Section I Introduction - Philadelphia University materi… · is produced per day? Answer 0.2 *...

Section I Introduction

1

Environmental Engineering (0670343)

Very Important Note: Most of the figures and tables are downloaded from google

and various text books

System Units:

1- International system (SI)

is the modern form of the metric system.

It is the world's most widely used system of units, both in everyday commerce

and in science.

The SI was developed in 1960 from the metre-kilogram-second (mks) system,

rather than the centimetre-gram-second (cgs) system which, in turn, had many

variants.

2- Non SI unit like U.S. and UK

In the United States, industrial use of SI is increasing, but popular use is still limited.

In the United Kingdom, conversion to metric units is official policy but not yet

complete. Those countries that still recognize non-SI units (e.g. the U.S. and UK)

have redefined their traditional non-SI units in terms of SI units.

Basic SI Units

Type Name Symbol

length metre m

mass kilogram kg

time second s

electric current ampere A

temperature kelvin K

amount of substance mole mol

luminous intensity candela cd

http://www.convertunits.com/type/length

http://www.convertunits.com/info/metre

http://www.convertunits.com/type/weight

http://www.convertunits.com/info/kilogram

http://www.convertunits.com/type/time

http://www.convertunits.com/info/second

http://www.convertunits.com/type/electric+current

http://www.convertunits.com/info/ampere

http://www.convertunits.com/type/temperature

http://www.convertunits.com/info/kelvin

http://www.convertunits.com/type/amount+of+substance

http://www.convertunits.com/info/mole

http://www.convertunits.com/type/luminous+intensity

http://www.convertunits.com/info/candela


2

SI prefixes

Factor Prefix Symbol

1024

1E24 yotta Y

1021

1E21 zetta Z

1018

1E18 exa E

1015

1E15 peta P

1012

1E12 tera T

109 1E9 giga G

106 1E6 mega M

103 1E3 kilo k

102 1E2 hecto h

101 1E1 deca da

101 1E1 deka da

10-1

1E-1 deci d

10-2

1E-2 centi c

10-3

1E-3 milli m

10-6

1E-6 micro µ

10-9

1E-9 nano n

10-12

1E-12 pico p

10-15

1E-15 femto f

10-18

1E-18 atto a

10-21

1E-21 zepto z

10-24

1E-24 yocto y

What to remember:

1ft = 12inch

1 inch = 2.54 cm

1 ft = 30.48 cm


3

1 US gallon = 3.78L

1 mile = 1609 m

I pound = 454g

Example:

Some employees at GE wash the PCB tainted floor with organic solvent (TCE) and

the discharge enters a holding tank that is 25 m x 25 m x 5 ft and is full with water.

The volume of solvent is 3 L and the concentration of PCBs in the solvent is 10 ppm.

What is the final concentration of PCB in mg/l in the holding tank?

Solution:

Tank Volume = 25*25*5*0.3048 = 937.5m3

Added solvent quantity (gramm) = 3L*10 mg/L = 30mg = 0.03g

The final concentration (mg/l) = 30/(937500+3) = mg/l

Reactor and Material balance:

Law of matter conservation: the matter can neither be created or destroyed

Law of Energy conservation: the energy can neither be created or destroyed but it

transformed form one type of energy to another

Material Balance:

- Is a key tools in achieving a quantitative understanding of the behavior of

environmental systems.

- It provide us with a tool for modeling the production, transport, and fate of

pollutants in the environment

Main equation:

Accumulation = input – output

- The control volume: it is a imaginary line around the process flowchart or part

of it in which the inputs, outputs and accumulations are determined to ease the

problem solution.


4

Example: a family brings approximately 50 kg of consumer good a week (food,

magazine , appliances…) of this amount 60% is consumed as food . half of the food is

used for biological maintenance and ultimately released as CO2 and the reminder is

discharged to sewer. The green recycle is about 25% of the solid waste that is

generated. Approximately 1kg is accumulated at home. Estimate the solid waste

amount that placed at curb each week??

consumer good food

(input)

Solid waste

Consumed food = 0.6 * 50 = 30 kg

Input = output (food) + output (solid waste) + accumulation

50 = 30 + solid waste + 1 total amount of generated solid waste = 19 kg

25% of the solid waste is recycled = 0.25 * 19 = 4.75 kg is recycled

Thus the solid waste send to curb side collection weekly = 19- 4.75 = 14.25 kg.

Time factor:

Mass rate of input = mass rate of output + mass rate of accumulation

𝑑𝑚𝑎𝑠𝑠𝑖𝑛𝑝𝑢𝑡

𝑑𝑡=

𝑑𝑚𝑎𝑠𝑠𝑜𝑢𝑡𝑝𝑢𝑡

𝑑𝑡+

𝑑𝑚𝑎𝑠𝑠𝑎𝑐𝑢𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛

𝑑𝑡

Example:

a tank with volume 0.35 m3 has a leak, the inlet tap flows water at rate 2.32 L/min

while the leak (outlet) is running at 0.32 L/min of water. how long it takes to fill the

tank? How much water has been wasted?

𝑑(𝑖𝑛)

𝑑𝑡=

𝑑(𝑜𝑢𝑡)

𝑑𝑡+

𝑑(𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛)

𝑑𝑡

Accumulation


5


𝑑𝑡= 2.32 – 0.32 = 2L/min

The tank volume = 0.35 m3 = 350 L

The time needed to fill the tank = 350 𝐿

2 𝐿/𝑚𝑖𝑛= 175 𝑚𝑖𝑛

The amount of wasted water = 175 min * 0.32 l/min = 56 L.

Efficiency = (𝒊𝒏−𝒐𝒖𝒕

𝒊𝒏) × 𝟏𝟎𝟎%

Example:

a waste incinerator has an air pollution control bag filter to collect the particulate

matter. The bag filter consists of 424 cloth bags arranged in parallel.where 1/424 of

the flow goes through each bag. The gas flow rate in and out of the filter is 47m3/sec.

and the concentration of the particulate entering the filter is 15g/m3. Under normal

operation the bag filter meets the authority discharge requirement of 24mg/m3

?estimate the removed particle in kg? estimate the efficiency of the bag filter? if due

to emergency one cloth bag is damaged and getout of the service, dose the effluent

comply with the standard?

Air inFlow Q= 47m3/sec out flow = 47m

3/sec

Cin= 15g/m3

Cout= 24mg/m3

Particulate removed by filter??

𝑑(𝑖𝑛)

𝑑(𝑣𝑜𝑙𝑢𝑚𝑒)=

𝑑(𝑜𝑢𝑡)

𝑑(𝑣𝑜𝑙𝑢𝑚𝑒)+


𝑑(𝑣𝑜𝑙𝑢𝑚𝑒)

Note: the accumulated particulate is discharged as waste.

Concentration of inlet particulate =

15000 𝑚𝑔

𝑚3×

47 𝑚3

𝑠𝑒𝑐= 705,000 mg/sec


6

Outlet concentration =

24 𝑚𝑔

𝑚3×

47 𝑚3

𝑠𝑒𝑐= 1800 mg/sec

1- Removed particle = 705,000 – 1,800 = 703,872mg/sec = 0.7038 kg/sec

2- Efficiency = (15000 – 24 / 15000)*100% = 99.84%

Or

Efficiency = (705000 - 1800 / 705000 )*100% = 0.9984

3- Since the flow is distributed over the bag clothes

Q =(1

424) 47𝑚3/sec

Bypass

Air in-Flow Q= (423

424) *47m

3/sec out flow = (

423

424) 47m

3/sec

Cin= 15g/m3

Cout= ??

Particulate removed by filter??

- Then Concentration of Bypass = 15000 *1

424 * 47 = 1,662.7 mg/sec

- The inflow = 15000 𝑚𝑔

𝑚3 ×423

424× 47

𝑚3

𝑠𝑒𝑐 = 703,337.2 𝑚𝑔/𝑠𝑒𝑐

- The efficiency of the bag filter is constant = 99.84%

- Thus the removed particulate by filter = 0.9984 × 703,337.2 𝑚𝑔/ sec =

702,211.9

- Then the amount of particulate left the system = in – out

- = 703,337.2 𝑚𝑔/ sec − 702,211.9 = 1,125.2 𝑚𝑔/𝑠𝑒𝑐

Thus the total outlet concentration = out from bypass + filter outflow

= 1,662.7 mg/sec + 1,125.2 𝑚𝑔/ sec = 2,787.9 𝑚𝑔/𝑠𝑒𝑐


7

The overall concentration in the effluent = 2,787.9 𝑚𝑔/𝑠𝑒𝑐

47 𝑚3 /𝑠𝑒𝑐= 59.3 𝑚𝑔/𝑚3

The new efficiency = 115000−59

15000× 100% = 99.61%

Example:

A wastewater treatment plant with an output of 38400m3/day discharges the liquid

effluent with a BOD of 20mg/L into a river. If the BOD of the river upstream of the

discharge point is 0.2mg/l, at a minimum flow of 20m3/s, compare the BOD of the

river downstream of the discharge, assuming complete mixing?

Discharge amount = 20𝑚𝑔

𝐿× 38400

𝑚3

𝑑𝑎𝑦∗ 1000

𝐿

𝑚3 = 768 kg/day = 8888mg/sec

River flow = 0.2g/m3 *

20 𝑚3

𝑠𝑒𝑐 = 4g/sec

Plant discharge = 38400 m3/day = 0.44 m

3/sec

Thus the final concentration downstream = 4000𝑚𝑔

𝑠𝑒𝑐 + 8888

𝑚𝑔

𝑠𝑒𝑐/(20,000 + 440) l/sec

= 0.64mg/l

Example:

A slurry containing 20 percent by weight of limestone (CaCO3) is processes to

separate pure dry limestone from water. If feed rate of the slurry is 2000kg/h, how

much CaCO3 is produced per day?

Answer 0.2 * 2000 = 400 kg/h = 400*24 = 9,600 kg/day

HW#1:

1. Each day 3780 m3 of wastewater is treated at a municipal wastewater

treatment plant. The influent contains 220 mg/L of suspended solids. The

clarified water has a suspended solids concentration of 5mg/L. Determine the

mass of sludge produced daily from the clarifier and write down the mass

balance of the clarifier.

Answer (215g/m3

*3780m3/day = 812.7 kg/day)


8

Section II Reactors

9

Reactor and steady state:

in most of environmental system a transformation occurs within the system and by

product is formed (CO2; bio-mass formed…)

Accumulation rate = input rate – output rate ±𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒

𝑑𝑀

𝑑𝑡=

𝑑(𝑖𝑛)

𝑑𝑡−

𝑑(𝑜𝑢𝑡)

𝑑𝑡 + 𝑟 …… (**)

Where r is the rate of reaction that is a complex function of pressure, temperature and

reactants and products concentrations.

Analysis performance of reactor type

Reactor Types:

1- Batch reactor

2- Continuous flow stirred tank reactor (CSTR)

3- Plug flow reactor (PFR)

1- Batch reactor :

- There is no in or out (it is like domestic mixer) fill in the material ,

- Mix and allow time for reaction to occur then discharge.

- Most experiment done in such reactor, because it is cheap, easy to operate and

build

𝑑𝑀

𝑑𝑡=

𝑑(𝑖𝑛)

𝑑𝑡−

𝑑(𝑜𝑢𝑡)

𝑑𝑡 + 𝑟

influent Effluent

Qin

CAin

Q out

C Aout

V (m3)

Reactor

Section II Reactors

10

In this reactor 𝑑(𝑖𝑛)

𝑑𝑡=

𝑑(𝑜𝑢𝑡)

𝑑𝑡= 0 , and M = C.V

Thus the main reaction equation is (**)

𝑑𝑀

𝑑𝑡= 𝑉

𝑑𝐶

𝑑𝑡= 𝑟 𝑉

𝑑𝐶

𝑑𝑡= −𝑘𝐶𝑉

𝑑𝐶

𝑑𝑡= −𝑘𝐶 𝑙𝑛

𝐶𝑜𝑢𝑡

𝐶𝑖𝑛 = −𝑘t

Example:

a contaminated soil is to be excavated and treated in a completely mixed

aerated lagoon. How long time it is needed to treat the soil. If the

following data obtained from lab scale batch reactor, assume first order

reaction. Estiamte the reaction constant k? and the required time to

achieve reduction of 99% of the original concentration?

Tiem (day) Waste concentration (mg/l)

1 280

16 132

Solution:

During 15 days (16-1) the waste concentration decreased from 280-132

Using the (**) equation

𝑑𝑀

𝑑𝑡= 𝑟 = −𝑘𝐶

𝑑𝐶

𝑑𝑡= −𝑘𝐶

𝐶𝑡

𝐶𝑜= 𝑒−𝑘 (15)

Section II Reactors

11

𝑙𝑛 132

280= −15 × 𝑘

K = 0.0501 d-1

thus to achieve 99% reduction: 𝐶𝑡

𝐶𝑜 = 0.01

then ln(0.1) = 0.0501 (t) t = 91.9 day = 92 day

What dose it mean??

No material in or out

Section II Reactors

12

2- CSTR: or equilization basin

𝑑𝑀

𝑑𝑡=

𝑑(𝑖𝑛)

𝑑𝑡−

𝑑(𝑜𝑢𝑡)

𝑑𝑡 + 𝑟

Where M = CV

Then the equation (**) become

𝑉𝑑𝐶

𝑑𝑡= 𝐶𝑖𝑛𝑄𝑖𝑛 − 𝐶𝑜𝑢𝑡𝑄𝑜𝑢𝑡 − 𝑘𝐶𝑜𝑢𝑡𝑉

For completely mix Qin = Qout, and divide by tnak volme (𝑉)

Residence time (to)= V/Q

𝑑𝐶

𝑑𝑡=

𝐶𝑖𝑛

𝑡𝑜−

𝐶𝑜𝑢𝑡

𝑡𝑜− 𝑘𝐶𝑜𝑢𝑡

for stedy state dc/dt = 0

Section II Reactors

13

𝐶𝑖𝑛

𝑡𝑜=

𝐶𝑜𝑢𝑡

𝑡𝑜+ 𝑘𝐶𝑜𝑢𝑡

𝐶𝑖𝑛

𝑡𝑜= 𝐶𝑜𝑢𝑡 (

1

𝑡𝑜 + 𝑘)

𝐶𝑜𝑢𝑡 = 𝐶𝑖𝑛

1 + 𝑘𝑡𝑜

Section II Reactors

14

Section II Reactors

15

Example:

A sewerage lagoon with surface area = 10 ha and water depth of 1m. The lagoon

receive daily waste water = 430 m3/day with pollutant concentration = 180 ppm. The

pollutant degradation in the lagoon is according the first order reaction with K = 0.7

day-1

find the concentration of the effluent ?

Solution: Assume

1- The lagoon is completely mixed

2- Their is no water losses

3- And steady state

𝑑𝑀

𝑑𝑡=

𝑑(𝑖𝑛)

𝑑𝑡−

𝑑(𝑜𝑢𝑡)

𝑑𝑡 + 𝑟

𝑉𝑑𝐶

𝑑𝑡= 𝐶𝑖𝑛𝑄𝑖𝑛 − 𝐶𝑜𝑢𝑡𝑄𝑜𝑢𝑡 − 𝑘𝐶𝑜𝑢𝑡𝑉

For completely mix Qin = Qout, and divide by tnak volme (𝑉)

Residence time (to)= V/Q

𝑑𝐶

𝑑𝑡=

𝐶𝑖𝑛

𝑡𝑜−

𝐶𝑜𝑢𝑡

𝑡𝑜− 𝑘𝐶𝑜𝑢𝑡

Section II Reactors

16

for stedy state dc/dt = 0

𝐶𝑜𝑢𝑡 = 𝐶𝑖𝑛

1 + 𝑘𝑡𝑜

Tank volume = 10 *104 (m

2) * 1m = 10,000 m

3

residence time at the tank (to) = V/Q = 100,000/(430) = 232.558 day

Cout = 180 / (1+ 0.7*232.558) = 1.09 mg/l

Example 2 : CSTR

Before entering the underground utility to do a maintenance, the air sample was

analyzed and found to contain 29 mg/m3 of H2S. The maximum allowable exposer

limit is 14 mg/m3. Thus the worker ventilate vault (القبو) using an air blower that has a

flow of 10 m3/min. if the vault volume is 160 m

3 estimate how long time it is required

to lower H2S concentration to the allowable limit?

Assume the vault behaves as CSTR and the H2S is non-reactive?

The general MB equation is

𝑑𝑀

𝑑𝑡=

𝑑(𝑖𝑛)

𝑑𝑡−

𝑑(𝑜𝑢𝑡)

𝑑𝑡 + 𝑟

But since no inlet and no reaction (non reactive)

𝑑𝑀

𝑑𝑡= −

𝑑(𝑜𝑢𝑡)

𝑑𝑡

V𝑑𝑐

𝑑𝑡= −𝑄𝑜𝑢𝑡𝐶𝑜𝑢𝑡

But the residence time τ = V/Q

Ct = Co Exp(-t/τ)

where

Section II Reactors

17

Ct : is the concentration at any time

Co : is the initial concentration

We need to calculate the residence time

τ = 160 𝑚3

10 𝑚3/𝑚𝑖𝑛 = 16 min

Now the required time is

14 𝑚𝑔/𝑚3

29𝑚𝑔

𝑚3

= exp (-t/ 16)

Thus the required time t = 11.6 min

Note: the H2S threshold is 0.18 mg/m3 thus the vault will keep have a very strong

odor (ad smell) even after 12 min of blowing

Concentration

(ppm)

Symptoms/Effects

0.00011-

0.00033

Typical background concentrations

0.01-1.5 Odor threshold (when rotten egg smell is first noticeable to some). Odor becomes

more offensive at 3-5 ppm. Above 30 ppm, odor described as sweet or sickeningly

sweet.

2-5 Prolonged exposure may cause nausea, tearing of the eyes, headaches or loss of

sleep. Airway problems (bronchial constriction) in some asthma patients.

20 Possible fatigue, loss of appetite, headache, irritability, poor memory, dizziness.

50-100 Slight conjunctivitis ("gas eye") and respiratory tract irritation after 1 hour. May

cause digestive upset and loss of appetite.

100 Coughing, eye irritation, loss of smell after 2-15 minutes (olfactory fatigue).

Altered breathing, drowsiness after 15-30 minutes. Throat irritation after 1 hour.

Gradual increase in severity of symptoms over several hours. Death may occur

after 48 hours.

Section II Reactors

18

100-150 Loss of smell (olfactory fatigue or paralysis).

200-300 Marked conjunctivitis and respiratory tract irritation after 1 hour. Pulmonary

edema may occur from prolonged exposure.

500-700 Staggering, collapse in 5 minutes. Serious damage to the eyes in 30 minutes.

Death after 30-60 minutes.

700-1000 Rapid unconsciousness, "knockdown" or immediate collapse within 1 to 2 breaths,

breathing stops death within minutes.

1000-2000 Nearly instant death

Source : OSHA (US department of labor)

3- PFR:

it can be represented as a pipe or long river.

- Biological treatment of domestic waste water (WW) usually done in long

narrow tank that may modeled as PFR

𝑑𝑀

𝑑𝑡=

𝑑(𝑖𝑛)

𝑑𝑡−

𝑑(𝑜𝑢𝑡)

𝑑𝑡 + 𝑟 …… (**)

But because no mass excange accure during the flow in the pipe

d(in) and d(out) = 0

Thus the equation (**) became as batch reactor

𝑑𝑀

𝑑𝑡= 𝑟 𝑉

𝑑𝐶

𝑑𝑡= −𝑘𝐶𝑉

𝑑𝐶

𝑑𝑡= −𝑘𝐶 𝑙𝑛

𝐶𝑜𝑢𝑡

𝐶𝑖𝑛= −𝑘t

Another form of the equation:

t= is the travel time and equalt L/U

where

L: length of plug-flow segment

Section II Reactors

19

U: is the Linear velocity, m/s

In (Cout/Cin)=-K L/U= -k V/Q

Example:

(exam): a waste water plant needs to disinfect its effluent , the treated water contains

4.5 × 105 fecal coliform colony forming unit (CFU) per liter. But the maximum

allowable concentration is 2000 CFU per liter. If the pipe is used to carry the water,

determine the required pipe length if the water flow inside the pipe by 0.75m/sec. and

the reaction rate constant for fecal coliform destruction is 0.23 min-1

Solution: assume that the pipe behave a s steady state plug flow reactor and the

reaction rate is first order:

ln𝐶𝑜𝑢𝑡

𝐶𝑖𝑛 = −𝑘𝑡 ln

2000

4.5 × 105= −(0.23)𝑡

Required residence time (t) = 23.5 min = 1,412.9 sec

But t = 𝑣𝑜𝑙𝑢𝑚𝑒 (𝑉)

𝑄 =

𝐴𝑟𝑒𝑎∗𝐿

𝑣∗𝐴𝑟𝑒𝑎= 𝐿/𝑣

0.75 m/sec * 1412.9 = 1059.7 = 1060 m

Section II Reactors

20

General Material Balance equation for first order reaction rates:

influent Effluent

Qin

CAin

Q

out

C Aout

V (m3)

Reactor

Section II Reactors

21

VdC/dt =ΣCinQin − ΣCoutQout±kCV

C = concentration in the control volume (river/stream/reactor) [=] mg/L

V = volume of control volume [=] L, m3, ft

3

Qin = flowrate of inlet streams [=]m3/s, L/s, cfs, MGD

Qout = summation of all outlet streams

[=]m3/s, L/s, cfs, MGD

Cin = concentration in each inlet stream [= ] mg/L

Cout = concentration in each inlet stream [= ] mg/L

k = 1st order reaction rate constant (will be given) [=] 1/s

t = time [=] sec, mi

Section III Water Chemistry

22

Water Quality: Introduction

Very Important Note: Most of the figures and tables are downloaded from google

and various text books

It is a branch of Civil engineering deals with study, design and construct of drinking

and waste water plant and their network. It consists of

a- Water supply engineering: it deals with drinking water (collection, treatment,

storage and distribution)

b- Waste water engineering: deals with domestic and industrial waste water

(collection, treatment and disposal)

c- Environmental Engineering: deals with pollutant and their reduction, treatment

and disposal.

water uses are

Drinking, removing, or diluting wastes, agriculture, producing and goods

manufacturing and energy production

Required water quantity depends on various factors:

I. climatic conditions,

II. Extreme weather

III. lifestyle and city size

IV. Culture and tradition: and activities within the community.

V. Diet and population characteristics

VI. technology,

VII. and wealth


23

Source of water:

Chemistry: Review:

Molecular weight: is the mass of a molecule. It is a sum of the combining element

[g/mole].

Equivalent weight: is the molecular weight divided by the number of positive or

negative electrical charges that resulting from compound dissolution [g/eq. weight]

Molarity (M): [No. of mole/L]: is the number of mole per litter of solution

Molality [mole/kg]is the number of mole per kg of material

Normality (N) [g equ./l]: is the number of equivalent weight per litter of solution

It is N = M*n

ppm : part per million and equals to mg/l

Fe2 (SO4)3 has a MW = 400 [g/mole]

-since its dissolution produces 6 charge

thus its eq. weight = 400/6 = 66.7 [g/eq. weight]

https://en.wikipedia.org/wiki/Molecule


24

Examples:

(1) H2SO4 + 2OH-= 2H2O + SO4

2-

Molecular weight of sulfuric acid (H2SO4) = (2*1) + (1*32) + (4*16) = 98.07 g/mol

Equivalent weight of sulfuric acid (H2SO4) = 49.03 g/equivalent of H+

Reasoning:

98.07 g/mol H2SO4 * (1 mole H2SO4 / 2 equivalents of H+) = 49.03 g/equivalent of H

+

In this example, the magnitude of the equivalent weight of sulfuric acid is HALF of

that of the molecular weight. This is because according to the balanced chemical

reaction, one mole of sulfuric acid reacts with TWO equivalents of hydroxide.

Example 2:

NH4OH + H+ = H2O + NH4

+

Molecular weight of ammonium hydroxide (NH4OH) = 35.00 g/mol

Equivalent weight of ammonium hydroxide (NH4OH) = 35.00 g/equivalent of OH-

Reasoning:

35.00g/mol of NH4OH * (1 mole NH4OH / 1 equivalent of OH-) = 35.00 g/equivalent

of OH-

In this example, the magnitude of the equivalent weight of ammonium hydroxide is

the same as the molecular weight.This is beacuse according to the balanced chemical

reaction, one mole of ammonium hydroxide reacts with ONE equivalent of H+.

Example 3:

Calculate the equivalent weights of HNO3 (MW = 63) and Ga(OH)3

(MW = 121) in the following three reactions:

- (a) 3HNO3 + Ga(OH)3 → 3H2O + Ga(NO3)3

EW(HNO3) = 63/1 = 63g

EW[Ga(OH)3] = 121/3 = 40.3g


25

All 3 OH- enter the reaction. Note the 3 water formed and that no unused OH

- are

found in the salt Ga(NO3)3.

- (b) HNO3 + Ga(OH)3 → H2O + Ga(OH)2(NO3)

EW(HNO3) = 63/1 = 63g

EW(Ga(OH)3] = 121/1 = 121g

Since only 1 H+ is available to react with the 3 OH

- from the Ga(OH)3, only 1 OH

-

reacts. Note also that only one water forms and two OH- are left unused in the salt

Ga(OH)2NO3.

- (c) 2HNO3 + Ga(OH)3 → 2H2O + Ga(OH)(NO3)2

EW(HNO3) = 63/1 = 63g

EW[Ga(OH)3] = 121/2 = 60.5g

Two OH- are used by the 2H

+ provided by the 2 moles of HNO3. Note also that 2

moles of water form and 1 OH- is left unused in the product Ga(OH)(NO3)2.

Example:

500 mg of anhydrous Ca(HCO3)2 is dissolved in 750 ml of water. What is the molar

concentration, normality and the concentration in ppm as expressed as CaCO3?

Mw of Ca(HCO3)2 = 162gm/mole

The equivalent weight = 162/2 = 81g/equivelant

Concentration of Ca(HCO3)2 in the solution = 500/0.75 =666.67mg/l (ppm)

No of mole = 500/162 = 3.08 mmole

Morality concentration = 3.08/0.75 = 4.12 m mole/L

Normality concentration = 500/(81*0.75) = 8.24 N

666.67mg

L

81𝑔

𝑒𝑞𝑢𝑖𝑣𝑒𝑙𝑎𝑛𝑡

= 8.24𝑚𝑒𝑞𝑢𝑖

𝐿

Thus the concentration as CaCO3 = 8.24*50 = 412mg/l as CaCO3


26

Example:

what mass of CO2 would be produced if (100g) of butane (C4H10) is completely

oxidized into CO2 and water?

2C4H10+ 13O2 8CO2 + 10H2O

Solution : from the balanced chemical equation

Each 2 mole of butan produces 8 mole of CO2

- Determine the number of mole of butane

No of mole = weight/MW = 100/58 = 1.724 molw

2 mole of CH4 8 mole of CO2

1.72 of CH4 X

Thus X= 1.72*8 / 2 = 6.8 mole of CO2 is produced

MW of CO2 = 44 g/mole

Thus the amount of CO2 is 6.8 * 44 = 303 g of CO2

Example: Commercial Sulphuric acid H2SO4 is often purchased as 93% solution. Find the

concentration (mg/l),the molarity and normality of the solution. If the specific gravity

of H2SO4 = 1.839.

Solution:

If H2SO4concentration is 100% , thus its itsspecific gravity will be 1.839

but since the concentration is only 93%

thus the solution S. gravity = 0.93* 1.839 = 1.710

MWH2SO4 = 2*1 + 1*32+4*16 = 98g/mole

Assume one litter of the solution:

Thus it contain 1710 g/l

Thus the molarity = 1710

𝑔

𝑙

98𝑔

𝑚𝑜𝑙𝑒

= 17.45𝑚𝑜𝑙𝑒

𝑙 𝑜𝑟 17.45 𝑀

Normality : MWH2SO4 = 98/2 = 48

N= 1710/48 = 34.9 eq/l


27

Example: find the weight of sodium bicarbonate [NaHCO3] that is required to prepare 1M

solution?

MWNaHCO3= 84g/mole

1mole/l= mass

84𝑔/𝑚𝑜𝑙𝑒 = mass = 84g/l = 84,000mg/l


28

Water

The source of water supply may be generally classified as

(a) Surface waters: (i) Rivers (ii) Lakes (iii) Impounding reservoirs

(b) Ground waters: (i) Springs (ii) Infiltration galleries (iii) wells.

GENERAL CHARACTERISTICS OF SURFACE WATERS:

1. More suspended impurities are present in surface waters.

2. Surface water may contain more pathogenic bacteria as pathogenic

bacteria are generally embodied in suspended impurities.

3. Surface water generally has color due to presence of Iron-oxide,

mineralogical compounds, suspended particles…. Etc

GENERAL CHARACTERISTICS OF GROUND WATERS:

1. Ground water contain large amount of the dissolved impurities.

2. It is generally free from suspended matter as it gets strained during its passage

through the porous underground strata.

3. It is soft or hard as it comes into contact with the Geological formations.

4. The Bacterial content is usually low due to the straining of suspended

impurities.

5. It may have taste and odour due to presence of organic matter dissolved during

passage through underground strata.

Water quality is studied for the following reasons:

1- Determine the degree of pollution

2- Determine the required step for W.T. P (water treatment plant)

3- Proper design for treatment unit

4- Check the effluent of WTP in reference with the specified standards.

Characteristic of water:

1- Physical characteristics

2- Chemical characteristics

3- Microbiological characteristics

4- Radiological characteristics


29

Quality assessment of Harvested Rainwater for domestic use in Jordan

- The water samples were from Zarqa, Irbid, Jerash and Ajloun

- Water demand (available ) for Jordanian in 2008 is 160 m3/capita / year and is

expected to reach 90 m3/capita / year by 2025

Sources of Rainwater Harvesting In Jordan

1- Roof top of houses and institutions (schools, hospitals…etc)

2- Yard and garden

3- Street

Rainwater Harvesting In Jordan For

1- Drinking and domestic purposes (washing, bathing, cooking)

2- Irrigation and livestock

The Harvested Water Is Stored In:

1- Concrete tank : made from (bricks, reinforced concrete

Usually near the houses and collect the water from the roof

2- Rock tank with cement interior :

Usually in gardens and yards and collect the water street and yards

Quality of Harvested Rainwater depends on:

1- Location (in which governorate)

2- Catchment area

3- Public sewer system availability or use a septic tank

Effect of location on harvested rain water quality:

In Zarqa : the water was polluted by air pollution in the city that emitted from

chimneys (chemical pollutant such as iron (Fe),lead (pb), chromium (Cr), high

alkalinity and hardness)


30

It was free from Chemical Oxygen Demand (COD) and Nitrate (NO3)

- For other governorate the main pollutants were Nitrate and biological

contaminants due to agricultural activities in Irbid, Ajloun and Jerash.

Effect of catchment area:

1- Roof top catchment:

-usually the main pollutants are from air and chimneys

-organic pollutant such as (COD, NO3, Ammonia) usually exist in yard and garden

catchment

2- Rock tank: since it collects the water from yard the main pollutants were

COD, Nitrate and biological

Effect of tank types:

1- Concrete tank: since it is included with houses , thus it collects the water from

the roof top.

2- Rock tank: it collects the water from yard thus it has same catchment area

effects.

Effect of public sewer availability:

- If public sewer is available: less biological contamination is expected

- If No public system : means dependant on septic tanks there is high risk of

biological contamination due to low maintenance and monitoring levels.

General conclusion: the harvested rainwater is suitable only for irrigation due

to effect of biological and pollutants contaminations.

Ammonia and nitrate (NO3) is sign for fertilizer and livestock pollution.


31

A- Physical characteristics: it is associated with water appearance and includes

Parameter Range for Drinking Water (DW)

Temperature 15:20 oC

Turbidity ≤ 1 𝑁𝑇𝑈 (Nephlometric turbidity unit)

Color Colorless

Taste and Odor No odor

B- Chemical Characteristics:

- can be evident be their reaction like hardness , PH residual elements

- it includes also toxic substances such as organic matters, Nitrate (NO3),

Cyanides (CN) and heavy metals.

- It controlled by maximum allowable pollutants (organic, non-organic)

concentrations

Parameter Maximum Range for Drinking Water

(DW)

PH 6.5-8.5

Acidity

Alkalinity

Dissolved Oxygen

Hardness >300 mg/l

Chlorides >250 mg/l

sulphates 150-200 mg/l

Nitrates >45 mg/l

Iron 0.1 – 0.3 mg/l

Solids - Total dissolved Solid (TDS) >500*

- suspended solids

*if there is no other source it my increased to 1500 mg/l

C- Microbiological characteristics

- Mainly caused by microorganism activities like bacteria, Viruses, protozoa

- E-coli Fecal Coliform and total coliform tests are commonly used to

determine the water pollution with bacteria

- Their presence in water indicates pollution with human or animal fecal.

- Controlled by safe drinking water standards and wastewater effluent standard

- Usually the result presented by presence absence test, most probable

number (MPN)


32

- For Filtered water Pseudomonas aeruginosa is tested , because it comes as a

result of unclean old filter.

D- Radiological characteristics

- The water should not contain any radioactive materials, that cause chronic or

cancer diseases

Solids: Types of impurities in water:

1- Suspended Solids (SS):

It is the large particle in water with diameter between (10-1

: 10-3

mm) such as

clay, salt, sand.

Note: ground water is usually free from SS? Why

-It cause the water turbidity

- Can be removed by filtration or sedimentation

-Turbidity measurement principle:


33

2- Colloidal matters: its smaller particle size usually (10-3

: 10-6

mm) and can’t

be settled alone

- Can be removed by very high force centrifugation or using an additive

(coagulant)

3- Dissolved solids (D.S): it is the salts and dissolved organic matter

- Usually cause water problem like calcium and potassium carbonate and

bicarbonate

- Can be removed by distillation, adsorption , ion –exchange or liquid extraction

What is water conductivity? Ability of water to conduct

Electrical current

Max .value 5000 µS/cm

TDS [ppm] = 1.56 of

conductivity [µS/cm]

Why? Sugar and salt


34

Exercise: At home: Bring a cup of water and dissolve the following:

1- One spoon salt (NaCl)

2- Single spoon of Sugar (C6 H12 O6)

3- Single spoon of very fine spice (any type)

4- Single spoon of fine sand

Describe to which W characteristic each one belong with justification?

SS Testing summary:

1- Dry known weight of water sample at 103 oC to 105

oC to drive off water in the

sample.

2- The residue is cooled, weighed, and dried again at 550oC to drive off volatile

solids in the sample.

3- The total, fixed, and volatile solids are determined by comparing the mass of the

sample before and after each drying step. Where

Laboratory procedure to determine TS and TVS Source: UMT

Total Suspended solid (SS) = ws

V

Volatile solid = 𝑤𝑠−𝑤𝑓

𝑉

Note: for water the each 1000ml= 1000g


35

Example:

If the following data is obtained from a 100ml of water sample ,calculate TSS and

VS?

filter mass = 1.5413g

Mass after drying at 105 oC= 1.5541g

Mass after igniting at 550oC= 1.5519g

Solution:

Suspended solid (SS) (𝑤𝑠= 1.5541−1.5413

100/1000= 0.128 g/l = 128mg/l

Volatile solid = 𝑤𝑠−𝑤𝑓

𝑉 =

1.5541−1.5519

100/1000= 0.022 g/l = 22mg/l

Fixed solid (𝑤𝑓) = 1.5519−1.5413

100/1000= 0.106 g/l = 106mg/l

Dissolved Oxygen (DO):

- The source of D.O in water is photosynthesis and aeration

- It is one of important parameters to measure the water quality.

- It gives pleasure taste to water

- As the temp D.O

- If the D.O concentration decreases to less than 4mg/l all fish die

- If the D.O concentration is less than 2 mg/l all organism dies and the water is

called septic water

- Best D.O concentration is between 8-10 ppm. Optimum is 9ppm.

- The maximum naturally accrued is 14 mg/l.


36

Biochemical Oxygen demand (BOD) :

- is the quantity of oxygen that is used by microorganism to stabilize the wastewater,

Usually measured after 5 days.

- a BOD test can be used to measure waste loadings to treatment plants, plant

efficiency and the effects of a discharge on a receiving stream, and to control the

plant process.

- It is indicator for the required aeration amount.

- The main equation describes the process is:

DO + organic matter CO2 + biological growth

- Drinking water usually has a BOD of less than 1 mg/L

- Ordinary domestic sewage may have a BOD of 200 mg/L.

-Any effluent to be discharged into natural bodies of water should have BOD less than

30 mg/L.

Test Summary:

1- The sample is filled in an airtight bottle and incubated at 20 oC for 5 days.

2- The dissolved oxygen (DO) content of the sample is determined before and

after five days of incubation at 20°C

3- and the BOD is calculated from the difference between initial and final DO.

The initial DO is determined shortly after the dilution is made; all oxygen uptake


37

occurring after this measurement is included in the BOD measurement

Calculations:

BOD5 mg/l = (Initial DO - DO5) x Dilution Factor

Dilution Factor =Bottle Volume (300 ml)

𝑠𝑎𝑚𝑝𝑙𝑒 𝑉𝑜𝑙𝑢𝑚𝑒

BOD At any time:

BODt = L(1-10-kt

)…. Or BODt = L(1-exp-kt

)….

Where:

BODt : BOD at any time

L: ultimate bio-oxygen demand

k: oxygen decay constant, [day-1

]

t: time, [days]

However: The BOD reaction rate constant (K) is dependent on the following:

1. The nature of the waste:

- degradability of the organic matter. For example , Simple sugar

and starches are rapidly degraded

2. The ability of the organisms in the system to utilize the waste


38

3. The temperature

- the water temperature may vary from place to place for the same river; hence, the

BOD rate constant is adjusted to the temperature of receiving water using

following relationship:

KT = K20 θ (T-20)

Where

T = temperature of interest, oC

KT = BOD rate constant at the temperature of interest, day-1

K20 = BOD rate constant determined at 20oC, day

-1

θ = temperature coefficient. This has a value of 1.056 in general and 1.047 for higher

temperature greater than 20oC

BOD incubator

Example:

Determine ultimate BOD for a wastewater having 5 day BOD at 20oC as 160 mg/L.

Assume reaction rate constant as 0.23 per day (base exp).

Solution

BOD5 = Lo ( 1 – exp-k.t

)

160 = Lo (1 – exp-5 x 0.23

)

Therefore, Lo = 234.1 mg/L


39

Example:

A BOD test is done by pipiting 5 ml of waste water into 300 ml testing bottle. If the

initial DO was 8.4 mg/l and the DO after 5-days of incubation at 20 oC was 3.7mg/l,

calculate the BOD and estimate the 20-days BOD value assuming the reaction decay

constant k = 0.1 day-1

. (USE base 10)

Solution:

Dilution factor = 300/5 = 60

BOD5 = (8.4 – 3.7)*60 = 282 mg/l ….*

b- to determine the BOD after 20 days:

from (*) calculate L

282 = L (1-10−0.1 ×5 ) L = 412 mg/l

Thus BOD20 = 412 (1-10−0.1 ×20 ) BOD20 = 407.8 mg/l

Try the solution using Base (Exp ) but in this case the K = 0.23 day-1

Example:

The wastewater is being discharged into a river that has a temperature of 15oC. The BOD

rate constant determined in the laboratory for this mixed water is 0.12 per day. What

fraction of maximum oxygen consumption will occur in first four days? (Base Exp)

Solution

Determine the BOD rate constant at the river water temperature:

K15 = K20 (1.056) (T-20)

= 0.12 (1.056) (15-20)

= 0.091 per day .....

note 1: per day mean day-1...

note 2: as the temp decrease the reaction rate (K) decreases

Using this value of K to find the fraction of maximum oxygen consumption in four days:

BOD4 = Lo (1 – e-0.091x4

)

Therefore, BOD4 / Lo = 0.305


40

Chemical oxygen demand (COD):

-Measure the amount of organic compounds in water that can be oxidized by strong

oxidant like mixture of sulfuric and chromic acids.

-Most applications of COD determine the amount of organic pollutants found in

surface water.

- It indicates of the strength (degree of pollution) of industrial WW that are not

biodegradable

- the test is faster than BOD test.

https://en.wikipedia.org/wiki/Organic_compound

https://en.wikipedia.org/wiki/Water

https://en.wikipedia.org/wiki/Organic_compound

https://en.wikipedia.org/wiki/Pollutant

https://en.wikipedia.org/wiki/Surface_water


41

Water PH

- Very important parameter that affects treatment processes, especially

coagulation, anddisinfection

- any unusual change may reflect a major event

H2O H+ + OH

-1 ,

A- kw = [H+] [OH

-1];;;; kw = 1 × 10−14

B- PH = -log[H+] [H

+] = 10

-PH

Where the [PH] is in Molar [mole/L]

Alkalinity :

- The source of alkalinity in natural water is bicarbonates that formed in reactions in

the soils through which the water percolates.

- It measures the capacity of water to neutralize acids (buffer capacity) in other

words its inherent resistance to pH change).

- river alkalinity values of up to 400 mg/l CaCO3

- it is determined by titration against standardized sulfuric acid (H2SO4)

- source of alkalinity is bicarbonate (main) carbonate and hydroxide

CO2 + H2O H2CO3 (carbonic acid)

H2CO3 H+ + HCO3

-1 Bicarbonate ka1 = 4.3 × 10−7

HCO3-1

H+ + CO3

-2 carbonate ka2 = 4.7 × 10−11

C- ka1 = 4.3 × 10−7 = [𝐻+] [𝐻𝐶𝑂3−1]

[H2CO3]

D- ka2 = 4.7 × 10−11 = [𝐻+] [𝐶𝑂3−2]

[𝐻𝐶𝑂3−1]

Alkalinity = [HCO3-1

]+2[CO3-2

] + [OH-1

] - [H+] …. Mole/l

-in a typical alkalinity problem , it is required to determine [H+]; [HCO3

-1]; [OH

-1];

[H2CO3]; [CO3-2

]


42

Example:

a water has alkalinity of 250mg/l as CaCO3 and a PH equals 7.5, determine [H+];

[HCO3-1

]; [OH-1

]; [H2CO3]; [CO3-2

].

Solution:

1- PH = 7.5 = -log [H+] …

thus[H+] = 10−7.5 = 3.16 × 10−8 𝑚𝑜𝑙𝑒/𝑙

thus [H+] = 3.16 × 10−8 × 1 × 103 = 3.16 × 10−5 𝑚𝑔/𝑙

2- Kw = 10-14

= [H+][OH

-1] .. thus [OH

-1] =

10−14

3.16 × 10−8 = 3.16 × 10−7 𝑚𝑜𝑙𝑒/𝑙

[OH-1

] = 3.16 × 10−7 𝑚𝑜𝑙𝑒

𝑙 ×

17𝑔

𝑚𝑜𝑙𝑒= 5.37 × 10−6 𝑔

𝑙= 5.37 × 10−3 𝑚𝑔/𝑙

3- Determine the equivalent number of mole :

No of Eq. mole = weight/ eq. weight

250 × 10−3𝑔/𝑙

50𝑔/𝑒𝑞. 𝑚𝑜𝑙𝑒= 5 × 10−3 𝑚𝑒𝑞/𝑙

Thus


43

CaCO3 Ca + CO3-2

- Every 1 mole of CaCO3 produces 1 mole of CO3-2

- 100g/mole of CaCO3 produces 60 g/ mole of CO3-2

- thus 250mg/mole of CaCO3 will produce = 250*60/100 = 150 mg of CO3-2

[CO3-2

] = 5 × 10−3 𝑒𝑞.𝑚𝑜𝑙𝑒

𝑙

Alkalinity = 5 × 10−3 𝑒𝑞.𝑚𝑜𝑙𝑒

𝑙= [HCO3

-1]+2[CO3

-2] + [OH

-1] - [H

+]…..A

ka1 = 4.3 × 10−7 = [𝐻+] [𝐻𝐶𝑂3−1]

[H2CO3]

ka2 = 4.7 × 10−11 = [𝐻+] [𝐶𝑂3−2]

[𝐻𝐶𝑂3−1]…. Thus

[𝐶𝑂3−2] = 4.7× 10−11

3.16× 10−7 𝑚𝑜𝑙𝑒/𝑙* [𝐻𝐶𝑂3−1] =1.48 × 10−4[𝐻𝐶𝑂3−1]

Substitute in A

5 × 10−3 𝑒𝑞.𝑚𝑜𝑙𝑒

𝑙= [HCO3

-1]+2×1.48 × 10−4]+[OH

-1] - [H

+]

[HCO3-1

]= 4.99× 10−3𝑞.𝑚𝑜𝑙𝑒

𝑙

Hardness:

- It is the capacity of a water to destroy the lather of soap.

- Causes economical losses.

- hardness mainly caused by the calcium (Ca+2

) and magnesium (Mg+2

)ions

- It is expressed as mg/l CaCO3

CaCO3 [mg/l] Type of water

Up to 50 ppm as CaCO3 Soft water

51- 100 ppm as CaCO3 Moderately soft

101-150 ppm as CaCO3 Slightly hard water

151-250 ppm as CaCO3 Moderately hard water

> 250 ppm as CaCO3 Hard water

There is a differences among countries in term of accepted water quality . in other

words, different water standards…..why?

Nitrate in Germany… blue baby …..


44

Total Hardness: is the expression of the results of direct measurement (principally of calcium and

magnesium) expressed as mg/l CaCO3

Calcium Hardness: is the expression of the results of the measurement of calcium only, as mg/l CaCO3.

Magnesium Hardness:

is the difference between total hardness and calcium hardness is taken as the

magnesium hardness .

There are other types of hardness like Carbonate Hardness( is removed by heating)

, Non-carbonate Hardness and Permanent Hardness.

Ca+2

+ CO3 CaCO3

Mg+2

+ OH- Mg[OH]2

Molecular weight of CaCO3 = 1*40 + 12 + 3*16 = 100 g/mole

Equivalent weight = Mw/charge = 100/2 = 50 g/eq. weight

Example:

the result of water analysis show the following for cations :calcium 35.8; magnesium

9.9; sodium 4.6 and potassium 3.9 ppm. While the anions were chloride 7.1; HCO3-1

131; SO4-2

26.4, draw a milliequivalent – per litter bar graph and express alkalinity

and hardness in unit of mg/l as CaCO3?

element Conc. [mg/l] Equ. weight M eq/l

Major cations Ca+2

35.8 20 1.79 Mg

+2 9.9 12.2 0.81

Na+ 4.6 23 0.20

K+ 3.9 39.1 0.10

Total cation 2.9

Major anions HCO3-1

131 61 2.15 SO4

-2 26.4 48 0.55

Cl-

7.1 35.5 0.20 Total anion 2.9

Note: usually total anions equal total cations; remember the arrangement

Thus we have:

1.79 meq/l of Ca(HCO3)2

0.36 meq/l of Mg(HCO3)2

0.45 meq/l of MgSO4

Draw milieq./l bar

graph


45

0.1 meq/l of KCl

0.1 meq/l of Na2 SO4

0.1 meq/l of NaCl

thus total carbonate hardness = 1.79 +0.36 = 2.15 meq/l

2.15 meq/l * 50 mg/meq = 107.5 mg/l as CaCO3

- total non-carbonate hardness

= 0.45 * 50 = 22.5 mg/l as CaCO3

Example 2:

element Conc. [mg/l] Equ. weight m eq/l

Major cations Ca+2

29 20 1.45 Mg

+2 16.4 12.2 1.34

Na+ 23 23 1.00

K+ 17.5 39.1 0.45

Total cation 4.24

Major anions HCO3-1

171 61 2.81 SO4

-2 36 48 0.75

Cl-

24 35.5 0.68 Total anion 4.24

1.45meq/l of Ca(HCO3)2

1.34meq/l of Mg(HCO3)2

0.02 meq/l of Na(HCO3)

0.75meq/l of Na2 SO4

0.23 meq/l of NaCl

0.45 meq/l of KCl

Total hardness = 1.45 + 1.34 = 2.79

2.79* 50 = 139.5mg/l as CaCO3

Hardness is determined (tested) in Lab by tetrametric with EDTA

1.79 Ca++ 0.81 Mg++ 0.2 Na+ 0.1 K+

2.15 HCO3- 0.55SO4

-2 0.2 Cl

-


46

Residual Chlorine :


47

Heavy Metals:

Volatile Organic Matter (VOC):

- Emitted mainly from chemical and petrochemical industries, which include

most solvents such as thinner cleaners,lubricants, and liquid fuels.

- The most commonly reported VOC’c are:

1- formaldehyde (CH2O),

2- tetrachloroethylene (C2Cl4),

3- benzene (C6H6),

4- toluene (C7H8),

5- xylene (C8H10)

6- and acetaldehyde (C2H4O).

- VOC’s are hazardous to environment because they severely affect

- both human health and environment such as global warming;

- destroy the ozone layer

- and photochemical smog [1-3].

- Especially, benzene and toluene are known to cause cancer in animals.

- Technologies used to control VOCs can be subdivided into two major sections:

I) Decomposition:

This method involves thermal oxidation and/or bio-filtration methods to reduce

the effects of emissions by decomposing VOCs.


48

II) Recovery: Absorption, adsorption, and condensation are common recovery

technologies.

various technologies for VOC removal have been investigated in recent years, with

treatments now including:

1- thermal oxidation,

2- bio-filtration,

3- absorption,

4- adsorption,

5- and condensation

Extra Examples:

Example 1:

A water treatment plant produces 100kg of dry weight of sludge daily, this sludge is

removed with 10m3

of water. What is the percentage solid concentration in the

effluent? And what is the ppm concentration assuming the water density is

1000kg/m3?

Sludge % concentration= 100 kg dry sludge

100 𝑘𝑔 𝑠𝑙𝑢𝑑𝑔𝑒+10,000 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟 × 100% = 0.99%

Sludge concentration (ppm) = 100 kg dry sludge×106

10,000 𝐿𝑖𝑡𝑒𝑟 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 10,000 ppm

Example 2:

Commercial sulfuric acid (H2SO4) is often produced as 93% W/V? find the

concentration of H2SO4 [ ppm], and the molarity and normality of the solution. If the

specific gravity of H2SO4 is 1.839

Solution:

Since the S.G = 1.839 then

1L of H2SO4 = 1.839 kg /L= 1839g/L


49

But each litter contains only 93% of H2SO4

then H2SO4 in 1L = 0.93 * 1839g/L = 1,710g/L = 1.710 * 106

ppm (mg/L) #1

#2:

MWH2SO4 = 98g/mole

Then the molarity = 1,710g/L

98g/mole = 17.45 mole/L or 17.45 M

#3 the Normality

Equiv Weight of H2SO4 = 98/2 = 48g/equiv

Then the normality = 1,710g/L

48g/equiv = 34.9 equiv./L or N

Example3

If the solubility constant of AlPO4 dissolution in water is (Ks) = 10-20

what is the

concentration of such solution that is in equilibrium with water?

AlPO4Al+3

+ PO4-3

for every mole of AlPO4 produces one mole ofAl+3

and one mople of PO4 -3

(Ks) = 10-20

= [Al+3

] [PO4 -3

]

10-20

= X2

…. Then X = [PO4 -3

] = 10-10

mole/l

Since MWPO4 -3 = 95g/mole

Thus the final [PO4 -3

] in mg/l = 95g/mole * 10-10

mole/l = 9.5 * 10-6

mg/l

Example4:

If 100 mg of H2SO4is added to1L of water what is the final PH?

Solution


50

[H2SO4] = 100mg/L

98 g/mole×

1𝑔

1000 𝑚𝑔 = 1.02 ×10

-3mole/L

The reaction is:

H2SO4 2H+

+ SO4 -2

Every mole of H2SO4 produces 2 mole of H+

then 1.02 ×10-3

mole/L needs = 2* 1.02*10-3

= 2.04*10-3

mole of H+

PH = -log [H+] = 2.69

Section IV Water Treatment - Introduction

51

The main purpose of water treatment is:

- To protect both human and environmental health,

- To save water by using the treated

- To protect the ground water resources form pollution.

Water source could be

River

Lakes and moorland (swamp)

Groundwater; most likely better quality than surface water

Rainwater ++/ Dam

Raw water quality varies with the source, and if the source is surface water,

the quality vary seasonally, particularly with flooding, upland and lowland, etc


52

Sources of Water pollution:


53

Section V Water Treatment Plant-conventional Treatment

54

Conventional water treatment:

is a combination of coagulation-Flocculation, sedimentation, filtration and

disinfection process.

1- Primary treatment involves : pumping, screening and grit removal

- Screening: aims to remove large objects, such as stones or sticks, that could

plug lines or block tank inlets.

- grit chamber- slows down the flow to allow grit to fall out

- sedimentation tank (settling tank or clarifier): settleable solids settle out and

are pumped away, while oils float to the top and are skimmed off

The typical functions of each unit operations are given in the following table:

Functions of Water Treatment Units Unit treatment

Function (removal)

Aeration, chemicals use Colour, Odour, Taste

Screening Floating matter

Chemical methods Iron, Manganese, etc.

Softening Hardness

Sedimentation Suspended matter

Coagulation Suspended matter, a part of colloidal matter and bacteria

Filtration Remaining colloidal dissolved matter, bacteria

Disinfection Pathogenic bacteria, Organic matter and Reducing


55

Aeration:

- It removes odor, color and taste due to reducing the concentration of volatile

gasses like H2S (hydrogen sulfide) , and algae and other related organisms.

- It oxidizes iron and manganese,

- increasing [DO], removes CO2 , CH4 and other flammable gasses, reduces

corrosion.

- Its work principle is based on the fact that the atmospheric oxygen will replace

the volatile gasses in water , while it will escape into atmosphere.

- The replacement will continue till reaching the equilibrium depending on the

partial pressure of each specific gasses.

Type of aeration:

1- Gravity aerators:

In this type water is allowed to fall by gravity, such that large area of water will

exposed to atmosphere.

2- Fountain (Spray) aerators

A special nozzle is used to produce a fine spray. Each nozzle is 2.5- 4cm diameter.

Discharging about 18-36 L/h

Nozzle spacing should be such that each meter cube of water (m3) has aerator area of

0.03 – 0.09 m2 for one hour.

3- Injection or Diffused aerator :

- It consists of tank with perforated pipes, tube or diffuser plates, fixed at the

bottom to release fine air bubbles from compressor unit.

- The tank depth is 3-4m and the tank width is 1.5 times of tank depth.

- If the depth is more the diffusers must be placed at 3-4m below the water

surface


56

- Aeration time is 10-30 min and 0.2-0.4 litter of air is required per litter of

water.

- Increase the diffuser depth increases the aeration rate and efficiency

- Decrease the orifice size increases the aeration rate and efficiency

4- Mechanical aerator:

Mixing paddle as in flocculation are used.


57

The puddle may be submerged or at surface.

Section VII Water Treatment - Coagulation

58

Settling (Sedimentation)

it is a type of solid liquid separation process in which a suspension is separated into two phases:

a- Clarified supernatant leaving the top of the sedimentation tank (overflow).

b- Concentrated sludge leaving the bottom of the sedimentation tank (underflow).

Purpose of Settling

To remove coarse dispersed phase. To remove coagulated and flocculated impurities.

To remove precipitated impurities after chemical treatment. To settle the sludge (biomass) after activated sludge process /

tricking filters.

Principle of Settling

Suspended solids present in water having specific gravity

greater than that of water (1) tend to settle down by gravity as soon as the turbulence is retarded by offering storage.

Basin in which the flow is retarded is called settling tank. Theoretical average time for which the water is detained in

the settling tank is called the detention period. Sedimentation tanks are circular or rectangular.

Sedimentation types:

It exist, based on characteristics of particles. 1- Discrete or type 1 settling particles whose size, shape, and specific gravity do not

change over time (has a density (ρ) = 2000 – 2200 kg/m3).

2- Flocculating particles or type 2 settling; particles that change size, shape and

perhaps specific gravity over time(has a density (ρ) = 1030 – 1070 kg/m3).

3- Hindered settling or type 3 settling: blanket sedimentation occurs at Lime

softening sedimentation and Sludge thickeners in water treatment

4- Compressed settling or type 4 settling occurs Sludge thickeners in water treatment


59

Applications in Water Treatment:

1. grit removal

2. suspended solids removal in primary clarifier

Place of Sedimentation in various WW plant:

Or


60

For treating hard water to removes flocculated solids. The sedimentation tank comes after the flocculation tank.


61


62


63


64

Sedimentation process:

It is a physical treatment that allows for particles having specific gravity higher than

water to settle under its own weight.

Factor affecting the sedimentation:

1- Water prosperities:

a) Sed. Decreases when water viscosity increase

b) Sed. Decreases when water density increase

c) Sed. Increase when water temp. Increase.

2- Suspended solids concentration, size and shape

3- Detention time: sed. Efficiency increase by increasing the water retention at

the tank , but after time the precipitation is decreased sharply thus the time

should be determined correctly. (2-4hr)

4- Flow velocity: decrease the flow velocity increase the sed. The max.

Allowable velocity in the tank is 0.3 m/sec

5- Tank shape: circular tanks are more efficient for sedimentation

Ideal sedimentation tank:

1- The flow is laminar

2- There are no dead zones

3- The horizontal velocity is constant

4- Good arrangement of inlet and outlet weirs.

Type of sedimentation:

1- Plain sedimentation: no chemical is used

2- Chemical sedimentation: chemical is used to enhance the efficiency

Critical Settling Velocity & settling Velocity (Overflow rate)

- Settling velocity Vs or as called overflow rate


65

Particles move horizontally with the fluid (all particles have the same horizontal

velocity)

Particles move vertically with terminal settling velocity(different for particles

with different size, shape and density)

All particle with Vs,o>Vcwill be completely settled.

All particle with Vs,o<Vcwill be removed in the ratio Vp/Vc


66

Available length = Circumference

of Circleمحيط الدائرة


67

In plain sedimentation : no chemical is used the precipitation just is done by gravity

Example:

Design the plain rectangular sedimentation tank for a water purification of an hourly

output 5000m3 if the retention time is 4hr, then get the amount if the sludge in the S.S

if the raw wastewater has SS concentration 80ppm and the sludge specific gravity is

1.1?

Solution:

Assume Surface .Load .Rate (S.L.R) = 30 m3/m

2/day

Thus the minimum total volume = 4 * 5000 = 20,000 m3 = n.w.l.d

Thus total tank area = 5000 * 24 / 30 = 4000 m2

=No. of tank * length* width of tank

4000m2 = n.l.w

Then the water depth = 20,000/ 4,000 = 5m.

Assume number of tank = 8 then each tank area = 500 m2

then the tank width = =√500

4 = 11.2 make the width = 10m

then the length = 50m ,


68

then each tank volume = 10*50*5 = 2,500 m3

actual total volume = 2,500 * 8 = 20,000 m3

Check:

Actual Retention time = 20,000/5000 = 4hr .ok

Actual SLR = (5000/4000)*24 = 30 m3/m

2/d . ok

Hydraulic .Line (H.L) = 5000 ×24

8 ×50 = 300 m3

/m/d .ok

Sludge Amount:

S.S = 80 mg/l ; assume Recovery .Rate (R.R) = 80%

Then the amount of sludge per day = Y 9ton/day)

= 5000 × 24 (m3/d)

×80 ×0.8

106 = 7.68 ton/day

Assume the S.S concentration in the sludge = 3%

Thus the total amount of sludge (water + SS) = 7.68/0.03 = 256 ton/day

Thus the total sludge volume = 256/1.1 = 232 m3/day


69

Coagulation Flocculent Sedimentation ( type 2 and 3):

- The design procedure for sedimentation tanks of type 2 and 3 are

the same as type 1.

- The difference is mainly in the overflow rate (vs)???

- The following table gives the design criteria of these two types.

Typical sedimentation tanks design dimensions :

A- Rectangular Tank

- depth: 3-5 m

- length: 15-90 m

- width: 3-24 m

B- Circular Tank

- depth: 3-5 m

- diameter: 4-60 m

C- Retention time 2-4 hr

D- Tank bottom slope 1-2%in rectangular tank; 2-4% in circular tank

E- The weir (الهدارات) length should be 1/7 of total tank length

F- Surface load rate = Q/ (total surface area)= 𝑄

𝑛.𝐴= over flow rate = 20-40m

3/m

2/d

Where n = number of tanks

G- Hydraulic load (H.L) on outlet weir = weir loading rate = over load =

H.L =𝑄/(𝑛. 𝐿𝑤) = 150 − 300 m3/𝑚/𝑑

H- Horizontal velocity ≤ 0.3 𝑚/𝑚𝑖𝑛

sizes change shape change specific gravity change


70

Example:

a circular clarifier (sedimentation tank) will be used to treat 3000m3hourly, assume

the surface area of the tank is 2400 m2

estimate the tank dimensions?

Solution :

Water depth =

Assume tank diameter = 30m thus number of tank =

No of tank = n× 𝜋 × 302

4= 2400

Thus n = 3.4 take 3 tanks

Modify the required diameter :

3× 𝜋 × 𝑑

2

4= 2400 d = 32m

S.L.R (over flow) = 3000×24

3×𝜋 × 322

4

= 29.8 m3/m

2/d within the range (20:40)

H.L= Q/Lw = 3000×24

3×𝜋 ×32 = 238 m

3/m/d within the range (150:300) .OK


71

Coagulation :

Particles smaller than 10 µm are difficult to remove by simple settling or by filtration.

It is a chemical addition process of coagulant, it consists two steps

Source: Terry L. Engelhardt, HACH

5- Coagulation is a process to neutralize the particles charges

6- form a gelatinous mass (floc) to trap (or bridge) particles thus forming a mass large

enough to settle or be trapped in the filter

Aims of Coagulation:

7- Reduce retention time

8- Increase filtration rate


72

9- Improve water quality

10- Increase sedimentation efficiency (90 – 96%)

Coagulation used chemicals:

A- Alum (Aluminiumsulphate )

B- Iron based salts

1- Ferric Sulfate (Fe2SO

4 · 9 H2O )

2- Ferric chloride (FeCl

3)

C- Sodium Aluminate, NaAlO2

D- Polyelectrolytes or resin: is generally low molecular weight (<500,000) it may be

anionic or cationic

Factors affecting coagulation:

1- Water PH

2- Water temperature

3- Coagulant doase

4- Coagulant type

5- Mixing type

Coagulation process:

1- Coagulant addition: wet or dry

2- Mixing

3- Flocculation,

4- Sedimentation

Coagulant feeding:

a- Dry feeding:

- the coagulant is added as powder to the water

- Main limitations are: hard to control the optimum doase, more workers and the

powder is sensitive to humidity.

b- Wet feeding:

- a solution from the coagulant is prepared in a separate tank

- High dispersion rate, and easy to control

- The solution is prepared by rabid mix (flash) mix


73

The Jar Test is used to determine the optimum Dose

1- it is the most basic test for control of coagulation/flocculation/filtration

How to determine the optimum coagulant and dosage using the jar test:

1- do the basic analysis for water (PH, Alkalinity, Hardness, Temp)

2- in each jar add different coagulant (choose the doase between 20:30mg/l)

3- flash mix for 30 sec (100:300rpm)

4- gentle mix for 10 min (20:30 rpm)

5- allow for sedimentation (30:40 min)

6- get removal efficiency for each vessel (depth of sedimentation solids)

7- determine the best coagulant

8- repeat the steps from 2-7 using the best coagulant in step (2) add different

dosage (10:60 ppm) now determine the optimum doase

Note: optimum PH is determined by repeating the steps 2-7.

Design of mixing tank:

Tank volume = 𝑸×.𝑺

𝑪×𝜸×.𝒏×.𝟏𝟎𝟔

where


74

S:coagulant dose

C: Coagulant concentration

𝜸: solution spesific gravity usually (1.02 − 1.07)

n: number of tanks

usually the tank is circular with dimensions is

d= 1:2m

n≥ 2

Example: Jar test

two sets of jar test on raw water with turbidity = 15TUand an HCO3- concentration =

50ppm expressed as CaCO3, given the below data find the optimum PH, coagulant

dose, and the theoretical amount of alkalinity that can be consumed at the optimal

dose?

First test:

1 2 3 4 5 6

PH 5 5.5 6 6.5 7 7.5

Alum dose (ppm) 10 10 10 10 10 10

Turbidity (TU) 11 7 5.5 5.7 8 13

Second test:

1 2 3 4 5 6

PH 6 6 6 6 6 6

Alum dose (ppm) 5 7 10 12 15 20

Turbidity (TU) 14 9.5 5 4.5 6 13

Thus from the result the optimum PH is 6.25 and the optimum doase = 12.5mg/l

Al2(SO

4)3 · 14 H

2O + 6HCO

3

-1 2Al(OH)3

+ 3SO4+ 6CO2

+ 11H2O

From the equation:

1 mole alum 6 mole HCO3

-1


75

thus

594 g/mole 6*61 g/mole

12.5 of alum X g/mole of carbonic acid

X = carbonic acid mass consumed = 12.5 ×6×61×10−3

594𝑔/𝑚𝑜𝑙𝑒= 7.7 × 10−3 𝑔/𝑙 = 7.7 mg/l

To represent the consumed mass by CaCO3

7.7 ×50

61= 6.31 mg/l as CaCO3

Coagulant Mixing:

This strep aims to disperse the coagulant solution in the sedimentation tank

Type of mixing:

1- coagulant injection to suction pipes of Low .Lifting .Pump

2- hydraulic mixing :

- using hydraulic weirs that causes water jump or using the venturi pipe

- low coast and easy to operate and maintain

3- Mechanical Mixing:

- Using a mixer with 150:200 r.p.m


76

Mechanical mixing

Design of mixing tank: it can be square or circular

Retention time = 5:60 sec

D= 2:4m

n≥ 1


77

Velocity Gradient calculation:

It is used to measure the degree of mixing

G= √𝑷

µ×𝑽

G: velocity gradient = velocity/distance [sec-1

] usually 300:500sec-1

P: motor power in watt

µ: 𝑤𝑎𝑡𝑒𝑟𝑣𝑒𝑠𝑐𝑜𝑠𝑖𝑡𝑦 (N.sec/m2) depends on water temp. (1.014 𝑋 10

-3)

V: mixing tank volume

B-Flocculation or gentle mixing tank:

-It is used to complete the reaction and interaction between the SS and the coagulant

solution ,

-The SS aggumulerate to form a flocks(increase the size and the weight)

Type of Flocculation tanks:

1- Hydraulic type: it is a tank with vertical or horizontal

Maze Flocculation basin , source (google image)

2- Mechanical mixing: it uses paddle to mix the water


78

Flocculation tank design

Time retention = 20: 30 min

D = 2:4m

L = 3:5m

n≥ 2

Example:

If Q =60000m3

daily, and the required alum dose = 40ppm calculate the

1- The required Alum solution tank and alum doase (dosage) if the alum

concentration in thank is 10%

2- The volume of flash mixing tank, if retention time is 30sec

3- The dimension of rectangular flocculation tank (L=3b)if retention time is

20min

Solution:

Total required volume = 𝑸×.𝑺

𝑪×𝜸×.𝒏×.𝟏𝟎𝟔

= 𝟔𝟎,𝟎𝟎𝟎 𝒎×𝟒𝟎

𝟎.𝟏×𝟏.𝟎𝟏×.𝟏𝟎𝟔= 24m3

assume 3 tanks then each tank has a volume of 8m3

assume water depth d= 2 then

tank area = 8/2 = 4m2

thus the tank is square length = 2m , n = 3, and water depth = 2

step by step solution:

daily alum dose required = 60000 * 40 * 10-6

= 2.4 ton/day of alum

but alum concentration = 0.1 then the required mass = 2.4/0.1 = 24 tons

the required volume = mass/density ... assume the solution density = 1ton/1.01ton/m3

thus the required volume = 24m3


79

Rate of dosing:

Dosing rate = required alum volume / time

24m3

/day 1000

24 ×60 = 16.67l/min

2-For flash mixing tank:

Q = 60000 𝑚3

𝑑𝑎𝑦 =

60000

24 ×60 = 41.67 m

3/min

V= Q * retention time = 41.67 * 30/60 = 20.835 m3

If the tank is circular (one tank)

V = π/4 * φ2

* d ... assume φ= d

Then d= 3m and tank diameter = 3m

3-for flocculation tank

V = 41.67 m3/min * 20 min = 833.4 m

3

Assume two tanks and water depth in each tank = 3m

Required area of each tank = 833.4 / (2*3) = 139 m2

thus b = 6.8m and l = 20.4 m

thus the final design: 2 tanks of l=20.4m and b=6.8m and water depth 3m


80

Development of coagulation-flocculation and sedimentation system:

A- At the first each process was done in separated tank as the following picture

B- Improvement was done to use one tank for flocculation part and sedimentation

part


81

C- One circular tank is used and called clariflocculater

It is very efficient , the SS removing efficiency may reach 96%

Retention time in the inner tank is 20:30 min...for flocculation

Retention time in the outer tank 2:3hr..... for sedimentation


82

Clariflucculator basic design :

For outer part:

To = 2:3 hr (total time for flocculation and sedimentation)

do = 3:5m

φo≤ 35𝑚

For inner part:

Ti = 20:30min

di = do - (0.5:1m)

ni = no

φi= (0.3: 0.5 )𝜑𝑜

Check:

Lw

Surface Load Rate (S.L.R) = 𝑄

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 =

𝑄

𝑛.𝜋

4 .(𝜑𝑜

2−φi2) = 25:40m

3/m

2/day


83

Hydraulic Load Rate (H.L.R) = 𝑄

𝑤𝑒𝑖𝑟 𝑙𝑒𝑛𝑔𝑡ℎ (𝑙𝑤 ) =

𝑄

𝑛.𝜋 .(𝜑𝑜)

≤ 300m3/m/day

Note: if mention retention time in sed. Tank = 2.5hr

And retention time in floc. Tank =30min

Thus the To = 2.5+0.5 = 3hr

Ti = 0.5hr

Treated water by coagulation can be measured by:

1- Zeta potential

2- Streaming Current : it is an on-line measurement of how well charge

neutralization has occurred

Example:

Design clarifluccolator tanks for water treatment plant of a daily flow (Q) = 48000m3

And working hours is 16hrs per day?

Solution:

Q = 48000 𝑚3 /𝑑𝑎𝑦

16ℎ𝑟/𝑑𝑎𝑦 = 3000m

3/hr

Outer tank design:

Retention time To = 3hr +0.5 = 3.5hr

Thus the required volume = 3000 *3.5 = 10,500 m3

assume water depth in the tank (d) = 3m

thus the required total surface area = 10,500m3/3m = 3500m

2

total area=3500 m2

= n ×𝜋

4× 𝜑𝑜

2assume 𝜑𝑜 = 35m

thus n = 3.6 use 4 tanks

thus𝜑𝑜 = 33.4𝑚


84

Design of inner tank:

Ti = 0.5hr

Required volume = 3000 * 0.5 = 1500m3

Di = d0 – (0.5:1m) = 3 -0.75 = 2.25m

Thus the required surface area = 1500/2.25 = 666.67m2

use 4 tanks ...

666.67 m2

= 4 ×𝜋

4× 𝜑𝑖

2

thus the required 𝜑𝑖 = 14.6m

Check:

S.L.R = 𝑄

𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎=

3000𝑚3

ℎ𝑟×24ℎ𝑟

𝑛.𝜋4

.(33.42

−14.62

) = 25.4m

3/m

2/day

H.L.R = 𝑄

𝑤𝑒𝑖𝑟 𝑙𝑒𝑛𝑔𝑡ℎ (𝑙𝑤 ) =

3000∗24

𝑛.𝜋 .(33.4) 171.5 m3

/m/day

Vh =

𝑄

𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 =

3000/60

𝑛.𝜋 .(14.6)∗3 = 0.09 m/min

φi= (0.3: 0.5 )𝜑𝑜 𝑡ℎ𝑢𝑠

φi

𝜑𝑜 =

14.6

33.4 = 0.44

Example:

For an existing water treatment plant with 4 clarifluculator with the following

dimension find the Max population that can be served

φo= 32m φi = 14m

do = 3.5m di= 3m

if Wc = 300L/c/day and water pump = 20hr

Solution:

- In such question the answer dose not consider the tank volume only but the

discharge from each tank (SLR;HLR, and Vh) then we determine the minimum


85

discharge which is equal to the maximum allowable discharge for the

treatment plant.

For outer part:

Tank volume or capacity = Q.T

To find the total retention time we estimate the sedimentation time (2hr) and the

flocculation time (20min) … thus T = 2.33 hr (which allow for max Q)

Q = 4×

𝜋

4 ×322 ×3.5

2.33= 4832 m3

/hr

From S.L.R

S.L.R = 𝑄 𝑝𝑒𝑟 𝑑𝑎𝑦

𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 = 25: 40 …take 40 to get the max. flow

Then Q2 = (40/24) × 4 ×𝜋

4 × (322 − 142 ) = 4335.4 m

3/hr

-note: S.L.R is converted to hr by dividing on 24 but not 20 (pump regime)

From H.L.R

H.L.R = 𝑄 𝑝𝑒𝑟 𝑑𝑎𝑦

𝑛.𝑙 𝑤𝑒𝑖𝑟= 300 m3

/m/day

Thus Q3 = (300/24) × 4 × 𝜋 × 32 = 5026.5 m3/hr

From Vh

Vh = 𝑄

𝑋.𝐴 = 0.3 m/min

Q4 = (0.3 *60) × ( 4 × 𝜋 × 14) × 3.5 = 11,875 m3/hr

- X.A is perimeter (circumference )of cylinder at the face of the internal tank

multiplied by water depth in the outer tank.

For inner part:

c = T.Q

from capacity: T = 0.33 hr

Q5 = 4×

𝜋

4 ×142 ×3

0.33= 5597.7 m

3/hr

-So maximum hourly flow from the plant is the minimum Q value =4335.4 m3/hr


86

- Maximum daily flow = 4335.4 m3/hr * 20 hr = 86,708 m

3/day

Thus maximum served population can be =𝑄𝑑𝑎𝑖𝑙𝑦

𝑊𝑐

= 86,708 m3/day

0.3𝑚3

𝑐/𝑑𝑎𝑦

= 289,027 capita

Section VIII

87

Water intake:

- Aims to safely withdrawing water from the source

- Its location is predetermined under specific pool levels

Factors Governing Location of Intake:

- The site should be near the treatment plant so that the cost of conveying water

is less.

- The intake must be located in the purer zone of the source to draw best quality

water from the source, thereby reducing load on the treatment plant.

- The intake must never be located at the downstream or in the vicinity of the

point of disposal of wastewater.

- The intake must be located at a place from where it can draw water even

during the driest period of the year.

- The intake site should remain easily accessible during floods and should not

get flooded.

Types of water intake:

1- Pipe intake:

- It is best type, good water quality

- Used when water level is changing

- Used for long distance water transmission

Section VIII

88

2- Shore intake:

-The water intake is placed directly near the water source like river bank

-The pumped waster quality is low due to pollution risk

-Good for industrial and agricultural water uses

-Low discharge capacity

Section VIII

89

3- Submerged (deep) intake:

- Used in narrow and deep water streams

- when the water cover is polluted

- When the water stream is used for sailing

Section VIII

90

4- Tower intake:

- Used for large water surfaces (lakes , dams)

- When water level is varied

- Water quality is varied with water depth

Water intake design depend mainly on water consumption

Section VIII

91

Filtration

- The resultant water after sedimentation will not be pure, and may contain

some very fine suspended particles and bacteria in it.

- To remove or to reduce the

a- remaining impurities

b- Algae

c- Iron and manganese salts

d- Color and taste of water

- The water is filtered through the beds of fine granular material, such as sand,

etc.

Filtration rates for granular media filtration are typically expressed at the number of

gallons filtered per square foot of filter surface area per min, gal/ft2/min or m3/m

2/hr

It is called filtration rate or Flux

Filter Materials

1- Gravel:

- It supports the sand layer, uniformly distribute the effluents

- It permits the filtered water to move freely to the under drains, and allows the

wash (backwash) water to move uniformly upwards.

Section VIII

92

2- Sand:

It is either fine or coarse, is generally used as filter media.

- The size of the sand is measured and expressed by the term called effective size.

The effective size, i.e. D10 may be defined as the size of the sieve in mm through

which ten percent of the sample of sand by weight will pass.

- The uniformity in size or degree of variations in sizes is represented by uniformity

coefficient. The uniformity coefficient, i.e. (D60/D10)

3- Other materials:

Such as Anthracite that is made from anthracite, which is a type of coal-stone that

burns without smoke or flames.

It is cheaper and has been able to give a high rate of filtration, remove odor .

Section VIII

93

Types of Filter:

Slow sand filter: (SSF)

- it consists of fine sand, supported by gravel.

- it captures particles near the surface of the bed and are usually cleaned by

scraping away the top layer of sand that contains the particles.

- The filtration is done on the layer (sticky, gelatinous film)

- The layer takes several weeks to form can consist of bacteria, fungi, protozoa,

alga, and microscopic aquatic organisms, once fully

- water turbidity should less than <50 NTU

- filtration rate = 0.05 gal/ft2/min

Section VIII

94

Construction of SSF

a- 1.2 : 1.5 m of water depth at the top

b- 0.9 : 1.5 m layers of fine sand with effective size (D10 = 0.25:0.35mm) and

uniformity coefficient (Cu = 1.75)

c- 0.3:0.6 m of well graded Gravel

d- Under-drainage system : which is group of pipes that collect the filtered water

into main pip

SSF operation and maintenance:

1- Prepare the filter layers (gravel, sand) then fill with water for 2-3 weeks to

form the dirty skin (stick layer)

2- The produced water during the preparation period is not usable.

3- Pass the water to be filtered for a period of 2-5 months ,

4- When significant the pressure drop is happened. Stop water inlet and clean the

filter

5- Filter cleaning: by removing the top sand and stick layers and replace the sand

Section VIII

95

Design Data:

Rate of Filtration (R.O.F) = 5:8 m3/m

2/day

No of tank ≥ 2 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 + 𝑛𝑐𝑙𝑒𝑎𝑛𝑖𝑛𝑔

Area of one filter = 1,000 : 2,000 m2

L=(1:1.25)b and both L, b are ≤ 50m

Operation 2:5 months

Cleaning (removing skin) 1: 15 days depends on shift

Preparation (forming dirty skin) 1:3 weeks

Example:

Design SSF for a water treatment plant working 16 hr/day if the design flow is

32,000 m3/day, if the rate of filtration (R.O.F) is 6 m

3/m

2/day?

Since the working hours is 16 per day then every hour the flow is

32,000 m3/day

16 hr/day = 2,000 m

3/hr

Then the surface area (S.A)

S.A = 2,000

m3

hr∗24hr/day

6 m3/m2/day = 8,000 m

2

Note: when using R.O.F always use the day as 24 hrs nevertheless the

filtration working hours

Assume filter area is 2,000 m2 .. thus we need 4 filters

No of filters = total area/ area of each filter = 8,000/2,000 = 4 filters

Since L = 1.25 b then

Then 2,000 = 1.25 b2

.. thus b = 40 m

Then L = 50m

Thus the final desing is 4 filters... (40*50m)

Note: For stand by during cleaning use 2 filters with same dimensions

Section VIII

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Rapid-sand filter (R.S.F):

- it consists of larger sand grains supported by gravel and capture particles

throughout the bed.

- Most popular nowadays

- They are cleaned by backwashing water through the bed to 'lift out' the

particles.

- Filtration rate 2 gal/ft2/min., a dual media filter with a rate of 4-6 gal/ft

2/min.

multimedia filter with a rate of 7-10 gal/ft2/min

- Can be

e- single media (usually sand)

f- Double media (sand and anthracite)

g- Multimedia (sand, anthracite , granular activated carbon, and ilmenite)

- Advantages: higher loading rate;

- Water need pre-treatment to stabilize particle charges

Operation procedures:

1- Preparation phase:

a- Open the backwash water from bottom to release the escaped air

b- Open the inlet valve (from top) for 15 min to form a stick layer

2- Filtration phase:

a- Open the waste valve and open the effluent valve and use the water

b- This phase continue for 12:24 hrs depending on water quality

3- Wash phase:

a- After filtration the voids (فجوات) will be filled and there is a pressure head

loss, and the effluent water is not clear

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97

b- Close the effluent valve and open the waste valve (5 min)

c- Open pressure air valve (bottom) for 5 min

d- Open water back wash valve (water is pressurized) for 10 min

Design data:

R.O.F = 120 : 200 m3/ m

2/ day

Area of one filter = 40:60 m2

L= (1:1.25) b

Note: No stand by filter is required

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98

Quantity of wash water:

Rate of washing (R.O.W) =5:6 R.O.F

Time of washing using compressed water is 8:10 min

Thus the required water quantity for washing in a day =

= No of washing time* (R.O.W) * time of washing * total area of filters

= No of washing time * (5:6 R.O.F) * (8:10min)* area of one filter* No of filters.

Example:

Design SSF for a water treatment plant working 16 hr/day if the design flow is

32,000 m3/day, if the rate of filtration (R.O.F) is 120 m

3/m

2/day? If L = 8m?

Daily flow (Qd) = 32,000 m3/day

16 hr/day = 2,000 m

3/hr

Total Surface area = 2,000*24 / 120 = 400 m2

If L= 8m thus b = 8/1.25 = 6.25 m

Thus the single filter area = 8*6.25 = 50 m2

Then no of filters = 400 /50 = 8 filters

If the filters are washed once a day estimate the required water quantity for filters

wash?

R.O.W = 5 ROF = 5 ∗ 120 𝑚3/𝑚2/𝑑𝑎𝑦 = 600 m3/m

2/day

Amount of water = 600 * 400 *10min /(24hr*60min) = 1,666.6 m3/ day

Section VIII

99

Example:

An existing WTP with 10 Rapid sand filters (and two stand by) each filter dimension

is 8*6m with R.O.F = 150 m3/m

2/day. Check if this plant can serve the future

400,000 capita who has maximum daily water consumption rate = 250L/capiat/day?

Current water capacity (Qc) = 150 * 10 * 8*6 = 72,000 m3/day

Required water in the future (Qf) = 250 L

capita.day∗

1m3

1000L∗ 400,000 = 100,000 m3

/day

Thus the water deficit (shortage) = 100,000 – 72,000 = 28,000 m3/day

How many additional filters are required?

Additional Total area required = 𝟐𝟖,𝟎𝟎𝟎 m3/day

150𝑚3

𝑚2.𝑑𝑎𝑦

= 186.7 m2

Additional number of filters = 186.7 / (8*6) = 3.8 = 4 filters

Pressure filters (in Vessel)

- Same as rapid filter but the filter media is in Vessel and under a high pressure

caused by Air

- The R.O.F = 170:480 m3/m

2/day

- They consist of two or more layers of different granular materials, with

different densities. Usually, anthracite coal, sand, and gravel are used.

- The different layers combined may provide more versatile collection than a

single sand layer.

- Because of the differences in densities, the layers stay neatly separated, even

after backwashing

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100

Filter backwash:

-Rapid filters, pressure filter – require periodic cleaning to remove accumulated

solids.

-The pressure (head loss) increase with time thus the filter need to backwash

During the cleaning cycle, or backwash, the flow is reversed and a flow rate of

13-20 gals/ft2/min is forced back through the bed to remove accumulated solids

Section VIII

101

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