Handbook_on_Material_and_Energy_Balance_Calculat1

Chapter 7 Energy and the First Law of Thermodynamics 425

The Calculator tool can also be used when a heat content value is known, but not the
temperature. You may enter either the heat content or enthalpy of formation of a species, and the
Calculator tool will calculate the temperature.

Quite often the Ητ-Η29% equation for a substance requires five or even six terms to adequately
represent the accuracy of the table data. For example the Нт-Н29% equation for MgCl2 has five
terms. A multi-term HT-H29z equation can be somewhat cumbersome to use in heat balance
calculations, especially if several species are involved. But if the temperature span of interest is
shorter, and/or the required accuracy is less than for the FREED equations, a three-term Ят-Я298
equation may be adequate. For example if Cp is adequately represented by a linear temperature
equation, this integrates to a quadratic HT-H29s equation, which is much easier to use for unknown
Г problems that a five-term equation. FREED's Graphics tool can be used to determine if simpler
equations are appropriate, and if so, Excel's Trendline tool can be used to develop the simpler
equation.

Suppose we need thermal data for S02 over a temperature span of 700 - 1100 K. FREED's
Graphics tool can generate a Cp chart as shown in the left side of Figure 7.7*. If a linear
approximation to Cp vs. T is acceptable, then FREED's Graphics tool can be used to plot Ят-Я298
vs. T, and the data fitted to a quadratic equation with Trendline, as shown in the rightmost chart.
The value of R2 = 1.00000 indicates an excellent fit to the textbox equation, showing that the
assumption of linear Cp gives an acceptable quadratic heat content equation.

Ят-Я298 for S02(g) = 0.00129Г2 + 10.43IT7- 3497 (cal/mol)

It's sometimes more convenient to have the heat content equation in Celsius temperature.
FREED will generate kelvin or Celsius temperature charts, in calories or joules per mole.

Plot: Cp vs. T for S02 Plot: HT-H298 vs. T for S 0 2
10000
11.5
600 700 800 900 1000 1100 1200 H 8000 у = 0.00129X2 + 10.431X - 3497
T(K) R2 = 1.00000

сSм 7000 ' Щ*
X 6000
t 5000

4000 1150
650 750 850 950 1050
T(K)

Figure 7.7 Heat content from ExcePs Trendline tool on charts from FREED's Graphics tool.

A similar procedure can be used if a substance is cooled from or heated from a specific
temperature. Suppose S02 was involved in a process where heat was transferred to it at an initial
S02 temperature of 450 °C such that the final S02 temperature was between 1000 °C and 1400 °C.
For this application it is convenient to have an equation for HrH450. FREED's Table tool can be
used to generate a set of enthalpy data that can be used to compute such an equation. First, the
table data is reduced to the essential values as shown in the leftmost two columns (next page).
Next the difference between the 450 °C and higher temperatures is computed and converted to kJ
(rightmost data column). Finally, Excel's Chart and Trendline tool are used to obtain the desired
equation.

Obviously this procedure is applicable only when the temperature span and/or the desired
accuracy permit. And for substances where Cp changes very little with temperature, a linear
enthalpy change equation may be suitable.

* The format of the FREED charts have been altered to better present them for print.

426 Chapter 7 Energy and the First Law of Thermodynamics

t, °C J/mol kJ/mol _ 55 Enthalpy of S 0 2 above 450 °C
450 Eо 50 |Ht-H450 = 1.71E-06t2 + 0.05286t ■ 2 4 . 7 7 ^ > |
1000 Ht-H25 Ht-H450
1050 R2 = 1.000
1100 19743 0.0
1150 49547 29.8 1300 1400
1200 52362 32.6
1250 55188 35.4
1300 58025 38.3
1350 60870 41.1
1400 63724 44.0
66586 46.8
69455 49.7
72331 52.6

7.8 Effect of Temperature on Heat of Reaction

Figures 7.4 and 7.6 showed that AHform of a substance varies somewhat with temperature, as
does АЯех (Figure 7.5). For the general chemical reaction:

p? + qQ + ► rR + sS + - - - [7.23]

there will not only be a change in enthalpy (i.e., AHrx) but also a change in the heat capacity,
designated here ACP, as given by:

AC, = [ r ( Q R + j(C,)s · · · ] - [ р ( ф Р + q{Cp)Q ■■■]

where the Cp terms here are the molar heat capacities. The variation in heat of reaction with
temperature is given by Kirchhoff s equation, which states that the rate of variation of the heat of
reaction with temperature, at constant pressure, is equal to the change in heat capacity
accompanying the reaction.

Гд(АЯгх) ACr [7.24]
L ST \p

This equation is applicable to both chemical and physical processes, with an important
limitation for physical processes, as discussed in section 7.8.2.

7.8.1 Application of Kirchhoff s Equation to Chemical Reactions

Process calculations often require values of ДЯГХ as a function of temperature. If we know
AHYX at one temperature, we can use Kirchhoff s equation and Cp data to calculate it at any other
temperature. Between the limits T\ and T2:

AH2-AHl = ^2ACpdT [7.25]

Equation [7.15] shows that the integral of Cp is ΗΊ2-ΗΊ\. Therefore, we can derive a general

expression for the heat of a reaction as a function of temperature if we know it at one temperature.
If the temperature of known A#rx is T\, then:

A#rx at T2 - Δ#ΓΧ at Tx = Ση(Ητ2-ΗΊι)ρτοάηοίζ - Ση(Ητ2-ΗτiWtants [7.26]

It's common to select 298.15 as Tu in which case FREED's multi-term heat content equations
can be used directly. Over a limited temperature range, a three-term (quadratic) heat content
equation might give acceptable accuracy, as discussed in the text near Figure 7.7. For generic
chemical Equation [7.23] and a quadratic heat content equation, Equation [7.26] becomes:

Atfrx at T- AHrx at 298.15 = r[AR + BR(7) + CR(7*)] + s[As + Bs(7) + Cs(7*)] [7.27]
-p[Ap + Bp(7) + CP(7^)] - <7[AQ + BQ(7) + CQ(7^)]

Chapter 7 Energy and the First Law of Thermodynamics 427

where А, В and С are the equation parameters for the #Т-Я298 equation. The first term on the
right-hand side of Equation [7.27] refers to the reaction products, while the second term refers to

the reactants. A more general version of [7.27] is:

Atfrx at T- Δ#ΓΧ at 298.15 - n[AA + ΔΒ(7) + AC(7^)] [7.28]

where n refers to the appropriate number of moles of each reaction constituent, and Δ refers to the
product - reactant difference in heat content equation coefficient. Equation [7.28] is valid only in
a temperature range that does not contain a phase change of any reactant or product.

Kirchhoff s equation is shown conceptually by setting up a cyclic process. Since Я is a state
function, ΣΑΗ for any number of individual steps around a cyclic process = 0. Consider the

reaction for the chlorination of nickel oxide at 298.15 К and higher temperatures.

NiO(c) + 2HCl(g) -> NiCl2(c) + H20(g) [7.29]

There are two ways to carry out the reaction. First, carry it out at an elevated temperature T2.
Equivalently, cool the reactants from T2 to 298.15 K, carry the reaction out at 298.15 K, and heat
the products to T2. Figure 7.8 shows these equivalent steps.

Step #4: ЛЯ0^ at T2 Figure 7.8 Diagram of
Kirchhoff s equation for
NiO(c ) + 2HCl(g ) -> NiCl2(c ) + H20(g ) calculating the heat of reaction
at any temperature. The value
lift of AHrx at T2 (step #4) is equal
to the sum of the heat effects of
Step#l: Step#3: steps 1, 2 and3.

Cool reactants Heat products

to 298.15 К X X to72

NiO(c ) + 2HCl(g ) -> NiCl2(c ) + H20(g )

Step#2:Atf°rxat298.15K

FREED's Reaction tool can be used (for example at T2 = 700 K) to calculate the sum of steps

2 and 3 (-77 380 J), step 4, (-122 800 J) and step 1 (-45 420 J). The agreement is perfect. The
Reaction tool also calculates and charts the AHrx as a function of temperature.

EXAMPLE 7.4 — Standard Heat of the Water-Gas Shift Reaction from 800 to 1500 K.

A hydrogen production process includes a unit where the WGR (see Equation [6.59] and
Figure 6.32) controls the product gas composition. The heat balance requires an analytical
expression for the heat of the WGR as a function of temperature.

Data. Values of Ητ-Η29% for the two reactants (CO and H20) and the two products (C02 and H2)
were obtained from FREED, and converted to units of kJ/mol. Figure 7.9 shows a plot of these
values. Table 7.1 shows the quadratic equation heat content parameters for each gas, based on
Excel's Trendline tool fit. R2 >0.9999 for all cases.

Table 7.1 Standard heat of formation from the elements, and equation parameters for heat content
above 800 К as developed from Trendline fit of FREED data. #T-#298 = A + ВГ+ СГ2, kJ/mol.

AВ С MPform, 298.15 К

co2 -16.07 0.04480 4.710 x IO"6 -393.51

H20 -9.48 0.02969 5.795 x IO"6 -241.81

CO -9.22 0.02867 2.259 х IO"6 -110.53

H2 -7.67 0.02643 1.924 х IO"6 0

428 Chapter 7 Energy and the First Law of Thermodynamics

Figure 7.9 Heat content above 298.15 К for the four gas species involved in the WGR.

Solution. The first step is to calculate A#°rx at 298.15 К from values of ΔΗ°ΐονΏΙ of the species at
298.15 K.

CO(g) + H20(g) - C02(g) + H2(g) [7.30]

A#°rx at 298.15 К = -393.51 +241.81 + 110.53 =-41.17 kJ [7.31]

Next calculate ΔΑ, ΔΒ, and AC so we can use Equation [7.27]. The arithmetic is easy
because n = 1 throughout.

ΔΑ = -16.07 - 7.67 + 9.22 + 9.48 = -5.04

ΔΒ = 0.04480 + 0.02643 - 0.02867 - 0.02969 = 0.01287

AC = 4.710 x 10"6+1.924 x 10"6-5.795 x 10"6-2.259 x 10"б=1420х 10"6

Finally include the АЯ°ГХ at 298.15 К and collect terms according to Equation [7.28]

A#°rx WGR, kJ = -46.21 +0.01287Г- 1.420 x 10"6(7^) [7.32]

where T is any value between 800 and 1500 K. A//°rx for the WGR is exothermic for all
temperatures in this range, but less so as temperature increases.

Assignment. Use FREED's Reaction tool to determine АЯ°ГХ of the WGR between 800 and 1500
K, convert to kJ, and use the Trendline tool to develop a quadratic equation to fit the results.
Calculate the difference between the АЯ°ГХ obtained from Equation [7.23].

7.8.2 Heat of Transformation for Non-Standard and Non-Physical States

The usual thermodynamic tables express enthalpy changes for substances in their equilibrium
(i.e., standard) state. However, phase transformations often take place at pressures other than 1
atm, or materials may persist in a certain form at temperatures below the standard transition
temperature. For example, a liquid may evaporate above or below the normal boiling point, in
which case the heat of vaporization will be different than the standard heat of vaporization at the
normal boiling point and 1 atm pressure.

In section 7.71 we showed how to apply Kirchhoff s equation to calculate the heat of reaction
at any temperature from АЯ°ГХ at one temperature, plus the change in heat content of the products
minus that of the reactants, as summarized by Equations [7.27] and [7.28]. There is no trouble
with this application because we can maintain a constant pressure (at 1 atm or otherwise) for all
temperatures, as implied by the use of Cp.

Chapter 7 Energy and the First Law of Thermodynamics 429

Where a substance exists in two phases that change with temperature, then the application of
Kirchhoff s equation depends on whether the pressure is maintained constant during the phase
change (say at 1 atm), or whether it is the equilibrium value. In the case of vaporization, a rigorous
application of Kirchhoff s equation would take into account the difference in volume of the liquid
and vapor phases, which is a function of temperature. But since the volume of the condensed
phase is so much smaller than the volume of the vapor phase, and if we assume ideal behavior of
the gas phase, Equation [7.33] may therefore be integrated as before without consideration of the
change in volume taking place during vaporization. This only holds true if we are far from the
critical point.

[ dT = ACr [7.33]

\p

Suppose MgCl2(/) was being vaporized at a pressure greater than one atm, which requires the
temperature to be above the normal boiling point of 1643 K. The AHV of MgCl2(/) is a (weak)

function of temperature, being 178,209 J/mol at the normal boiling point (Figure 7.6). We can
calculate the AHV of MgCl2(/) at different temperatures as outlined in Example 7.4, or we can use

FREED's tables (with some extrapolation) for MgCl2(c,/,g) and MgCl2(g). We simply take the
difference in the heat of formation of MgCl2(/) and MgCl2(g) using equations for AH{ of

MgCl2(/,g) from Figure 7.6. These are the equations designated Del(Hf) in the figure. The
extrapolation of the AHf MgCl2(/) equation above the normal boiling point is acceptable so long as
we recognize the inherent danger of extrapolating too far. The two AH{ equations are:

A#f MgCl2(/) = 29.2775(7) + 0.000845(7^) - 109 828/7+245.164Г4- 1.0484 x 10"7(7^) - 772 230 [7.34]

Atff MgCl2(g) = -1.3703(7) + 0.000845(P) - 109 828/T+ 245.1647^ - 1.0484 x 10"7(7^) - 543 666 [7.35]

The difference between these two equations (Equation [7.35] - [7.34]) is the AHV of MgCl2(/j.
The result is a linear equation because the heat content equations are linear for the two phases.

AHV of MgCl2(/) (J/mol) - -30.648Γ+ 228 560 [7.36]

Alternatively, we could have used FREED's Reaction tool to develop a table of AHV of
MgCl2(/) for the reaction of MgCl2(/) to MgCl2(g).

FREED also has data for the non-physical standard state of supercooled elemental gases.
Figure 7.10 shows a portion of the FREED table for Zn(g) at 1 atm. Below the normal boiling
point of 1180 K, the data is for the non-physical standard state of zinc gas at one atm pressure.
The values in the dHf column are the enthalpies for the reaction:

Zn(c,/,g)^Zn(g)

This data is useful for calculating the vapor pressure of Zn(g) in equilibrium with the
condensed phases below 1180 K. Such an equation for liquid zinc can be extrapolated above or
below the normal; melting point to estimate the/?Zn for superheated or supercooled liquid zinc.

430 Chapter 7 Energy and the First Law of Thermodynamics

Zn, (g) GFW: 65.39 Density 0.00267
(Zinc) dH298°: 130416 S298°: 160.879

Comments: Zn: melts at 692.73 and boils at 1180 K.

Unit: (K, g-mol, J)

T (K) Cp HT-H298 S° dHf

298.15 20.786 0 160.879 130416

300.00 20.786 39 161.008 130408 Figure 7.10 Thermodynamic data
500.00 171.626 129296 table for Zn(g) at 1 atm. The
692.73 20.786 4197 178.403 127822 standard state for zinc vapor is non-
physical below 1180 K.
692.73 20.786 8203 178.403 120500
700.00 178.620 120423
900.00 (Melting Pt) 183.843 118304
1100.00 188.015 116185
1180.00 20.786 8203 189.474 115338

1180.00 20.786 8354 189.474
1300.00 191.487
20.786 12511

20.786 16668

20.786 18331

(Boiling Pt)

20.786 18331 0
0
20.786 20825

EXAMPLE 7.5 — Supercooling Liquid Tin.

Clean metals and alloys can be supercooled tens of degrees below their melting points, and
even lower for short time periods. Calculate the heat of fusion of supercooled liquid tin from the
normal melting point of 505.12 К down to 300 K.

Data. The FREED database gives the following equations for the sensible heat of tin:

Solid Tin: #т-Я298 = 20.786(7) +0.009799(P) - 319 78/Г- 6962 (J/mol)

Liquid Tin: #T-#298 = 28.098(7) +0.000085(P) - 183 242/Г- 848 (J/mol)

Solution. The difference between these two equations (liquid - solid) at the normal melting point
of 505.12 К is the AH°f of Sn (7029 J/mol). Using the modification of Kirchhoff s equation
discussed earlier in connection with the MgCl2 example, the difference between the two equations
was calculated at 25° intervals from the normal melting point to 300 K. Figure 7.11 shows the
results. An increase in AHf of Sn occurs for the first 80° of supercooling, but then the trend
reverses, which is thermodynamically impossible. This shows the danger in extrapolating
equations having higher-order terms outside the range of the data used to derive the equation.

AHfOfTin

300 350 400 450 500
T,K

Figure 7.11 Δ//0^ of supercooled tin calculated by extrapolation of higher-temperature data.

Chapter 7 Energy and the First Law of Thermodynamics 431

Assignment. Liquid tin is supercooled to 450 К in an insulated container. A nucleus of solid tin
forms, and the mass begins to solidify. The AH^ adds heat to the mass and its temperature rapidly

increases to 505.12 K. What fraction of the tin remains liquid?

7.9 The Properties of Steam and Compressed Air

Steam is a ubiquitous substance in modern industrial processes. More energy is consumed in
the generation of steam than in any other single process. In materials production and processing,
steam may be a reactant or a product of the process. Some pyrometallurgical processes produce
extra heat, which is used to make high-pressure steam in a boiler. Recall the discussion of the zinc
roaster example in Chapter 6, where steam tubes extracted heat from the bed of the roaster, and the
roaster offgas was cooled in a waste-heat boiler. Steam is treated in FREED as an ideal gas at 1
atm.

Compressed air is also in common use in materials processing. Compressed air can be an
effective way to store and transport energy, and has even been proposed as a propulsion source for
automobiles. Air is treated in FREED as an ideal mixture of N2 and 0 2 having the empirical
formula N1.5sO0.42, which represents one mole of air. A more accurate formulation for C02-free air
was given in Chapter 1 as:

φΝ2 = 0.7812; φ02 = 0.2096; <pAr = 0.0092

If so desired, an ideal-gas thermodynamic table for this more accurate composition of air
could be derived from the FREED tables for each species, and added to the FREED database.
However, if this degree of accuracy is desired, it is probably better to resort to an equation of state
derived from P-V-7-#data, as will be shown in section 7.9.2.

7.9.1 Properties of Steam

Several terms were introduced earlier to describe the H20 system. At temperatures below the
vapor-liquid equilibrium (VLE) curve, H20 is an undercooled liquid. On the VLE curve where
water and steam are in equilibrium, water is a saturated liquid and steam is a saturated vapor. At
temperatures above the VLE curve, H20 is a superheated vapor. We defined H20 as the species,
water as the liquid phase, and steam as the vapor phase.

Many of the properties of steam were discussed in Chapter 2 in terms of equations of state for
steam as an ideal and a non-ideal gas, and the phase diagram for steam was shown as Figure 2.3
from 250 К to 650 K. The thermodynamic properties of water and steam at 1 atm are part of the
FREED database, where steam is assumed to behave ideally. This data is reasonably accurate at
pressures up to two or three atm, but as was shown in Example 2.3, steam at higher pressures
becomes increasingly non-ideal. However, the van der Waals equation of state may give
acceptable accuracy over a range of pressures between 1 and 15 bar if the values of a and b are
statistically determined over the same range (please see Section 4.10.1). Owing to the importance
of steam, a special database has been prepared containing information on the volume, enthalpy,
internal energy, and entropy of steam and water from the ice point to temperatures well above the
critical temperature of 374 °C. This database is called a steam table, and is often included in
thermodynamics texts for chemical and mechanical engineers, and in various industrial handbooks.
Steam table information may be downloaded from the internet (NIST 2007), and various steam
table computer programs can be found by searching the internet. Two simple programs suitable
for student use (ChemicaLogic 2003, Archon 2001) are discussed in more detail in the Appendix.
The SI units used in a steam table involve Pa or bars as pressure units, and kJ/kg for specific
enthalpy. The enthalpy reference temperature is liquid water at 0 °C. Most computerized steam
tables allow other units.

The use of non-ideal steam properties in this text will be limited to temperatures between 100
- 250 °C and pressures between 100 - 2500 kPa (~1 - 25 atm). In many cases, the properties of
saturated steam and the enthalpy of vaporization are the only values required for an energy

432 Chapter 7 Energy and the First Law of Thermodynamics

balance. A regression analysis of steam table data indicates that the important properties of water
and steam in the above range can be expressed to an accuracy of three significant figures by two-
or three-term equations, where t designates °C, and P is in bar. The subscript "sat" refers to
conditions on the VLE curve. These equations are of satisfactory accuracy for most purposes of
this text. The values from these equations are not equivalent to corresponding values from the
FREED database (Equations [2.15] and [2.16]) because FREED treats steam as an ideal gas at 1
atm throughout the entire temperature range, while the steam table data is based on real steam.
The reference temperature is also different.

Log(pH2Osat) (bar) - 48.70/t + 4.748(log t) - 9.980* [7.37]
Volume of water (L/kg) - 3.81 x W\t2) + 3.61 x 1(Г5(0 + 1.0006 [7.38]

Log(volume of steamsat) (L/kg) - -24 290112 + 586.5// - 0.24 [7.39]

Enthalpy of water (above 0 °C) (kJ/kg) = 9.70 x 10'V) + 4.083(0 [7.40]

Enthalpy of vaporizationsat (kJ/kg) = -4.456 x 10~V) - 1.975(0 + 2502 [7.41]
Cp of superheated steam (kJ/(deg · kg) - 0.066(P, bar) - 0.00311 + 2.46 [7.42]

Cv of superheated steam (kJ/(deg · kg) = 0.035(P, bar) - 0.0017/ + 1.77 [7.43]

An isobaric process illustrates the application of the above equations. Suppose we require the
amount of heat absorbed by one kg of water initially at 0 °C when heated at a pressure of 10 bar to
produce superheated steam 50 °C above the VLE temperature. Equation [7.37] is used first (with
Excel's Goal Seek tool) to calculate a tSdX of 453 К (180 °C). The enthalpy increase of water when
heated to 180 °C is calculated from Equation [7.40] as 860 kJ. The enthalpy of vaporization from
Equation [7.41] is 2002 kJ. The isobaric heat capacity of superheated steam at 10 bar is given by
Equation [7.42] which when integrated over a Δ/ of 50° gives an enthalpy change of 124 kJ
required to superheat the steam. The total enthalpy change is the sum of these three values, or
2986 kJ for the stated mass of one kg. Given the accuracy of the equations, the value is best
expressed as 2990 kJ/kg.

7.9.2 Properties of Compressed Air

The thermodynamic properties of C02-free air have been tabulated by NIST (Lemmon 2000),
and are included as the file AirPVT.xls in the Air folder on the Handbook CD. For most material
processing applications, the pressure seldom exceeds 50 bar and the temperature seldom exceeds
2000 K. In this range, the van der Waals equation of state is applicable and used in this text for
making P-V-T calculations for air. Section 2.3 introduced the vdW equation, and presented one
version of it as Equation [2.10]. The vdW equation has two parameters, a and b9 that can be
calculated by a statistical evaluation of the P-V-T data in a specified range. However, in order to
derive the best values for a and b, the vdW equation should be recast in terms of the
compressibility z, as described by Equation [2.9], and the density of air:

z= _i _^P_ [7.44]
l-bp RT

where p is the density in mol/L, and R is the gas constant 0.0831447 L · bar/(mol · K). The units
for a are L2 · bar/mol2 and the units for b are L/mol.

Figure 7.12 shows the compressibility of air in the above range as calculated from the NIST
data. Clearly air does not deviate markedly from ideal behavior except at higher pressures. An
Excel statistical procedure (discussed in Section 3.4.3) was used to determine values of a and b
that best represent the P-V-T properties of air according to Equation [7.44]. Values of a and b
were sought that minimized the sum of the squares of the differences (SSD) in the actual and
calculated values of z. The details are outlined in worksheet AirPVT.xls on the Handbook CD

For atm. subtract 0.006

Chapter 7 Energy and the First Law of Thermodynamics 433

along with the NIST air data. Equation [7.45] shows the vdW equation of state for dry C02-free
air, valid between 250 to 2000 К and up to 50 bar.

1 1.28028p [745]
Z " 1-0.042542p 0.083145Г

A statistical evaluation of the isobaric enthalpy (J/mol) of real air (P in bar) above 298.15 К
was made using Excel's Regression tool. Equation [7.46] gives the expression.

ΗΊ "я298.15 =46.972Γ-0.000174Γ2 - 8 3 5 ' 6 1 ^ / -911.9л/г + 1 0 . 9 P - 3 2 0 9 % + 4558 Г7·46]

Equation [7.46] is an isobaric equation, so it should only be used to calculate the high
temperature heat content as a function of temperature for a given pressure. The standard state
(where Ητ-Η29$ = 0) is therefore a function of the stated pressure. If we define the standard state to
always be 298.15 К and 1 bar, then a slightly different equation results:

Ητ -#298.15 =46.9727 -0.000175Г2 - 8 3 5 > 6 1 % -911.9л/г+ 4 . 8 P - 3 2 0 9 % + 4560 Г7·47]

Figure 7.12 Compressibility of dry C02-free air between 1 and 50 bar, and 250 - 2000 K.

7.9.3 Temperature Change for Free Expansion of a Gas

In Section 7.4, we stated that the enthalpy of an ideal gas is a function of temperature only,
not of pressure (or volume). When any gas expands adiabatically and reversibly it does work on
the surroundings, its temperature drops, and there is a change in enthalpy. It also does work when
it expands isothermally and reversibly; heat must be supplied to keep the temperature constant and
there is no enthalpy change since there is no temperature change,.

In some operations, a pressurized gas is allowed to expand freely (so-called Joule expansion)
without doing any work. In the particular case of the free expansion of an ideal gas, there is no
change in temperature or enthalpy. For real gases, free expansion usually results in a change of
temperature, expressed as the Joule-Thompson coefficient μίΤ.. At moderate pressures, μίΤ. for
most gases is positive, meaning that real gases undergo cooling on free expansion. This
phenomena is utilized in the liquefaction of air and other gases.

Compressed air and superheated steam are sometimes used to inject substances into reactors,
or to atomize fuel oil at the entry point of a burner. Steam expands freely, and owing to the change

434 Chapter 7 Energy and the First Law of Thermodynamics

in temperature upon expansion, the heat balance for the process may be affected. Here we
consider the change in temperature for the adiabatic expansion of compressed air and superheated
steam. The units are one mole of gas, temperature in K, pressure in bar, and energy in J.

For compressed air, we make use of Equation [7.47], which calculates the enthalpy of air at
any temperature and pressure relative to STP of 298.15 К and 1 bar. Suppose the initial air
temperature was 310 К and the initial pressure was 10 bar. An adiabatic change in pressure to 1
atm (1.013 bar) requires a change in temperature to supply the necessary enthalpy. At 310 К and
1.013 bar, the enthalpy of one mole of air is 348 J, and at 310 К and 10 bar, the enthalpy is 298 J.
Free expansion requires 50 J, which is supplied by cooling the air. Cp for air is about 29.2 J/mol-
deg between 295 - 340 K, so the air temperature drops about 1.7 degree. Figure 7.13 shows the
results of similar calculations for a range of initial air temperatures and pressures.

о ΔΤ for Free Expansion of Compressed Air

2 —0—300 К
- a - 320к
+Ш3-* -6 - - Δ - - 340 К
ФEaL- -8
АТзоо = -0.202P + 0.20
Ф ΔΤ320 = -0.179Ρ + 0.18
дТ34о = -0.159Р + 0.16
-10
10 20 30 40
initial pressure, bar 50

Figure 7.13 Temperature change resulting from free expansion of air to 1 atm pressure as a
function of initial air temperature and pressure. Calculations based on Equation [7.47].

The situation for steam is more complex than for air because steam is farther from ideal. The
equations given earlier for the properties of saturated steam are not of sufficient accuracy for
calculating the enthalpy of saturated or supersaturated steam above the standard state of water at
0.01° C, so we must use one of the on-line steam software programs described in Sections 2.5 or
2.7. The table below shows the enthalpy of saturated steam (kJ/kg) from 1.013 to 50 bar, the
enthalpy of supersaturated steam at 1.013 bar, and the saturation temperature. An interesting
characteristic of saturated steam is that its maximum enthalpy is at about 33 bar.

Psat, bar 1.013 3 6 10 17 23 30 40
100 133.5 158.8 179.9 204.3 219.6 233.9 250.4
77sat, °C 2676 2725 2756 2777 2794 2801 2803 2801
2676 2744 2794 2836 2884 2914 2942 2975
#sat-steam, kJ/kg
Hsteam, at 1.013 bar & Tsat

To get an idea of the temperature reached by free adiabatic expansion of saturated steam to
steam at 1.013 bar, consider the data at Psat of 17 bar, where //sat = 2794. Inspection of the data
for steam at 1.013 bar shows an enthalpy of 2794 at about 159° С Therefore, when saturated
steam at 204 °C undergoes free expansion to 1.013 bar, it cools to 159 °C, a drop of 45°. More
detailed calculations were made using the SteamTab program for the expansion of saturated and
supersaturated (/sat + 25°) steam. The results are displayed in Figure 7.14. A notable result is
that the free expansion of saturated steam initially between 210° and 250° С results in about the
same final temperature: 162° С This is because the enthalpy of saturated steam is almost constant
between 210° and 250° С

Chapter 7 Energy and the First Law of Thermodynamics 435

T Resulting from Free Expansion of Steam

300 —0—Tsat _^--Δ-^
—O—Tsatad
ίo° 250 - - Δ - Tsat+25° \^tsr

% 200 -X--Tsat+25°,ad

Ж

10 20 30 40 50
initial pressure, bar

Figure 7.14 Temperature of steam after free adiabatic expansion of saturated and supersaturated
steam to 1 atm (1.013 bar). Line labeled Tsat refers to the temperature and pressure of the (initial)
saturated steam, while the line labeled Tsat+25° refers to the temperature of the (initially)
supersaturated steam (25° above Tsat). The line labeled Tsat,ad refers to the final steam
temperature after free adiabatic expansion from saturation. The line labeled Tsat+25°,ad refers to
the temperature of freely expanded supersaturated steam.

7.9.4. Cooling by Steam Venting

Some hydrometallurgical processes employ autoclaves so they can use pressurized aqueous
solutions for leaching. Most sulfide leaching processes are carried out at temperatures below 260°
C, although some processes take place near or even above the critical temperature of water (374°
C). Once the desired extent of reaction has occurred, the autoclave is vented to release steam.
Water evaporates rapidly until the /?H20 approaches 1 atm, the temperature drops, and the
autoclave contents are discharged into a separation unit. Proper design of a pressure leaching
process requires knowing how much water will evaporate during steam venting. This requires data
on the heat capacity of water and its heat of vaporization. Either of the on-line steam programs
will provide this information. The data are plotted in Figure 7.15 from 100° С to 250° C, with
linear approximations to the data obtained by using the Trendline tool.

Properties of Superheated Water

2300 3>^^AHvap 4.9

2200 Η of vap = -3.63t + 2650 DJ 4.8 3)
ö) 2100
D" r -* 4.7 Ф
4.6 "?
2 2000 O)
XXD
hi ^-*τro 1900 4.5 i

> ет-а- CJKHO'I ^^ . 4.4 ■*

< 1800 π Cp = 0.0040t + 3.74 ^ ^ ^ V ^ w -11 4.3 d
' 4.2
1700 О

——ι 1 ' —г————ι 1

1600100 120 140 160 180 200 220 240 4.1

temperature, °C

Figure 7.15 Heat capacity and heat of vaporization of superheated water. Text box equations
represent linear approximations to data obtained from a steam table calculation program.

436 Chapter 7 Energy and the First Law of Thermodynamics

One way to structure the calculations is to write a differential equation expressing the change
in heat content vs. mass of water evaporating, and substitute the appropriate text box equations to
express heat content as a function of temperature. But owing to the variation in ΔΗν£φ and Cp with
temperature, the variables cannot be easily separated, so this method is impractical. Another
method is to divide the process into small steps, and make a heat balance for each step. The results
of that step are used as a starting point for the next step, and so on. We designate this method as
iterative gas-phase removal.

Suppose we start with 100 kg of water in an autoclave initially at 240° С We assume that the
mass of steam in the autoclave is small compared to the mass of the water, and adiabatic
conditions. Our first step vents 1 kg of steam at 240° C, which requires 1779 kJ. The heat balance
on the autoclave contents is:

-1779 = 4.7(100)(Tf- 240); Tf = 236.2° С

We now have 99 kg of water at 236.2° C, and can repeat the above calculation (but with
slightly different values of A//vap and Cp). This was done by writing the formulae into a worksheet,
and filling across the columns until the final temperature went below 100° С At that point, the
vapor pressure of water has reached the ambient pressure (assumed to be 1 atm), and steam venting
ceases. The results are shown in Figure 7.16. About 25% of the water evaporates in bringing the
temperature to 100 °C.

There may be other factors to consider for an actual pressure leach process. First, the
autoclave may contain insoluble material, whose Cp must be factored into the heat balance.
Second, the walls of the autoclave act as a heat reservoir to the process. Third, the ЛЯуар from a
solution will be larger than from pure water. And fourth, the pH20 will be less for an aqueous
solution. If these factors can be evaluated, they should be included in the calculation.

Evaporative Cooling of Superheated Water

100

95 I ^ O - л
xσ.> 90
Φъ.
85
ГО
«о(£Я^
80
(0 75
Eга 70 ^^^^^^—
-^^^^^^—1

240 220 200 180 160 140 120 100

water temperature, °C

Figure 7.16 Decrease of superheated water temperature by evaporation of steam. Results were
obtained by iterative gas-phase removal in 1 kg increments. The temperature decrease for each
increment of 1 kg water vaporized increases as the temperature drops. This is for two reasons.
First, because the ЛЯуар increases with decreasing temperature, and second, the mass of remaining
water is less.

7.9.5 Enthalpy of Psychrometry

Earlier chapters contained several sections dealing with the vapor pressure of water, the
humidification of air, and the condensation of water when air is cooled below the dew point. Some
calculations were made by assuming ideal gas behavior and using Equations [2.15] - [2.16] to
calculate /?H20. Alternatively, the PsyCalc 98 program (Linric 1998) was used to make a material

Chapter 7 Energy and the First Law of Thermodynamics 431

balance on a clay dryer (see Example 2.7). PsyCalc also calculates the enthalpy change for
changing the moisture content of air, or changing the temperature of moist air. Like the steam
tables, the basis temperature for enthalpy calculations is STP and the reference state for H20 is
liquid water at 0°C. The enthalpy values are based on one kg of dry air plus whatever water vapor
the air contains.

PsyCalc shows that at the reference state, the/?H20 is 0.611 kPa (0.00603 atm). One kg of
dry air contains 3.79 g H20, and φΗ20 = 0.00603 (listed as 6033 PPMv). The enthalpy is listed as
9.4 kJ/kg dry air, which indicates that the heat of vaporization of 3.79 g H20 at 0 °C requires 9.4 kJ
of energy (remember that dry air has zero enthalpy at the basis temperature). The enthalpy of the
same moist gas at 30 °C is 39.8 kJ, which is a combination of the heat content of one kg of dry air,
plus the heat content of 3.79 g of H20(g). The enthalpy change when 1003.8 g of moist air is
raised from 0 °C to 30 °C is the difference between these two enthalpy values, or 30.4 kJ. With a
couple of entries, PsyCalc can also give data to calculate the enthalpy requirement for heating each
constituent of the gas (dry air and steam). When using PsyCalc, remember that the mass term kg
refers to the mass of dry air.

7.10 The Use of FREED in Making Heat Balances

The previous sections introduced the essential principles involved in applying the first law to
an energy balance. In the remaining Chapters, these principles will be applied to a variety of
materials processes. Usually, the enthalpy data will come from FREED, so before leaving this
Chapter, we illustrate how to use the FREED database to make simple heat balance calculations.
Four numerical examples are presented. First, a calculation of the isobaric heat of combustion of
CO using preheated air. Next, the temperature change when steam is compressed adiabatically.
Third, the enthalpy change when carbon reduces nickel oxide isothermally and isochorically.
Finally, we use FREED's Reaction tool to calculate the product temperature for an adiabatic
reaction between CO and 02.

Before proceeding to the examples, it's worthwhile to note that most heat balances use 298.15
К as the basis temperature. Reactants are cooled to (or heated to) 298.15 K, all chemical reactions
are carried out at 298.15 K, and the products are heated (or cooled) to their final temperature. This
is the method FlowBal uses. There are exceptions to this methodology, but they are few.

EXAMPLE 7.6 — Combustion of CO with Preheated Air.

Calculate the enthalpy change for the combustion of one mole of CO with ten moles of air at
P=\ atm. The CO enters the burner at 298 К and the air enters at temperatures up to 700 K.

Data. FREED data for CO, C02 and air (assumed to be 21 %02, rest N2) are shown below.

Nl.580.42,.(g) GFW: 28.8504 CO,(g) GFW: 28.0104
(Air) dH298°: 0
(Carbon Monoxide) dH298°: -110,529

Unit: i(K, g-mole,J} = C02,(g) GFW: 44.0098
(Carbon Dioxide) dH298°: -393,509
T (K) Cp HT-H298

298.15 29.177 0

300.00 29.165 55
400.00 29.242 2966
500.00 29.901 5922
600.00 30.649 8949
700.00 31.366 12051

Solution. The first step in this example is a material balance. Ten moles of air bring in 7.9 moles

of N2 and 2.1 moles of 02. The material balance requires a stoichiometric chemical equation for
the combustion of one mole of CO:

438 Chapter 7 Energy and the First Law of Thermodynamics

CO(g) + y202(g) - C02(g)

This reaction has a large value of ATeq, which can be verified from FREED. This means that
proceeds spontaneously until the limiting reactant is completely consumed (i.e., it is chemically
irreversible). The reaction for the combustion of one mole of CO requires Vi mole of 02, which
leaves 1.6 moles of 0 2 unreacted. Since air is in excess of the stoichiometrically-required amount,
CO is the limiting reactant, and is completely consumed. The products of combustion are thus
C02, N2, and 02. The product contains 7.9 moles of N2, 1.6 moles of 02, and 1 mole of C02.

The next step is to pick a basis temperature, here 298 K. The first enthalpy change is for
cooling air to 298 K. The second enthalpy change is for the chemical reaction at 298 K. The
enthalpy change for cooling ten moles of air from an elevated temperature to 298 К is the same
value, but opposite in sign to the values in the data table. For example, the value of Ητ-Η29% for
one mole of air at 500 К is 5922 J, so cooling ten moles of air from 500 to 298 К results in an
enthalpy change of-59 220 J.

The A#°rx is obtained by subtracting the AH°{ of the product (C02) from the A#°f of the
reactant (CO). (Since 0 2 is an element, its AH°f = 0 by definition). In addition, air has a A#°f of
zero, so the reaction enthalpy change involves only CO and C02. It's good practice in heat
balance exercises to write the AH°f values next to each chemical reaction:

CO(g) + i/202(g)-C02(g)

АЯ°ГХ - Atf°f C02 - AH°{ CO = -393 509 + 110 529 = -282 980 J

The negative number for ΑΗ°τχ means CO oxidation is an exothermic reaction. The net
overall enthalpy change for the process is the sum of the enthalpy changes for the two steps:
cooling the air to 298 K, and carrying out the reaction at 298 K.

An Excel worksheet was set up to make these calculations for the combustion of one mole of
CO entering at 298 К with ten moles of air entering at different temperatures. The results are
shown in Figure 7.17. Excel's Trendline Tool was used to develop an equation representing the
results, as shown in the text box.

Enthalpy Change for Process Figure 7.17 Overall heat effect for the
cooling of ten moles of air to 298 К and
x« -350 the combustion of one mole of CO to C02
-400 at 298 K. Preheating the air to 700 К
increases the available enthalpy from
combustion by 40 %.

400 500 600

air temperature, К

Assignment. Calculate the required air preheat temperature to produce a 10 % increase in the
overall enthalpy change for the process.

EXAMPLE 7.7 — Adiabatic Compression of Steam.

100 L of steam, initially at 1 atm and 400 K, undergoes reversible adiabatic compression until
the temperature reaches 650 K. Calculate the volume and pressure as a function of temperature.

Data. The example requires knowledge of Cp for steam as a function of temperature because
adiabatic compression means that the gas temperature will increase. FREED's Graphics tool was

* N2 is also not involved because it takes no part of the chemical reaction, and in any event, its A#°f is zero
by definition.

Chapter 7 Energy and the First Law of Thermodynamics 439

used to obtain a plot of Cp vs. T9 as shown in Figure 7.18. A Cp vs. T equation was developed for
steam as an ideal gas using Excel's Trendline tool. Owing to the deviation of steam from ideal
behavior, a linear equation for Cp was deemed adequate.

Equation [7.16] is applicable in this case, but we need Cv instead of Cp. For an ideal gas, Cv is
obtained by subtracting the gas constant R (here, 1.99 cai deg-1 тоГ1) from the Cp equation for
steam: Cv = 0.0024(7) + 5.24. Also, Equation [7.16] was derived assuming that the heat capacity
was constant. Making a correction for the variation in Cv with temperature, the following equation
is obtained:

5.241η(Γ2/400) + 0.0024(72 - 400) = -Rln(T2/100)

Terms were collected, and the equation was written as a formula in an Excel worksheet. The
volume was calculated at 50° temperature intervals, and the pressure calculated from the ideal gas
law. The results were plotted in Figure 7.19.

9.20 Cpvs. TforH20 Figure 7.18 Diagram of Cp vs.
9.00 H T for steam in units of cai deg4
8.80 i it тоГ1. A linear equation
8.60
8.40 Cp = 0.0024T + 7.23 ^ i developed from Excel's
o 8.20 R2 = 0.984
8.00 Trendline tool is shown in the
7.80 ^0^^^ text box. The trendline
equation is shown as a dashed
300 400 T(K) 500 line.

600 70d

120 Adiabatic Compression of Steam 10 Figure 7.19 Pressure and
100 volume relationships for the
—Volume 8Φ I ε adiabatic compression of 100
jB 80 liters of steam initially at 400 К
- - - - Pressure at 1 atm. As the pressure
increases from 1 to 7.89 atm,
^^^ the temperature rises to 650 К
and the volume drops to 20.6
о 60 ^^ *Φ 43 liters.
3 40 Φ
о * * " ^"
> 20 \ m " m 2|

1

400 450 500 550 600 650
Τ,Κ

Assignment. Calculate and plot the work done on the system during the adiabatic compression.
Calculate the amount of heat that must be removed from the system if cooled isochorically to the
dew point.

EXAMPLE 7.8 — Enthalpy Change During Reduction of NiO with C.

One mole of NiO and one mole of С are mixed, pressed into a pellet, and heated to 400 K.
The warm pellet is then dropped into a furnace at 1000 К in the presence of a very small amount of
inert gas. The pellet heats rapidly to 1000 К and NiO is reduced. The gas flowing from the
reaction zone contained <pCO = 0.71, balance C02. Calculate the amount of heat required to bring
the reactants to 1000 К and reduce all of the NiO.

Data. Relevant portions of the FREED data tables are shown below. There are two standard
reactions for the carbothermic reduction of NiO (next page). The AH°r: at 1000 К is shown for
each reaction on a basis of one mol of gas produced.

440 Chapter 7 Energy and the First Law of Thermodynamics

C(c) + NiO(c) -»■ CO(g) + Ni(c); ΑΗ°α = -111 966 + 234 931 = 122 965 J [7.48]
C(c) + 2NiO(c) -> C02(g) + 2Ni(c); Δ#°ΓΧ = -394 577 + 469 862 = 75 285 J [7.49]

C, (graph) C02,(g)
(Carbon) (Carbon Dioxide)

Unit: (K, g-mol, J) Unit: (K, g-mol, J)
T (К) НТ-Н298
dHf T (K) dHf
0
298.15 0 298.15 -393,508
0
400.00 1048 0 1000.00 -394,577

1000.00 11,781

NiO,(c,l) CO,(g)
(Carbon Monoxide)
(Nickel Monoxide)

Unit: (K, g-mol, J) Unit: (K, g-mol, J)

T (К) НТ-Н298 dHf T (K) dHf

298.15 0 -239,742 298.15 -110,530

400.00 4897 -239,132 1000.00 -111,966

1000.00 37,916 -234,931

Solution. A material balance on the reduction process indicates that the overall reaction produces
0.225 moles of C02 and 0.550 moles of CO. Carbon is present in excess, so NiO is the limiting
reactant, and 0.225 moles of С remain unreacted. The process material and heat balance is set up
as a ledger for input and output amounts. The heat effect for each part of the process is readily
calculated. As a contrast, the heat balance basis temperature here is 1000 К instead of the usual
298.15 K.

Species Moles in ЯЮ00-Я400 Moles out АЯ°гх, J
0
С 1.000 10,733 0.225
NiO 1.000 33,019 0 234,931
-61,581
CO 0 — 0.550 -88,780
0 0.225
co2 —

Sum of siensible heat, J: 43,752
SumofA#°rx, J: 84,570
128,322
Overall Enthalpy Change, J:

The values of i/iooo-#4oo are based on the amounts of each species put in, and the АЯ°ГХ for the
amount of each species involved in the overall reaction. The process is clearly endothermic, with
an overall enthalpy change of+128 322 J. Alternatively, the same result for A//°rx can be obtained
by multiplying Equation [7.48] by 0.550 and Equation [7.49] by 0.225.

Assignment. Show that the overall enthalpy change is the same if the heat balance basis
temperature is changed to 298.15 K. 1) cool the reactants to 298 K, 2) carry out the chemical
reaction at 298 K, and 3) heat the products to 1000 K.

EXAMPLE 7.9 — Temperature Change of an Adiabatic Reaction.

One mole of CO and ten moles of air, initially at 298 K, are ignited and react in an insulated
isobaric chamber. Calculate the product temperature using FREED's Reaction tool.

Data. Example 7.6 described the overall reaction stoichiometry for the oxidation of CO. Since
oxygen is present in excess, the reaction is spontaneous and complete. One mole of C02 is
formed, plus 1.6 moles of 0 2 and 7.9 moles of N2.

Chapter 7 Energy and the First Law of Thermodynamics 441

Solution. The initial worksheet array created by the Reaction tool is shown below.

Reaction: CO 02 N2 ==> C02 N2 02
Descriptor: (g) (g) (g) (g) (g) (g)
Amount: 1 2.1 7.9 1 7.9 1.6
Unit: m m m m m m

The Reaction tool can create a table of reaction properties, a chart of values vs. temperature,
and make a specific calculation. Here we seek the adiabatic reaction temperature, where the
overall heat (defined as the overall enthalpy change) is zero. In other words, the reaction heat plus
the sensible heat of the products = 0. We open the Calculate window, enter a value of 0 in the
upper left window, select Celsius for temperature, click on Overall Heat, and FREED iterates to
find a result, as shown. Notice that the energy units (J or cai) have no effect on the result.

Calculator: Unit: Cai LogKr Overall H
\ Input (heat) Iteration 8.671 0.0
T(C) dGr
0.0 0.0 -44321
Solutions T(C) 843.9
dHr
1 843.9 -67437

The final product temperature is 844 °C. You can check this value on the chart.

Assignment. Use the Reaction tool to calculate the adiabatic reaction temperature for air amounts
between 6 and 12 moles. Fit the results to a linear equation. Use the Reaction tool to calculate the
adiabatic reaction temperature for 10 moles of air as a function of air input temperature.

7.11 Heat of Solution

When two substances are mixed and dissolve in each other, the solution temperature can be
notably different from (often hotter than) the initial temperature of the substances. This enthalpy
change is called the heat of solution, designated AHsoin, or in some texts, as the heat of mixing,
designated AHmix or Δ//1^. An equivalent term is the integral heat of mixing, by which is meant the
sum of the heat effects of dissolution of the individual components. The use of the "integral" term
is to distinguish between the overall heat effect of forming a solution from the partial heat
contributed by each substance. The chemical equation for the formation of one mole of solution
from pure components is:

nA(c,l) + (l-w)B(c,/) -> AxB(i-x)(c,0 [7.50]

In Section 2.7.1, Raoult's Law was used to describe the properties of ideal solutions, for
which the activity coefficient γ is one. Very few condensed solutions obey Raoult's Law, but
some come close enough to use it as a useful approximation. A consequence of Raoult's Law is
that the ideal AHS0\n = 0.

7.11.1 Formation of Non-ideal Metallic Solutions

In Section 2.7.2, we discussed certain properties of non-ideal solutions, particularly metallic
solutions, and pointed out that the regular solution concept may be applicable to systems where the
properties of the end members are not too different. The regular solution uses a single parameter a
(independent of temperature) to describe the variation in activity coefficient with composition.
The same parameter is also used in defining the AHsoin for a regular solution:

A#soin = axAxB [7.51]

If Δί/soin is exothermic (i.e., a is negative), the bond energy between solute A and В is greater
than the average bond energies of the pure A and В components. This is noted by a tendency for
solute atoms to "cluster" and form ordered phases or even intermetallic compounds. If Mfso\n is
endothermic, the solution tends toward immiscibility.

442 Chapter 7 Energy and the First Law of Thermodynamics

Equation [7.51] suggests a parabolic shape of Δ#80ιη vs. mole fraction. However, even if the
A//soin vs. mole fraction curve is essentially parabolic, that does not mean the solution is regular.
Also, the value of a derived from activity coefficient data may be different than the value derived
from A//soin data. Consider the Cd-Mg system, which was the subject of Example 2.11. Figure
7.20 shows the molar A//soin for liquid Cd-Mg alloys, which is very close to a true parabola. The
average value of a is -5300 cal/mol, but the corresponding value of a derived from activity
coefficient measurements at 923 К is -4100 cal/mol. Notwithstanding this difference, it is
sometimes necessary to express the variation of A//soin in the regular solution format, with a
derived from the enthalpy (calorimetrie) values.

0 ii AHsoin for Cd-Mg Alloys at 923 К г. xMg A//soln a
0—
i 0
0.1 -460 -5111
-200 0.2 -841 -5256
-1000 -5333
о -400 AHsoln = -5300(xCd)(xMg) 0.25 -1123 -5348
I -600 —r 0.3 -1293 -5388
w \ /P 0.4 -1341 -5364
-800 0.5 -1271 -5296
Ü л-^ 0.6 -1102 -5248
0.7 -980 -5227
< -1000 -842 -5263
0.75 -494 -5489
-1200 0.8
0.9 0—
-1400
1
0.2 0.4 0.6 0.8
xMg

Figure 7.20 Molar heat of solution to form liquid Cd-Mg alloys at 923 K.

It's important to note the state of aggregation of the substances before and after dissolution.
Two solid phases may dissolve to form a liquid solution, and two liquid phases may form a solid
solution. The overall heat effect for the formation of a solution must encompass any phase
transitions during dissolution. For example, suppose we wished to calculate the overall heat effect
when one mole of Cd(/) at 700 К was added to one mole of Mg(c) at 700 К to form a liquid
solution at 700 K. The reaction is:

1.5Cd(/) + Mg(c) ^2.5Cd0.6Mgo.4(/) [7.52]

The first heat effect term is the fusion of one mole of Mg(c) at 700 K, which is listed in
FREED's table for Mg(/) as 2140 cai. The second heat effect term is the A//soin to form 2.5 moles
of liquid solution, which is 2.5(-1293) = -3232 cai, which for a regular solution is independent of
temperature. The sum of these two steps gives the overall heat effect, -1092 cai.

Section 2.7.2 stated that deviations from regular behavior are common, but it may still be
acceptable to use Equation [7.51] over a limited composition range. If the solution is strongly non-
regular, an empirical equation may be satisfactory. Consider for example the AHsoin of carbon

(graphite) in liquid iron. The data in Figure 7.21 shows that a simple quadratic equation for A//soin
of carbon is perfectly adequate.

7.11.2 Polymeric Solutions

Materials processes often involve a slag, matte, or molten salt. The nature and solution
properties of these melts can be rather complex, and so falls outside the scope of this Handbook.
The physical chemistry of high-temperature solutions has been described in detail (Richardson
1974, Sano 1997, Turkdogan 1980). There is no freely-available database of AHsoin for polymeric
substances. To a first approximation, the Cp of solution data can be made by assuming that the Cp
of a solution is the weighted average of Cp of the constituent species.

Chapter 7 Energy and the First Law of Thermodynamics 443

2000 Δ Η δ θ Ι η for Fe-C Alloys at 1873 К xC M/soln a
0
_о 1600 О H of soln Poly. (H of soln) 0.02 0 5714
E 0.04 112 5990
ΐо 1200 AHsoln = 10770(xC2) + 5200(xC) 0.06 230 6294
I 800 R2 = 1 0.08 355 6617
а<: 0.1 487 6978
0.12 628 7367
400 0.14 778 7774
0.16 936 8214
0.04 0.08 0.12 0.16 0.2 0.18 1104 8692
xC 0.2 1283 9206
0.211 1473 9503
1582

Figure 7.21 Molar heat of solution of carbon (graphite) and liquid iron at 1873 К up to the limit
of carbon solubility.

If this sort of approximation is not appropriate, solution compilations are available from
commercial programs such as those cited in General References. Data from one of the commercial
programs can be configured for use in Handbook-type heat balances by correlating the enthalpy
data with the composition of a solution of interest, and expressing the result as an "overall" Cp.
It's handy to have a single Cp value that includes the overall enthalpy change for heating the
reactants and forming the solution, even though such a practice lacks rigor.

For example, suppose we needed to make a heat balance on a copper smelting flash furnace.
We would need enthalpy data for the slag and matte phases, both of which are non-ideal solutions.
FactSage's Equilib has the necessary information. Calculations were made on a series of mattes
having wCu2S between 65 % and 80 %. The reactants were pure Cu2S(c) and FeS(c) at 300 K, and
the product was a matte at 1570 K. Equilib calculated the overall heat for bringing 100 g of
mixture from 300 to 1570 К plus dissolution to form a matte. The Cp results are shown in Figure
7.22. The overall AH was fitted to a linear equation, as was the Cp of the matte. The heat capacity
value is the more useful of the heat terms since it can be used for a ΔΤ other than near 1570 K.

Overall AH, J/g - -5.083(%Cu) + 1220 [7.531

Average Heat Capacity of Matte 85
0.76

0.75 —Q — wCu2S ^ Φ
63
§? 0.74 , JO ΰ
75 *
Ψ 0.73 cT * * ^ or * 70 ю
ОS. 0.72
^> 3СЧ
0.71 Ü
0.70 I \Cp of matte = -0.0040(%Cu in matte) + 0.96 65 £
' 60
51 I I I II

53 55 57 59 61
wCu in matte, %

Figure 7.22 Average value of Cp of copper matte as a function of matte grade over the range 295
- 1600 K. The value can be used to calculate the amount of heat required to bring the constituent
solids (Cu2S & FeS) from any typical ambient temperatures to any typical copper smelting
temperatures and to form a solution (a matte). The unit for Cp is J/(g · deg).

444 Chapter 7 Energy and the First Law of Thermodynamics

7.11.3 Aqueous Solutions

When a strong electrolyte dissolves in water, energy is required to move the ions far apart and
work is done against mutual attractions. On the other hand, energy is produced by ion hydration.
The net result can be an exothermic or endothermic event. For example, we are all familiar with
the evolution of heat when a strong electrolyte such as KOH(c) is dissolved in water. In metallic
or polymeric systems, we found it useful to use the enthalpy change when two substances form one
mole of solution. For aqueous solutions, it is often more useful to consider the enthalpy change
when one mole of salt dissolves in several moles of water (or when one mole of a gas like S03 or
HCl dissolves to form an acid solution). The enthalpy change depends on the final composition of
the solution, so it is necessary to define a standard state for the solute. Section 6.10 showed that
the standard state concentration for aqueous solutions was a one molai solution, or one mole of
solute per 1 kg (55.51 moles) of water. The SI symbol for molality is b. However, since aqueous
solutions tend to deviate strongly from ideality, thermodynamic properties are based on a reference
state of infinite dilution.

Another convention used in electrolyte thermodynamics is to reference all thermodynamic
properties against those of the hydronium ion, usually expressed as R+(aq). This is because we
cannot determine the properties of any one ionic species by itself, so we define the change in
properties of an ion relative to H+(aq). This is done by setting the value of Кщ for the formation
of H+(aq) as exactly one. Similarly, the value of AH°iorm of H+(aq) from the element is set as zero.

The variety of chemical changes taking place when a strong electrolyte dissolves in water
requires certain assumptions in calculating the enthalpy change for an aqueous process from
thermodynamic data. If the final solution is very dilute, we can neglect the enthalpy change for the
solvent (water), and assume the overall enthalpy change is caused only by the formation of the
ionic species. This allows us to use tabulated data for the enthalpy of formation of ions. The
procedure is similar to using tabulated data for Кщ of formation of ions to calculate their molality
during an ionic reaction. We must be aware that the results are valid only for very dilute solutions.
Otherwise we need to apply complex non-ideal solution models, or have access to laboratory or
plant data on the actual system of interest. A second requirement is to know which ionic species
are present (and their amounts) during a process. We saw in Chapter 6 that the ionization of
sulfuric acid produced both S042~(aq) and HS04~(aq) ions in different proportions, depending on
concentration. In this Handbook, you will either be told which ionic species are present in
sufficient amounts to affect the material and heat balance, or given help to calculate them.

We illustrate the use of thermodynamic data to calculate the temperature rise caused by the
adiabatic dissolution of two different strong electrolytes at 50 °C in water at 50 °C to produce 1
molai solutions. First, KC1, which has an endothermic heat of dissolution, and next NiCl2, which is
exothermic. Both substances form two ions, so the products of dissolution are unambiguous. For
consistency, we take all of the necessary data from the HSC program (although other programs
such as FactSage are equally useful).

Specie NiCl2(c) KCl(c) Ni2+(a^) K+(aq) C\-{aq)

M/°form, kcal/mole at 50 °C -72.9 -104.3 -13.1 -60.3 ^10.8

KCl(c) -> K\aq) + CY{aq)\ AH°rx = 3200 cai at 50 °C [7.54]

NiCl2(c) -> m2+(aq) + 2 C\'(aq); Atf°rx = -21 800 cai at 50 °C [7.55]

Considering the reliability of the data away from its reference state of infinite dilution, we
may safely ignore the small contribution of the heat capacity of the ion species. The Cp of water is
18 cai тоГ1 deg-1 or 1.00 kcal k g 1 deg-1. Therefore, dissolution of KC1 to form a 1 molai solution

causes a 3.2 °C temperature decrease to 46.8 °C. Formation of a 1 molai solution of NiCl2 causes a
temperature increase of 21.8 °C to 71.8 °C. The temperature rise would be half that for the

formation of a 0.5 molai solution. Similar calculations at 18 °C give Af/°soin values in excellent
agreement with 18 °C handbook data. Recall, however, that the results obtained by

Chapter 7 Energy and the First Law of Thermodynamics 445

thermodynamic calculations are based on a reference state of infinite dilution, and can be
increasingly in error at higher concentrations.

We use a similar approach when calculating the АЯ°ГХ for ionic reactions. Suppose we wish
to calculate the adiabatic temperature change when NiO dissolves in hydrochloric acid. The initial
acid has 2.5 molai HC1 at 30 °C and one mole of NiO at 30 °C is added. The NiO completely
dissolves, but [СГ] remains the same. As before, we take data from HSC.

Specie NiO(c) H2O(/) Ni2+(a<?) U\aq)

A//°form, kcal/mole at 30 °C -57.3 -68.3 -13.0 0.0

The chemical reaction is then:

NiO(c) + 2H+(a<?) -» Ni2+(a<?) + H20(/); АЯ0™ - -24 000 cai at 30 °C [7.56]

Neglecting the heat capacity of the solute ions, the temperature increase is 24 °, to give a final
temperature of 54 °C.

7.12 Summary

The relationship between the work done by or on a system requires the introduction of the
concept of internal energy U. The function C/is a state variable (or state property) of a system, and
because of this, the change in U does not depend on the way the system undergoes the change.
The relationship between the change in internal energy as a function of the work done and heat
absorbed is formalized as the first law of thermodynamics.

The total energy of a process system has three components: kinetic energy, potential energy,
and internal energy. The first law formalizes the conversion of energy from one form to another.
In a closed system, energy can be transferred between the system and its surroundings as work or
heat. Application of the first law allows an energy balance to be made of any process, but places
no restriction on the process direction or even whether or not the process can or might occur.
Concepts introduced in earlier chapters (extent of reaction, yield, and equilibrium constant) are
used to deal with these issues.

The amount of work and heat for a process are functions of the path. The four common paths
that simulate common process conditions are isothermal, isochoric, isobaric, and adiabatic.
Calculations on isobaric processes are facilitated by the use of the state variable H (enthalpy), and
the heat capacity (Cp). The enthalpy changes associated with temperature changes, phase
transformations and chemical reactions are the basis for system heat balances.

For an ideal gas the values of U and H are functions only of temperature. Cp and Cv are
independent of pressure and volume, and Cp - Cv = R. For an isothermal process, U remains
constant, so the heat which enters or leaves the gas is equal to the work done by or on the gas.

Absolute values of U and H cannot be known (but they can be arbitrarily assigned a value).
The enthalpy (called the "high temperature heat content") is usually set equal to zero at 298.15 K,
which is the basis temperature for most thermodynamic databases. The database used in this text is
FREED, and is included on the Handbook CD. Enthalpy data for pure substances is available in
FREED in table format for any desired temperature interval, and contains equations for various
thermodynamic functions. FREED has a number of computational tools that help in making heat
balances. Several other commercially-available computerized thermodynamic databases are also
available (Wikipedia 2010) and as cited in General References.

The enthalpy (sensible heat, or high-temperature heat content) of a substance increases
smoothly with temperature, but with discontinuities when the substance undergoes a phase change.
If the reference temperature is 298.15 K, the term used to describe the enthalpy above this
temperature is designated ΗΊ-Η29%. The enthalpy change for a chemical reaction is a (weak)
function of temperature, with discontinuities at temperatures where phase changes take place in the

446 Chapter 7 Energy and the First Law of Thermodynamics

constituent elements or the substances. The enthalpy change for phase transitions or chemical
reactions is designated AH. The enthalpy change for an isothermal reaction when all substances
are in their normal standard state is designated Δ#°ΓΧ, and is the value tabulated in FREED.

FREED's reference temperature (and that of most thermochemical databases) is 298.15 K.
This encourages adoption of 298.15 К as the basis temperature for making heat balance
calculations. Reactants are cooled (or heated) to 298.15, all chemical reactions are carried out at
298.15 K, and the products are then heated to (or cooled to) their final temperature. To conform to
the first law, (i.e., for heat balance closure), the sum of all heat terms (including heat loss to or gain
from the surroundings) must equal zero.

Hess's law makes it possible to calculate the AHrx at any temperature by algebraically
combining ЛЯГХ for the various chemical reactions at 298.15 К with heat content data for reactants
and products. Kirchhoff s equation makes it possible to calculate the enthalpy of phase changes or
reactions at temperatures other than the normal standard state transformation temperature listed in
the database. FREED's Reaction tool incorporates the arithmetic of Kirchhoff s equation, and
includes a Calculator feature which finds the temperature and other thermodynamic functions for
any given thermodynamic function.

Specialized databases are available for certain substances, such as steam at or above the
saturation temperature (a steam table). Data from a steam table program was used as a basis for
developing equations to calculate the volume of water and steam at the saturation condition, the
enthalpy of water and steam, and the heat of water vaporization. Two steam calculation programs
and a psychrometric program are available on-line. Over a limited range of T and P, the van der
Waals equation was found to be a satisfactory equation of state for compressed air. The vdW a
and b parameters were calculated by a statistical treatment of the NIST compressed air data. The
NIST data and the calculational procedures are on worksheet AirPVT.xls on the Handbook CD.

An enthalpy effect occurs when two or more substances form a non-ideal solution, and is
called the heat of solution, designated AHS0\n, which may be positive or negative. When the
solution tends to be regular, a single parameter a may be sufficient to describe the AHS0\n over an
extended range of composition. Reference data for A#Soin is not as readily available as enthalpy
data for pure substances, but can be found in selected handbooks or in commercially-available
databases.

The enthalpy changes taking place during aqueous ionic reactions can be calculated from the
enthalpy of formation of the ionic species from the elements. The best source of such data is from
a commercially-available thermodynamic database program. Typical handbook data is based on a
reference state of infinite dilution, and can lead to significant error for heat balances on non-dilute
solutions. The heat balance requires a correct identification of the nature and molality of the ionic
species present.

References and Further Reading

Archon Engineering, Steam Tables, [Online], http://www.archoneng.com/steam.html, 2001.

Barthel, Josef M.G., Krienke, Hartmut, and Kunz, Werner, Physical Chemistry of Electrolyte
Solutions: Modern Aspects. Springer, 1998.

ChemicaLogic Corporation, SteamTab Companion, [Online].
http://www.chemicalogic.com/steamtab/companion/default.htm. November 2003.

DeHoff, Robert, Thermodynamics in Materials Science, 2nd Edition, CRC Press, 2006.

Engineers Edge, First Law of Thermodynamics,
http://www.engineersedge.com/thermodynamics/first_law.htm

Chapter 7 Energy and the First Law of Thermodynamics 447

Fernandez-Prini, Roberto, High-Temperature Aqueous Solutions: Thermodynamic Properties.
CRC Press, 1992.

Gale, William F, and Totemeier, Terry C, Eds., Smithells Metals Reference Book, 8th Edition 2004,
Elsevier.

Gaskell, David, Introduction to the Thermodynamics of Materials, 5th Edition, Taylor & Francis,
2008.

Han, Kenneth N., Fundamentals of Aqueous Metallurgy. SME, 2002.

Hultgren, Ralph H, Raymond L. Orr, Philip D. Anderson, and Kenneth K. Kelley, Selected Values
of the Thermodynamic Properties of Binary Alloys, Metals Park, Ohio, American Society for
Metals [1973]

Linric Company, PsyCalc 98, [Online], http://www.linric.com/webpsysi.htm and
http://linricsoftw.webl27.discountasp.net/webpsycalc.aspx. 1998.

NIST Fluid Properties, 2007. http://webbook.nist.gov/chemistry/fluid/
Sano, Nobuo, Lu, Wei-Kao, and Riboud, Paul V, Advanced Physical Chemistry for Process
Metallurgy, Academic Press; 1st Edition, 1997.

Wikipedia contributors, "Thermodynamic Databases for Pure Substances", "Enthalpy", Hess's
Law", "Regular solution".
Wikipedia, The Free Encyclopedia, 2010, http://en.wikipedia.org/wiki/Main_Page.

Wright, Margaret Robson, An Introduction to Aqueous Electrolyte Solutions, John Wiley, 2007.

Zaytsev, Ivan D, and Aseyev, Georgiy G. Properties of Aqueous Solutions of Electrolytes, CRC
Press, 1992.

Zemaitis, Joseph F., Jr., Clark, Diane M., Rafal, Marshall, and Scrivner, Noel С Handbook of
Aqueous Electrolyte Thermodynamics: Theory & Application. AIChE, 1988.

Exercises

7.1. Two moles of 0 2 are contained in a volume of 50 liters. The gas undergoes a reversible
isochoric process taking it from an initial pressure of 1.83 bar to 3.37 bar. Calculate the value of Q
and AU. Use data from Example 7.1.

7.2. One mole of N2 at STP is subject to a process involving three steps, each carried out
reversibly:

a) an isobaric doubling of the volume, followed by
b) an isothermal tripling of the pressure, followed by
c) an adiabatic expansion to 1 atm.

Calculate the heat and work effects for each step, using the procedure of Example 7.1, and
assuming ideal behavior.

7.3. 2280 J of heat is added in an isobaric process to one mole of N2 initially at STP. How many J
of work does the gas do during expansion?

7.4. A lead bullet having a mass of 20 g is traveling at 450 m/s, and it impacts a fixed block of
lead weighing 1.00 kg. The block is initially at 25 °C and the bullet is initially at 30 °C. Calculate
the change in U for the system, and the temperature of the combined block/bullet system after
impact, assuming negligible transfer of heat to the surroundings. For a small temperature change,
the Cp of lead is 6.42 cai deg"1 mol"1.

448 Chapter 7 Energy and the First Law of Thermodynamics

7.5. Use Kirchhoff s law to determine the heat of vaporization of one kg of water at 20 degree
intervals from 300 К to 400 K, and compare the results to those obtained from FREED's Reaction
tool and one of the steam table programs.

7.6. Fifteen kJ of heat is added to one mole of MgCl2(c) initially at 500 K. Calculate the final
temperature of the MgCl2(c).

7.7. Use FREED data (at 50° intervals) to calculate the parameters for a vapor pressure equation
for liquid Mg, valid between the melting point and the normal boiling point. Use Excel's
Regression tool to determine which equation format gives the best data fit:

log(p) = a/77 + Ъ/f + с log(p) = а/Г + b(7) + с

7.8. Copper is being melted in an electric furnace. Eighteen tonnes of copper initially at 25 °C is
placed in the furnace, and is to be heated to 25 ° above its melting point in one hour. How many
kW of power should be applied to the charge for the purpose of melting the copper?

7.9. Solid magnetite is being transported pneumatically from a hopper to a reaction vessel. One
kg of air is capable of transporting 6 kg of magnetite. The air enters the transport pipe at 47 °C and
the magnetite enters at 31 °C. The gas and solid reach the same temperature within a few seconds
of contact. What is the temperature?

7.10. Foundry sand consists of a mass fraction of 73 %Si02, 18 %ZrSi04, and 9 %Fe304.
Calculate the heat capacity of foundry sand (units: cai deg"1 kg"1) over the temperature range 25 -
200 ° C, and fit the Cp data to a linear equation.

7.11. 100 lb/hr of compressed air is required for atomizing fuel oil for a burner. Air enters the
compressor at 80 °F and 2 psig, and exits at 65 psig. How many Btu must be removed from the
compressor to maintain a compressed air exit temperature of 140 °F?

7.12. Pure CO is stable at 1400 K, but as the temperature drops, it tends to decompose to C02 and
С If cooled slow enough to let the reaction proceed to equilibrium, at 800 К 88 % of the CO has
decomposed. Calculate the enthalpy change for the process for one mole of CO in going from
1400 К to 800 К and reacting as specified.

7.13. A mixture of two moles of Ag and one mole of S were pressed into a pellet at 298 К and
surrounded by ceramic foam fiber insulation. A corner of the pellet was heated by a tiny electric
spark, whereupon the entire pellet reacted to form Ag2S and reached a temperature of 588 K.
Within experimental error, was the process adiabatic?

7.14. Superheated steam is used to heat water from 0 °C to 40 °C. What mass of superheated
steam at 200 °C and 10 bar pressure should be injected into 100 kg of water to accomplish the
desired goal? Make calculations using the text equations, and check the answer using one of the
steam calculation programs.

7.15. A fuel oil has a composition of 87.5 %C and 12.5 %H, and produces 44,100 kJ of heat when
one kg is burned to C02(g) and H20(/) at 298 K. What is the adiabatic flame temperature when the
oil is burned with twice the stoichiometric air?

7.16. The enthalpy of air at 1 atm between 260 and 400 К is shown below, from NIST tables. Use
appropriate Excel statistical tools to calculate a suitable Hj-H29s.\5 enthalpy equation for air from
this data, and compare it to FREED data for air.

T 260 280 300 320 340 360 380 400
H above 0 K, J/mol 7531.1 8113.7 8696.5 9279.8 9863.6 10448 11034 11621

7.17. One kg of dry air initially at 27 °C is exposed to 100 g of water at 27 °C. Water evaporates
to the air until water vapor saturation. P=\ atm. How much heat must be added to the system to
keep the temperature at 27 °C? Work this out with PsyCalc and with FREED data (use water
vapor equation from Chapter 2).

Chapter 7 Energy and the First Law of Thermodynamics 449

7.18. 40 grams of water vapor are added to one kg of dry air initially at 60 °C and 1 atm. In the
process, the temperature drops to 52 °C. Use the PsyCalc program to calculate the enthalpy
change at 3° intervals as the moist air cools to the dpt. Calculate the enthalpy change for cooling
the dry air, and the enthalpy change for cooling the water vapor, and compare to the enthalpy
change calculated from the FREED database.

7.19. Use FREED's Reaction tool to calculate the enthalpy change for the reduction of one mole

of Mo02 with five moles of H2 as a function of temperature between 1000 and 1600 K. The H2
enters at 298 К and the Mo02 enters at the reaction temperature. Plot the results and use the
Trendline function to find an equation that reproduces the results to ±20 J.

7.20. Use the Reaction tool to calculate the adiabatic reaction temperature for the reaction between
6 moles of air and one mole of ammonia as a function of the air temperature between 298.15 and
1000 K. Plot the results and use Trendline to fit the results to a quadratic equation.

7.21. Write equation for the reduction of wustite by CO, and another by reduction with H2. Use
FREED's Reaction feature to calculate the ЛЯ°ГХ for each reaction at 50 degree intervals between
1000 К and 1200 K. A process engineer wants to reduce wustite by a mixture of CO and H2 such
that the net isothermal process heat is zero. What fraction of the wustite should be reduced by CO

in order to attain the goal? Plot the results and use Excel's Trendline feature to obtain a linear

equation to fit the result.

7.22. Calculate the amount of heat required cast Ag-Cu liquid alloys from 1400 К to 350 К as a
function of wCu between 0.3 and 0.7. Fit the data from the chart to a regular solution model to
calculate a, and use this value in your calculations.

1200 T - AHsoin for Ag-Cu Liquid Alloys

1000 0.2 0.4 0.6 0.8

| 800 xCu

8 eoo

< 400

200

0
0

7.23. An acid solution from copper cementation has pH = 2.58, [Fe2+] = 0.500, [S042~] = 0.403,
and [HS04~] = 0.197 at 40 °C. It is neutralized by the addition of limestone (at 40 °C), which
precipitates all of the iron as siderite, and virtually all of the sulfur-containing ions as gypsum.

C02(g) is evolved, and the final solution has [HC03~] = 0.011 with a pH of 6.05. Write one or
more balanced equations for the process, and calculate the enthalpy change required to maintain

the system at 40 °C. Neglect the small amount of water that evaporates and leaves with the C02.
Data for the ionic species is shown below. Use FREED for the non-ionic species.

Species HS04"(tf<7) S042~(aq) ¥e2+(aq) НСОз~{aq)

A//°form, kcal/mole at 40 °C -212.3 -218.7 -22.1 -165.3

CHAPTER 8

Enthalpy Balances in Non-Reactive Systems

The previous chapter presented the basis for process energy balance calculations, namely, the law
of conservation of energy as expressed in first law terms. The energy required to increase the
temperature of a substance was expressed in terms of the change in enthalpy in going from a
reference temperature to some other temperature. The energy effect of phase transformations and
chemical reactions was also presented in terms of a change in enthalpy. In this chapter, all of these
effects will be brought together for the enthalpy analysis of metallurgical and materials processes
where no chemical reactions occur.

As in the case for material balances, you must consider the purpose of an energy balance
before formulating the problem. The purpose may be to compare several complete processes to
see which requires the lowest total energy from the universe. Or, the purpose may be to compare
alternative modes of operation of an existing process to find a way to use less energy. Other
objectives might be to develop a control system for a process, or to make a conceptual design for
an entirely new process for which no plant data is available. Each purpose requires different data
and procedures. For example, the objective of lowering the total energy requirement to produce a
cast iron engine block would require knowing the electrical energy to move materials around the
plant, the amount of energy required to heat the work space, the energy of melting iron, the energy
for reprocessing foundry sand, the energy consumed by pollution control equipment, etc. On the
other hand, developing a control strategy for heat-treating a gear requires knowing the
effectiveness of electrical energy transfer to the gear, the energy involved in production of a
carburizing gas, the relationship between part size and annealing time, etc. Different objectives
require different system boundaries.

8.1 Combined Material and Heat (System) Balances

Our treatment of material balances followed a progressive course, starting with non-reactive
systems. Similarly, our treatment of heat balances will start with non-reactive systems and follow
a progressive course to reactive systems in the next chapter. The tools made available from the
previous chapters will be used here but with some modifications. It's very common to make a
material and a heat balance for a process; we'll call this a system balance.

The most common type of energy balance on non-reactive systems is the "heat" balance, in
which only the thermal energy (heat) into and out of a process is accounted for. It should properly
be referred to as a thermal energy balance, or an enthalpy balance, but it's called a heat balance in
common usage. As we did for a material balance, we will often use a ledger and various Excel
tools. Referring to Chapter 7, limiting consideration to thermal energy, and assuming no heat loss
(i.e. HL = 0), the heat put into a system will equal the heat taken out plus any accumulation.

In a steady-state system, no heat accumulates. Also, most processes are (or are nearly)
isobaric, so enthalpy rather than internal energy is the measure of heat energy. If the system is not
isolated, HL* Φ 0, so a term must be included for heat gained from or lost to the surroundings. The
equation adopted for use in this Handbook is that the algebraic sum of all heat terms will equal
zero, as shown by Equation [8.1], where the symbol H refers to the sum of each type of heat effect.

H of instreams + H of reactions + H to or from surroundings + H of outstreams = 0 [8.1]

Henceforth, we will dispense with the use of Q to designate heat, and use H instead, which implies an
isobaric process.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 451

Chapter 7 showed how to use FREED data for the heat content and heat of transformation of
all common substances. Heat entering from the surroundings must be specified, or it can be
calculated from a heat balance. Heat transferred to the surroundings is designated as heat loss (HL)
and is a positive number, but values for HL are difficult to obtain because they are both process-
specific and not easily measured. As a result, the uncertainty in HL is the greatest source of error in
a heat balance. If the HL isn't specified or can't be estimated, it can usually be calculated from the
heat balance. Heat loss can be expressed in terms of a rate per unit of time, or a percentage of heat
produced for a process. In rough plant heat balances, you may see statements like "10% of the
combustion heat is lost".

A heat balance calculation involves summing heat terms for the various actions taking place
in the system. Generally the heat terms are summed around each process device, then the device
heats are summed to get the overall process heat. There is no universal way to designate the sum
of heat effects around a device. In this Handbook, it's called the sum of heat effects, or the net
device (or net process) heat. For lack of a better symbol, we'll use H, and describe it in its context
of use. Making an accurate heat balance for an actual process is no easy thing; we have not only
the uncertainties of mass flow, composition, and time variations we discussed in connection with
material balances, but also the additional uncertainties of measuring the temperature and finding
accurate thermodynamic data for complex substances. When we sum the enthalpy changes for a
process having a known material balance, we almost always find that Equation [8.1] Φ zero. In
that case, we declare that the "heat balance doesn't close". If the overall H is on the order of
expected process uncertainties, or consistent with the number of significant figures used for stream
properties, we may ignore its non-zero value and consider that the heat balance does close. If the
overall H is unacceptably far from zero, some stream property or the HL must be changed. This is
the usual practice for making a heat balance on a conceptual process where plant data is absent. In
most cases, you either ignore the heat loss, make a reasonable guesstimate, or make multiple heat
balances to study the effect of heat loss.

A major decision is choosing a basis temperature for making the heat balance. FREED uses
298.15 К (25 °C) as its standard reference temperature, as do other databases, but the steam tables
use 273.15 К as their reference temperature. Except when using the steam tables, 298.15 К is
commonly used as a heat balance basis temperature. This is true even if none of the streams is at
298.15 K, and none of the reactions occurs at 298.15 K. However, a 298.15 К basis temperature
may not result in the simplest arithmetic. If most process changes are taking place at an elevated
temperature, then that temperature may be the arithmetically simplest basis temperature. Often the
choice of a basis temperature is arbitrary. Because of Kirchhoff s law, the basis temperature
choice will not change the overall result.

Before starting out on heat balance examples, it's helpful to recognize the two main types of
composite material/heat balance situations. First, if the material balance can be solved separately
from the heat balance, the balances are described as being uncoupled, and the material balance
should be solved first. Once the stream properties are known, the heat balance calculations are
simpler. Second, if the material balance cannot be solved separately, the balances are coupled, and
the composite material and heat balances must be solved together. If a heat balance on an
uncoupled system does not close, the discrepancy is usually caused by some error in the heat loss,
or some other exchange of heat to or from the surroundings. The term system balance (on a device
or process) means a coupled heat and material balance for a device or a multi-device process.

It's helpful to categorize the various types of heat balance calculations. There is no clear
boundary between the listed types, so they are offered as typical rather than definite categories.
They're listed in general order of arithmetic difficulty.

• The stream flows and compositions are stated, apparent, or are easily calculated, so the
material balance is known or can be calculated independent of any heat effects. The stream
temperatures are stated. You are asked to calculate a heat balance, which may include
calculating the heat loss or the amount of heat input required to maintain the system at a given
temperature. This is an uncoupled situation.

452 Chapter 8 Enthalpy Balances in Non-Reactive Systems

• Some stream flows and/or compositions are unknown, but stream temperatures are
known. The main task is to make a material balance using the heat balance as a constraint.
This is a coupled situation.

• The stream flows and compositions are stated or are apparent, but one or more
temperatures are unknown. The main task is to make a heat balance to solve for the unknown
temperatures. This is an uncoupled situation.

• Finally, complex systems where there is a mixture of unknown stream flows,
compositions, temperatures, and heat loss. This is a coupled situation.

A DOF calculation for a non-reactive system balance is quite straightforward. A heat balance
problem can be considered as just a material balance problem with one additional constraint, a heat
balance. There are several additional variables such as stream temperature, heat losses, and
enthalpy/temperature relationships.

Let's start with a heat balance for the humidification of air. Water is sprayed into 100
mol/min of warm dry air to produce cool humidified air. Figure 8.1 shows a sketch of the
flowsheet. The context of the problem suggests that 320 К is the arithmetically simplest basis
temperature for a heat balance. The system pressure isn't needed if ideal gas behavior is assumed
and dew point and relative humidity aren't involved. The objective of this heat balance is to
determine the heat loss, if any, from the humidifier device.

Water 1 ^ι 1 3 Cool moist air
(20 moles, 320 K) V 3 = ?,320K)
x'
Air humidifier
Warm dry air __ 2__ I 4 ^ Water
(100 moles, 390 K) " 1 (16 moles, 320 K)

Figure 8.1 Flowsheet for the humidification of warn dry air for one minute of operation. Solid
lines depict liquid streams, and dashed lines are gas streams.

A DOF analysis shows that the material balance is correctly specified, so it can be solved
independently of the heat balance; the system is uncoupled. This system is thus the simplest of the
abovementioned categories. The water enters and leaves at the same temperature, so the heat
balance involves only the air and the A//vap of water (at 320 K). FREED data showed that Cp for
air was nearly constant at 29.1 J mol-1 deg-1 in the temperature range of interest. FREED's
Reaction tool shows that the A//vap of water is a function of temperature, as shown in Table 8.1.
The 320 К basis-temperature A//vap of water is displayed in bold font.

Table 8.1 Enthalpy of vaporization of water as a function of temperature. 400

Г(К) 298.15 300 320 340 360 380 39 752
2207
ΔΗναρ, J/mol 44 019 43 942 43 108 42 275 41440 40 600
ΔΗναρ, kJ/kg 2443 2439 2393 2347 2300 2254

The material balance shows that four moles of water evaporate, so F* = 104 moles. The heat
balance equation steps for the basis time of one minute are as follows:

1. Cool 100 moles of air from 390 К to 320 K: Hx = Н320-Нш = Cp(AT)n
= 29.1(-70)100 = -203 700J.
2. Evaporate 4 moles of water at 320 K: H2 = 43 108(4) - 172 400 J.
3. Sum of heat effects = zero: H\ + H2 + HL = 0

The heat loss is then 31 300 J/min, or -31 kJ/min.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 453

Enthalpy is a state property and therefore independent of path, so any basis temperature is
valid for the heat balance. The selected basis temperature of 320 К involved the least arithmetic.
However, there are advantages in using 298.15 К (or 25 °C) as a basis temperature.

The use of 25 °C as a basis temperature is illustrated for a different version of the above
example. Again, cold water is used to cool hot air and at the same time, humidify it. The
flowsheet is the same as that shown in Figure 8.1, but with mass units. The heat loss was
measured to be 7360 kJ/min, and P = 0.96 atm.

Stream number SI S2 S3 S4
Temp, °C 26 680 60 50

Flow, kg/min 244 66 ? ?

The materials balance DOF = 1 for this system, so it is underspecified. However, the heat
balance adds another equation, so the system balance is correctly specified. The material and heat
balances are now coupled, and both must be solved together. Since there is no clear choice for a
basis temperature, we pick 298.15 К (25 °C). The procedure is to cool all instreams to 25 °C,
evaporate W kg of water at 25 °C, and heat all outstreams. Figure 8.2 shows a sketch for this
procedure, and Table 8.2 shows heat data from FREED. For large known temperature changes,
you should use the heat content, which may be a non-linear function of temperature. For small
temperature changes Cp may be convenient because it's nearly constant over small temperature
intervals.

Air, 680 °C

^25"^680

Moist air Heat loss

Water, H25-H26 r Air, Steam, Water,
26 °C 60 °C 50 fC
60 °C

ф ф Evap'n, 25 °C J ^60"^25 ^60~^25 I H50-H25

AH' vap

Figure 8.2 Diagram of heating and cooling steps for spray cooling of hot air. The basis

temperature for the process is 25 °C.

Table 8.2 Heat data for substances involved in the process. Heat content data in units of kJ/kg,

i i i i L ^ p vaia tu n id ait/ JVJ jvg, ut/g, .

Heat term A#°vap H2O(0 at 25 °C Я68О-Я25 dry air CP H20(g) cp H2O(/) Cp dry air
700 1.869 4.184 1.009
Value 2440

The system balance steps are:

1. Cool 66 kg of dry air from 680 °C to 25 °C.
#1 = #25-#68o = 66(-700) = -46 200 kJ.

2. Cool 244 kg of cold water from 26 °C to 25 °C.
#2 = #25-#26 = 244(-l)(4.184) = -1020 kJ.

3. Vaporize W kg of water at 25 °C.
Hi = A#°vap = W(2440) kJ.

4. Heat 66 kg of dry air from 25 °C to 60 °C.
Ht =tf6o-#25= 66(1.009)(60 - 25) = 2330 kJ.

454 Chapter 8 Enthalpy Balances in Non-Reactive Systems

5. Heating W kg of steam from 25 °C to 60 °C.
# s - #6o-#25 = W(1.869)(35) = W(65.4) kJ.

6. Heat (244 - W) kg of water from 25 °C to 50 °C.
H6 = H50-H25 = (244 - W)(4.184)(25) = 25 520 - 104.6W kJ.

7. # 7 = #L = 7360kJ
8. ΣΗ = 0 = -46 200 - 1020 + 25 520 + 7360 + W(2440) + 65.4 -104.6).

Solving, W = 5.00 kg water. S3 is then 71 kg/min and S4 is 239 kg/min. The/?H20 in S3 is
0.103 atm, so the %RH in S3 is about 52 %. The/?H20 in equilibrium with the S4 water (at 50 °C)
is about 0.122 atm, so by any comparison, the humidified air is not saturated with water vapor.

EXAMPLE 8.1 — Heat Balancefor Melting Aluminum.

A continuous aluminum melting furnace receives ingots via a conveyor system and discharges
molten aluminum. Figure 8.3 shows a sketch of the process. Volumes are STP. A slight negative
pressure on the furnace draws some air in at the ingot-feeding door. The furnace is heated by a
natural gas burner with CH4 as the fuel and 118 % of stoichiometric air, and produces a flame
temperature of 1800 °C. The furnace design is such that the exit combustion gas goes directly to
the stack at 800 °C. Make system balance for the melting furnace, and calculate the heat loss.

Data. Portions of FREED's data tables are reproduced in Figure 8.4.

CH4 Hot combustion gas Stack gas
(32 m3/min, 25 °C) * (800 °C)
J1800°C]_
Aluminum ingot
Air Burner Aluminum < (445 kg/min, 25 °C)
melting
(118%stoich,25°C) Air
(40 m3/min, 25 °C)
Molten aluminum furnace 4
(445 kg/min, 700 °C)

Figure 8.3 Depiction of aluminum melting furnace.

Solution. The first requirement is to decide if the material and heat balances are coupled. The
DOF for the material balance = 0, so they are not. Therefore, we start by making a material
balance on the furnace. The required air at 118 % of stoichiometric is 359.6 m3/min (STP). The
ledger below shows the results. The N2 and 0 2 content of the air leaking in at the ingot feeding
port is simply added to the N2 and 0 2 content of the combustion gas.

Material Balance on Gas Streams, m3/min, STP

C02 H20 o2 N2 Total

Combustion gas 32.0 64.0 11.5 284.1 391.6
Leak air — — 8.4 31.6 40

Total stack gas 32.0 64.0 19.9 315.7 431.6

A basis temperature of 25 °C is appropriate because two streams enter at 25 °C. The
aluminum and air contribute no heat to the system because they enter at the standard reference
temperature of 25 °C. The only source of heat for the furnace is the heat content of the combustion
gas. If HL - 0, all of this heat is transferred to the outlet streams (aluminum and stack gas). Owing
to the uncertainty of HL and the accuracy of typical flow and temperature measurements, the STP
molar volume was rounded off to 22.4 mVkmol.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 455

Al,(c,l,g) GFW = 26.98 C02, (g) H20, (g)

(Aluminum) (Carbon Dioxide) (Hydrogen Oxide)

Unit: (C, g-mole, Cai) Unit: (C, g-mole, Cai) Unit: (C, g-mole, Cai)
—Τ~(ϋ) Cp HT-H2<)8
-77ΓΓ -UT HT-H298 cP нт-нт

25.00 5.820 0 25.00 8.884 0 25.00 8.025 0
800.00 800.00 10.079 6943
660.46 7.963 4310 1800.00 13.170 8943 1800.00 12.347 18263

dH 2580 H800-H25 = 14.477 22909

660.46 7.590 6890 H1800-H25 = H800-H25 = H1800-H25 =

700.00 7.590 7190 8943/22.4 = 22,909/22.4 = 6943/22.4 = 18,263/22.4 =
399 kcal/m3 1023 kcal/m3 310 kcal/m3 815 kcal/m3
H700-H25 =
7190/26.98 = N2, (g) 02, (g)
266.5 kcal/kg
(Nitrogen dimer) (Oxygen dimer)

Unit: (C, g-mole, Cai) Unit: (C, g-mole, Cai)

T (C) Cp HT-H298 T (C) Cp HT-H298

25.00 6.961 0 25.00 7.021 0

800.00 7.905 5710 800.00 8.401 6038
1800.00 8.640 14043 1800.00 9.059 14807

H800-H25 = Hl800-H25 = H800-H25 = H1800-H25 =

5710/22.4 = 14,043/22.4 = 6038/22.4 = 14,807/22.4 =
255 kcal/m3 627 kcal/m3 270 kcal/m3 661 kcal/m3

Figure 8.4 FREED data for the substances involved in melting aluminum. Note units.

There are three steps in the heat balance. First, cool the combustion gas from 1800 °C to 25
°C. Second, heat the aluminum to 700 °C. Third, heat the stack gas to 800 °C. You should sketch
these steps in a diagram as was done in Figure 8.2. The steps below outline the heat balance
calculations.

Step #1. Cool the combustion gas to 25 °C.
# ! =Я25-Я18оо = 32.0(-1023) + 64.0(-815)+ 11.52(-661) + 284.Ц-627)
# i = -270 600 kcal/min.

Step #2. Heat the aluminum to 700 °C.
#2 = Я700-Я25 = 445(266.5) = +118 600 kcal/min.

Step #3. Heat the stack gas to 800 °C.
# з = Я8оо-Я25 = 32.0(399) + 64.0(310) + 19.92(270) + 315.7(255)
H3 = 118 500 kcal/min.

Step #4. Sum all heat terms.
Σ # = -270 600 + 118 600 + 118 500 + HL = 0. Σ # - 3 3 500 kcal/min + HL

The sum of the above energy terms gives HL = 33 500 kcal/min. The system is transferring (i.e.
losing) 33 500 kcal/min to the surroundings.

Assignment. Suppose that the feed rate of aluminum ingots was increased to 450 kg/min, while all
other operating conditions (including the heat loss) remained the same. Calculate the temperature
of the molten aluminum.

EXAMPLE 8.2 — Heat Balancefor Spray Cooling of Hot Air.

The hot air cooling system described by Figure 8.1 cooled 66 kg/min of hot air initially at 680
°C with 244 kg of water initially at 26 °C. The spray cooler was re-designed for extremely good
gas/liquid contact so as to cause both outstreams to have the same temperature, and the moist air to
be water vapor saturated. Calculate the system balance.

456 Chapter 8 Enthalpy Balances in Non-Reactive Systems
Data. Table 8.2 gives most of the necessary heat data for the process. The use of Cp is convenient
and acceptable since the Table 8.2 value was obtained over the same temperature range used here,
and Cp is only a weak function of temperature. However, we need information on the mass of
water vapor in vapor-saturated air as a function of temperature when P = 0.96 atm. The PsyCalc
program is used for this because it calculates the grams of water vapor per kg of dry air as a
function of ambient conditions, and we know the mass of dry air in S3. Four data points were
obtained from PsyCalc between 48 °C and 57 °C at 97.2 kPa, and revised as kg steam/kg dry air.
Figure 8.5 shows the relationship.

Figure 8.5 Mass of steam in one kg of dry air as a function of temperature from the PsyCalc
program. Text box equation developed by Excel's Trendline tool.
Solution. We have two unknowns in this Example instead of just one as in the earlier spray
cooling example. DOF for the material balance = +1, so the material and heat balances are
coupled and we must solve them together as a system balance. Let W = kg of water evaporated
and / be the temperature of S3 and S4 in °C.

The system balance equation steps are listed below, in heat units of kJ.
1. Cool 66 kg of dry air from 680 °C to 25 °C.

# ! - Я25-#б8о = 66(-700) = -46 200.
2. Cool 244 kg of cold water from 26 °C to 25 °C.

#2 = Я25-Я26 = 244(-l)(4.184) - -1020.
3. Vaporize W kg of water at 25 °C.

# з = Atf°vap = W(2440).
4. Heat 66 kg of dry air from 25 °C to t.

#4 = HrH15 = 66(1.009)0 - 25).
5. Heating W kg of steam from 25 °C to t.

# 5 = #r#25 = W(1.869)(i-25).
6. Heat (244 - W) kg of water from 25 °C to t.

H6 = HrH25 = (244 - W)(4.184)(/ - 25).
7. H7 = HL = 7360
8. W = 66(0.0001778/2 - 0.01261/ + 0.277)
9. Σ # = 0

Chapter 8 Enthalpy Balances in Non-Reactive Systems 457

Rather than go to the trouble of collecting terms and trying to solve a quadratic equation, it's
easier to insert the above equations into Solver and let it find the value of W and t. Solver requires
initial estimates for W and /; here, W was set at 7 and t at 50. These estimates are called
placeholder values, arbitrarily chosen as reasonable, that Solver will change to find a solution to
the equation set. Two equations were written, one for step 8 and one for step 9. Solver found a
solution:

W = 5.7 kg and / = 49.1 °C

Assignment. Try using SuperSolver to find / and W as a function of the incoming water
temperature.

Example 8.3 shows the approach for a one-unknown system that's simple enough not to need
Goal Seek or Solver.

EXAMPLE 8.3 — Fog Cooling of Ceramic Parts.

Ceramic insulators made from a magnesium aluminum silicate called cordierite are sintered
and then cooled prior to packaging. The cooler uses fog generated in an air/water mixing nozzle.
The mass ratio of air/water to the nozzle mixer is 10. Figure 8.6 shows a sketch of the cooling
system. Calculate the mass of air and water required to cool the cordierite to 450 K.

FA=FE + FF Heat loss: F-E ■_ ? kg/min
ΓΑ=1200Κ 3000 kcal/min E TE=300 К

Hot gas Air

Hot parts Water
FB = 600 kg/min
ΓΒ=1250Κ FF = ? kg/min
Cool 7F = 300 К
Tu = 450 К parts

Figure 8.6 Sketch of the cooling system for cordierite parts.

Data. Enthalpy data for cordierite, water (l,g) and air are from FREED, as shown in the table
excerpts below. Note units of cal/mol.

Mg2AI4Si5018 GFW: 584.953 H20(l,g) GFW: 18.01528 N1.580.42, (g) GFW: 28.8504

(Mg-AI Silicate) (Hydrogen Oxide) (Air)
Unit: (K, g-mole, Cai) Unit: (K, g-mole, Cai) Unit: (K, g-mole, Cai)

T (K) Cp HT-H298 T (К) Cp HT-H298 T (К) Cp HT-H298

298.15 108.101 0 298.15 18.018 0 298.15 6.974 0
300.00 108.467 200 300.00
450.00 134.773| 18573 300.00 18.011 33 400.00 6.971 13|
1200.00 171.407 137466 1200.00
1250.00 172.823 373.70 18.168 1362 6.989 709
146071
8.141 6803

dH 9767

373.70 8.115 11130
1200.00 10.434 18766

Solution. Scale the mass flow by 1/1000 to conform more closely to FREED's units, so the flow of
cordierite is 1.026 mol/min, and the mole ratio of air/water is 6.244. The amount of water is set as
W, and the amount of air as 6.244W. There are four unknowns: the flows and enthalpies of
streams E and F. The material and heat balances are coupled, so we must make a system balance.

458 Chapter 8 Enthalpy Balances in Non-Reactive Systems

The steps in making the calculation are to cool the cordierite from 1250 to 450 K, and heat the
water and air from 300 to 1200 K. The sum of the heat effects (including the heat loss) is set equal
to 0, and W is calculated. The enthalpy values (calories) used in the calculation are noted in the
above tables in boxed entries.

1. Cool cordierite. Η450-Ηί250= 1.026(18 573 - 146 071) - - 1 3 0 800.

2. Heat water: Ншо-Нш = W(18 766 - 33) = 18 730W.

3. Heat air: НШо-Нш = 6.244W(6803 - 13) - 42 400W.

4. Heat loss: 3000.

Solving, W = 2.091. The mass of water is thus 37.66 g and the mass of air 376.6 g.
Converting back to the original basis, the cooler requires 37.7 kg/min water and 377 kg/min air.
The mass of fog is 415 kg/min.

Assignment. Set up an Excel worksheet to calculate the mass of fog over a range of air/water mass
ratios between 12 and 8, and plot the results.

8.2 Heat Balance for Adiabatic Processes

In an adiabatic process (which may be open, closed, or isolated), HL= 0 so no heat is
exchanged with the surroundings. In a practical sense, HL is never zero, but it may be very small
in comparison to the heat content of streams and the AH of transformations/reactions. In this text,
the term adiabatic is used to describe systems where HL is smaller than the number of significant
figures used in describing the stream and transformation enthalpy. The concept of an adiabatic
process was discussed in several sections of Chapter 7.

Consider the simple case of humidifying dry air by spraying water into it. We considered this
case early in the Chapter by noting that mixing 20 moles of water at 320 К and 100 moles of dry
air at 390 K, and evaporating four moles of water at 320 K, resulted in a heat loss of 31 300 J/min.
This loss closed the heat balance and kept the temperature at 320 K. For an adiabatic process, HL
would be zero, in which case, the outstreams temperature would have to be above 320 K, or the
instream air temperature would be below 390 K. In the latter case, the question is: what is the
instream air temperature such that it has 31 300 J less heat than in the example? In other words,
what is the air temperature such that ЯТ-Я320 = 1724 J/mol?

The first example used Cp of air, so we can do the same here. The relevant equation is:

HT-H320 - 1724 = СДАТ) = 29.1(ΔΤ)

ΔΤ is then 59°, so the incoming air must be at 379 К for the cooler to operate adiabatically.
There are other ways of finding this temperature, such as using FREED's Calculator tool or
interpolating in FREED's table for air. Your method choice will depend a lot on how comfortable
you are with FREED's various tools.

Sometimes a single value for the Cp of a substance does not give sufficiently accurate results.
This happens when calculations have to be made over different temperature ranges, especially
when Cp changes rapidly with temperature. FREED's graphics tool tells you at a glance how
rapidly Cp changes with temperature, and gives a data table that can be used to derive an
expression for Cp vs. T. For example, suppose a combustion process involved a CH4 instream
temperature that varied between 50 °C and 250 °C. Is it satisfactory to use an average value of Cp
for CH4 over this temperature range? Figure 8.7 shows the chart produced by FREED's Graphics
tool over that temperature range. Clearly, Cp for CH4 has too large a variation to be considered a
constant over the 200° temperature range.

When the extent of Cp variation is unacceptably large, it's better to avoid using Cp, and use a
heat content equation instead. When a linear equation is acceptable for Cp, a quadratic equation is
acceptable for the heat content. However, the temperature range of interest suggests two possible

Chapter 8 Enthalpy Balances in Non-Reactive Systems 459

basis temperatures, 25 °C (which is the reference temperature for the FREED database) or 50 °C.
Figure 8.8 shows heat content data for CH4 between 50 °C and 250 °C for both basis temperatures,
along with quadratic equations developed by Trendline.

Plot:Cpvs.TforCH4

5.00E+01
4.80E+01
4.60E+01
3 4.40E+01
ü 4.20E+01
4.00E+01
3.80E+01
3.60E+01

100 150 200 250

T(degC)

Figure 8.7 Chart from FREED's Graphics tool showing the variation in molar Cp for CH4 between
50 °C and 250 °C. The Trendline tool was used to develop a linear relationship between the

variables. Actual chart slightly edited for display purposes.

10000 Heat Content vs. t for CH4 200 250
_ 9000 Hi-His = 0.0273Г + 33.68/ - 848
I 8000 /Л-/У50 = 0.0283/2 + 33.34/ -1733
^5 7000
•g 6000 50 100 150
3 5000 temp, deg. С
О 4000
^ 3000
S 2000
^ 1000

0
0

Figure 8.8 Chart from FREED's Graphics tool showing the variation in molar heat content for
CH4 for basis temperatures of 25 °C and 50 °C. Boxed equations were developed by using the
Trendline tool. Actual chart slightly edited for display purposes.

The acceptability of each equation's fit can be estimated by calculating H25-H25 for the 25 °C
basis temperature equation, and Н50-Н50 for the 50 °C basis temperature equation; both equations
should give zero at their basis temperature. The error for the 25 °C basis temperature equation is
11 J, while that for the 50 °C basis temperature equation is 5 J. This is a very small error
compared to the magnitude of other typical heat balance error sources. In conclusion, either basis
temperature is equally acceptable.

EXAMPLE 8.4 — Atomization of a Molten Metal

Copper powder is prepared by atomizing molten copper with nitrogen. The gas impinges on a
stream of molten copper and breaks it into tiny droplets. Sufficient N2 must be used to cool the Cu
to <140 °C so that it will not oxidize when discharged. Figure 8.9 shows a sketch of the molten

460 Chapter 8 Enthalpy Balances in Non-Reactive Systems
metal atomizer. The heat loss through the walls of the powder collection vessel is negligible.
Experience showed that the N2 gas leaving the powder reception vessel is 35° cooler than the
temperature of the powdered copper discharged. The average flow rate of molten Cu is 30 kg/min.
Make a system balance to calculate: a) the flow rate of N2 required at the average flow rate of Cu
to attain a powder temperature of 120 °C, and b) the temperature of the Cu and N2 for the N2 flow
rate calculated in part a), when the Cu flow fluctuates from 25 to 35 kg/min as the level in the
holding crucible varies.

Figure 8.9 Sketch of atomization process for making copper powder. The powder is collected in a
sealed vessel and discharged periodically.
Data. FREED was the source of all data. The relevant data is introduced in the solution:
Solution, a) Here the material and heat balances are coupled. The average flowrate of Cu is 30
kg/min, or 0.4721 kmol Cu/min. The amount of N2 required is unknown, so a material balance
cannot be made. Since the material and heat balance are coupled, the problem requires a system
balance. The problem statement indicates a variation in the Cu discharge temperature, so the basis
temperature is selected as 25 °C. The steps in making the system balance are listed below, along
with enthalpy values obtained from FREED's Calculator tool.

1. Cool Cu to 25 °C. #115o-tf25 = -44 790kJ/kmol.
2. Heat Cu to 120 °C. Hm - H25 = 2350 kJ/kmol.
3. Heat N2 from 25 to 85 °C. #85 - #25 = 1736 kJ/kmol.
The net enthalpy change for one kmol of Cu is the sum of the values for step 1 and step 2.
For a basis of one minute of operation (0.4721 kmol Cu/min), the H\2o - HU50 for Cu = -20 040 kJ.
The enthalpy change for N2 is 1736 kJ/kmol. Since H L = 0, the enthalpy change of Cu + the
enthalpy change of N2 = 0:

-20 040+17360?N2) = 0
Solving, 11.54 kmol N2/min are required (or 259 m3 N2/min).
b) This too is a coupled system because temperature is unknown. Here the flow of N2
remains at 11.54 kmol/min, while the flow of Cu varies between 0.3934 and 0.5507 kmol Cu/min.
The arithmetic is complicated because the Cu and N2 temperatures aren't known, and the FREED
enthalpy equations (like those in most other databases) list the enthalpy as a function of
temperature, not the reverse. Two general approaches can be employed when faced with an
unknown-temperature calculation. Both approaches will be used in part b) of this example to
illustrate their differences.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 461

First, a graphical approach is illustrated. Several equally spaced values of Cu discharge
temperature are selected at the N2 flow of 11.54 kmol/min and the mass of Cu that will close the
heat and material balance is calculated. The results are plotted and the desired results read from
the graph. This approach works by restating the problem to simplify the arithmetic, and then
plotting the results to get the answer. Excel's Trendline tool can be used to obtain an equation
relating the variables. The following steps are used:

1. Select Cu discharge temperatures (^Cu) of 90, 110, 130, and 150 °C. This requires N2
discharge temperatures (^Ν2) to be 55, 75, 95, and 115 °C.

2. Calculate the molar enthalpy change for Cu in heating from 25 °C to tdCxx.

3. Add -44 790 kJ to the above value to obtain the net molar enthalpy change for Cu. This
is the value for Htacu-Hn50.

4. Calculate the molar enthalpy change for N2 in heating to 4Ν2, and multiply by 11.54 to
get the total enthalpy change for N2.

5. Set up the balance equation to solve for the amount of Cu.

6. Plot the results.

As was done in part a), FREED's Calculator tool was used to obtain enthalpy change values
for steps 2 and 4 above. Table 8.3 shows the results.

Table 8.3 Heat content (Ш-Н25) for copper and nitrogen, kJ.

/dN2, °C 65 75 95 105
One kg-mol N2 1156 1447 2025 2315
11.54kg-molN2 13 340 16 698 23 369 26 715

tdCu, °C 100 110 130 140
One kg-mol Cu 1850 2100 2603 2856

A sample calculation for tdCu = 110 °C is: 16 698 +rcCu(2100- 44 790) = 0

Solving, the amount of Cu = 0.3911 kmol, or 24.96 kg Cu. Figure 8.10 shows results for each
value in the table above, and a chart of the results between 25 and 35 kg Cu/min. As expected, an
increase in the flowrate of Cu results in an increase in /dCu, but even at the upper limit of 35
kg/min, tdCu does not exceed the maximum permissible value of 140 °C.

Atomizing Copper with 11.54 kmol N2
42

с kg Cu/min = 0.518(tdCu) - 32.1 ^^^^^ ^^^^^
| 38

Ì 5 34 ^*^0 idCu, °C n Cu kg Cu
Ό

N 30 100 0.3107 19.7

*о-» 26 110 0.3912 24.9
CO 130 0.5539 35.2
22 140 0.6371 40.5
δ L •^^^

18

К )0 110 120 130 Л*Ю

Cu discharge temperature, °C

Figure 8.10 Results of coupled material/heat balance calculations on the atomization of copper.
Nitrogen enters at 25 °C and exits 35 °C below the copper powder discharge temperature (VdCu).
Text box equation obtained by using Excel's Trendline tool. As expected, the trendline passes
through the point calculated in part a) of 30 kg/min Cu and a tdCu of 120 °C.

462 Chapter 8 Enthalpy Balances in Non-Reactive Systems

The second approach is to use FREED's analytical equations for the heat content of Cu and
N2 and solve the equations for a value of T that closes the heat balance. Since the enthalpy
equations listed in FREED are in degrees K, it's common to express the temperature in K. The

equations for in setting up the material/heat balance are:

Molar heat content of liquid Cu at 1150 °C: 44 790 kJ/kmol.
Molar heat content of solid Cu: HT-H29s = 96.991(7) - 2.6723 x 10"2(P) - 9.7064 x 105/Γ-
2400VT + 7.3382 x 10~6(7^) +17 958 (kJ/kmol).

Molar heat content of N2: #τ-#298 = 38.167(7) + 1.6458 x 10"3(P) - 5.0733 x 105/Г-
541.1 Vr - 2.3407 x 10"7(7^) - 477 (kJ/kmol).

The overall enthalpy change for Cu and N2 requires that the molar heat content equations be
multiplied by the number of moles of Cu and N2. Also, remember that the T in the solid Cu
equation is TdCu and the Tin the N2 equation is rdN2, which is 35° less than rdCu. The combined
system balance equation for 25 kg/min Cu (0.3934 kmol/min; bold value in the equation) is shown

in Equation [8.2].

11.54[38.167(Г-35) + 1.6458 x 10~3(Г-35)2 - 5.0733 x 105/(Г-35) - 541.1 VrÖ5 - 2.3407 x 10"

7(Г-35)3 - 477] + 0.3934[96.991(7) - 2.6723 x 10"2(P) - 9.7064 x 105/Г- 2400V7 + 7.3382 x

10~6(7^) +17 958 - 44 790] = 0. [8.2]

At first glance, Equation [8.2] looks very difficult to solve. However, Excel's Goal Seek tool
made the solution quite easy. The value of T (= TdCu) was found to be 383 K, or 110 °C, which is
the same result obtained from the text box equation on Figure 8.10. The equation was also solved
for 35 kg Cu/min by inserting 0.5507 (kmol/min) in place of 0.3934 in the equation. The result
was 403 K, or 130 °C, again in agreement with the graphical approach.

There are pros and cons for each method. The graphical method usually has less-complex
arithmetic and is recommended for simpler problems where a high degree of accuracy is not
warranted. The graphical approach also has the advantage of showing trends and rates of change.
The equation method is often the only one suitable where the problem has a large number of
species and multiple reactors. And sometimes a (simpler) three-term Ητ-Η2η equation can be
developed from FREED data that gives adequate accuracy.

Assignment. Molten Cu is water atomized to produce a slurry of water and powder. The process is
adiabatic and continuous. Calculate the flowrate of water required to atomize 30 kg/min of molten
Cu to produce a slurry at 75 °C. The Cu is initially at 1175 °C and the water initially at 30 °C.

8.3 Psychrometric Calculations

Many processes involve a change in the moisture content of a gas stream. In the gasification
of coal, steam is added to industrial oxygen to increase the amount of hydrogen produced. In iron
ore reduction, water must be removed from the spent reduction gas so it can be recycled. The
general term used for altering the water vapor content of a gas is humidification or
dehumidification. Adjustment of the moisture content and temperature of air is called air
conditioning. The general subject is called psychometrics (Wikipedia 2010). A gas can be
humidified by adding steam, or by spraying water into it. A gas can be dehumidified by cooling it
below the dew point to condense water. In either case, heat must be added or removed to attain the
desired objective. The PsyCalc 98 program was mentioned earlier as a helpful tool for calculating
the enthalpy change for heating and cooling moist air, but it is not suitable for use with the variety
of other gases found in materials processing.

Dehumidification of a gas requires removal of heat, not only to cool the gas, but to condense
water (with an enthalpy change opposite that of A//vap). The amount of heat that must be removed
from a gas until it reaches the dew point is a function of the initial gas temperature and its
composition. After reaching the dew point, heat must be removed to condense moisture and cool

Chapter 8 Enthalpy Balances in Non-Reactive Systems 463

the dehumidified gas and water as the pH20 changes along the VLE line. The thermal effect of
dehumidification along the VLE line is somewhat complicated because both gas cooling and water
condensation occur simultaneously, and the mass of the dehumidified gas and water change
continuously as dehumidification proceeds. The heat and material balances are therefore coupled.

EXAMPLE 8.5 — Dehumidifying Spent Gas from an Iron Ore Reducing Furnace.

Spent gas from an iron ore reducing furnace is treated to remove H20 and C02. The H20 is
removed first by cooling the gas to condense some of the steam, then removing most of the C02 by
amine extraction. (See Example 4.15). The composition of the spent gas is:

<pCO2 = 0.18; <pCO = 0.16; <pH2O = 0.36; <pH2 = 0.30

The spent gas exits the furnace at 380 °C and enters the dehumidifier at 365 °C and 1.88 bar.
Make a system balance to calculate the amount of heat removed per m3 (STP) of spent gas as it
cools to 25 °C, and the mass of water condensed. Neglect heat loss.

Data. FREED and Equation [2.16] are the source of all data used in this example. The relevant
heat content data derived from FREED is shown on the next page.

Solution. As the spent gas cools to the dew point, no water condenses, so the only enthalpy change
is for cooling the gas. Once the gas reaches the dew point (i.e., the VLE temperature), further
cooling results in the condensation of steam to water, so not only is the amount and heat content of
the gas and water changing, there is also a heat of condensation. One procedure for calculating the
heat balance of a continuously-changing process is to divide it into a number of temperature
intervals, and calculate the enthalpy change for each interval. Steps of similar <pH20 are chosen
below the dew point such that approximately equal amounts of water are condensed in each step.
Calculations are made down to 25 °C. The results are plotted and fitted to an analytical
expression. Bar is the pressure unit used in this example. The basis temperature is 25 °C (298 K).
The advantage of carrying out the condensation for all steps at 298 К is that the enthalpy of
condensation of steam to water (-Ai/vap) is the same value for each step: -44 020 J/mol. Please see
Table 8.1, which shows the relevant FREED tables for H20(/,g).

The initial />H20 in the spent gas is 1.88(0.36) = 0.6768 bar. You can use Equation [2.16] to
calculate that the dew point for the spent gas is 362 K, and when cooled to 298 K, the pH20 is
0.0316 bar. Thus, <pH20 goes from 0.36 in the original gas to 0.0316/1.88 - 0.0168 at 298 K. The
following table shows the results of a series of calculations on the/?H20 and <jpH20 as the gas cools
below the dew point. P = 1.88 bar.

T, К 360 355 349 342 334 325 315 298.15
/?H20, bar 0.0316
0.6208 0.5100 0.3993 0.2963 0.2070 0.1350 0.0812 0.0168
<pH20 0.3302 0.2713 0.2124 0.1576 0.1101 0.0718 0.0432

A material balance is required to calculate the amount of water condensed at each of the
above temperatures. Starting with 1 m3 (STP) spent gas (44.60 g-mol), the amount of steam
originally present is 16.056 moles. The amount of steam present (and by difference the amount of
water condensed) after cooling to some temperature below the dew point can be calculated from
Equation [8.3]:

<рП20 = иН20 [8.3]
44.6(1- 0.36) + иН20

where яН20 is the amount of steam remaining in the gas after condensation. At 355 К for
example, nH20 = 10.616 moles. By difference, 5.44 moles of water condense as the gas cools
from the dew point of 362 К to 355 K.

The heat balance for each temperature requires three steps. Where the gas is cooled below the
dew point, the first step is cooling the entire spent gas to 298.15 K. The second step is condensing

464 Chapter 8 Enthalpy Balances in Non-Reactive Systems

out an amount of water (at 298.15 K) as calculated by use of Equation [8.3]. The third step is
heating the remaining gas to the temperature used in making the Equation [8.3] calculations. The
overall heat effect is the sum of these three steps. However, before making the heat balance, we
need to recall how to develop heat content equations that are more manageable than those listed at
the bottom of FREED's tables. Those equations have several terms in order to fit the heat content
data over a span of 2000° or more, and as shown in Equation 8.2, their use complicates the
arithmetic. FREED can help develop simpler equations that accurately represent Нт-Н2^ over a
shorter temperature range. Usually a three-term equation is adequate for gases over a 600 К
temperature span. A two-term Нт-Н29% is adequate for water because Cp for water is nearly
constant.

The procedure in developing the simpler HT-H2gs equations is to use FREED's Graphics tool
to chart Ят-Я298 vs. T. Excel's Trendline tool then fits a quadratic equation to the data for the
gases, and a linear equation for water. (This same procedure was used to develop a Cp equation for
CH4; see Figure 8.7). The molar heat content coefficients for the four gases and water are as
follows, with units of J/g-mol:

ЯТ-Я298 = AT2 + ВТ + С; J/mol

AВС

H2(g) 0.00038 28.853 -8641
C02(g) 0.01707 27.458 -9713
CO(g) 0.00161 28.050 -8489

H20(g) 0.00477 30.396 -9480

H2O(0: — 75.42 -22 490

These coefficients (plus the Af/°condensation of steam at 298 К of -44 020 J/(g_1 тоГ1) are used
for all heat balance calculations. A sample calculation for the case of the spent gas cooling to 500
К (no water condensation because 500 К is above the dpt of 362 K) is shown first, followed by a
sample calculation for cooling to 355 К (5.44 moles of water condensed).

The ledger below shows that the total enthalpy change for cooling 1 m3 (STP) spent gas from
638 to 500 К is -214.8 kJ. Remember — no water condenses in cooling to 500 K.

Ledger for Cooling 1 m3 Spent Reducing Gas from 638 to 500K

Gas H20(g) H2(g) COfe) C02(g) Totals
16.056 13.38 7.14 8.03 44.6
Amount of gas cooled, mol -190.3 -132.8 -71.8 -513.4
H for cooling gas from 638 to 298 K, kJ 34.5 24.8 13.4 -118.5
-155.8 -108.0 -58.4 19.8 92.5
H for heating gas to 500 K, kJ -98.7 -420.9
ΣΗ, kJ

The next-page ledger shows the total enthalpy change for cooling 1 m3 (STP) spent gas from
638 to 355 K. ΣΗ = -657.7 kJ. Of that change, -239.5 kJ (36%) is for the condensation of 5.44
moles of water between 362 and 355 K.

Figure 8.11 shows the results of calculations at several temperatures between 638 and 298 K.
Considering the accuracy needed, the enthalpy change down to the dew point of 362 К is
adequately expressed by a linear equation:

#T-#638 = 1.523(7) - 974 (kJ/m3, STP) [8.4]

However, there is no simple mathematical expression that adequately represents the enthalpy
change below the dew point, or the amount of water condensed.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 465

Ledger for Cooling 1 m3 Spent Reduction Gas from 638 to 355 К

Gas H20(g) H2(g) CO(g) C02(g) H2O(0 Totals
Amount of gas cooled, mol 16.056 13.38 7.14 8.03 0 44.6
H for cooling gas from 638 to 298 K, kJ -190.3 -132.8 -71.8 -513.4
Condense 5.44 moles H20 at 298 K, kJ -118.5 — -239.5
Amount of species at 298.15 K, mol 10.616 13.38 7.14 44.6
//for heating species to 355 K, kJ 20.3 22.1 11.9 8.03 -239.5 95.2
-170.0 -110.7 -59.9 17.6 5.44 -657.7
L#,kJ -100.9 23.3

-216.2

200

-400 ω

-600 гсОо)

Q.

-800
С

1000 ω

1200

Figure 8.11 Graphical depiction of the changes taking place during the cooling of 1 m3 (STP)
stack gas initially at 638 К and containing φΗ20 of 0.36 (16.056 moles of H20). Pressure = 1.88
bar throughout. The enthalpy to the dew point (362 K) is for simply cooling the stack gas. The
enthalpy change below the dew point is the sum of the enthalpy change for condensation of water
and stack gas cooling, which accounts for the sharp change in slope at 362 K.

Assignment. Calculate the enthalpy change for cooling the stack gas from 638 to 320 К according
to the following path: First, cool the gas to 320 K. Second, condense water at 320 К according to
the material balance for the gas being at its dew point at 320 K. Be sure you use the AiPcondensation
value at 320 К (see Table 8.1).

Often it's necessary to make a system balance on a process where several different things are
going on in one device. For example consider the drying of a wet powder in a horizontal rotary
dryer. The temperature of the solid in such a device does not vary uniformly along its length.
First, the wet solid heats up, with very little water evaporating until the mass comes close to 100
°C. Next, heat is transferred to the solid to evaporate water, during which the solid temperature
remains near 100 °C. Finally, the dry solid is heated from near 100 °C to the exit temperature.
While the three regions overlap somewhat in the dryer, it's useful from a conceptual standpoint to
consider the dryer as having three zones. A heat balance for each zone gives a good understanding
of the heat balance on such a dryer, although an overall heat balance is adequate for a general
picture of the process.

A combustion stack gas can be used as a source of heat for drying. If the powder material has
a tendency to oxidize, the system should be designed to exclude as much air as possible from
leaking into the dryer. For greatest efficiency, the gas and solid should flow counter-currently

466 Chapter 8 Enthalpy Balances in Non-Reactive Systems

through the dryer, and it is necessary to keep the gas temperature above the dpt at all times. Please
review the clay dryer material balance that was given in Example 2.7.

EXAMPLE 8.6 — Using Stack Gas to Dry Cadmium Powder.

Cadmium powder is prepared by water-atomizing liquid Cd. The powder can undergo
unacceptable oxidation if exposed to air above 150 °C. Calculate the quantity of combustion stack
gas required to dry one tonne of wet Cd powder containing wH20 = 0.18. The wet powder enters
the dryer at 50 °C and the stack gas temperature fluctuates between 140 - 160 °C. The
composition of the stack gas is: <pC02 = 0.18, φΗ20 = 0.35, <pN2 = 0.50, φ02 = 0.03. HL is
estimated to be 100 000 kJ/tonne wet Cd.

Data. Considering the narrow range of temperatures for the different substances, and the
uncertainty of heat loss, it is sufficiently accurate to use an average value of Cp for each species.
Also, mass and degrees Celsius are chosen for units. The program MMV-C was used to convert
the stack gas composition to mass units, and FREED was used to obtain Cp data except for the
АЯуар of H20(/), which was taken from the steam table calculator.

Stack gas composition C02 H20 N2 o2

Voi % 18.0 35.0 50.0 3.0

Mass % 27.1 21.6 48.0 3.3

Species c o 2 H20(g) N2 o2 Cd H20(/)
Cp, kJ kg-'deg"1
0.92 1.95 1.03 0.93 0.24 4.2

A#vap H20 at 100 °C - 2260 kJ/kg

Solution. The first step is to sketch the process flowsheet, with the three different zones of the
dryer separated by function. The sketch implies three separate devices, but actually there are three
zones in one device. The zone lengths may be quite different. The heat and material balances are
coupled, so we must make a system balance.

m"= 1000 kg wH2O = 0.15 820 kg : 820 kg
wH2O = 0.18 tB=100°C wH2O = 0
ivH2O = 0
tA = 50 °C tc = 100°C tD=130°C

Figure 8.12 Flowsheet for drying powdered cadmium. Dryer is one device, depicted as having
three zones. The flow of solid or wet materials is indicated by a solid line, while gas flow is
indicated by a dashed line.

The next step is to calculate the mass of stream B, which by a mass balance around the
preheater is 965 kg. So 35 kg H20(/) evaporate in the preheater and 145 kg evaporate in the
evaporator unit.

Sometimes it's helpful to make an overall system balance for a process before getting into the
details of each device. This gives an idea of the mass of stack gas as a function of the dryer offgas
temperature (7H). Start by calculating Cp for the stack gas, based on the values of Cp for each
individual species. The following equation shows the calculation, where the Cp is listed first, and
the mass fraction second.

Cp stack gas, = 0.92(0.271) + 1.95(0.216) + 1.03(0.480) + 0.93(0.033) = 1.20 kJ/(kg · deg).

Chapter 8 Enthalpy Balances in Non-Reactive Systems 467

The overall heat balance involves calculating the heat effect (in kJ) of each part of the
process, using the integrated version of Equation [7.15]: Ht2 - Нц = Cpm(t2 - t\) = CpmAt. A
sample calculation is made for a stack gas inlet temperature of 150 °C.

Step #1. Heat 820 kg Cd to P\ Hx2 - Нц = 0.24(820)0° - 50)

Step #2. Heat 180 kg water to 100 °C: tft2 -Щ = 4.20(180)(100 - 50) - 37 800 kJ

Step #3. Evaporate 180 kg of water: Atfvap = 2260(180) = 406 800 kJ
Step #4. Heat 180 kg of steam to tH: Hi2-Htx = 1.95(1Щ(Р - 100)
Step #5. Heat X kg of stack gas to tH: Ht2-Htx = 1.20(X)(/H - 150)

Step #6. Heat loss = 100 000 kJ.

The sum of the above heat effects must equal zero. The unknowns are tD, tH, and X (the mass
of stack gas). If we specify any one of the above three, we can explore the effect of one of the
other two on the other. The above equations were written into an Excel worksheet with P = 130
°C. Values of tR between 90 and 110 °C were inserted, and SuperGoalSeek was used to calculate
X. The results are shown in the following table.

Л°с 90 95 100 105 110

Mass of stack gas, kg 7734 8463 9339 10 409 11747

As expected, the lower the value of tH, the less stack gas required. It's desirable to minimize
the amount of stack gas to minimize dust carryover.

The overall system balance does not tell us which of the above conditions actually exist in the
system. That requires a system balance around each device, which requires estimated heat losses
for each. Since the unit temperatures decrease in the direction of stack gas flow, we arbitrarily
assign a heat loss of 37 000 kJ for the post-heater, 33 000 kJ for the evaporator, and 30 000 kJ for
the preheater.

The unknown temperatures in the system are the three stack gas temperatures, /F, tG and /H.
The stack gas mass is also unknown, but the result of the system balance can be used to develop an
equation relating the stack gas mass and tH. If the values in the above table are plotted, then
Excel's Trendline tool can fit polynomial equations to the data. A quadratic equation was deemed
adequate:

Mass of stack gas, kg = X = 4.03 l(tHf - 606.7(/H) + 29 700

System balance equations around each device are as follows.

Preheater: Heat Cd 50°: 820(0.24)(50) - 9840 kJ

Heat water 50°: 180(4.2)(50) = 37 800 kJ

Evaporate 35 kg water at 100 °C: 35(2260) = 79 100
Cool X kg of stack gas: X(l.2)(tH-tG)

Heat loss: 30 000kJ

Evaporator: Evaporate 145 kg water at 100 °C: 145(2260) - 327 700 kJ
Cool X kg of stack gas: X(1.2)(iG-/F)
Heat loss: 33 000Ы

Postheater: Heat Cd 30°: 820(0.24)(30) = 5904
Cool X kg of stack gas: X(1.2)(iF - 150)

Heat loss: 37 000kJ

Terms in the above equations were collected and recast in a convenient form for Solver,
which found a solution as follows:

tH = 100.7 °C; tG = 114.5 °C; f = 146.2 °C; X - 9474 kg.

468 Chapter 8 Enthalpy Balances in Non-Reactive Systems

These results are only slightly affected by minor adjustments in the distribution of heat loss.
The stack gas exit temperature (tH) is above the normal boiling point of water, hence there is no
tendency for water vapor to condense in the preheater section. Once the system material and heat
balance is set up properly, you can examine the effect of several factors that were fixed values in
the problem statement.

Assignment. Examine the effect of increasing the stack gas inlet temperature (tE) by 10° to see how
much less stack gas can be used.

8.4 Energy Efficiency

Example 8.1, the melting of aluminum, showed that not all of the energy input to a system is
used productively. In fact, some energy never gets into the system because it is lost in transit.
Some energy is lost from the system to the surroundings (or is deliberately removed), and some
energy is incompletely transferred from one substance to another. Inefficiencies in the use or
transfer of energy in a process can be a large part of the operating cost. The ratio of the energy that
is used productively by a process, divided by the total energy consumed, is called the energy
efficiency of the process. Where heat is the only energy term involved, the ratio is sometimes
called the thermal efficiency. In Example 8.1, the rate of heat entering the melting furnace in the
hot combustion gas was 270 600 kcal/min, while the heat output in the molten aluminum was 118
600 kcal/min. This is an energy efficiency of 44%.

Thermal efficiency can be increased in various ways. First, adding insulation can cut down
on heat losses. Second, increasing the throughput of material can cut the heat loss per unit of
material processed. Third, modifying the method of producing heat might increase its utilization.
Finally, recovered heat from an output stream can be directed back (i.e., recycled) to the process.
Each of these techniques can be applied to the aluminum melting furnace example.

Adding insulation to a furnace is an obvious way to increase thermal efficiency. The heat loss
for Example 8.1 was 33 450 kcal/min. The furnace probably has some insulation, but if more
insulation could cut the heat loss to 20 000 kcal/min, 13 450 kcal/min more would be made
available to the process. If all of this heat were recovered by the aluminum, the melting rate would
increase to 495 kg/min and the energy efficiency would increase to 49%.

Increasing the throughput of the system is another way to increase thermal efficiency,
providing that the additional heat doesn't just end up making the stack gas hotter. For example,
adding a stirring device to the molten aluminum bath might increase the heat transfer coefficient
from the combustion gas to the bulk aluminum while keeping the stack gas temperature at 800 °C
but at a higher gas throughput rate. The mass flow of aluminum could then be increased such that
the molten aluminum still exits at 700 °C. Suppose that the burner gases and leak air increase 10
%, while the heat loss remains the same at 33 450 kcal/min. The increased mass of aluminum is
calculated by the following material and heat balance.

+10% Material/Heat Balance on Gas Streams, m /min, STP

Gas species C02 H20 o2 N2 Total

Combustion gas 35.2 70.4 12.67 312.51 430.78
— — 9.24 34.76 44
Leak air

Total stack gas 35.2 70.4 21.9 347.3 474.8

#25-#i800 of combustion gas = -297 440 kcal/min

Я1800-Я25 of stack gas = 130 190 kcal/min

Heat loss = 33 450 kcal/min

The net result is that 133 800 kcal/min is available to heat the aluminum, so the throughput
rate of aluminum increases by 13% to 502 kg/min. The thermal efficiency is 45%, not much better
than the original case.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 469

Lowering the amount of excess air to the burner will decrease the amount of heat leaving the
system as sensible heat in the stack gases. The combustion gas temperature will increase, but the
amount of heat entering in the combustion gas will remain the same because the same amount of
CH4 is burned. If this heat is effectively transferred to the aluminum, more can be melted. The
result of the materials and heat balance calculation for 110 % of stoichiometric air is shown in the
ledger below.

110% Stoichiometric Air Material/Heat Balance on Gas Streams. m3/min, STP

Gas species C02 н2о o2 N2 Total
Combustion gas 32.0 64.0 6.4 264.8 367.2
— — 9.24 34.76 44
Leak air

Total stack gas 32.0 64.0 15.6 299.6 411.2

Я25-Я1800 of combustion gas = -297 440 kcal/min
Я1800-Я25 of stack gas = 130 190 kcal/min
Heat loss - 33 450 kcal/min

The net result is that 123 850 kcal/min is available to heat aluminum, so the throughput rate of
aluminum can be increased 5% to 465 kg/min. The thermal efficiency is increased to 46%, again a
small but not insignificant amount.

These balances show that slight modifications to a process can increase the thermal efficiency,
but no single approach has a large effect. In practice, these and other approaches would receive
serious consideration and if justified by the economics, be implemented. One of the approaches
not explored above is to recover heat from one of the output streams and direct it back into the
process. This approach has the potential for large increases in energy efficiency, and is discussed
in detail in the next section.

8.5 Recovery and Recycling of Heat

The aluminum melting furnace example showed that about 40 % of the heat content of the
combustion gas leaves the system as sensible heat in the stack gas. There is thus considerable
potential for improving the thermal efficiency of the process if some of that heat could be returned
to the system. This might be done in two ways. First, the stack gases could be redirected to pass
over the incoming aluminum ingots to increase their temperature before they enter the furnace.
Second, the stack gases could be redirected to a device where they can exchange heat with (i.e.,
preheat) the combustion air. These two approaches are examples of heat exchange, which will be
discussed in detail in the next sections.

8.5.1 Heat Exchange Between Fluids

The recovery of heat by transferring some of the sensible heat in one stream to another is
common practice. The device for doing this is called a heat exchanger, or sometimes a
recuperator, and in the simplest case for gas streams consists of adjacent metallic conduits
(Wikipedia 2010). The two gas streams may move concurrently or counter-currently, but since the
latter arrangement is much more efficient, we will emphasize counter-current heat exchange. A
common type of heat exchanger for gases is the tube-in-shell type, where the hot gas surrounds a
series of tubes containing the gas to be heated. The objective is to have a high surface area to
improve heat transfer efficiency. A sketch of a tube-in-shell heat exchanger is shown in Figure
8.13 for the exchange (recovery) of heat from a stack gas. The tube-in-shell design is just one of
many fluid-to-fluid (gas or liquid) heat exchangers.

470 Chapter 8 Enthalpy Balances in Non-Reactive Systems
Hot stack gas I

Hot Cold
combustion air combustion air

Warm stack gas Ψ
Figure 8.13 Sketch of a counter-current flow tube-in-shell heat exchanger.

The exterior of a heat exchanger is generally insulated to minimize heat loss to the
surroundings, and may approach adiabatic operation. When operating in steady-state, the change
in enthalpy of the hot stream equals the change in enthalpy of the cold stream. The magnitude of
the heat transferred between streams is designated hstr, usually expressed in thermal units per unit
time (such as kcal/min). For steady-state operation, it's simpler to define a unit time, and replace h
with H. When the heat capacity of each stream is (approximately) constant and independent of
temperature, then for any heat exchanger:

Я hx ——.Т*1Т hot —_ TT cold [8.5]
str Γ1
s t r

V z ^ o t / o hot/τ-ί hot rp hot\ v zrcoolldd^ coolldd//Tτ7 cold rp cokk [8.6]
2J" tp (iin -iout )-2J" Cp (7C ^ in )

where F = stream quantity and Cp = heat capacity in appropriate units. Here we use H to

designates the total enthalpy of a stream above some basis temperature (usually 298 K). The
superscripts hot and cold refer to the hot and cold stream, and the subscripts jn and out refer to the
entering and leaving streams. The symbol YJFCP represents the total heat capacity per unit time of
the flowing streams, and is called the thermal mass.

Equation [8.6] is an approximation because Cp is taken to be a constant. This approximation
is valid only when Tm - Tmt is small, or Cp is a weak function of temperature. For accurate
calculations, it is necessary to replace [8.6] by a relation where the heat capacities are temperature
functions:

TT _ fL'm v^1 77hot^ hot /γιhot _ rTout^i y^cold ^, cold 7j, cold [8.7]

To avoid integrating the Cp expression we can use the equivalent enthalpy expression as
defined by Equations [7.15] and [7.21].

Heat exchanger characteristics are best illustrated by noting the temperature profile of the gas
streams along the axis of the exchanger. The "hot" end can always be identified, being that at
which the stream having the highest temperature enters the heat exchanger. The temperature
difference between two streams at a section x through the exchanger is defined as:

ATx = (Τχαοί - T- xcoclmda\ ) [8.8]

In a counter-current operation:

at the hot end: AT*01 = (rinhot- 1rpoutcold\)

at the cold end: ΔΓ°ω = (routho1 - Tinccoo,lad)\

The steady-state heat transfer can be related to A, the exchange area, and U, the heat transfer
coefficient (enthalpy per unit area) as follows:

# h x = UAAT™* [8.9]

Chapter 8 Enthalpy Balances in Non-Reactive Systems 471

where Δ7^Μ is the so-called log mean temperature driving force which is valid when U is constant.
By definition:

ΔΓ LM ΔA7nnhot - AA iT cold [8.10]

cold

If one value of Aris not more than twice the other, then Δ7^Μ values are 4 % or less than the
arithmetic mean of АГоШ and A7*ot.

The stream temperature along the length of a co-current heat exchanger is such that the two
exit streams approach each other's temperature at the exit (cold) end. The stream temperature
profile along the length of a counter-current heat exchanger is more complex. Figure 8.14 shows
this profile for two fluid streams (an undefined "gas", and air) along a counter-current heat
exchanger. The arrangement in the left-hand sketch is for ^ o t C / o t = ^oldC,cold (i.e., the thermal
mass is the same for both streams). The right-hand arrangement is for 1^о1Ср ot > ^oldC/old, which
means that the gas entering the hot end (the "hot" stream) has a larger thermal mass than the air
entering the cold end (the "cold" stream). When ^ o t C / o t - FoldC/old, then Δ7 is the same all
along the axis of the exchanger. This condition is called a balanced heat exchange condition. On
the other hand, when ^ o t C / o t > FoldC/old, then A7*ot < ΔΓοΜ This is an unbalanced heat
exchange condition, where here the thermal mass of the hot stream is greater than the thermal mass
of the cold stream.

Hot end Cold end Hot end Cold end
hot
4 *" hot
t
t f
T
T cold^^^

Figure 8.14 Temperature profiles along the length of balanced and unbalanced counter-current
heat exchangers transferring heat from hot gas to cold air.

For a balanced exchanger and a given value of Hhx, the stream temperature difference AT
along the length decreases with increasing A and U, approaching zero as either A or U increase
without limit. For an unbalanced exchanger and a given value of Я, the value of AT is smaller at
the point where the stream with the greatest thermal mass enters, and approaches zero at that point
as either A or U increase without limit. For an unbalanced exchanger, the stream with the greatest
thermal mass undergoes a smaller change in temperature than the other stream.

A very important principle of any heat exchanger is that at no point along the axis of a heat
exchanger can the temperature of the initially-cold stream exceed that of the initially-hot stream.

EXAMPLE 8.7 — Heat Exchange in a Waste Heat Boiler.

Offgas from a zinc roasting furnace enters the shell of a counter-current tube-in-shell heat
exchanger at 880 °C, and transfers heat to the water/steam fluid in the tubes. When the heat
exchanger tubes are clean, the roaster gas exits at 400 °C. As dust accumulates on the tubes, the
heat transfer coefficient U decreases and the roaster gas exit temperature increases. Boiler water
enters the heat exchanger tubes at 88 °C and 7.8 kg/sec and exits as superheated steam at 60 bar.
The flow rate and composition of the roaster gas is:

33.0m7secSTP; <pN2 = 82.0%; <pSO2=14.0%; φΟ2 = 4.0%

This may not be true if phase changes or chemical reactions are taking place in one or both streams. If this
happens, the device is really a reactor, not a heat exchanger.

472 Chapter 8 Enthalpy Balances in Non-Reactive Systems

Calculate the exit temperature of the superheated steam as a function of the roaster gas exit
temperature. At what roaster gas exit temperature will the steam reach the VLE temperature?
Neglect heat losses from the waste heat boiler.

Data. We need data for the enthalpy change when water at 88 °C is converted to superheated
steam and for the enthalpy change when roaster gas is cooled from 880 °C to its exit temperature.
For steam data, we can use either of the steam property calculators to find that water at 88 °C has
an enthalpy (above 0 °C) of 89 kcal/kg. The VLE temperature for H20 at 60 bar is 276 °C, so
steam must exit above that temperature. Enthalpy data was taken from the calculator at 40 °C
intervals above 280 °C, and 89 kcal/kg was subtracted to account for the enthalpy brought in by the
hot water. Figure 8.15 shows the enthalpy difference between one kg of water at 88 °C and steam
at various superheat temperatures. The Trendline tool was used to develop a quadratic expression
for the enthalpy/temperature relationship. H is expressed as a positive number, meaning that it
represents the amount of heat absorbed by one kg of the water/steam fluid in going from water at
88 °C to superheated steam at 60 bar.

Enthalpy above 88 °C of Sht'd Steam at 60 bar

800 J85.5
"763.1
750 "717.6 '740.6
_^e^694.5
CD
^^^670.6
cu
_^645.2 H = -0.00043r + 1.003f + 338
Ά 317.0 R2 = 0.9991

« 650
600 <JT581.4

550

280 320 360 400 440 480 520 560 600
f,°C

Figure 8.15 Enthalpy of superheated steam at 60 bar above a reference temperature of °C.
Saturation temperature is 276 °C.

FREED was the source of enthalpy data for the three roaster gas species in terms of cal/mol,
which were converted to kcal/m3 (STP). The volume fraction of each gas was multiplied by its
volumetric enthalpy to convert to a basis of one m3 (STP) of roaster gas. Figure 8.16 shows the
results. H is represented by a negative number because it represents the heat given up by one STP

m roaster gas as it cools from 880 °C to the its exit temperature rout °. H is adequately expressed
by a linear equation.

-120 Enthalpy of Roaster Gas (Hsso'^t)
H =0.364/ -326
-130
420 440 460 480 500 520
Q. -140 544,-128
540
co -150

CO
E
CO -160
Ü
.*
-170
a: -180

-190

400

Figure 8.16 Ht-Hsoo of roaster gas above various reference temperatures.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 473

Solution. The application of Equation [8.6] in enthalpy terms expresses the relationship between
the heat transferred from the roaster gas to the water/steam fluid. An Excel worksheet was set up
to make calculations for six different roaster gas exit temperatures. Remember that the roaster gas
is the "hot" stream and the water/steam fluid is the "cold" stream. A sample calculation at /outhot =
400 °C for one second of operation is shown below. Owing to our practice of writing heat
balances with the sign of the heat effect term properly designated, the sum of the two equations
written below must equal zero, i.e., H ° + Hrcold . 0.

Roaster Gas: Я110' = ^ot(tfouthot-) = 33(0.364 x 400 - 326) = -5953 kcal/sec.

Steam: cM = FM(H outcold) = 7.8[-0.00043(Wcold)2 + 1.003(/outcold)+ 338]
H

SuperGoalSeek found solutions at six different roaster gas exit temperatures from 400 to 525
°C as shown in the Figure 8.17. The results show that the thermal mass of the roaster gas is about
triple that of the steam. For example, a 25 degree change in roaster gas exit temperature results in
a 70 degree change in steam temperature.

Waste Heat Boiler Operation

250 425 450 475 500 4000
400 525
roaster gas exit t, °C

Figure 8.17 Steam temperatures as a function of roaster gas temperature for WHB operation.

As dust builds up on the heat exchanger tubes, the rate of heat transfer to the steam decreases.
The roaster gas exit temperature increases and the steam exit temperature decreases. To assure that
only steam exits the boiler tubes, the steam exit temperature must be several degrees above the
VLE of 276 °C, or at least 280 °C. Inserting this value into the text box equation on the figure, the
roaster gas exit temperature should not drop below about 515 °C. At that point, the roaster gas
transfers about 4100 kcal/sec to the steam. The danger of incomplete water vaporization in the
boiler tubes can be alleviated by decreasing the steam pressure or the water flow until the boiler
tubes can be cleaned.

Assignment. Calculate the steam exit temperature for a roaster gas exit temperature of 420 °C and
a steam exit pressure of 80 bar.

Another way to recover and reuse heat from a furnace gas is to use a regenerator, which
alternately absorbs heat from one fluid and transfers it to another. Regenerators are most often
used in furnaces which can be fired from the ends. Two regenerators are used in sequence; one is
being heated while the other is in use. A regenerator consists of a large mass of porous brickwork,
plus a valve system to switch offgas and combustion air between each one. During one cycle, the
hot offgas is passed through one regenerator while combustion air is passed through the other one.
In a second cycle the gas flows are switched, and the previously-heated regenerator transfers its
heat to the combustion air. This is an unsteady-state process, so the flame and combustion air
temperature varies from cycle to cycle. Regenerators (called stoves) are commonly used to heat air
for the iron blast furnace.

474 Chapter 8 Enthalpy Balances in Non-Reactive Systems

8.5.2 Heat Exchange between Solids and Fluids

Many processes require that a solid be heated before it enters a reactor, or cooled before it can
be discharged. In so-called pebble-bed heat exchangers, a moving bed of ceramic solids is used to
exchange heat with a counter-current flow of fluid, usually a gas. This type of device can be used
to remove heat from a dirty or corrosive gas that would adversely affect a simpler heat exchanger.
The high solid surface area results in a high heat transfer coefficient. Considerations similar to
fluid/fluid heat exchange apply when one of the streams is a solid.

Fluid/solid heat exchange is important in the operation of a moving-bed shaft furnace, such as
used in the production of DRI (see for example Figure 6.46). In the top half of the furnace, the hot
spent reducing gas transfers heat to the descending iron oxide pellets, thereby heating them to the
reduction temperature. In another application, the offgas from an electric arc furnace is passed
over incoming scrap in a long tunnel, thus preheating the scrap before it enters the electric furnace.

EXAMPLE 8.8 — Preheating HCl in a Pebble-Bed Vertical Shaft Heat Exchanger.

A pebble-bed heat exchanger uses 60 kg/min of mullite pebbles to heat 40 to 50 kg/min of
HCl(g) from 300 to 900 K. The pebble flow has a much larger thermal mass than the gas flow.
Owing to efficient heat transfer, the HCl exits the top of the heat exchanger shaft only 6 ° cooler
than the incoming pebble temperature. Calculate the pebble exit temperature as a function of the
mass flowrate of HCl being heated. Heat loss from the heat exchanger is estimated at 2400 kJ/min.

Data. FREED is the data source for both mullite (Al6Si20i3) and HCl. The enthalpy required to
heat one mol of HCl from 300 to 900 is 17 870 J, or 490 kJ/kg. FREED's graphics tool plotted the
value of ΗΊ-Η29% from 500 to 900 K, after which Excel's Trendline tool derived the following
quadratic equation (FREED's enthalpy units were converted to kJ/kg):

ЯТ-Я298, kJ/kg mullite = 0.0002047^ + 0.832Γ- 284

Гш1101 for the mullite is 906 K, and ffinhot = 637 kJ/kg, so it's necessary to recast the above
equation to express the enthalpy change of mullite in cooling from 906 К to 7Outhot:

#outhot - #inhot = #outhot - #960 = # h o t = 0.000204(routhot)2 + 0.832(routhot) - 921 kJ/kg
The value of Hhot is a negative number because mullite is giving up heat to the gas.

Solution. Note that mullite is the "hot" stream and HCl is the "cold" stream. As before, the
procedure is to sum the enthalpy terms for heating HCl, cooling mullite, and heat loss, and solve
the equation (using Goal Seek) for the Tou^ot that will give H = zero. For 40 kg/min HCl entering,
the exit temperature of mullite is 583 K. The system balance calculation was repeated for different
amounts of HCl entering. A statistical analysis of the results showed a linear relationship between
the variables:

Pebble exit temperature (deg K) - -7.748(mass flowrate of HCl heated to 900 K) + 893

Assignment. Calculate the mass of mullite required per kg of HCl if the mullite exit temperature is
to be controlled at 550 K.

8.5.3 Application of Heat Recovery Techniques to Aluminum Melting

In Section 8.4 we calculated the heat balance for an aluminum melting process, and noted that
a considerable amount of heat left the system as sensible heat in the stack gas. One way to
increase the thermal efficiency of a process is to use the stack gas to either preheat the combustion
air or preheat the ingots. In either case, heat is transferred back to the system, thereby increasing
its thermal efficiency. We start by calculating the effect of preheating the ingots, then comparing
the increased efficiency so obtained to that obtained by preheating the combustion air.