Isentropic Compression of an N 2 /CO 2 Ideal-Gas Mixture: 04-BS-10, May 2015, Q7

Question 7: 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. Mole fractions $y_{N_2}=0.80$, $y_{CO_2}=0.20$. Inlet: $T_1=1000$ K, $P_1=100$ kPa. Exit: $P_2=500$ kPa. Constant specific heats evaluated at 300 K: $c_{p,N_2}=1.039$, $c_{v,N_2}=0.743$ kJ/kg·K ($M_{N_2}=28.013$); $c_{p,CO_2}=0.846$, $c_{v,CO_2}=0.657$ kJ/kg·K ($M_{CO_2}=44.01$).

Find. $w_{in}$ [kJ/kg mixture].

Approach

Convert the given mole fractions to mass fractions via each species' molar mass, mass-weight the constant specific heats to get mixture $c_p$, $c_v$, and $k$, then apply the standard constant-$k$ isentropic temperature-pressure relation to find the exit temperature. Work input per unit mass of mixture follows from the mixture's own $c_p$.

Mixture molar mass and mass fractions. $$M_{mix}=y_{N_2}M_{N_2}+y_{CO_2}M_{CO_2}=0.80\times28.013+0.20\times44.01=\boxed{31.212\ \text{kg/kmol}}.$$ $$mf_{N_2}=\frac{y_{N_2}M_{N_2}}{M_{mix}}=0.7180,\qquad mf_{CO_2}=\frac{y_{CO_2}M_{CO_2}}{M_{mix}}=0.2820.$$
Mixture specific heats and $k$. $$c_{p,mix}=mf_{N_2}c_{p,N_2}+mf_{CO_2}c_{p,CO_2}=0.7180\times1.039+0.2820\times0.846=\boxed{0.9846\ \text{kJ/kg}\cdot\text{K}}.$$ $$c_{v,mix}=mf_{N_2}c_{v,N_2}+mf_{CO_2}c_{v,CO_2}=0.7188\ \text{kJ/kg}\cdot\text{K},\qquad k_{mix}=\frac{c_{p,mix}}{c_{v,mix}}=\boxed{1.3698}.$$
Isentropic exit temperature. Because the mixture composition (and hence the Gibbs mixing-entropy term) is unchanged across the compressor, the isentropic relation applies directly with the mixture's own constant $k$: $$T_2=T_1\left(\frac{P_2}{P_1}\right)^{(k_{mix}-1)/k_{mix}}=1000\times(5)^{0.2699}=\boxed{1544.24\ \text{K}}.$$
Work input per unit mass of mixture. $$w_{in}=c_{p,mix}(T_2-T_1)=0.9846\times(1544.24-1000)=\boxed{535.85\ \text{kJ/kg mixture}}.$$

Quantity	Result
$c_{p,mix}$	0.9846 kJ/kg·K
$k_{mix}$	1.3698
$T_2$	1544.24 K
$w_{in}$	535.85 kJ/kg mixture

Concept & Theory

Topic. Constant-specific-heat isentropic compression of a fixed-composition ideal-gas mixture.

Key relations. Mass-fraction mixing rule for extensive-per-mass properties: $c_{p,mix}=\sum mf_i\,c_{p,i}$ (and likewise for $c_v$). Constant-$k$ isentropic relation: $T_2/T_1=(P_2/P_1)^{(k-1)/k}$.

Why this works. For an ideal-gas mixture of fixed composition, the Gibbs mixing-entropy contribution is identical at inlet and exit (same mole fractions throughout) and cancels exactly from the isentropic condition — this is why the mixture can be treated with a single "effective" $k_{mix}$ exactly as if it were a pure substance, provided the composition never changes.

Common pitfall. Mixing on a mole-fraction basis directly into $c_p$ (i.e. computing $\sum y_i c_{p,i}$) instead of converting to mass fractions first — $c_p$ is a per-unit- mass property, so the correct mixing rule requires mass, not mole, fractions.

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

Question 7 of 9: Isentropic Compression of an N 2 /CO 2 Ideal-Gas Mixture

Question 7: Isentropic Compression of an N2/CO2 Ideal-Gas Mixture (15 marks)

Approach

Question 7: Isentropic Compression of an N₂/CO₂ Ideal-Gas Mixture (15 marks)