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Determination of viscosity by viscometer

Short answer

A viscometer determines dynamic viscosity using Newton's law τ = μ du/dy. In a rotating-cylinder type, torque on a cylinder in a fluid-filled gap yields μ = T t / (2πωR³h). The capillary-tube type compares flow times of test and reference liquids: μ₁/μ₂ = ρ₁t₁/(ρ₂t₂).

Explain the determination of viscosity by a viscometer.

A viscometer measures the dynamic viscosity μ\mu of a fluid from a measured relation between shear stress and velocity gradient, using Newton's law τ=μ dudy\tau=\mu\,\dfrac{du}{dy}.

Rotating-cylinder (Couette) viscometer

The fluid fills the thin gap tt between a fixed outer cylinder and an inner cylinder (radius RR, immersed height hh) turned at angular speed ω\omega. The torque TT on the inner cylinder is measured.

   fixed outer cylinder
   |   |  fluid gap t |   |
   |   |<->|  inner   |   |
   |   |   | cylinder |   |
   |   |   | (omega)  |   |

For a thin gap the velocity profile is linear, so

τ=μωRt,T=τ (2πRh) R\tau=\mu\frac{\omega R}{t},\qquad T=\tau\,(2\pi R h)\,R μ=T t2πωR3h\mu=\frac{T\,t}{2\pi\omega R^{3}h}

Capillary-tube (Ostwald) viscometer

The time t1t_1 for a fixed volume of the test liquid to flow through a capillary under its own head is compared with the time t2t_2 for a reference liquid (water) of known viscosity:

μ1μ2=ρ1t1ρ2t2\frac{\mu_1}{\mu_2}=\frac{\rho_1 t_1}{\rho_2 t_2}

For laminar flow (Hagen-Poiseuille), μ=π Δp r48QL\mu=\dfrac{\pi\,\Delta p\,r^{4}}{8QL}.

Falling-sphere viscometer

A small sphere of diameter dd and density ρs\rho_s falls at terminal velocity vv in the liquid (density ρ\rho). Balancing weight, buoyancy and Stokes drag 3πμdv3\pi\mu d v:

μ=d2(ρs−ρ) g18 v\mu=\frac{d^{2}(\rho_s-\rho)\,g}{18\,v}

Other types

The Saybolt (efflux) viscometer gives kinematic viscosity from the time taken for 60 mL of oil to flow out through a standard orifice.

✓ Answer checked by Claude Opus 5.5 (high effort)

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