2. (24 marks total) A process for gasifying coal produces a fuel gas with a comp

Question

2. (24 marks total) A process for gasifying coal produces a fuel gas with a composition of 13 25% Co, and 62% H2 by volume. (a) (4 marks) Calculate the specific volume of this gas at 25 °C and 101.3 kPa in m/kg. (b) (5 marks) Determine the amount of oxygen required for stoichiometric combustion of this fuel in kmol O2 / kmol fuel. (c) (5 marks) The products of stoichiometric combustion in dry air are cooled to 40°C at a total pressure of 101.3 kPa. Determine the amount of product water condensed as liquid in kmol water /kmol fuel. (d) (10 marks) This fuel is now burned with excess air. If the products of combustion contain 11% CO, by volume on a dry basis, calculate the percent excess air. Assume that the air is dry % CH 4,

Explanation / Answer

From gas law, PV= nRT

Where n= no of moles = mass/molar mass

Basis : 1 mole of gas , it contains 0.13 moles of CH4, 0.25 moles of CO and 0.62 moles of H2 ( since at constant pressure, volume% = mole%)

For molar mass= molar mass of CH4* mole fraction of CH4+molar mass of CO* mole fraction of CO+ molar mass of H2* mole fraction

=0.13*16+0.25*28+2*0.62= 10.32 g/mole

For 1 mole , the gas law equation becomes

PV= RT, V= RT/P,when R= 0.0821*10-3 m3atm/mole.K and T= 25+273= 298K

P= 101.3 Kpa= 101.3/101.3 atm =1 atm (101.3 Kpa= 1atm)

V= 0.0821*10-3*298/1=0.0244 m3

Hence specific volume= 0.0244/10.32*10-3 m3/kg=2.346 m3/kg           

Basis : 1 kmole of fuel, moles : CH4=0.13, CO= 0.25 and H2=0.62

Combustion reactions of these fuels as

CH4+ 2O2---àCO2+2H2O (1) , CO+0.5O2---àCO2 (2) and H2+0.5O2-àH2O (3)

1 mole of CH4 requires 2moles of oxygen from reaction-1, 1 mole of CO requires 0.5 moles of oxygen from reaction-2 and 1 mole of CH4 requires 0.5 moles of O2 to produce 1 mole of H2O

Moles of O2 requried ( total)=0.13*2+0.5*0.25+0.62*0.5= 0.26+0.125+0.31=0.695 kmoles

Hence kmoles of oxygen required/ kmole of fuel =0.695/1=0.695 kmoles

Products of combustion

CO2= 0.13 moles ( from combustion of CH4), CO2=0.25 moles from combustion of CO and 0.62 moles of H2O

Products of combustion

CO2= 0.13+0.25=0.38 and H2O=0.62 moles (reaction-3)+2*0.13 ( from reactiom-1)=0.62+0.26=0.88 moles

Air contains 21% O2 and 79% N2, moles of N2 added from air

0.79*(0.695/0.21)=2.61 moles      

Moles of CO2 +N2= 2.61+0.38= 2.99 kmoles

at 40 deg.c, vapor pressure of water=55.3 Torr= (55.3/760)*101.3 Kpa =7.4 Kpa

water condenses when partial pressure of water vapor = vapor pressure of liqudi at the given temperature

partial pressure of water vapor =7.4Kpa and parital pressure of CO2+N2= 101.3-7.4= 93.9 Kpa

mole of water vapor present/ mole of CO2 and N2 present = partial pressure of water vapor/ partial pressure of CO2+N2

since moles of CO2 and N2 together remain constant

moles of water vapor/2.99= 7.4/93.9

moles of water vapor present = 0.24 moles

moles of water condensed= moles of water initially present – moles of water present at 40 deg.c= 0.88-0.24=0.64 k moles

this much water is condensed per kmole of fuel.

hence water condensed/ kmole of fuel =0.64/1

stoichiometric requirement of oxygen = 0.695

moles of air to be supplied =0.695/0.21=3.31

let x =% of excess air

air supplied = (1+x/100)*3.31

prodcuts contains

N2 present = 2.61*(1+0.01x)

O2 = O2 supplied-O2 reacted = 3.31*(1+x/100)*0.21-0.695= 0.00695x

CO2= 0.38 moles

Total moles on dry basis = 2.61+2.61*0.01x+0.00695x+0.38 =2.99+0.03305x

Mole fraction of CO2 in the product = 0.38/(2.99+0.03305x)= 0.11 ( given)

0.38= 0.11*(2.99+0.03305x)

0.38=0.11*2.99 +0.11*0.03305x

Hence x=14.05%

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