Find the question you need
Choose a route or read from the beginning. All teaching chapters stay open; the grouped index below is expandable.
Set the physical model
Chapters 1–2. Separate stress, yield and translational motion.
Derive the moving regions
Chapters 3–5. Locate the plug and calculate its velocity.
Calculate and compare
Chapters 6–8. Integrate flow and change pressure or pipe radius.
Practise and continue
Chapters 11–12. Use a new threshold case and connected reading.
Full chapter index · Practice and worked checks · Rheology subject root
An unyielded plug inside a flowing material can move. In the ideal pipe-flow problem, material near the centre translates without local shear while an outer annulus deforms around it. Calling the plug unyielded describes deformation, not zero velocity relative to the laboratory.
This guide derives that distinction for a fictional Bingham material in a straight circular pipe. The numbers are educational models, not pump settings or material specifications. Start with the Rheology root for stress, shear rate, yield behaviour and measurement limits.
Chapter index
Set the physical model · Chapters 1–2
1. State the geometry and flow assumptions
2. Derive the stress distribution from force balance
Derive the moving regions · Chapters 3–5
3. Locate the unyielded core
4. Derive the velocity in the annulus
5. Work a moving-plug example
Calculate and compare · Chapters 6–8
6. Integrate the total volume flow
7. Change pressure without assuming proportional flow
8. Change radius and inspect the threshold too
Test the model · Chapters 9–10
9. Connect the calculation to measurement evidence
10. Repair the common mistaken moves
Practise and continue · Chapters 11–12
11. Practise with a new pressure and viscosity
12. Choose a connected flow route
CHAPTER 1 OF 12 · Set the physical model
1. State the geometry and flow assumptions
Take a horizontal circular pipe of radius R and length L, with a positive pressure drop ΔP along the direction of flow. Assume steady, incompressible, fully developed, axisymmetric laminar flow. The wall is stationary and no slip applies. Material properties are uniform and do not evolve with time in this model.
The material has an ideal yield stress τ_y and plastic viscosity μ_p. Above yield, shear-stress magnitude and shear-rate magnitude obey τ = τ_y + μ_pγ̇. Below yield, the ideal model permits stress without shear deformation.
Real pastes can exhibit slip, thixotropy, elasticity, temperature changes or a different flow curve. Those effects are not silently included in the Bingham parameters. First solve the clearly defined model, then examine whether it describes the material and experiment of interest.
CHAPTER 2 OF 12 · Set the physical model
2. Derive the stress distribution from force balance
Consider a coaxial cylinder of material of radius r. Pressure force across its ends is ΔPπr². Its side area is 2πrL, so balancing axial pressure force and shear force gives shear-stress magnitude τ(r) = ΔPr/(2L).
At the wall, τ_w = ΔPR/(2L). At the centre, stress magnitude is zero by symmetry. Stress is therefore not uniform across the pipe, even though the driving pressure gradient is uniform in the fully developed model.
This distribution follows from momentum balance, rather than from assuming a Newtonian velocity profile. The material law is needed next to connect stress with deformation. Keeping balance and constitutive response separate makes the derivation transferable to other adequately specified flow models.
Where τ(r) < τ_y, the ideal Bingham material has zero shear rate. If τ_w > τ_y, the boundary is r_p = 2Lτ_y/ΔP = Rτ_y/τ_w. The region 0 ≤ r < r_p is the plug; the region r_p < r ≤ R is the yielded annulus.
Define a = r_p/R = τ_y/τ_w. This is a radius fraction, not an area fraction. The plug occupies fraction a² of the pipe’s cross-sectional area.
If τ_w ≤ τ_y, there is no yielded annulus in this model. With the stationary no-slip wall and no other driving motion, the ideal steady solution has no flow. The threshold pressure drop is ΔP_y = 2Lτ_y/R. At the threshold itself, the yielded annulus has zero thickness and the model flow rate is zero.
Let u(r) be axial velocity. It decreases towards the stationary wall, so shear-rate magnitude is −du/dr. In the yielded annulus, μ_p(−du/dr) = ΔPr/(2L) − τ_y.
Integrate from r to R using u(R) = 0. The result is u(r) = ΔP[R² − r² − 2r_p(R − r)]/(4μ_pL), for r_p ≤ r ≤ R.
At r = r_p, the derivative is zero and the velocity is u_p = ΔP(R − r_p)²/(4μ_pL). Inside the plug, du/dr = 0, so every radius has this same axial velocity u_p.
Zero shear rate therefore means neighbouring layers within the core do not slide past one another. It does not require their common velocity to vanish. The annulus supplies the deformation connecting that moving core to the fixed wall.
Use fictional R = 0.010 m, L = 1.0 m, τ_y = 10 Pa and μ_p = 1.0 Pa·s. At ΔP = 4,000 Pa, wall stress is 4,000 × 0.010/2 = 20 Pa.
Then a = 10/20 = 0.50 and r_p = 0.005 m. The core occupies a² = 0.25, or 25%, of cross-sectional area. Its velocity is 4,000 × 0.005²/4 = 0.025 m/s.
| Position | Stress magnitude | Velocity under the model |
|---|---|---|
| Centre, r = 0 | 0 Pa | 0.025 m/s |
| Plug boundary, r = 0.005 m | 10 Pa | 0.025 m/s |
| Wall, r = 0.010 m | 20 Pa | 0 |
The centre has zero shear stress yet moves downstream. A statement that every location below yield must remain stationary would contradict this valid translational motion. The correct distinction is between motion and deformation.
Volume flow is Q = 2π∫u(r)r dr across the radius. Include a constant-velocity core from zero to r_p and the derived annular profile from r_p to R. Omitting the core because it has zero shear rate would omit real transported volume.
For 0 ≤ a < 1, integration gives Q = Q₀[1 − 4a/3 + a⁴/3], where Q₀ = πR⁴ΔP/(8μ_pL). Q₀ is the Newtonian reference flow using the same plastic viscosity and driving pressure. It is not the actual Bingham flow unless the yield contribution vanishes.
For a = 0.50, the factor is 1 − 2/3 + 1/48 = 17/48 ≈ 0.35417. Q₀ is about 15.708 mL/s, so Q is about 5.563 mL/s. Mean speed Q/(πR²) is about 0.01771 m/s, below the 0.025 m/s core speed.
The plug itself transports πr_p²u_p ≈ 1.963 mL/s, about 35.3% of the total. Its 25% area share and 35.3% flow share differ because the outer annulus includes slower material approaching the stationary wall.
CHAPTER 7 OF 12 · Calculate and compare
7. Change pressure without assuming proportional flow
For the same material and pipe, the threshold is ΔP_y = 2,000 Pa. Halving the main example’s pressure reaches that threshold and gives zero model flow, rather than half the original flow.
| Pressure drop | Wall stress | Radius fraction a | Q/Q₀ | Actual flow Q |
|---|---|---|---|---|
| 2,000 Pa | 10 Pa | 1.00 | 0 | 0 |
| 4,000 Pa | 20 Pa | 0.50 | About 0.35417 | About 5.563 mL/s |
| 8,000 Pa | 40 Pa | 0.25 | About 0.66797 | About 20.985 mL/s |
Doubling pressure from 4,000 to 8,000 Pa increases flow by about 3.77 times. The Newtonian reference doubles, while the dimensionless yield penalty also becomes smaller. The plug radius decreases and more of the cross-section shears.
This nonlinear response follows from the model’s yield term. It does not establish that a real material will retain identical properties, boundary behaviour and flow regime under a large pressure change.
CHAPTER 8 OF 12 · Calculate and compare
8. Change radius and inspect the threshold too
Newtonian reference flow contains R⁴, but the Bingham correction also depends on radius through a = 2Lτ_y/(ΔPR). Changing radius therefore changes both the reference scale and the yielded fraction.
Halve R from 0.010 to 0.005 m while retaining ΔP = 4,000 Pa and the other parameters. Wall stress becomes 10 Pa, exactly the model yield stress. The new threshold is 4,000 Pa, so the predicted steady no-slip flow is zero.
Simply dividing the original flow by sixteen would miss this threshold crossing. Before applying a familiar power law, check whether the changed geometry moves the problem into another regime. The pipe radius and diameter must also remain distinct; confusing them changes the force balance and the fourth-power reference substantially.
CHAPTER 9 OF 12 · Test the model
9. Connect the calculation to measurement evidence
A pressure–flow experiment needs suitable pressure locations, pipe dimensions, flow measurement, temperature, preparation and observation time. Fully developed flow excludes entrance-region complications. No slip must be justified rather than assumed because it makes the algebra convenient.
Wall slip can allow material translation without the no-slip threshold behaviour derived here. Time-dependent rebuilding or creep can also make a single ideal yield value inadequate. A smooth pressure–flow curve does not uniquely prove a Bingham mechanism.
The rotational rheometry guide examines a different measurement geometry. Parameters inferred there need an appropriate comparison before being used to predict a pipe experiment. The viscosity terminology guide helps preserve units and quantity definitions.
Is the plug a stationary blockage?
Not in this yielded-annulus solution. It moves at u_p without internal shear. A wholly unyielded cross-section attached to a stationary no-slip wall is a different case.
Is plastic viscosity the same as apparent viscosity?
No. In positive steady Bingham shear, apparent viscosity is τ/γ̇ = μ_p + τ_y/γ̇. It depends on rate. Substituting an unqualified apparent viscosity into a Newtonian pipe equation changes the model.
Can I use the same profile for a flat channel?
No. Geometry changes stress balance and the integration measure. Derive the appropriate channel relation instead of replacing radius with half-gap in the final pipe formula.
Does this calculation establish safe pumping conditions?
No. It describes an ideal flow field. Real equipment, material history, pressure limits, transients and application-specific requirements need their own evidence. The teaching values are not operating instructions.
CHAPTER 11 OF 12 · Practise and continue
11. Practise with a new pressure and viscosity
Keep the original pipe and material but use ΔP = 6,000 Pa. Wall stress is 30 Pa and a = 1/3. The plug radius is about 0.00333 m, and its area fraction is 1/9, about 11.1%.
The flow factor is 1 − 4/9 + 1/243 = 136/243 ≈ 0.55967. Q₀ is about 23.562 mL/s, so Q is about 13.187 mL/s. Plug speed is 6,000(0.010 − 0.00333…)²/4 ≈ 0.06667 m/s.
Now double only plastic viscosity to 2.0 Pa·s, preserving the ideal yield stress and driving pressure. Plug radius stays unchanged because the stress balance and yield condition do not contain μ_p. Velocities and flow halve. Explain this separation before calculating: stress selects the yielded regions, while plastic viscosity controls their rate response in this model.
Foundations and measurements
Return to Rheology for the subject map. Continue with rotational rheometry, rheology terminology and viscosity units. Use each measurement’s geometry and history when interpreting its parameters.
Authoritative further study
Cambridge’s non-Newtonian-fluid mechanics notes on yield problems discuss Bingham modelling, rigid regions and pressure thresholds. They also examine more complicated geometries, where the simple pipe account cannot be copied unchanged. The derivation and numerical pipe cases here are original educational examples rather than reported measurements from those notes.
Learning transfer
Connect to The Gold Standard of Education and How X Works. The transferable sequence is force balance, material law, boundary conditions, integration and a check of which assumptions survive the changed problem.
Sources checked 21 October 2026. Worked examples are educational models with the assumptions stated above.