Heat Balance in Pyrometallurgical Processes

Transcription

1 Heat Balance in Pyrometallurgical Processes

2 The heat balance shows the important sources of heat energy and their relative contribution to the total energy usage in a process The heat balance in general accounts for heat quantities in two categories, input and output, whose total must be identical Careful study of the heat balance often discloses possible lines of improvement in the process, especially improvements leading to saving in fuel In a metallurgical process, modifications like change in compositions of input materials, in amount of fuel, in rate of treatment or in process temperature are often necessary Understanding of the probable effects of such modifications on the heat balance helps the engineer in preparing for operating difficulties resulting from the changes Heat Input + Retained Heat Heat Output + Heat Loss

3 Heat Input Items Sensible heats of input materials Heats evolved in exothermic reactions Heat supplied from outside of the system Heat Output Items Sensible heats of output materials Heats absorbed in endothermic reactions Heats absorbed in bringing low T input materials to reference temperature and state Heat loss to the surroundings Heat input is equal to heat output in steady state processes In autogeneous processes like roasting of zinc, the heat evolved in the metallurgical reactions and the sensible heats in the input materials account for the heat input In non-autogeneous processes like ironmaking in reverberatory furnace, the heat is supplied wholly or in part by heat evolved from combustion of fuel, heat supplied electrically or by other means from outside the system Thermal efficiency = Useful heat output(total heat input heat losses) Total heat input

4 Sankey diagrams help visualization of the distribution of energy in a process or an entire plant General energy balance Energy balance in an electric arc furnace

5 Procedure in Calculating a Heat Balance 1. Work out the complete stoichiometry of the reactions and materials balance 2. Denote the temperatures at which all materials enter and leave the system 3. Fix and specify the basis of the heat balance (quantity throughout the process), reference temperature and reference state 4. Calculate the sensible heat for each input and output material 5. Calculate heats of reaction for the quantities of all the chemical reactions 6. Calculate if present, heats required to bring input materials up to the reference states 7. Calculate if present, heat supplied electrically or by other means from the surroundings 8. List and add input and output items, finding heat loss or deficiency by the difference

6 The consideration of which set of reactions should be used to account for the overall chemical change of the process is important I - Fe 2 O 3 + 3C = 2Fe + 3CO ΔH 298 = calories/mole Fe 2 O 3 II - Fe 2 O 3 = 2Fe + 3/2O 2 ΔH 298 = calories/mole Fe 2 O 3 3/2O 2 + 3C = 3CO ΔH 298 = calories/mole Fe 2 O 3 III - Fe 2 O 3 + 3CO = 2Fe + 3CO 2 ΔH 298 = calories/mole Fe 2 O 3 3CO 2 + 3C = 6CO ΔH 298 = calories/mole Fe 2 O 3 All three sets of reactions represent the same change in state, the net contribution to the heat balance is calories/mole Fe 2 O 3 The heat output or heat consumption accompanying reaction I is represented as a single item in the heat output side of heat balance The largest heat output and heat input are obtained by choosing set II, each heat input and output being calories larger than those of set I For all three sets, the heat loss to the surrounding is the same

7 Heat Input Heat Output ΔH for chemical change Reaction set I cal cal Reaction set II cal cal cal Reaction set III 6390 cal cal cal Thermodynamically, all three methods are equally correct However the choice of the set affects the outlook of the heat balance Using the combination of reactions which gives the largest input and output totals gives greater emphasis on reaction heats compared to other items The combination giving the largest input and output makes the heat loss appear in a smaller proportion of the total heat input so that a higher thermal efficiency than actual is obtained Thermal efficiency = Useful heat output(total heat input heat losses) Total heat input The set of reactions must represent real heat evolutions or absorptions in the process as much as possible

8 ΔH Cooling of flue gas Flue gas is passed through a waste heat boiler which cools it from 1121 C to 293 C Calculate the heat given up by the flue gas 2170 m 3 /min Hot flue 1121 C, 1394 K 74% N 2 14% CO 2 Boiler 10% H 2 O 2% O Cool flue gas 293 C, 566 K Basis: 2170 m 3 Reference temperature = 25 C CO2 H2O N2 O T (K)

9 P 1 V 1 /T 1 = P 2 V 2 /T 2, P 1 =P 2 = 1 atm, V 2 = V 1 T 2 /T 1 = 2170*273/1394= 425 m 3 425*0.74/22.4 = kg-mole/min N 2 Heat Input for N 2 Heat Output for N 2 Sensible heat for 1121 to 25 C Sensible heat for 25 to 293 C *(H H 298 ) = * = kcal 14.04*(H 566 -H 298 ) = 14.04*1884 = kcal 425*0.14/22.4 = 2.66 kg-mole/min CO 2 Heat Input for CO 2 Heat Output for CO 2 Sensible heat for 1121 to 25 C Sensible heat for 25 to 293 C -2.66*(H H 298 ) = -2.66*13277 = kcal 2.66*(H 566 -H 298 ) = 2.66*2711 = 7211 kcal 425*0.1/22.4 = 1.90 kg-mole/min H 2 O Heat Input for H 2 O Sensible heat for 1121 to 25 C -1.90*(H H 298 ) = -1.90*10320 = kcal Heat Output for H 2 O Sensible heat for 25 to 293 C 1.90*(H 566 -H 298 ) = 1.90*2219 = 4216 kcal 425*0.021/22.4 = 0.38 kg-mole/min O 2 Heat Input for O 2 Heat Output for O 2 Sensible heat for 1121 to 25 C Sensible heat for 25 to 293 C -0.38*(H H 298 ) = -0.38*8783 = kcal 0.38*(H 566 -H 298 ) = 0.38*1953 = 742 kcal Total Input = kcal Total Output = kcal Heat given up by the flue gas = = kcal/min

10 Example - Limestone of 84% CaCO 3, 8% MgCO 3, 8% H 2 O is calcined in a rotary kiln. Gaseous fuel is combusted with stoichiometric air to supply the required heat. The limestone, fuel and air are supplied at 298 K, lime is discharged at 1173 K and gases leave at 473 K. Calculate the energy required to calcine 1000 kg of limestone. Reactions: CaCO 3 = CaO + CO 2 ΔH= Kcal/kg-mole C p CaO = x10 3 T 6.945x10 5 /T 2 MgCO 3 = MgO + CO 2 ΔH= Kcal/kg-mole C p MgO = x10 3 T x10 5 /T 2 CO + ½O 2 = CO 2 ΔH= Kcal/kg-mole C p CO 2 = T Material balance gives calcined products and off-gases as 8.4 kg-moles CaO kg-moles MgO kg-moles CO kg-moles H 2 O 8.4 kg-moles CaO is produced by consuming 8.4*42750 = Kcal kg-moles MgO is produced by consuming 0.952*24250 =23086 Kcal Total heats of decomposition of the two reactions = Kcal The sensible heat in calcined products H = C P CaO dt = = Kcal C P MgO dt

11 8.4 kg-moles CaO is produced by consuming 8.4*42750 = Kcal kg-moles MgO is produced by consuming 0.952*24250 =23086 Kcal Total heats of decomposition of the two reactions = Kcal The sensible heat in calcined products H = C P CaO dt = = Kcal C P MgO dt The sensible heat in CO 2 H = The heat content in H 2 O H = C P H 2 O l = Kcal = = Kcal Total heat requirement to calcine 1000 kg of limestone C P CO 2 dt dt H m C P H 2 O v dt H = = Kcal

12 Example - Limestone is calcined in a rotary kiln. Gaseous fuel at 900 K is combusted with stoichiometric air to supply the required heat. The limestone and air are supplied at 298 K, lime is discharged at 1173 K and gases leave at 473 K. Calculate the amount of gaseous fuel required to obtain 10 kg-moles CaO CaCO 3 = CaO + CO 2 ΔH= Kcal/kg-mole C p CaO = x10 3 T 6.945x10 5 /T 2 MgCO 3 = MgO + CO 2 ΔH= Kcal/kg-mole C p MgO = x10 3 T x10 5 /T 2 CO + ½O 2 = CO 2 ΔH= Kcal/kg-mole C p CO 2 = T Rational Analysis wt%, v% CaCO 3 MgCO 3 H 2 O CO 2 O 2 CO N 2 Limestone Fuel The calorific value of the fuel or quantity of heat produced by combustion of 1 kg-mole of fuel: 1 kg-mole of fuel contains kg-mole CO and kg-mole O 2 The heat of combustion for kg-mole CO = 0.166*(-67900) = Kcal Heat Input Heat Output Decomposition heat of limestone (10*ΔH CaO +1.13*ΔH MgO ) Sensible heat in the fuel (X* H ) Calorific value of the fuel ( X) Sensible heat in CaO (10* H ) Sensible heat in MgO (1.133* H ) Sensible heat in off-gas (( X)* H CO 2 + Material balance gives (0.74 X X)* H N 2 + CO 2 = X CaO = * H H 2 O) N 2 = 0.74 X X MgO = H 2 O = 5.29

13 ΔH Heat Input Sensible heat in the fuel (X* H ) Calorific value of the fuel ( X) Heat Output Decomposition heat of limestone (10*ΔHCaO+1.13ΔHMgO) Sensible heat in CaO (10* H CaO) Sensible heat in MgO (1.133* H MgO) Sensible heat in off-gas (( X)* H CO 2 + (0.74 X X)* H N * H H 2 O) CaCO 3 = CaO + CO 2 ΔH= Kcal/kg-mole C p CaO = x10 3 T 6.945x10 5 /T 2 MgCO 3 = MgO + CO 2 ΔH= Kcal/kg-mole C p MgO = x10 3 T x10 5 /T CO2 H2O N2 CaO MgO T (K)

14 Alumina calcination 200 kg Fuel 298 K 84% C 16% H Al(OH) K 55% Al 2 O 3 45% H 2 O Rotary Kiln 1700 K 1.2 Air 298 K 1000 kg Al 2 O K Reactions: 2 Al(OH) 3 = Al 2 O 3 + 3H 2 O ΔH= Kcal/kg-mole H 1000 H 298 (Al 2 O 3 ) =18710 Kcal/kg.mol C + O 2 = CO 2 ΔH= Kcal/kg-mole 2H + ½ O 2 = H 2 O ΔH= Kcal/kg-mole Calculate the volume of gases leaving the kiln per 1000 kg alumina Off-gas 800 K

15 m Al Al(OH) K 1700 K V g Off-gas 800 K 55% Al 2 O 3 X CO2 45% H 2 O X H2O 200 kg Fuel Rotary Kiln X O2 298 K X N2 84% C V A Air 298 K 1000 kg Al 16% H 2 O K Reaction: 2 Al(OH) 3 = Al 2 O 3 + 3H 2 O ΔH= Kcal/kg-mole H 1000 H 298 (Al 2 O 3 ) =18710 Kcal/kg.mol C + O 2 = CO 2 ΔH= Kcal/kg-mole 2H + ½ O 2 = H 2 O ΔH= Kcal/kg-mole Perform the material balance

16 m Al Al(OH) K 1700 K 4014 m 3 Off-gas 800 K 55% Al 2 O 3 7.8% CO 2 45% H 2 O 2.5% O kg Fuel Rotary Kiln 55.4% N K 34.3 H 2 O 84% C V A Air 298 K 1000 kg Al 16% H 2 O K Reaction: 2 Al(OH) 3 = Al 2 O 3 + 3H 2 O ΔH= Kcal/kg-mole H 1000 H 298 (Al 2 O 3 ) =18710 Kcal/kg.mol C + O 2 = CO 2 ΔH= Kcal/kg-mole 2H + ½ O 2 = H 2 O ΔH= Kcal/kg-mole Perform the heat balance Heat Input Heat of combustion Heat Output Heat of alumina calcination Sensible heat in alumina Sensible heat in off-gas

17 Heat Balance in Roasting Oxidation reactions occurring during roasting processes to convert metal sulphides to oxides are highly exothermic Fuel is seldomly used and the sulphide ores that enter the furnace convert to oxides and flue gases which take the liberated heat as their sensible heat Products are raised from 298 to an elevated temperature which can be calculated by making the heat balance Heat Input = Heat Output = (H T -H 298 ) products Since there is no fuel in the roasting process, the final temperature of the products can not be controlled by arrangement of the amount of combusting material The temperature attained by the products is calculated in a similar way to the calculation of flame temperature

18 Flame temperature The maximum temperature the gaseous products can reach upon proceeding of an exothermic reaction is called the flame temperature The furnace is considered as adiabatic for no heat loss to the surroundings and maximum flame temperature T aa( s)( T1 ) bb( g)( T2 ) H cc( g)( T3 ) dd( g)( T3 ) H3 aa( s)(298) bb( g)(298) cc( l)(298) dd( g)(298) H 1 H 2 H 3 H 4 H 5 H 4 c T m( C ) 298 C P( C( l)) dt H v( C) a Tflame T m( C ) C P( C( g )) dt c( H Tflame H 298 ) C( g ) H 5 d T Tflame 298 C P( D( g )) dt d( H Tflame H 298 ) D( g )

19 Roasting furnace analysis Zinc concentrate of the following composition is roasted in a fluidized bed reactor with stoichiometric amount of air. During roasting 80% of total iron charged forms ZnO.Fe 2 O 3 Find the bed temperature when 10% heat input is lost to the surroundings Rational Analysis wt% Material ZnS FeS 2 PbS SiO 2 H 2 O Zinc concentrate Basis 1000 kg of zinc concentrate 75% ZnS 18% FeS 2 3% PbS 3% SiO 2 1% H 2 O Air Reactions ZnS + 3/2O 2 = ZnO + SO 2 2FeS /2O 2 = Fe 2 O 3 + 4SO 2 PbS + 2O 2 = PbSO 4 Fluidized Bed Reactor Flue gases SO 2, N 2, H 2 O Roast product ZnO. Fe 2 O 3 ZnO Fe 2 O 3 SiO 2 PbSO 4

20 Basis 1000 kg of zinc K 75% ZnS 18% FeS 2 3% PbS 3% SiO 2 1% H 2 O Fluidized Bed Reactor Flue gases SO 2, N 2, H 2 O Air Reactions ZnS + 3/2O 2 = ZnO + SO 2 2FeS /2O 2 = Fe 2 O 3 + 4SO 2 PbS + 2O 2 = PbSO 4 Roast ZnO. Fe 2 O 3 ZnO Fe 2 O 3 SiO 2 PbSO 4 Material balance gives: Concentrate content Roast product content Gas content ZnS kg-moles ZnS + 3/2O 2 = ZnO + SO 2 ZnO kg-moles SO mol FeS kg-moles 2FeS /2O 2 = Fe 2 O 3 + 4SO 2 Fe 2 O moles SO mol PbS kg-moles PbS + 2O 2 = PbSO 4 PbSO moles SiO kg-moles SiO kg-moles H 2 O kg-moles H 2 O kg-moles Oxygen consumed 7.732*3/ *11/ *2 = kg-moles Nitrogen in the off-gas /0.21*0.79 = kg-moles

21 Some ZnO is tied up with Fe 2 O 3 as ZnO.Fe 2 O 3 Fe tied up with ZnO = 0.8* = kg-moles, kg-moles Fe 2 O 3 in mol ZnO.Fe 2 O 3 Free Fe 2 O 3 in roast product = ( ) = kg-moles kg-moles ZnO in mol ZnO.Fe 2 O 3 Free ZnO in roast product = = kg-moles Roast product content Flue gas content ZnO kg-moles SO kg-moles Fe 2 O kg-moles N kg-moles ZnO.Fe 2 O kg-moles H 2 O kg-moles PbSO kg-moles SiO kg-moles Heat balance Heat Input Heat Output Sensible heats of input materials (0) Sensible heats of output materials Heats evolved in exothermic reactions Heats absorbed in endothermic reactions (0) Heat loss to the surroundings (10% heat input)

22 Heat Input Heat Output Sensible heats of input materials (0) Sensible heats of output materials Heats evolved in exothermic reactions Heats absorbed in endothermic reactions (0) Heat loss to the surroundings (10% heat input) Heats evolved in reactions ZnS + 3/2O 2 = ZnO + SO 2 ΔH= kcal/kg-mole ZnO 2FeS /2O 2 = Fe 2 O 3 + 4SO 2 ΔH= kcal/kg-mole Fe 2 O 3 PbS + 2O 2 = PbSO 4 ΔH= kcal/kg-mole PbSO 4 ZnO + Fe 2 O 3 = ZnO.Fe 2 O 3 ΔH= kcal/kg-mole ZnO.Fe 2 O 3 Total heat liberated = ( *6.914)+( *1.0225)+( *0.126)+(-4750*0.818) = kcal 10% of heat input is lost to the surroundings = kcal Heat available for raising the temperature of the products = kcal Sensible heats of output materials ZnO 6.914*H T -H 298 = 6.914*(11.71T+0.61*10-3 T *10 5 /T-4277) Fe 2 O *H T -H 298 = 0.204*(31.75T+0.88*10-3 T ) ZnO.Fe 2 O *H T -H 298 = 0.818*(27.78T+8.86*10-3 T ) PbSO *H T -H 298 = 0.126*(10.96T+15.5*10-3 T *10 5 /T-3327) SiO 2 0.5*H T -H 298 = 0.5*(14.41T+0.97*10-3 T ) H 2 O 0.555*H T -H 373 = 0.555*(7.30T+1.23*10-3 T ) *ΔH m =0.555*(11170) SO *H T -H 298 = *(11.04T+0.94*10-3 T *10 5 /T-3992) *H T -H 298 = *(6.83T+0.45*10-3 T *10 5 /T-2117) N 2

23 Sensible heats of output materials ZnO 6.914*H T -H 298 = 6.914*(11.71T+0.61*10-3 T *10 5 /T-4277) Fe 2 O *H T -H 298 = 0.204*(31.75T+0.88*10-3 T ) ZnO.Fe 2 O *H T -H 298 = 0.818*(27.78T+8.86*10-3 T ) PbSO *H T -H 298 = 0.126*(10.96T+15.5*10-3 T *10 5 /T-3327) SiO 2 0.5*H T -H 298 = 0.5*(14.41T+0.97*10-3 T ) H 2 O 0.555*H T -H 373 = 0.555*(7.30T+1.23*10-3 T ) *ΔH m =0.555*(11170) SO *H T -H 298 = *(11.04T+0.94*10-3 T *10 5 /T-3992) N *H T -H 298 = *(6.83T+0.45*10-3 T *10 5 /T-2117) Total Heat Output = T+49.83*10-3 T Heat available for raising the temperature of the products = kcal T+49.83*10-3 T = *10-3 T T = 0 T= b± b2 4ac 2a a= 49.83*10-3, b= , c= T = K

24 The rate of roasting reactions are low due to the solid state of the sulphide concentrates Extra oxygen is supplied by the excess air which considerably increases the rate of oxidation Consider the same process with the use of 20% excess air The flue gas content SO kg-moles N kg-moles *0.2 = kg-moles O *0.2/0.79*0.21 = 3.49 kg-moles H 2 O kg-moles Increase in the total heat output *0.2*H T -H 298 (N 2 ) *H T -H 298 (O 2 ) = T *10-3 T Total heat output = T *10-3 T Available heat = kcal 56.55*10-3 T T =0 T = 1545 K If 40% excess air was used to increase the oxidation rate, Increase in the total heat output *0.4*H T -H 298 (N 2 ) *H T -H 298 (O 2 ) = 2*(118.73T *10-3 T ) So, 63.36*10-3 T T =0 T=1407 K Materials for the construction of the furnace may change due only to use of excess air

25 Heat Balance in Ironmaking Consider an iron blast furnace charged with iron ore, limestone and coke for 20-day period to produce 507 tons pig iron per day Rational Analysis wt% Material Fe 2 O 3 SiO 2 MnO Al 2 O 3 H 2 O C CaO CO 2 Limestone Coke Air is blown through tuyeres at 704 C, the charge, flux and coke are at 25 C, gases leave at 149 C Moisture of 292 kg in one charge per ton of pig iron is added The ultimate analysis of the pig iron gives 93% Fe, 3.9% C, 1.1% Si, 1.8% Mn, 0.22%P The rational analysis of flue gases on dry basis gives 60% N 2, 21.3% CO, 16.5% CO 2, 2.2% H 2 Basis 1000 kg of pig iron Blast furnace C Ore Limestone 410 kg Blast furnace Slag 486 kg Coke 764 kg 704 C Pig iron 1000 kg

26 Heat Balance in Ironmaking

27 Basis 1000 kg of pig iron Ore 292 kg H 2 O Limestone 410 kg 43.75% CO 2 Coke 764 kg 95.2% C, 4.8% H 2 O 704 C Reactions Fe 2 O 3 + 3CO = 2Fe + 3CO 2 CaCO 3 = CaO + CO 2 C + 1/2O 2 = CO Blast furnace SiO 2 + 2C = Si + 2CO MnO + C = Mn + CO CO 2 + C = 2CO Blast furnace 149 C 60% N 2, 21.3% CO, 16.5% CO 2, 2.2% H 2 H 2 Slag 486 kg Pig iron 1000 kg 93% Fe, 3.9% C, 1.1% Si, 1.8% Mn, 0.22%P P 2 O 5 + 5C = 2P + 5 CO H 2 O + C = CO + H 2 CO 2 + C = 2CO C = C H 2 O(l) = H 2 O(g) Material Balance C balance is solved to find the quantity of top gases C in coke + C in limestone = C in pig iron + C in top gas Dry weight of C in coke = 764*95.2/100 = kg C in limestone = 410*43.75/100*12/44 = kg C in pig iron = 1000*3.9/100 = 39 kg C in top gas = = kg = 64.8 kg-atom

28 C in top gas = = kg = 64.8 kg-atom The top gas contains 16.5 CO 2 and 21.3% CO on dry basis Dry top gas contains in total = 64.8/( ) = kg-moles Therefore on a dry basis the top gas contains: CO 2 = *16.5/100 = kg-moles CO = *21.3/100 = kg-moles H 2 = *2.2/100 = 3.77 kg-moles N 2 = *59.9/100 = kg-moles In addition the top gas contains H 2 O from moisture in the charge and coke H 2 O = 292/ *4.8/100 = = kg-moles N 2 balance is solved to find the quantity of air blast N 2 in air = N 2 in top gas = kg-atoms O 2 in air = *21/79 = 27.3 kg-atoms Total air blast = kg-atoms

29 Heat balance diagram Air 704 C Pig iron T Slag T Top C I II Fe 2 O 3 + 3CO = 2Fe + 3CO 2 IV V 25 C III SiO 2 + 2C = Si + 2CO I Sensible heat in the air blast II Heat evolution by the exothermic reactions III Heat absorption by the endothermic reactions IV Sensible heat in pig iron V Sensible heat in slag VI Sensible heat in top gas I + II = III + IV + V + VI + heat loss

30 Material and heat balances in the reactions Fe 2 O 3 + 3CO = 2Fe + 3CO 2 Exothermic Rxn, ΔH= kcal/kg-mole Kg-atom Fe in pig iron = 2/3 kg-mole CO = 2/3 kg-mole CO 2 produced Total heat evolved = -6300*24.91*1/3 = kcal Kg-mole CO = 3/2*100*(92.99/100)/56 = kg-moles SiO 2 + 2C = Si + 2CO Kg-atom C = kg-mole CO = 2*kg-atom Si in pig iron = 2*1000*(1.1/100)/28 = 0.79 kg-moles CaCO 3 = CaO + CO 2 Kg-mole CO 2 = kg-mole CaCO 3 = 410*(43.75/100)/44 = 4.08 kg-moles MnO + C = Mn + CO Kg-atom C = kg-mole CO = kg-atom Mn in pig iron = 1000*(1.8/100)/55 = 0.33 kg-moles C + 1/2O 2 = CO Kg-atom C = kg-mole CO produced = 2*kg-mole O 2 consumed =2*27.3 = 54.6 kg-moles P 2 O 5 + 5C = 2P + 5 CO Kg-atom C = kg-mole CO = 5/2*kg-atom P in pig iron =5/2*1000*(2.1/100)/31 = 0.17 kg-moles Endothermic Rxn, ΔH= kcal/kg-mole Heat absorbed by reduction = *0.79/2 = kcal Endothermic Rxn, ΔH= kcal/kg-mole Heat absorbed by decomposition = 42500*4.08 = kca Endothermic Rxn, ΔH= kcal/kg-mole Heat absorbed by reduction = 62400*0.33 = kcal Exothermic Rxn, ΔH= kcal/kg-mole C Total heat evolved = *54.6 = kcal Endothermic Rxn, ΔH= kcal/kg-mole Heat absorbed by reduction = *0.17/5 = 7208 kcal CO 2 + C = 2CO Endothermic Rxn, ΔH= kcal/kg-mole Kg-atom CO = 2*kg-atom C = 2*kg-atom CO 2 Heat absorbed by conversion = 38000*3.5 = kcal Kg-atom C = kg-atom in coke kg-atom in C in pig iron kg-atom C consumed in all reactions = / *(3.9/100)/12 ( ) = 3.5 kg-atom 3.5 kg-moles CO 2 consumed, 7 kg-moles CO produced

31 H 2 O + C = CO + H 2 Kg-mole H 2 O = kg-atom H 2 in the top gas = 3.77 kg-moles H 2 O(l) = H 2 O(g) Kg-mole H 2 O = kg-moles Endothermic Rxn, ΔH= kcal/kg-mole Heat absorbed by decomposition = 28200*3.77 = kca Endothermic Rxn, ΔH v = 9756 kcal/kg-mole Heat absorbed by evaporation = 9756*52.89 = kcal Total heat evolution by 2 exothermic reactions Total heat absorption by 7 endothermic reactions C + 1/2O 2 = CO kcal SiO 2 + 2C = Si + 2CO kcal Fe 2 O 3 + 3CO = 2Fe + 3CO kcal CaCO 3 = CaO + CO kcal MnO + C = Mn + CO kcal P 2 O 5 + 5C = 2P + 5 CO 7208 kcal CO 2 + C = 2CO kcal H 2 O + C = CO + H kcal H 2 O(l) = H 2 O(g) kcal kcal kcal There is excess heat of kcal that is available to take the products to a higher temperature There is also heat evolution from the sensible heat in the air blast at 704 C: O 2 N 2 H 977 -H 298 = kcal/kg-mole H 977 -H 298 = kcal/kg-mole Sensible heat = *27.3 = kcal Sensible heat = * = kcal Total heat input = Heat evolution from reactions + Sensible heat in the air blast = kcal = kcal

32 Sensible heat in the products: Sensible heat in pig iron: It is estimated that 1 kg of pig iron carries about 284 kcal of sensible heat from the furnace H T -H 298 = 284 kcal/kg Heat absorption = 284*1000 = kcal Sensible heat in slag: It is estimated that 1 kg of slag carries about 500 kcal of sensible heat from the furnace H T -H 298 = 500 kcal/kg Heat absorption = 500*486 = kcal Sensible heat in top gas: Top gas leaves the furnace at 149 C CO 2 = kg-moles H 422 -H 298 = kcal/kg-mole Heat absorbed = *28.29 = CO = kg-moles H 422 -H 298 = kcal/kg-mole Heat absorbed = 865.1*36.51= H 2 = 3.77 kg-moles H 422 -H 298 = kcal/kg-mole Heat absorbed = 859.0*3.77= N 2 = kg-moles H 422 -H 298 = kcal/kg-mole Heat absorbed = 865.1*102.69= H 2 O = kg-moles H 422 -H 298 = kcal/kg-mole Heat absorbed = *52.89=53292 Total sensible heat in top gas = kcal Total sensible heat in the products = kcal Total heat absorption from the reactions = kcal Total heat output = kcal Heat loss = Total heat input Total heat output = = kcal Thermal efficiency = ( Total heat input Heat loss )/ Total heat input) = 75.51%

33 Iron Smelting Analysis An iron blast furnace produces pig iron of composition 94.2% Fe, 3.5% C, 1.5% Si, 0.8% Mn. The burden of blast furnace consists of the following materials. Limestone is pure CaCO 3 and is converted to CaO that makes up 36% of the slag. The analysis of exit gas is given as 26% CO, 13% CO 2 and 61% N 2 ; assume no loss of iron in slag. Rational Analysis wt% Material Fe 2 O 3 SiO 2 Al 2 O 3 MnO C Iron ore Coke Reactions: CaCO 3 = CaO + CO 2, C + ½ O 2 = CO, Fe 2 O 3 + 3CO = 2Fe + 3CO 2, SiO 2 + 2CO = Si + 2CO 2 Iron ore Flux: CaCO kg Coke Iron blast furnace Flue 600 K (CO, N 2, CO 2 ) 1700 K (Al 2 O 3, SiO 2, CaO, MnO) 1000 kg Pig K (xcu 2 S.yFeS) Air

34 Iron Smelting Analysis Iron ore 76% Fe 2 O 3 14% SiO 2 9% Al 2 O 3 1% MnO Flux: CaCO kg Coke 88% C 12% SiO 2 Iron blast furnace Flue 600 K 26% CO, 61% N 2, 13% CO K (Al 2 O 3, SiO 2, 36% CaO, MnO) 1000 kg Pig K 94.2% Fe 3.5% C 1.5% Si 0.8% Mn Air Reactions: CaCO 3 = CaO + CO 2, C + ½ O 2 = CO, Fe 2 O 3 + 3CO = 2Fe + 3CO 2, SiO 2 + 2CO = Si + 2CO Slag composition 2 Basis 1000 kg Pig iron 36% CaO Material balance kg SiO 2 Fe in ore = Fe in pig iron 159 kg Al 2 O kg MnO (76/100)*(112/160)*X = 1000*(94.2/100) = 942 kg kg = 64% Iron ore = X = 1771 kg CaO amount = kg Si in slag = Si in ore + Si in coke - Si in pig iron Slag amount = 804 kg = 1771*(14/100)*(28/60) *(12/100)*(28/60) 1000*(1.5/100) = kg SiO 2 in slag = *(60/28) = kg Mn in slag = Mn in ore Mn in pig iron = 1771*(1/100)*(55/71) 1000*(0.8/100) = 5.72 kg MnO in slag = 5.72*(71/55) = 7.38 kg Al 2 O 3 in slag = Al 2 O 3 in ore = 1771*(9/100) = 159 kg

35 Iron Smelting Analysis Iron ore 76% Fe 2 O 3 14% SiO 2 9% Al 2 O 3 1% MnO Flux: CaCO kg Coke 88% C 12% SiO 2 Iron blast furnace Air Reactions: CaCO 3 = CaO + CO 2, C + ½ O 2 = CO, Fe 2 O 3 + 3CO = 2Fe + 3CO 2, SiO 2 + 2CO = Si + 2CO 2 Basis 1000 kg Pig iron Material balance CaO in limestone = CaO in slag = kg CaCO 3 in limestone = 298.5*(100/56) = 533 kg C in flux + C in coke = C in pig iron + C in flue gases C in flue gases = 533/ *(88/100)/ *(3.5/100)/12 = 83.1 kg-atoms Total flue gas = 83.1*(100/39) = kg-moles = 213.1*22.4*(600/298) = 9611 m 3 N 2 in flue gases = 213.1*(61/100) = 130 kg-atoms N 2 in air = 130 kg-moles O 2 in air = 130*(21/79) = kg-moles Flue 600 K 26% CO, 61% N 2, 13% CO kg 1700 K kg CaO, kg SiO kg Al 2 O 3, 7.38 kg MnO 1000 kg Pig K 94.2% Fe 3.5% C 1.5% Si 0.8% Mn

36 CaCO 3 = CaO + CO 2 Fe 2 O 3 + 3CO = 2Fe + 3CO 2 SiO + 2CO = Si + 2CO Iron Smelting Analysis Iron ore 76% Fe 2 O 3 14% SiO 2 9% Al 2 O 3 1% MnO Flux: CaCO kg Coke 88% C 12% SiO 2 Iron blast furnace 9611 m 3 Flue 600 K 26% CO, 13% CO 2, 130 kg-atoms N kg 1700 K kg CaO, kg SiO kg Al 2 O 3, 7.38 kg MnO 1000 kg Pig K 94.2% Fe 3.5% C 1.5% Si 0.8% Mn 3686 m 3 Air Reactions: CaCO 3 = CaO + CO 2, C + ½ O 2 = CO, Fe 2 O 3 + 3CO = 2Fe + 3CO 2, SiO 2 + 2CO = Si + 2CO 2 Basis 1000 kg Pig iron Heat balance Pig iron 1700 K Slag 1700 K Top K Reference T 25 C C + ½ O 2 = CO

37 Iron Smelting Analysis Iron ore 76% Fe 2 O 3 14% SiO 2 9% Al 2 O 3 1% MnO Flux: CaCO kg Coke 88% C 12% SiO 2 Iron blast furnace 3686 m 3 Air Reactions: CaCO 3 = CaO + CO 2, C + ½ O 2 = CO, Fe 2 O 3 + 3CO = 2Fe + 3CO 2, SiO 2 + 2CO = Si + 2CO 2 Basis 1000 kg Pig iron Heat balance Exothermic reactions C + ½ O 2 = CO ΔH = kcal/kg-mole CO C in coke = 81 kg-atoms Heat of formation of 81 kg-moles CO = *81 = kcal Endothermic reactions CaCO 3 = CaO + CO 2 ΔH = kcal/kg-mole CaCO 3 CaO in slag = 289.5/56 = 5.17 kg-moles Heat of decomposition of 5.17 kg-moles CaCO 3 = 42500*5.17 = kcal 9611 m 3 Flue 600 K 26% CO, 13% CO 2, 130 kg-atoms N kg 1700 K kg CaO, kg SiO kg Al 2 O 3, 7.38 kg MnO 1000 kg Pig K 94.2% Fe 3.5% C 1.5% Si 0.8% Mn

38 Iron Smelting Analysis Iron ore 76% Fe 2 O 3 14% SiO 2 9% Al 2 O 3 1% MnO Flux: CaCO kg Coke 88% C 12% SiO 2 Iron blast furnace 3686 m 3 Air Endothermic reactions Fe 2 O 3 + 3CO = 2Fe + 3CO 2 ΔH = kcal/kg-mole Fe 2 O 3 Fe 2 O 3 in ore = 1771*(76/100) = 1346 kg = 8.41 kg-moles Heat of reduction of 8.41 kg-moles Fe 2 O 3 = *8.41 = kcal 9611 m 3 Flue 600 K 26% CO, 13% CO 2, 130 kg-atoms N kg 1700 K kg CaO, kg SiO kg Al 2 O 3, 7.38 kg MnO 1000 kg Pig K 94.2% Fe 3.5% C 1.5% Si 0.8% Mn SiO 2 + 2CO = Si + 2CO 2 ΔH = kcal/kg-mole SiO 2 Si in pig iron = 15 kg = 0.54 kg-atoms Heat of reduction of 0.54 kg-moles SiO 2 = *0.54 = kcal CO 2 + C = 2CO ΔH = kcal/kg-mole CO 2 reduced = CO 2 produced CO 2 in flue gases = = 3.78 kg-moles Heat of reduction of 3.78 kg-moles CO 2 = 41600*3.78 = kcal

39 Iron Smelting Analysis Iron ore 76% Fe 2 O 3 14% SiO 2 9% Al 2 O 3 1% MnO Flux: CaCO kg Coke 88% C 12% SiO 2 Iron blast furnace 3686 m 3 Air Heat outputs Heat content of pig iron at 1700 K = Sensible heat of elements + Heat of mixing Heat content of slag = Sensible heat of oxides Sensible heat in flue gases m 3 Flue 600 K 26% CO, 13% CO 2, 130 kg-atoms N kg 1700 K kg CaO, kg SiO kg Al 2 O 3, 7.38 kg MnO 1000 kg Pig K 94.2% Fe 3.5% C 1.5% Si 0.8% Mn Pig iron ΔH m = 7500 kcal/ton Fe C Mn Si SiO2 Al2O3 MnO CaO CO

40 Iron Smelting Analysis Iron ore 76% Fe 2 O 3 14% SiO 2 9% Al 2 O 3 1% MnO Flux: CaCO kg Coke 88% C 12% SiO 2 Iron blast furnace 3686 m 3 Air Heat balance Heat Input Heat evolution from exothermic reactions kcal kcal Heat content in pig iron kcal Heat content in slag kcal Sensible heat in flue gases kcal A heat deficit of kcal 9611 m 3 Flue 600 K 26% CO, 13% CO 2, 130 kg-atoms N kg 1700 K kg CaO, kg SiO kg Al 2 O 3, 7.38 kg MnO 1000 kg Pig K 94.2% Fe 3.5% C 1.5% Si 0.8% Mn Heat Output Heat evolution from endothermic reactions