Question 12.13: DETERMINATION OF HEADLOSS, POWER, AND LOADING OF A BAFFLED C...
DETERMINATION OF HEADLOSS, POWER, AND LOADING OF A BAFFLED CHANNEL
Water zigzags through a baffled channel basin at a velocity of 0.5 ft/s (0.15 m/s) and speeds up to 1.5 ft/s (0.46 m/s) in the slots.
There are 19 around-the-end baffles. The flocculation basin is 10 ft × 30 ft × 60 ft (3.05 m × 9.14 m × 18.29 m). Estimate
1. The loss of head, neglecting normal channel friction
2. The power dissipated
3. The G and Gt_d values for a flow of 6.46 MGD (10.0 ft³/s = 0.28 m³/s), with a displacement time of 30 min
4. The channel loading
Assume a water temperature of 50°F (10°C), that is, 𝜇 = 2.74 × 10^{−5} lb force-s/ft² (1.31 × 10^{−3} N-s/m²) .
1 (US Customary System):
1. The loss of head
h = nν^{2}_{1} ∕ 2g + (n − 1)ν^{2}_{2} ∕ 2g
h = 20 × 0.5²∕2g + 19 × 1.5²∕2g = 0.74 ft.
2. The power dissipated
P = Q𝜌gh
P = 10 ft³∕s × 62.4 lb∕ft³ × 0.74 ft
= 462 ft-lb/s = 0.84 hp = 626 W.
3. The G and Gt_{d} values
V = 10 ft × 30 ft × 60 ft = 18,000 ft³.
G = \sqrt[2]{P/\mu V}
G = [462∕(2.74 × 10^{−5} × 18,000)]^{1∕2}
= 30.6 ∕ s.
Gt_{d} = 30.6∕s × 30 min × 60 s∕min = 5.5 × 10^{4}
4. The channel loading, flow per unit volume.
Q∕V = 6.46 × 10^{6} gpd∕(18,000) ft³
= 360 gpd ∕ ft³.
2 (SI System):
1. The loss of head
h = nν^{2}_{1}/2g + (n − 1)ν^{2}_{2}/2g
h = 20 × (0.15)²∕2 × 9.81 + 19 × (0.46)²∕2 × 9.81
= 0.23 m.
2. The power dissipated
P = Q𝜌gh
P = 0.28m³∕s × 9,804N∕m³ × 0.23m
= 627 N-m/s=0.627kW = 627 W.
3. The G and Gt_{d} values
V = 3.05 m × 9.14 m × 18.29 m = 510 m³.
G = \sqrt[2]{P/\mu V}
G = [(627 N-m∕s)∕(1.31 × 10^{−3} N-s∕m²)(510 m³)]^{1∕2}
= 30.6 ∕ s.
Gt_{d} = 30.6∕ s × 30 min × 60 s∕min = 5.5 × 10^{4}.
4. The channel loading, flow per unit volume
Q ∕ V = [(0.28 m³∕s)∕(510 m³)](60 s∕min)(60 min ∕ h)(24 h ∕ day)
= 47.4 m³ / day / m³.