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23-Chem-A1 Process Balances and Chemical Thermodynamics · May 2014

Question 6 of 8: Joule–Thomson Throttling of Oxygen

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Exams, May 2014 — 04-Chem-A1 Process Balances and Chemical Thermodynamics, three-hour, open book, any non-communicating calculator. The paper is in four parts: Part A (Q1–Q2, 20%), Part B (Q3–Q4, 30%), Part C (Q5–Q6, 20%) and Part D (Q7–Q8, 30%); a candidate answers ONE question from each part. For completeness all eight questions are fully worked below.

Reference texts: Smith, Van Ness & Abbott, Introduction to Chemical Engineering Thermodynamics (8th ed.) for the energy balances, residual properties, VLE and reaction equilibrium; Felder & Rousseau, Elementary Principles of Chemical Processes (4th ed.) for the material and energy balances; steam properties from the ASME/NIST steam tables (Cengel & Boles appendix). SI throughout; ideal-gas constant \(R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}\).

Question 6: Joule–Thomson Throttling of Oxygen (Part C — 20%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. O\(_2\), \(T_1 = 15.5\ ^\circ\mathrm{C} = 288.65\) K, \(P_1 = 13.8\) MPa (138 bar) → \(P_2 = 1.38\) MPa (13.8 bar). Steady flow, adiabatic (\(Q=0\)), no shaft work, negligible \(\Delta\)KE/PE. Critical constants \(T_c=154.6\) K, \(P_c=50.43\) bar, \(\omega=0.022\); ideal-gas \(C_p/R = 3.639 + 0.506\times10^{-3}T - 0.227\times10^{5}T^{-2}\).

Find. The outlet temperature \(T_2\) (Joule–Thomson expansion).

throttle valve O₂ in 13.8 MPa, 15.5 C O₂ out 1.38 MPa, T₂=? adiabatic, well-insulated → H = const (isenthalpic)
A well-insulated throttle valve: oxygen expands from 13.8 MPa to 1.38 MPa with no heat or work, so the enthalpy is constant (isenthalpic). Because the inlet is far above the ideal-gas region, the real-gas enthalpy departure makes the temperature fall (positive Joule–Thomson coefficient).

Approach

A throttle is isenthalpic. Split the enthalpy into ideal-gas (temperature) and residual (pressure) parts; setting the total enthalpy change to zero gives one equation for \(T_2\), solved by iteration with PR-EOS residual enthalpies.

  1. Isenthalpic condition. With \(Q=W_s=0\) and negligible kinetic/potential terms, the energy balance is \(H_2 = H_1\), i.e. $$\big[H^{ig}(T_2)-H^{ig}(T_1)\big] + \big[H_2^{R}(T_2,P_2) - H_1^{R}(T_1,P_1)\big] = 0.$$
  2. Inlet residual enthalpy (138 bar). From PR-EOS at 288.65 K, 138 bar (reduced \(T_r=1.87,\ P_r=2.74\)): $$H_1^{R} = -1223.8\ \mathrm{J/mol}.$$ The high-pressure inlet is strongly non-ideal; the low-pressure outlet is nearly ideal (\(H_2^R\) only a couple hundred J/mol).
  3. Ideal-gas enthalpy change. \(\displaystyle H^{ig}(T_2)-H^{ig}(T_1)=\int_{T_1}^{T_2}C_p^{ig}\,dT\), with \(C_p^{ig}\approx 29.4\) J/mol·K near ambient. The cooling must supply the difference in residual enthalpy.
  4. Iterate for \(T_2\). Solving \(\int_{T_1}^{T_2}C_p^{ig}dT + H_2^{R}(T_2,13.8\,\text{bar}) - (-1223.8) = 0\) gives \(H_2^{R}\approx -173\) J/mol at the solution and $$T_2 = 252.2\ \mathrm{K}.$$

    \(T_2 \approx 252\) K = −21 °C, a drop of about 36 K.

QuantityValue
\(H_1^{R}\) (138 bar)−1224 J/mol
\(H_2^{R}\) (13.8 bar, \(T_2\))−173 J/mol
Outlet temperature \(T_2\)252 K (−21 °C)
Joule–Thomson cooling≈ 36 K (≈ 0.29 K/bar)
Check / assumptions

Assumptions stated per the rubric: steady flow, adiabatic valve, no shaft work, negligible kinetic and potential energy, oxygen enthalpy from PR-EOS residuals on a constant ideal-gas \(C_p\). The result (cooling of ≈36 K) is consistent with the measured Joule–Thomson coefficient of O\(_2\) (≈0.2–0.3 K/bar at these conditions). A generalized departure-chart evaluation gives the same outlet temperature to within ±2 K.