Working Out Shaft Torque in a Stirred Tank with a Cylindrical Shape

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SUMMARY

The discussion focuses on calculating shaft torque in a stirred tank containing yield stress non-Newtonian (viscoplastic) fluids. The total torque is defined by the equation Γ = (π/2)τYHC DC² + (π/6)τYDC³, where the first term is derived from the curved area of the impeller, while the second term accounts for the flat surface area. The contributor successfully computes the first term but struggles with the second term, mistakenly deriving D²/4 instead of D²/6. The solution involves considering the flat surface as concentric ring zones to accurately sum individual torque contributions.

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  • Understanding of non-Newtonian fluid dynamics
  • Familiarity with torque calculations in mechanical systems
  • Knowledge of shear stress and yield stress concepts
  • Experience with cylindrical geometry in engineering applications
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  • Study the principles of non-Newtonian fluid behavior in stirred tanks
  • Learn about torque calculations involving concentric ring zones
  • Explore the application of shear stress in mechanical engineering
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Mechanical engineers, chemical engineers, and researchers working with non-Newtonian fluids in stirred tank applications will benefit from this discussion.

Tom Hardy
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Here is some background to the problem (in a stirred tank):

"With yield stress non-Newtonian (viscoplastic) fluids, it is possible to generate an agitated volume around the impeller, defined as a cavern, surrounded by a stagnant region where the shear stress is insufficient to overcome the apparent yield stress of the fluid."

Sometimes you can get a cylindrical cavern around the impeller, see the below image.

YeL2m.png


"By performing a force balance between the applied torque, Γ and the shear stress acting on the surface of a cylinder, we can define the boundary by setting the shear stress equal to the yield stress τ = τY. The total torque is given by:
$$\Gamma = \frac{\pi}{2} \tau_{y}H_{C}D_{C}^2+\frac{\pi}{6}\tau_{y}D_{C}^3$$

I just can't get the second term. The first term I can get by doing:
$$\Gamma_{1}=\tau_y \cdot Area_{Curved} \cdot \frac{D}{2} = \pi \cdot \frac{D^2}{2} \cdot H_{c} \cdot \tau_{y}$$

This gets me the first term...but the second term I just can't get, this is what I'm doing:

$$\Gamma_{2}=\tau_{y} \cdot Area_{Faces} \cdot \frac{D}{2} =\tau_{y} \cdot 2 \pi \cdot \frac{D^2}{4} \cdot \frac{D}{2} = \tau_{y} \cdot \pi \cdot \frac{D^3}{4} $$

Argh, so I'm getting D^2/4 instead of D^2/6 for the second term and I just can't work it out, if anyone can help I'd appreciate it. Thanks.
 
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Consider the flat surface as a series of concentric ring zones . Each zone has a different shear area and a different radial distance from axis . Total torque is worked out by summing the individual torque contributions from each zone .
 
Last edited:

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