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?
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.
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.
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%.
Last Modified on: 11-Sep-2014
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