5
$\begingroup$

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. Cylindrical Cavern around impeller

"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.

$\endgroup$
  • $\begingroup$ From where did you get the equation for Γ? Are you sure the numerator for the second part of the equation is 6? $\endgroup$ – Fred Apr 3 '16 at 11:50
  • $\begingroup$ Yes, it's from my university notes and the professor confirmed that it was right $\endgroup$ – MathsIsHard Apr 3 '16 at 16:00
0
$\begingroup$

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:

$$ \int\int\tau_y\cdot r\, dA $$

where the double integral is over the area where the force acts. Be careful to pick the appropriate differential area for a constant z surface in cylindrical coordinates. I did the calculation and the answer matches the one you state.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.