Linde Hampson Liquefaction Calculator
Cryogenic cycle thermodynamics: Calculate liquid yield fraction, compressor specific work, Joule-Thomson throttling enthalpy drop, and Carnot figure of merit.
Cycle Operating Parameters
Liquefaction Performance Summary
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Frequently Asked Questions
How does the Linde-Hampson cycle produce liquid cryogens without an expander?
The simple Linde-Hampson system relies entirely on the Joule-Thomson (JT) effect. High-pressure gas (150-200 bar) passes through a counter-current heat exchanger where it is cooled by cold return vapor from the liquid separator. It then expands isenthalpically through a JT throttle valve. Because the gas operates below its maximum inversion temperature, isenthalpic expansion causes temperature reduction, condensing a fraction (y) into saturated liquid at ~1 bar.
What is the typical liquid yield fraction for simple air liquefaction?
For an ambient feed air stream at 295 K compressed to 200 bar, the simple Linde-Hampson cycle achieves a liquid yield (y) between 8% and 10% per pass. The remaining 90-92% recirculates back through the heat exchanger to pre-cool the incoming feed. By adding auxiliary ammonia or Freon pre-cooling down to ~220 K, the liquid yield can increase to 18-20%.
Why is the Figure of Merit (FOM) relatively low compared to Claude cycles?
The Figure of Merit (FOM) for a simple Linde-Hampson cycle is typically only 7% to 15%. This thermodynamic irreversibility stems from large entropy generation during the isenthalpic throttle valve expansion, where no external work is recovered. Modern air separation units (ASUs) utilize Claude or Kapitza cycles with cryogenic turbo-expanders that expand gas isentropically, recovering shaft work and boosting FOM above 35-45%.