Propane gas is use as the fuel in a combustion furnace. The propane is fed into
ID: 474483 • Letter: P
Question
Propane gas is use as the fuel in a combustion furnace. The propane is fed into the furnace at 25 degree C and 1.1 atm (absolute) at a flowrate of 1000 mol/s and burned with 130% excess air at 250 degree C and 1.2 atm (absolute). The products from the combustion reaction leave the furnace at 900 degree C and 1 atm (absolute). What is the volumetric flowrate of the air fed to the furnace in (L/s)? What is the flowrate in SCFM? Assuming 100% combustion of the propane and that no carbon monoxide is formed, calculate the volumetric flowrate of the combustion products in L/s. Using the criteria from 5.2 in the book, verify that the ideal gas equation of state is acceptable for calculating the volumetric flowrates determined in parts a and b.Explanation / Answer
The combustion of propane is C3H8+ 5O2-->3CO2 + 4H2O
Basis : 1000 mole/s. Moles of water according to the stoichiometry of the reaction =5*1300= 6500 moles/s. Air supplied = 130% excess. Air supplied =2.3*6500 =14950 moles/s
For Calculating the flow rate of air, n= 14950 moles/s, P= 1.2 atm, T =250+273= 523K
From ideal gas equation, V= nRT/P, R =0.0821 L.atm/mole.s
V= 14950*0.0821*523/1.2 =534940 L/s
For converting this to standard condition, apply P1V1/T1= P2V2/T2
Where P2, V2 and T2 refer to standard conditions ( i.e P2=1 atm, T2= 273K)
V2= P1V1T2/(P2T1)= 1.2*534940*273/(1*523) =335079 L/s = 335.079m3/s (SCFM)
Combustion products : Excess air= 6500*1.3= 8450 moles/S, CO2 formed = 3*1000=3000 moles/ and H2O= 4*1000= 4000moles/s
Total moles of products = 8450+4000+3000= 15450 moles/s. These gases are leaving at 900deg.c and 1 atm pressure. P= 1 atm, T= 900+273= 1173K,
V=15450*0.0821*1173/1 = 1487886 L/s.
Ideal gas law equation is applicable at low pressure and high enough temperature. For the present conditions, the temperature and pressure are high and low enough to apply ideal gas law equation.
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