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}, S_{1}, S_{2},..., S_{n})

Where N_{P} = Power Number = P/(n^{3}D_{a}^{5}ρ)

N_{Re} = Reynolds number = nD_{a}^{2}ρ/m

N_{Fr} = Froude Number = n^{2}D_{a}/g

And S_{1}, S_{2},..., 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.

Calculations:

For the given problem,

For the case 1:

μ_{1} = 2 cP

n_{1} = 100

N_{Re,1} = 50000

N_{Re,1} = n_{1}D_{a}^{2}ρ/μ_{1} = 50000

Therefore, D_{a}^{2}ρ = 50000 x 2 / 100 = 1000

For the case 2:

μ_{2} = 4 cP

n_{2} = 200

N_{Re,2} = (D_{a}^{2}ρ)n_{2}/μ_{1 }= 1000 x 200/4 = 50000

Here N_{Re} is more than 10000. Therefore, N_{P} = constant = K

N_{P} = P/(n^{3}D_{a}^{5}ρ) = K

i.e., P = Kn^{3}D_{a}^{5}ρ

since D_{a} and ρ are same for the two cases,

we can group KD_{a}^{5}ρ as a constant, say M.

i.e., P = Mn^{3}

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

P_{2}/P_{1} = n_{2}^{3}/n_{1}^{3}

= 200^{3}/100^{3} = 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%.

Last Modified on: 04-Feb-2022

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