1. Fuels for motor vehicles other than gasoline are being evaluated because they

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. ) 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. ) 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. )

Explanation / Answer

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 %

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