Analysis of a Cone-and-Plate Rheometer The problem concerns the analysis of a cone-and-plate rheometer, an instrument developed and perfected in the 1950s and 1960s by Prof. Karl Weissenberg, for measuring the viscosity of liquids, and also known as the “Weissenberg rheogoniometer.”^4 A cross sectio

Question

Analysis of a Cone-and-Plate Rheometer

The problem concerns the analysis of a cone-and-plate rheometer, an instrument developed and perfected in the 1950s and 1960s by Prof. Karl Weissenberg, for measuring the viscosity of liquids, and also known as the "Weissenberg rheogoniometer." [latex]{ }^{4}[/latex] A cross section of the essential features is shown in Fig. E6.9, in which the liquid sample is held by surface tension in the narrow opening between a rotating lower circular plate, of radius [latex]R[/latex], and an upper cone, making an angle of [latex]\beta[/latex] with the vertical axis. The plate is rotated steadily in the [latex]\phi[/latex] direction with an angular velocity [latex]\omega[/latex], causing the liquid in the gap to move in concentric circles with a velocity [latex]v_{\phi}[/latex]. (In practice, the tip of the cone is slightly truncated to avoid friction with the plate.) Observe that the flow is of the Couette type.

The top of the upper shaft --- which acts like a torsion bar---is clamped rigidly. However, viscous friction will twist the cone and the lower portions of the upper shaft very slightly; the amount of motion can be detected by a light arm at the extremity of which is a transducer, consisting of a small piece of steel, attached to the arm, and surrounded by a coil of wire; by monitoring the inductance of the coil, the small angle of twist can be obtained; a knowledge of the elastic properties of the shaft then enables the restraining torque [latex]T[/latex] to be obtained. From the analysis given below, it is then possible to deduce the viscosity of the sample. The instrument is so sensitive that if no liquid is present, it is capable of determining the viscosity of the air in the gap!

[latex]{ }^{4}[/latex] Professor Weissenberg (1893-1976) once related to the author that he (Prof. Weissenberg) was attending an instrument trade show in London. There, the rheogoniometer was being demonstrated by a young salesman who was unaware of Prof. Weissenberg's identity. Upon inquiring how the instrument worked, the salesman replied: "I'm sorry, sir, but it's quite complicated, and I don't think you will be able to understand it."

The problem is best solved using spherical coordinates, because the surfaces of the cone and plate are then described by constant values of the angle [latex] heta[/latex], namely [latex]\beta[/latex] and [latex]\pi / 2[/latex], respectively.

Accepted Answer

Assumptions and the continuity equation. The following realistic assumptions are made:

There is only one nonzero velocity component, namely that in the [latex]\phi[/latex] direction, [latex]v_{\phi}[/latex]. Thus, [latex]v_{r}=v_{ heta}=0[/latex].
Gravity acts vertically downward, so that [latex]g_{\phi}=0[/latex].
We do not need to know how the pressure varies in the liquid. Therefore, the [latex]r[/latex] and [latex] heta[/latex] momentum balances, which would supply this information, are not required.

The analysis starts once more from the continuity equation, [latex](5.50)[/latex] :

[latex]\frac{\partial ho}{\partial t}+\frac{1}{r^{2}} \frac{\partial\left( ho r^{2} v_{r} ight)}{\partial r}+\frac{1}{r \sin heta} \frac{\partial\left( ho v_{ heta} \sin heta ight)}{\partial heta}+\frac{1}{r \sin heta} \frac{\partial\left( ho v_{\phi} ight)}{\partial \phi}=0,[/latex] (5.50)

which, for constant density and [latex]v_{r}=v_{ heta}=0[/latex] reduces to:

[latex]\frac{\partial v_{\phi}}{\partial \phi}=0,[/latex] (E6.9.1)

verifying that [latex]v_{\phi}[/latex] is independent of the angular location [latex]\phi[/latex], so we are correct in examining just one representative cross section, as shown in Fig. E6.9.

Momentum balances. There are again three momentum balances, one for each of the [latex]r, heta[/latex], and [latex]\phi[/latex] directions. From the third assumption above, the first two such balances are of no significant interest, leaving, from Eqn. (5.77), just that in the [latex]\phi[/latex] direction:

In terms of velocities, for a Newtonian fluid with constant [latex] ho[/latex] and [latex]\mu[/latex] :

[latex]\begin{aligned} ho\left(\frac{\partial v_{r}}{\partial t} ight.&\left.+v_{r} \frac{\partial v_{r}}{\partial r}+\frac{v_{ heta}}{r} \frac{\partial v_{r}}{\partial heta}+\frac{v_{\phi}}{r \sin heta} \frac{\partial v_{r}}{\partial \phi}-\frac{v_{ heta}^{2}+v_{\phi}^{2}}{r} ight) \=&-\frac{\partial p}{\partial r}+\mu\left[\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial v_{r}}{\partial r} ight)+\frac{1}{r^{2} \sin heta} \frac{\partial}{\partial heta}\left(\sin heta \frac{\partial v_{r}}{\partial heta} ight) ight.\&\left.+\frac{1}{r^{2} \sin ^{2} heta} \frac{\partial^{2} v_{r}}{\partial \phi^{2}}-\frac{2 v_{r}}{r^{2}}-\frac{2}{r^{2}} \frac{\partial v_{ heta}}{\partial heta}-\frac{2 v_{ heta} \cot heta}{r^{2}}-\frac{2}{r^{2} \sin heta} \frac{\partial v_{\phi}}{\partial \phi} ight]+ ho g_{r},\ ho\left(\frac{\partial v_{ heta}}{\partial t} ight.&\left.+v_{r} \frac{\partial v_{ heta}}{\partial r}+\frac{v_{ heta}}{r} \frac{\partial v_{ heta}}{\partial heta}+\frac{v_{\phi}}{r \sin heta} \frac{\partial v_{ heta}}{\partial \phi}+\frac{v_{r} v_{ heta}}{r}-\frac{v_{\phi}^{2} \cot heta}{r} ight) \=&-\frac{1}{r} \frac{\partial p}{\partial heta}+\mu\left[\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial v_{ heta}}{\partial r} ight)+\frac{1}{r^{2} \sin heta} \frac{\partial}{\partial heta}\left(\sin heta \frac{\partial v_{ heta}}{\partial heta} ight) ight.\ &\left.+\frac{1}{r^{2} \sin ^{2} heta} \frac{\partial^{2} v_{ heta}}{\partial \phi^{2}}+\frac{2}{r^{2}} \frac{\partial v_{r}}{\partial heta}-\frac{v_{ heta}}{r^{2} \sin ^{2} heta}-\frac{2 \cos heta}{r^{2} \sin ^{2} heta} \frac{\partial v_{\phi}}{\partial \phi} ight]+ ho g_{ heta}, \ ho\left(\frac{\partial v_{\phi}}{\partial t} ight.&\left.+v_{r} \frac{\partial v_{\phi}}{\partial r}+\frac{v_{ heta}}{r} \frac{\partial v_{\phi}}{\partial heta}+\frac{v_{\phi}}{r \sin heta} \frac{\partial v_{\phi}}{\partial \phi}+\frac{v_{\phi} v_{r}}{r}+\frac{v_{ heta} v_{\phi} \cot heta}{r} ight) \=&-\frac{1}{r \sin heta} \frac{\partial p}{\partial \phi}+\mu\left[\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial v_{\phi}}{\partial r} ight)+\frac{1}{r^{2} \sin heta} \frac{\partial}{\partial heta}\left(\sin heta \frac{\partial v_{\phi}}{\partial heta} ight) ight.\ &\left.+\frac{1}{r^{2} \sin ^{2} heta} \frac{\partial^{2} v_{\phi}}{\partial \phi^{2}}-\frac{v_{\phi}}{r^{2} \sin ^{2} heta}+\frac{2}{r^{2} \sin heta} \frac{\partial v_{r}}{\partial \phi}+\frac{2 \cos heta}{r^{2} \sin ^{2} heta} \frac{\partial v_{ heta}}{\partial \phi} ight]+ ho g_{\phi} . (5.77)\end{aligned}[/latex]

[latex]\begin{aligned} ho\left(\frac{\partial v_{\phi}}{\partial t} ight.&\left.+v_{r} \frac{\partial v_{\phi}}{\partial r}+\frac{v_{ heta}}{r} \frac{\partial v_{\phi}}{\partial heta}+\frac{v_{\phi}}{r \sin heta} \frac{\partial v_{\phi}}{\partial \phi}+\frac{v_{\phi} v_{r}}{r}+\frac{v_{ heta} v_{\phi} \cot heta}{r} ight) \=&-\frac{1}{r \sin heta} \frac{\partial p}{\partial \phi}+\mu\left[\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial v_{\phi}}{\partial r} ight)+\frac{1}{r^{2} \sin heta} \frac{\partial}{\partial heta}\left(\sin heta \frac{\partial v_{\phi}}{\partial heta} ight) ight.\&\left.+\frac{1}{r^{2} \sin ^{2} heta} \frac{\partial^{2} v_{\phi}}{\partial \phi^{2}}-\frac{v_{\phi}}{r^{2} \sin ^{2} heta}+\frac{2}{r^{2} \sin heta} \frac{\partial v_{r}}{\partial \phi}+\frac{2 \cos heta}{r^{2} \sin ^{2} heta} \frac{\partial v_{ heta}}{\partial \phi} ight]+ ho g_{\phi} . (E6.9.2) \end{aligned}[/latex]

With [latex]v_{r}=v_{ heta}=0[/latex] (assumption 1[latex]), \partial v_{\phi} / \partial \phi=0[/latex] [from Eqn. (E6.9.1)], and [latex]g_{\phi}=0[/latex] (assumption 2), the momentum balance simplifies to:

[latex]\frac{\partial}{\partial r}\left(r^{2} \frac{\partial v_{\phi}}{\partial r} ight)+\frac{1}{\sin heta} \frac{\partial}{\partial heta}\left(\sin heta \frac{\partial v_{\phi}}{\partial heta} ight)-\frac{v_{\phi}}{\sin ^{2} heta}=0.[/latex] (E6.9.3)

Determination of the velocity profile. First, seek an expression for the velocity in the [latex]\phi[/latex] direction, which is expected to be proportional to both the distance [latex]r[/latex] from the origin and the angular velocity [latex]\omega[/latex] of the lower plate. However, its variation with the coordinate [latex] heta[/latex] is something that has to be discovered. Therefore, postulate a solution of the form:

[latex]v_{\phi}=r \omega f( heta),[/latex] (E6.9.4)

in which the function [latex]f( heta)[/latex] is to be determined. Substitution of [latex]v_{\phi}[/latex] from Eqn. (E6.9.4) into Eqn. (E6.9.3), and using [latex]f^{\prime}[/latex] and [latex]f^{\prime \prime}[/latex] to denote the first and second total derivatives of [latex]f[/latex] with respect to [latex] heta[/latex], gives:

[latex]\begin{aligned}f^{\prime \prime} &+f^{\prime} \cot heta+f\left(1-\cot ^{2} heta ight) \& \equiv \frac{1}{\sin ^{2} heta} \frac{d}{d heta}\left(f^{\prime} \sin ^{2} heta-f \sin heta \cos heta ight) \&=\frac{1}{\sin ^{2} heta} \frac{d}{d heta}\left[\sin ^{3} heta \frac{d}{d heta}\left(\frac{f}{\sin heta} ight) ight]=0. (E6.9.5)\end{aligned}[/latex]

The reader is encouraged, as always, to check the missing algebraic and trigonometric steps, although they are rather tricky here! [latex]^{5}[/latex]

By multiplying Eqn. (E6.9.5) through by [latex]\sin ^{2} heta[/latex], it follows after integration that the quantity in brackets is a constant, here represented as [latex]-2 c_{1}[/latex], the reason for the "-2" being that it will cancel with a similar factor later on:

[latex]\sin ^{3} heta \frac{d}{d heta}\left(\frac{f}{\sin heta} ight)=-2 c_{1}.[/latex] (E6.9.6)

Separation of variables and indefinite integration (without specified limits) yields:

[latex]\int d\left(\frac{f}{\sin heta} ight)=\frac{f}{\sin heta}=-2 c_{1} \int \frac{d heta}{\sin ^{3} heta}.[/latex] (E6.9.7)

[latex]^{5}[/latex] Although our approach is significantly different from that given on page 98 et seq. of R.B. Bird, W.E. Stewart, and E.N. Lightfoot, Transport Phenomena, Wiley & Sons, New York, 1960, we are indebted to these authors for the helpful hint they gave in establishing the equivalency expressed in our Eqn. (E6.9.5).

To proceed further, we need the following two standard indefinite integrals and one trigonometric identity [latex]{ }^{6}[/latex] :

[latex]\begin{aligned}\int \frac{d heta}{\sin ^{3} heta} &=-\frac{1}{2} \frac{\cot heta}{\sin heta}+\frac{1}{2} \int \frac{d heta}{\sin heta}, (E6.9.8) \ \int \frac{d heta}{\sin heta} &=\ln \left( an \frac{ heta}{2} ight)=\frac{1}{2} \ln \left( an ^{2} \frac{ heta}{2} ight), (E6.9.9) \ an ^{2} \frac{ heta}{2} &=\frac{1-\cos heta}{1+\cos heta} . (E6.9.10)\end{aligned}[/latex]

Armed with these, Eqn. (E6.9.7) leads to the following expression for [latex]f[/latex] :

[latex]f=c_{1}\left[\cot heta+\frac{1}{2}\left(\ln \frac{1+\cos heta}{1-\cos heta} ight) \sin heta ight]+c_{2},[/latex] (E6.9.11)

in which [latex]c_{2}[/latex] is a second constant of integration.

Implementation of the boundary conditions. The constants [latex]c_{1}[/latex] and [latex]c_{2}[/latex] are found by imposing the two boundary conditions:

At the lower plate, where [latex] heta=\pi / 2[/latex] and the expression in parentheses in Eqn. (E6.9.11) is zero, so that [latex]f=c_{2}[/latex], the velocity is simply the radius times the angular velocity:

[latex]v_{\phi} \equiv r \omega f=r \omega c_{2}=r \omega, \quad ext { or } \quad c_{2}=1.[/latex] (E6.9.12)

At the surface of the cone, where [latex] heta=\beta[/latex], the velocity [latex]v_{\phi}=r \omega f[/latex] is zero. Hence [latex]f=0[/latex], and Eqn. (E6.9.11) leads to:

[latex]c_{1}=-c_{2} g(\beta)=-g(\beta)[/latex], where [latex]\frac{1}{g(\beta)}=\cot \beta+\frac{1}{2}\left(\ln \frac{1+\cos \beta}{1-\cos \beta} ight) \sin \beta .[/latex] (E6.9.13)

Substitution of these expressions for [latex]c_{1}[/latex] and [latex]c_{2}[/latex] into Eqn. (E6.9.11), and noting that [latex]v_{\phi}=r \omega f[/latex], gives the final (!) expression for the velocity:

[latex]v_{\phi}=r \omega\left[1-\frac{\cot heta+\frac{1}{2}\left(\ln \frac{1+\cos heta}{1-\cos heta} ight) \sin heta}{\cot \beta+\frac{1}{2}\left(\ln \frac{1+\cos \beta}{1-\cos \beta} ight) \sin \beta} ight].[/latex] (E6.9.14)

As a partial check on the result, note that Eqn. (E6.9.14) reduces to [latex]v_{\phi}=r \omega[/latex] when [latex] heta=\pi / 2[/latex] and to [latex]v_{\phi}=0[/latex] when [latex] heta=\beta[/latex].

[latex]^{6}[/latex] From pages 87 (integrals) and 72 (trigonometric identity) of J.H. Perry, ed., Chemical Engineers'Handbook, 3rd ed., McGraw-Hill, New York, [latex]1950 .[/latex]

Shear stress and torque. Recall that the primary goal of this investigation is to determine the torque [latex]T[/latex] needed to hold the cone stationary. The relevant shear stress exerted by the liquid on the surface of the cone is [latex] au_{ heta \phi}[/latex]---that exerted on the under surface of the cone (of constant first subscript, [latex] heta=\beta[/latex] ) in the positive [latex]\phi[/latex] direction (refer again to Fig. [latex]5.12[/latex] for the sign convention and notation for stresses). Since this direction is the same as that of the rotation of the lower plate, we expect that [latex] au_{ heta \phi}[/latex] will prove to be positive, thus indicating that the liquid is trying to turn the cone in the same direction in which the lower plate is rotated.

From the second of Eqn. (5.65), the relation for this shear stress is:

[latex]\begin{aligned} au_{ heta \phi} = au_{\phi heta}=\mu\left(\frac{\sin heta}{r} \frac{\partial}{\partial heta}\left(\frac{v_{\phi}}{\sin heta} ight)+\frac{1}{r \sin heta} \frac{\partial v_{ heta}}{\partial \phi} ight), (5.65) \end{aligned}[/latex]

[latex] au_{ heta \phi}=\mu\left[\frac{\sin heta}{r} \frac{\partial}{\partial heta}\left(\frac{v_{\phi}}{\sin heta} ight)+\frac{1}{r \sin heta} \frac{\partial v_{ heta}}{\partial \phi} ight] .[/latex] (E6.9.15)

Since [latex]v_{ heta}=0[/latex], and recalling Eqns. (E6.9.4), (E6.9.6), and (E6.9.13), the shear stress on the cone becomes:

[latex]\left( au_{ heta \phi} ight)_{ heta=\beta}=\mu\left[\frac{\sin heta}{r} \frac{\partial}{\partial heta}\left(\frac{r \omega f}{\sin heta} ight) ight]_{ heta=\beta}=-\left(\frac{2 c_{1} \omega \mu}{\sin ^{2} heta} ight)_{ heta=\beta}=\frac{2 \omega \mu g(\beta)}{\sin ^{2} \beta} .[/latex] (E6.9.16)

One importance of this result is that it is independent of [latex]r[/latex], giving a constant stress and strain throughout the liquid, a significant simplification when deciphering the experimental results for non-Newtonian fluids (see Chapter 11). In effect, the increased velocity differences between the plate and cone at the greater values of [latex]r[/latex] are offset in exact proportion by the larger distances separating them.

The torque exerted by the liquid on the cone (in the positive [latex]\phi[/latex] direction) is obtained as follows. The surface area of the cone between radii [latex]r[/latex] and [latex]r+d r[/latex] is [latex]2 \pi r \sin \beta d r[/latex] and is located at a lever arm of [latex]r \sin \beta[/latex] from the axis of symmetry. Multiplication by the shear stress and integration gives:

[latex]T=\int_{0}^{R / \sin \beta} \underbrace{(2 \pi r \sin \beta d r)}_{ ext {Area }} \underbrace{r \sin \beta}_{\begin{array}{c} ext { Lever } \ ext { arm }\end{array}} \underbrace{\left( au_{ heta \phi} ight)_{ heta=\beta}}_{ ext {Stress }}[/latex], (E6.9.17)

Substitution of [latex]\left( au_{ heta \phi} ight)_{ heta=\beta}[/latex] from Eqn. (E6.9.16) and integration gives the torque as:

[latex]T=\frac{4 \pi \omega \mu g(\beta) R^{3}}{3 \sin ^{3} \beta} .[/latex] (E6.9.18)

Problem [latex]6.15[/latex] reaches essentially the same conclusion much more quickly for the case of a small angle between the cone and the plate. The torque for holding the cone stationary has the same value but is, of course, in the negative [latex]\phi[/latex] direction.

Since [latex]R[/latex] and [latex]g(\beta)[/latex] can be determined from the radius and the angle [latex]\beta[/latex] of the cone in conjunction with Eqn. (E6.9.13), the viscosity [latex]\mu[/latex] of the liquid can finally be determined.

Question 6.9: Analysis of a Cone-and-Plate Rheometer The problem concerns ...

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