Question

1. Fuels for motor vehicles other than gasoline are being evaluated because they generate lower levels of pollutants than does gasoline. Compressed propane(C3H8) has been suggested as a source of power for vehicles. Supposed that in a test 30 kg of C3H8 is burned with 400 kg ofair(21mole%O2,therestN2)toproduce44kgofCO2 and12kgofCO.Inorderto calculate the percent excess air, follow the below steps.

(a) What are the molecular weights of C3H8, air(21 mole % O2, the rest N2), and O2? Hence what is the moles of air entering the system?

(b) Percentage of excess air is based on the complete combustion. Write down the stoichio- metric equation of relevant reaction. Check whether the reaction equation is correctly balanced.

(c) Calculate the theoretical air (moles of air required for complete combustion) and percent excess air.

2. A stream of air (21 mole % O2, the rest N2) flowing at a rate of 10.0 kg/h is mixed with a stream of CO2. The CO2 enters the mixer at a rate of 20.0 m3/h at 150? and 1.5 bar. What is the mole percentage of CO2 in the product stream. ( hint: you can assume the inlet stream of CO2 as an ideal gas. )

Answer 1

1. a. Molecular Weight of C3H8 = 44 kg/kmol

Molecular Weight of Air = mol fraction of O2*Mol Wt of O2 + mol fraction of N2*Mol Wt of N2

=0.21*32+0.79*28 = 28.84 kg/kmol

Mol Wt of O2 = 32 kg//kmol

Amount of air entering into the system = 400 kg

Amount of air entering into the system in kmol = 400/28.84 = 13.87 kmol

1.b. Stoichio- metric equation for Complete Comustion of propane

C3H8 + 5 O2 = 3 CO2 + 4 H2O or

C3H8 + 5 O2 + 18.8 N2 = 3 CO2 + 4 H2O + 18.8 N2

1.C. From stiochimetric equation, for 1 mol of propane 5 mol of O2 required.

30 kg of propane entering into the system

Amount of propane entered in kmol = 30/44 =0.68 kmol.

For complete combustion, therotical O2 required = 0.68*5 = 3.4 kmol

hence the theoretical air required = 3.4/0.21 = 16.23 kmol of air or 468.18 kg of Air.

percent Excess Air = (Actual Air - stoichiometric air)/stoichiometric air*100 = (400-468.18)/468.18*100 = -14.5 %

2. Amount of Air stream = 10 kg/h

Mol wt of Air = 28.84

Amount of air in kmol = 20/28.84 = 0.345 kmol

CO2 assume as an ideal gas

As per ideal gas law PV=nRT

n= (1.5 *20 )/(423.15 * 8.314*10^-5) = 852.74 gmol

Total mol in product stream = 0.8527 +  0.345 = 1.1977 kmol

Mol percentage of CO2 in product stream = 0.8527 / 1.1977 = 71.2 %