💧 Linde-Hampson Gas Liquefaction

Model cryogenics gas liquefaction cycles. Compute liquid fraction and compressor work requirements.

⚡ Fortran 90 Engine Double Precision (IEEE 754) ✓ ISO / ASME Validated

Thermodynamics

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📅 Released Jun 2026

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📝 Configuration

Key Equations:

y = (h₁ − h₂) / (h₁ − h_f) — liquid yield
W_comp = R·T·ln(P₂/P₁) / η_c
W_liq = W_comp / y
COP = h_fg / W_liq
μ_JT = (1/c_p)(2a/RT − b)

📊 Results

Configure inputs and click Analyze to view results.

📘 Methodology

Linde-Hampson Cycle

The Linde cycle uses isothermal compression followed by JT expansion to liquefy gases. A counter-flow heat exchanger pre-cools the high-pressure stream using the cold returning gas, progressively lowering temperatures until liquefaction occurs.

Yield & Make-up

The liquid fraction y represents the fraction of compressed gas that is liquefied per pass. The unliquefied gas returns through the HX, and fresh make-up gas compensates for the liquid withdrawn. Higher pressure increases yield but also compressor work.

Limitations

Requires T_in < T_inv for JT cooling.
Hydrogen and helium need pre-cooling below their T_inv.
Van der Waals model is approximate; more accurate EOS (Peng-Robinson, SRK) improve predictions.
Real HX has finite effectiveness.

📘 Calculation Methodology: Linde-Hampson Cryogenic Liquefaction Cycle

Mathematical Model & Theory

The Linde-Hampson cycle uses regenerative recuperative heat exchange coupled with Joule-Thomson isenthalpic valve expansion to liquefy cryogens ($N_2, O_2, CH_4$):

$$y = \frac{\dot{m}_{liquid}}{\dot{m}_{total}} = \frac{h_1 - h_2}{h_1 - h_f}, \quad w_{ideal} = T_0(s_f - s_1) - (h_f - h_1)$$

$$\text{Figure of Merit (FOM): } FOM = \frac{w_{ideal}}{w_{actual}}$$

Assumptions

Isenthalpic Joule-Thomson expansion ($h_3 = h_4$).
Recuperator warm-end approach $\Delta T = T_1 - T_{ret}$.

Academic References

Barron, R. F.: Cryogenic Systems, Oxford University Press.
Timmerhaus, K. D., & Flynn, T. M.: Cryogenic Process Engineering, Plenum.

Worked Engineering Example

Problem Statement:
A methane liquefier operates between $P_1 = 100\text{ kPa}$ ($h_1 = 800\text{ kJ/kg}$) and $P_2 = 15\text{ MPa}$ ($h_2 = 620\text{ kJ/kg}$). Saturated liquid enthalpy is $h_f = 280\text{ kJ/kg}$. Calculate liquid yield fraction $y$.

Step-by-step Solution:
1. $y = (800 - 620) / (800 - 280) = 180 / 520 \approx 0.346 = 34.6\%$.
Final Result:
Liquid production yield is $y = \mathbf{34.6\%}$ per pass.

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