Service call: a rooftop storage tank that fills slower than it used to, and a booster set in the basement that fills it through 90 m of 100 mm commercial steel riser. The only instrument on site is a clamp-on ultrasonic meter, so the tech straps its transducers to a clean straight length of the riser, gels the contact faces, and waits for the signal to lock: the meter reads a steady 2.00 m/s in the pipe. A tape drop settles the static side of the story — the tank water level stands 12 m above the pump. The rest comes off the reference card: take the water at 1,000 kg/m³ and 1.00 mPa·s, and the roughness of commercial steel as 0.045 mm. From one velocity reading and a tape measure, the booster's whole operating point can be reconstructed.
Every number in this problem is editable — change any value below and the whole chain recalculates.
Given
v = 2 m/s — Measured pipe velocity
D = 100 mm — Inside diameter
L = 90 m — Riser length
ρ = 1,000 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
(a)the flow the riser is carrying
(b)the Reynolds number of the flow
(c)the Darcy friction factor
(d)the friction head lost over the 90 m
(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=4πD2v
Your values, in your units
Q=4π⋅(100mm)2⋅(2m/s)
Answer
Q=942.48L/min
Carried onward at full precision, not this rounded figure.
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=μρvD
Your values, in your units
Re=(1mPa⋅s)(1,000kg/m3)(2m/s)(100mm)
Converted to base units
Re=(0.001Pa⋅s)(1,000kg/m3)(2m/s)(100mm)
Answer
Re=200,000
Carried onward at full precision, not this rounded figure.
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.
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%.
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=2gv2
Your values, in your units
hv=2(9.80665m/s2)(2m/s)2
Answer
hv=203.94mm
Carried onward at full precision, not this rounded figure.
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+hv
3.426 mcarried from step 4
203.94 mmcarried from step 5
Your values, in your units
TDH=(12m)+(3.42598m)+(0.203943m)
Answer
TDH=15.63m
Carried onward at full precision, not this rounded figure.
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.
