Flow Behaviour Index and Consistency Coefficient from Narrow-Gap Viscometer Data: 22-Agric-A6 Physical Properties of Biological Materials and Food Products, May 2016, Q3

22-Agric-A6 Physical Properties of Biological Materials and Food Products · May 2016

Question 3: Flow Behaviour Index and Consistency Coefficient from Narrow-Gap Viscometer Data (20 marks)

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.

Viscometer geometry and readings
Quantity	Symbol	Value
Full-scale spring torque	$T_{fs}$	7187 dyn·cm
Inner cylinder radius (OD/2)	$r_i$	0.5 cm
Outer cylinder radius (ID/2)	$r_o$	0.75 cm
Cylinder height	$L$	6 cm

Derived shear stress and shear rate at each speed
N (rpm)	% FS	Torque, dyn·cm	$\dot\gamma$ (s$^{-1}$)	$\tau$ (dyn/cm$^2$)
2	15	1078.1	0.419	114.4
4	26	1868.6	0.838	198.3
10	53	3809.1	2.094	404.2
20	93	6683.9	4.189	709.2

Find. The flow behaviour (power-law) index $n$ and the consistency coefficient $b$ (commonly written $K$) of the power-law model $\tau = b\,\dot\gamma^{\,n}$.

Approach. Reduce every torque/speed pair to a shear stress and a shear rate using the narrow-gap Couette approximation, then fit a straight line to $\ln\tau$ versus $\ln\dot\gamma$: the slope is $n$ and the antilog of the intercept is $b$.

Check: with $r_o/r_i = 1.5$, this viscometer's gap is wider than the ~10% radius-ratio limit usually quoted for the narrow-gap approximation to be exact. The approximation below is nonetheless the intended (and standard textbook) solution route for this style of problem; a rigorous wide-gap treatment would instead integrate the shear-rate profile across the annulus (Krieger–Elrod method).

Convert torque and speed to shear stress and shear rate. Torque at each reading is $T = T_{fs}\times(\%\text{FS}/100)$. The shear stress on the inner (bob) cylinder wall is $$\tau = \frac{T}{2\pi r_i^2 L} = \frac{T}{2\pi(0.5)^2(6)} = \frac{T}{9.425}.$$ The narrow-gap shear rate, with the outer cylinder stationary and angular speed $\omega = 2\pi N/60$, is $$\dot\gamma = \frac{\omega\, r_i}{r_o - r_i} = \frac{\omega(0.5)}{0.25} = 2\omega.$$ Applying these to all four readings gives the $\tau$–$\dot\gamma$ table above.
Linearize and fit. Taking logarithms of $\tau = b\dot\gamma^{\,n}$ gives $\ln\tau = \ln b + n\ln\dot\gamma$, a straight line on the log-log plot shown below. A least-squares fit through the four points gives $$n = 0.79, \qquad \ln b = 5.43 \;\Rightarrow\; b = 227\ \text{dyn}\cdot\text{s}^{n}/\text{cm}^2.$$
State the result. $$\boxed{n \approx 0.79,\qquad b \approx 227\ \text{dyn}\cdot\text{s}^{n}/\text{cm}^2\;(\approx 22.7\ \text{Pa}\cdot\text{s}^{n})}$$ Since $n<1$, the food product is pseudoplastic (shear-thinning).

Log-log plot of shear stress versus shear rate; the four viscometer readings fall on a straight line of slope $n = 0.79$, confirming power-law (pseudoplastic) behaviour over this shear-rate range.

Final results
Quantity	Value
Flow behaviour index, $n$	0.79 (pseudoplastic)
Consistency coefficient, $b$	227 dyn·s$^n$/cm$^2$ (22.7 Pa·s$^n$)

Concept & Theory

Topic: Narrow-gap (Couette) rotational viscometry and power-law fluid characterization.

Key relations: power-law model $\tau = K\dot\gamma^{\,n}$; narrow-gap shear rate $\dot\gamma = \omega r_i/(r_o-r_i)$; shear stress on the bob $\tau = T/(2\pi r_i^2 L)$.

Why this works: a rotational viscometer measures torque, which is an integral of shear stress over the wetted cylinder surface; the narrow-gap approximation converts that integral quantity back into a local shear stress and shear rate pair by assuming the velocity gradient across the annulus is uniform, which lets every torque–speed reading be treated as a single point on the fluid's own $\tau$-$\dot\gamma$ curve. Because a power law is linear in log-log space, ordinary linear regression on the logarithms recovers both power-law constants from as few as two points, with more points improving the fit.

Common pitfall: using the outer-cylinder radius (or the mean radius) instead of the inner (bob) radius in the stress formula, or forgetting to convert rpm to rad/s before forming $\dot\gamma$.

Source: Steffe, Rheological Methods in Food Process Engineering, concentric-cylinder viscometry chapter; Rao et al., Engineering Properties of Foods.

Quantity	Symbol	Value
Full-scale spring torque	\(T_{fs}\)	7187 dyn·cm
Inner cylinder radius (OD/2)	\(r_i\)	0.5 cm
Outer cylinder radius (ID/2)	\(r_o\)	0.75 cm
Cylinder height	\(L\)	6 cm

Quantity	Value
Flow behaviour index, \(n\)	0.79 (pseudoplastic)
Consistency coefficient, \(b\)	227 dyn·s\(^n\)/cm\(^2\) (22.7 Pa·s\(^n\))

Question 3 of 9: Flow Behaviour Index and Consistency Coefficient from Narrow-Gap Viscometer Data

Question 3: Flow Behaviour Index and Consistency Coefficient from Narrow-Gap Viscometer Data (20 marks)