Question

The volumetric analysis of mixture of gases is 30 percent oxygen, 40 percent nitrogen, 10 percent carbon dioxide, and 20 percent methane. This mixture is heated from 20°C to 200°C while flowing through a tube in which the pressure is maintained at 150 kPa. Determine the heat transfer to the mixture per unit mass of the mixture.

Answer 1

Answer:$Q=234kJ/kg$

Explanation:

Hello,

In this case, the transferred heat to the mixture is given by the addition among each compound's specific enthalpies:

$Q=h_{O_2}+h_{N_2}+h_{CO_2}+h_{CH_4}$

Now, each specific enthalpy is given by the following equation:

$h_i=\int\limits^{T_2}_{T_1} {Cp_i(T)} \, dT$

As long as there is a large difference between the initial and the final temperature. In such a way, the shown below polynomial for the Cp which is given kJ/kmol*K is considered and subsequently integrated:

$Cp(T)=a+bT+cT^2+dT^3\\\int\limits^{T_2}_{T_1} {Cp(T)} \, dT =a(T_2-T_1)+\frac{b}{2}(T_2^2-T_1^2)+\frac{c}{3}(T_2^3-T_1^3)+\frac{d}{4}(T_2^4-T_1^4)$

Now, the specific enthalpy is computed for each component considering the temperatures in Kelvin:

$h_{O_2}=25.48(473.15K-293.15K)+\frac{1.520x10^{-2}}{2}(473.15K^2-293.15K^2)+\frac{-0.7155x10^{-5}}{3}(473.15K^3-293.15K^3)+\frac{1.312x10^{-9}}{4}(473.15K^4-293.15K^4)=5456.2kJ/kmol$

$h_{N_2}=28.9(473.15K-293.15K)+\frac{-0.1571x10^{-2}}{2}(473.15K^2-293.15K^2)+\frac{0.8081x10^{-5}}{3}(473.15K^3-293.15K^3)+\frac{-2.873x10^{-9}}{4}(473.15K^4-293.15K^4)=5280.4kJ/kmol$

$h_{CO_2}=22.26(473.15K-293.15K)+\frac{5.981x10^{-2}}{2}(473.15K^2-293.15K^2)+\frac{-3.501x10^{-5}}{3}(473.15K^3-293.15K^3)+\frac{7.469x10^{-9}}{4}(473.15K^4-293.15K^4)=7269.4kJ/kmol$

$h_{CH_4}=19.89(473.15K-293.15K)+\frac{5.024x10^{-2}}{2}(473.15K^2-293.15K^2)+\frac{1.269x10^{-5}}{3}(473.15K^3-293.15K^3)+\frac{-11.01x10^{-9}}{4}(473.15K^4-293.15K^4)=7269kJ/kmol$

Now, we convert them per unit of mass as:

$h_{O_2}=5456.2kJ/kmol*\frac{1kmol}{32kg}=170.5kJ/kg\\h_{N_2}=5280.4kJ/kmol*\frac{1kmol}{28kg}=188.6kJ/kg\\h_{CO_2}=7269.4kJ/kmol*\frac{1kmol}{44kg}=165.2kJ/kg\\h_{CH_4}=7269kJ/kmol*\frac{1kmol}{16kg}=454.3kJ/kg$

Finally, by considering the volumetric percentages, we compute the heat transferred to mixture:

$Q=0.3*170.5kJ/kg+0.4*188.6kJ/kg+0.1*165.2kJ/kg+0.2*454.3kJ/kg\\Q=234kJ/kg$

Best regards.