Calculate pressure drop from flow rate

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Pipe Head Loss Calculator

The Pipe Head Loss Calculator carries out the whole chain, from velocity through the Colebrook iteration to the total pressure drop, in one pass, and splits the result into major and minor losses. The embed below is loaded with this example so each hand step can be read against its output.

Problem

Water at $20^{\circ}$ C ( $ρ = 998$ kg/m $^{3}$ , $ν = 1.0 \times 1 0^{- 6}$ m $^{2}$ /s) flows at $Q = 4$ L/s through $L = 80$ m of $D = 50$ mm commercial steel pipe with roughness $ε = 0.046$ mm. The line has two standard $9 0^{\circ}$ elbows, $K = 1.5$ each, so the total fitting coefficient is $K = 3$ . Find the pressure drop.

Because the flow is given, no iteration over the flow is needed. The only implicit step is the friction factor itself.

Step 1: velocity and Reynolds number

Velocity comes from continuity and sets the flow regime. Everything after this depends on whether the flow is laminar or turbulent.

1Area: $A = π (0.05)^{2} /4 = 1.963 \times 1 0^{- 3}$ m $^{2}$ .
2Velocity: $v = Q / A = 0.004/1.963 \times 1 0^{- 3} = 2.037$ m/s.
3Reynolds number: $R e = v D / ν = 2.037 (0.05) /1 0^{- 6} = 101, 900$ . That is far above $4000$ , so the flow is turbulent and Colebrook-White applies.

Step 2: friction factor

The relative roughness is $ε / D = 0.046/50 = 0.00092$ . Colebrook-White, $\frac{1}{f} = - 2 lo g_{10} (\frac{ε}{3.7 D} + \frac{2.51}{R e f})$ , is implicit in $f$ . Starting from the explicit Swamee-Jain estimate $f = 0.0220$ and iterating converges to $f = 0.02186$ in a few passes. The Moody chart would read about $0.022$ at these values.

Step 3: major and minor losses

The dynamic pressure is $\frac{ρ v ^{2}}{2} = \frac{998 ( 2.037 ) ^{2}}{2} = 2071$ Pa. Both loss terms are multiples of it.

1Major loss: $Δ P_{f} = f \frac{L}{D} \frac{ρ v ^{2}}{2} = 0.02186 (1600) (2071) = 72, 440$ Pa.
2Minor loss: $Δ P_{m} = K \frac{ρ v ^{2}}{2} = 3 (2071) = 6, 213$ Pa.
3Total: $Δ P = 72, 440 + 6, 213 = 78, 650$ Pa, or $78.7$ kPa, which is $11.4$ psi.
4As a head of water: $h = Δ P / (ρ g) = 78, 650/ (998 \cdot 9.81) = 8.03$ m. A pump serving this line must supply at least that, plus any static lift.

Reading the result

Friction in the straight pipe accounts for $92%$ of the drop, because the run is $1600$ diameters long. Two elbows contribute $8%$ . In a short manifold with many fittings the proportions would reverse. The embedded Pipe Head Loss Calculator shows the same split as separate major and minor head lines.

Common Pitfall

The Darcy friction factor and the Fanning friction factor differ by a factor of four: $f_{D a rcy} = 4 f_{F annin g}$ . A Fanning value near $0.0055$ put into the Darcy-Weisbach equation as written gives a pressure drop four times too small. Charts and tables label which one they use, and the equation here, with $L / D$ and $ρ v^{2} /2$ , takes the Darcy factor.

Try a Variation

Double the flow to $Q = 8$ L/s and, before computing, predict the new pressure drop. If it doubled with the flow it would be about $157$ kPa. Both loss terms contain $v^{2}$ , so the real result is close to four times the original. Run the new flow through the Pipe Head Loss Calculator and see how far the drop is from a doubling, and why $f$ falls slightly as the flow rises.

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Worked ExampleEngineeringPipe Head Loss Calculator

Calculate flow rate from pressure and pipe diameter

Pressure and diameter alone do not fix a flow rate. What the pressure is doing decides the method: a known velocity gives $Q = A v$ , an orifice turns a pressure difference into flow, and a long pipe loses pressure to friction and has to be solved by iteration.

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