A fuel gas containing 45.00 mole% methane and the balance ethane is burned compl

Question

A fuel gas containing 45.00 mole% methane and the balance ethane is burned completely with pure oxygen at 25.00 degree C, and the products are cooled to 25.00 degree C. Suppose the reactor is continuous. Take a basis of calculation of 1.000 mol/s of the fuel gas, assume some value for the percent excess oxygen fed to the reactor (the value you choose will not affect the results), and calculate - Q(kW), the rate at which heat must be transferred from the reactor if the water vapor condenses before leaving the reactor and if the water remains as a vapor. Now suppose the combustion takes place in a constant-volume batch reactor. Take a basis of calculation 1.000 mol of the fuel gas charged into the reactor, assume any percent excess oxygen, and calculate -Q(kJ) for the cases of liquid water and water vapor as products.

Explanation / Answer

For contonuous reactor

total feed= 1 mol/s

methane= 45%, XCH4= 0.45, nCH4 = 0.45*1= 0.45 mol/s

ethane= 55%, XC2H6= 0.55, nC2H6= 0.55*1 = 0.55 mol/s

Complete combustion reaction can be written as for case where water condenses before leaving the reactor

CH4 + C2H6 + 11/2 O2 = 3 CO2 (g) + 5 H2O (l)

1 mol CH4 = 1 mol C2H6 = 11/2 mol O2 = 3 mol CO2 = 5 mol H2O .................as per stoichimetry

0.45 mol/s CH4 = 0.45 mol/s C2H6= 2.475 mol/s O2 = 1.35 mol/s CO2 = 2.25 mol/s H2O (l) .......as per given condition

Standard heat of formation for these gases are as follows

For CH4, -74.9 kJ/mol, For C2H6 -83.7 kJ/mol, For CO2 -393.509, For H2O(l) -282.8 kJ/mol, For O2 0 kJ/mol

Then

Heat of reaction = (1.35*-393.509 + 2.25*-285.8) - (0.45*-74.9 + 0.45*-83.7 + 2.475 * 0)

Heat of reaction = -1102.93 kJ/s

Heat of combustion Q = -heat of reaction = 1102.93 kJ/s = 1102.93 kW

Hence -Q= -1102.93 kW of heat needs to be transfred from reactor if water condenses before leaving the reactor.

Now,

Complete combustion reaction can be written as for case where water remains as vapor in the reactor

CH4 + C2H6 + 11/2 O2 = 3CO2 (g) + 5H2O (g)

1 mol CH4 = 1 mol C2H6 = 11/2 mol O2 = 3 mol CO2 = 5 mol H2O .................as per stoichimetry

0.45 mol/s CH4 = 0.45 mol/s C2H6= 2.475 mol/s O2 = 1.35 mol/s CO2 = 2.25 mol/s H2O (g) .......as per given condition

Standard heat of formation for these gases are as follows

For CH4, -74.9 kJ/mol, For C2H6 -83.7 kJ/mol, For CO2 -393.509, For H2O(g) -241.818 kJ/mol, For O2 0 kJ/mol

Then

Heat of reaction = (1.35*-393.509 + 2.25*-241.818) - (0.45*-74.9 + 0.45*-83.7 + 2.475 * 0)

Heat of reaction = -1003.96 kJ/s

Heat of combustion Q = -heat of reaction = 1003.96 kJ/s = 1003.96 kW

Hence -Q= -1003.96 kW of heat needs to be transfred from reactor if water remains as vapor in the reactor.

For vessel of constant volume

feed of 1 mol,

No change total feed or feed compostion for continuous reactor and constant volume vessel except flow. Hence there will be no change in the total heat of combustion value except in the first case it is heat of combustion per second (kJ/s) and in second it is total heat of combustion (kJ)

no change in the the Q value in both cases

-Q= -1102.93 kJ for liquid water product

and -Q= -1003.96 kJ for water vapor product

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