含义Meaning
气体经节流阀等焓膨胀时温度的变化。μ_JT > 0 时膨胀降温（多数气体室温下如此，是空气液化和制冷的原理）；高于转化温度时升温（室温下 H₂、He）。理想气体 μ_JT = 0。
Temperature change on isenthalpic throttling. μ_JT > 0 means cooling on expansion (most gases at room temperature — the basis of gas liquefaction and refrigeration); above the inversion temperature gases warm (H₂, He at room temperature). For ideal gases μ_JT = 0.
符号说明Symbols
| 符号Symbol | 含义 Meaning | 单位 Unit |
|---|---|---|
| \(\mu_\mathrm{JT}\) | 焦耳–汤姆孙系数Joule–Thomson coefficient | K·Pa⁻¹ |
| \(\alpha\) | 体膨胀系数thermal expansion coefficient | K⁻¹ |
| \(T_\text{inv}\) | 转化温度inversion temperature | K |
推导Derivation
绝热节流：上游做功 \(p_1V_1\)，下游 \(p_2V_2\)，\(U_2 - U_1 = p_1V_1 - p_2V_2\) ⇒ \(H_1 = H_2\)。
Throttling conserves enthalpy.
\((\partial T/\partial p)_H = -(\partial H/\partial p)_T/C_p\)，用麦克斯韦关系 \((\partial H/\partial p)_T = V - T(\partial V/\partial T)_p\)。
Use the Maxwell relation for (∂H/∂p)_T.
理想气体 \(T\alpha = 1\) ⇒ μ = 0；范德华气体低压近似 \(\mu_\mathrm{JT} \approx (2a/RT - b)/C_{p,m}\)。
Ideal: μ = 0; van der Waals: μ ≈ (2a/RT − b)/Cp.
适用条件与局限Assumptions & limitations
范德华估算的转化温度偏高约一倍（N₂ 预测 850 K，实测约 621 K）。
The van der Waals estimate is roughly twice too high.
例题Worked example
N₂ 在 298 K 时 μ_JT ≈ 0.27 K/bar：从 200 bar 节流到 1 bar 约降温 50 K（林德法液化空气）。
N₂ at 298 K: ~0.27 K/bar; throttling 200 → 1 bar cools by ~50 K (Linde process).