Pipe friction & head loss: practice problems with step-by-step solutions

1What head loss is

Bernoulli’s equation assumes the fluid flows without friction. In a real pipe the water rubs against the wall and against itself, and some of its energy turns into heat.

Engineers measure this lost energy as a height of fluid, called head loss, in metres. A head loss of 3 m means the water lost as much energy as it would need to climb 3 m.

You can see it with two thin tubes (standpipes) on a horizontal pipe. The water rises less high in the downstream tube. The difference is the head loss between them.

Head loss is why long pipelines need pumps, and why a garden hose gives a weaker jet when it is very long.

A head loss converts to a pressure drop with Δp = ρ · g · h_f. For water, 1 m of head is about 9.81 kPa.

2Reynolds number: laminar or turbulent

Slow flow in a thin pipe moves in smooth layers. This is laminar flow. Fast flow in a wide pipe swirls and mixes. This is turbulent flow.

The Reynolds number tells you which one you have. It compares the push of the moving fluid with its stickiness (viscosity):

• V: the mean velocity in m/s, V = Q/A.
• D: the inside diameter in m.
• ν: the kinematic viscosity in m²/s (about 1.0 × 10⁻⁶ m²/s for water at 20 °C), or μ: the dynamic viscosity in Pa·s, used with the density ρ.

Re has no unit. Below about 2000 the flow is laminar. Above about 4000 it is turbulent. Most water pipes are turbulent.

3Friction loss and minor losses

The friction loss along a straight pipe is given by the Darcy–Weisbach equation:

Here f is the Darcy friction factor (given in the question, typically 0.015 to 0.03) and L is the pipe length. The term V²/(2g) is the velocity head.

The loss grows with the square of the velocity. Doubling the flow through the same pipe gives four times the friction loss.

Fittings such as bends, valves, the entrance from a tank and the exit into a tank cause extra minor losses. Each has a loss coefficient K:

Typical values: a square-edged entrance K = 0.5, a 90° bend about 0.3, the exit into a tank K = 1.0. The total loss is h_L = h_f + h_m.

4The method step by step

1. Convert units: diameters to m, flow rates to m³/s (L/s ÷ 1000).
2. Find the area A = π·D²/4 and the mean velocity V = Q/A.
3. Calculate the velocity head V²/(2g). Almost every formula in this topic uses it.
4. For the regime, find Re = V·D/ν. For the friction loss, h_f = f·(L/D)·V²/(2g).
5. Add minor losses with ΣK·V²/(2g). Convert to a pressure drop with Δp = ρ·g·h if asked.

L/D is often a large number (1000 or more). Calculate it on its own first so you can see that it is sensible.

5A worked example

Follow every step. Try to predict the next one before you open it.

6Common mistakes

• Forgetting the 2g, or the square on V, in the velocity head V²/(2g).
• Leaving the diameter in mm, which makes L/D a thousand times too small.
• Using the flow rate Q in place of the velocity V. Convert with V = Q/A first.
• Counting the exit loss (K = 1.0) twice: once as a minor loss and once as the velocity head of a jet into a tank.
• Giving a head in metres when a pressure is asked, or the other way round.

7How exam questions look

• Easy: the Reynolds number. From the velocity or the flow rate, the diameter and the viscosity (ν, or ρ and μ), find Re and say whether the flow is laminar or turbulent.
• Medium: Darcy–Weisbach. Find the friction head loss h_f or the pressure drop Δp for a given pipe, flow and friction factor.
• Hard: minor losses and total head. A pipe from a reservoir with an entrance, bends and a valve. Find the reservoir height needed for a flow, or the flow between two reservoirs.

For a pipe from a reservoir that ends in a free jet, the jet still carries its velocity head: H = (1 + f·L/D + ΣK)·V²/(2g).

Between two open reservoirs, the height difference is used up by the losses alone: Δz = (f·L/D + ΣK)·V²/(2g), with the exit K = 1.0 in ΣK. Solve for V, then Q = A·V.