Compound Molecular weight (g/mol) Solubility

Transcription

1 Q1. a. 5kg Alum (Al2(SO4)3.18H2O) added t 4m 3 f water. Sludge prduced in kg/m 3? Reactin in slutin that prduces precipitate: 3Ca(HCO3)2 + Al2(SO4)3.18H2O 2Al(OH)3 + 3CaSO4 + 6CO2 + 18H2O 2Al(OH)3 is insluble and settles as precipitate Mlecular Weight Ca(HCO3)2 = 1g/mle (as CaCO3) Mlecular Weight Al2(SO4)3.18H2O = 666.5g/mle Mlecular weight f 2Al(OH)3 = 28g/mle One mle f Al2(SO4)3.18H2O results in 2 mles f Al(OH) g f Al2(SO4)3.18H2O result in 2 78 = 156 g f Al(OH)3 1g f Al2(SO4)3.18H2O will result in 156/666.5 g f Al(OH)3 5kg Al2(SO4)3.18H2O will result in 156 5/666.5g f AL(OH)3 5kg Al2(SO4)3.18H2O will result in g = 11.72kg f AL(OH)3 S, abut 11.72kg sludge will be prduced per 4m kg/4 = 2.925kg/1 3 m 3 b. 5kg Ferrus Sulfate (FeSO4.7H2O) and 5 kg Lime (2Ca(OH)2) are added t 4m 3 f water. Reactins in water and resultant FeSO4.7H2O + Ca(HCO3)2 Fe(HCO3)2 + CaSO4 + 7H2O Fe(HCO3)2 + 2Ca(OH)2 Fe(OH)2 + 2CaCO3 + 2H2O Fe(OH)2 +1/4O2 + 1/2H2O Fe(OH)3 Cmpund Mlecular weight (g/ml) Slubility

2 FeSO4.7H2O 278 Sluble Ca(HCO3)2 1 (as CaCO3) Sluble Fe(HCO3)2 178 Sluble Ca(OH)2 56 Slightly sluble Fe(OH) Very Slightly Sluble CaCO3 1 Smewhat Sluble Fe(OH) Insluble CaSO4 136 Sluble H2O 18 X CO2 44 Sluble O2 32 Sluble Reactin Set 1: FeSO4.7H2O + Ca(HCO3)2 Fe(HCO3)2 + CaSO4 + 7H2O 278g f FeSO4.7H2O results in 178 g Fe(HCO3)2 5 kg f FeSO4.7H2O results in 178 5/278 g f Fe(HCO3)2 5 kg f FeSO4.7H2O results in 32.1 kg f Fe(HCO3)2 Reactin Set 2: Fe(HCO3)2 + 2Ca(OH)2 Fe(OH)2 + 2CaCO3 + 2H2O Nw, 178g f Fe(HCO3)2 results in 89.9 g Fe(OH) kg f Fe(HCO3)2 results in /178g f Fe(OH) kg f Fe(HCO3)2 results in kg f Fe(OH)2 In additin, 178g f Fe(HCO3)2 results in 2 g CaCO kg f Fe(HCO3)2 results in 2 321/178g f CaCO kg f Fe(HCO3)2 results in kg f CaCO3 Amunt f Lime cnsumed 178g f Fe(HCO3)2 cnsume 2 56g f Lime (Ca(OH)2) We have 32.1kg f Fe(HCO3)2, 32.1kg f cnsume /178g f Lime (Ca(OH)2) 32.1kg f cnsume 2.14kg f Lime (Ca(OH)2) Amunt f Lime nt cnsumed: = 29.76kg

3 This additinal lime will frm CaCO3 accrding t the belw reactin Ca(OH)2 + Ca(HCO3)2 2CaCO3 + 2H2O 56g f Lime (Ca(OH)2) frms 2 g f CaCO3 We have 29.76kg f Lime 29.76kg f lime will results in /56 g f CaCO kg f lime will results in kg f CaCO3 Fe(OH)2 +1/4O2 + 1/2H2O Fe(OH)3 Nw, 89.9g f Fe(OH)2 results in 16.9 g Fe(OH)3 Since, we have kg f Fe(OH) kg f Fe(HCO3)2 results in /89.9g f Fe(OH) kg f Fe(HCO3)2 results in kg f Fe(OH)3 Nw, we have in 4m 3 water, we have ( ) f CaCO3, and 19.23kg Fe(OH)3 But 15 4 = 6 kg f CaCO3 is nt sluble (given) and will nt result in sludge. Hence, we ttal sludge in 4m 3 f water = = kg Hence, sludge prduced = kg/1 3 m 3 f water

4 Q2. Half reactin 1: H2S S + 2H + + 2e - Oxidatin Ptential E = +.14 vlts Half reactin 1: Cl2 +2e - 2Cl - Reductin Ptential E = vlts Ttal reactin: H2S+Cl2 S + 2HCl - Over all ptential Over all ptential = E reactin = E reductin E reactin E reactin E xidatin = = vlts Because the E reactin is psitive, the reactin will prceed as written, i.e., twards right.

5 lg inactivatin CE 28 Water & Wastewater Engineering Assignment 7 Slutin Spring 218 Q Time A Time B Time C Time D Linear (Time A) Linear (Time B) Linear (Time C) Linear (Time D) -6 Time (min) Lg(Nt/N) vs time T estimate inactivatin rate cnstant, lg(nt/n) = -kt Fr sample A, let s take last pint (the plt is linear, s we can take any pint thrugh which the straight-line passes) -5 = -(k/min) (1) k =.5/min Fr Sample B -5 = -k 3.1 k=.17/min Fr Sample C -3 = -k 12.6 k=.24/min Fr Sample D -2 = -k 5.9 k=.34/min

6 Lgnt CE 28 Water & Wastewater Engineering Assignment 7 Slutin Spring 218 Q4. a. First, we estimate lgn using variatin f Nt with time at the minimum Cl dse, 1 mg/l lgnt Series2 Linear (lgnt) Linear (lgnt) Time Plt f Lg Nt v/s time Assuming a linear fit, lgnt =lgn -k t, we get, i.e., the intercept f the straight line, is lgn lgnt-lgn = -k t (lgnt-lgn)/t = -ks Nw, lets use first and last pint n the plt abve (4-lgN)/15 = -k (2.698 lgn)/6 = -k k~.3/min And lgn = 4.45 N ~ 28 Nte: The answer t this questin will vary depending n the value yu estimate fr N Plts f lg(nt/n) v/s CT are prvided belw:

7 lgnt/n lgnt/n lgnt/n CE 28 Water & Wastewater Engineering Assignment 7 Slutin Spring T15 Linear (T15) CT CT T3 Linear (T3) CT T6 Linear (T6) The plts data in all the three plts shws apprximately linear variatin. S, in equatin 12-7 f the bk, lg N t N = k C n t The cefficient f dilutin n can be taken as 1. Hence, we have

8 lg N t N = k Ct Since, the plts are lg N t N v/s Ct, k can be taken as the slpe f the lines Fr sample 1, (Time = 15) The line crss the pint at CT = = -k 12 k = 2.72/12 =.23/mg-min/L Fr sample 2, (Time = 3) The line crss the pint at CT = = -k 24 k = 3.17/24=.13/mg-min/L Fr sample 3, (Time = 6) The line crss the pint at CT = = -k 36 k = 3.37/36=.9/mg-min/L b. Vlume required fr 6-min cntact time at average winter flw Average winter flw Q = 26m3/d Detentin time τ = 6 min τ = Q = τq = 6min 26m 3 /d = m 3 Sample 1 Fr residual clifrm cunt 2MPN/1, hw much CT is needed[this was nt asked in the questin] Lg(Nt/N) = -k CT CT = Lg(2/28)/-k

9 CT ~ 93 mg-min/l Cntact time T = 6 min C= 93/6 = 1.55 mg/l (Chlrine dsage. Actually, we may need a little higher t keep sme residual Cl) Similarly, t have 1MPN/1 residual clifrm cunt, CT = Lg (1/28)/-k CT ~ 63 mg-min/l C = 63/6 C = 1.5 mg/l Similarly, yu ll need t calculate dsages fr ther samples. c. Yearly Chlrine dsage Average flw ver a year = (2 + 26)/2 = 23 m3/d Chlrine mass required fr a year fr Sample 1 at average lading at residual clifrm 2MPN/1mL, Cl2, required = Cl2 dsage Q 365 = 1.55 mg/l 23 m 3 /d 365 d/y = kg/y Similarly, fr 1MPN/1mL residual clifrm, Cl2, required = kg/y Similarly, yu ll need t calculate chlrine requirements fr ther samples.

10 CHLORINE RESIDUAL, MG/L CE 28 Water & Wastewater Engineering Assignment 7 Slutin Spring 218 Q5. Plt f chlrinatin data Sample B CHLORINE DOSAGE, MG/L Frm the plt, I take breakpint chlrinatin apprximately as 9mg/L The design dsage t btain a residual f 1mg/L: Cmbined residual at breakpint ~.8mg/L We want free residual 1mg/L ttal residual =1.8mg/L, which ccurs at Chlrine dsage f abut 1.15mg/L Design dsage = 1.15mg/L