Department of Civil Engineering, IIT Delhi CEL 212: Environmental Engineering (Second Semester 2011-2012) Minor I- February 7 th , 2012 Solution (Total Points = 50; Duration = 60 minutes )

Name______Entry no.______Group No______Note: Assume missing data (if any) and mention the same. Be precise in all open-ended questions. Q1. Determine the settling velocity of a spherical particle with diameter of 200 micron and a specific gravity of 2.65 in water at 20°C? Comment on settling behaviour of this type of particles. µ=1.001×10 -3 N.s/m 2 at 20°C. [8+2=10 points]

Solution: Iteration 1: Assume laminar flow and apply Stokes law: 2 3 -6 2 2 -3 2 Vt= (9.81m/s )*(2650-1000) (kg/m )*(200*10 ) m / [18*1.001×10 N.s/m ] =0.0359 m/sec

Now calculate Reynolds Number. Reynolds number =Re= {(1000 kg/m 3) *(0.0359m/s)* (200*10 -6m)}/ [1.001×10 -3 N.s/m 2] = 7.17 (>1 and <10000 => transitional flow)

Iteration 2: 0.5 Now calculate C D = (24/Re) + {3/ (Re) } +0.34 = 4.806 Now use general formula to calculate Vt (for non-laminar flow conditions) 2 -6 Vt = (4/3*9.81)* (2650-1000)*(200*10 m)/ [4.806*1000] =0.0299 m/sec

Re= {(1000 kg/m 3) *(0.0299m/s)* (200*10 -6m)}/ [1.001×10 -3 N.s/m 2] = 5.98 (>1 and <10000 => transitional flow)

Iteration 3: Now calculate C D = 5.574 2 -6 Vt = (4/3*9.81)* (2650-1000)*(200*10 m)/ [5.574*1000] Vt =0.02783 m/sec

So final settling velocity =0.02783 m/sec

This type of particles follow type-1 settling where particles behave as discrete particles and their shape and size do not change during their settling.

Q2. For the following samples, calculate hydroxide, carbonate, and bicarbonate alkalinity by the procedure (Alkalinity and pH measurements). The sample size is 100 mL, N/10 sulfuric acid is used as the titrant and the water temperature is 25°C. [4+4+4=12 points] Sample pH Total mL titrant to reach end point Phenolphthalein Bromcresol green A 11 10 15.5

Answer: pH 11 => All three types of alkalinity; pOH=14-pH = 14-11=3; [OH-] =10 -pOH = 10 -3 moles/L

1 Number of hydroxide equivalents = (17g/mole*10 -3 mole/L)/ (17g/equivalent) = 10 -3 equivalents/L = number of equivalents of CaCO 3

-3 Hydroxide alkalinity = (equivalent weight of CaCO 3 =50 g/equivalent)*(10 equivalents of CaCO 3/L) -3 3 =50*10 *10 = 50 mg/L CaCO 3

As titration was done using 0.01N H 2SO 4, so phenolphthalein alkalinity = (X mL of 0.01N H 2SO 4) *(1000/sample volume in mL) mg/L CaCO 3 Given 100 mL sample volume: So, phenolphthalein alkalinity = (50)*(1000/100) = 500 mg/L CaCO 3 Similarly, total alkalinity = (100)*(1000/100) = 1000 mg/L CaCO 3

Carbonate alkalinity = 2*[Phenolphthalein alkalinity-hydroxide alkalinity] =2*[500-0.5] =2*[499.5] =999 mg/L CaCO 3

Bicarbonate alkalinity = Total alkalinity - [Carbonate alkalinity + hydroxide alkalinity] =1000-[999+0.5] = 0.5 mg/L CaCO 3

Q3. A city water supply is obtained from a deep aquifer. The water has uniform quality, which is clear and free of organics; hardness is in excess of 300 mg/L and consists of both 2+ calcium and magnesium. Dissolved CO 2 is approximately 15 mg/L and iron (Fe ) is about 1 mg/L. Other dissolved constituents are below problem levels. Draw a schematic diagram of a treatment plant that will render this water potable. Identify each unit and briefly state its purpose. Show points of chemical addition and identify the chemicals. [3+3+2=8 points]

Answer: Processes should be given to remove things creating problem. [Aeration->Coagulation-flocculation-Adsorption] combination can remove carbon dioxide gas, hardness and ferrous ions. As groundwater does not have pathogen (indicated through water quality information), no disinfection unit is necessary here. Adsorption can also remove pathogens if there any.

Q4. The IIT Delhi wastewater (pH7) has calcium ions (50 mg/L), magnesium (10 mg/L) ions, sodium (14.8 mg/L), bicarbonate ions (130 mg/L), 20 mg/L sulfate ions, and 100 mg/L carbonate ions. Calculate total hardness, total alkalinity, and carbonate hardness values? Comment on carbonate hardness and total alkalinity values. [4*4=16 points]

Answer: Only cations with bivalency contribute to hardness. 2+ 2+ Hardness (in mg/L) as CaCO 3 = M (in mg/L) *[(50)/ (Equivalent weight of M )] Cation Concentration Equivalent weight Hardness (in mg/L) as CaCO 3 (mg/L) (g/mole) Ca 2+ 50 20 =(50) *[(50)/(20)]=125 Mg 2+ 10 12.2 =(10) *[(50)/(12.2)]=41

2 2+ 2+ Total hardness = Hardness due to Ca ions and Mg ions. = 125+41= 166 mg/L as CaCO 3 (answer)

Total alkalinity calculation Anion Concentration Equivalent weight Alkalinity (in mg/L) as CaCO 3 (mg/L) (g/mole) 2- CO 3 100 =(12+3*16)/2= 30 =(100) *[(50)/(30)]=166.7 - HCO 3 130 =(1+12+3*16)/1= 61 =(130) *[(50)/(61)]=106.56 Total =166.7+106.56 =273.26 alkalinity

Sulfate-based hardness Anion Concentration Equivalent weight Hardness (in mg/L) as CaCO 3 (mg/L) (g/mole) 2- SO 4 20 =(32+4*16)/2= 48 =(20) *[(50)/(48)]=20.83

Maximum value of non-carbonate hardness = 20.83 mg/L as CaCO 3

As the maximum value of carbonate hardness could be equal to total hardness is 166 mg/L as CaCO 3 (assuming that calcium and magnesium ions do not react with sulfate ions).

Given that maximum value of non-carbonate hardness = 20.83 mg/L as CaCO 3 So minimum value of carbonate hardness = (166 mg/L as CaCO 3 )– ( 20.83 mg/L as CaCO 3) = 145.17 mg/L as CaCO 3

Note that this value is higher than alkalinity present in the sample.

Q5. Define following terms briefly (<100 words): (i) Flocculating particles, (ii) Critical DO deficit, (iii) oxygen regeneration rate and (iv) indicator organisms. [1*4=4 points] Answer: These things have been discussed in class and also explained in lecture notes.

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