Question

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Answer in units of ft. 011 (part 3 of 3) 5.0 points Using Table 6 (p. 37), what is the typical range of mixing intensity for low-turbidity 006 (part 4 of 6) 5.0 points moreno (m5773) HW04 Mixing walker - (18385) flocculation? 3. (a) 2.78 hp, (b) 1.42 hp, (e) 0.443 hp 1. G . 20 - 100 -1 4. (a) 3.21 hp, (b) 1.11 hp, (c) 0.443 hp 2. G = 10-80- 5. (a) 2.38 hp, (b) 1.11 hp, (c) 0.443 hp 3. G 20 - 70:- 6. (a) 3.21 hp, (b) 2.38 hp, (e) 1.11 hp 4. G 10-50-1 7. (a) 2.78 hp, (b) 2.38 hp, (c) 0.443 hp 012 (part 1 of 8) 5.0 points Following Example 9 12), for a plante siyon flow of 68 MGD and 8 p ale proces trains, what is the volume (ft) of each flocou lation basin with a hydraulic residence time of 26 min? Answer in units of a 018 (part 7 of 8) 5.0 points Calculate the diameter (ft) of the axial Im- peller of each compartment using a D/T ratio of 0.43 Answer in units of ft. 013 (part 2 of 8) 5.0 points What is the volume ( ) of each of 3 floccula- tion compartiments per train? Answer in units of A 019 (part of 8) 6.0 points Based on previously calculated motor powers which choice is the set of rotational speeds for the three compartments with an axial power number of 0.20 6 2.4 lb/ 014 (part 3 of 8) 5.0 points What is the surface area (R) of each com- partment of the water depth is 12 ft? Answer in units of it. 015 (part 4 of 8) 5.0 points What is the length, L (ft), of each side of a tee Snare compartment? Answer in units of ft. 1. (a) 15.372 rpm, (b) 10.1 rpm, (c) 4.3324 rpm 2. (a) 12.6 rpm, (b) 10.1 rpm, (c) 7.14 rpm 3. (a) 11.2644 rpm, (b) 8.1103 rpm, (c) 4.30824 rpm 1. (a) 12.6 rpm, (b) 11.2614 rpm, (c) 8.1103 pm 5. (a) 15.372 rpm. (b) 8.1103 rpm, (c) 4.32424 rpm 6. (a) 15.372 rpm, (b) 11.2640 rpm, (c) 7.14 pm 7. (a) 12.6 rpm, (b) 10.1 rpm, (c) 8.1103 rpm 016 (part 5 of 8) 5.0 points What is the equivalent diameter, T (ft) of each compartment? Answer in units of ft. 017 (part 6 of 8) 5.0 points Which choice show the correct motor power (hp) required for each flocculation compart- went with a mixing efficiency of 80 percent, and a mixing intensity of (a) 70 . b) 50 (c) 306 =3.2 x 10-5 1. (A) 3.21 hp, (b) 2.38 hp, (c) 0.511 hp 2. (a) 2.78 hp, (b) 1.12 hp, (e) 0.511 hp

Answer 1

Ans 11) Option 3 is correct because for low turbidity , value of G ranges from 20 - 70 s^-1

Ans 12) Number of trains (n) = 8

Discharge = 68 x 10⁶ gallon /day or 47222.22 gallon/min

Residence Time = 26 min

Volume (V) = Discharge x time

= 47222.22 x 26

= 1227777.77 gallon or 164130 ft³ ( 1 gallon = 0.1336 ft³)

Volume of each flocculation basin = V / ( n- 1) = 164130 / 7 = 23447 ft³

Ans 13) Volume of each compartment = 23447/3 = 7816 ft³

Ans 14) Area = 7816 / 12 = 651.5 ft²

Ans 15) Length = Width

=> Length = Area^1/2 = 651.5^1/2 = 25.5 ft

Ans 16) Equivalent diameter , T = (4 A/$\pi$)^1/2

=> T = ( 4 x 651.5 / 3.14)^1/2

=> T = 28.8 ft

Ans 17) P = G² x $\mu$ x Volume of each compartment

   P₁ = 70² x 3.2 x 10^-5 x 7816

=> P₁ = 1225.4 x (0.00181/0.8) = 2.78 hp

  P₂ = 50² x 3.2 x 10^-5 x 7816

=> P₂ = 625.28 x (0.00181/0.8) = 1.42 hp

P₃ = 30² x 3.2 x 10^-5 x 7816

=> P₃ = 225.1 x (0.00181/0.8) = 0.511 hp

Hence correct answer is option 2

Ans 18) D = 0.43 T = 0.43 (28.8) = 12.38 ft

Ans 19) We know,

P = N_p$\rho$ N³ D⁵

where, P = Power (Watts)

Np = Power number

  $\rho$ = density ( kg/m³) = 62.4 lb/ft³ or 1000 kg/m³

N = rotational speed (rpm)

D = Diameter of impeller(m) = 12.38 ft or 3.775 m ( 1 ft = 0.3048 m)

Power (P1) = 2.78 hp or 2074 W

Power (P2) = 1.42 hp or 1059 W

Power (P3) = 0.511 hp or 381.2 W

Putting values,

a) 2074 = 0.28 (1000) N₁³ (3.775)⁵

=> N₁³ = 0.0096

=> N₁ = 0.21 rev per second

=> N₁ = 0.21 rev/sec x 60 sec / min = 12.6 rpm       

b) 1059 = 0.28 (1000) N₂³ (3.775)⁵

=> N₂³ = 0.00495

=> N₂ = 0.17 rev/sec or 10.1 rpm

c) 381.2 = 0.28 (1000) N₃³ (3.775)⁵

=> N₃³ = 0.00177

=> N₃ = 0.121 rev/sec or 7.14 rpm  

Hence, option 2 is correct