P1 V2 P2 n2 y2 pane (C3H8) is burned in a combustion chamber at a flow rate of 1

Question

P1 V2 P2 n2 y2 pane (C3H8) is burned in a combustion chamber at a flow rate of 10.00 m3/min with 30.0% excess air. If the propane and air streams enter the chamber at 17.00C and 1.00 atm (absolute), and the products leave the chamber at 275.00C and atm the flow rate of product gas from the 100% conversion and ideal behavior. solve for the flow rate, first label the flow diagram below. Label any quantities as "unknown" if they are not given or implied above and "none" if the quantity is zero. Do not leave any blank spaces. Variable names are given for reference later. m3/min CRH 18 m3/min product OC °C T3 atm atm mol/min product m3/min air Combustion Chamber mol/min CO2 n4 atm mol/min H20 n5 mol/min air mol/min O2 n6 mol O2/ mol air mol/min N2 17 mol N2/mol air unknown 1.00 0.79 10.00 275.0 n8 mol/min C3H 4.60 17.0 0.21 none

Explanation / Answer

use basis of 1 minute

at T = 17 degC = 290.15 K
and
p = 1.0 atm = 101325 Pa
and
V = 10m^3

pV = nRT

so,
101325*10 = n*8.3145*290.15

n = 420 mol

for complete combustion 1 mol propane needs 5 mol O2

so 420 mol propane needs 5*420 = 2100 mol O2

and we have 30% excess air

so O2 actually fed = 1.3*2100 = 2730 moles

and N2 that comes along with 2730 moles of O2 = (79/21)*2730 = 10270 moles

total moles of air fed = 10270+2730 = 13000 moles

so, 10270 moles N2 and 630 moles O2 remain un-reacted.

upon combustion 1 mol propane gives 3 mol CO2 and 4 mol H2O vapor
or 1 mol propane gives 7 mol gaseous products

so 420 mol propane will give 420 = 1960 mol total gaseous products

so in product total gaseous products =
(moles of CO2+moles of steam) + (moles of O2 and N2 unreacted)
= 1960+10270+630 = 12860 moles

at exit side
p = 4.6 atm = 466095 Pa
T = 275 degC = 548.15 K
n = 12860 moles

so, V =(nRT/p) = 125.75 m^3

so answers for part

(a) 420 mol/min

(b) 13000 mol/min

(c) 12860 mol/min

(d) 125.75 m^3/min

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