ThermoFluidCalc Report

Throttling & Joule-Thomson β€” 2026-09-15 13:29:33
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πŸ’¨ Joule-Thomson Throttling

Analyze isenthalpic expansion and throttling of real gases. Compute temperature changes and inversion curves.

⚑ Fortran 90 Engine Double Precision (IEEE 754) βœ“ ISO / ASME Validated
Joule-Thomson Throttling Thermodynamics
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⚑ Solves 200
πŸ’Ύ Downloads 434 πŸ“¦ Fortran Code 11.4 KB
πŸ“… Released Jun 2026
⏱️ Latency < 1 ms
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πŸ“ Configuration

πŸ§ͺ Gas Selection
βš—οΈ Custom Gas Properties
🌑️ Operating Conditions
Must be less than Pin
Key Equations:

Isenthalpic: hin = hout
ΞΌJT = (1/cp)(2a/RT βˆ’ b) [Van der Waals]
Tout = Tin + ΞΌJTΒ·Ξ”P
Tinv = 2a/(Rb) [max inversion temp]
Ideal gas: ΞΌJT = 0

πŸ“Š Results

Configure inputs and click Analyze to view results.

πŸ“˜ Methodology

Isenthalpic Process

Throttling through a valve is isenthalpic (hin = hout) at steady state with negligible kinetic and potential energy changes. For a real gas, this causes a temperature change due to intermolecular forces.

Joule-Thomson Effect

The JT coefficient ΞΌJT = (βˆ‚T/βˆ‚P)h determines whether throttling causes cooling (ΞΌ>0) or heating (ΞΌ<0). Most gases cool at room temperature; hydrogen and helium heat up because their inversion temperatures are very low.

Inversion Curve

The inversion temperature is where ΞΌJT = 0. Below Tinv, the gas cools upon expansion. For the Van der Waals model: Tinv,max = 2a/(Rb). Gas liquefaction (Linde process) requires operating below Tinv.

πŸ“˜ Calculation Methodology: Joule-Thomson Expansion & Inversion Temperature

Mathematical Model & Theory

Isenthalpic throttling through a porous plug or restriction causes temperature change quantified by the Joule-Thomson coefficient $\mu_{JT}$. Cooling occurs when $\mu_{JT} > 0$ below the inversion curve:

$$\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_h = \frac{1}{c_p}\left[ T\left(\frac{\partial v}{\partial T}\right)_P - v \right] = \frac{V}{c_p}(T \beta - 1)$$
$$\text{Maximum Inversion Temperature (van der Waals): } T_{inv,max} = \frac{2a}{R b}$$

Assumptions

  • Steady isenthalpic expansion ($h_1 = h_2$) with negligible kinetic energy change.
  • Real gas equation of state behavior.

Academic References

  1. Moran, M. J. et al.: Engineering Thermodynamics, Ch. 11.
  2. Perry's Chemical Engineers' Handbook: Thermodynamic Properties.

Worked Engineering Example

Problem Statement:
Nitrogen at $T = 300\text{ K}$ has $\mu_{JT} = 0.22\text{ K/bar}$. Calculate temperature drop when throttled from $100\text{ bar}$ to $20\text{ bar}$.

Step-by-step Solution:
1. $\Delta P = 20 - 100 = -80\text{ bar}$.
2. $\Delta T = \mu_{JT} \times \Delta P = 0.22 \times (-80) = -17.6\text{ K}$.
3. $T_{exit} = 300 - 17.6 = 282.4\text{ K}$ ($9.25^\circ\text{C}$).
Final Result:
Throttling cools nitrogen by $\mathbf{17.6\text{ K}}$ to an exit temperature of $\mathbf{9.25^\circ\text{C}}$.