Question 12.11: DESIGN OF IN-LINE MECHANICAL BLENDERS Choose a typical comme...

1 (US Customary System):

Select a residence time of 0.5 s in blender (check on typical residence time of a commercial blender) from Section 12.3.3.
0.5 water hp is required for each MGD of water flow.

P = required power input = (0.5 hp∕1 MGD)(1.6 MGD)
= 0.8 water hp
= 0.8 hp × (550 ft-lb∕s)∕(1 hp)
= 440 ft-lb∕s

t_{d} = 0.5 s
Q = 1.6 MGD = 1.6 MGD× (1.547 cfs∕MGD) = 2.4752 ft³∕s
V = water volume to which power is supplied at 0.5 s
= Qt_d = (2.4752 ft³∕s)(0.5 s) = 1.2376 ft³

At 8°C

𝜇 = viscosity = 1.387 centipoises at 8°C
= 1.387 centipoises × (2.088 × 10^{−5} lb-s∕ft²)∕(1 centipoises)
= 2.90 × 10^{−5} lb-s∕ft² at 8°C

G = (P∕𝜇V)^{0.5}
= {(440 ft-lb∕s)∕[(2.9 × 10^{−5} lb-s∕ft²)(1.2376 ft³)]}^{0.5}
= 3,500 s^{−1}

At 18°C

𝜇 = 2.2 × 10^{−5} lb-s∕ft² at 18°C
G = {(440 ft-lb∕s)∕[(2.2 × 10^{−5} lb-s∕ft²)(1.2376 ft³)]}^{0.5}
= 4,000 s^{−1}

The range of G values lies within the recommended values of 3,000–5,000 s^{−1}

2 (SI System):

Select a residence time of 0.5 s in blender (check on typical residence time of a commercial blender) from Section 12.3.3.
0.37 kW is required for 43.8 L/s of water flow. 6.0560 MLD = 6.0560 × 10^{6}L/1,440 × 60 s = 70.1 L/s

P = required power input = (0.37 kW∕43.8 L∕s)(70.1 L∕s)
= 0.592 kW
= 581.61 N-m∕s or watts
t_{d} = 0.5 s
Q = 6.0560 MLD = 6.0560 × 10³ m³∕(1,440 × 60 s)
= 0.07 m³∕s
V = water volume to which power is supplied at 0.5 s
= Qt_d
= (0.07 m³∕s)(0.5 s) = 0.035 m³

At 8°C

𝜇 = viscosity = 1.387 centipoises at 8°C
= 1.387 × 0.001 Pa-s or N-s∕m²
= 1.387 × 10^{−3} N-s∕m²
G = (P∕𝜇V)^{0.5}
= {(581.61 N-m∕s)∕[(1.387 × 10^{−3} N-s∕m²)(0.035 m³)]}^{0.5}
= 3,500 s^{−1}

At 18°C

𝜇 = 1.060 × 10^{−3}  N-s∕m²
G = {(581.61 N-m∕s)∕[(1.06 × 10^{−3} N-s∕m²)∕(0.035 m³)]}^{0.5}
= 4,000

The range of G values lies within the recommended values of 3,000–5,000 s^{−1}.