In the event of an explosive combustion of vapor at atmospheric pressure, the va
ID: 483991 • Letter: I
Question
In the event of an explosive combustion of vapor at atmospheric pressure, the vapor cloud can be modeled as adiabatic because the combustion occurs so rapidly. The vapor cloud expands rapidly due to the increase in moles due to combustion, but also due to the adiabatic temperature rise. Estimate the volume increase of a 22°C, 1 m3 mixture of propane and a stoichiometric quantity of air that burns explosively to completion. Estimate the temperature rise. Thermodynamics Problem (Thermodynamics of Processes)
Explanation / Answer
The reaction is:
C3H8 + 5O2 = 3CO2 + 4H2O
taking stoiciometric quantities
1 mole of propane requires 5 mole of oxygen to give 3 moles of carbon dioxide and 4 moles of water.
Propane added
PV = nRT (ideal gas equation)
P = pressure = 1 atm
V = volume = 1 m3
R = gas constant
T = temperature = 22 + 273 = 295 K
Hence, n = number of moles = PV/RT
n = 105*1/(8.314*295)
n = 40.77 moles
Hence,40.77 moles of propane will require 5*40.77 = 203.862 moles of oxygen
Air contains 21% oxygen and 79% nitrogen.
Hence,
stoichiometric amount of air required = 203.862 /0.21 = 970.775 moles of air is required.
volume of air required = 970.775 * 22.4 = 21745.36 L = 21.745 m3
Hence, initial moles added = 40.77 propane + 970.775 moles of air = 1011.545 moles
total volume initially added = 1 propane + 21.745 air = 22.745 m3 mixture
Now, after the reaction, assuming 100% conversion
moles of carbon dioxide obtaines = 3 * 40.77 = 122.31 moles
volume of carbon dioxide = 122.31 * 22.4 = 2739.744 L = 2.7397 m3
moles of water formed = 40.77 * 4 = 163.08 moles
volume = 163.08 * 22.4 = 3652.992 L = 3.6529 m3
Volume of nitrogen unreacted = 970.775 * 0.79 * 22.4 = 17178.8334 L
= 17.1788 m3
Total final volume = 2.7397 CO2 + 3.6529 H2O + 17.1788
Total volume in the end = 23.5714 m3
volume increase = 23.5714 - 22.745 =0.8264 m3
Since it is an adiabatic operation, the temperature and volume relationship is:
(T2 / T1) = (V1 / V2)k-1
T1 = 295 K
V1 = 22.745
V2 = 23.5714
k = 1.4 (assuming ideal gas, since specific heats at constant volume and pressure is not given in question)
Substitute to get T2 = 299.241 K = 26.09 celsius