Question

A flat blade turbine having six blades is centrally installed in a vertical tank. The tank is 6 ft (1.83 m) in diameter; the turbine is 2 ft. (0.61 m) in diameter and is positioned 2 ft (0.61 m) from the bottom of the tank. The turbine blades are 5 in. (127 mm) wide. The tank is filled to a depth of 6 ft (1.83 m) with a solution of 50 percent caustic soda, at 150°F (65.6°C), which has a viscosity of 12 cP and a density of 93.5 lb/ft^3 (1498 kg/m^3). The turbine is operated at 90 r/min. the tank is baffled. What power will be required to operate the mixer? (Given KL=403680).

(a) 2.47 kW

(b) 6.05 kW

(c) 3.21 kW

(d) 4.00 kW

The question was asked during an online exam.

Question is taken from Power for Agitation in section Agitation and Mixing of Liquids of Food Processing Unit Operations

Answer 1

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).