N₂/CO₂ Ideal-Gas Mixture, Isentropic Compression: 04-BS-10, December 2017, Q4

Question 4: N₂/CO₂ Ideal-Gas Mixture, Isentropic 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. Ideal-gas mixture, $y_{N_2}=0.80$, $y_{CO_2}=0.20$ (mole basis). Inlet $T_1=600$ K, $P_1=100$ kPa. Isentropic compression to $P_2=500$ kPa. Molar masses $M_{N_2}=28.013$, $M_{CO_2}=44.01$ kg/kmol.

Find. Work input per unit mass of mixture, $w_{in}$ [kJ/kg].

Approach

Because the mixture is only 2 components at fixed mole fractions, the molar entropy change of the mixture on compression is the mole-fraction-weighted sum of each species' own molar entropy change (Gibbs' theorem for ideal-gas mixtures: each component behaves as if it alone occupied the container at its own partial pressure, and the $R\ln(y_i)$ mixing terms cancel identically between the inlet and outlet states since composition doesn't change). Root-solve the isentropic condition for $T_2$ using each species' variable-cp (temperature-dependent) molar entropy function, then get the molar enthalpy change the same way and convert to a per-unit-mass mixture result via the mixture's molar mass.

Mixture molar mass. $$M_{mix}=y_{N_2}M_{N_2}+y_{CO_2}M_{CO_2}=0.80\times28.013+0.20\times44.01=22.410+8.802=\boxed{31.212\ \text{kg/kmol}}.$$
Isentropic condition ⇒ exit temperature. Composition is unchanged across the compressor, so the mixing entropy terms cancel and the isentropic condition reduces to $$y_{N_2}\big[\bar s^\circ_{N_2}(T_2)-\bar s^\circ_{N_2}(T_1)\big]+y_{CO_2}\big[\bar s^\circ_{CO_2}(T_2)-\bar s^\circ_{CO_2}(T_1)\big]=R_u\ln\frac{P_2}{P_1}.$$ Right-hand side: $R_u\ln(500/100)=8.314\times1.6094=13.381$ kJ/(kmol·K). Root-solving (both species' molar entropy evaluated via high-accuracy ideal-gas-limit properties) gives $$\boxed{T_2=881.29\ \text{K}}.$$
Molar enthalpy change of each species. $N_2$: $\bar h(600\,\text{K})=17{,}564$, $\bar h(881.29\,\text{K})=26{,}295$ kJ/kmol, $\Delta\bar h_{N_2}=8731$ kJ/kmol. $CO_2$: $\bar h(600\,\text{K})=35{,}205$, $\bar h(881.29\,\text{K})=49{,}348$ kJ/kmol, $\Delta\bar h_{CO_2}=14{,}143$ kJ/kmol.
Work input per unit mass of mixture. Steady-flow, negligible ΔKE/PE: $w_{in}=\Delta h_{mix}$ (per unit mass). Mole-weighted molar enthalpy change, divided by $M_{mix}$: $$\Delta \bar h_{mix}=y_{N_2}\Delta\bar h_{N_2}+y_{CO_2}\Delta\bar h_{CO_2}=0.80\times8731+0.20\times14{,}143=6985+2829=9813\ \text{kJ/kmol}.$$ $$w_{in}=\frac{\Delta\bar h_{mix}}{M_{mix}}=\frac{9813}{31.212}=\boxed{314.41\ \text{kJ/kg}}.$$

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 with variable specific heats.

Key relations. Amagat/Gibbs ideal-gas mixture rule: mixture molar mass $M_{mix}=\sum y_iM_i$; mixture isentropic condition reduces to $\sum y_i\Delta\bar s^\circ_i(T)=R_u\ln(P_2/P_1)$ when composition is fixed.

Why this works. For an ideal-gas mixture, each species' entropy is evaluated at its own partial pressure $y_iP$; since $y_i$ is identical at states 1 and 2 (no separation or reaction occurs), the $-R\ln y_i$ mixing term appears identically in $s_i(T_1,y_iP_1)$ and $s_i(T_2,y_iP_2)$ and cancels out of the difference, leaving only the pure-species $\bar s^\circ_i(T)$ terms plus the overall $R_u\ln(P_2/P_1)$ — this is why the mixture isentropic condition looks identical in form to a pure-substance one, just mole-fraction-weighted.

Common pitfall. Using a constant, room-temperature $k$ (specific heat ratio) for this mixture instead of accounting for the strong temperature dependence of CO₂'s specific heat between 600 K and 880 K — CO₂'s $c_p$ rises roughly 15% over that span, so a constant-$k$ shortcut understates both $T_2$ and $w_{in}$.

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

Question 4 of 9: N₂/CO₂ Ideal-Gas Mixture, Isentropic Compression

Question 4: N₂/CO₂ Ideal-Gas Mixture, Isentropic Compression (15 marks)

Approach