That is 8.343873 ft of head in this liquid. Straight pipe accounts for 2.995263 psi; fittings account for 0.614621 psi.
1. Calculator Inputs
A horizontal water line carries 50 US gpm through 150 ft of NPS 2 Schedule 40 pipe. Find the loss through the pipe and a group of fittings with a specified total K of 4. Use the actual 2.067 in inside diameter.
| Input | Enter this value |
|---|---|
| Solve for | Pressure drop, ΔP |
| Pressure unit | psi |
| Flow | 50 US gpm |
| Inside diameter | 2.067 in |
| Pipe length | 150 ft |
| Absolute roughness | 0.00015 ft, or 0.0018 in |
| Density | 62.3 lbm/ft³ |
| Dynamic viscosity | 1.00 cP |
| Friction factor | Automatic from Re and roughness |
| Fittings | One “Custom fitting / equipment (enter K)”; quantity 1; K = 4 |
| Fitting velocity basis | Same 2.067 in bore as the pipe |
Remove any preloaded fittings first. K = 4 is an explicit example assumption, not a universal value for four elbows. There is no elevation rise, equipment loss, or imposed outlet-pressure requirement in this example.
2. Work through the math
Convert flow, diameter and viscosity
The US gallon contains 231 in³. Keep diameter in feet when calculating area in ft².
Find area and velocity
Find Reynolds number and relative roughness
This is turbulent flow. The automatic method uses the Darcy friction factor from Colebrook–White.
Solve the Colebrook equation
Start at f = 0.020 and repeat the second expression. These are Darcy factors, not Fanning factors.
| Iteration | Darcy f |
|---|---|
| 1 | 0.022546346 |
| 2 | 0.022374855 |
| 3 | 0.022385567 |
| 4 | 0.022384894 |
| 5 | 0.022384937 |
| 6 | 0.022384934 |
Calculate straight-pipe and fitting head losses
Convert head to pressure
At standard gravity, a density of 62.3 lbm/ft³ corresponds to a specific weight γ of 62.3 lbf/ft³. The factor 144 converts ft² to in².
For direct calculations with mass density in lbm/ft³, include gc = 32.17404856 lbm·ft/(lbf·s²): ΔP = (fL/D + K)ρV²/(2gc × 144). Do not use lbm as though it were lbf.
3. Compare your answers
| Checkpoint | Reference answer | Your check |
|---|---|---|
| Area | 0.0233 ft² | ________ |
| Velocity | 4.78 ft/s (1.46 m/s) | ________ |
| Reynolds number | 76,300 | ________ |
| Darcy f | 0.0224 | ________ |
| Straight-pipe loss | 3 psi | ________ |
| Fitting loss | 0.615 psi | ________ |
| Total pressure loss | 3.61 psi | ________ |
| Head loss | 8.34 ft (2.54 m) | ________ |
Live calculator check, October 19, 2026: 3.61 psi total, 3 psi pipe, and 0.615 psi fittings. The calculator displays velocity and head in SI even when pressure is in psi; the equivalent US values are shown above. Compare within 0.001 psi and 0.01 ft when hand rounding.
Quick sanity check: with the same diameter, flow, roughness and fittings, doubling only the straight length doubles the pipe loss, not the fitting loss. The new total would be 6.61 psi.
4. Assumptions and sources
This is a full, circular pipe containing a steady, single-phase Newtonian liquid. Density, viscosity, bore and roughness are constant. The specified values are illustrative water properties near room temperature. No elevation term or pump head is hidden in the friction result.
If your answer differs, check actual bore versus nominal size, Darcy versus Fanning factor, cP conversion, and accidental extra fittings. A factor-of-four discrepancy often points to the friction-factor convention.
Method reference: DOE Fundamentals Handbook, fluid flow. Site references: pipe dimensions, roughness, and Darcy versus Fanning. The numerical case was calculated separately from the website code and then compared with the live calculator. Agreement checks this case, not every possible input.
Keep checking
Browse all worked examples or choose the right pipe calculator.