Question

Chemical Engineering Question

100 mol/minute of propane gas is burned completely in a continuous reactor with air to form carbon dioxide and water. Develop an Excel spreadsheet to calculate the adiabatic flame temperature if the gasses are fed at 30oC, 1atm for the following cases:

% Excess Air = 0%, +10%, +25%, +50%, +75%, +100%, +150%, +200%

Use heat capacity data from Table B.2 for all components in the combustion reaction.Turn in a hard copy of the Excel spreadsheet as well as a plot of adiabatic flame temperature vs.% excess air.

Answer 1

propane (C3H8) combustion can be represened as

C3H8+ 5O2-----> 3CO2+ 4H2O

the probelm can be divided into two parts. one mass balance and the second energy balance

mass balance

1 mole of C3H8 requires 5 moles of oxygen and gives 3 moles of CO2 and 4 moles of H2O. moles of air = 1/0.21 ( since air contains 21% O2 and 79% N2) = 4.76 moles. ( this data can be used for calculation of air flow rate requireemnt for any moles of feed)

excess air to be supplied = 4.76*(1+x/100), where x = % excess air

N2= 4.76*(1+x/100)*0.79

O2 = 4.76*(1+x/100)*0.21

the products of combustion contains 3 moles of CO2, 4 moles of H2O, 4.76*(1+x/100)*0.79 and oxygen= 4.76*(1+x/100)*0.21- 5.   The mass balance on spread shee it shown below.

							Products of combustion
Moles of propane ( moles/min)	oxygen	Air   required ( mol/min)	% excess air	air to be upplied	N2( Mole/min)	O2 (in air supplied)	CO2	H2O	N2	O2
100	500	2380.952381	0	2380.952381	1880.952381	500	300	400	1880.952	0
100	500	2380.952381	10	2619.047619	2069.047619	550	300	400	2069.048	50
100	500	2380.952381	25	2976.190476	2351.190476	625	300	400	2351.19	125
100	500	2380.952381	50	3571.428571	2821.428571	750	300	400	2821.429	250
100	500	2380.952381	75	4166.666667	3291.666667	875	300	400	3291.667	375
100	500	2380.952381	100	4761.904762	3761.904762	1000	300	400	3761.905	500
100	500	2380.952381	150	5952.380952	4702.380952	1250	300	400	4702.381	750
100	500	2380.952381	200	7142.857143	5642.857143	1500	300	400	5642.857	1000

Energy balance

heat trasnferred= enthalpy of products + heat of reaction - enthalpy of reactants

since reactants are at 25 deg.c and 1 atm, enthalpy of products =0

for adiabatic operation , heat transferred = 0

enthalpy of prodcucts =- heat of reaction

heat of reaction = -2220.1Kj/mol

So heat of reaction for 100 mol/min = -2220.1*100 Kj/min=2220100 Kj/min=2220100/60 Kj/sec= 37002 Kj/sec= 37002 KW

CP/R data : CO2 : 5.457+1.045*10-3 T, O2= 3.639+0.506*10-3*T, N2= 3.280+0.593*10-3 T , H2O= 3.470+1.450*10-3 T

enthalpy change= number of moles* integral of CpdT ( between T and 298)=37002 KW