The correct choice is (a) 2.47 kW
The explanation is: First calculate Reynold’s number. We have Da = 2 ft, n =\(\frac{90}{60}\) = 1.5 r/s, μ= 12 X 6.27 X 10^-4 = 8.06 X 10^-3 lb/ft-s, ρ=93.5 lb/ft^3 and g = 32.17 ft/s^2.
NRe = \(\frac{Da^2n\rho}{μ}\) = \(\frac{2^2X1.5X93.5}{8.06 X 10^{-3}}\) = 69,600
We know that at low Reynold’s number for both baffled and unbaffled tanks, the relationship is given by
Np =\(\frac{K_L}{N_{Re}}\) = \(\frac{403680}{69600}\) = 5.8
P = \(\frac{5.8 X 93.5 X 1.5^3 X 2^2}{32.17}\)=1821 ft-lb/s
The power requirement is 1821/550 = 2.47 kW (3.31 hp).