Couette Shear Stress, Viscous Heating & Rotational Viscometer Torque

Distinguish planar shear from cylindrical flow

For steady laminar Newtonian flow between parallel plates with no pressure gradient, constant dynamic viscosity and no slip, the velocity gradient is U/h. Here U is signed relative wall velocity and h is the positive gap. The planar formulas do not by themselves describe an eccentric loaded journal bearing or every commercial viscometer spindle.

For the cylindrical comparison, the inner cylinder has radius R and surface velocity U; the outer cylinder is stationary at radius R+h. Both cylinders are concentric. Assume the ideal purely azimuthal laminar solution, constant viscosity and negligible end effects. The worksheet does not check stability, Taylor vortices, heating-induced viscosity changes or instrument calibration.

Shear rate in B8

Inputs B2:B4 are viscosity in Pa s, signed speed in m/s and gap in metres. The initial rate is 5000 s⁻¹. Zero speed is allowed; missing data, nonpositive viscosity and nonpositive gap are rejected.

=IF(COUNT(B2:B4)<>3,"Enter numeric mu, U and gap",IF(OR(B2<=0,B4<=0),"Viscosity and gap must be positive",B3/B4))

Planar shear stress in B9

The initial stress is 600 Pa. Its sign follows the chosen velocity-gradient direction. Drag exerted on the moving wall opposes its motion; do not confuse that wall-traction direction with the signed quantity reported here.

=IF(ISNUMBER(B8),B2*B8,B8)

Planar viscous dissipation in B10

The rate of conversion of mechanical energy into heat per unit fluid volume is mu × (U/h)². It is 3000000 W/m³ for the supplied example, remains nonnegative when motion reverses, and becomes zero at rest. It is not a temperature rise; a thermal model is needed for that. This uniform planar dissipation must not be assumed uniform across a finite annular gap.

=IF(ISNUMBER(B8),B2*B8^2,B8)

Narrow-gap driving torque in B11

R in B5 and immersed length L in B6 are in metres. Multiplying the planar shear estimate by 2πR²L gives approximately 0.18849556 N m. This is a narrow-gap approximation, even though multiplying actual wall shear by area and lever arm is exact. The approximation requires h small compared with R.

=IF(ISNUMBER(B9),IF(COUNT(B5:B6)<>2,"Enter numeric radius and length",IF(OR(B5<=0,B6<=0),"Radius and length must be positive",2*PI()*B5^2*B6*B9)),B9)

Ideal annular driving torque in B12

For the stationary outer cylinder, Ω=U/R and the ideal driving torque is 4πmu L Ω R²(R+h)² / ((R+h)²−R²). The formula below uses h(2R+h) for the denominator to avoid subtracting nearly equal squared radii. The fluid’s resisting torque on the inner cylinder has the opposite sign.

This exact solution of the stated ideal model gives about 0.19416909 N m with the initial inputs, roughly 3% above the planar approximation. Reversing U reverses both torque signs. The ratio tends to one as h/R tends to zero; a larger gap makes the distinction important. Neither value includes bearing friction, spindle-end torque or turbulence.

=IF(ISNUMBER(B9),IF(COUNT(B5:B6)<>2,"Enter numeric radius and length",IF(OR(B5<=0,B6<=0),"Radius and length must be positive",4*PI()*B2*B3*B6*B5*(B5+B4)^2/(B4*(2*B5+B4)))),B9)

Further reading

• Richard Fitzpatrick, University of Texas — Taylor-Couette flow
• The Feynman Lectures on Physics II, Chapter41 — Viscosity and Couette flow