12 L/min of liquid benzene are to be burned with 40% excess air to produce steam
ID: 995710 • Letter: 1
Question
12 L/min of liquid benzene are to be burned with 40% excess air to produce steam (60 bar and 500 degrees C) from liquid water (60 bar and 20 degrees C. Before entering the combustor, the air is pre-heated from 25 degrees C to 200 degrees C.
a) How much heat must be supplied to preheat the air?
b) Suppose benzene is completely burned out with 5% of the burned benzene from CO and the flue gas leaves the combustor at 1200 degrees C. Determine the steam production rate in m3/min (assume all heat generated from the combustion is used to produce steam, i.e., no heat loss from the combustor to the surroundings.)
Explanation / Answer
Flow rate of Benzene= 12 L/min =12*1000cc/min and density= 0.876 g/cc
mass flow rate of Benzene = Volumetirc flow rate* density = 12000*0.876=10512g/min
Molecular weight of Benzene= 78, moles of Benzene= 10512/78=134.7692 moles/min
The reaction between Benzene and air is given by C6H6+7.5O2---> 6CO2 +3H2O
From the reaction, 1 moles of Benzene requires 7.5 moles of oxygen
134.7692 moles/min requires 134.7692*7.5=1010.769 moles/min of air
Ais is supplied 40% in excess, hence moles of air= 1.4*1010.769 =1415.1 moles/min
Molecular weight of air= 29, mass of air= 1415.1*29=41038 gms/min
The specific heat of air =1.003 j/g,deg.c and hence heat to be supplied= mass flow rate* specific heat* temperature difference= 41038*1.003*(250-25)=9261251 joules/min
b) Benzene incomplete combustion requires
C6H6+ 4.5O2----> 6CO +3H2O
Enthalpy change of reaction = 6* enthalpy change of CO + 3* enthalpy change of water vapor- 1* Enthalpy change of Benzene = 6*(-110.5) + 3*(-241.82)+49.2Kj/mol=-1336.26 Kj/mol
1mole os Benzene produces 1336.26 Kj of energy
134.7692*5/100 moles of Benzene ( since 5% of Benzene is burned to produce CO)= 6.74 moles/min
6.74 moles/min gives 6.74*1336.26 Kj=9006.392 Kj, steam generated= 3*6.74 moles/min=20.22 moles/min
This heat energy is used to completely burn rest of the Benzene
Rest of the Benzene= 134.7692-6.74=128.092
The reaction is C6H6 + 15/2O2 -> 3H2O + 6CO2
moles of Benzene utilized= 128.092, enthalpy of combustion of benzene =-6534 Kj
Heat avialble= 9006.392 Kj/min, So moles of Benzene burned = 9006.392/6534=1.378 Moles/min
Moles of Steam produced= 1.378*3 moles/min=4.134 moles/min
total steam produced =4.134+20.22 =24.354 moles/min= 24.354*29 gm/min=706g/min=0.706 kg/min
at STP density of steam is18/22.4= 0.8 kg/m3
at 1200 deg.c( assumig a pressure of 1 atm), density = 0.8*273.15/(1200+273.15)=0.15 kg/m3
volumetrif flow rate of steam =0.706/0.15 =4.71 kg/min
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.