Agitation - Power Required

An agitated baffle vessel is being used to prepare a uniform solution of viscosity 2 cP, running the agitator at 100 rpm, so as to obtain a Reynolds number of 50,000. If the contents of the vessel are replaced by a solution of viscosity 4 cP, and the agitator rpm is increased to 200, by how much will the power requirement change?

Theory:

For agitated vessel, the following dimensional relationship is applicable.

N_P = ψ (N_Re, N_Fr, S₁, S₂,..., S_n)

Where N_P = Power Number = P/(n³D_a⁵ρ)

N_Re = Reynolds number = nD_a²ρ/m

N_Fr = Froude Number = n²D_a/g

And S₁, S₂,..., S_n are shape factors.

n = rotation per unit time of agitator

D_a = dia of impeller

ρ = density of fluid

μ = viscosity of fluid

If the shape factors are remaining constant, then

N_P = ψ (N_Re, N_Fr)

For baffled vessel, N_P is a function of only N_Re provided the shape factors are remaining at constant value.

i.e.,

N_P = ψ (N_Re)

For various impeller configurations, and system geometry, N_P vs. N_Re chart is available to estimate the power required.

From the charts available, it could be seen that, for NRe > 10000, N_P is independent of N_Re, and remains at a constant value.

i.e., N_P = constant, (for N_Re > 10000), and viscosity is not a factor.

For the given problem,

For the case 1:

μ₁ = 2 cP

n₁ = 100

N_Re,1 = 50000

N_Re,1 = n₁D_a²ρ/μ₁ = 50000

Therefore, D_a²ρ = 50000 x 2 / 100 = 1000

For the case 2:

μ₂ = 4 cP

n₂ = 200

N_Re,2 = (D_a²ρ)n₂/μ₁ = 1000 x 200/4 = 50000

Here N_Re is more than 10000. Therefore, N_P = constant = K

N_P = P/(n³D_a⁵ρ) = K

i.e., P = Kn³D_a⁵ρ

since D_a and ρ are same for the two cases,

we can group KD_a⁵ρ as a constant, say M.

i.e., P = Mn³

The ratio of power required for case 2 to 1 is,

P₂/P₁ = n₂³/n₁³

= 200³/100³ = 8.

The power required for the second case will be 8 times that of the first case. In other words, the power requirement will rise by, 100 x (8-1)/1 = 700%.