1. Energy and Heat in Reacting Systems

2. In the setup of a process and choice of raw materials, availability of fuel or low cost energy are important factors An energy balance of the process showing input and output of heat and other forms of energy similar to the materials balance is necessary Principle of conservation energy that stems from the first law of thermodynamics is used for setting up an energy balance: Whenever a quantity of one kind of energy is produced, an exactly equal amount of other kinds must be used up

3. The law of conservation of energy is stated in the form of an energy balance or energy equation for a process occurring within a system: System before the processSystem after the process The process is referred to as thermodynamic change of state A system is in a definite state when all of its properties are defined Temperature, pressure, concentration of components, kinds of components, states of phase T, P XN π Initial state (State 1) Final state (State 2)

4. Kinds of energy stored within a body or system: Internal energy – Energy stored within a system by virtue of the relative motions, forces and arrangements of the atoms or molecules in the system Temperature is an indication of internal energy which reduces when some of the internal energy is withdrawn as heat Pressure also represent internal energy which reduces when some of the internal energy is withdrawn as expansion work Part of the internal energy of a system may be withdrawn as heat or work by a chemical reaction occurring in the system Kinetic energy – Energy possessed by a body by virtue of its relative motion It is particularly important for the flow of gases and liquids Potential energy – Energy possessed by a body by virtue of its position and the force of gravity

7. Like internal energy, the change in enthalpy during a process depends only on the initial and final states, not on the path Therefore heat absorption or evolution of practical processes can be evaluated from the heat content data before and after the process Enthalpy data are readily available for the following simple thermodynamic changes in state, at constant pressure: Temperature changes in pure substances Phase changes in pure substances Formation of compounds from the elements at STP Formation and dilution of solutions Enthalpy changes associated with solution of various oxides in each other, as in slags, can be estimated or neglected

9. Changes in state of phases As a solid is heated to its melting point, additional heat must be supplied to melt it The heat required for melting at constant pressure = the increase in heat content from the solid to the liquid = ΔH fusion An equal quantity of heat is liberated during solidification, ΔH solidification = -ΔH fusion Similarly, heat effects accompanying vaporization = the increase in heat content from the liquid to vapour = ΔH vaporization Also, heat effects accompanying allotropic changes in solids = the increase in heat content from one allotrope to the other = ΔH transformation These ΔH values are tabulated for 1 atm pressure and they vary with temperature e.g. Heat of vaporization of water at 100 C is 542 cal/g, it is 583 cal/g at 25 C S L ΔH’ m ΔHmΔHm ΔH’’ m T’’ m TmTm T’ m H T -H 298

11. @ 298 K

12. ΔH for a high temperature process can be calculated in the same way as Hess’ law is used calculate the heats of fomation: The process is represented schematically as ΔH T = ΔH 298 + ΣΔH (cooling reactants) + ΣΔH (heating products) If there are no changes in the states of phase of the reactants or products, ΔH T = ΔH 298 T °C aA bB cC dD T °C Base temperature I II III

13. Example – Find the net heat available or required when the following reaction takes place at 800 K SubstanceΔH o 298 (kJ/mole)C P (J/mole K) CaO(s) -634.3 49.62+4.52*10 -3 *T-6.95*10 5 *T -2 CO 2 (g) -393.5 44.14+9.04*10 -3 *T-8.54*10 5 *T -2 CaCO 3 (s) -1206.7 104.52+21.92*10 -3 *T-25.94*10 5 *T -2 = -1206.7 - 634.3 + 393.5 = -178.9 kJ = 10.76 + 8.36*10 -3 *T-10.45*10 5 *T -2 J/K J

14. Alternatively ΔH T can be calculated from Hess’ law 12 3 4

15. It is usually possible to obtain reasonably complete data on the initial and final states for most of the complex processes that consists of the amounts of components, temperature, and states of phase In calculating ΔH for a complex process, a schematic diagram facilitates analysis of the problem and is helpful in avoiding errors For the conversion of Cu 2 S to Cu, the process is represented schematically as The steps shown do not correspond to the way the process is carried out in practice but this ideal process has the same thermodynamic change in state as the actual process ΔH (conversion process) = ΔH I + ΔH II + ΔH III + ΔH IV (-) (-) (+) (+) Air 25 °C 1200 °C M.Pt. Liquid Cu 2 SWaste gases 1250 °C Liquid Cu 1300 °C M.Pt. Base temperatureCu 2 S (l) + O 2 (g) → 2Cu (l) + SO 2 (g) I II IIIIV

16. Hess’ law states that enthalpy change accompanying a chemical reaction is the same whether it takes place in one or several stages since enthalpy is a state function AΔHB XYZ Reaction Enthalpy change A  X ΔH(1) X  Y ΔH(2) Y  Z ΔH(3) Z  B ΔH(4) A  B ΔH Calculation of ΔH for a complex process involves algebraic addition of ΔH values for the following 3 kinds of steps: 1.Cooling all input substances from actual temperatures and states to the base temperature and references states at the base temperature 2.Carrying out the reaction at the base temperature (tabulated ΔH) 3.Heating all reaction products and output materials from the base temperature and reference states to actual final temperature and states 1 23 4

17. Non-isothermal complex processes 2145

18. Example – A furnace that is designed to melt silver/copper scrap is to be fired with propane and air. The propane vapor mixes with dry air at 298 K. Flue gases are expected to exit the furnace at 1505 K under steady state conditions. How long will a 45.5 kg container of propane maintain the furnace temperature if heat is conducted through the brickwork at the rate of 10000 kJ/hour ? 298 1 2 1505 Substance H T -H 298 (J/mole) ΔH o 298 (kJ/mole) C P (J/mole K) C 3 H 8 (g) -103.55 CO 2 (g)-16476+44.25*T+0.0044*T 2 +8.62*10 5 T -2 -393.5 44.14+9.04*10 -3 *T-8.54*10 5 *T -2 H 2 O(l) -285.85 75.47 H 2 O(g)34660+30.01*T+0.00536*T 2 -0.33*10 5 T -2 -241.95 30.01+10.72*10 -3 *T+0.33*10 5 *T -2 N 2 (g)-8502+27.88*T+0.00213*T 2 -- 27.88+4.27*10 -3 *T ΔH v (H 2 O) = 40897 J/mole Air: 21% O 2 + 79% N 2

19. Heat balance comprises of heat input which is equal to heat output plus heat accumulation The benefits of heat balance: Calculating the retained heat If the furnace is to work at a particular temperature, a certain amount of heat has to be retained inside to increase the temperature of the product Calculating the heat deficit If the calculated heat output is larger than the calculated heat input, there is a heat deficit which should be compensated by supplying an extra amount of thermal energy from an outside source The sources can be the combustion of fuel or electricity Once you decide for combustion of fuel, then you have to decide on which type of fuel among solid, liquid, gasesous, to use Having decided on the type of fuel, you must make sure that sufficient quantity of this fuel is available in the reserves

20. The ability to calculate the furnace temperature the temperature attained by the products inside the furnace is important because the furnace should be constructed of the materials that are able to sustain that particular high temperature without creep or fusing At low operating temperatures, there are many materials available for the design of the furnace Once 900 or 1000 degree celsius reached, the choices are only limited to refractory materials Refractory materials Materials that can withstand high temperatures, corrosion from liquids and abrasion of hot gases Silica (SiO 2 )Melts at 1724 C Temperatures attained in Metallurgical processes Alumina (Al 2 O 3 )2050 CCopper smelting1000-1100 C Aluminosilicate (xAl 2 O 3.ySiO 2 ) 1600-1820 CZinc retorts1400-1600 C Lime (CaO)Bessemer converter1600 C Magnesia (MgO)2165 COxygen converter1850 C Forsterite (2MgO.SiO 2 )Tuyeres in iron blast furnace1900 C Dolomite (MgO.CaO)Electric arc temperature3600 C Hematite (Fe 2 O 3 ) or Magnetite (Fe 3 O 4 ) Electric arc furnace1800 C Chromite (FeO.Cr 2 O 3 )2050-2200 C Carbon (Graphite)3600 C Metals (Water cooled) Carbides (silicon carbide)2700 C

21. Acid refractories absorb oxygen ions when dissolved in a basic melt e.g. SiO 2 + 2O 2- = SiO 4 4- Siliceous materials that consist of silica and are low in metallic oxides and alkalies Natural rock, quartzite sand, silica brick Aluminosilicates that consist of chemically combined silica in alumina Free silica should not be present as they lower the melting point Natural rock, fireclay, firebrick Basic refractories provide oxygen ions when dissolved in a melt e.g. MgO = Mg 2+ + O 2- Aluminum oxides Bauxite or bauxite brick, electrically fused bauxite Calcium and magnesium oxide Magnesia, lime, dolomite Neutral refractories are not attacked by acidic or basic oxides and are used to replace basic refractories where the corrosive action is strong Aluminosilicates are sometimes classified as neutral refractories, but they exhibit an acid reaction in contact with basic slags Carbonaceous refractoriesMetals Graphite, carbon bricksFe, Cu, Mo, Ni, Pt, Os, Ta, Ti, W, V and Zr Artificial refractories Others Zirconium carbide, silicon carbideForsterite, concrete, serpentine Chromite

22. Properties of refractories Thermal conductivity: Must be low to minimize heat losses from walls Coefficient of thermal expansion: Must be low to avoid expansion when heated up to the operating temperature Thermal shock resistance: Must be high to avoid expansion and contraction when exposed to repeated heating and cooling. All refractories are generally heated and cooled very slowly Porosity: Should be minimized to improve the strength, thermal shock resistance except in the case of insulating refractories that are used in the outer walls to prevent heat losses Resistance to chemical attack: Chemical attack results from the contact of acid and basic refractories with slag or dust Acidic refractories should be in contact with acid slag that is high in silica Basic refractories should be in contact with basic slag that is high in CaO or MgO Most of the oxide or silicate refractories are fully oxidized so that they will not be affected by oxygen Graphite and silicon carbide oxidize and burn at high temperatures Softening point: The temperature at which the refractory is plastically deformed under load More important criterion for the selection of refractories than the melting point