Please help with part (c), I have added my answers to part (a) and (b). The lowe
ID: 1007359 • Letter: P
Question
Please help with part (c), I have added my answers to part (a) and (b).
The lower flammability limit (LFL) and the upper flammability limit (UFL) of C3H8 in air at 1 atm are 2.3 mole% and 9.5 mole%. If the percentage of C3H8 is below the LFL, the mixture is said to be too lean; if it is above the UFL, the mixture is too rich to ignite.
(a) Which would be safer to release into the atmosphere, a fuel-air mixture that is too lean or too rich to ignite? Explain.
ANS: It's more dangeroud the release a mixture too rich to ignite because it could get diluted and fall between the limits of flammability.
(b) If 4.3 mole% C3H8 in air enters a furnace at a rate of 150mol C3H8/s, what is the minimum molar flow rate of the diluting air to get the propane concentration down to LFL?
ANS: 150/0.0203 = 7389.16
Molar flowrate of air = 7389.16 - 150 = 7239.16
(c) The actual diluting air molar flow rate is specified to be 130% of the minimum value. Assuming the fuel mixture (4.03 mol% C3H8) enters the furnace at the same rate as in part (b) at 125 degrees C and 131 kPa and the diluting air enters at 25 degrees C and 110kPa, calculate the ratio (m3 diluting air)/(m3 fuel gas) and the mole percent of C3H8 in the diluted mixture.
Explanation / Answer
Given 130% of min mol flow rate , for diluting mixture (130/100)*7239.16=9410.908 mole/sec
PV=nRT , 110*10^3*V=9410.908*8.314*298
V=211.96 m3
For fuel mixture,
4.03/100 *x=150 ,x = 3722.08 mole/sec (mola flow rate of air)
PV =nRT, 131*10^3*V=3722.08*8.314*398
V= 94.017 m3
Ratio = 211.96/94.017=2.2544
Mole percent of c3h8/100 *9410.908=150
Mole % =1.59%