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Pipe friction from a clamp-on velocity reading

Piping · Reynolds number, friction factor and system head

A booster set fills a rooftop storage tank through 90 m of 100 mm commercial steel riser, and the only instrument on site is a clamp-on ultrasonic meter, which reads a steady 2.00 m/s in the pipe. The tank water level stands 12 m above the pump. Take the water at 1000 kg/m³ and 1.00 mPa·s, and the roughness of commercial steel as 0.045 mm.

Given
  • v = 2 m/s — Measured pipe velocity
  • D = 100 mm — Inside diameter
  • L = 90 m — Riser length
  • ρ = 1000 kg/m³ — Water density
  • μ = 1 mPa·s — Water viscosity
  • ε = 0.045 mm — Commercial steel roughness
  • h_s = 12 m — Static lift to the tank level
Determine
  1. (a)the flow the riser is carrying
  2. (b)the Reynolds number of the flow
  3. (c)the Darcy friction factor
  4. (d)the friction head lost over the 90 m
  5. (e)the total head the booster pump is developing
Step 1 of 6(a) · solve for Flow rate

The meter reads velocity but everything is bought and billed in flow, so turn 2.0 m/s through the 78.5 cm² bore into 15.7 L/s — about 249 gpm — first. The classic slip is using the pipe's nominal size: a 100 mm nominal steel pipe's true bore differs by schedule, and Q goes with the square of it.

Rearranged for Q
Q=πD24vQ = \tfrac{\pi D^{2}}{4} v
Your values, in your units
Q=π⋅(100 mm)24⋅(2 m/s)Q = \tfrac{\pi \cdot \left( 100\ \text{mm} \right)^{2}}{4} \cdot \left( 2\ \text{m/s} \right)
Answer
Q=942.48 L/minQ = 942.48\ \text{L/min}

Carried onward at full precision, not this rounded figure.

Open the Pipe Velocity from Flow and Diameter solver →

Step 2 of 6(b) · solve for Reynolds number

Re = ρvD/μ = 200,000 — deep into turbulence, which is what licenses the turbulent friction-factor correlation in the next step. Check this before reaching for any friction formula: below about 2,300 the whole machinery changes to f = 64/Re, and roughness stops mattering entirely.

Rearranged for Re
Re=ρvDμRe = \tfrac{\rho v D}{\mu}
Your values, in your units
Re=(1,000 kg/m3) (2 m/s) (100 mm)(1 mPa⋅s)Re = \tfrac{\left( 1{,}000\ \text{kg/m}^{3} \right) \, \left( 2\ \text{m/s} \right) \, \left( 100\ \text{mm} \right)}{\left( 1\ \text{mPa}{\cdot}\text{s} \right)}
Converted to base units
Re=(1,000 kg/m3) (2 m/s) (100 mm)(0.001 Pa⋅s)Re = \tfrac{\left( 1{,}000\ \text{kg/m}^{3} \right) \, \left( 2\ \text{m/s} \right) \, \left( 100\ \text{mm} \right)}{\left( 0.001\ \text{Pa}{\cdot}\text{s} \right)}
Answer
Re=200,000Re = 200{,}000

Carried onward at full precision, not this rounded figure.

Open the Reynolds Number solver →

Step 3 of 6(c) · solve for Darcy friction factor

Swamee–Jain is the explicit stand-in for the Colebrook equation — no iteration, within about 1% — and it needs both the relative roughness (0.045/100 = 0.00045) and the Reynolds number from step 2. Watch the ratio: ε and D must be in the SAME unit before dividing, or the factor comes out wild.

Rearranged for f
f=0.25[log⁡10 ⁣(ε3.7D+5.74Re0.9)]2f = \tfrac{0.25}{\left[\log_{10}\!\left(\tfrac{\varepsilon}{3.7D} + \tfrac{5.74}{Re^{0.9}}\right)\right]^{2}}
200,000 —carried from step 2
Your values, in your units
f=0.25[log⁡10 ⁣((0.045 mm)3.7⋅(100 mm)+5.74(200,000)0.9)]2f = \tfrac{0.25}{\left[\log_{10}\!\left(\tfrac{\left( 0.045\ \text{mm} \right)}{3.7 \cdot \left( 100\ \text{mm} \right)} + \tfrac{5.74}{\left( 200{,}000 \right)^{0.9}}\right)\right]^{2}}
Answer
f=18.665 per millef = 18.665\ \text{per mille}

Carried onward at full precision, not this rounded figure.

Open the Swamee–Jain Friction Factor solver →

Step 4 of 6(d) · solve for Friction head loss

Darcy–Weisbach assembles the loss: f × (L/D) × v²/2g ≈ 3.43 m over the 90 m run — about 3.8 m per 100 m, a heavy but plausible number at 2 m/s. Note the loss grew from the measured velocity squared: had the reading been 2.5 m/s the loss would be 56% larger, not 25%.

Rearranged for h_f
hf=fLDv22gh_f = f \tfrac{L}{D} \tfrac{v^{2}}{2g}
18.665 per millecarried from step 3
Your values, in your units
hf=(0.0186652)⋅(90 m)(100 mm)⋅(2 m/s)22(9.80665 m/s2)h_f = \left( 0.0186652 \right) \cdot \tfrac{\left( 90\ \text{m} \right)}{\left( 100\ \text{mm} \right)} \cdot \tfrac{\left( 2\ \text{m/s} \right)^{2}}{2\left( 9.80665\ \text{m/s}^{2} \right)}
Answer
hf=3.426 mh_f = 3.426\ \text{m}

Carried onward at full precision, not this rounded figure.

Open the Darcy–Weisbach Head Loss solver →

Step 5 of 6(e) · solve for Velocity head

The last head component is the kinetic energy the water keeps: v²/2g = 0.204 m. It is small next to friction here, and the temptation is to drop it — resist, because on short high-velocity runs it is the biggest term after static, and the habit of computing it costs one line.

Rearranged for h_v
hv=v22gh_v = \tfrac{v^{2}}{2g}
Your values, in your units
hv=(2 m/s)22(9.80665 m/s2)h_v = \tfrac{\left( 2\ \text{m/s} \right)^{2}}{2\left( 9.80665\ \text{m/s}^{2} \right)}
Answer
hv=203.94 mmh_v = 203.94\ \text{mm}

Carried onward at full precision, not this rounded figure.

Open the Velocity Head (h = v²/2g) solver →

Step 6 of 6 · solve for Total dynamic head

Static plus friction plus velocity head: 12 + 3.426 + 0.204 ≈ 15.63 m. This is the point on the pump curve the booster is actually operating at, reconstructed from one ultrasonic reading and a tape measure.

Rearranged for TDH
TDH=hs+hf+hvTDH = h_s + h_f + h_v
3.426 mcarried from step 4
203.94 mmcarried from step 5
Your values, in your units
TDH=(12 m)+(3.42598 m)+(0.203943 m)TDH = \left( 12\ \text{m} \right) + \left( 3.42598\ \text{m} \right) + \left( 0.203943\ \text{m} \right)
Answer
TDH=15.63 mTDH = 15.63\ \text{m}

Carried onward at full precision, not this rounded figure.

Open the Total Dynamic Head solver →

The riser carries about 15.7 L/s at Re = 200,000, the friction factor is 0.0187, the run loses 3.43 m to friction, and with 12 m of static lift the booster is developing about 15.6 m of total head.

Why this order

This chain is the diagnostic a service tech actually runs when the only trustworthy number on site is a clamp-on velocity. The order is forced by dependency: flow is just geometry, but the friction factor needs the Reynolds number first, because f is not a property of the pipe — it is a property of the pipe AND the flow through it. Re = 200,000 says turbulent, which licenses Swamee–Jain; the same pipe at a trickle would be laminar and take f = 64/Re instead, where the roughness that dominates here would not matter at all. Darcy–Weisbach then assembles head loss from the factor, and the last step adds the three head components the way every pump curve expects them. The cross-check on the friction factor is to invert it: 10^(−0.5/√f) − 5.74/Re^0.9 should hand back ε/3.7D, and it does.

The two reliable wrecks are both ratios. Relative roughness ε/D must be formed from a single unit — 0.045 mm over 100 mm is 0.00045, but 0.045 mm over 0.1 m read carelessly is 0.45, and the factor that comes back describes a pipe lined with gravel. And the L/D in Darcy–Weisbach is 900 here, a pure number, so a diameter left in millimetres while the length sits in metres inflates the loss a thousandfold — an error so large it usually gets caught, unlike the roughness one, which quietly doubles the answer. Worth keeping for the field: at 2 m/s in clean commercial steel, friction lands near 4 m per 100 m of run; a reading far off that, with this chain in hand, is how you find the throttled valve or the fouled strainer nobody logged.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.