Cryogenic Air Separation & Gas Liquefaction, Joule-Thomson & Linde-Hampson (1895) Cycle
Joule-Thomson Cooling & Linde-Hampson Liquefaction Formula
Calculate Isenthalpic Joule-Thomson Temperature Drop (K), Warm-End Isothermal Enthalpy Deficit (kJ/kg), Linde-Hampson Liquid Yield Fraction (%), and Liquid Nitrogen Production (kg/h).
Throttling 200-bar nitrogen drops 36.8 K across a single Joule-Thomson valve—yet without Carl von Linde’s counterflow recuperator, 0% condenses until warm-end enthalpy deficit locks in 11.87% liquid yield!
When compressed real gas (`Nitrogen` or `Air` below its `621` K inversion curve—unlike `300` K `H_2` or `He`, which *heat up* upon throttling unless pre-cooled below `202` K and `40` K!) expands adiabatically through a porous plug or needle valve (`dh = 0`), intermolecular van der Waals attraction produces **Joule-Thomson Cooling `Delta_T_{JT} = mu_{JT} * (P_2 - P_1)`** (`K` across Mean J-T Coefficient `mu_{JT}` K/bar)! **Carl von Linde and William Hampson (`1895`)** turned that modest single-pass drop into continuous `77.4` K liquefaction by routing un-liquefied cold flash vapor (`1 - y`) back through a counterflow recuperator! Applying a control-volume energy balance around the cold box (`Warm Low-Pressure Exit Enthalpy h_1` kJ/kg, `Warm High-Pressure Inlet Enthalpy h_2` kJ/kg, and `Saturated Liquid Enthalpy h_f` kJ/kg) proves **Liquid Yield Fraction `y = ((h_1 - h_2) / (h_1 - h_f)) * 100%`** depends ENTIRELY on the **Warm-End Isothermal Real-Gas Enthalpy Deficit `(h_1 - h_2)`** compressed into `P_2`!
Build this Cryogenic Cold-Box Balance Sheet—Compressor_Discharge `P_2` = 200 bar (B2), Separator_Return `P_1` = 1.013 bar (B3), Mean_`mu_{JT}` = 0.185 K/bar (B4), Low-Pressure_Warm_Gas `h_1` = 301.5 kJ/kg (B5), `200`-bar_Warm_Gas `h_2` = 269.2 kJ/kg (B6), `77.4`-K_Saturated_`LN_2` `h_f` = 29.4 kJ/kg (B7), and Recycle_Compressor_Flow `m_{dot}` = 140 kg/h (B8). Ask the in-app assistant for a first-law control-volume hint before writing the cell formulas: Initial J-T Valve Drop equals `36.81` K (E2), Warm Isothermal Enthalpy Deficit `h_1 - h_2` equals `32.30` kJ/kg (E3), Steady-State Linde-Hampson Liquid Yield equals **`11.87%`** (E4), producing **`16.62` kg/h** of `LN_2` (E5)!
| Linde-Hampson N2 Liquefier KPI (300 K Warm End: h1=301.5, hf=29.4 kJ/kg, m=140 kg/h) | Formula | Verified Output |
|---|---|---|
| Single-Pass J-T Temp Drop (K) | Mean_JT_K_bar * (High_P2_bar - Low_P1_bar) | 36.81 K @ 200 bar (19.74 K @ 95 bar) |
| Isothermal Enthalpy Deficit (kJ/kg) | LowP_Warm_h1_kJ_kg - HighP_Warm_h2_kJ_kg | 32.30 kJ/kg (16.90 kJ/kg @ 95 bar) |
| Linde-Hampson Liquid Yield (%) | ((h1 - h2) / (h1 - Sat_Liq_hf)) * 100 | 11.87% yield (6.21% @ 95 bar sag!) |
| Liquid LN2 Production (kg/h) | Compressor_kg_hr * ((h1 - h2) / (h1 - hf)) | 16.62 kg/h (8.70 kg/h halved output!) |
Linde-Hampson (1895) Cold-Box Control-Volume Enthalpy & J-T Liquefaction Formula
Validate `High_Pressure P_2` in `[20, 400]` bar, `Low_Pressure P_1` in `[0.8, 10]` bar (`with P_2 > P_1 + 15`), `Mean mu_{JT}` in `[0.02, 1.50]` K/bar, `Enthalpies h_f < h_2 < h_1` (`kJ/kg`), and `Mass_Flow > 0` kg/h!
=LET(ph_b, B2, pl_b, B3, mj_k, B4, hl_k, B5, hh_k, B6, hf_k, B7, mf_h, B8, yx_f, (hl_k - hh_k) / (hl_k - hf_k), IF(OR(ph_b < 20, ph_b > 400, pl_b < 0.8, pl_b > 10, ph_b <= pl_b + 15, mj_k < 0.02, mj_k > 1.5, hf_k >= hh_k, hh_k >= hl_k, mf_h <= 0), "Invalid Linde Input", ROUND(mf_h * yx_f, 2)))Derated Final Compressor Stage (`P_2 = 95` bar -> `h_2 = 284.6` kJ/kg) Yield-Halving Check
Bypass a leaking `4`th-stage piston ring so discharge pressure drops to `B2 = 95.0` bar (`mu_{JT}` B4 = 0.210 K/bar, `h_2` rises to `B6 = 284.6` kJ/kg): Warm Enthalpy Deficit E3 shrinks from `32.30` down to **`16.90` kJ/kg**, cutting Liquid Yield E4 nearly in half to **`6.21%`** (`8.70` kg/h `LN_2` in E5)—proving why industrial plants add Claude expansion turbines or `NH_3` pre-cooling!
=IF(ISTEXT(E5), E5, ROUND(((B5 - B6) / (B5 - B7)) * 100, 2))Try the example
Use the original values or change them to test your understanding.