Agitation - Power Required

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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.

NP = y (NRe, NFr, S1, S2,..., Sn)

Where NP = Power Number = P/(n3Da5r)

NRe = Reynolds number = nDa2r/m

NFr = Froude Number = n2Da/g

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

n = rotation per unit time of agitator

Da = dia of impeller

r = density of fluid

m = viscosity of fluid

If the shape factors are remaining constant, then

NP = y (NRe, NFr)

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

i.e.,

NP = y (NRe)

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

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

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

Calculations:

For the given problem,

For the case 1:

m1 = 2 cP

n1 = 100

NRe,1 = 50000

NRe,1 = n1Da2r/m1 = 50000

Therefore, Da2r = 50000 x 2 / 100 = 1000

For the case 2:

m2 = 4 cP

n2 = 200

NRe,2 = (Da2r) n2/m1 = 1000 x 200/4 = 50000

Here NRe is more than 10000. Therefore, NP = constant = K

NP = P/(n3Da5r) = K

i.e., P = Kn3Da5r

since Da and r are same for the two cases,

we can group KDa5r 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%.


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Last Modified on: 11-Sep-2014

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