4, (11 marks total) A natural gas consists of 90% CH, and 10% N2 by volume. Afte

Question

4, (11 marks total) A natural gas consists of 90% CH, and 10% N2 by volume. After it is burned in a furnace, the products of combustion contain 9% CO2 by volume on a dry basis. The total pressure is 101 kPa, and the natural gas and air both enter the furnace at 25°C. (a) (4 marks) Determine the percent excess air. (b) (3 marks) The products leave the furnace at 40°C. Determine the composition of the products after condensation in kmolVkmol fuel. (c) (4 marks) Calculate the heat transfer to the furnace in MJ/kmol fuel.

Explanation / Answer

Basis : 1 mole of natural gas containing 90% CH4 and 10% N2. Moles of CH4= 0.9 and moles of N2=0.1

The combustion of methane is CH4+ 2O2---->CO2+2H2O

1 mole of CH4 requires 2 moles of oxygen for complete combustion

0.9 moles of CH4 requires 2*0.9= 1.8 moles of oxygen for complete combustion. Air contains 21% O2 and 79% N2, moles of air = 1.8/0.21=8.57, let x= % of excess air.

So moles of air supplied =8.57*(1+x/100)= 8.57+0.0857x

Moles of N2 in air = 0.79*(8.57+0.0857x)= 6.77+0.068x

Total moles of N2= 6.77+0.068x+0.1= 6.87+0.068x

Moles of CO2 formed =0.9 moles, moles of oxygen remaining = 0.21*(8.57+0.0857x)-1.8= 0.018x

Products ( dry basis): CO2=0.9, O2=0.018x and N2= 6.87+0.068x

Total moles ( on dry basis )= 0.9+0.018x+6.87+0.068x= 7.77+0.086x

Mole % of CO2= 100*(0.9/(7.77+0.086x)= 9

10/(7.77+0.086x)= 1

10= 7.77+0.086x, 0.086x= 10-7.77, x= 25.93%

Products on wet basis ( moles): CO2=0.9, O2=0.018*25.93=0.468, H2O= 2*0.9=1.8, N2=6.87+0.086*25.93=9.1, moles of dry gas = 0.9+0.468+9.1=10.468

At 40 deg.c, vapor pressure of water =55.3mm Hg= 55.3/760 atm ( 1atm= 760 mm Hg)= 0.073 atm

At dew point of 40 deg.c, partial pressure of water vapor = vapor pressure of liquid =0.073 atm

Hence moles of water vapor / moles of dry gas = partial pressure of water vapor/partial pressure of dry gas

Moles of water vapor/10.468= 0.073/(1-0.073)

Moles of water vapor at 40 deg.c= 10.468*0.073/(1-0.073)= 0.82

Moles of water vapor condensed/mole of fuel = 1.8-0.82= 0.98

Kmoles of water vapor condensed/kmoels of fuel =0.98

Rate of heat transfer= enthalpy of products+ heat of reaction- enthalpy of reactants

Since feed is entering at 25 deg.c, taking reference temperature is 25 deg.c, enthalpy of reactants=0

Standard heat of formation of CH4= -67 Kj/mole,

But moles of CH4= 0.9, standard heat=-67*0.9 KJ=-60.3 KJ

Enthalpy of products ( since products are at 40 deg.c and feed is at 25 deg.c, it is reasonable to assume the specific heat of gases to be independent of temperature and an average specific heat can be assumed.

Products at 40 deg.c: H2O(g)= 0.98 moles, CO2=0.9 moles, N2= 9.1 and O2=0.468

1Kcal= 4184 joules, 1 cal =4.184 joules

Cp data : CO2=50 j/moles.deg.c , N2= 31.41 J/mole. Deg.c, O2= 33 J/mole.K and H2O= 33.1 J/mole.deg.c

Enthalpy of product = 0.98*33.1*(40-25)+0.9*50*(40-25)+9.1*31.41*(40-25)+0.468*33*(40-25)=5681 joules

Heat transferred= 5681-60.3*1000=-54619 Joules= -54.619 Kj

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