Pipe pressure loss: from conservation laws to Poiseuille flow

The question and its assumptions

Reynolds number compares inertia and viscosity; it does not supply the pressure needed to drive a specified flow through a pipe. That question also needs length, flow rate and wall conditions. Consider steady, incompressible flow of a Newtonian fluid with constant viscosity in a straight, horizontal, rigid circular pipe. The velocity is fully developed, laminar and subject to no slip at the wall.

Entrance effects, bends, valves, elevation changes and machines are excluded here. The result describes a qualifying pipe segment, rather than a complete installation. Use the inside diameter in the Re calculator when checking the flow conditions.

Variables and two different lengths

Steady mass conservation gives ṁ = ρAv. With constant density, Q = Av and A = πD²/4. Reducing diameter at fixed Q changes the speed as well: a diameter-only edit at fixed speed is a different comparison.

Integrating the viscous momentum equation

Set R = D/2. The axial momentum equation reduces to 0 = −dp/dx + μ(1/r)d/dr(r du/dr). Integrate twice using a finite gradient at the axis and u(R) = 0.

Area averaging gives v = Q/(πR²) = ΔpR²/(8μℓ). The centerline speed is twice the mean. Replacing R by D/2 yields Δp = 32μℓv/D². This is a solution under the stated assumptions, rather than an empirical turbulent-flow fit.

Connecting pressure loss to the energy balance

For an incompressible flow segment, write p₁/(ρg) + α₁v₁²/(2g) + z₁ + hpump = p₂/(ρg) + α₂v₂²/(2g) + z₂ + hturbine + hL. Here α denotes the kinetic-energy correction factor. It is 2 for a fully developed parabolic laminar profile, rather than the uniform-profile value of 1. Equal diameter, profile and elevation make those terms cancel, leaving hL = Δp/(ρg).

Compare the Darcy definition Δp = fD(ℓ/D)(ρv²/2) with the viscous solution to obtain fD = 64/Re. A Fanning factor is one quarter of a Darcy factor. Mixing them creates a factor-of-four error. Omitting hL from a same-diameter horizontal energy balance would incorrectly imply zero pressure drop.

An independently constructed worked example

Use illustrative constants ρ = 1000 kg/m³, μ = 0.001 Pa·s, D = 0.01 m, ℓ = 2 m, v = 0.1 m/s and g = 9.81 m/s². These are not a quoted property table.

Dissipated mechanical power is not the electrical input to a pump. Other losses and pump efficiency would need to be included.

Sensitivity and limits

At fixed D and v, doubling ℓ doubles pressure loss without changing Re. At fixed Q and ℓ, halving D increases the required pressure drop by a factor of 16, but also quadruples v and doubles Re. Recheck whether the new flow remains laminar and developed before using the same solution.

• Convert millimeters to meters before entering diameter.
• Add entrance, fitting and elevation contributions to the appropriate balance.
• Do not extend this solution directly to non-Newtonian flow, turbulent flow or slip boundaries.