CHAPTER 2-Water and Wastewater Analysis (Part 2)

8/17/2019 CHAPTER 2-Water and Wastewater Analysis (Part 2)

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- Stream water hardness is the totalconcentration of cations, specifically calcium(Ca 2+ ),magnesium (Mg 2+ ), iron (Fe 2+ ),

manganese (Mn2+

) in the water.- Water rich in these cations is said to be

‘hard’. Stream water hardness reflects thegeology of the catchment area.

- Sometimes it also provides a measure of theinfluence of human activity


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For instance, acid mine drainage oftenresults in the release of iron into a stream.The iron produces extraordinarily high

hardness is a useful water quality indicator.Hardness is a reflection of the amount ofcalcium and magnesium entering thestream through the weathering of rock

such as limestone (CaCO 3).


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- Many people object to water containing harness greater than 150 mg/Las CaCO 3, suppliers of public water have considered to soften thewater (remove hardness) to < 120 mg/L as CaCO 3.

- Total Hardness (TH) = Ca 2+ + Mg 2+

- Total Hardness (TH):- 1. Carbonate Hardness (CH) - Combination of Ca and Mg ion with

CO 32-, or HCO -3 ion. Also called as temporary hardness because can be removed by heating.

- 2. Non-carbonate Hardness (NCH) - associated with other anions,Cl -, SO 42-, or NO -3 . Called as permanent hardness . Defined as thetotal hardness in excess of the alkalinity. If the alkalinity ≥ totalhardness, there is no NCH.

- TH = CH + NCH


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Source: combination of Ca dan Mg ionswith CO 3 2- ion or HCO 3- .

Carbonate components can be removed

through softening such as boiling,addition of lime

Water hardness are removed as thecarbonate components in water

precipitate.


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Mg(HCO 3)2

Ca 2+

Ca(HCO3)

2 CaCO 3MgCO

3

Mg 2+ HCO 3- CO 32- +


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Source: Combination of Ca & Mg ions with Cl - , SO 4 2- , or NO - 3 ion.

Non-Carbonate ion cannot be removed through boiling, but itcan be removed through soda-ash softening or lime-soda

softening.Hardness of element in mg/L as CaCO 3 mg/L element X = conc. Of X mg/L) 50 mg CaCO 3 /meq)

as CaCO 3 (EW of X mg/meq))

Total Hardness CaCO 3 : = Ca 2+ + Mg 2+ the predominantContributors)


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MgCl 2

+

Ca(NO 3)2 CaSO4 Mg(NO 3)2

Ca 2+

CaCl 2 CaSO 4

Mg 2+ Cl - SO 42- NO 3-


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Unit of hardness: mg/L as CaCO 3 .Hardness classification:soft 50mg/l asCaCO 3 moderately hard 50-150mg/l asCaCO 3 hard 150-300 mg/l as CaCO 3 very hard >300 mg/l as CaCO 3


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Find the equivalent weight (EW) of each of thefollowing: Ca 2+ , CO 3 2- ,CaCO 3 .

Solution:Equivalent weight was defined as

EW = atomic or molecular weight / n(valence)units: grams/equivalent (g/eq) or miligrams/miliequivalent(mg/meq)

For calcium, n=2 (valence or oxidation state in water). Atomicweight = 40.08, therefore the EW is thenEW = 40.08/2 = 20.04 g/eq or mg/meq


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For carbonate ion (CO 3 2- ), the oxidation state of 2- is used for nsince the base CO 3 2- can potentially accept 2 H + . Themolecular weight is 60.01. Therefore,EW = 60.01/2 = 30 g/eq or mg/meq

In CaCO 3 , n=2 since it would take 2H + to replace the cation(Ca 2+ ) to form carbonic acid, H 2CO 3 . the MW is 100.Therefore,EW = 100/2 = 50 g/eq or mg/meq


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Constituent mg/L MW EW mg/L as CaCO 3

Ca2+ 95.2

Mg2+ 13.5

Na+ 25.8

Cl- 67.8

SO2-4 76.0

NO -3 15.3

Al 2(SO4)3 18.4


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A sample of groundwater has 120 mg/L of Ca 2+ and 20 mg/L of Mg 2+ . Express it hardness in unitof mg/L as CaCO 3 .

Solution:recalled: Mg/L of X = concentration of X mg/L) 50 mg /meq CaCO 3 )

as CaCO3 (equivalent weight of X mg/meq))

1. Convert Ca 2+ and Mg 2+ to mg/L as CaCO 3 Ca 2+ : MW = 40, n=2, EW=40/2 =

Mg 2+ : MW = 24.3, n=2, EW=24/2 =


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Now, find the mg/L as CaCO 3 of cations◦ Ca 2+ =◦ Mg 2+ =

Total Carbonate Hardness = Ca 2+ + Mg 2+

==


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Is measured to determine the ability of astream to resist changes in pH.Alkalinity results from the dissolution ofcalcium carbonate (CaC0 3) from limestonebedrock which is eroded during the naturalprocesses of weatheringAlkalinity values of 20 -200 ppm are commonin freshwater ecosystems. Alkalinity levelsbelow 10 ppm indicate poorly bufferedstreams.These stream are the least capable of resistingchanges in pH, therefore they are mostsusceptible to problems which occur as a resultof acidic pollutants


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Is a measure of the acid-neutralizing capacity of water tothe equivalence point of carbonate or bicarbonate.

This ability to neutralize acid, or H+ ions, is particularly importantin regions affected by acid rain.

Carbonate species which contribute to alkalinity are as thefollowing:◦ Hydroxyl ( OH -) , Ion Carbonate ( CO 3 2- ), Ion Bicarbonate HCO 3 - ), Carbon

dioxide CO 2)

Alkalinity mol/L) = [HCO 3 - ] + 2[CO 3 2- ] + [OH - ] [H+ ]

Alkalinity eq/L or mg/L as CaCO 3) = HCO 3 - ) + CO 3 2- ) + OH - ) H+ )

For nearly neutral water, the concentrations of H + and OH - are negligible.

http://en.wikipedia.org/wiki/Equivalence_pointhttp://en.wikipedia.org/wiki/Equivalence_point


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Nitrogen is an essential nutrient that isrequired by all plants and animals for theformation of amino acids.

In its molecular form, nitrogen cannot beused by most aquatic plants, therefore itmust be converted to another form.One such form is ammonia (NH 3). Ammonia

may be taken up by plants or oxidized bybacteria into nitrate (NO 3 - ) or nitrite (NO 2 - ).Of these two forms, nitrate is usually themost important.


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It is a measure of the quantity of oxygenused by microorganisms (eg.aerobicbacteria) in the oxidation of organic matter.In other words: BOD measures the changein dissolved oxygen concentration causedby the microorganisms as they degrade theorganic matter.

High BOD is an indication of poor waterquality


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BOD dalam sample yang dicairkan dikira sebagai:

Persamaan BOD tanpa pembenihan air kawalan.(Unseeded dilution water)

P

DO DO BOD f i

sampel pencairan faktor P

DO final akhir DO DO

DOinitial awal DO DO

f

i

)(

)(


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Persamaan BOD dengan pembenihan air kawalan. (Seeded dilutionwater)

P

f BO BO DO DO BOD

f i f i )]()[(

sampel pencairan faktor P

kawalanair akhir DO BO

kawalanair awal DO BO

DO final akhir DO DO

DOinitial awal DO DO

f

i

f

i

)(

)(


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Latihan 3

Dengan menggunakan persamaan BOD, kira nilaiBOD suatu sample yang mempunyai nilai-nilaiberikut:DO i = 7.0mg/l DO f = 6.3mg/lBO i = 7.1mg/l BO f = 7.0mg/l

P = 1/10 f = 1.0

BOD =


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Proses BOD adalah satu proses yang amat perlahan dansecara teorinya memerlukan masa yang infiniti umtukmelengkapkan proses.Dalam masa 20 hari, pengoksidaan organic berkarbon(carbonaceous organic) adalah sekitar 95-99% lengkap.

Dalam masa 5 hari, pengoksidaan adalah sekitar 60-70%lengkap.20 C diambil kerana ia adalah nilai suhu purata air sungaiberiklim sederhana. Tambahan pula, ia mudah dicapaimenggunakan incubator.Nilai BOD yang berbeza didapati bagi setiap suhupengeraman yang berbeza kerana kadar tindakbalas biokimiaadalah bergantung kepada suhu.

stabil produk O H

CObiomassnewbarubiojisimnutrienOorganik Bahan

2

22 )(


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Tindakbalas kinetik BOD ditunjukkan sebagai tindakbalastertib pertama seperti berikut:

t t L

dt

dL

t t kL

dt

dL

Lt = jumlah BOD yang tinggal pada masa t

Lo = BOD muktamadk = pemalar kadar tindakbalas, hari -1, (base e) K = k /2.303, pemalar kadar tindakbalas, hari -1 (base 10)

Pers. 1


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Kamirkan pers. 1

t

o

L

Lt

t

dt k L

dLt

o

kt L t o

L

Lt ln Pers. 2

kt L L ot lnln

kt L L

o

t

ln

Kt kt

o

t e L L

10 Pers. 3


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y, jumlah BOD yang dikeluarkan pada bila-bila masa t

BODt, y t = L o – Lt= L o – Lo e -kt

= L o (1 – e -kt )

BOD 5 hari :

y5 = L o –

L5= L o (1-e -5k )

kt ot e L L

Pers. 4


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Dengan menggunakan persamaan:

nilai pemalar tindakbalas k boleh dianggarkanpada suhu-suhu yang berlainan.

kT = k 20 θ (T-20)

Di mana = 1.047


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Latihan 4Tentukan BOD 1 hari dan BOD akhir (ultimate BOD, L o ) bagi satusample air sisa. Diberi BOD5,20C air sisa tersebut adalah 200 mg/ldan k = 0.23 d-1

Tentukan BOD ultimateL t = L o e -kty 5 = L o – L 5

= L o (1 - e -kt )200 = L o (1 - e -5(0.23) )

= L o (1- 0.316)L o =

Tentukan BOD 1 , L 1L t = L o e -kt

L 1 = 293 (e -0.23(1 ))= 293 (0.795)

L 1 = 233 mg/l Y = L o - L 1

= (293-233) mg/l=


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According to Thomas,(1-e -kt ) = kt (1+kt/6 )-3 Therefore

BODt=L o(1-e -kt ) BODt=L o(kt )[1+(1/6) kt ]-3 .....(1)

By rearranging terms & taking the cube root of both sides,equation (1) can transformed to:

(t /BOD t)1/3 =( k Lo)-1/3 + ( k )2/3 /6(L o)1/3 (t ) ....(2)

A plot of ( t /BOD t)1/3 versus t is linear. The intercept isdefined as:

A = ( k Lo)-1/3 ….( 3)


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a slope is defined by:B = ( k )2/3 /6(L o)1/3 ….( 4)

Recalled , y = c + m x x =t

C =(kLo )-1/3

b

a m=a/b = (k 2/3 /6Lo1/3 )


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Solving L o1/3 in Eq.(3) substituting into Eq. (4) and solving fork yields:

k= 6 (B/A) …( 5)Likewise, substituting Eq. (5) into Eq.(3) and solving for L o

yields:Lo = 1/6(A) 2(B) …( 6)


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The following data were obtained from anexperiment to determine the BOD rate constantand ultimate BOD for an untreated wastewater:

Time day) 2 4 6 8 10BOD(mg/L)

125 200 220 230 237


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Solution

Calculate values of (t/BOD t)1/3 for each day.

Plot (t/BOD t)1/3 versus t on graph paper and drawthe line of best fit by eye.

Time day) 2 4 6 8 10

BOD (mg/L) 125 200 220 230 237(t/BOD t)1/3 0.252 0.271 0.301 0.326 0.348


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Time ( t)

A= 0.224

0 2 4 6 8 10

0.1

0.2

0.3

0.4

0.5

B= 0.0124


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Determine the intercept (A) and slope (B)from the plot.

A = 0.224B = (0.348-0.224)/(10-0)

= 0.0124Calculate k and L o using the followingformula

k = 6 (B/A) Lo = 1/ 6(A) 2(B)


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Therefore:k = 6(B/A)

= 6(0.0125/0.224)

= 0.335 / dayLo = 1/6(A) 2(B)= 266 mg/L

CHAPTER 2-Water and Wastewater Analysis (Part 2)

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Transcript of CHAPTER 2-Water and Wastewater Analysis (Part 2)