N₂/O₂ Mixture — Adiabatic Compression: 04-BS-10, December 2018, Q4

Question 4: N₂/O₂ Mixture — Adiabatic Compression (15 marks)

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. $m=2$ kg, mass fractions $mf_{N_2}=0.70$, $mf_{O_2}=0.30$. $T_1=7\ ^\circ$C (280.15 K), $P_1=100$ kPa. $T_2=177\ ^\circ$C (450.15 K), $P_2=500$ kPa. Adiabatic ($Q=0$), closed system, negligible KE/PE.

Find. (a) $W_{in}$ [kJ]; (b) $S_{gen}$ [kJ/K].

Approach

With $Q=0$ for a closed system, the first law gives the work input directly from the mixture's internal-energy rise; no assumption of reversibility is made or needed. Each pure component's $u$ and $s^\circ$ are evaluated at the real (variable-property) ideal-gas states via its equation of state, mass-fraction-weighted, and the mixing (Gibbs) entropy term is identical at inlet and exit because the composition does not change through a compressor — it therefore cancels from $S_{gen}$.

First law (part a): $Q=0$ closed system. Mass-fraction-weighted internal energy change (real variable-property ideal-gas values, not a single constant $c_v$): $$\Delta u_{mix}=mf_{N_2}(u_{2,N_2}-u_{1,N_2})+mf_{O_2}(u_{2,O_2}-u_{1,O_2})=123.28\ \text{kJ/kg}$$ $$W_{in}=m\,\Delta u_{mix}=2\times123.28=\boxed{246.56\ \text{kJ}}.$$
Entropy change per component. Each pure component's entropy change combines its own temperature and partial-pressure terms (partial pressure ratio equals the total pressure ratio since composition is unchanged): $$\Delta s_i=s_i^\circ(T_2)-s_i^\circ(T_1)-R_i\ln\frac{P_2}{P_1},\qquad R_{N_2}=0.2968,\ R_{O_2}=0.2598\ \text{kJ/kg}\cdot\text{K}.$$
Mixture entropy generation (part b). The Gibbs mixing-entropy term is identical at state 1 and state 2 (same mole fractions throughout) and cancels exactly, leaving the mass-weighted sum of the pure-component changes; since $Q=0$, $S_{gen}=\Delta S_{mix}$ directly: $$\Delta s_{mix}=mf_{N_2}\Delta s_{N_2}+mf_{O_2}\Delta s_{O_2}=0.01927\ \text{kJ/kg}\cdot\text{K}$$ $$S_{gen}=m\,\Delta s_{mix}=2\times0.01927=\boxed{0.03854\ \text{kJ/K}}.$$
Sign check. $S_{gen}>0$ as required by the second law: the given end states correspond to an actual (irreversible) adiabatic compression, since the isentropic exit temperature for this pressure ratio ($T_{2s}\approx443$ K) is below the stated $T_2=450.15$ K.

Quantity	Result
(a) $W_{in}$	246.56 kJ
(b) $S_{gen}$	0.03854 kJ/K

Concept & Theory

Topic. Adiabatic (but not isentropic) compression of a fixed-composition ideal-gas mixture, worked from directly-given end states rather than an efficiency.

Key relations. Mixture property = mass-fraction-weighted sum of pure-component properties evaluated at the mixture's actual $T,P$: $\Delta u_{mix}=\sum mf_i\Delta u_i$, $\Delta s_{mix}=\sum mf_i\Delta s_i$ (Gibbs mixing term cancels for fixed composition). Adiabatic closed system: $S_{gen}=\Delta S$ directly (no $Q/T$ term).

Why this works. Ideal-gas mixture enthalpy/internal energy have no mixing term at all (components don't interact), and the composition-dependent mixing-entropy term is a function of mole fractions alone — since those don't change through the compressor, the term is identical at inlet and exit and subtracts out of $\Delta S_{mix}$ even though it is never computed explicitly.

Common pitfall. Using a single constant $c_v$ for "the mixture" without first mass-weighting the two components' individual (and here, markedly different-temperature) property values — over a 170 K span the variable-property route used here differs from a constant room-temperature $c_v$ estimate by several percent.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 13 (Gas Mixtures).

Question 4 of 9: N₂/O₂ Mixture — Adiabatic Compression

Question 4: N₂/O₂ Mixture — Adiabatic Compression (15 marks)

Approach