Unit 03 · Lesson 08 of 09

Why does pressure fall along a pipe?

A small diameter can demand a large pressure.

Explore the model. Derive the result. Check its limits.

What you’ll be able to do

Derive the laminar flow rate.
Use a common pressure-loss form.
Turbulence introduces roughness.

Before you touch the controls

At fixed volume flow in laminar motion, what happens to pressure drop when diameter halves?

An experiment you can change

Let diameter reshape the required pressure

Predict the direction of change, move one control, then explain the response using the equation.

Diameter0.02 mVolume flow0.02 L/sRoughness0.001 mm

V=\frac{4\dot V}{\pi D^2},\quad \mathrm{Re}=1273.2,\quad f_D=0.050265,\quad \Delta p=(0.050265)\frac{10}{0.02}\frac{1000(0.063662)^2}{2\times1000}=0.05093\,\mathrm{kPa}

In laminar flow, substituting 64/Re yields Δp = 128 μ L volume-flow/(π D⁴). Roughness does not enter that exact laminar result.

Model assumptions and units

Water ρ = 1000 kg/m³, μ = 0.001 Pa s; straight 10 m circular pipe, fully developed flow. Darcy factor, not Fanning factor. Haaland approximation used only above Re = 4000; transition is bracketed.

\Delta p=f_D\frac LD\frac{\rho V^2}{2},\quad f_D=\frac{64}{\mathrm{Re}}\ (\text{laminar}),\quad \frac1{\sqrt{f_D}}=-1.8\log_{10}\!\left[\left(\frac{\varepsilon/D}{3.7}\right)^{1.11}+\frac{6.9}{\mathrm{Re}}\right]

Derive the laminar flow rate.

Integrate the parabolic velocity profile over concentric rings. With $R=D/2$ this gives the Hagen–Poiseuille relation. It applies to fully developed laminar Newtonian flow in a straight circular pipe.

\dot V=\int_0^R u(r)2\pi r\,dr=\frac{\pi R^4\Delta p}{8\mu L}=\frac{\pi D^4\Delta p}{128\mu L}

Use a common pressure-loss form.

Darcy–Weisbach expresses pressure drop using the Darcy friction factor, pipe length/diameter, and mean-speed dynamic pressure. Substituting laminar flow gives $f_D=64/\mathrm{Re}$ . Fanning factors are one quarter of Darcy factors.

\Delta p=f_D\frac LD\frac{\rho V^2}{2},\quad f_D=64/\mathrm{Re}

Turbulence introduces roughness.

For turbulent fully developed flow, friction factor depends on Reynolds number and relative roughness. The displayed Haaland formula approximates the implicit Colebrook relation; logarithms are base ten. It is not a transition-region law.

\frac1{\sqrt{f_D}}\approx-1.8\log_{10}\left[\left(\frac{\varepsilon/D}{3.7}\right)^{1.11}+\frac{6.9}{\mathrm{Re}}\right]

State what is omitted.

The model includes straight-pipe friction only. Entrances, valves, bends, exits, developing profiles, and varying properties can need additional terms. Check regime again whenever diameter or flow changes.

h_L=f_D\frac LD\frac{V^2}{2g}+\sum K\frac{V^2}{2g}

A worked example

A laminar water line.

Take $\dot V=0.020\,\mathrm{L/s}=2\times10^{-5}\,\mathrm{m^3/s}$ , $D=0.020\,\mathrm m$ , $L=10\,\mathrm m$ , and $\mu=0.001\,\mathrm{Pa\,s}$ .
Mean speed is $0.06366\,\mathrm{m/s}$ and Reynolds number is about 1273, so laminar theory applies.
Pressure drop is $128(0.001)(10)(2\times10^{-5})/[\pi(0.020)^4]=50.93\,\mathrm{Pa}$ . Halving diameter would multiply the laminar expression by 16, but also doubles Re and may invalidate the regime assumption.

Make the idea your own

Calculate. Compare. Test the assumptions.

Work out your answer, check it, then explore the visual solution. The answer checker uses a 1.5% rounding tolerance; for model-limit questions it checks the displayed model result.

1 · Calculate

A laminar pressure calculation

For water in a 10 m pipe, $D = 0.020$ m and $flow = 0.020$ L/s, find pressure drop in kPa. Roughness is 0.001 mm.

Hint

Find mean speed and Re first; then use 64/Re or Poiseuille.

Explore the visual solution

Diameter0.02 mVolume flow0.02 L/sRoughness0.001 mm

V=\frac{4\dot V}{\pi D^2},\quad \mathrm{Re}=1273.2,\quad f_D=0.050265,\quad \Delta p=(0.050265)\frac{10}{0.02}\frac{1000(0.063662)^2}{2\times1000}=0.05093\,\mathrm{kPa}

In laminar flow, substituting 64/Re yields Δp = 128 μ L volume-flow/(π D⁴). Roughness does not enter that exact laminar result.

Model assumptions and units

Water ρ = 1000 kg/m³, μ = 0.001 Pa s; straight 10 m circular pipe, fully developed flow. Darcy factor, not Fanning factor. Haaland approximation used only above Re = 4000; transition is bracketed.

\Delta p=f_D\frac LD\frac{\rho V^2}{2},\quad f_D=\frac{64}{\mathrm{Re}}\ (\text{laminar}),\quad \frac1{\sqrt{f_D}}=-1.8\log_{10}\!\left[\left(\frac{\varepsilon/D}{3.7}\right)^{1.11}+\frac{6.9}{\mathrm{Re}}\right]

Controls explore variations. The steps below retain the original question’s values.

Water ρ = 1000 kg/m³, μ = 0.001 Pa s; straight 10 m circular pipe, fully developed flow. Darcy factor, not Fanning factor. Haaland approximation used only above Re = 4000; transition is bracketed.

\Delta p=f_D\frac LD\frac{\rho V^2}{2},\quad f_D=\frac{64}{\mathrm{Re}}\ (\text{laminar}),\quad \frac1{\sqrt{f_D}}=-1.8\log_{10}\!\left[\left(\frac{\varepsilon/D}{3.7}\right)^{1.11}+\frac{6.9}{\mathrm{Re}}\right]

2 · Compare

Recheck after narrowing

Use $D = 0.010$ m with the same 0.020 L/s flow and roughness. Inspect Reynolds number and report the displayed comparison estimate.

Hint

At fixed flow halving diameter doubles Re; it enters transition here.

Explore the visual solution

Diameter0.01 mVolume flow0.02 L/sRoughness0.001 mm

V=\frac{4\dot V}{\pi D^2},\quad \mathrm{Re}=2546.5,\quad f_D=0.046885,\quad \Delta p=(0.046885)\frac{10}{0.01}\frac{1000(0.25465)^2}{2\times1000}=1.5201\,\mathrm{kPa}

Transition: shown pressure drop is the larger of laminar and turbulent estimates, a comparison of two estimates rather than a validated transition correlation.

Model assumptions and units

Water ρ = 1000 kg/m³, μ = 0.001 Pa s; straight 10 m circular pipe, fully developed flow. Darcy factor, not Fanning factor. Haaland approximation used only above Re = 4000; transition is bracketed.

\Delta p=f_D\frac LD\frac{\rho V^2}{2},\quad f_D=\frac{64}{\mathrm{Re}}\ (\text{laminar}),\quad \frac1{\sqrt{f_D}}=-1.8\log_{10}\!\left[\left(\frac{\varepsilon/D}{3.7}\right)^{1.11}+\frac{6.9}{\mathrm{Re}}\right]

Controls explore variations. The steps below retain the original question’s values.

Water ρ = 1000 kg/m³, μ = 0.001 Pa s; straight 10 m circular pipe, fully developed flow. Darcy factor, not Fanning factor. Haaland approximation used only above Re = 4000; transition is bracketed.

\Delta p=f_D\frac LD\frac{\rho V^2}{2},\quad f_D=\frac{64}{\mathrm{Re}}\ (\text{laminar}),\quad \frac1{\sqrt{f_D}}=-1.8\log_{10}\!\left[\left(\frac{\varepsilon/D}{3.7}\right)^{1.11}+\frac{6.9}{\mathrm{Re}}\right]

3 · Challenge the model

Rough turbulent pipe

Use $D = 0.005$ m, $flow = 0.10$ L/s, $roughness = 0.050$ mm, and length 10 m. Find pressure drop using the displayed Haaland Darcy factor.

Hint

Relative roughness is ε/D after converting both lengths to meters.

Explore the visual solution

Diameter0.005 mVolume flow0.1 L/sRoughness0.05 mm

V=\frac{4\dot V}{\pi D^2},\quad \mathrm{Re}=25465,\quad f_D=0.040098,\quad \Delta p=(0.040098)\frac{10}{0.005}\frac{1000(5.093)^2}{2\times1000}=1040.1\,\mathrm{kPa}

In laminar flow, substituting 64/Re yields Δp = 128 μ L volume-flow/(π D⁴). Roughness does not enter that exact laminar result.

Model assumptions and units

Water ρ = 1000 kg/m³, μ = 0.001 Pa s; straight 10 m circular pipe, fully developed flow. Darcy factor, not Fanning factor. Haaland approximation used only above Re = 4000; transition is bracketed.

\Delta p=f_D\frac LD\frac{\rho V^2}{2},\quad f_D=\frac{64}{\mathrm{Re}}\ (\text{laminar}),\quad \frac1{\sqrt{f_D}}=-1.8\log_{10}\!\left[\left(\frac{\varepsilon/D}{3.7}\right)^{1.11}+\frac{6.9}{\mathrm{Re}}\right]

Controls explore variations. The steps below retain the original question’s values.

Water ρ = 1000 kg/m³, μ = 0.001 Pa s; straight 10 m circular pipe, fully developed flow. Darcy factor, not Fanning factor. Haaland approximation used only above Re = 4000; transition is bracketed.

\Delta p=f_D\frac LD\frac{\rho V^2}{2},\quad f_D=\frac{64}{\mathrm{Re}}\ (\text{laminar}),\quad \frac1{\sqrt{f_D}}=-1.8\log_{10}\!\left[\left(\frac{\varepsilon/D}{3.7}\right)^{1.11}+\frac{6.9}{\mathrm{Re}}\right]

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