Energy and Heat in Reacting Systems

MME 4517 Materials and Energy

Balance

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

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 process System 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 X N π

Initial state(State 1)

Final state(State 2)

𝐹=2+ (𝑁−1 ) (𝜋 )− (𝜋−1 ) (𝑁 )=2−𝜋+𝑁

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in the systemTemperature is an indication of internal energy which reduces when some of the internal energy is withdrawn as heatPressure also represent internal energy which reduces when some of the internal energy is withdrawn as expansion workPart of the internal energy of a system may be withdrawn as heat or work by a chemical reaction occurring in the system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

Transient kinds of energy:

Heat – The kind of energy that passes from one body to another solely as a result of a difference in temperature

Mechanical work – Work is done when a force acts on a body and a displacement of the body occurs in the direction of application of the force –or– Work is done when a pressure difference between the system and the surroundings causes expansion or compression of the system

or

Electrical work – Work is done by the passage of an electric current in s direct current circuit

The first law of thermodynamics for a process occurring in a system is:

where and are the internal energies of the system in state 1 and state 2, is the absorbed heat by the system from surroundings and is the work done by the system on the surroundings

Almost all processes in extractive metallurgy follow constant pressure paths, at about 1 atmHeat content of a system for these constant pressure processes is defined as:

for any thermodynamic change in state,

For a constant pressure process in which all work done by the system on the surroundings is the work of expansion,

and where is the heat absorbed by the system in changing from state 1 to state 2 through a constant pressure path in which only expansion work is done

If changes in state at constant pressure involve work other than expansion, like electrical work in electric furnace,

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

The effect of temperature on the heat content of a system can be given by graphs, tables,

empirical equations,

and indirectly by the heat capacity

The engineer uses heat content data most frequently for determining sensible heats () and the heat quantities evolved or absorbed when the temperature of a substance is changed between 2 known levels

calories/mole sensible heat

∆𝐻=𝐻𝑇 2−𝐻𝑇 1=(𝐻 𝑇 2−𝐻298 )−(𝐻𝑇 1−𝐻 298)

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 = ΔHfusion

An equal quantity of heat is liberated during solidification, ΔHsolidification = -ΔHfusion

Similarly, heat effects accompanying vaporization = the increase in heat content from the liquid to vapour = ΔHvaporization

Also, heat effects accompanying allotropic changes in solids = the increase in heat content from one allotrope to the other = ΔHtransformation

These ΔH values are tabulated for 1 atm pressure and they vary with temperaturee.g. Heat of vaporization of water at 100 C is 542 cal/g, it is 583 cal/g at 25 C

S

LΔH’mΔHmΔH’’m

T’’m Tm T’m

HT-H

298

Heat of formation of a compound from its elements may be liberated or absorbed

Quantity of heat absorbed (QP) = heat content of the system resulting from the reaction (ΔH), if the formation reaction is carried out at constant pressure

Quantity of ΔH is fixed when the quantities and the thermodynamic properties (P, V, T) of the reacting elements, and the quantities and the thermodynamic properties of the product are fixed

state 1 state 2 (P1, V1, T1) (P1, V2, T2)

Heat of formation of compounds are tabulated for T= 298 K and P= 1 atmΔHH2O

f= -241.8 kJ/mole

Signs of ΔH for constant pressure processes

Positive Negative Heat absorption from surroundings Heat evolved to surroundingsMelting FreezingVaporization Condensation Most dissociation reactions Most reactions of formation from elements

The ΔH for a reaction is equal to the algebraic sum of the ΔH values for the individual reactions when the main reaction is a combination of two or more individual reactions

- -

Hess’ law is a useful method for calculating the unknown enthalpy change of a reaction using known reactions combination of which forms the original one Example – Calculate the standard enthalpy of formation of solid Fe3O4 from the following enthalpy data

ΔH˚298 = -264500 J ΔH˚298 = -292500 J ΔH˚298 = -230650 J ΔH= 6 ΔH1+3 ΔH2- ΔH3

ΔH/2=-1117240 J/mole

@ 298 K

32243

322

2

32/12

2/12

2/1

OFeOOFe

OFeOFeO

FeOOFe

43223 OFeOFe

24332

322

2

2/123

32/36

636

OOFeOFe

OFeOFeO

FeOOFe

432

432

23

246

OFeOFe

OFeOFe

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

ΔHT= ΔH298 + ΣΔH(cooling reactants)+ ΣΔH(heating products) If there are no changes in the states of phase of the reactants or products, ΔHT= ΔH298

T °C aA bB cC dD

T °C

Base temperature

III

III

)()()()( TdDTcCTbBTaA TH

PP

CT

H

dT

dHb

dT

dHa

dT

dHd

dT

dHc

dT

Hd BADC

.)(.)( reactPprodPP CCC

PBPAPDPC CCCC P

P

CT

H

T

PT

T

P

H

H

dTCHH

dTCHdT

298298

298298

)()()()( 298 TdDTcCTbBTaA H

Example – Find the net heat available or required when the following reaction takes place at 800 K

Substance ΔHo298 (kJ/mole) CP (J/mole K)

CaO(s) -634.3 49.62+4.52*10-3 *T-6.95*105*T-2

CO2(g) -393.5 44.14+9.04*10-3 *T-8.54*105*T-2

CaCO3(s) -1206.7 104.52+21.92*10-3 *T-25.94*105*T-2

= -1206.7 - 634.3 + 393.5 = -178.9 kJ

= 10.76 + 8.36*10-3*T-10.45*105*T-2 J/K

)()()( 22982983298298 COHCaOHCaCOHH oooo

)()()( 32 sCaCOgCOsCaO

)()()( 23 COPCaOPCaCOPP CCCC

800

298

253800 )10*45.1010*36.876.10(178900 dTTTH o

1733955505178900800 oH J

Alternatively ΔHT can be calculated from Hess’ law

)298()298()298(

)800()800()800(

32

32

298

800

CaCOCOCaO

CaCOCOCaO

H

H

1 2

3

4

298

800 )(1 dTCH CaOP

298

800 )(2 2dTCH COP

oHH 2983

800

298 )(4 3dTCH CaCOP

4321 HHHHHT

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 Cu2S 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) = ΔHI + ΔHII + ΔHIII + ΔHIV

(-) (-) (+) (+)

Air 25 °C

1200 °C

M.Pt.

Liquid Cu2S Waste gases1250 °C

Liquid Cu1300 °C

M.Pt.

Base temperatureCu2S (l) + O2 (g) → 2Cu (l) + SO2 (g)

III

III IV

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

A ΔH B

X Y Z

Reaction Enthalpy change AX ΔH(1) XY ΔH(2) YZ ΔH(3) ZB ΔH(4) AB 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 temperature2. 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

2 34

Non-isothermal complex processes

))(())(())(())(( 3321 TsdDTlcCTlbBTsaA TH

)298)(()298)(()298)(()298)(( 3 sdDscCsbBsaA H

21 4 5

)(2983298 ))((5

)(2983))(()(298 ))((4

.)(298.)(2982983

)(2982

298

))(()())((2

298

)(298))((1

54321

)(

)(

)(

)(

3

3

)(

)(

)(

)(

2

11

sDT

T

sDP

lCT

T

T lCPCm

T

sCP

oreact

oprod

o

lBTT sBPBm

T

T lBP

T sATsAP

T

HHddTCdH

HHcdTCHdTCcH

HHHH

HHbdTCHdTCbH

HHadTCaH

HHHHHH

Cm

Cm

Bm

Bm

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)(8.18)(4)(3

)(8.18)(4)(3)(8.18)(5)(

222

2222283

gNgOHgCO

gNgOHgCOgNgOgHC H

12

298 1505

Substance HT-H298 (J/mole) ΔHo298 (kJ/mole) CP (J/mole K)

C3H8(g) -103.55CO2(g) -16476+44.25*T+0.0044*T2+8.62*105 T-2 -393.5 44.14+9.04*10-3 *T-8.54*105*T-2

H2O(l) -285.85 75.47H2O(g) 34660+30.01*T+0.00536*T2-0.33*105 T-2 -241.95 30.01+10.72*10-3 *T+0.33*105*T-2

N2(g) -8502+27.88*T+0.00213*T2 -- 27.88+4.27*10-3 *TΔHv(H2O) = 40897 J/mole

Air: 21% O2 + 79% N2

Heat balance comprises of heat input which is equal to heat output plus heat accumulation

The benefits of heat balance:

Calculating the retained heatIf 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 deficitIf 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

The ability to calculate the furnace temperaturethe 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 furnaceOnce 900 or 1000 degree celsius reached, the choices are only limited to refractory materials

Refractory materialsMaterials that can withstand high temperatures, corrosion from liquids and abrasion of hot gases Silica (SiO2) Melts at 1724 C Temperatures attained in Metallurgical processesAlumina (Al2O3) 2050 C Copper smelting 1000-1100 CAluminosilicate (xAl2O3.ySiO2) 1600-1820 C Zinc retorts 1400-1600 CLime (CaO) Bessemer converter 1600 CMagnesia (MgO) 2165 C Oxygen converter 1850 CForsterite (2MgO.SiO2) Tuyeres in iron blast furnace 1900 CDolomite (MgO.CaO) Electric arc temperature 3600 CHematite (Fe2O3) or Magnetite (Fe3O4) Electric arc furnace 1800 CChromite (FeO.Cr2O3) 2050-2200 CCarbon (Graphite) 3600 CMetals (Water cooled)Carbides (silicon carbide) 2700 C

Acid refractories absorb oxygen ions when dissolved in a basic melte.g. SiO2 + 2O2- = SiO4

4-

Siliceous materials that consist of silica and are low in metallic oxides and alkalies• Natural rock, quartzite sand, silica brickAluminosilicates that consist of chemically combined silica in aluminaFree silica should not be present as they lower the melting point• Natural rock, fireclay, firebrick

Basic refractories provide oxygen ions when dissolved in a melte.g. MgO = Mg2+ + O2-

Aluminum oxides• Bauxite or bauxite brick, electrically fused bauxiteCalcium 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 strongAluminosilicates are sometimes classified as neutral refractories, but they exhibit an acid reaction in contact with basic slagsCarbonaceous refractories Metals• Graphite, carbon bricks Fe, Cu, Mo, Ni, Pt, Os, Ta, Ti, W, V and ZrArtificial refractories Others• Zirconium carbide, silicon carbide Forsterite, concrete, serpentineChromite

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 dustAcidic refractories should be in contact with acid slag that is high in silicaBasic refractories should be in contact with basic slag that is high in CaO or MgOMost of the oxide or silicate refractories are fully oxidized so that they will not be affected by oxygenGraphite and silicon carbide oxidize and burn at high temperatures

Softening point: The temperature at which the refractory is plastically deformed under loadMore important criterion for the selection of refractories than the melting point