Answer to Question #181513 in Mechanical Engineering for Mohamed sadique

Question

In an engine cylinder a gas has a volumetric analysis of 18% CO2, 12.5 % O2, and 74.5 % N2. The temperature at the beginning of expansion is 950°C and the gas mixture expands reversibly through a volume ratio of 8: 1, according to the law PV1.2 = C. Calculate per kg of gas (i) the work done (ii) the heat flow (iii) the change of entropy per kg of mixture. The values of Cp for the constituents CO2, O2, and N2 are 1.235 kJ/kgK, 1.088 kJ/kgK and 1.172 kJ/kgK respectively.

Answer 1

Percentage of mass (m) =

% of CO2=0.18

% of O2 = 0.13

% of N2 = 0.68

Specific heat at constant pressure $=0.18×C p ​ 1+0.13×C p ​ 2+0.68×C p ​ 3$

$C_p=1.17KJ/Kg K$

Gas constant -

$R=8.314( \frac{ 0.44}{ 0.187} ​ + \frac{ 0.44}{ 0.131} ​ + \frac{0.44}{ 0.682} ​ )$

$R=0.2718KJ/kg/K$

We know that,

$C p ​ −C v ​ =R$

$C v ​ =1.1725−0.271$

$C v ​ =0.9$

$i) W=\frac{R(T_1-T_2)}{n-1}$

$\frac{T_2}{T_1}=(\frac{v_1}{v_2})^{n-1}=(\frac{1}{8})^{1.2-1} =0.659$

$W=566.8KJ/kg$

$Q=ΔU+W=U 2 ​ −U 1 ​ +W$

$=0.9×10 3 (806−1223)+566.8×10 3$

$Q=191.41kJ/kg$

$ii) S=R\log(\frac{v_2}{v_1}) =0.2718\times 10^3 \log 8$

$S=0.565KJ/kg K$

$S=C _v log( \frac{ T_ 2}{T_ 1} ​ ​ )$

$=0.9\times 10^3\log(\frac{122.3}{805.9}) = 0.376$

Change in entropy

$(S_2-S_1)=0.19kJ/kg K$