16 Energy Balances on Reactive Process
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Manolito E. Bambase Jr Assistant Professor, Department of Chemical Engineering CEAT, University of the Philippines, Los Banos, Laguna, Philippines
16-1. Roasting of Iron Pyrite Ore
An iron pyrite ore containing 85.0% FeS2 and 15.0% gangue (inert dirt, rock, etc.) is roasted with an amunt equal to 200% excess air according to the reaction 4 FeS2 + 11 O2 ======> 2 Fe2O3 + 8 SO2 in order to produce SO2. All the gangue plus the Fe2O3 and unreacted FeS2 end up in the solid waste product (cinder). Analysis shows the cinder contains 4.0% FeS2. Determine the heat transfer per kilogram of ore to keep the product stream at 250C if the entering stream streams are at 250C. Assume that pressure is constant at 1 atm. 1
16-1. Roasting of Iron Pyrite Ore
Ore, F = 100 kg
Gas Products, P mol
0.85 kg FeS2/kg F 0.15 kg gangue/kg F
n1 mol SO2 n2 mol O2 n3 mol N2
Reactor Air, A mol
Cinder, C kg
0.21 mol O2/mol A 0.79 mol N2/mol A 200% excess
m1 kg gangue m2 kg Fe2O3 m3 kg FeS2
4 FeS2 + 11 O2 ======> 2 Fe2O3 + 8 SO2 2
16-1. Roasting of Iron Pyrite Ore
The heat transfer required is calculated form the energy balance equation. Assuming DK = DP = WS = 0, then Q = DH = SHout – SHin The total enthalpy at the inlet is: SHin = (Hgangue + HFeS2 + HN2 + HO2)1 The total enthalpy at the outlet is: SHout = (HSO2 + HO2 + HN2 + Hgangue + HFe2O3 + HFeS2)2
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16-1. Roasting of Iron Pyrite Ore
Hence, Q = (HSO2 + HO2 + HN2 + Hgangue + HFe2O3 + HFeS2)2 – (Hgangue + HFeS2 + HN2 + HO2)1 Since the amount of N2 and gangue are the same at the inlet and out let and the there are no changes in T and P conditions at the inlet and outlet, then (Hgangue)1 = (Hgangue)2 and (HN2)1 = (HN2)2 And the heat transfer equation becomes, Q = (HSO2 + HO2 + HFe2O3 + HFeS2)2 – (HFeS2 + HO2)1 4
16-1. Roasting of Iron Pyrite Ore
From Table F.1, (Ĥfo)FeS2 = – 177.9 kJ/mol (Ĥfo)Fe2O3 = – 822.156 kJ/mol (Ĥfo)SO2 = – 296.90 kJ/mol (Ĥfo)O2 = 0.00 kJ/mol These are the specific enthalpies at 250C. The actual enthalpy at the given temperature condition is: T ˆ ˆ H H f CP dT TR
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16-1. Roasting of Iron Pyrite Ore
Since the actual temperature is also at 250C, then Ĥ = Ĥfo The total enthalpy of each component at the inlet and outlet is determined as: H = nĤ = nĤfo The molar amount of each component can be obtained thru material balances.
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16-1. Roasting of Iron Pyrite Ore
From excess air information: Total O2 in = 5.8437 kmol Total N2 in = 21.983 kmol Gangue Balance: N2 Balance:
m1 = 15.0 kg n3 = 21.983 kmol
S Balance (kmol) Fe Balance (kmol) O2 Balance (kmol)
2(85.0/120.0) = n1 + (m3/120.0)(2) (85.0/120.0) = (m2/159.02)2 + (m3/120.0) 5.8437 = n2 + n1 + (m2/159.02)(1.5)
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16-1. Roasting of Iron Pyrite Ore
Also, m3/(15.0 + m2 + m3) = 0.04 Solving the last 4 equations simultaneously, Fe2O3: FeS2:
m2 = 54.63 kg ====> 0.342 kmol m3 = 2.90 kg ====> 0.0242 kmol
SO2: O2:
n1 = 1.368 kmol n2 = 3.938 kmol
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16-1. Roasting of Iron Pyrite Ore
Solving for the total enthalpy of the inlet: SHin = (85.0/120.0 kmol)(–177.9 kJ/mol)(1000 mol/kmol) + (5.8437)(0) SHin = – 126,007 kJ
Solving for the total enthalpy of the outlet: SHout = (1.368)(–296.90)(1000) + (0.0242)(–177.90)(1000) + (1.368)(–296.90)(1000) SHout = – 691,641 kJ 9
16-1. Roasting of Iron Pyrite Ore
Solving for Q: Q = – 691,641 – ( – 126,007) = – 565,634 kJ per 100 kg of ore Per kg of ore: Q = – 5,656 kJ per kg of ore The negative sign indicates that meat must be removed during the process in order to maintain the temperature at 250C.
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Example 16-2. Combustion of Carbon Monoxide
Carbon monoxide at 500F is completely burned at 2 atm pressure with 50% excess air that is at 10000F. The products of combustion leave the combustion chamber at 8000F. Calculate the heat evolved from the combustion chamber expressed as Btu/lbm CO entering. 1 lbmol CO, 500F Air, 10000F n1 lbmol O2/lbmol n2 lbmol N2/lbmol 50% excess
Flue Gas, 8000F Combustion Chamber
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n3 lbmol CO2 n4 lbmol O2 n5 lbmolN2
Example 16-2. Combustion of Carbon Monoxide
With DK = DP = WS = 0, the heat evolved is calculated as: Q = DH = SHout – SHin Q = (HCO2 + HN2 + HO2)2 – (HCO + HN2 + HO2)1
Calculate total enthalpy of each component: Hi = niĤi = ni(Ĥfo + CPidT) The standard heat of formation can be obtained from Table F.1.
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Example 16-2. Combustion of Carbon Monoxide
The CP can be expressed as a function of temperature: CP = a + bT + gT2 Values of the constants for CO, CO2, O2, and N2 can be obtained from Table E.1. The molar amount of each component can be obtained thru material balance. n1 = ____ lbmol O2 ; n2 = ____ lbmol N2 n3 = ____ lbmol CO2; n4 = ____ lbmol O2 n5 = ____ lbmol N2 13
Example 16-2. Combustion of Carbon Monoxide
Standard heat of formation (Table F.1): (Ĥfo)CO = __________ Btu/lbmol (Ĥfo)CO2 = __________ Btu/lbmol (Ĥfo)N2 = (Ĥfo)O2 = 0 Heat capacity equations (Table E.1): (T is in 0F and CP is Btu/lbmol-0F)
(CP)CO2 (CP)CO (CP)N2 (CP)O2
= _____ + __________ T ___________ T2 = _____ + __________ T ___________ T2 = _____ + __________ T ___________ T2 = _____ + __________ T ___________ T2 14
Example 16-2. Combustion of Carbon Monoxide
Solving for the specific enthalpy: Products CO 2 :
___ ˆ H CO2 169,179 CP )CO2 dT _________ Btu / lbmol ___
N2 :
___ ˆ H N2 (0) CP ) N2 dT _________ Btu / lbmol
\O 2 :
___ ˆ H CO2 (0) CP )O2 dT _________ Btu / lbmol
___
___
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Example 16-2. Combustion of Carbon Monoxide
Reactants: CO :
N2 :
\O 2 :
___ ˆ H CO 47,515 CP )CO dT _________ Btu / lbmol ___
___ ˆ H N2 (0) T CP ) N2 dT _________ Btu / lbmol ___
___ ˆ H CO2 (0) CP )O2 dT _________ Btu / lbmol ___
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Example 16-2. Combustion of Carbon Monoxide
Solving for the total enthalpy of the products and reactants: SHout = - 143,743 Btu SHin = -22,170 Btu And the heat transfer is: Q = SHout – SHin = - 121,570 Btu per 1 lbmol CO Q = - 121,570 Btu/lbmol CO (1 lbmol CO/28 lbm CO) Q = - 4,342 Btu/lbm CO
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