Question 12.14: DETERMINATION OF THE VELOCITY DIFFERENTIAL, POWER, DETENTION...
DETERMINATION OF THE VELOCITY DIFFERENTIAL, POWER, DETENTION TIME, VELOCITY GRADIENT, AND LOADING OF A FLOCCULATOR
A flocculator designed to treat 20 MGD (75.7 MLD = 0.88 m³/s) is 100 ft long, 40 ft wide, and 15 ft deep (30.48 m × 12.19 m × 4.57 m). It is equipped with 12 in. (305 mm) paddles supported parallel to and moved by four 40 ft (12.19 m) long horizontal shafts that rotate at a speed of 2.5 rpm. The center line of the paddles is 6.0 ft (1.83 m) from the shaft, which is at mid-depth of the tank. Two paddles are mounted on each shaft, one opposite the other. If the mean velocity of the water is approximately one-fourth the velocity of the paddles and their drag coefficient is 1.8, find
1. The velocity differential between the paddles and the water
2. The useful power input and the energy consumption
3. The detention time
4. The value of G and the product Gt_{d}
5. The flocculator loading
Assume a water temperature of 50°F (10°C), 𝜇 = 2.74 × 10^{−5} (lb force)(s)/ft² (1.31 × 10^{−3} N-s/m²).
1 (US Customary System):
1. The paddle velocity is
ν_{i} = 2𝜋rn = 2𝜋 × 6(2.5∕60) = 1.57 ft/s, and
the velocity differential is
ν= (1 − k) ν_{i}
= (1 − 0.25)1.57
= 1.18 ft∕s.
2. Because the area of the paddles is A = 40 × 2 × 4 × 1 = 320 ft² and the coefficient of drag C_{D} = 1.8, the useful power input, by Eq. (12.6), is
P = 1/2C_{D}ρAν³
P = 0.5 × 1.8(62.4∕32.2)(320)(1.18)³
= 918 ft-lb∕s
= 918∕550 hp
= 1.67 hp or 1.24 kW.
The energy consumption per million gallons, therefore, is 1.67 × 24/20 = 2.0 hph/MG, or 1.5 kWh/MG (0.39 kWh/ML) treated. For electrical drive, there must be added the energy required to overcome mechanical friction and to provide for electrical losses in the lines and motor. (In practice, flocculators consume 2–6 kWh/MG treated or 0.53–1.6 kWh/ML treated.)
3. Because the volume of the tank is 40 × 100 × 15 = 6 × 10^{4} ft³, the detention period is
t_{d} = V∕Q
= 6 × 10^{4} ft³ × 7.48 gal∕ft³ × 24 h∕day
× 60 min∕h∕(20 × 10^{6})MGD
= 32.5 min.
4. The value of G is
G = \sqrt[2]{P/\mu V}
G = [918∕(2.74 × 10^{−5} × 6 × 10^{4})]^{1∕2}
= 23.7 ft∕s∕ft.
Gt_{d} = 23.7∕s × 32.5 min × 60 s∕min
= 4.64 × 10^{4}.
5. The flocculator loading is
Q∕V = 20 × 10^{6}/(6 × 10^{4})
= 333 gpd∕ft³
= 44.5 ft³∕d∕ft³.
2 (SI System):
1. The paddle velocity is
ν_{i} = 2𝜋rn = 2𝜋 × 1.83(2.5∕60) = 0.48 m/s, and
the velocity differential is
ν = (1 − k)ν_{i}
= (1 − 0.25)(0.48)
= 0.36 m∕s.
2. Area of the paddles is A = (12.19 × 0.30 m² /paddle) (4 shafts)(2 paddles/shaft) = 30 m², the useful power input, by Eq. (12.6), is
P = 1/2C_{D}ρAν³
P = 0.5 × 1.8[(9,804 N∕m³)∕(9.81 m∕s²)](30 m²)(0.36 m∕s)³
= 1,260 N-m∕s
= 1.25 kW.
Energy consumption per ML = 1.25 × 24/75.7 ML = 0.40 kWh/ML treated. For electrical drive, there must be added the energy required to overcome mechanical friction and to provide for electrical losses in the lines and motor.
3. Volume of tank = 30.48 × 12.19 × 4.57 = 1,700 m³
The detention period is t_{d} = V∕Q
= [(1,700 m³)∕(0.88 m³∕s)]∕(60 s∕min)
= 32.2 min.
4. The value of G is
G = \sqrt[2]{P/\mu V}
G = [1,260∕(1.31 × 10^{−3} × 1,700)]^{1∕2}
= 23.8∕s.
Gt_{d} = (23.8∕s)(32.2 min × 60 s∕min)
= 4.60 × 10^{4}.
5. The flocculator loading is
Q∕V = 75.7 × 10^{6} L∕day∕(1,700 m³)
= 44, 500 L ∕ day ∕ m³ or 44.5 m³ ∕ day ∕ m³.