Tests on a small scale tank 0.3 m diameter (Rushton impeller, diameter 0.1 m) have shown that a blending process between two miscible liquids (aqueous solutions, properties approximately the same as water, i.e. viscosity 1 mN s/m², density 1000 kg/m³) is satisfactorily completed after 1 minute using

Question

Tests on a small scale tank 0.3 m diameter (Rushton impeller, diameter 0.1 m) have shown that a blending process between two miscible liquids (aqueous solutions, properties approximately the same as water, i.e. viscosity 1 mN s/m², density 1000 kg/m³) is satisfactorily completed after 1 minute using an impeller speed of 250 rev/min. It is decided to scale up the process to a tank of 2.5 m diameter using the criterion of constant tip-speed.
(a) What speed should be chosen for the larger impeller?
(b) What power will be required?
(c) What will be the blend time in the large tank?

Accepted Answer

a) In the small scale tank, the 0.1 m diameter impeller is rotated at 250 rev/min or:

[latex](250 / 60)=4.17 Hz[/latex]

The tip speed is then: [latex]\pi D N=(\pi imes 0.1 imes 4.17)=1.31 m / s[/latex]

If this is the same in the large scale tank, where [latex]D=(2.5 / 3)=0.83[/latex] m, then:

[latex]1.31=(\pi imes 0.83 imes N)[/latex]

from which the speed of rotation to the larger impeller,

[latex]N=\underline{\underline{0.346 Hz }}[/latex] or [latex]\underline{\underline{20.8 ext{rev / min}}}[/latex]

b) In the large scale tank: N = 0.346 Hz, D = 0.83 m, [latex] ho=1000 kg / m ^3[/latex] and [latex]\mu=1 imes 10^{-3} Ns / m ^2[/latex].

Thus, [latex]R e=D^2 N ho / \mu=\left(0.83^2 imes 0.346 imes 1000 ight) /\left(1 imes 10^{-3} ight)=238,360[/latex].

From Fig. 7.6, for a propeller mixer, the Power number, [latex]N_p=0.6[/latex].

Thus: [latex]0.6= P / ho N^3 D^5[/latex]

and: [latex]P =0.6 ho N^3 D^5=\left(0.6 imes 1000 imes 0.346^3 imes 0.83^5 ight)[/latex]

[latex]=\underline{\underline{9.8 W }}[/latex]

c) In the smaller tank: [latex]R e=D^2 N ho / \mu=\left(0.1^2 imes 4.17 imes 1000 ight) /\left(1 imes 10^{-3} ight)=41700[/latex]

In Equation 7.22, the dimensionless mixing time is:

[latex] heta_m=N t_m=k R e[/latex]

or for [latex]t_m=60 s :(4.17 imes 10)=k imes 41700[/latex]

and: k = 0.0060

Thus in the larger tank: [latex]N t_m=0.0060 Re[/latex]

or: [latex]0.346 t_m=(0.0060 imes 238,360)[/latex]

and: [latex]t_m=4140 s[/latex] or [latex]\underline{\underline{1.15 min }}[/latex]

Question 7.P.7: Tests on a small scale tank 0.3 m diameter (Rushton impeller...

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