Upload
lucinda-eaton
View
221
Download
2
Tags:
Embed Size (px)
Citation preview
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in the systemTemperature is an indication of internal energy which reduces when some of the internal energy is withdrawn as heatPressure also represent internal energy which reduces when some of the internal energy is withdrawn as expansion workPart of the internal energy of a system may be withdrawn as heat or work by a chemical reaction occurring in the system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