6. A flat-blade turbine agitator with disk having six blade is installed in a ta

Question

6. A flat-blade turbine agitator with disk having six blade is installed in a tank. The tank diameter is 1.83 m, the turbine diameter is 0.61 m and the width is 0.122 mm. The tanlk contains four baffles, each having a density of 929 kg/m3 It is desired to design a small pilot unit with a vessel volume of 2.o L so that effects of various process variables on the system can be studued in the laboratory. The rate of the mass transfer appear to be important in this system. So the scale-doen should be on this basis. Design the new system specifying sizes, rpm and power.

Explanation / Answer

Tank Diameter = 1.83 m

Since, Tank length is not given.

Assuming L/D = 1.2

Therefore, Tank length = 1.2*1.83 = 2.19 m

Tank Volume = 3.14/4*d^2*l = 3.14/4*1.83^2*2.19 = 5.75 m3

Turbine Diameter/Tank diameter = .61/1.83 = 0.33

For Pilot reactor,

Volume = 2 litre = 2*10^-3 m3

L/D = 1.2

Therefore, 2*10^-3 = 3.14/4*D^2*1.2*D

D=0.12 m

L = 0.14 m

Now for mass transfer rate to be same, Power/Volume ratio in each reactor need to same

Power = Power no.*density*N^3*D^5

N = rotational speed of agitator, rpm

Power number is same for same mass

Power/Volume of tank =Power/Volume of Pilot reactor

=>(Ntank)^3*(tank Diameter)^5/(5.75) = (Npilot)^3*(pilot Diameter)^5/(2*10^-3)

=>Ntank/Npilot = (5.75/(2*10^-3)*(.12/1.83)^5)^.33 =0.15

Assuming Ntank = 100 rpm

=> NPilot=100/.15 = 666 rpm

for same mass transfer, Tip velociy of both the reactors are equal

Tip velocity = Agitator diameter*N/2

=>0.61*100 = 666*(Pilot agitator diameter)

=>Pilot agitator diameter = 0.09 m

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