Ass Ans

 

Assignment For Chapter 5 1.  How much alkalinity (as CaC03) is consumed by a dose of 60 mg/L m g/L of each of the following coagulants: Al2(S04)3 . 14H2O, Fe2(S04)3, and FeCl3? If the water alkality of 100mg/l as CaCo3.Determine the additional Lime required

Solution For alum , Al2(so4)3.14H2O 2-

-

Al2(SO4)3.14H2O + 6HCO3 ↔ 2Al(OH)3↓ + 6CO2 + 14H2O + 3SO4

Molecular mass of Alum, Al 2(SO4)3.14H2O =(27*2)+(32*3)+16*12+18*14=594mg Molecular mass of bicarbonate, 6HCO3=6*((1+12+(16*3)) =366mg -

594mg of Al2(SO4)3.14H2O consume →366 mg of HCO3

                    

60 mg mg of Al2(SO4)3.14H2O will consume →

 

Alkalinity as CaCo3 = Eq wt of CaCo3 = -

Eq wt of HCo3  =

 HCO3

= 50

= 61

Alkanlinity as CaCo3 =

 =30.3mg/L as . CaCo3

For Fe2(SO4)3 Fe2(SO4)3 + 3Ca(HCO3) ↔ 2Fe(OH)3 +3CaSo4 +6CO2 Molecular mass Fe2(SO4)3 =(56*2)+(32*3)+(16*12) =400 mg Molecular mass 3Ca(HCO3) =3*((40+1+12+(16*3)) =303 mg 400 mg of Fe2(SO4)3 consume → 303 mg of Ca(HCO3) 60 mg of Fe2(SO4)3 will consume

→

    

 =45.45 mg. Ca(HCO3)

 

                 

Alkanlinity as CaCo3 = Eq wt of CaCo3 =

 

= 50

Eq wt of Ca(HCo3) =

= 50.5

Alkanlinity as CaCo3 =

 

For FeCl3

 =45mg/L as . CaCo 3

2FeCl3 + 3Ca(OH)2 ↔ 3CaCl2 +2Fe(OH)2  Molecular mass 2FeCl3 =2*(56+(35.5*3))=325 mg Molecular mass 3Ca(OH)2=3*((40+(16*2)+(1*2))=222mg 325 mg of FeCl3 consumes →222 mg of Ca(OH) 2 60 mg mg of FeCl3 will consumes→

                    

=40.98 mg Ca(OH)2  

Alkanlinity as CaCo3 = Eq wt of CaCo3 =

Eq wt of Ca(OH)2 =

Alkanlinity as CaCo3 =

= 50

= 37

 =55.14 mg/L as . CaCo3

 

  2. Design a square rapid mixing basin for a water treatment plant with a design flow of 8000m 3/d. The basin depth is equal to 1.25 times the width. The design velocity gradient will be 1000 s-l at 15°C and the detention time is to be 30 s. Determine the basin dimensions, and the input power required. 3. A water treatment plant is processing a flow of 42000 m3/d. The flocculation system consists consists of three identical parallel units. A side view of one of the units is shown in the sketch below. The dimensions of each flocculation unit are 4 m wide by 4 m deep by 15.5 m long. The inlet baffle is located 0.4 m from the front wall. Each unit has three sets of paddles mounted on a horizontal shaft. There are three sets of paddles attached to the shaft. The first set of paddles has four paddles on each arm (total of eight paddles) with centers located at 1.6, 1.4, 1.2, and 1 m from the shaft. The second set has three paddles on each arm with centers located at 1.6, 1.4, and 1.2 m from the shaft, and the third set has two paddles per arm with centers located at 1.6 and 1.4 m from the shaft. Each paddle is 0.1 m wide and 3.2 m long. The first paddle set is 0.8 m from the inlet baffle and the paddle set are separated by a distance of 0.8gradient m. At temperatures 20 and 25°C is the rotational speedfor ofeach the shaft to achieve mean velocity of 30 s-l in aofunit? What is what the local velocity gradient paddle set? CD=a 1.8, k = 0.25. Solution

Rotational speed of the shaft V=15.5*4*4 =248 m3 Power per unit From table at T=20    C, μ=16.624*1 μ=16.624*10 0-4 N.s/m2  P=G2μV  =302*16.624*10-4*248 =371.05Nm/s From table at T=25    C, μ=8.996*10-4 N.s/m2 P=G2μV  =302*8.996*10-4*248 =200.86Nm/s Angular velocity of plate Ap=0.1*4*18=7.2 m2

 

From table at T=20    C,  C, ᵨ =999.78Kg/m3 W =

     √  



=0.176m/sec

Actual speed =0.176/0.75 =0.235 m/s ɯ =(0.235m/s )*(rev/3.8*π m)*(60s/min) 

=2.26 rev/min W =

     √  



=0.266m/sec

Actual speed =0.266/0.75 =0.355 m/s ɯ =(0.355m/s )*(rev/3.8*π m)*(60s/min) 

=1.78 rev/min Velocity gradient for each paddle set

From table at T=2    C,  C, ᵨ =999.78Kg/m3 μ=16.624*10-4 N.s/m2

Paddle with 4 blades Ap= 0.1*4.5*4 =1.8 m2  V=380.7 m3

         

3 =1.8*1.8*999.78*0.3263 =56Nm/s

P=Cd*Ap*ᵨ

2 

2

 

μ

G=9.42 /s Paddle with 6 blades

 

Ap= 0.1*4.5*6 =2.7 m2  V=380.7 m3

 

          

3 =1.8*2.7*999.78*0.3263 =84

P=Cd*Ap*ᵨ 2

2

 

μ

G=11.52 /s

Paddle with 8 blades Ap= 0.1*4.5*8 =3.6 m2  V=380.7 m3

 

P=Cd*Ap*ᵨ 2

3 =18*3.6*999.78*0.3263 =112

Nm/s

2

       

 

μ

G=13.30 /s

From table at T=25    C,  C, ᵨ =996.955 Kg/m3 μ=8.996*10-4 N.s/m2 Paddle with 4 blades Ap= 0.1*4.5*4 =1.8 m2  V=380.7 m3

 

P=Cd*Ap*ᵨ 2

3 =1.8*1.8*996.955*0.2663 =30.4

Nm/s

2

       

 

μ

G=9.42 /s Paddle with 6 blades Ap= 0.1*4.5*6 =2.7 m2  V=380.7 m3

Nm/s

 

 

P=Cd*Ap*ᵨ 2

3 =1.8*2.7*996.955*0.2263 =45.56

Nm/s

2

        μ

 

G=11.53 /s Paddle with 8 blades Ap= 0.1*4.5*8 =3.6 m2  V=380.7 m3

 

P=Cd*Ap*ᵨ 2

3 =1.8*3.6*996.955*0.2263 =60.79

Nm/s

2

       

 

μ

G=13.32 /s From the above velocity gradient is the same for different water temperature

4. A flocculation b basin asin designed to treat 50,000 m 3 /d of water is 21 m long, 15 m wide, and 3.60 m deep. The paddle-wheel units consist of four horizontal shafts that rotate at 4 rpm. The shafts are located perpendicular to the direction of flow at mid-depth of the basin. Each shaft is equipped with four paddle wheels 3 m in diameter and each wheel has four blades 3.30 m long and 150 mm wide with two blades located on each side of the t he wheel. The blades are 300 mm apart. Assume the water velocity to be 30% of the velocity of the paddles and that the water temperature is 10 o C. Determine: 1. The power input to the water 2. The v velocity elocity gradient gradient

 

18.8 m 0.8 m 2 AAiT Department of Civil Engineering Water Treatment 3. The retention time 4. The Gt value 5. The population of a town is 100,000 and the average per capita demand is 135 litres/day/capita. Design the coagulation and sedimentation tank for supplying water to the town. The maximum demand may be taken as 1.5 times the average demand. Detentio n period is 5 hours for settling tank and 30 minutes for flocculent chamber. Flow rate is 900 litres/hour/m 2 of plan area. 6. The water works of a city treats 50 million litres/day of water in a coagulation sedimentation tank. The quantity of alum consumed is 15 mg/litres of water. w ater. If the alkalinity of raw water is equivalent to 1.0 mg/litre of CaCO 3 , determine the quantity of alum and quick line (containing 80% CaO) required per month by water works.