Question

A heater burns normal octane (n-C₈H₁₈) using 20.0 percent excess air. Combustion is complete. The flue gas leaves the stack at a pressure of 100 kPa and a temperature of 160⁰C.

a) Calculate the complete flue gas analysis.

b) What is the volume of the flue gas in cubic meter per kg mol of n-octane?

Answer 1

C8H18 + 25/2O2 = 8CO2 + 9H2O

assume 100 mol of C8H18

then, we need 25/2 --> 1250 mol of O2

but we need 20% excess so 1250*1.2 = 1500 mol of O2

for air---> 0.21 mol of air is O2

so

1500 mol of O2 -- >1500/0.21 = 7142.85 mol of air

so:

a)

mol of C8H18 left --> 0

mol of O2 left --> 1500 -1250 = 250 mol of O2 left

mol of N2 left = 7142.85*0.79 = 5642.85 mol of N2

mol of CO2 produced = 1:8 ratio so --> 8*100 = 800 mol of CO2 produced

mol of H2O produced = 1:9 ratio so --> 9*100 = 900 mol of H2O produced

Total mol = 250 + 5642.85 + 800 +900 = 7592.85 mol

x-N2 = 5642.85 /7592.85 = 0.7431

x-H2O = 900 / 7592.85 = 0.11853

x-CO2 = 800 /7592.85 = 0.105362

x-O2 = 900/ 7592.85 = 0.1185

b)

total mol = 100 mol, so this is actuall 0.1 kmol

we must multiply by 10*7592.85 = 75928.5 mol

volume

PV = nRT

V = nRT/P = 7592.85 *0.082*(160) / (1) = 99618.192 L --> 99618.192/(1000) m3 = 99.618 m3 per kmol