W_1)\times10^6}{V}$$

so $TS = \text{fixed} + \text{volatile}$.

Suspended and dissolved solids

A measured volume of sample is filtered through a weighed glass-fibre filter paper (Whatman GF/C). The filter with the residue is dried at 103-105°C and weighed. The weight gain gives the suspended solids (SS). Then dissolved solids = total solids − suspended solids.

Settleable solids

One litre of well-mixed sample is poured into an Imhoff cone and allowed to stand for 1 hour (after 45 minutes the sides of the cone are gently scraped or rotated to release the trapped matter). The volume of the solids settled at the bottom is read in mL/L. Settleable solids can also be found by weight:

$$\text{Settleable solids}=SS_{\text{raw sample}}-SS_{\text{supernatant after 1 h settling}}$$

Non-settleable solids

These are the solids which do not settle in 1 hour (colloids and very fine suspended matter):

$$\text{Non-settleable solids}=\text{Suspended solids}-\text{Settleable solids}$$

(or equal to the suspended solids in the supernatant after settling; some books also include dissolved solids in this term.)

 Sample
   |
   +-- evaporate 103 C ---> Total solids (TS)
   |        '-- ignite 550 C -> Volatile / Fixed
   +-- filter ---> Suspended (SS) / Dissolved (DS)
   '-- Imhoff cone 1 h --> Settleable / Non-settleable

Physical characteristics of wastewater

1. Colour: fresh sewage is light brownish-grey. As it becomes stale (septic), it turns dark grey and finally black because of sulphides. Colour tells the age and condition of the wastewater.
2. Odour: fresh sewage has a faint, musty smell. Septic sewage smells of hydrogen sulphide (rotten egg) and other gases like mercaptans. It is measured by the threshold odour number.
3. Temperature: usually 3-5°C higher than the water supply (about 20-25°C in Nepal). Temperature affects bacterial activity, DO solubility and the rate of decomposition. It is measured with a thermometer.
4. Turbidity: due to suspended and colloidal matter (clay, silt, organic matter); high in raw sewage. Measured in NTU with a nephelometer.
5. Solids: total solids are 500-1000 mg/L (suspended, dissolved, settleable).
6. Specific gravity is about 1.0 (nearly that of water).

Decomposition of wastewater

Organic matter in wastewater is broken down by bacteria. There are two ways.

1. Aerobic decomposition (in the presence of free oxygen) Aerobic bacteria oxidise organic matter using dissolved oxygen:

$$\text{Organic matter}+O_2\xrightarrow{\text{aerobic bacteria}}CO_2+H_2O+NO_3^-+SO_4^{2-}+\text{new cells}$$

Nitrogen goes to nitrates, sulphur to sulphates, and carbon to CO₂. It is rapid, odourless and the products are stable. It occurs in streams, oxidation ponds, trickling filters and activated sludge.

2. Anaerobic decomposition (in the absence of oxygen) When oxygen is exhausted, anaerobic bacteria work in two stages:

• Acid formation: complex organic matter is changed to organic acids and alcohols.
• Methane formation: these acids are converted to methane, carbon dioxide, ammonia and hydrogen sulphide:

$$\text{Organic matter}\xrightarrow{\text{anaerobic bacteria}}CH_4+CO_2+NH_3+H_2S+\text{new cells}$$

It is slow, and gives foul smell and a black colour, but methane is useful as biogas. It happens in septic tanks, sludge digesters and the bottom deposits of ponds.

3. Facultative decomposition: facultative bacteria work either with or without oxygen (as in facultative ponds).