Question 14.10: Burning of a hydrocarbon at 25°C and 1 atm pressure takes pl......
Burning of a hydrocarbon at 25°C and 1 atm pressure takes place adiabatically. Temperature of the products is 980 K. Find out the irreversibility if entropies of the reactants and products are given as:
Sum of entropies of reactants at 25°C = 18,782.01 kJ/kg mol K
Sum of entropies of products at 980°C = 22,481.68 kJ/kg mol K
Given: (1) Combustion reactants: Air and a hydrocarbon, (2) Initial condition: 1 atm pressure and 298K temperature. 3). Sum of entropies of reactants at 25°C = 18,782.01 kJ/kg mol K and 4). Sum of entropies of products at 980°C = 22,481.68 kJ/kg mol K
Find out: irreversibility
Properties: Molar mass of C, H_2 and O_2 and air are: 12 kg/kmol, 2 kg/kmol, 32 kg/kmol and 29 kg/kmol respectively.
Assumption: (1) Combustion is complete, (2) Air and combustion gases are ideal (3) Kinetic and potential energies are negligible
Analysis:
\sum{N_p} S_p = 22,481.68 kJ/kg mol K and \sum{N_R S_R} = 18782.01 kJ/kg mol K .
Knowing that- I = T_0 S_{gen} , S_{gen} = \Delta S_{sys} + \frac{Q_{sur}}{T_0}
Since, process is adiabatic, Q_{sur} = 0
So, S_{gen} = \Delta S_{sys} \Longrightarrow S_{gen} = ( \sum{N_pS_p} – \sum{N_R S_R} )
⇒ S_{gen} = 22,481.68 – 18,782.01= 3,699.67 kJ/kg mol K.
I = T_0 S_{gen} = 298 \times 3,699.67 = 1,102,501.66 kJ/kg mol