Transcript MME 4517

Energy and Heat in Reacting Systems
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
Initial state
(State 1)
Final state
(State 2)
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
𝐹 =2+ 𝑁−1 𝜋 − 𝜋−1 𝑁 =2− 𝜋+𝑁
π
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energy is withdrawn as heat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withdrawn as expansion work
Part of the internal energy of a system may be withdrawn as heat or work by a chemical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
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
𝑊 = 𝐹∆𝑙 or 𝑊 = 𝑃∆𝑉
Electrical work – Work is done by the passage of an electric current in s direct current circuit
𝑊 = 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 ∗ 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 ∗ 𝑡𝑖𝑚𝑒 = 𝑉 ∗ 𝐼 ∗ 𝑡
The first law of thermodynamics for a process occurring in a system is:
∆𝑈 = 𝑈2 − 𝑈1 = 𝑄 − 𝑊
where 𝑈1 and 𝑈2 are the internal energies of the system in state 1 and state 2, 𝑄 is the
absorbed heat by the system from surroundings and 𝑊 is the work done by the system on the
surroundings
Almost all processes in extractive metallurgy follow constant pressure paths, at about 1 atm
Heat content of a system for these constant pressure processes is defined as:
𝐻 = 𝑈 + 𝑃𝑉
for any thermodynamic change in state, 𝜕𝐻 = 𝜕𝑈 + 𝑃𝑑𝑉
For a constant pressure process in which all work done by the system on the surroundings is
the work of expansion,
𝑉2
𝜕𝑊 = 𝑃
𝑑𝑉 ,
𝑊 = 𝑃 𝑉2 − 𝑉1
𝑉1
and ∆𝐻 = 𝑄𝑃 where 𝑄𝑃 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,
∆𝐻 = 𝑄𝑃 − 𝑊 ′
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,
𝐷
𝑇
∆𝐻 = 𝐻𝑇 − 𝐻298 = 𝐴 + 𝐵𝑇 + 𝐶𝑇 2 + calories/mole
sensible heat
∆𝐻 = 𝐻𝑇2 − 𝐻𝑇1 = 𝐻𝑇2 − 𝐻298 − (𝐻𝑇1 − 𝐻298 )
and indirectly by the heat capacity
𝜕𝐻
𝜕𝑇 𝑃
The engineer uses heat content data most frequently for determining sensible heats
(𝐻𝑇 − 𝐻298 ) and the heat quantities evolved or absorbed when the temperature of a substance
is changed between 2 known levels
𝐶𝑃 =
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
HT-H298
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
ΔH’’m
ΔHm
L
ΔH’m
S
T’’m
Tm
T’m
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
1
𝐻2 +
𝑂 = 𝐻2 𝑂
2 2
state 1
state 2
(P1, V1, T1)
(P1, V2, T2)
Heat of formation of compounds are tabulated for T= 298 K and P= 1 atm
ΔHH2Of= -241.8 kJ/mole
Signs of ΔH for constant pressure processes
Positive
Heat absorption from surroundings
Melting
Vaporization
Most dissociation reactions
Negative
Heat evolved to surroundings
Freezing
Condensation
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
∆𝐻 =
∆𝐻𝑓 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 −
∆𝐻𝑓 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
@ 298 K
Fe3 O4 + 4CO = 3Fe + 4CO2
4C + 4O2 = 4CO2
∆𝐻𝑓 = 4 ∗ −94050 𝑘𝑖𝑙𝑜𝑐𝑎𝑙𝑜𝑟𝑖𝑒𝑠
- 3Fe + 2O2 = Fe3 O4
∆𝐻𝑓 = −267000 𝑘𝑖𝑙𝑜𝑐𝑎𝑙𝑜𝑟𝑖𝑒𝑠
- 4𝐶 + 2𝑂2 = 4𝐶𝑂
∆𝐻𝑓 = 4 ∗ −26420 𝑘𝑖𝑙𝑜𝑐𝑎𝑙𝑜𝑟𝑖𝑒𝑠
∆𝐻 = 4 ∗ −94050 − −267000 − 4 ∗ −26420 = −3520 𝑘𝑖𝑙𝑜𝑐𝑎𝑙𝑜𝑟𝑖𝑒𝑠
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
3Fe  2O2  Fe3O4
6 Fe  3O2  6 FeO
ΔH˚298 = -264500 J 6 FeO  3 / 2O2  3Fe2O3
Fe  1 / 2O2  FeO
3Fe2O3  2 Fe3O4  1 / 2O2
ΔH˚
=
-292500
J
2 FeO  1 / 2O2  Fe2O3
298
2 Fe3O4  1 / 2O2  3Fe2O3 ΔH˚298 = -230650 J 6 Fe  4O2  2 Fe3O4 ΔH= 6 ΔH1+3 ΔH2- ΔH3
3Fe  2O2  Fe3O4 ΔH/2=-1117240 J/mole
ΔH for a high temperature process can be calculated in the same way as Hess’ law is used
calculate the heats of fomation:
HT
aA(T )  bB(T ) 

 cC (T )  dD(T )
dH C
dH
dH D
dH A
dH B
c
d
a
b
dT
dT
dT
dT
dT
 H 

  CP

T

P
 H 

  CP  CPC  CPD  CPA  CPB

T

P
CP   CP( prod.)  CP( react.)

H T
T
dH   C P dT
H 298
298
T
H T  H 298   C P dT
298
The process is represented schematically as
aA bB
I
cC dD
T °C
II
aA(T )  bB(T )  cC (T )  dD(T )
H 298
T °C
III
Base temperature
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
Example – Find the net heat available or required when the following reaction takes
place at 800 K
CaO(s)  CO2 ( g )  CaCO3 (s)
ΔHo298 (kJ/mole)
CP (J/mole K)
-634.3
49.62+4.52*10-3 *T-6.95*105*T-2
-393.5
44.14+9.04*10-3 *T-8.54*105*T-2
-1206.7
104.52+21.92*10-3 *T-25.94*105*T-2
Substance
CaO(s)
CO2(g)
CaCO3(s)
o
o
o
o
H 298
 H 298
(CaCO3 )  H 298
(CaO)  H 298
(CO2 )
= -1206.7 - 634.3 + 393.5 = -178.9 kJ
CP  CP (CaCO3 )  CP (CaO )  CP (CO2 )
= 10.76 + 8.36*10-3*T-10.45*105*T-2 J/K
H
o
800
800
 178900   (10.76  8.36 *103 T  10.45 *105 T 2 )dT
298
o
H 800
 178900  5505  173395 J
Alternatively ΔHT can be calculated from Hess’ law
H 800
CaO(800)  CO2 (800) 
 CaCO3 (800)
1
2
4
H 298
CaO(298)  CO2 (298) 
 CaCO3 (298)
3
298
H1   CP (CaO) dT
800
298
H 2   CP (CO2 ) dT
800
o
H 3  H 298
800
H 4   CP (CaCO3 ) dT
298
HT  H1  H 2  H 3  H 4
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
Liquid Cu2S
1200 °C
Liquid Cu
Waste gases
1300 °C
1250 °C
M.Pt.
I
Air 25 °C
II
Cu2S (l) + O2 (g) → 2Cu (l) + SO2 (g)
III
M.Pt.
IV
Base temperature
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
(-) (-)
(+) (+)
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
Y
4
Z
1
X
Reaction
AX
XY
YZ
ZB
AB
2
3
Enthalpy change
ΔH(1)
ΔH(2)
ΔH(3)
ΔH(4)
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
Non-isothermal complex processes
HT
aA( s)(T1 )  bB(l )(T2 ) 

 cC (l )(T3 )  dD( s)(T3 )
1
2
4
5
H3
aA(s)( 298)  bB( s)(298) 

cC (s)( 298)  dD( s)(298)
H T  H1  H 2  H 3  H 4  H 5
H1  a 
298
T1
C P ( A( s )) dT  a ( H T1  H 298 ) A( s )
Tm ( B )
298

H 2  b 
C P ( B (l )) dT  H m ( B )   C P ( B ( s )) dT   b( H T 2  H 298 ) B (l )
 T2

Tm ( B )
o
o
o
H 3  H 298
  H 298
( prod .)   H 298 ( react .)
Tm ( C )
T3

H 4  c 
C P (C ( s )) dT  H m (C )   C P (C (l )) dT   c( H T 3  H 298 ) C (l )
 298

Tm ( C )
T3
H 5  d  C P ( D ( s )) dT  d ( H T 3  H 298 ) D ( s )
298
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 ?
H
C3 H 8 ( g )  5O2 ( g )  18.8 N 2 ( g ) 
3CO2 ( g )  4 H 2O( g )  18.8 N 2 ( g )
298
1505
1
2
3CO2 ( g )  4 H 2O( g )  18.8 N 2 ( g )
298
Substance
C3H8(g)
CO2(g)
H2O(l)
H2O(g)
N2(g)
ΔHv(H2O) =
HT-H298 (J/mole)
-16476+44.25*T+0.0044*T2+8.62*105 T-2
34660+30.01*T+0.00536*T2-0.33*105 T-2
-8502+27.88*T+0.00213*T2
40897 J/mole
Air: 21% O2 + 79% N2
ΔHo298 (kJ/mole)
-103.55
-393.5
-285.85
-241.95
--
CP (J/mole K)
44.14+9.04*10-3 *T-8.54*105*T-2
75.47
30.01+10.72*10-3 *T+0.33*105*T-2
27.88+4.27*10-3 *T
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
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 (SiO2)
Melts at 1724 C
Temperatures attained in Metallurgical processes
Alumina (Al2O3)
2050 C
Copper smelting
1000-1100 C
Aluminosilicate (xAl2O3.ySiO2) 1600-1820 C
Zinc retorts
1400-1600 C
Lime (CaO)
Bessemer converter
1600 C
Magnesia (MgO)
2165 C
Oxygen converter
1850 C
Forsterite (2MgO.SiO2)
Tuyeres in iron blast furnace 1900 C
Dolomite (MgO.CaO)
Electric arc temperature
3600 C
Hematite (Fe2O3) or Magnetite (Fe3O4)
Electric arc furnace
1800 C
Chromite (FeO.Cr2O3)
2050-2200 C
Carbon (Graphite)
3600 C
Metals (Water cooled)
Carbides (silicon carbide)
2700 C
Acid refractories absorb oxygen ions when dissolved in a basic melt
e.g.
SiO2 + 2O2- = SiO44Siliceous 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 = Mg2+ + O2Aluminum 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 refractories
Metals
• Graphite, carbon bricks
Fe, Cu, Mo, Ni, Pt, Os, Ta, Ti, W, V and Zr
Artificial refractories
Others
• Zirconium carbide, silicon carbide
Forsterite, concrete, serpentine
Chromite
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