Fluid mechanics: theory and checks

Pipe pressure loss: from conservation laws to Poiseuille flow

Connect mass and energy balances to a viscous solution; Reynolds number alone does not determine pressure loss.

Pressure-loss derivation and examples rechecked: 2026-10-22

The question and its assumptions

Re compares inertia and viscosity but does not give the pressure needed to drive a flow; that needs length, flow rate and wall conditions too. The derivation below assumes steady, incompressible flow of a Newtonian fluid with constant viscosity in a straight, horizontal, rigid circular pipe, with fully developed laminar flow and no slip at the wall.

Entrance effects, bends, valves, elevation changes and machines are excluded: the result describes one pipe segment, not an installation. Use the inside diameter in the Re calculator when checking the flow conditions. Example 3 leaves the laminar assumption and uses the Darcy–Weisbach equation with a Colebrook–White friction factor.

Variables and two different lengths

The diameter D and axial length ℓ play different roles.
VariableSI unitMeaning
ρkg/m³Density
μPa·sDynamic viscosity
D, r, ℓmInside diameter, radial coordinate, axial length
v, u(r)m/sMean speed and local axial speed
Qm³/sVolume flow rate
Δp, τwPaPressure drop and wall shear magnitude
g, hLm/s², mGravity and head loss
fD, ε–, mDarcy friction factor, absolute wall roughness

Steady mass conservation gives ṁ = ρAv. With constant density, Q = Av and A = πD²/4. At fixed Q a smaller diameter also changes the speed, so editing only D at fixed v 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.

u(r) = Δp(R² − r²)/(4μℓ)
Q = 2π∫₀ᴿu(r)r dr = πR⁴Δp/(8μℓ)

Area averaging gives v = Q/(πR²) = ΔpR²/(8μℓ), and the centerline speed is twice the mean. With R = D/2 this becomes Δp = 32μℓv/D². It is an exact solution under the stated assumptions, not an empirical fit.

Connecting pressure loss to the energy balance

For an incompressible segment, p₁/(ρg) + α₁v₁²/(2g) + z₁ + hpump = p₂/(ρg) + α₂v₂²/(2g) + z₂ + hturbine + hL, where α is the kinetic-energy correction factor: 2 for the parabolic laminar profile, not the uniform-profile value of 1. With equal diameter, profile and elevation those terms cancel, leaving hL = Δp/(ρg).

Comparing the Darcy definition Δp = fD(ℓ/D)(ρv²/2) with the viscous solution gives fD = 64/Re. A Fanning factor is one quarter of a Darcy factor, so mixing them creates a factor-of-four error. The wall shear stress follows from a force balance on the segment: τw = ΔpD/(4ℓ) = (fD/8)ρv². Omitting hL from a same-diameter horizontal balance would wrongly imply zero pressure drop.

Three pressure-loss calculations

Example 1. Laminar segment with rounded constants

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

Re = 1000; Q = πD²v/4 = 7.85398 × 10⁻⁶ m³/s

Δp = 32μℓv/D² = 64 Pa

Independent check: fD = 0.064, ℓ/D = 200 and ρv²/2 = 5 Pa → Δp = 64 Pa

hL = 0.006524 m; τw = ΔpD/(4ℓ) = 0.08 Pa

Mechanical dissipation in this segment: ΔpQ = 0.00050265 W

Interpretation. Dissipated power is not pump electrical input. The entrance length also matters: with Le/D = 0.065 Re from the MIT notes, Le ≈ 0.65 m, about a third of the 2 m segment. The developing region has higher wall shear, so 64 Pa is a lower bound if the pipe starts at a reservoir or a fitting.

Example 2. The same pipe with IAEA water at 60 °C

Given. IAEA Table 3.3, saturated water at 60 °C, used as low-pressure liquid: ρ = 983.2 kg/m³, μ = 4.677 × 10⁻⁴ Pa·s. Same assumed D = 0.01 m, ℓ = 2 m and v = 0.1 m/s.

Re = 983.2 × 0.1 × 0.01 / (4.677 × 10⁻⁴) ≈ 2102.2

If laminar: Δp = 32 × 4.677 × 10⁻⁴ × 2 × 0.1 / 0.01² ≈ 29.93 Pa, a ratio of 0.4677 to Example 1

τw ≈ 0.0374 Pa; ΔpQ ≈ 0.000235 W

Le = 0.065 × 2102.2 × 0.01 ≈ 1.37 m

Interpretation. In laminar flow Δp is proportional to μ, so warming the water more than halves the loss. But the result sits on a judgment boundary: Re ≈ 2102 is just above the 2100 that the WIPP model uses as its laminar limit, and below the 2300 of the MIT notes and the 2320 of the TU Delft text. The entrance length is also about two thirds of the pipe. Report 29.93 Pa with that caveat (laminar likely, long developing region, loss probably higher) and measure if it matters.

Example 3. Turbulent water flow with a Colebrook–White friction factor

Given. IAEA water at 20 °C: ρ = 998.2 kg/m³, μ = 1.0000 × 10⁻³ Pa·s. Assumed: D = 0.05 m, v = 1.5 m/s, ℓ = 10 m, g = 9.81 m/s². Two wall cases: hydraulically smooth (ε/D = 0) and an assumed relative roughness ε/D = 0.001.

Re = 998.2 × 1.5 × 0.05 / 0.001 = 74,865; ℓ/D = 200; ρv²/2 ≈ 1123 Pa

Colebrook–White, 1/√fD = −2 log₁₀(2.51/(Re√fD) + 0.27 ε/D), iterated from fD = 0.02: smooth fD ≈ 0.01913 (four iterations reach 0.0191); rough fD ≈ 0.0228

Blasius check (smooth, 2320 < Re < 10⁵): 0.3164 Re−0.25 ≈ 0.01913

Smooth: Δp = 0.01913 × 200 × 1123 ≈ 4.30 kPa; hL ≈ 0.439 m; τw ≈ 5.37 Pa

Rough: Δp ≈ 5.13 kPa

Q = π × 0.05² × 1.5 / 4 ≈ 2.945 × 10⁻³ m³/s; ΔpQ ≈ 12.7 W (smooth), 15.1 W (rough)

Wrongly using 64/Re: fD = 0.000855, Δp = 192 Pa, about 22.4 times too small

Interpretation. At this Re the friction factor no longer depends on μ alone, and roughness adds about 19% to the loss even at ε/D = 0.001. The explicit Swamee–Jain approximation listed by the TU Delft text gives 0.0190 and 0.0230, within about 1% of the iterated values, which is adequate for most sizing. Applying 32μℓv/D² without checking Re is wrong here by more than an order of magnitude. Real roughness values should come from the manufacturer or a handbook; 0.001 is an assumption.

Sensitivity and limits

At fixed D and v, doubling ℓ doubles pressure loss without changing Re. At fixed Q and ℓ in laminar flow, halving D raises Δp 16 times but also quadruples v and doubles Re, so check again whether the flow stays laminar and developed. In turbulent flow Δp grows roughly with v² (the MIT notes state that turbulent pressure changes vary as the square of mean velocity for a wide range of geometries), so the laminar scaling no longer holds.

Questions about pipe pressure loss and friction factors

How can I tell whether a friction factor is Darcy or Fanning?

For laminar flow in a circular pipe, a Darcy factor equals 64/Re and a Fanning factor 16/Re. A Moody-type chart that reads about 0.02 at Re 10⁵ in a smooth pipe is Darcy. If Δp comes out four times too large or small, the conventions have been mixed.

When is Δp = 32μℓv/D² acceptable?

For steady, fully developed laminar flow of a Newtonian fluid in a rigid circular pipe. Check Re against the threshold your reference uses (about 2100 to 2320), and check that the entrance length 0.065 Re D is short compared with the pipe.

Do I need to iterate the Colebrook–White equation?

It is implicit in fD, so a direct solution needs iteration; starting from 0.02, a few passes converge to four digits, as in Example 3. Explicit fits such as Swamee–Jain or, for smooth pipes below Re 10⁵, Blasius usually agree within about 1%.

Is ΔpQ the power my pump needs?

No. ΔpQ is the mechanical power dissipated in the segment. Pump shaft power adds the other segments, fittings and elevation, then divides by pump efficiency; motor and drive losses come on top.

Why is the kinetic-energy factor 2 for laminar flow?

The parabolic profile carries more kinetic energy than a uniform profile with the same mean speed; the cube of u(r) averaged over the area gives exactly twice v³. It cancels between two sections of the same pipe, but not across a change of diameter or flow regime.

References