Question 7.P.7: Tests on a small scale tank 0.3 m diameter (Rushton impeller...
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?
a) In the small scale tank, the 0.1 m diameter impeller is rotated at 250 rev/min or:
(250 / 60)=4.17 Hz
The tip speed is then: \pi D N=(\pi \times 0.1 \times 4.17)=1.31 m / s
If this is the same in the large scale tank, where D=(2.5 / 3)=0.83 m, then:
1.31=(\pi \times 0.83 \times N)
from which the speed of rotation to the larger impeller,
N=\underline{\underline{0.346 Hz }} or \underline{\underline{20.8 \text{rev / min}}}
b) In the large scale tank: N = 0.346 Hz, D = 0.83 m, \rho=1000 kg / m ^3 and \mu=1 \times 10^{-3} Ns / m ^2.
Thus, R e=D^2 N \rho / \mu=\left(0.83^2 \times 0.346 \times 1000\right) /\left(1 \times 10^{-3}\right)=238,360.
From Fig. 7.6, for a propeller mixer, the Power number, N_p=0.6.
Thus: 0.6= P / \rho N^3 D^5
and: P =0.6 \rho N^3 D^5=\left(0.6 \times 1000 \times 0.346^3 \times 0.83^5\right)
=\underline{\underline{9.8 W }}
c) In the smaller tank: R e=D^2 N \rho / \mu=\left(0.1^2 \times 4.17 \times 1000\right) /\left(1 \times 10^{-3}\right)=41700
In Equation 7.22, the dimensionless mixing time is:
\theta_m=N t_m=k R e
or for t_m=60 s :(4.17 \times 10)=k \times 41700
and: k = 0.0060
Thus in the larger tank: N t_m=0.0060 Re
or: 0.346 t_m=(0.0060 \times 238,360)
and: t_m=4140 s or \underline{\underline{1.15 min }}