Introduction & Context

A helical ribbon impeller is the standard geometry for laminar blending of highly viscous, non-Newtonian fluids in cylindrical vessels. The ribbon’s large surface area and close wall clearance generate a strong axial circulation loop that eliminates stagnant zones and promotes uniform temperature and composition. This calculation sheet predicts the mechanical response (power draw and torque) from the fluid rheology and operating speed, enabling correct motor sizing, shaft design, and scale-up from pilot to industrial scale. Typical applications include polymerisation reactors, fermentation broths, food pastes, and pharmaceutical gels.

Methodology & Formulas

Tank and impeller geometry
The impeller diameter \(D\) is fixed by the tank diameter \(T\) through the diameter ratio
\[ D = \left(\frac{D}{T}\right)\,T \] The ribbon pitch is taken equal to \(D\) (single-start). Validity limits are:

Parameter	Minimum	Maximum
\(D/T\)	0.90	0.95

Shear-rate estimation
For a close-clearance ribbon, the average shear rate is proportional to the rotational speed:
\[ \dot\gamma = k_{s}\,N \] with \(k_{s}=30\) for a single-start ribbon of pitch equal to diameter.
Apparent viscosity for power-law fluid
The fluid follows the Ostwald–de Waele model
\[ \tau = K\,\dot\gamma^{\,n} \] hence the apparent viscosity at the estimated shear rate is
\[ \mu_{\text{a}} = K\,\dot\gamma^{\,n-1} \]

Reynolds number
The impeller Reynolds number is defined as
\[ Re = \frac{\rho\,N\,D^{2}}{\mu_{\text{a}}} \] Correlation accuracy is guaranteed only in the following regime:

Flow regime	Re range
laminar (strict)	0.01–10
extended laminar	10–30

Power and torque
In the laminar regime, the power number is inversely proportional to Reynolds:
\[ N_{p} = \frac{350}{Re} \] The power draw \(P\) and shaft torque \(M\) are then
\[ P = N_{p}\;\rho\,N^{3}D^{5}, \qquad M = \frac{P}{2\pi N} \]

All algebraic symbols retain their customary units: \(N\) in s⁻¹, \(D\) in m, \(\rho\) in kg m⁻³, \(K\) in Pa sⁿ, \(P\) in W, \(M\) in N m.

Use a helical ribbon when the dynamic viscosity exceeds 10 Pa·s and the Reynolds number drops below 100. At these conditions, the ribbon’s close-clearance design and large surface area maintain a uniform velocity profile across the entire vessel diameter, eliminating stagnant zones common with anchor or paddle impellers.

Diameter ratio D/T = 0.90–0.98 to sweep the wall and minimise bypass
Pitch P = 0.5–1.0 D to balance axial pumping and viscous drag
Width W = 0.08–0.12 D for structural stiffness without excessive power
Clearance c ≤ 5 mm or 0.005 T, whichever is larger, to prevent product build-up

For a single-start helical ribbon in the laminar regime, the dimensionless power number Po ≈ 300–350 when calculated with Newtonian reference fluids. Use this constant in P = Po·ρN³D⁵ for early design; validate with pilot trials because non-Newtonian fluids can raise Po by 20–40 %.

Retrofit is possible if the vessel has an ASME dished head and a top-entering agitator drive rated for the increased torque. Check that the straight-shell height is at least 0.8 D to accommodate the long ribbon; otherwise, add a stub tube or select a double-flight ribbon to shorten the element length.

Worked Example – Sizing a Helical-Ribbon Impeller for a Shear-Sensitive Paste

A food plant needs to blend a tomato-based vegetable paste (power-law fluid) in a 600 L dished-bottom vessel. The process engineer must check whether a 30 rpm helical-ribbon impeller can deliver full-tank motion without exceeding 110 W shaft power.

Vessel inside diameter, T = 0.600 m
Fill height, H = 0.600 m
Fluid density, ρ = 1400 kg m⁻³
Consistency index, K = 25 Pa sⁿ
Flow behaviour index, n = 0.55
Impeller speed, N = 30 rpm (0.5 s⁻¹)
Impeller-to-tank ratio, D/T = 0.90 (within the 0.90–0.95 range for helical ribbons)
Pitch-to-diameter ratio, P/D = 1.0 (single-flight ribbon)
Number of flights, N_s = 0.5 (one ribbon)

Calculate impeller diameter: D = 0.90 × 0.600 m = 0.540 m.
Estimate average shear rate for a helical ribbon: \[ \dot\gamma = k_s\,N = 30 \times 0.5 = 15\ \text{s}^{-1}. \]
Compute apparent viscosity at this shear rate: \[ \mu_a = K\,\dot\gamma^{\,n-1} = 25 \times 15^{0.55-1} = 7.39\ \text{Pa s}. \]
Determine Reynolds number: \[ Re = \frac{\rho N D^2}{\mu_a} = \frac{1400 \times 0.5 \times 0.54^2}{7.39} = 27.6. \]
Because Re < 30, the flow is still in the laminar regime; use the correlation \[ N_p = 350\,Re^{-1} = 350 / 27.6 = 12.67. \]
Calculate power: \[ P = N_p\,\rho\,N^3 D^5 = 12.67 \times 1400 \times (0.5)^3 \times (0.540)^5 = 101.8\ \text{W}. \]
Convert to torque: \[ M = \frac{P}{2\pi N} = \frac{101.8}{2\pi \times 0.5} = 32.4\ \text{N m}. \]

Final Answer: The 30 rpm helical-ribbon impeller draws 102 W and delivers 32.4 N m torque—within the 110 W limit—confirming adequate laminar blending for the paste.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle