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`

Step-by-step 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)+(b)/(2)(T_2^2-T_1^2)+(c)/(3)(T_2^3-T_1^3)+(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)+(1.520x10^(-2))/(2)(473.15K^2-293.15K^2)+(-0.7155x10^(-5))/(3)(473.15K^3-293.15K^3)+(1.312x10^(-9))/(4)(473.15K^4-293.15K^4)=5456.2kJ/kmol`

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

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

`h_(CH_4)=19.89(473.15K-293.15K)+(5.024x10^(-2))/(2)(473.15K^2-293.15K^2)+(1.269x10^(-5))/(3)(473.15K^3-293.15K^3)+(-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*(1kmol)/(32kg)=170.5kJ/kg\\h_(N_2)=5280.4kJ/kmol*(1kmol)/(28kg)=188.6kJ/kg\\h_(CO_2)=7269.4kJ/kmol*(1kmol)/(44kg)=165.2kJ/kg\\h_(CH_4)=7269kJ/kmol*(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.