Measurement of Flow and Viscoelastic Properties

Abstract

Techniques for measuring rheological properties of fluid foods are discussed in this chapter. First, the traditional measuring systems are considered, followed by microrheological techniques using micrometer-scale probes.

Appendices

Appendix 1

Analysis of Flow in a Concentric Cylinder Geometry

One can derive applicable equations for shear rate starting from either the general equations of conservation of mass (continuity) and conservation of momentum (motion) or by conducting balances of mass or momentum on a differential shell. Numerous examples of using the conservation equations or shell balances can be found in Bird et al. (1960) and other texts on transport phenomena. For the concentric cylinder geometry shown in Fig. 3A.1, it will be assumed that the inner cylinder is rotating and the outer cylinder is stationary. Interested readers can derive the applicable equations for the case of outer cylinder rotating using appropriate boundary conditions.

First, from a simple force balance, the total torque (M) on the inner cylinder is

$${{\sigma }_{r}}\theta \,\,\times \,2\pi {{r}_{i}}h\,\,\times \,{{r}_{i}}=M$$

(3A.1)

In terms of torque per unit height,$\text{T}=\frac{M}{h}$ Eq. 3A.1 becomes

$${{\sigma }_{\text{r}}}\theta =\frac{T}{2\pi r_{\text{i}}^{2}}$$

(3A.2)

To describe the flow between the inner and outer cylinders, we use cylindrical coordinates (r, θ, z) and note that the fluid moves in a circular motion; the velocities in the radial and the axial directions are zero: v _θ = rΩ, v _r = 0, v _z = 0, and due to symmetry ∂/∂θ = 0.

Also, at steady state ∂ρ/∂t = 0. The equation of continuity in cylindrical coordinates is

$$\frac{\partial \rho }{\partial t}+\frac{1}{r}\frac{\partial }{\partial r}(\rho r{{v}_{r}})+\frac{1}{r}\frac{\partial }{\partial \theta }(\rho {{v}_{\theta }})+\frac{\partial }{\partial z}(\rho {{v}_{z}})=0$$

All terms in the above equation are zero.

The r-component of the equation of motion is

$$\begin{aligned} & \rho \left( \frac{\partial {{v}_{r}}}{\partial t}+{{v}_{r}}\frac{\partial {{v}_{r}}}{\partial r}+\frac{{{v}_{\theta }}}{r}\frac{\partial {{v}_{r}}}{\partial \theta }\frac{v_{\theta }^{2}}{r}+{{v}_{z}}\frac{\partial {{v}_{r}}}{\partial z} \right) \\ & =-\frac{\partial p}{{{\partial }_{r}}}+\left( \frac{1}{r}\frac{\partial }{{{\partial }_{r}}}(r{{\sigma }_{rr}})+\frac{1}{r}\frac{\partial {{\sigma }_{r\theta }}}{\partial \theta }\frac{\sigma \theta \theta }{r}+\frac{\partial {{\sigma }_{rz}}}{\partial z} \right)+\rho {{g}_{r}} \\ \end{aligned}$$

The r-component of the equation of motion reduces to

$$-\rho \frac{v_{\theta }^{2}}{r}=\frac{1}{r}\frac{\partial }{{{\partial }_{r}}}(r{{\sigma }_{rr}})-\frac{\sigma \theta \theta }{r}$$

(3A.3)

Equation 3A.3 deals with the normal stresses σ_rr and σ_θθ. For the purpose of illustration, we consider the θ-component of the equation of motion in detail and note in particular that v_θ = ƒ (r)

θ-component

$$\begin{aligned} & \rho \left( \frac{\partial v\theta }{\partial t}+{{v}_{r}}\frac{\partial v\theta }{\partial r}+\frac{v\theta }{r}\frac{\partial v\theta }{\partial \theta }+\frac{{{v}_{r}}v\theta }{r}+{{v}_{z}}\frac{\partial v\theta }{\partial z} \right) \\ & =-\frac{1}{r}\frac{\partial p}{\partial \theta }+\left( \frac{1}{{{r}^{2}}}\frac{\partial }{\partial r}\left( {{r}^{2}}{{\sigma }_{r}}\theta \right)+\frac{1}{r}\frac{\partial {{\sigma }_{\theta \theta }}}{\partial \theta }+\frac{\partial {{\sigma }_{\theta }}_{z}}{\partial z} \right)+\rho {{g}_{\theta }} \\ \end{aligned}$$

(3A.4)

Under steady flow condition ∂v_θ /∂t = 0, v_r = 0, ∂v_θ /∂θ = 0, and ∂v_θ /∂z = 0, the LHS of Eq. 3A.3 reduces to zero. On the right-hand side, ∂p/∂θ = 0 because there is no pressure gradient in the θ-direction and ρg_θ = 0 because there is no θ-component of gravity. The ∂σ_θθ /∂θ and ∂σ_θz /∂z = 0 because there are no shear gradients in the θ and z directions. There is no pressure gradient in the θ direction. We, thus, have for the θ-component

$$\frac{\partial \left( {{r}^{2}}{{\sigma }_{r}}\theta \right)}{\partial r}=0$$

(3A.5)

The z-component reduces to

$$0=-\frac{\partial P}{\partial z}+\rho {{g}_{z}}$$

(3A.6)

Equation 3A.6 simply describes the hydrostatic pressure in the gap between the two cylinders. Differentiation of Eq. 3A.5, which contains the shear stress of interest, σ_rθ , results in

$${{r}^{\text{2}}}\text{d}{{\sigma }_{r}}_{\theta }+2r{{\sigma }_{r}}_{\theta }\text{d}r=0$$

(3A.7)

Equation 3A.7 can be rearranged to:

$$\text{d}r=-\frac{r}{2}\frac{\text{d}{{\sigma }_{r}}_{\theta }}{{{\sigma }_{r}}_{\theta }}$$

(3A.8)

The boundary condition at the inner cylinder rotating at an angular velocity Ω is v _θ= r _iΩ at r = r _i and at the stationary outer cylinder is v_θ = 0 at r = r_o; both expressions are based again on the assumption of no slip condition at a solid–fluid interface. The velocity distribution is obtained from

$$ \frac{{{v_\theta }}}{r} = \int\limits_{{r_i}}^r {\frac{{{\rm{d}}({v_\theta }/r)}}{{{\rm{d}}r}}} {\rm{d}}r $$

(3A.9)

Substituting from Eq. 3.8 for dr in Eq. 3A.9 and using the appropriate integration limits

$$ \frac{{{v_\theta }}}{r} = \int\limits_{{\sigma _i}}^{{\sigma _{r\theta }}} {} \left( {\frac{{r{\rm{d}}({v_\theta }/r}}{{{\rm{d}}r}}} \right)\frac{{{\rm{d}}{\sigma _{r\theta }}}}{{{\rm{2}}{\sigma _{r\theta }}}} = \int\limits_{{\sigma _{_{r\theta }}}}^{{\sigma _i}} {} \left( {\frac{{r{\rm{d}}({v_\theta }/r)}}{{{\rm{d}}r}}} \right)\frac{{{\rm{d}}{\sigma _{r\theta }}}}{{{\rm{2}}{\sigma _{r\theta }}}} $$

(3A.10)

By using the Leibnitz rule and noting to differentiate with respect to σ _rθ , and with the boundary conditions: σ _rθ  = σ_i at r = r _i and σ _rθ  = σ_o at r = r _o

$$ \frac{{{\rm{d}}({v_\theta }/r)}}{{{{\rm{d}}_{\sigma {\rm{i}}}}}} = \frac{1}{{{{\rm{2}}_{\sigma {\rm{i}}}}}}\left[ {{{\left( {r\frac{{{\rm{d}}({v_\theta }/r}}{{{\rm{d}}r}}} \right)}_{\rm{i}}} - {{\left( {r\frac{{{\rm{d}}({v_\theta }/r)}}{{{\rm{d}}r}}} \right)}_{\rm{o}}}} \right] $$

(3A.11)

Noting that Ω = (v _θ /r) and $\dot \gamma$ = (r(d(v _θ /r)/dr)), Eq. 3A.9 can be written as

$$ {\rm{2}}{\sigma _{\rm{i}}}\frac{{{\rm{d}}{\Omega _{\rm{i}}}}}{{{\rm{d}}{\sigma _{\rm{i}}}}} = .{\gamma _{\rm{i}}}--.{\gamma _{\rm{o}}} $$

(3A.12)

Therefore, while it is relatively easy to calculate the shear stress at the surface of the rotating cylinder from Eq. 3A.2, one can only derive an expression for the difference in shear rates at the surfaces of the inner and outer cylinders from the basic equations of flow. Additional work is required to calculate the corresponding shear rate ${\dot \gamma _1}$ and there have been several approaches to determine it. One approach has been to apply infinite series solution to the differential equation in 3A.12.

A popular method of calculating the shear rate at the surface of the rotating cylinder is to assume that the test fluid follows the simple power law model

$$ \sigma=- K{\left[ {r\frac{{{\rm{d}}({v_\theta }/r)}}{{{\rm{d}}r}}} \right]^n} $$

(3A.13)

Equation 3A.12 can be solved after recalling Eq. 3A.2 for shear stress

$$ \frac{{{V_\theta }}}{r} =- {\left( {\frac{T}{{{\rm{2}}nK}}} \right)^{{\rm{1}}/n}}\frac{n}{{\rm{2}}}{r^{ - {\rm{2}}/n}} + C $$

(3A..14)

Using the boundary conditions v_θ = 0 at r = r_o and v_θ = r _iΩ at r = r_i, one can obtain the equation given earlier, Eq. 3.7

$$ .\gamma= \frac{{2\Omega r_i^{\rm{2}}}}{{\left[ {1 - {{\left( {\frac{{{r_i}}}{{{r_o}}}} \right)}^{\rm{2}}}} \right]}}\left\{ {\frac{{1 - {{(\frac{{{r_i}}}{{{r_o}}})}^{\rm{2}}}}}{{n{{\left[ {1--\frac{{{r_i}}}{{{r_0}}}} \right]}^{{\rm{2}}/n}}}}} \right\} $$

(3A.15)

When n = 1 (Newtonian fluid), the Margules equation for shear rate is obtained

$$ .{\gamma _{\rm{i}}} = \frac{{{\rm{2}}\Omega r_i^2}}{{[ {1 - {{({r_i}/{r_o})}^2}} ]}} $$

Fig. 3A.1

Decision tree for using a concentric cylinder geometry

Examining the limiting form of the second part of Eq. 3.7 as r_i → r_o

$$ {\rm{Limit}}\left\{ {\frac{{1 - {{({r_i}/{r_o})}^{\rm{2}}}}}{{n[ {1 - {{({r_i}/{r_o})}^{\rm{2}}}} ]}}} \right\} = _{n \to {r_o}}^{\lim }\left( {\frac{{{\rm{2}}{r_i}/{r_o}}}{{ - n({\rm{2}}/n){{({r_i}{r_o})}^{({\rm{2}}/n) - 1}}}}} \right) = 1 $$

Therefore, when ri → ro, the shear rate of a non-Newtonian fluid tends to that of a Newtonian fluid. In Table 3.1, values of the correction factor in parenthesis in Eq. 3.7 are given for several values of the flow behavior index, n, and the ratio (r_i/r_o) of the concentric cylinders, and they confirm that: (1) they are small when r _i → r _o, and (2) they may be large when the fluids are substantially non-Newtonian and when (r_i/r_o)  0.95.

Although the concentric cylinder geometry is relatively easy to use in rheological studies, some of its limitations should be recognized as shown in Fig. 3.44.

Appendix 2

Analysis of Steady Laminar Fully Developed Flow in a Pipe

The shell balance method will be used to examine steady laminar flow of a fluid in a pipe. For the geometrical system illustrated in Fig. 3B.1 and for steady laminar fully developed flow of a fluid, a shell momentum balance can be conducted (Bird et al. 1960; Geankoplis 1983) using the cylindrical coordinates, r, θ, and z. The momentum balance is conducted on a control volume shell at a radius r with dimensions Δr and Δz.

• 1. The balance of two opposing forces is

$$pA{{|}_{z}}-pA{{|}_{z}}{{+}_{\Delta z}}=P(\text{2}\pi r\Delta r){{|}_{z}}-p(\text{2}\pi r\Delta r){{|}_{z}}{{+}_{\Delta z}}$$

(3B.1)

• 2. The shear or drag force at radius r = σ _rz (2πrΔz) which can be considered to be the momentum flow into the cylindrical surface of the shell. The net efflux of momentum is the difference between the magnitudes the momentum out and momentum in

$${{\sigma }_{rz}}(\text{2}\pi r\Delta z){{|}_{r+\Delta r}}(\text{OUT})-{{\sigma }_{rz}}(\text{2}\pi r\Delta z){{|}_{r}}(\text{IN})$$

(3B.2)

• 3. The momentum flux across the annular surface at z and z + Δz is zero because the axial velocity v _z is of the same magnitude at z and z + Δz.

• 4. Recognizing that the sum of forces acting on control volume = rate of momentum out of control volume – rate of momentum into control volume

$$\begin{aligned}& p(\text{2}\pi r\Delta r){{|}_{z}}-p(\text{2}\pi r\Delta r){{|}_{z+\Delta z}}={{\sigma }_{rz}}(\text{2}\pi r\Delta z){{|}_{r+\Delta r}}(\text{OUT}) \\ & -{{\sigma }_{rz}}(\text{2}\pi r\Delta r){{|}_{r}}(\text{IN}) \\ \end{aligned}$$

(3B.3)

The above equation can be simplified to

$$r\Delta r(p{{|}_{z}}-p{{|}_{z}}{{+}_{\Delta z}})=\Delta z\left[ (r{{\sigma }_{rz}}){{|}_{r}}{{+}_{\Delta r}}-(r{{\sigma }_{rz}}){{|}_{r}} \right]$$

(3B.4)

Fig. 3B.1

Schematic diagram for force balance in laminar capillary flow

Dividing both sides by (ΔrΔz), results in

$$\frac{r\left( p{{|}_{z}}-p{{|}_{z}}{{+}_{\Delta z}} \right)}{\Delta z}=\frac{\left( r{{\sigma }_{rz}} \right){{|}_{r+\Delta r}}-\left( r{{\sigma }_{rz}} \right){{|}_{r}}}{\Delta r}$$

(3B.5)

As Δr and Δz → 0, the above equation becomes

$$r\frac{\text{d}p}{\text{d}z}=\frac{\text{d}}{\text{d}r}(r{{\sigma }_{rz}})$$

(3A.6)

After integration, one gets

$$r{{\sigma }_{rz}}=-\frac{{{r}^{2}}}{2}\frac{\text{d}p}{\text{dz}}+c1$$

(3B.7)

which can be rearranged to

$${{\sigma }_{rz}}=-\frac{r}{2}\frac{\text{d}p}{\text{dz}}+\frac{c1}{r}$$

(3B.8)

By using the condition σ_rz = 0 at r = 0 (center line), the integration constant c ₁ = 0, so that the stress distribution across the radius of the pipe is

$${{\sigma }_{rz}}=-\frac{r}{2}\frac{\text{d}p}{\text{dz}}$$

(3B.9)

that for the wall of the pipe becomes: σ_w = –[(r _o/2; dp/dz)], because at r = r _o, σ_rz = σ_w. When the pressure drop, Δp, is measured for fully developed flow over a length, L, the equation for the shear stress at the wall is

$${{\sigma }_{\text{w}}}=\frac{\text{D}\Delta p}{\text{4}L}$$

(3B.10)

We will derive an expression for the volumetric flow rate, Q, and then another for the shear rate. Because the axial velocity, v _z, is dependent on the radial position

$$Q=\int\limits_{_{0}}^{_{ro}}{2\pi r{{v}_{z}}\text{d}r=}\int\limits_{_{o}}^{_{ro}}{{{v}_{z}}\text{(}r\text{)}\,\text{d}\,\text{(}{{r}^{2}}\text{)}=}\,\pi \int\limits_{_{0}}^{_{ro}}{{{v}_{z}}\text{2}r\,\text{d}r+\pi \int\limits_{_{{{v}_{z}}\,\text{at}r=0}}^{_{{{v}_{z}}\,\text{at}r=ro}}{{{r}^{2}}\,\text{d}{{v}_{z}}}}$$

(3B.11)

Assuming a no slip boundary condition, that is, v _z = 0 at r = r _o, the right-hand side of the above equation can be written as

$$\pi =\int\limits_{_{o}}^{_{ro}}{{{v}_{z}}\text{2}r\,\text{d}r\,-\pi }\int\limits_{_{\,o}}^{_{{{v}_{z}}}}{{{r}^{2}}\,\text{d}{{v}_{z}}}$$

(3B.12)

The above equation can be integrated and noting that v _z = 0 at r = r _o, and r = (r _o/σ_w)σ_rz, one can obtain the expression

$$Q=\left( -\frac{\pi r_{\text{o}}^{3}}{\sigma _{\text{w}}^{3}}\,\, \right)\int\limits_{_{0}}^{_{{{\sigma }_{\text{w}}}}}{\sigma _{rz}^{2}\left( -\frac{\text{d}{{v}_{z}}}{\text{d}r}\,\, \right)d{{\sigma }_{rz}}}$$

(3B.13)

Multiplying both sides of the above equation by 4 and rearranging, we get

$$\frac{4Q}{\pi r_{\text{o}}^{3}}\,\left( \frac{4}{\sigma _{\text{w}}^{3}}\,\, \right)\int\limits_{_{0}}^{_{{{\sigma }_{\text{w}}}}}{\sigma _{rz}^{2}\left( -\frac{\text{d}{{v}_{z}}}{\text{d}r}\,\, \right)d{{\sigma }_{rz}}}$$

(3B.14)

The above equation can be used for deriving a general solution for tube flow and specific expressions for the volumetric flow rates of fluids exhibiting different rheological behaviors. For a Newtonian fluid, noting that the shear rate, $-(\text{d}{{v}_{z}}/\text{d}r)=\frac{{{\sigma }_{rz}}}{\eta },$ it can be shown that

$$\frac{4Q}{\pi r_{\text{o}}^{3}}=\,\frac{{{\sigma }_{\text{w}}}}{\eta }=\,\frac{1}{\eta }\frac{D\Delta p}{4L}$$

(3B.15)

or that the viscosity of a Newtonian fluid

$$\eta =\frac{(D\Delta p/4L)}{(8{{\overset{\mathbf{--}}{\mathop{v}}\,}_{z}}/D)}\text{,}\,\text{that}\,\text{is,}\,\text{ }\!\!\eta\!\!\text{ }\sim \frac{1}{(8{{\overset{\mathbf{--}}{\mathop{v}}\,}_{z}})}\sim \text{flow}\,\text{time}$$

(3B.16)

Equation 3B.16 is the basis for calculation of viscosity of a Newtonian fluid using glass capillary viscometer. It should also be recognized that $(4Q/\pi r_{\text{o}}^{3})=(32Q/\pi {{D}^{3}})$ gives the shear rate for Newtonian fluids but not for non-Newtonian fluids and it is called pseudoshear rate. Additional steps are required to obtain an expression for the true shear rate.

Differentiating both sides of Eq. 3B.14 and setting the limits

$$\frac{\text{d }\!\![\!\!\text{ }\sigma _{\text{w}}^{3}(4Q/\pi r_{\text{o}}^{3})}{\text{d}{{\sigma }_{\text{w}}}}=4\sigma _{\text{w}}^{2}\left( -\frac{\text{d}{{v}_{z}}}{\text{d}r} \right)$$

(3B.17)

Further differentiating the left-hand side and rearranging, one obtains the general solution for the true shear rate in tube flow

$$\left( \frac{\text{d}{{v}_{z}}}{\text{d}r} \right)=\left( \frac{3}{4} \right)\,\frac{4Q}{\pi r_{\text{o}}^{3}}+\frac{{{\sigma }_{\text{w}}}}{4}\,\frac{\text{d(4Q/}\pi r_{\text{o}}^{3}\text{)}}{\text{d}{{\sigma }_{\text{w}}}}$$

(3B.18)

Equation 3B.18 is known as the Weissenberg–Rabinowitsch–Mooney (WRM) equation in honor of the three rheologists who have worked on this problem. An alternate equation can be derived for fluids obeying the power law model between shear stress and the pseudoshear rate

$$\sigma =K'\,{{(4Q/\pi r_{\text{o}}^{3})}^{n'}}$$

(3B.19)

$$\left( -\frac{\text{d}{{v}_{z}}}{\text{d}r} \right)=\frac{3n'+1}{4n'}\,\left( \frac{4Q}{\pi r_{\text{o}}^{3}} \right)$$

(3B.20)

To use the WRM equation, the steps involved are: (1) using a tube/pipe/capillary of known diameter (D) and length (L), several sets of volumetric flow rate (Q) versus pressure drop (ΔP) data are obtained under isothermal fully developed and no slip at the pipe wall conditions.

(2) The quantities $(4Q/\pi {{r}^{3}}_{0})$ are calculated, plotted, and a smooth curve fitted.

(3) At each value of $(4Q/\pi {{r}^{3}}_{0})$ the value of: $\text{d}(4Q/\pi {{r}^{3}}_{0})/\text{d}{{\sigma }_{\text{w}}}$ is determined and the true shear rate calculated using Eq. 3B.18. Alternatively, for many foods, plots of the quantities: log $(4Q/\pi {{r}^{3}}_{0}),$ and log (D Δp/4 L) are straight lines with slopes n′ that can be used in Eq. 3B.20 to obtain the true shear rate.

Fig. 3B.2

Decision tree for using a capillary/tube viscometer

(4) Generally, for a given fluid, values of n′ and of n of the power law model: $\sigma = K'{(4Q/\pi r_0^3)^{n'}}$ will be the same. However, the value of K′ in is related to the consistency index K by the equation

$$K' = K{\left( {\frac{{3n + 1}}{{4n}}} \right)^n}$$

(3B.21)

As an example, for n = 0.3, K′ = 1.15 K.

From Eq. 3B.19, an expression can be derived for a pipe-flow apparent viscosity (η _ap) based on the diameter D and the average axial velocity in the tube $({\bar \nu _z})$

$${\eta _{{\rm{ap}}}} = K'{\left( {\frac{{8{{\mathop v\limits^{{\bf{ - - }}} }_z}}}{D}} \right)^{n' - 1}}$$

(3B.22)

When n = n′, together with Eqs. 3B.21 and 3B.22, one can obtain the generalized Reynolds number (GRe)

$$\text{GRe}=\frac{{{D}^{n}}\overset{\mathbf{--}}{\mathop{v}}\,_{_{z}}^{2-n}\rho }{{{8}^{(n-1)}}K}{{\left( \frac{4n}{3n+1} \right)}^{n}}$$

(3B.23)

Some of the considerations in selecting a capillary/tube viscometer for viscosity measurement are shown in Fig. 3B.2.

Appendix 3

Analysis of Flow in a Cone-Plate Geometry

We assume that the cone of radius r_o and cone angle θ_o is on top over the plate rotating with an angular velocity Ω while the bottom plate is stationary as shown in Fig. 3.7. We consider the equation of motion in spherical coordinates r,θh,ϕ. For steady fully developed flow, the velocity components v_θ = 0 and v_r = 0, and v _ϕ is a function of r and θ. In addition, we consider pressure variations (body forces) to be negligible and that the value of θ _o is very small,  < 0.1 rad. The r-component reduces

$$ \rho \frac{{v_\phi ^{\rm{2}}}}{r}\phi= \frac{1}{{{r^{\rm{2}}}}}\frac{\partial }{{\partial r}}({r^{\rm{2}}}{\sigma _{rr}}) - \frac{{{\sigma _{\theta \theta }} + {\sigma _{\phi \phi }}}}{R} $$

(3C.1)

The θ-component reduces to

$$ \frac{1}{{r\sin \theta }}\frac{\partial }{{\partial \theta }}({\sigma _{\theta \theta }}\sin \theta ) - \frac{{{\rm{cot}}\theta }}{r}{\sigma _\phi }_\phi= 0 $$

(3C.2)

The ϕ-component of the equation of motion reduces to

$$ \frac{1}{r}\frac{{\partial {\sigma _{\theta \phi }}}}{{\partial \theta }} + \frac{{{\rm{2cot}}\theta }}{r}{\sigma _\theta }_\phi= 0 $$

(3C.3)

We note that the ϕ-component contains the shear stress σ _θϕ . The boundary conditions are: v _ϕ  = 0 at θ = π/2 because the plate is stationary and v_ϕ = rΩ cos θ _o at θ = (π/2) – θ _o. First, we will derive an expression for the shear stress by conducting a simple torque balance on the plate

$$M = \int\limits_0^{2\pi } {\int\limits_0^{{r_o}} {{r^2}} {\sigma _{\theta \phi }}{\rm{d}}r{\rm{d}}\phi {\rm{ = }}\frac{{2\pi r_0^3}}{3}} {\sigma _{\theta \phi }}|\pi /2$$

(3C.4)

Integration of Eq. 3C.3 results in the expression: σ _θϕ |_(π/2) = σ _θϕ (θ) sin² θ. Therefore,

$$ {\sigma _{\theta \phi }}(\theta ) = \frac{{3M}}{{2\pi r_o^3{{\sin }^2}\theta }} $$

(3C.5)

When ${\theta _o} < 0.1\,{\rm{rad,}}\,{\rm{si}}{{\rm{n}}^{\rm{2}}}\left( {\frac{\pi }{2}{\theta _o}} \right) \sim 1,$ so that

$$ {\sigma _{\theta \phi }} = \frac{{{\rm{3}}M}}{{{\rm{2}}\pi r_o^3}} $$

(3C.6)

Defining T_cn as the torque per unit area, Eq. 3C.4 can be written in terms of the cone diameter D

$$ {\sigma _{\theta \phi }} = \frac{{3{T_{cn}}}}{D} $$

(3C.7)

From Eqs. 3C.5 and 3C.6, it follows that the shear stress in a cone-plate geometry is essentially uniform.

To obtain an expression for shear rate, we can simply say that for low values of the cone angle θ_o we need not distinguish between sin θ_o and θ_o (Whorlow 1980) when the shear rate at radius r will be

$$ .{\gamma _r} = \frac{{r\Omega }}{{r{\theta _o}}} = \frac{\Omega }{{{\theta _o}}} $$

(3C.8)

Alternatively, we can consider the velocity distribution (Brodkey 1967) for v_ϕ

$$ \frac{{{v_\phi }}}{{r\,\sin \theta }} = \int\limits_{\theta= \pi /2}^\theta{\frac{{{\rm{d(}}v\phi /r\sin \theta )}}{{{\rm{d}}\theta }}} {\rm{d}}\theta$$

(3C.9)

We can also substitute for from Eq. $3{\rm{C}}{\rm{.3,}}d\theta= \frac{1}{2}\frac{{\sin \theta }}{{\cos \theta }}\frac{1}{{{\sigma _{\theta \phi }}}}$ and also switch limits: σ_θϕ at θ = 0 and σ_plate at θ = π/2. For a Newtonian fluid, we can substitute in 3C.9 the relationship between shear rate and shear stress and get 3C.11

$$ {\sigma _{\theta \phi }} = \eta \left[ {\sin \theta \frac{{{\rm{d(}}{{\rm{v}}_\phi }/r\sin \theta )}}{{{\rm{d}}\theta }}} \right] $$

(3C.10)

$$ \frac{{{v_\phi }}}{{r\sin \theta }} = \int\limits_\theta ^{\pi /{\rm{2}}} {\frac{1}{\eta }} \frac{{{\sigma _{\theta \phi }}}}{{\sin \theta }}{\rm{d}}\theta$$

(3C.11)

After substituting for σ_θϕ in 3C.11 and performing the integration results in

$$ \frac{{{v_\phi }}}{r} = \frac{{{\rm{3}}T}}{{{\rm{4}}\eta {r_o}}}\left[ {\cot {\theta _o} - 1{\rm{n}}\left( {\tan \frac{\theta }{2}} \right)\sin \theta } \right] $$

(3C.12)

Equation 3C.12 can be written for the boundary condition at $\theta= \frac{\pi }{2}--{\theta _0}:$

$$ \Omega \cos {\theta _o} = \frac{{{\rm{3}}T}}{{{\rm{4}}\eta {r_o}}}\left[ {\tan {\theta _o} - \cos {\theta _o}1{\rm{n}}\left( {\frac{{1 - \tan {\theta _o}}}{{1 + \tan {\theta _o}}}} \right)} \right] $$

(3C.13)

For low values of the cone angle θ_o, the above equation becomes

$$ \begin{array}{l}\Omega= \frac{{{\rm{3}}T}}{{{\rm{4}}\eta {r_o}}}({\theta _o} + {\theta _o}) = \frac{{3T{\theta _o}}}{{\eta D}}\\\end{array} $$

$$ {\rm{Therefore}},{\sigma _{\theta \phi }} \cong \eta \frac{\Omega }{{{\theta _o}}}. $$