Water pollution: a very short introduction

WATER
POLLUTION
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Precipitation on sea

The
Hydrologic
Cycle

458x10 km
3

3

Precipitation on land
119x103 km3
Evapotranspiration

Evaporation

72x103 km3

505x103 km3
Runoff
47x103 km3

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Water Resources
Resources
Water
Conventional
Conventional

Surface water
water
Surface

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Groundwater
Groundwater

Unconventional
Unconventional

Desalination
Desalination

Treated
Treated
wastewater
wastewater


Water Resources
About 2.2 billion people in developing countries lack access to safe drinking
water.

About 2.7 billion people in developing countries lack access to safe sanitation
services.

Water Use
Use
Water
Domestic
Domestic
Industrial
Industrial
Agricultural
Agricultural

Percent of withdrawal

100

Domestic
Industrial
Agricultural

80
60
40
20
0
Low

Mid

High

Income level

Water withdrawals by sector in low, mid and high income
countries (Environmental Science: A Global Concern, W.P. Cunningham
and B.W. Saigo, 3rd. Ed. Wm. Brown Pub. © 1995)

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Water Usage
Household Water Usage

Liters

Standard toilet, per flush

10-30

Shower head, per minute

20-30

Dishwasher, per load

50-120

Washing car with running water, 20 minute

400-800

Uncovered 60m2 pool, per day

100-400

Agricultural Items
One egg

150

Glass of milk

380

One kg of rice

4240

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Water pollution
pollution
Water

Physical
Physical
Radiological
Radiological

•Thermal
•Solids

Chemical
Chemical

Biological
Biological

•Hardness
•Pathogens
•Heavy metals
•Nutrients
•Pesticides
•Oxygen demanding waste
•Volatile organic compounds

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•Radio-isotopes


•Thermal Pollution
Causes a drop in the dissolved oxygen due to higher metabolic rate
and lower DO solubility at higher temperature.

•Solids
Could be suspended (causing turbidity) or dissolved causing
salinity).

Total dissolved solids (TDS) in water is the sum of all cations
and anions present expressed in mg/L.
Water use
Drinking

TDS, mg/L
<500

Irrigation
Animals

<2100
Up to 10,000

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•Hardness
Sum of divalent cations especially Ca and Mg.
Causes scale formation
Hardness is commonly expressed in meq/L or mg/L as CaCO3
Hardness
Hardness

Hardness
classification
of water

Description

meq/L

mg/L as CaCO3

Soft

<1

<50

Moderately
hard

1-3

50-150

Hard

3-6

150-300

Very hard
>6
>300
mg
Concentration of X ( mg / L) × 50 (mg CaCO 3 / meq)
of X as CaCO 3 =
L

Equivalent weight of X (mg / meq)

Equivalent weight =
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Atomic or molecular weight
equivalent number


Symbol

Atomic or
molecular weight

Equivalent
number

Equivalent weight
(mg/meq)

Al3+

27

3

9

Ca2+

40.1

2

20

1

1

1

Mg2+

24.3

2

12.2

K+

39.1

1

39.1

Na+

23

1

23

HCO3-

61

1

61

CO3 2-

60

2

30

35.5

1

35.5

NO3-

62

1

62

PO4 3-

95

3

31.7

SO42-

96

2

48

CaCO

100

2

50

H+

Cl-

3
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•Heavy Metals
Cations with specific gravity >4.
Examples: Hg, Pb, Cd, Cr, Co, Fe, Mn, Zn, etc.
Most heavy metals are toxic at low levels.
Some are useful as nutrients at low level but toxic at high level.

•Oxygen Demanding Waste
Substance that oxidize in water (most organics).
Cause a drop in the dissolved oxygen.

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•Pesticides
Classification based on intended use: insecticides, herbicides,
fungicides, etc.
Classification based on chemical structure: organochloroines
(ex. DDT), organophosphate, and carbamates.

•Nutrients
Include N, P, S, C, Ca, K, Fe, etc.
Allow algal growth

•Volatile Organic Compounds (VOCs)
Chlorinated compound, PCE, TCE, DCE, vinyl chloride, TCA
Aromatic hydrocarbons, BTEX
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Biochemical Oxygen Demand (BOD)
BOD: The amount of O2 needed by the microorganisms
to oxidize organic wastes.
Organic matter + O2

Microorganisms

CO2 + H2O +……….

BOD5: Amount of oxygen consumed in 5 days per liter of solution.
BOD
BOD

CBOD
CBOD

Carbonaceousoxygen
oxygendemand
demand
Carbonaceous

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NBOD
NBOD

Nitrogeneousoxygen
oxygendemand
demand
Nitrogeneous


Case 1: No Dilution
BOD5 = DOi - DO5

5 days

Dissolved oxygen
initial

at 20 oC

DOi

DO5

Sample

Sample

Dissolved oxygen after
5 days

Case 2: Diluted sample with blank free from organics
5 days
at 20 oC

DOi

DO5

Diluted
sample

Diluted
sample

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DOi − DO 5
BOD5 =
P
Vsample
P=
Vsample + Vdilution

Dilution factor


Case 3: Diluted sample with blank that has organics
5 days

5 days

at 20 oC

at 20 oC

DOi

DO5

DOi

DO5

Diluted
sample

Diluted
sample

Dilution
water

Dilution
water

BOD5 =

(DOi − DO5 ) diluted sample − (DOi − DO5 ) dilution (1 − P)

How to select P?

Two conditions have to be met:
(i) DO5 > 2 mg/L
(ii) DOi-DO5>2 mg/L
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P


Example 5.2
A test bottle containing dilution water has its DO level drop by 1.0
mg/L in a five-day test. A 300-ml BOD bottle filled with 15 ml of
wastewater and the rest dilution water experiences a drop of 7.2
mg/L in the same time period. What is the BOD5 of the waste?

Solution
P=

Vsample
Vsample + Vdilution

BOD5 =
BOD5 =

=

15
= 0.05
15 + 285

(DO i − DO5 ) diluted sample − (DOi − DO5 ) dilution (1 − P)

P
(7.2) − (1.0)(1 − 0.05)
= 125 mg / L
0.05

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Problem 5.4 with modification
A BOD test is to be run on a sample of wastewater that has a BOD5
of approximately 250 mg/L. If the initial DO of a mix of dilution
water and wastewater is 8 mg/L. What is the best dilution factor to
be used in order to determine the exact BOD5 of the sample?

Solution
Assume dilution water is free of organics

BOD5 =

 Two conditions have to be met in determining P:
(i) DO5 > 2 mg/L

(ii) DOi-DO5>2 mg/L

BOD5 =

DOi − DO5
8−2
⇒ 250 =
⇒ P = 1 / 41.7

P
P

BOD5 =

DOi − DO5
2
⇒ 250 = ⇒ P = 1 / 125
P
P

The best P value is the average of the two values
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DO i − DO5
P

P=1/83


Example 6.4
A groundwater sample has 100 mg/L Ca and 10 mg/L Mg. Classify
the water from a hardness point-of-view.

Solution
mg
Concentration of X (mg / L) × 50 (mg CaCO 3 / meq)
of X as CaCO 3 =
L
Equivalent weight of X (mg / meq)

mg
100 (mg / L) × 50 (mg CaCO 3 / meq)
of Ca as CaCO 3 =
= 250
L
20 (mg / meq)
mg
10 (mg / L) × 50 (mg CaCO 3 / meq)
of Mg as CaCO 3 =
= 41
L
12.2 (mg / meq)

So, the total hardness of the water is 291 mg/L as CaCO3 and the
water is classified as hard.
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