Question 7 of 9: Isentropic Compression of an N₂/CO₂ Mixture

Question 7: Isentropic Compression of an N₂/CO₂ 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: $y_{N_2}=0.80$, $y_{CO_2}=0.20$. $T_1=1000$ K, $P_1=100$ kPa, $P_2=500$ kPa. Constant specific heats evaluated at 300 K.

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

Approach

Convert mole fractions to mass fractions using each species' molar mass, mass-weight the individual $c_p$ values (at 300 K) into a mixture $c_p$, derive the mixture gas constant from the mixture's average molar mass, and apply the standard constant-$k$ isentropic relation for $T_2$. The mixing (Gibbs) entropy term is identical at inlet and exit since composition does not change through the compressor, so it drops out of the isentropic condition entirely.

1. Mixture composition (mass basis). With $M_{N_2}=28.013$, $M_{CO_2}=44.01$ g/mol: $$m_{N_2}=0.80\times28.013=22.410,\qquad m_{CO_2}=0.20\times44.01=8.802\quad(\text{per kmol mixture}).$$ $$M_{mix}=22.410+8.802=31.212\ \text{kg/kmol},\qquad mf_{N_2}=0.7180,\quad mf_{CO_2}=0.2820.$$
2. Mixture properties. At 300 K: $c_{p,N_2}=1.0414$, $c_{p,CO_2}=0.8526$ kJ/kg·K. $$c_{p,mix}=mf_{N_2}\,c_{p,N_2}+mf_{CO_2}\,c_{p,CO_2}=0.7180\times1.0414+0.2820\times0.8526 =\boxed{0.9881\ \text{kJ/kg}\cdot\text{K}}.$$ $$R_{mix}=\frac{R_u}{M_{mix}}=\frac{8.314}{31.212}=0.2664\ \text{kJ/kg}\cdot\text{K},\qquad k_{mix}=\frac{c_{p,mix}}{c_{p,mix}-R_{mix}}=\frac{0.9881}{0.7217}=1.3691.$$
3. Isentropic exit temperature. $$T_2=T_1\left(\frac{P_2}{P_1}\right)^{(k_{mix}-1)/k_{mix}}=1000\times5^{0.2695}=\boxed{1543.19\ \text{K}}.$$
4. Work input. $$w_{in}=c_{p,mix}(T_2-T_1)=0.9881\times(1543.19-1000)=\boxed{536.74\ \text{kJ/kg mixture}}.$$