Problem 9.60 Fuel Gas Combustion A fuel gas containing 5.000 mole% methane and t

Question

Problem 9.60 Fuel Gas Combustion A fuel gas containing 5.000 mole% methane and the balance ethane is burned completely with pure oxygen at 25.000 and the products are cooled to 25.0° C Physical Property Tables ? Continuous Reactor 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-Qlkw), 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. State of water (kw) liquid vapor

Explanation / Answer

consider

1 MOLE OF gas mixture-

methane ( CH4 ) =0.05 Moles/s

& ethane (C2H6) =0.95 Moles/s

temperature - T =273 +25 =298 k

CH4 ( g) +2O2 ( g) = CO2 ( g) +2H2O ( Vapor)

, Hc ( 298 ) = -802 kj/mol

1 MOL + 2 MOL = 1 MOL + 2 MOLES

0.05MOL + 2 *0.05 = 0.05 MOL + 2*0.05 MOLES

0.05 MOLES + 0.1 MOLES O2 = 0.05 MOLES +0.1 MOLES

C2H6 ( g) + 7/2 O2 (g) = 2 CO2 ( g) + 3H2O ( vapor) ,

Hc = -1428.0 kj /mol

1 mol + 3.5 moles = 2 moles + 3 moles

1*0.95 mol + 3.5 *0.95moles = 2*0.95 moles + 3*0.95 moles

total moles of Oxygen ( O2 ) = 0.1 + 3.325 =3.425MOLES

SUPPOSE 20 % EXCESS O2 ( OXYGEN ) IS TAKEN TO BURN THE FUEL MIXTURE COMPLETELY

TOTAL O2 = 3.425 *((100 +20)/100) = 4.11

Mass OF OXYGEN = 4.11* 32 =131.52 g

total heat transfer = 0.05*( -802 ) +( - 1428) *0.95 = -1396.7kJ/mol.s

1 J /s = 1 watt

1 kJ /s = 1 kw

1 kJ = 1 kw -s

Q = -1396.7 kW - S

HEAT OF PRODUCTS MUST BE BROUGHT BACK TO STANDARD CONDITION OF 25oC ( 298 k ) and 1 atm

- H = - Q ( HEAT OF COMBUSTION AT 25oC and 1 atm pressure) H = -1396.7 KW/mol

& if water is condense the extra heat need to remove that is equal to heat need remove in above case & latent heat of vaporisation of water molecules

the molar heat of vaporisation of water at 298 K= 44.01 kJ/mol

as total water molecules = 2.95 moles

heat of vaporisation = 2.95*44.01 =129.8295kJ/mol need to remove extra

so for liquid -Q=1396.7+129.8295 =1526.5295kW

& for gas -Q= 1396.7 kW

b)Closed Vessel at constant Volume

for isochoric systems Q=U =H-PV

volume can be calculated by ideal gas law

PV=nRT

n= (0.15[total moles produces by combution of 0.05 mole of methane]+4.75[total moles produces by combution of 0.95 mole of ethane]+(4.11-3.425)[excess oxygen])=5.585 moles

101325*V = 5.585*8.314*298.16

V=0.136636m3

for vapor phase Q=-1396.7-101.325*0.136636 =-1410.544kJ

similarly for liquid

calculate volume dont include moles of water produce while calculating volume as they are condensed

n=0.05 [moles of CO2 by combution of 0.05 mol methane]+2*0.95[moles of CO2 by burning 0.95 mole of ethane ]+4.11-3.425 [excess oxygen]=2.635moles

101325*V=2.635*8.314*298.16

V=0.064464m3

Q = -1526.5295-PV

Q= -1526.5295-101.325*0.064464=-1533.0613kJ

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