For a turbine agitator installed in a vertical tank speed is 1.5 rps. Diameter of tank is 1.8 m and the diameter of turbine is 0.6 m. The density of liquid is 1120 kg/m3 and viscosity is 120 Ns/m2. If the power Number is given by Np = 65/NRe. Calculate the power required for agitation.

</div> > Diameter of tank, D_T = 1.8 m > > Speed, N = 1.55 rps > > Diameter of turbine agitator, D_a = 0.6 m > > Density of liquid, ρ = 1120 kg/m³ > > Viscosity of liquid, μ = 120 Ns/m² = 120 kg/(m-s) > > ∴ Reynold’s number, N_p = ${\D_a^2 \Nρ}/{μ} = {(0.6)^2 \× 1.5 \× 1120}/{120} = 5.04$
> > Power number, N_p = ${65}/{N_{Re}} = {65}/{5.04}$ = 12.9 > > ∴ Power required, P = N_p ρ N³ D_a⁵ = 12.9 × 1120 × (1.5)³ × (0.6)⁵
> > = 3791 watt = 3.79 kW