Question 4 of 9: Isentropic Compression of an N₂/CO₂ Mixture

Question 4: 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. Mixture: 80% N$_2$ + 20% CO$_2$ by mole. Inlet: $P_1=100$ kPa, $T_1=1000$ K. Exit: $P_2=500$ kPa, isentropic. Constant specific heats evaluated at 300 K.

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

Approach

Convert the given mole fractions to mass fractions using each species' molar mass, then form the mixture's mass-weighted constant-property $c_p$, gas constant, and specific-heat ratio $k$. Since the compression is isentropic and $k$ is treated as constant, use the ideal-gas isentropic relation directly on temperature and pressure, then get the specific work from $w=c_p\Delta T$.

1. Mixture composition (mass basis). Per kmol of mixture: $m_{N_2}=0.80\times 28.013=22.410$ kg, $m_{CO_2}=0.20\times44.01=8.802$ kg, so $M_{mix}=31.212$ kg/kmol, giving mass fractions $mf_{N_2}=0.7180$, $mf_{CO_2}=0.2820$.
2. Mixture properties at 300 K. $$c_{p,mix}=mf_{N_2}c_{p,N_2}+mf_{CO_2}c_{p,CO_2}=0.7180\times1.0414+0.2820\times0.8526=\boxed{0.9881\ \text{kJ/kg}\cdot\text{K}}.$$ $$R_{mix}=\frac{R_u}{M_{mix}}=\frac{8.314}{31.212}=0.2664\ \text{kJ/kg}\cdot\text{K},\qquad k_{mix}=\frac{c_{p,mix}}{c_{p,mix}-R_{mix}}=\boxed{1.3691}.$$
3. Isentropic exit temperature. $$T_2=T_1\left(\frac{P_2}{P_1}\right)^{(k_{mix}-1)/k_{mix}}=1000\times\left(\frac{500}{100}\right)^{0.2696}=\boxed{1543.2\ \text{K}}.$$
4. Work input per unit mass of mixture. For an isentropic, steady-flow compressor with negligible KE/PE change, $w_{in}=\Delta h=c_{p,mix}(T_2-T_1)$: $$w_{in}=0.9881\times(1543.2-1000)=\boxed{536.7\ \text{kJ/kg mixture}}.$$