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Hydraulics, open channels and fluid power
Pipe wall shear stress from frictional pressure loss
Average shear stress on the wall of a horizontal straight circular pipe, from the frictional pressure drop Δp over length L and diameter D.
- Ref
- HYD-07
- Rev
- v1
- Field
- Hydraulics, open channels and fluid power
- Result
- τw [Pa]
A force balance on a length L of straight horizontal pipe gives Δp π D²/4 = τw π D L, so the average wall shear stress is τw = Δp D/(4L). It holds for laminar and turbulent flow alike, since it comes from a force balance and not from a velocity profile. The pressure drop must be due to friction alone, with no change in height or speed.
Average shear stress on the wall of a horizontal straight circular pipe, from the frictional pressure drop Δp over length L and diameter D. A force balance gives it for laminar or turbulent flow.
Calculator
- Pressure gradient, Δp/L400 Pa/m
- Wall friction force over the length, τw π D L7.854 N
- Friction velocity for water, √(τw/1000)0.07071 m/s
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A drop of 4,000 Pa over 10 m in a 50 mm pipe gives a wall shear of 5 Pa.
- The same stress follows from τw = f ρ V²/8 with the Darcy friction factor, so a measured pressure drop gives f directly.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (Pa) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Δp Frictional pressure drop | 1 to 1e7 Pa | 0.00125 | 12,500 | 12,500 | +10% | +10 % |
| L Pipe length | 0.1 to 10,000 m | 0.005 | 500 | 500 | +10% | −9.09 % |
| D Pipe diameter | 0.002 to 5 m | 0.2 | 500 | 499.8 | +10% | +10 % |
Sweep of Frictional pressure drop
| Frictional pressure drop (Pa) | Wall shear stress (Pa) |
|---|---|
| 1 | 0.00125 |
| 7.499 | 0.009374 |
| 56.23 | 0.07029 |
| 421.7 | 0.5271 |
| 3,162 | 3.953 |
| 23,710 | 29.64 |
| 177,800 | 222.3 |
| 1,334,000 | 1,667 |
| 1e7 | 12,500 |
Sweep one variable
Variables and how each one contributes
Wall shear stress
What the equation gives you. Everything else feeds it.
Frictional pressure drop
The frictional pressure drop over the length, in Pa. Raises the wall shear in direct proportion. Raising it by 10 % (to 4,400) moves the result by +10 %, from 5 to 5.5 Pa.
Pipe diameter
The pipe internal diameter, in m. A larger pipe needs more shear to balance the same pressure force per length, so the shear rises in direct proportion. Raising it by 10 % (to 0.055) moves the result by +10 %, from 5 to 5.5 Pa.
Pipe length
The length over which the pressure drops, in m. A longer length spreads the pressure drop, so the shear falls in inverse proportion. Raising it by 10 % (to 11) moves the result by −9.09 %, from 5 to 4.545 Pa.
Worked example with the values above
- Pressure gradient: Δp/L = 4,000 / 10 = 400 Pa/m.
- Multiply by D/4 = 0.0125 m.
- τw = 5 Pa.
Assumptions and limits
- Steady, fully developed laminar flow of a Newtonian liquid with no slip at the walls.
- Smooth, straight geometry, and no body force other than gravity where stated.
Common mistakes
- Using these results for turbulent flow, which has a flatter profile.
- Entering the diameter where the radius is needed, which changes the answer by a factor of two or four.
- Mixing the sign of the pressure gradient: it is −dp/dx, positive when pressure falls along the flow.
Statistics and uncertainty
These laminar formulas are exact for their model, so the uncertainty comes from the inputs. Gap and film thickness enter with the third power, so a 1 % error in thickness gives about a 3 % error in flow.
Where it is used
- Lubrication films, coating, sealing gaps and thin-film runoff.
- Checking viscous pipe flow and wall shear from measured pressure drops.
Maths you need
- Viscosity: μ, the shear stress per unit velocity gradient
- No-slip condition: the liquid moves with the wall at the wall
- Reynolds number: the ratio of inertial to viscous forces