OpenSpecEngineering

Estimates the frictional pressure drop of a gas and a liquid flowing together in a pipe, from the pressure drop each phase would have if it flowed alone. It uses the Lockhart–Martinelli method [1] with Chisholm’s equations for the two-phase multipliers [2]. Process and pipeline engineers use it to size gas–liquid lines, such as wet-gas, condensate-return and reactor-outlet piping.

How the pressure drop is calculated

First work out two single-phase pressure drops for the same pipe and length: ΔP_L with only the liquid flowing, at its own mass flow, and ΔP_G with only the gas flowing. Any Darcy–Weisbach calculation gives them. The page then forms the Lockhart–Martinelli parameter and the multipliers:

X = √(ΔP_L / ΔP_G)
φ_L² = 1 + C/X + 1/X²,  φ_G² = 1 + C·X + X²
ΔP_TP = φ_L²·ΔP_L = φ_G²·ΔP_G

where φ_L² and φ_G² are the liquid and gas two-phase multipliers and ΔP_TP the two-phase frictional pressure drop. The Chisholm constant C depends on whether each phase, flowing alone, would be laminar or turbulent: 20 for both turbulent, 12 for laminar liquid and turbulent gas, 10 for turbulent liquid and laminar gas, and 5 for both laminar. Because φ_G² = X²·φ_L², the two forms always agree. The page shows both as a check.

Limits and common mistakes

• Each single-phase drop uses that phase’s own flow in the full pipe bore, not the total flow.
• Lockhart and Martinelli treated a phase as laminar below a single-phase Reynolds number of about 1000 and turbulent above about 2000.
• The result is friction only. Add the static head of the mixture in risers and the acceleration loss separately.
• The method is a correlation of air–liquid data. Expect scatter of ±30 % or more, and use it with more care for boiling or flashing flow [3].
• Gas density changes along a long line. Split the line into lengths with small pressure drops and work each in turn.

Worked example

A 30 m run of 2 in. pipe carries water and air together. On their own, the water flow would lose 3.2 kPa in this run and the air flow 11.5 kPa, and both would be turbulent. What is the two-phase friction loss?

1. Enter Pressure Drop if Liquid Flowed Alone 3.2 and Pressure Drop if Gas Flowed Alone 11.5.
2. Choose Flow Regime of Each Phase Turbulent–turbulent (C = 20).
3. Press Solve. The Solution shows: Lockhart–Martinelli parameter X: 0.527504
   Chisholm constant C: 20
   Liquid two-phase multiplier φL²: 42.5081
   Gas two-phase multiplier φG²: 11.8283
   Two-phase pressure drop ΔP_TP: 136.026 kPa
   Cross-check φG²·ΔP_G: 136.026 kPa

Check by hand. X = √(3.2/11.5) = 0.5275, so φ_L² = 1 + 20/0.5275 + 1/0.2783 = 1 + 37.91 + 3.59 = 42.51, and 42.51 × 3.2 ≈ 136 kPa. The mixture loses about nine times what the two phases would lose separately (14.7 kPa in total). That is typical: the liquid crowds the gas into a smaller area and the gas drives the liquid faster.