A fuel mixture of 30% Propylene and 70% Methyl Alcohol, on a molar basis, at 25
ID: 500070 • Letter: A
Question
A fuel mixture of 30% Propylene and 70% Methyl Alcohol, on a molar basis, at 25 degree C and 100 kPa is combusted with 200% theoretical air at 527 C and 100 kPa. For this steady-state process, the products of combustion exit the combustion chamber at 1000 degree C and 100 kPa. (a) Write the theoretical (or stoichiometric) chemical reaction; (b) Write the actual chemical reaction; (c) Determine the mole fractions of the actual products of combustion; (D) Determine the dew point temperature of the appropriate component in the products of combustion; (e) Determine the mass percentage of the products of combustion; and, (f) Determine the volume occupied by 2 kg of the product of combustion mixture.Explanation / Answer
The combustion of propylene is C3H6+ 4.5 O2------>3CO2 + 3H2O (1)
The combustion of methyl alcohol CH3OH + 1.5 O2 ----->CO2 + 2H2O (2)
Basis : 1 mole of mixture. It contains 0.3 moles of C3H6 and 0.7 moles of CH3OH.
moles of oxygen required for reaction -1= 4.5*0.3= 1.35 and for reaction -2 = 0.7*1.5= 10.5
Air contains 21% O2 and 79% N2, moles of air required= 1.35/0.21= 6.43 for reaction-1 and 10.5/0.21=50 for reaction-2
Air supplied is 200% excess, air supplied for reaction-1= 6.43*2= 12.86 and for reaction-2 = 50*2= 100
moles Oxygen in reaction-1= 12.86*0.21=2.7 , N2= 12.86*0.79= 10.16 and for reaction-2 = 100*0.21= 21, N2= 100*0.79=79
Oxygen consumed in reaction -1 = 2.7-1.35=1.35 , N2 =10.16, for reaction-2, O2= 21-10.5=10.5, N2 for reaction-1= 10.16 and for reaction-2 =79
Hence 0.3C3H6+1.35O2 +1.35O2 +10.16 N2 ------> 0.9CO2+0.9H2O +10.16 N2
for reaction-2, 0.7CH3OH +10.5O2 +10.5O2+79N2------->0.7CO2+1.4H2O+10.5O2+79N2
Products (Moles) : CO2= 0.9+0.7=1.6, H2O= 0.9+1.4 =2.3, N2= 10.16+79= 89.16, O2= 1.35+10.50= 11.85
total moles of products = 1.6+2.3+89.16+11.85=104.91
Composition of Products : CO2= 1.6/104.91=0.015, H2O= 2.3/104.91=0.022, N2= 89.16/104.91=0.85, O2= 11.85/104.91 =0.1129
partial pressure of water vapor = 0.022*100 = 2.2Kpa, 101.3 Kpa= 760mm , 2.2 Kpa= (2.2*760/101.3)= 16.5 mm Hg
Antoine constants for water , log Psat(mmHg)=8.07131-1730.63/(t+233.426)
Dew point is the temperature at which partial pressure = vapor pressure= 16.5 mm Hg
From Antoine equation, log(16.5)= 8.07131- 1730.63/(t+233.426)
6.85= 1730.63/(t+233.426)
t+233.426= 1730.63/6.85= 252.51
t= 252.51-233.426= 19.084 deg.c which is the dew point.
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