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

Question 5 of 8: Reversible Isothermal Compression of a CH₄/N₂ Mixture

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 5: Reversible Isothermal Compression of a CH₄/N₂ Mixture (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. \(y_{CH_4}=0.70,\ y_{N_2}=0.30\); \(T=250\) K (isothermal); \(P_1=10\) bar → \(P_2=100\) bar; inlet volumetric flow 0.2 m³/min. At 250 K the gas is well below the methane critical temperature region, so real-gas (residual) properties are needed. Treating the mixture as an ideal solution (Lewis–Randall), each residual property is the mole-fraction average of the pure-component residuals evaluated at the mixture \(T,P\).

Property (PR EOS, ideal soln)State 1 (10 bar)State 2 (100 bar)
Compressibility \(Z\)0.9700.750
Residual enthalpy \(H^R\) (J/mol)−201.5−2168.3
Residual entropy \(S^R\) (J/mol·K)−0.553−6.296

Find. (a) rate of heat transfer \(\dot Q\); (b) power \(\dot W\).

P (bar) V (molar) state 1: 10 bar state 2: 100 bar Reversible isothermal compression at 250 K
The reversible isothermal compression path at 250 K: molar volume shrinks roughly ten-fold from state 1 (10 bar) to state 2 (100 bar). Because the gas is non-ideal at 250 K, the enthalpy and entropy changes carry residual (departure) contributions computed from the Peng–Robinson equation of state.

Approach

Get the property changes from the ideal-gas part plus PR-EOS residuals; for a reversible step \(\dot Q = \dot n\,T\,\Delta S\); the steady-flow first law then gives the shaft power \(\dot W = \dot n(\Delta H)-\dot Q\). Molar flow follows from the real-gas density at the inlet.

  1. Entropy and enthalpy change (per mole). Isothermally the ideal-gas enthalpy change is zero and the ideal-gas entropy change is \(-R\ln(P_2/P_1)\). Adding residuals, $$\Delta S = -R\ln\frac{P_2}{P_1} + (S_2^R - S_1^R) = -8.314\ln 10 + (-6.296+0.553) = -19.14 - 5.74 = -24.89\ \tfrac{\mathrm{J}}{\mathrm{mol\,K}},$$ $$\Delta H = 0 + (H_2^R - H_1^R) = -2168.3 + 201.5 = -1966.9\ \mathrm{J/mol}.$$
  2. Molar flow from the inlet density. With \(Z_1=0.970\), $$\dot n = \frac{P_1 \dot V}{Z_1 R T} = \frac{(10\times10^{5})(0.2/60)}{0.970(8.314)(250)} = 1.654\ \mathrm{mol/s}.$$
  3. (a) Rate of heat transfer (reversible). $$\dot Q = \dot n\,T\,\Delta S = (1.654)(250)(-24.89) = -1.029\times10^{4}\ \mathrm{W}.$$

    \(\dot Q \approx -10.3\) kW (about 10.3 kW must be removed).

  4. (b) Power requirement. Steady-flow first law \(\dot n\,\Delta H = \dot Q - \dot W_s\) gives the shaft work; the compressor input power is \(|\dot W_s|\): $$\dot W_s = \dot Q - \dot n\,\Delta H = -10{,}289 - (1.654)(-1966.9) = -10{,}289 + 3253 = -7036\ \mathrm{W}.$$

    Power input \(|\dot W_s| \approx 7.0\) kW.

QuantityValue
\(\Delta H\) (per mole)−1967 J/mol
\(\Delta S\) (per mole)−24.89 J/mol·K
Molar flow \(\dot n\)1.654 mol/s
Heat transfer rate \(\dot Q\)−10.3 kW (removed)
Power requirement7.0 kW (input)
Check / method

Residual properties and \(Z\) were computed from the Peng–Robinson equation of state for pure CH\(_4\) and N\(_2\) at 250 K and each pressure, then mole-fraction averaged (ideal-solution / Lewis–Randall rule). Using generalized (Lee–Kesler) departure charts instead gives the same numbers to within chart-reading precision. The inlet volumetric flow is taken at the stated 10 bar, 250 K.