CHEMICAL ENGINEERING

108 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

(c) For the large tank, from Equation 7.13:

PDkN^3 D^5

or:  17. 0 ð 103 Dkð 2. 903 ð 15

from which: kD 697

Thus, for the smaller tank, assuming power/unit volume is constant:

volume of fluidD/ 4  0. 32 ð 0. 3 D 0 .021 m^3

and: power supplied,PD 0. 021 ð 0. 8 ð 103 

D17 W

Thus, for the smaller tank:

17 D 697 N^3 ð 0. 15

and: ND 13 .5Hzor 807 rev/min.