Isentropic Compression of an N₂/CO₂ Ideal-Gas Mixture: 04-BS-10, May 2016, 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. Molar composition: 80% N₂, 20% CO₂. Inlet: $P_1=100$ kPa, $T_1=1000$ K. Exit: $P_2=500$ kPa. Isentropic compression, constant specific heats evaluated at 300 K, ideal-gas mixture.

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

Approach

Convert the molar composition to mass fractions using each species' molar mass, then form the mixture's mass-weighted $c_p$ and gas constant $R_{mix}=R_u/M_{mix}$. With constant specific heats, the isentropic temperature ratio follows the usual $T_2/T_1=(P_2/P_1)^{(k-1)/k}$ relation for the mixture's own effective $k_{mix}$, and the specific work input is $c_{p,mix}\Delta T$.

Mass fractions. Per kmol of mixture: $m_{N_2}=0.8\times28.013=22.410$ kg, $m_{CO_2}=0.2\times44.01=8.802$ kg, $M_{mix}=31.212$ kg/kmol. $$mf_{N_2}=\frac{22.410}{31.212}=0.7180,\qquad mf_{CO_2}=\frac{8.802}{31.212}=0.2820.$$
Mixture properties at 300 K. $c_{p,N_2}=1.0414$ kJ/kg·K, $c_{p,CO_2}=0.8526$ kJ/kg·K. $$c_{p,mix}=0.7180\times1.0414+0.2820\times0.8526=0.7477+0.2404=\boxed{0.9881\ \text{kJ/kg}\cdot\text{K}}.$$ $$R_{mix}=\frac{R_u}{M_{mix}}=\frac{8.314}{31.212}=0.2664\ \text{kJ/kg}\cdot\text{K},\qquad k_{mix}=\frac{c_{p,mix}}{c_{p,mix}-R_{mix}}=\frac{0.9881}{0.7217}=1.3691.$$
Isentropic exit temperature. $$T_2=T_1\left(\frac{P_2}{P_1}\right)^{(k_{mix}-1)/k_{mix}} =1000\times5^{0.2696}=\boxed{1543.19\ \text{K}}.$$
Specific work input. $$w_{in}=c_{p,mix}(T_2-T_1)=0.9881\times(1543.19-1000)=0.9881\times543.19 =\boxed{536.74\ \text{kJ/kg mixture}}.$$

Quantity	Result
$c_{p,mix}$	0.9881 kJ/kg·K
$k_{mix}$	1.3691
$T_2$	1543.19 K
$w_{in}$	536.74 kJ/kg mixture

Concept & Theory

Topic. Isentropic compression of a fixed-composition ideal-gas mixture, constant specific heats.

Key relations. Mass-fraction-weighted mixture $c_p$: $c_{p,mix}=\sum mf_i c_{p,i}$. Mixture gas constant from the mixture molar mass: $R_{mix}=R_u/M_{mix}$. Isentropic temperature ratio: $T_2/T_1=(P_2/P_1)^{(k-1)/k}$.

Why this works. An ideal-gas mixture of fixed composition behaves as a single "pseudo-pure" ideal gas with mass-weighted properties, so every single-component isentropic-process relation carries over unchanged once $c_{p,mix}$, $R_{mix}$, and $k_{mix}$ are computed from the mole fractions and molar masses of the constituents.

Common pitfall. Weighting $c_p$ by mole fraction instead of mass fraction — $c_p$ in kJ/kg·K is a mass-specific property, so the mixture value must be a mass-fraction-weighted average, not a mole-fraction-weighted one (CO&sub2;'s larger molar mass gives it more mass-fraction weight than its 20% mole fraction alone would suggest).

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