In the given example, when propane burns in stiochiometric amount of air in clos

Question

In the given example, when propane burns in stiochiometric amount of air in closed chamber then the reaction occur

C3H8 + 6O2 -----------------> 3CO2 + 4H2O + O2

Molar mass of propane = 44

Molar mass of oxygen = 6 X 16 = 96

Molar mass of CO2 = 3 X 44 = 132

Molar mass of water = 72

Initial pressure = 1 atm

Initial Temp = 298 K

Final Temp = 2000K

Now Initial pressure = Final Pressure + Pox

Pf = 1 X 1 / 7 = 0.143 atm

Pox = 1 X 6/7 =0.857 atm

Considering volume = 22.4 L or 0.022 m3

Mass of fuel = PVn/RT = 0.143 X 44 X 0.02 X 105/ 8314.4 X 298 = 0.125 X 105/ 2477691.2 = 0.0050 Kg

Mass of oxygen = 0.857 X 96 X 0.02 X 105/ 8314.4 X 298 = 1.64 X 105/ 2477691.2 = 0.066 Kg

Total mass = 0.0050 + 0.066 = 0.071 Kg

Mass of CO2 = 132/204 X 0.143 = 0.092 Kg

Mass of H2O = 72 / 204 X 0.143 = 0.050 Kg

Total mass = 0.092 + 0.050 = 0.142 Kg

P(CO2) = mRT/NV = 0.092 X 8314.4 X 2000 / 132 X 0.02 = 1529849.6/ 2.64 = 5.79 X 105 bar

Heat removed = Change in enthalpy

Explanation / Answer

propane gas burned completely with stochiometric amount of air in closed rigid combustion chamber. initial temperature and pressure for both substances are 25 C and 1 atm. The final temperature in the chamber is 2000K. Determine the final pressure, the heat removed, the entropy generation and the exergy destruction if the surroundings temperature is 25 C

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