Isentropic Compression of an N₂/CO₂ Ideal-Gas Mixture: 04-BS-10, December 2013, Q5

Question 5: Isentropic Compression of an N₂/CO₂ Ideal-Gas Mixture (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. Ideal-gas mixture, 80 mol% N₂ / 20 mol% CO₂. Inlet $P_1=100$ kPa, $T_1=600$ K; exit $P_2=500$ kPa. Isentropic (steady-flow) compression, negligible $\Delta$KE/PE.

Quantity	Value
$y_{N_2},\ y_{CO_2}$ (mole fr.)	0.80, 0.20
$P_1,\ T_1$	100 kPa, 600 K
$P_2$	500 kPa

Find. $w_{in}$, work input per unit mass of mixture.

Approach

Compute the mixture's apparent molar mass from the mole fractions, then find the exit temperature from the mixture isentropic condition (mole-fraction-weighted sum of each component's molar entropy change equals the mixture's $R\ln(P_2/P_1)$ term, solved by root-finding since $s^\circ_i(T)$ is nonlinear), and finally get the specific work input from the mole-fraction-weighted enthalpy rise divided by the mixture molar mass.

Mixture molar mass. $$M_{mix}=y_{N_2}M_{N_2}+y_{CO_2}M_{CO_2}=0.80\times28.013+0.20\times44.01=\boxed{31.212\ \text{kg/kmol}}.$$
Isentropic mixture condition ⇒ $T_2$. For an ideal-gas mixture, the molar entropy change of the mixture is the mole-fraction-weighted sum of each component's own entropy change (each component's partial pressure ratio equals the total pressure ratio for a fixed-composition mixture, so the usual $R\ln(P_2/P_1)$ term applies to the mixture as a whole): $$y_{N_2}\big[\bar s^\circ_{N_2}(T_2)-\bar s^\circ_{N_2}(T_1)\big]+y_{CO_2}\big[\bar s^\circ_{CO_2}(T_2)-\bar s^\circ_{CO_2}(T_1)\big]=\bar R\ln\!\frac{P_2}{P_1}.$$ Solving by root-finding: $$T_2=\boxed{881.29\ \text{K}}.$$
Work input per unit mass of mixture. The mixture's molar enthalpy rise is the mole-fraction-weighted sum of each component's molar $\Delta\bar h$; dividing by $M_{mix}$ converts to a per-kg-of-mixture basis (steady-flow energy balance, adiabatic/isentropic, $\Delta$KE/PE negligible: $w_{in}=\Delta h$): $$\Delta\bar h_{mix}=y_{N_2}\Delta\bar h_{N_2}(T_1\!\to\!T_2)+y_{CO_2}\Delta\bar h_{CO_2}(T_1\!\to\!T_2)$$ $$w_{in}=\frac{\Delta\bar h_{mix}}{M_{mix}}=\boxed{314.41\ \text{kJ/kg}}.$$

Quantity	Result
$M_{mix}$	31.212 kg/kmol
$T_2$	881.29 K
$w_{in}$	314.41 kJ/kg

Concept & Theory

Topic. Isentropic compression of a fixed-composition ideal-gas mixture, evaluated component-by-component with variable specific heats (Gibbs–Dalton model).

Key relations. Gibbs–Dalton: each component behaves as if it alone occupied the mixture volume at the mixture temperature, with partial pressure $y_iP$; mixture molar entropy and enthalpy are mole-fraction-weighted sums of the pure-component molar properties evaluated at the mixture $T$.

Why this works. Because every component shares the same $T$ and undergoes the same pressure ratio $P_2/P_1$ (partial pressures scale with total pressure at fixed composition), the mixture's isentropic path can be found by requiring the composition-weighted sum of the individual components' entropy generation to vanish — it reduces to a single scalar root-find in $T_2$, not a coupled system.

Common pitfall. Using a single "average" $c_p$ or $R$ for the mixture computed at only the inlet temperature, rather than integrating each component's actual (temperature-dependent) property functions across the full 600–881 K span, introduces a several-percent error at this temperature rise.

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

Question 5 of 9: Isentropic Compression of an N₂/CO₂ Ideal-Gas Mixture

Question 5: Isentropic Compression of an N₂/CO₂ Ideal-Gas Mixture (15 marks)

Approach