Interactive Lecture
Major and Minor Losses in Pipe Flow Systems
by Dr Arnaldo Delli Carri
When fluid flows through pipes, energy is lost due to friction between the fluid and pipe walls (major losses) and turbulence created by fittings, bends, and changes in flow area (minor losses). These losses directly set how large your pump needs to be — get them wrong and the system either doesn't flow or wastes energy.
Total head loss is the sum of major losses (friction in straight pipes) and minor losses (fittings, bends, valves). This lost energy must be supplied by pumps to maintain flow.
Major losses occur along the length of straight pipe due to viscous friction between the fluid and pipe walls. These are calculated using the Darcy-Weisbach equation.
Darcy-Weisbach Equation:
\[h_f = f \frac{L}{D} \frac{v^2}{2g}\]
Where:
The friction factor depends on the flow regime (characterised by the Reynolds number) and the relative roughness of the pipe wall:
Reynolds Number:
\[Re = \frac{\rho v D}{\mu} = \frac{v D}{\nu}\]
Where \(\mu\) is dynamic viscosity and \(\nu\) is kinematic viscosity.
Colebrook-White Equation (implicit, turbulent flow):
\[\frac{1}{\sqrt{f}} = -2 \log_{10}\left(\frac{\epsilon/D}{3.7} + \frac{2.51}{Re\sqrt{f}}\right)\]
This implicit equation is the basis of the Moody diagram. For practical calculations, the explicit Swamee-Jain approximation is often used:
\[f = \frac{0.25}{\left[\log_{10}\left(\frac{\epsilon/D}{3.7} + \frac{5.74}{Re^{0.9}}\right)\right]^2}\]
Minor losses occur at pipe fittings, bends, valves, and other components that disturb the flow. Despite their name, they can be significant in systems with many fittings.
Minor Loss Equation:
\[h_L = K \frac{v^2}{2g}\]
Where:
| Fitting Type | K-value |
|---|---|
| Pipe entrance (sharp-edged) | 0.5 |
| Pipe entrance (rounded) | 0.05 |
| Pipe exit | 1.0 |
| 90° elbow (standard) | 0.9 |
| 90° elbow (long radius) | 0.6 |
| 45° elbow | 0.4 |
| T-junction (flow through) | 0.6 |
| T-junction (side branch) | 1.8 |
| Gate valve (fully open) | 0.2 |
| Gate valve (half open) | 5.6 |
| Globe valve (fully open) | 10.0 |
| Check valve | 2.5 |
The total head loss in a piping system is:
\[h_{total} = h_f + \sum h_L = f \frac{L}{D} \frac{v^2}{2g} + \sum K \frac{v^2}{2g}\]
The Moody diagram (or Moody chart) is one of the most important tools in fluid mechanics. Published by Lewis Moody in 1944, it provides a graphical representation of the Darcy friction factor as a function of Reynolds number and relative pipe roughness \(\epsilon/D\).
The Moody diagram plots the Darcy friction factor against Reynolds number, with a separate curve for each relative roughness \(\epsilon/D\). A few things to notice:
To find the friction factor for a given pipe flow:
| Pipe Material | Roughness \(\epsilon\) (mm) |
|---|---|
| Drawn tubing (smooth) | 0.0015 |
| Commercial steel / Wrought iron | 0.046 |
| Galvanised iron | 0.15 |
| Cast iron | 0.26 |
| Concrete | 0.3 – 3.0 |
| Riveted steel | 0.9 – 9.0 |
Explore the Moody diagram below. Adjust the flow conditions using the sliders and watch the operating point move on the chart. Hover over the diagram to read friction factor values at any Reynolds number and roughness.
Design a piping system and observe how pipe properties, fittings, and valves affect total head loss. The simulation shows the Energy Grade Line (EGL) and Hydraulic Grade Line (HGL) along the pipe, with pressure gauges at each fitting.
| Concept | Equation | Description |
|---|---|---|
| Major Loss (Darcy-Weisbach) | \(h_f = f \frac{L}{D} \frac{v^2}{2g}\) | Friction loss in straight pipe |
| Minor Loss | \(h_L = K \frac{v^2}{2g}\) | Loss due to fittings and components |
| Reynolds Number | \(Re = \frac{\rho v D}{\mu}\) | Determines flow regime |
| Friction Factor (Laminar) | \(f = \frac{64}{Re}\) | For Re < 2300 |
| Colebrook-White (Turbulent) | \(\frac{1}{\sqrt{f}} = -2 \log\left(\frac{\epsilon/D}{3.7} + \frac{2.51}{Re\sqrt{f}}\right)\) | Implicit equation, basis of Moody diagram |
| Swamee-Jain (Turbulent) | \(f = \frac{0.25}{[\log(\frac{\epsilon/D}{3.7} + \frac{5.74}{Re^{0.9}})]^2}\) | Explicit friction factor approximation |
| Total Head Loss | \(h_{total} = h_f + \sum h_L\) | Sum of major and minor losses |
| Pressure Drop | \(\Delta P = \rho g h_{total}\) | Pressure loss due to head loss |
| Velocity from Flow Rate | \(v = \frac{4Q}{\pi D^2}\) | Average velocity in pipe |
Water flows at 3 m/s through a 150 mm diameter steel pipe that is 200 m long. The friction factor is 0.020. Calculate the head loss due to friction. (g = 9.81 m/s²)
Show Solution
Solution:
Given:
Apply Darcy-Weisbach equation:
\[h_f = f \frac{L}{D} \frac{v^2}{2g} = 0.020 \times \frac{200}{0.15} \times \frac{3^2}{2 \times 9.81}\]
\[h_f = 0.020 \times 1333.33 \times \frac{9}{19.62}\]
\[h_f = 0.020 \times 1333.33 \times 0.459 = 12.24 \text{ m}\]
Answer: The head loss due to friction is 12.24 m
Water flows through a pipe system at 2.5 m/s. The system contains two 90° elbows (K = 0.9 each), one gate valve fully open (K = 0.2), and one pipe exit (K = 1.0). Calculate the total minor head loss.
Show Solution
Solution:
Given:
Step 1: Calculate total K-value:
\[K_{total} = 2(0.9) + 0.2 + 1.0 = 1.8 + 0.2 + 1.0 = 3.0\]
Step 2: Apply minor loss equation:
\[h_L = K_{total} \frac{v^2}{2g} = 3.0 \times \frac{2.5^2}{2 \times 9.81}\]
\[h_L = 3.0 \times \frac{6.25}{19.62} = 3.0 \times 0.319 = 0.956 \text{ m}\]
Answer: The total minor head loss is 0.956 m (approximately 1.0 m)
Water (\(\rho = 998\) kg/m³, \(\mu = 1.002 \times 10^{-3}\) Pa·s) flows at 1.5 m/s through a 200 mm diameter cast iron pipe (\(\epsilon = 0.26\) mm). Use the Moody diagram (or the Swamee-Jain equation) to determine the friction factor, then calculate the head loss for a 500 m pipe length.
Show Solution
Solution:
Given:
Step 1: Calculate Reynolds number:
\[Re = \frac{\rho v D}{\mu} = \frac{998 \times 1.5 \times 0.2}{1.002 \times 10^{-3}} = 298,800\]
Step 2: Calculate relative roughness:
\[\frac{\epsilon}{D} = \frac{0.26}{200} = 0.0013\]
Step 3: Find friction factor (Swamee-Jain):
\[f = \frac{0.25}{\left[\log_{10}\left(\frac{0.0013}{3.7} + \frac{5.74}{298800^{0.9}}\right)\right]^2} = \frac{0.25}{\left[\log_{10}(4.19 \times 10^{-4})\right]^2} = 0.0219\]
Step 4: Calculate head loss:
\[h_f = f \frac{L}{D} \frac{v^2}{2g} = 0.0219 \times \frac{500}{0.2} \times \frac{1.5^2}{2 \times 9.81}\]
\[h_f = 0.0219 \times 2500 \times 0.1147 = 6.28 \text{ m}\]
Answer: \(f = 0.0219\), \(h_f = 6.28\) m
A water pipe system has the following specifications: L = 100 m, D = 100 mm, v = 2 m/s, f = 0.025. The system includes three 90° elbows (K = 0.9 each) and one gate valve (K = 0.2). Calculate: (a) major loss, (b) minor loss, (c) total head loss, (d) pressure drop. (\(\rho\) = 1000 kg/m³)
Show Solution
Solution:
Given: L = 100 m, D = 0.1 m, v = 2 m/s, f = 0.025, Ktotal = 3(0.9) + 0.2 = 2.9
(a) Major loss:
\[h_f = f \frac{L}{D} \frac{v^2}{2g} = 0.025 \times \frac{100}{0.1} \times \frac{4}{19.62}\]
\[h_f = 0.025 \times 1000 \times 0.204 = 5.10 \text{ m}\]
(b) Minor loss:
\[h_L = K_{total} \frac{v^2}{2g} = 2.9 \times \frac{4}{19.62} = 2.9 \times 0.204 = 0.591 \text{ m}\]
(c) Total head loss:
\[h_{total} = h_f + h_L = 5.10 + 0.591 = 5.69 \text{ m}\]
(d) Pressure drop:
\[\Delta P = \rho g h_{total} = 1000 \times 9.81 \times 5.69 = 55,830 \text{ Pa} = 55.8 \text{ kPa}\]
Answers: (a) 5.10 m, (b) 0.591 m, (c) 5.69 m, (d) 55.8 kPa