Non-Newtonian fluid pumping: power-law flow, pressure drop and NPSH

A sauce plant in Turkey extends a transfer line by 15 m to reach a new hot-filling hall. The rotary lobe pump seems to have headroom, so the change is signed off. On the first cold morning the pump trips on high pressure. By afternoon, with hot product, it runs, but rattles on the suction side. Two separate problems are at work: rheology and net positive suction head. Both are predictable with a few equations, and both sit at the centre of non-Newtonian fluid pumping.

What makes non-Newtonian fluid pumping different?

A non-Newtonian fluid is one whose viscosity changes with shear rate, and sometimes with time. Shear stress (τ, Pa) is the force per unit area that makes layers of fluid slide over each other; shear rate (γ̇, s⁻¹) is how quickly velocity changes from layer to layer. For a Newtonian fluid such as water, milk or clear juice, τ = μ·γ̇ and the viscosity μ is constant at a given temperature. For non-Newtonian foods the ratio τ/γ̇ varies, so it is called the apparent viscosity, η.

How do the power-law and Herschel-Bulkley models work?

The power-law model describes a shear-thinning food with two constants:

τ = K × γ̇ⁿ so η = K × γ̇⁽ⁿ⁻¹⁾

K is the consistency coefficient (Pa·sⁿ) and n the flow behaviour index (dimensionless): n below 1 means shear-thinning, n = 1 Newtonian. Fit them to rheometer data by plotting log τ against log γ̇; the slope is n and the intercept log K.

For a fruit preparation with K = 15 Pa·sⁿ and n = 0.35 (illustrative), the apparent viscosity is 15 Pa·s at 1 s⁻¹, 3.36 Pa·s at 10 s⁻¹ and 0.75 Pa·s at 100 s⁻¹: a twenty-fold range in one product.

Products with a yield stress, the minimum stress needed to start flow, follow the Herschel-Bulkley model:

τ = τ₀ + K × γ̇ⁿ

τ₀ is the yield stress (Pa); the power law (τ₀ = 0) and the Bingham plastic (n = 1) are special cases. A ketchup-like sauce with τ₀ = 30 Pa, K = 10 Pa·sⁿ and n = 0.4 has τ = 30 + 10 × 100⁰·⁴ = 93 Pa at 100 s⁻¹, an apparent viscosity of 0.93 Pa·s.

Two cautions apply. A model is valid only over the shear rates used to fit it, typically tens to hundreds of reciprocal seconds at a pipe wall. And K falls steeply as temperature rises, so data at 20 °C say little about a line running at 80 °C.

How do you calculate laminar pressure drop for a power-law fluid?

Find the true wall shear rate, convert it to wall shear stress, then apply a force balance. Laminar flow, smooth layered flow without eddies, is normal for viscous foods.

1. Mean velocity: V = Q ÷ (π × D²/4).
2. Nominal wall shear rate: 8V/D. This is the true value only for a Newtonian fluid.
3. Rabinowitsch-Mooney correction for a power-law fluid: γ̇w = [(3n + 1)/(4n)] × 8V/D.
4. Wall shear stress: τw = K × γ̇wⁿ.
5. Pressure drop: ΔP = 4 × τw × L ÷ D.
6. Check the regime with the Metzner-Reed Reynolds number.

Re_MR = ρ × V⁽²⁻ⁿ⁾ × Dⁿ ÷ [K × 8⁽ⁿ⁻¹⁾ × ((3n + 1)/(4n))ⁿ]

The Metzner-Reed (generalised) Reynolds number is defined so that laminar power-law flow obeys the Newtonian relation f = 16/Re for the Fanning friction factor, a dimensionless measure of wall friction. Flow stays laminar up to roughly 2,100, the exact limit varying a little with n.

Add static head (ρ·g·Δz, about 43 kPa for a 4 m lift) and any pressure the filler needs. Fitting losses are usually small beside straight-pipe friction for so viscous a product. The Intermediate Food Process Engineering course works through these steps from rheometer data to pump selection.

Why does a single viscosity figure give the wrong pressure drop?

Because the right figure depends on the wall shear rate and the shape of the velocity profile, and a single viscosity captures neither. Put three “typical” viscosities into the Newtonian laminar equation, ΔP = 32·μ·V·L/D²:

Shear-thinning also changes how a line responds. In laminar power-law flow, ΔP is proportional to Qⁿ and to 1/D⁽³ⁿ⁺¹⁾. Doubling the flow raises the pressure drop only 2⁰·³⁵ = 1.27 times (306 to 390 kPa), where a Newtonian liquid would double. Cutting the bore from 50 mm to 40 mm raises it 1.58 times (to 484 kPa), against 2.44 times for a Newtonian liquid.

Yield-stress products need a restart check: the minimum restart pressure is ΔP = 4·τ₀·L/D, so for τ₀ = 30 Pa in the same line, 4 × 30 × 40 ÷ 0.050 = 96 kPa, before allowing for cooled or restructured product. Temperature is the other big lever. Pressure drop is proportional to K, so if K doubles as product cools in an uninsulated line, the pressure drop doubles: the cold-morning trip in the opening scenario.

How do you calculate NPSH available?

Net positive suction head available (NPSHa) is the margin, in metres of liquid, by which the absolute pressure at the pump inlet exceeds the liquid’s vapour pressure. When pressure inside the pump falls below the vapour pressure, bubbles form and then collapse violently: cavitation, with noise like gravel, lost flow and pitted rotors.

NPSHa = h_atm + h_static − h_vap − h_f

Each term is a head in metres: h_atm = P_surface ÷ (ρ·g), using absolute pressure on the liquid surface (atmospheric for an open tank); h_static is the level above the pump inlet (negative for a suction lift); h_vap = P_vapour ÷ (ρ·g); h_f is suction friction. The supplier states the NPSH required (NPSHr); positive displacement pump makers often quote net positive inlet pressure (NPIPr) in kPa or bar instead. A margin of at least 0.5 to 1 m is common practice, more for hot duties.

Near the boiling point, h_atm and h_vap almost cancel, which is why pumps drawing from evaporators and vacuum vessels sit well below them.

Which pumps suit viscous non-Newtonian foods?

Positive displacement pumps suit most viscous, shear-sensitive and particulate foods; centrifugal pumps suit thin liquids. A positive displacement (PD) pump traps a fixed volume and pushes it forward, so flow follows speed, largely regardless of pressure. A centrifugal pump spins the liquid, so its flow depends on system resistance.

Above a few hundred mPa·s, a PD pump is usually the better choice. Never run one against a closed valve: fit a relief valve, bypass or pressure trip. Keep suction lines short, wide and flooded, and size the pump from the flow curve at process temperature. Food Technology for Industry Professionals covers rheology alongside heat transfer, thermal process lethality and hygienic design.

Frequently asked questions

What do K and n mean in the power-law model?

K is the consistency coefficient, in Pa·sⁿ, and sets how thick the product is overall. n is the flow behaviour index and describes how strongly viscosity changes with shear rate: below 1 means shear-thinning, and the lower n is, the stronger the effect. Measure both at process temperature, because K falls sharply as temperature rises.

Why use the Metzner-Reed Reynolds number?

The ordinary Reynolds number needs a single viscosity, which a power-law fluid does not have. The Metzner-Reed (generalised) Reynolds number combines density, velocity, diameter, K and n so that laminar power-law flow follows the Newtonian friction law, f = 16/Re for the Fanning friction factor. It shows whether flow is laminar, which it nearly always is for viscous foods.

Can a centrifugal pump handle viscous food products?

Only up to a point. As viscosity rises, a centrifugal pump delivers less head and flow and its efficiency falls, so manufacturers correct the water curve for viscosity. High impeller speeds can also thin delicate products such as stirred yoghurt. Above a few hundred mPa·s, or for products with pieces, a positive displacement pump is usually the better choice.

What causes cavitation in food pumps?

Cavitation happens when the pressure inside the pump falls below the liquid’s vapour pressure, so vapour bubbles form and then collapse as pressure recovers. Common causes are hot product, a low tank level, long or narrow suction lines, blocked strainers, high pump speed and high-altitude sites. Raising the level, enlarging the suction line, slowing the pump or cooling the product restores the margin.

How do you restart a line full of a yield-stress product?

The pump must push the product past its yield stress along the whole line before flow starts, which needs at least ΔP = 4·τ₀·L/D. Product that has cooled, or rebuilt structure at rest, needs more. Check that the pump and its relief setting can supply this pressure, keep the line insulated or trace-heated, and consider a warm water flush before long stops.

Next step. The Intermediate Food Process Engineering course takes you from rheometer data to pipe pressure drops, pump operating points and cavitation-free suction design. It ends with a proctored final assessment and an ASC certificate. You can also see all eleven food science and technology courses.

Sources. James F. Steffe, Rheological Methods in Food Process Engineering, 2nd edn (Freeman Press, 1996). R. Paul Singh, Dennis R. Heldman and Ferruh Erdogdu, Introduction to Food Engineering, 6th edn (Academic Press, 2024). Romeo T. Toledo, Rakesh K. Singh and Fanbin Kong, Fundamentals of Food Process Engineering, 4th edn (Springer, 2018). Zeki Berk, Food Process Engineering and Technology, 3rd edn (Academic Press, 2018).

This article is general guidance on pumping and pipe-flow calculations and is not a substitute for the applicable standard, your national legislation, pump supplier data or the advice of a qualified engineer.