A2 — Spinning cone viscometer torque: 23-Chem-B1 Transport Phenomena, May 2014, Q2

Question 2: A2 — Spinning cone viscometer torque

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An inverted cone (apex down) of rim radius $R=7\ \text{cm}=0.07\ \text{m}$ and half-angle $\theta=45^\circ$ (measured from the vertical axis — Fig. A1 draws the V as a right angle; since $\sin45^\circ=\cos45^\circ$ the torque is the same whether the printed “cone angle” is read from the axis or from the horizontal) rotates inside a matching conical seat. The fluid occupies a thin uniform gap $h=1.5\ \text{mm}=1.5\times10^{-3}\ \text{m}$ measured perpendicular to the conical surface; $\mu=1.5\ \text{Pa}\cdot\text{s}$; rotational speed $\dot{n}=1500\ \text{rpm}$.

Find. The torque $T$ needed to sustain rotation.

Fig. A2: Inverted-cone viscometer. The cone (half-angle $45^\circ$) rotates in a matching seat separated by a uniform fluid film of thickness $h$ measured normal to the conical surface. The surface speed grows linearly with radius, $v=\Omega r$, so shear stress and torque are dominated by the outer radii.

Approach. Build the torque as an integral over the conical surface: at radius $r$ the wall speed is $\Omega r$, the linear film gives shear stress $\tau=\mu\Omega r/h$, and each annular strip contributes $dT=\tau\,r\,dA$ with $dA=2\pi r\,dr/\sin\theta$.

Convert the rotational speed. $\Omega=\dfrac{2\pi\dot{n}}{60}=\dfrac{2\pi(1500)}{60}=157.08\ \text{rad/s}.$
Shear stress from the linear film. A point on the cone at radius $r$ moves azimuthally at $v=\Omega r$; with a linear profile across the normal gap $h$ (stationary seat), $$\tau=\mu\frac{v}{h}=\frac{\mu\Omega r}{h}.$$
Surface-area element of the cone. Moving a slant distance $ds$ changes the radius by $dr=ds\,\sin\theta$, so an annular strip at radius $r$ has area $dA=2\pi r\,ds=\dfrac{2\pi r}{\sin\theta}\,dr.$
Elemental torque. The azimuthal shear force $\tau\,dA$ acts at lever arm $r$: $$dT=\tau\,r\,dA=\frac{\mu\Omega r}{h}\,r\,\frac{2\pi r}{\sin\theta}\,dr=\frac{2\pi\mu\Omega}{h\sin\theta}\,r^3\,dr.$$
Integrate over the cone. $$T=\frac{2\pi\mu\Omega}{h\sin\theta}\int_0^R r^3\,dr=\frac{\pi\mu\Omega R^4}{2\,h\sin\theta}.$$
Insert the numbers. With $\sin45^\circ=0.7071$ and $R^4=(0.07)^4=2.401\times10^{-5}\ \text{m}^4$, $$T=\frac{\pi(1.5)(157.08)(2.401\times10^{-5})}{2(1.5\times10^{-3})(0.7071)}=\boxed{8.38\ \text{N}\cdot\text{m}}.$$

Quantity	Value
Angular speed $\Omega$	$157.08$ rad/s
Torque relation	$T=\dfrac{\pi\mu\Omega R^4}{2h\sin\theta}$
Required torque $T$	$8.38\ \text{N}\cdot\text{m}$

Concept & Theory

Topic. Newtonian shear on a rotating conical surface (rotational viscometry).

Key relations. Linear (Couette) film $\tau=\mu\Omega r/h$; conical area element $dA=2\pi r\,dr/\sin\theta$; torque $T=\int \tau\,r\,dA=\pi\mu\Omega R^4/(2h\sin\theta)$.

Why this works. The thin-gap assumption makes the velocity profile locally linear, so the local shear stress is simply viscosity times the surface speed divided by the gap. Because both the speed ($\Omega r$) and the lever arm ($r$) grow with radius, the integrand carries an $r^3$ weighting — the torque is set overwhelmingly by the fluid near the rim ($R^4$ dependence).

Common pitfall. Using the vertical gap instead of the true normal gap, or forgetting the $1/\sin\theta$ that stretches a radial increment into slant area — both distort the geometric factor. Also, always convert rpm to rad/s before computing shear.

Source. Bird, Stewart & Lightfoot, Ch. 2–3 (Newtonian shear stress, rotating surfaces); Welty et al., Ch. 8 (viscous flow and rotational viscometers).

Question 2 of 6: A2 — Spinning cone viscometer torque

Question 2: A2 — Spinning cone viscometer torque