Question 12.12: RAPID MIXING BASIN SIZING AND MECHANICAL MIXER DESIGN Calcul...
RAPID MIXING BASIN SIZING AND MECHANICAL MIXER DESIGN
Calculate the volume and dimension (L × W × H) of a square rapid mixing basin and design the mechanical mixer, assuming (a) maximum design flow = 100 MLD = 26.42 MGD; (b) number of mixing basins = 4; (c) number of mixing stages = 1; (d) mixing detention time = 30 seconds; (e) mixer velocity gradient (G) = 950/seconds; (f) depth-to-width ratio of the mixing basin = 1.5; (g) raw water temperature range = 5°C to 29°C; (h) efficiency of mechanical mixer’s gearbox = 90%; (i) power number of the impeller given by the manufacturer N_{p} = 2.75; and (j) Reynolds number equation for rapid mixers R = d² n𝜌/𝜇, where d = impeller diameter, m; n = impeller rotation speed, revolution per second, or rps; 𝜌 = mass density, kg/m³; and 𝜇 = dynamic viscosity = 1.518 × 10^{−3} N-s/m² at 5°C.
Four square rapid mixing basins can be properly sized using the maximum design flow and the required detention time. The Ten-States Standards specify that the mixing detention period should not be more than 30 seconds, with mixing equipment capable of imparting a minimum velocity gradient (G) of at least 750 fps/ft or m/s/m, or s^{−1}. The design engineer should determine the appropriate detention time and G value through a series of jar testing. The following are calculations for mixing basin sizing and mechanical mixer design.
N = 4 mixing values
Q = 100 MLD = 100,000 m³∕day = 69.44 m³∕min = 1.1574 m³∕s
Q∕N = (1.1574 m³∕s)∕4 = 0.2894 m³∕s
V for each mixing basin = (Q∕N)t_{d}
= (0.2894 m³∕s)(30 s) = 8.68 m³
Volume = L × W × H = W × W × 1.5 W = 1.5 W³ = 8.68 m³
W³ = 8.68∕1.5 = 5.7867
W = 5.7867^{0.333} = 1.8 m; L = W = 1.8 m;H = 1.5 × 1.8 m = 2.7 m
Dimension = L × W × H = 1.8 m × 1.8 m × 2.7 m
Design each mechanical mixer for a velocity gradient (G) of 950/s at a flow rate of 0.29 m³/s, which is one-fourth of the maximum design flow rate. The expected raw water temperature is in the range of 5°C to 29°C. The lowest temperature presents the critical condition in this mixer design.
𝜇 = 1.518 × 10^{−3} N-s∕m³ at 5°C
V = mixing basin volume = 8.68 m³
Rearranging Eq. (12.3) power equation.
P = G²V𝜇
P = (950∕s)²(8.68 m³)(1.518 × 10^{−3} N-s∕m²)
= 11,892 N-m∕s
= 11,892 W
= 11,892 kg-m²∕s³
= 11.89 kW
Since P is the power imparted to the water, the power of the drive (P_{d}) is calculated by dividing P by the efficiency of the gearbox, E_{g}, which is typically around 90%.
P_{d} = P∕E_{g}
= 11.89 kW∕0.9
= 13.21 kW
Selection of turbine diameter d = 0.5 × mixing basin width = 0.5 × 1.8 m = 0.90 m
Calculation of impeller rotation speed n, assuming N_{p} = 2.75 given by the manufacturer.
n = (P∕ρN_{p}d^{5})^{1∕3} (12.3b)
= [(11, 892 kg-m²∕s³)∕(1,000 kg∕m³)(2.75)(0.9 m)^{5}]^{1∕3}
= [7.3788]^{1∕3}
= 1.94 rps (or 117 rpm)
Find dynamic viscosity of water at 5°C.
𝜇 = 1.518 × 10^{−3} N-s∕m²
= 1.518 × 10^{−3}[(kg-m∕s²)s]∕m²
= 1.518 × 10^{−3} kg∕s-m
Here, 1 N = 1 kg-m/s².
Check Reynolds number for turbulent flow using a Reynolds number equation for rapid mixers.
ℝ = d²n𝜌∕𝜇 = d²n𝛾∕𝜇g
= (0.9 m)²(1.94 rps)(1,000 kg/m³)∕(1.518 × 10^{−3} kg/s-m)
= 1.0352 × 10^{6} > 10,000
Therefore, it is in the turbulent range. Okay.