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11 CHAPTER 6ENERGY BALANCES

6.0 INTRODUCTIONIn the previous chapter, the basis for process energy bal-.ance calculations was presented, namely, the Law of Conservationof Energy. The energy effects of various reactions, transformations, and changes of state were presented. In this chapter al lof these effects are brought together as metallurgical processplants and operations are analyzed.As in the case of material balances, the purpose of theenergy balance must be considered before the problem can be properly formulated. The purpose may be to compare several completeprocesses which produce the same product, to see which requiresless total energy from the universe. Or, i t may be to comparealternative modes of operation of an existing process, or i t maybe to develop information needed for control of a process. Eachpurpose requires differing details, and will require differentdata. Consider for example, the basic oxygen steelmaking process.If the process is being compared with the electric furnace steel-making process for purposes of total energy consumption comparison,the energy consumption of the process would have to include theelectrical energy required to t i l t the vessel, the electricalenergy to hoist the oxygen lance, the energy required to producethe oxygen from air, the energy to run the waste gas recovery system, etc.

electricity

DustCo 11 ector

"",.r:':----:--fluxes ~

Iel ectricity

water

l,as teGases

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MATERIAL AND ENERGY BALANCE CALCULATIONS IN METALLURGICAL PROCESSES

On the other hand, if an energy balance was being developed tobe used to control the process, many of these inputs would not berequired. In other words, we would draw the system boundary differently:

2 ~ ( Waste Gases (Hot)/' tl---,'< X luxesolten Pig Iron \_ I ~ Heat Losses

I\ 0 IX BOF J--...MOlten Steel

~ 1 0 1 t e n Slag ''''''----''1SystemBoundary

6.1 THE HEAT BALANCEThe most common type of energy balance on metallurgicalprocesses is the "heat" balance, in which only the thermal energy(heat) into and out of a process is accounted for. It should moreproperly be referred to as a thermal energy balance, but commonusage dictates that we call it a heat balance. All that is required to successfully develop a heat balance on a process is anaccountant's mind! Referring again to Chapter 5 and limitingconsideration to thermal energy:Heat IntoSystem Heat outof System + Heat Accumulatedin System

In a steady-state process, no heat is accumulated.

(6.1-1)

The biggest problem, and the place where more mistakes aremade than anywhere else, is in the choice of reference temperature.This will be considered in the following general discussion, butfirst the reader is reminded that heats of reaction can be calculated at any temperature according to Kirchoff's Law, since enthalpy is a state function . Also, since most metallurgical processesare constant pressure pr.ocesses, enthalpy, rather than internalenergy, is the measure of heat energy.

248

ENERGY BALANCES

Consider a process in which solid materials, fuels andpreheated ai r enter a process, react, and release process gasesand products at elevated temperatures, and loose energy by radiation and convection to the surroundings:In Out

So1 ds @298 K } {Gases @1800 KFue1@298 K - - - - - - - ~ Condensed Products@1600KAir @1000 KThe heat balance for such a process can be represented as follows,if the reference temperature is taken as 298K:

Gases, 1800 KProducts, 1600 K

t Ai r 1000 KTE ~ H R 298298 Fuel .. . ....-"-__ .... - - - - ' - - - - - ~ lossReactants

The enthalpy terms would be as follows, with the sign of the termbeing as indicated.Term Sign of Term

298

11 ,Cp,airdT1800

1Cp,gasesdT +16001= Cp,products dT +249

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Sum of the Heats of Reaction atLLlHR 298 = 298K to form the output products, and gases.

ILlHlos s = net heat loss to surroundings.The energy balance, then, is

Sign of Term+ if endothermic- if exothermic

+

(6.1-2)This is o n l y . c o r r ~ c t , h o ~ e v e r , if great care is taken to followthe proper dlrectlon of lntegration, from tail to head of thearrows on the diagram.

Now, c o n s i d ~ r the same process, but instead of calculatingthe heats reactlon at 298 K, assume, for example, that it ismore convenlent to calculate LLlHR at 1000 K (because, perhaps, theL I ~ f data reactants and products are available at 1000K.) Thedlagram wlth 1000 K as the r eference tempera ture would be:T Gases .1800 K

I 4 Products @ 1600 KLlH3 ILlHR 1000 LlH1000 A i r _ . - ~ - - - ~ - - - - - - - - - - - - - , - - - ~ ~ ~ ___ ______ - ~ 4 ~ ~ .. I- - LlHl ossI ILlHl LlH2

298 Fuel U ReactantsThe terms in this case would be

Term Sign of TermI 1000 ILlH = J Cp,Fuel dT1 +298I 1000

C ~ , r e a c t a n t s d TlH2 J +298I 1800 ILlH3 J C dTp,gases

1000+

25 0

ILlH loss

TermILlH =4

1600J IC dTp,products1000LLlHf,lOOOproducts LLlHf,lOOOreactants

ENERGY BALANCES

Sign of Term+

+ or -+

Again, as long as the integration direction follows the arrows, theenthalpy balance is :(6.1-3)

This time, however, note that there is no term involving the sensible heat contained in the incoming ai r and that none of the terms(except possibly LLlHR,lOOO) is a negative term.The argument can be carried further. The result would alwaysbe the same. To set up the balance properly, use the followingalgorithm:Step 1:Step 2:Step 3:

Decide at which temperature heats of reaction willbe calculated. This is the reference temperature.Draw a temperature scale.Place the reactants at their initial temperatureson the scale and draw an arrow from that temperature to the reference temperature.

Step 4: Repeat Step 3 for al l products leaving the system.Step 5: Calculate al l sensible heat terms by integratingheat capacity data in the direction of the arrowto the reference temperature.Step 6: Calculate the heats of reactions, solution, etc.,at the reference temperature.Step 7: Add al l of the terms obtained in previous steps.Step 8: The heat loss term is the result. A positivevalue means that there is a loss from the systemto the surroundings.

EXAMPLE 6.1-1: Show how the thermal energy balance can be obtainedfrom the First Law of Thermodynamics.Solution: The most general form of the First Law is given in

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Eq. (5.2-2). For a system at constant pressure, this formulationbecomes-2" n V.d(E + PV ) = oq - ow* + E omi (hi + __ + ~ z.)i=l 2gc gc 1

Since the thermal energy or heat balance consider s changesin enthalpy only, the changes in kinetic and potential energy ofthe system can be assumed to be equal to zero and*(E'+ PV') = d(U' + PV') = dH' = oq + Eomi hi - oW

If no electrical work is put into the system or produced by thesystem,

For a system at steady-state there is no change in the state function:dH' = 0

or , therefore, the heat put into or lost from the system,

oroq = (enthalpy of outputs)from system (enthalpy of inputs)to the system

On a molar basis, oq = -EoniHiTo show that this approach is equivalent to the heat balancegiven in Eq. (6.1-2), consider the example with a reference temperature of 1000 K. For the process,lIHi + lIH2 + lIH3 + lIH4 + ElIHR,lOOO = lIHioss

However,TRetlOOO

l I H ~ = f C ~ , f u e l d T = [ H ~ u e l @ ( T R e f = l O O O ) - H'fuel@(T in=298)]T n=298

252

,lIH3

ENERGY BALANCES!ef=lOOO ,= = H @T =1000Cp,reactants dT [ reactants ( Ref, )Tin = 298 -Hreactants@(Tin=298)]Tfout=1800, ,C dT= H @(T =p,gas [gas out 1800) - H @(T = 1000)]gas RefTRef=l 000

, " ,JTout=1600lIH = C dT=[H @(T =1600)-H d @(TR f=lOOO)]4 p,prod. prod. out pro. eTRetl 000andElIH = E lIHF @(TRef=lOOO) - E lIHF @(TRef = 1000)R,lOOO products reactants

E H @(TRef = 1000)- E H (TR f = 1000)products react ants e, ,Hproducts @ (TRef = 1000) + Hgas @ TRef = 1000), , ,-[H @(T =lOOO)+H @(TR f=lOOO)+H . (TRef=lOOO)]fuel Ref reactants e a1rFinally, combination of these equations yields

+ H -H @(T. )]~[ ants Ref) reactants 1n, ,+ [Hgas@(Tout) - ~ )+ [ H ~ r o d u c t s @ ( T o u t ) ~ s @ ( T R e f ) ]

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Since the inlet temperature of the ai r was the referencetempera tureI I I I I+ + + + r ~ H R , l O O O

= [H I @ (T ) + H' @ (T ) ]products out gases outI I I-[H @(T.) + H @(T. ) + H . @(T. ) ]fuel ln react ants ln al r ln

= (enthalpy of outputs from system)- (enthalpy of inputs to system)

Finally, the negative of the heat loss from the system mustcorrespond to the heat gained by the system

orI- ~ H l o s s = oq

oq = enthalpy of productsleaving system

enthalpy of inputsto systemwhich is equivalent to the statement of the First Law.

The previous example demonstrates that the First Law and theheat balance are equivalent. It also shows that the choice ofthe reference temperature is completely arbitrary since al l termsbased upon this value cancel out during the calculation. The heatloss term will be the same, regardless of what reference t e m p e r a ~~ u r e is chosen, if the problem is se t up properlX.EXAMPLE 6.1-2: A furnace for reheating billets, prior to theirbeing rolled, uses fuel oil and ai r to produce hot gases whichpass over the billets, transferring heat to both the billets andthe furnace walls. Some of the heat in the walls is re-radiatedto the billets and some travels through the walls and is lost tothe surrounding. The gases finally exit from the furnace at areduced temperature.

254

ENERGY BALANCES

AirAiro Fuel -==y __ _JLc - ODD o ~ o l - Billets InC"" --:c 0 0 0 0 - DODD 000 - 1 __ 1/ ~ . - - J )

\)Il't Gases to stack _."e'ts'0'

* h f 1 oil is 40,000 kJ/kg, 25%The net h e a t ~ n g value of t s ~ e u:ses analyze 4.6% 02. Theexcess ai r is supplled, and the wa dgTeave at 1400 K at a ratebillets enter the furnace at 298 K an at 1500 K. The oil isof 130,000 kg/hr. The waste gases ~ e : ~ : l Y Z e S 85% C, 14% Hand 1%supplied at a rate of 100 k 9 / h ~ ' t ~ ~ thermal efficiency. For purS. Compute the heat losses an heat capacity of flue gases =poses of the example, use a m ~ a n1.05 kJ/kgK. (=0.27 Btu /lb- F).f e temperature is clearly 298 K,Solution: The heat balance re ~ r e ~ ~ C h the heat of combustion ofsince that is the temperature aW l

the fuel is available.The diagram thus becomes: 1500 K1400 K

I~ H g a s e s

tT I~ H s t e e ls t e e l } . ~ ~ r ~ ~ ~ H ~ ' ~ 2 ~ 9 ~ ~ __________L-__________ai r ~ H l o s soil298

. h rmochemical energy liberated by* The net heating value l ~ h ~ h ~ u ~ l e t o H20(9) and C02 at.298 K. Thecomplete c?mbustion of 1 refers to the energy l l b ~ r a t e d bygross heatlng v a ~ u e of a fu(e) d CO at 298 K. The dlfferencecomplete combustlon to H20 t t . o ~ of water at 298 Kand theis obviously ~ h e ~ e a t of v ~ p o r l z ~ o ~ fuels made up or pure ele-heat of vaporlzatlon of wa ere the calorific power can bements, or a mixture of c o m p o u n d ~ ~ of the individual compounds,calculated from heats of c o m b u s . ~ o n W h i c h are indefinite in combut for fuels s ~ c h a S l ~ o a ~ ~ ~ r ~ ~ n ~ d values must be used.position, experlmenta y e

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The enthalpy terms are:L ~ H R , 2 9 8 = Combustion = -(100 kg)(40,000 kJ/kg)

= -4 x 106 kJI

steel = (130,000 kg)(103 g mol/55.85kg)(10,120 cal/g mol)

= 5.651 x 105 kJ (lkJ/4.186 x 103 cal)The enthalpy content (sensible h )after combustion calculatl'o ea t of the gases can be foundf . f' . ns are made in d . ,o ln lltrate al r (unmeasurable oth . )or er to flnd the amounterwlse. For 1 kg oil:

products(0.85 kg C)(44/12) 7 3.12 kg C02(0.14 kg H2)(18/2) 7 1.26 kg H20

19.01 kg SL(64/32) 7 ~ kg S021. 00 kg oi l 4.40 kg

wt. ~3.12 kg CO2 0.07091.26 kg H2O 0.070.02 kg S02 0.00030.87 kg O2 0.027215.10 kg N2 0.539320.37 kg gases 0.7077

required O22.34 kg1.120.013.47 kg O2

The 25% excessso that the

.YL!10.039.900.00043.8576.21

100. Since the oxygen analyzer i d' t 0ly lnfiltrate air presen t. Furth n lCf 4:6% O2, there is clear-analysis, on a volume basis . er ca atlons show that thisper mole of oi l enters the ~ y ~ ~ l l o c ~ ~ r lf 1 additional kg of ai rwaste ases must amount to 21 3 ~ m k us, the ~ o t a l amount of

I ases/k 011, and~ H f l u e gases = (2137 kg gas)(1.05 kJ/kgK)(1500 K-298 K)

2,697,100 kJ/hr.25 6

1

ENERGY BALANCES

The balance then is:I-4,000,000 + 565,100 + 2,697,100 = ~ H l o s s

I

~ H l o s s = 737,800 kJ/hr.The thermal efficiency of the process can be measured or describedin several ways. Since the requirement of the process is to heatsteel from 298K to 1400K, the efficiency has to be measuredagainst this requirement. However, the base of this efficiencycan vary.

For instance, in order that the steel reach 1400K, theenergy required to raise it to that temperature must be availableat temperatures above 1400K. The gases cannot transfer energy tothe metal unless they are hotter than the metal. In a countercurrent system such as in this Example, the gases could exit fromthe furnace at any temperature abgve 298K. Therefore, the potential available energy is 4.0 x 10 kJ, and the efficiency isEfficiency = enerqy absorbed by product x 100energy supplied

565,1004,000,000 x 10014.1%

On the other hand, had this been a batch type furnace, witha cold charge going into the furnace and staying there until theentire charge reaches 1400K, the energy available to be transfer-red to the metal would have been less than 4.0 x 106 kJ, becauseas the stock heated up, the waste gases could not transfer energyat temperatures below the stock temperature. This would causethe numerator in the efficiency equation to become smaller, in-creasing the apparent efficiency.Finally, the efficiency of the furnace could be measured bythe effectiveness of the insulation, in which case the efficiencywould be

Efficiency = n e r g Y E ~ ~ r ~ y H ~ ~ t Loss x 1004.0 x 106 - 0.74 x 106

4.0 x 10681.5%

6.1.1 SANKEY DIAGRAMSAn excellent technique for visualizing the distribution ofenergy in a process, or for that matter in an entire plant, isthe Sankey diagram. All incoming energy sources and all outgoing

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energy streams are placed on the diagram; the width of each streambeing proportional to the amount of energy contained in it . Forexample, the Sankey diagram for the furnace in Example 6.1-2 wouldbe as fo 11 ows :

ChemicalEnergy inFuel Oil,4.0 x 106kJ

ReheatingFurnace

Heat Losses, 0.74 x 106 kJ

Sensible Heat in Steel, 0.56 x 106 kJ

SensibleHeat inFlue Gases,2.7 x 106 kJ

It is immediately obvious that, in this case, a large amountof the energy put into the furnace is lost from this furnace assensible heat in the flue gases, and that the efficiency of heatingof the steel is low. Thus, it is easy to visualize where effortsought to be made to increase the energy efficiency of the operation.Redesign of the furnace, for example, to decrease heat losses orimprove heat transfer so that less fuel is required would be oneway, but even more obvious would be to utilize the heat in theflue gases. This might be accomplished by using the gases to produce steam in a boiler and then using the steam to produce electricity via a turbine. The use of waste-heat boilers is common inmany large process plant complexes.Another way to utilize the energy in the flue gas would beto transfer it s energy to the incoming combustion air, thus recirculating some of the energy and reducing the requirement forfuel oil. This is done by means of a recuperator, in which energyis Ireco!Jped", the French word for regained. Recuperators takemany forms, such as shell and tube heat exchangers, crossflow orcounter-current in arrangement, and they may be continuous orintermittent in operation. As fa r as energy efficiency is concerned, it is only the efficiency of heat exchange that is ofinterest here.

EXAMPLE 6.1-3 : Taking the situation in Example 6.1-2, examinewhat happens if the heat recuperator shown in Fig. 6.1-1 was puton the flue gas stream and i t was assumed that the recuperatorwas 50% efficient at heat exchange.Solution: Since the use of preheated ai r will decrease the requirement for chemical heat, the fuel requirements must be recalculated.Also, since less fuel is required, less combustion ai r will be

25 8

ENERGY BALANCES

required and fewer kg of waste gases will be available. The heatbalance is somewhat different now:

fT

I

lIH flue gas298 ~ H c o m b . I- - ~ - - - - - - - - - - ~ ~ ~ L - - - - - - - - - L - - - - - - - " l ' I H 1 0 S S

---_ D O O O O o - - O D O O O o - - D O ~ 1-0 0 _--___ - - -~ "- --- '\- ,"h''',d ' ; ' I : ~ ~ ~ ~ ~ e

__ ---' \ '- . '- - ' - ~ cold ai r---------.",.'"

recuperator

Fig. 6.1-1 Billet heating furnace with recuperator.

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lIH' is thesteelis the same, becauselIH'. is taken to beal rexchange efficiency.

same, and for the moment, assume that lIH10ssvirtual11 the same temperatures are involved.(0.5) lIHfl ' based on the assumed heatue gasLetting F be the number of kg of fuel oil.

Heat In Heat Out (m) (C p) (LIT)(0.5)(21.37F)(1.05)(1202)+40,000F=565,100+737,800+(21 .37F)(1 .05)(1202)13,485F+40,000F=1,302,900. + 26,971FF=49.14 kg fuel oi l,lIHfl =1,325,300 kJ/hrue gas,lIHrecouped=662,650 kJ/hr

lIHcombustion=1,965,500 kJ/hr.The Sankey diagram for this situation would look as follows:

RecoveredTherma1Energy

Chemical energyin fuel oil1.96 x 106 kJ

Heat losses/

0.74 x 106Reheating 106 Sensible Heat inFurnace ~ ; . ; . : " . . . . . ; . . ; . ~ ~ Steel

0.66 x 106 kJ Combustion AirSensible Heat inFlue Gases toStack

2O

il

ENERGY BALANCES

Now, total fuel consumption has been drastically decreased,from 100 kg oil/hr to 49.1 and the efficiency is up to 21%. Heatrecovery of this type should be practiced as often as possible,and usually is , consistent with local economic conditions.The Sankey diagram for processes involving chemical reactionsother than combustion of fuel also helps to highlight the potentialareas for fuel savings. Such a diagram, for the sinter plantillustrated in Fig. 4.3-2 is given in Fig. 6.1-2 . Notice firstthat the heat losses, as a percentage of the energy in , are small.This is typical of most gas-solid packed bed systems, such as

sinter plants and shaft furnaces. Notice also that the energy toevaporate moisture is not insignificant. Hydrated ores, such asFe203'H20 require extra energy to process. The only potentialenergy savings can be made by recouping heat from the waste gasesor the hot sinter itself. These amount to about 50% of the overallthermal output of the process. Some sinter plants do have sintercoolers in which the heat recovered from the hot sinter by ai r istransferred to the sinter on the machine via huge ducts that emptyjust above the sinter bed. Some fuel savings are effected in thismanner.One might be tempted to think that if these sensible heatstreams could be removed as outputs, the fuel required would bedecreased. They do look like "losses". However, this would be amistake, because they result from the fact that within the process,certain temperatures must be attained, and in order to reach thosetemperatures, the fuel is necessary in the first place. Furthermore, none of the streams leaving the system is at a temperatureanywhere near the maximum temperatures reached in the process it self, and may not be useful, except in neighboring processes orplants. This is a result of where the system boundary has beendrawn. In other words, the "qual ity" of the energy being lost istoo low for the energy to be useful. Thus, the use of Sankeydiagrams from the standpoint of understanding a process can bemisleading. More on the subject of energy guality is presentedin Section 6.4.

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Heat in flue ~ Idust, 7.7% I( r e t u r ~ e d to ' \ .Sinter mix) 1

Sintert1achi ne

f

Formation ofFe 304 , 0.8%(3FeO + -+ Fe304)

To evaporatemoisture, 15.7%

~ . ~ l ~ To d " ~ p o , ,' :. . "- carbonates, 14.1%Sensible ~ . (CaC0 3 -+ CaO + CO2)Heat in waste

""" 259:"'"ti" hee' j j . .. ( : : ~ ~ ~ ~ : : ' ~ ~ : : 0 3 ' "2091)of carbon insinter, 2.4%(unburned)

Sensibleheat insinter, 25.4%

Fig. 6.1-2 Sankey diagram of the heat balance onthe sinter plant depicted in Fig. 4.3-2.Energy values are as percent of totalenergy input and output. (Adapted fromAgglomeration of Iron Ores. D. F. Ball.et al . American Elsevier. New York.1973.)

262

ENERGY BALANCES

EXAMPLE 6.1-4: In Example 4.3-7. the material balance for a "heat"of basic oxygen steel was developed. Using the results of thatcalculation prepare a heat balance for the same "heat". Additionalpertinent data are on the figure below:

80 kg 02 as @273 K86 kBurnt Lime

@ 298 K

o

Infiltrate air@ 298 K

179 kg

I

264 kg"was te Gasf I @ 2000 K

Solution: The choice of a reference temperature requires thatwhich reactions are taking place and what the heats of thosereactions are be known. In this case the reactions areC + 02(g)

Si + 02(g)F e ( ~ ) + 1/2 02(g) + F e O ( ~ )

1.65 CaO(S) + Si02 + 1 65 C a O S i 0 2 ( ~ )

(a)(b)( c)(d)

where the underline. eg., Si , indicates that the element is dissolved in molten iron. From Chapter 5. we saw that since Eqs. (a),(b) and (c) involve two steps. eg S i ( ~ ) + 31,100 cal/mol

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AH f ,Si02 = - 217,700 cal/molthe sum of the two steps gives the net reaction and it s enthalpy:

AHR,(b) = - 186,600 cal/mol Si02In other words, we find the net heat effect by adding the heateffects of the two steps. This involves knowing the heat ofsolution of the elements in liquid iron. This data is available(Table 6.1-1) at 1600C (2912F, 1873 K) and since data for theother reactions are also available at 1600C, i t is convenient,this case, to choose 1600C as the reference temperature.

TABLE 6.1-1Heats of Sol ution of Elements in r10lten Ironat 1 wt.% Concentration and 1600C (1373 K)

-mReaction: H. ca1/g mol1C (gr) + C (1 % in Fe) + 5,100Si(q,) + Si (1% in Fe) - 31,100Mn + Mn (1 % in Fe) 0Cr( s) + Cr (1% in Fe) + 5,000A + AT (1% in Fe) - 10,3001/2 02(g) + o (1% in Fe) - 28,0001/2 S2(g) + S (1% in Fe) - 31.500Co( + Co (1% in Fe) 0C u ( ~ ) + Cu (1% in Fe) + 8,000N i ( ~ ) + Ni (1% in Fe) - 5,000At 1600C, the heat effects of reactions (a), (b), and (c) are:

(a )(b)( c)

C + 02(g) + C02(g)Si + 02(9) + S i 0 2 ( ~ )Fe + 1/2 02(g) + F e O ( ~ )

AH R 1873- 99,940 cal/mol-186,000 "- 55,780 "

In the case of reaction (d), this reaction is the closest that wecan come to estimating the heat of mixing of the slag from CaO,Si02, and FeO. The FeO - Si02 liquid system is nearly ideal froma thermodynamic standpoint, so the heat of mixing of those twoconstituents is small and will be ignored, but i t is known26 4

in

ENERGY BALANCES

(however inaccurately) that there is a heat evolution w h e n . C a O ( ~ )and S i 0 2 ( ~ ) are mixed to form a liquid. The heat effect ln thlScase will be approximated by:(d) 1.65CaO(s) + S i 0 2 ( ~ ) + 1 . 6 5 C a O S i 0 2 ( ~ ) ' AHR = -20,000 cal/molNow, draw the heat balance diagram:

walte ~ a s e s2000 AH R1873 AH6 I..;. - AH873 Steel Slag lossIAH4 Heat1620 Hot Metal LossesT,K

~ S c r a p . ~ C a O Infiltrate Air

The heat balance is then:I I I I I IAH1 + AH2 + AH3 + AH4 + AH5 + AH R1873 = - AH 10s s

This is the heat to bring the oxygen to 1873 K.

lBr3 Cp ,02dT " [H 1B73 - H273l02 x 103

= 35.0 x 106 cal.This is the heat needed to heat and melt scrap steel andbring it to 1873 K.

I 341 3AH2 = nscrap [H 1873 - H298 ]Fe = 55.85' [18550] x 10

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= 113.1 x 106 cal.This is the heat required to bring the lime to 1873 K. Sincethis lime is going to react with Si02(t ) to form the slag,and since the heat of fonnation, equation (d), is writtenin t e r m ~ of CaO(s), there is no need to add the heat offusion. 18738H; ""CaD J p.CaOdT " [H1873 - H298] x 103298

30.4 x 106 cal.The sensible heat increase in the liquid pig iron from it sinitial temperature to 1873 K.

1873 K= n J C dT=537506[16J[1873-1620JX1034 pi g iron p P1 g 1 on

1620 K= 56.6 x 106 cal.

The infiltrate air is heated from room temperature to 1873Kbefore i t reacts with CO in the waste gas duct.= nair Cp,air [1873-298J = [7.85J[1575J x 103= 76. x 106 cal.

r ~ H R 1873: Now that all of the reactants are at the reference' temperature, compute the amounts of reactants, theheats of reaction and add:Based on the waste gas analysis,

(0.4499 kg CO 2)(264 kg gas)(103 mol CO 2)(1.0 kg gas) (44 kg CO 2) = 2699 mol CO2 are formed.Based on the slag analysis,

FeO,and

* See footnote on page 210.26 6

ENERGY BALANCES

(172 kg slag)(0.149 kg Si02)(103 mol Si02)= 427 mol 5i02(1.0 kgslag)(60 kg 5i02)are formed.Heats of Reaction at 1873 K:

-99,940 cal(a) (2699 mol C) mol C02 formed(b) (427 mol 5i02) - l ~ ~ i o ~ ~ o ~ a l(c) (836 mol FeO) -55,780 calmol FeO

-269.7 x 106 cal= - 79.4 x 106 cal= - 46.6 x 106 cal

-20,000 cal = _ 8.54 x 106 cal(d) (427 mol 5i02) mol 1.65 CaO 5i02-404.2 x 106 cal

, the enthal py contained in the waste gases: This representsabove 1873 K.l I H ~ = n02[H2000-H1873]02 + nC02 [H2000-H1873]C02 + nN2 [H2000-H1873]N2

10.7 x 106 cal.Finally,

or

-lIH'l xlO-6=35.0 + 113.1 + 30.4 + 56.6 + 76.3 + 10.7 - 404.2oss

,loss

= + 322.1 - 404.2 .= - 82.1

82.1 x 106 calor 20.3% of heat input.

This amount of heat is lost via convection and/or r a d i ~ t i o n . t ~ thesurroundings, including the water-cooled hood, uP.to t e p01n bhere the 2000K gas temperature was measured, or 1S absorbed ythe refractory lining of the furnace.A 5ankey diagram for the BOF process with a reference temperature of 1873K would look as follows:

26 7

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lIHS SO--.!. -+ 1 2

BOFProcess

1IH'O2, 298 -+ 18731IH'~ Scrap, 298 -+ 1873

_ l IHp i g iron, 1620 -+ 18731-----0::::---

H' .......- : : : : ~ = = = - - - - - . L - .\ Fe -+ FeO .....lIHCaO'SiO 2

lIHWaste Gases, 1873 -+ 20001IH'losses

On the other hand, we could have chosen 298 K as the referencetemperature, in which case the heat balance diagram would havebeen:2000 Waste gases,1873 ,Steel 4 Slag

IHot Metal t.H1620 4 I gasesf llH 1s agIllHT K steel

_ EllH R298 I2 9 8 ~ ~ ~ - - - - ~ ~ ' - ~ ~ ~ ~ - - ~ ~ L - - - - - - - ~ - - - - - - ~ I l H 1 0 S S273- .nHO scrap CaO Infiltrate2 ai r

268

ENERGY BALANCES

and the Sankey diagram would be:

BOF

steel

Process I-_____ - _f'.Wgases

f'.WlossFigures 6.1-3 and 6.1-4 illustrate the material flows and a Sankeydiagram of the energy balance for an entire integrated steel plant,incorporating the preceding diagram for the BOF process.

Finally, another way to graphically present a heat balanceis by means of a bar graph. This is more useful for a batch process than for a continuous one. For the process in Example 6.1-4,such a diagram would be as follows, for the reference temperatureof 1873 C: Energy In Energy Out 100Fe -+ FeO Heat Losses 90

Si -+ Si02 Waste Gas,Suoerheat r-Infiltrate I-ai r heating 70 s-a>s::LLJ- 60 0Hot metal , +-'Heating - 50 s::a>u

C + CO2 CaO heating - s-40 a>a..I- 30Scrap heatingand melting l- 20I- 10

02 Sensible heatS l a g formation 'L-- 8.1%0269

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Row 1doOtrUte limettcne1451b 276lb C4 or24701> 46J581b1660 Ib 1- - i 4 4 i ! k ~ w h L - _ - - - - f - 1 ' - - - - 1 k : l ~ ~ I n > ; n P J a , ; ; ' f = ; : J I incl udes coke ovens27110 I.'

I Boiers I~ ~

I l J r ~ G e ~ . 1iSS . h IElectrcity I I II! II

01)' en

I and blast furnaces~ ~ ' 1 " .r - - - - - - - - ~ ~ ~ - - ~ ~ ir - - - - - - 1 ~ - _ t J ~ J ! 1

: 'Q 2 ei1--'1'--"'",,=='-=-1 oornlre lme{ ) 8 ~ J ~ D'- d ~ L - : - l C

Burnt lmeTaldolomlte

CO. 00,O ... n 2090 I.'

32kwh

AE

12380 It'

39 kwh

~ U ~ S ... Plan.

ICostinq Pi. IoS oS

iDj[ Sookinq P,t.J

'"1

I

J.W;lb

m scrapI " " ,2iu,2in 8.lleh

Fig. 6.1-3 Material flows in a primary steelmaking plant in whichcoke is produced in slot ovens, ore is reduced in ablast furnace, molten pig iron (B.F. metal) and scrapare charged to a BOF shop, and ingots are cast and hotrolled after soaking to even out temperature gradients.(Ref. The Effect of Various Steelmaking Processes onthe Energy Balances of Integrated Iron and Steel Works,Spec. Report 71, The Iron and Steel Inst., London, 1962)270

Fig. 6.1-4

541E I . c ' r ! C I ~ Y

2,aenzol Tar 28'44C.O.Ga.

H

ENERGY BALANCES

1-12

I Ca.,.nq Pi t ;1 8 ~ 7 ' b 7, i

1Ton-X. taille"All figures in ther.ns/ton billets AE=ancillary e"uipment

Sankey diagram for plant shown Fig. 6.1-3, with areference temperature of 298 K Joules (ref. Spec o Re- 8port 71, op cit.) 1 Therm - 100,000 BTU = 1.055 x 10joules.271

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