# Working Out Shaft Torque in a Stirred Tank with a Cylindrical Cavern Formation (using a non Newtonian fluid)

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. "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 D3/4 instead of D3/6 for the second term and I just can't work it out, if anyone can help I'd appreciate it.

• From where did you get the equation for Γ? Are you sure the numerator for the second part of the equation is 6? – Fred Apr 3 '16 at 11:50
• Yes, it's from my university notes and the professor confirmed that it was right – MathsIsHard Apr 3 '16 at 16:00

For your curved surfaces, the lever arm for the calculation of torque is a constant $D/2$. That is not true in the case of the cavern's faces. In that case the lever arm is a variable ($r$). To calculate the torque you would need to integrate the stress over the area times the lever arm. It should look something like this:
$$\int\int\tau_y\cdot r\, dA$$