Handbook_on_Material_and_Energy_Balance_Calculat1

Chapter 8 Enthalpy Balances in Non-Reactive Systems 475

A common device for transferring heat between a gas and a solid is a conveyor system
enclosed in a hood. The offgas is directed down the hood and flows counter-currently to the solid.
Figure 8.18 shows a sketch of such a system.

Stack gas Stack gas
(800 °C) (<800 °C)

Aluminum ingots Aluminum ingots
(<600 °C) (25 °C)

Figure 8.18 Counter-current flow heat exchanger for heat transfer between a gas and solid.

This type of heat exchanger can be incorporated in the aluminum melting process to improve
overall heat utilization. The goal is to transfer as much heat is to the incoming ingots as possible,
while making sure they stay below the melting point of aluminum (660 °C). The melting
flowsheet with heat exchanger is shown in Figure 8.19.

nCH4

(? m3/min, 25 °C)
Hot combustion gas Stack gas Stack gas
(1800 °C) (800 °C) (Tstout)

Air 1 Burner Aluminum Al HX
melting
(118%stoich, 25 °C)

Molten aluminum furnace Tingln Aluminum ingot
(445 kg/min, 700 °C) Air (445 kg/min, 25 °C)

(? m3/min, 25 °C)

Figure 8.19 Aluminum melting process incorporating a heat exchanger for heating ingot with
stack gas. The heat exchanger heats the ingot from 25 °C to /inIng while cooling the stack gas from
800 °C to /0UtSt. All gas quantities diminish as the /inIn8 increases. Volumes are STP.

The amount of heat transferred to the ingots can be varied by the design of the heat exchanger
(for example by varying the retention time). It may even be possible to bring the ingots to the
melting point, but a practical upper limit for tinlng is probably 600 °C to avoid any danger of
melting in the heat exchanger*. At any given location in the heat exchanger, the temperature of the
gas must be greater than the temperature of the solid. If HL = 0, all heat given up by the gas is
gained by the aluminum, and a heat balance can be set up to determine the other process
parameters. We start first with the heat balance for the overall system and then around the heat
exchanger. This requires heat content equations for the gas and aluminum like those used in earlier
examples. In this case, since we know that HL is small (but we don't know what it is), a heat
balance is necessarily approximate, and therefore, a two-term equation for Ht-H25 is adequate.
FREED's Graphics tool produced Ht-H25 equation coefficients for aluminum and air to 550 °C and
the stack gas components to 700 °C. FREED's molar units were converted to STP volume for gas
and mass for aluminum. Data for air is included for use in a later section. The basis temperature is
25 °C and the basis time is one minute. The mass of aluminum is kept at 445 kg, and as more heat
is transferred to the incoming ingot, less fuel is required, and hence the volume of the combustion
gas, in-leak air, and stack gas decrease. Methane is burned with 118 % stoichiometric air.

Hx-His = A + Bt fkcal/m3 for gas, kcal/kg for AL degrees Celsius)

Species 02: N2 н2о co2 Air Al(c)

A -18.3 -15.2 -22.9 ^10.7 -13.8 -15.1

В 0.356 0.334 0.409 0.541 0.334 0.259

We don't yet know which stream has the larger thermal mass: the stack gas or the aluminum.

476 Chapter 8 Enthalpy Balances in Non-Reactive Systems

In Example 8.1, the combustion gas flow was 391.6 m3 and brought in 270 600 kcal. The
enthalpy change on cooling the combustion gas to 25 °C is then -691 kcal/m3. The stack gas flow
was 431.6 m3 (remember 40 m3 air leaked in). Based on the above data and the stack gas
composition, the enthalpy change on heating the stack gas to some outlet temperature /0UtSt is:

#t-#25 stack gas (kcal/m3) - -18.4 + 0.3614(/outst) [8.11]

Four steps are required to make a system balance on the overall process. First, cool the
combustion gas to 25 °C. Second, heat the stack gas to t0UtSi· Third, heat 445 kg of aluminum ingot

to 700 °C. In steps one and two, the volume of the combustion and stack gas streams will be a

variable (called FcmbGas and KstkGas) related by a factor of 1.102. (Remember that the original
combustion gas input was 391.6 m3, and the stack gas output was 40 m3 more, hence the factor of

1.102). The air in-leak is assumed to decrease in proportion to the FcmbGas and FstkGas), but the

furnace heat loss remains 33 450 kcal. The sum of all enthalpy changes is set = 0, and FCmbGas and
^stkGas are calculated as a function of touts\

1. Cool combustion gas to 25 °C: Н25-Ншо = PcmbGas(-691) kcal

2. Heat stack gas to /0UtSt: HrH25 = U02(KCmbGas)(-18.4 + 0.3614 x /0UtSt) kcal

3. Heat and melt 445 kg of Al: Нш-Н25 = 445(266.5) = 118 600 kcal

The sum of these heat terms plus the heat loss is then set = 0.

152,050 \P · [ l J
VCmbGas = 7„ ,1,1-0.3983(i out st)Г

The fourth step is to make a system balance on the heat exchanger device. First, cool the
stack gas from 800 °C to /0UtSt· Second, heat the aluminum from 25 °C to /inIng. Neglect heat loss
from the heat exchanger.

1. Cool stack gas to /0UtSt: #t-#8oo = 1.102(FCmbGas)(18.4 - 0.3614 x 800 - 18.4 + 0.3614 x /outst) kcal/min
2. Heat aluminum ingots to /inIng: #25-#t = 445(-15.1 + 0.259 x /inIng) kcal/min
Now set the sum of these two heat terms equal to 0:

, Ino 6720-FCmbGas(-319 + 0.3983x out / ГЯ 1 i l

in " 115.3
The system balance equations [8.12] and [8.13] can be solved for various values of /0UtSt·

Figure 8.20 shows the results. An interesting result is that when the air preheat temperature is low
(i.e., t[nlng is high), the thermal mass of the stack gas is higher than that of the aluminum. As more

heat is transferred to the ingots, the thermal mass of the aluminum becomes greater than that of the
stack gas. As the heat exchanger becomes more efficient, the toutst decreases, /jnIng increases, and

the volume of required combustion gas decreases. The maximum theoretical heat exchanger
efficiency is where /0UtSt = 25 °C. However, heat transfer rates become vanishingly small as the

two temperatures approach each other, so in practice, one might expect the economic limit of heat
exchanger efficiency when the two streams are at least 75° apart. This is at /0UtSt ~ 100 °C. But to
of /inIng
avoid any danger of ingot melting in the heat exchanger, we set a limit about = 600 °C, which is
at a toutst - 120 °C and a combustion gas volume of about 225 m3, or 57 % of the volume

before adding the heat exchanger. For this condition, the thermal efficiency is about 78%. This

result shows that the addition of an ingot heat exchanger is an excellent way to improve the

thermal efficiency of the process, and much better than adding insulation or increasing throughput.

Another way to increase the thermal efficiency of a process is to use the stack gas to preheat
the combustion air. Heating the combustion air is similar to heating the ingots: heat is transferred
from the stack gas to the system, thereby increasing the thermal efficiency. A heat exchanger of
type of pictured in Figure 8.13 is commonly used to preheat combustion air. A heat balance can
show how much more aluminum (entering at 25 °C) can be melted if the combustion gas volume is

Chapter 8 Enthalpy Balances in Non-Reactive Systems 477

kept at 391.6 m3/min. Preheating the air will increase the temperature of the combustion gas.

However, for heat balance purposes alone , we needn't concern ourselves with the combustion gas

temperature since the amount of natural gas burned is the same, and contributes 270 600 kcal/min.
To this, add the sensible heat contributed by 359.6 m3/min of preheated air. The overall system

balance is expressed in terms of the additional mass of Al melted and the temperature of the
combustion air (7in ir)

Kg Al increase = -tft-tf9<p2o5fAir 359.6(-13.8 +0.334 x/inAir) [8.14]

#700-#25of A1 266.5

where the numerator is the heat content for 359.6 m3 of air, and the denominator is the heat content
for one kg of Al.

Ingot Heat Exchanger Parameters

E 600 Т JOT JOT ^ D ^ JOT - D ^ "3D
550
Ώ 500 —О—VCmbGas, mA3
450 — D — tingln, °C
О
>
ГС 400
σ> 350
300
о 250
сО) 200
150
100

750 650 550 450 350 250 150

stack gas exit temperature, °C

Figure 8.20 Results of system balance calculations for an aluminum melting furnace using stack
gas to preheat the ingot. The solid line represents the volume of combustion gas required as a
function of the temperature of the stack gas leaving the heat exchanger. The dashed line represents
the temperature of the aluminum ingots leaving the heat exchanger (and entering the melting
furnace). The mass of aluminum melted is constant at 445 kg/min.

The tm ir is related to ^out by noting that the enthalpy change of the stack gas equals the
enthalpy change of the air:

Stack gas: Нш-Нх = 431.6(-18.4 + 0.3614 x 800 + 18.4 - 0.3614 x /outst) kcal/min

Air: HrH25 = 359.6(-13.8 + 0.334 x /inAir) kcal/min

Setting the air and stack gas heats equal to each other:

124 700 - 156(70Utst) = -4960 + 120.1(7mAir) [8.15]

The heat balance Equations [8.14] and [8.15] can be solved for various values of/0UtSt- Figure
8.21 shows the results. As the efficiency of the heat exchanger increases, /0UtSt decreases and /inAir

increases, and more aluminum melts. The theoretical maximum efficiency of the heat exchanger
osictcuuartisonwhoecncuersithfierrst,/owutsht erneatcohuetsst 25 °C or /inAir reaches
is about 230 °C. Since 800 °C. The results show that the latter
the rate of heat transfer approaches zero as
Air approaches the stack gas inlet temperature of 800 °C, a practical maximum tm ir will be about

700 °C, which will allow an additional 300 kg/min of aluminum to be melted. This is a 67 %

increase in the melting rate without the air preheater in place. Maximum air preheating increases

the thermal efficiency of the process to about 73%.

As the combustion gas temperature goes up, the amount of NOx in the combustion gas also goes up.

478 Chapter 8 Enthalpy Balances in Non-Reactive Systems

Air Heat Exchanger Parameters

аEa 800 —O-kgAI JOT
—О—tAirPh, °C JüT~
700

600
к. 500
к.
400
<о
300

t-^^200

100

750 650 550 450 350 250 150
stack gas exit temperature, ° С

Figure 8.21 Results of system balance calculations on aluminum melting furnace as a function of
burner air preheat temperature. Burner uses 118 % stoichiometric air. The solid line represents the
increase in mass of Al that can be melted per minute as a function of the temperature of the stack
gas leaving the heat exchanger. The dashed line represents the temperature of the combustion air
leaving the heat exchanger (and entering the burner). The combustion gas volume was constant at
391.6 m3/min.

Environmental problems are caused by NOx in the very hot combustion gas produced by
preheating the combustion air. One way to lower the combustion gas temperature is to recycle
some of the stack gas back into the furnace alongside the combustion gas. Another way to
decrease the NOx level in very hot combustion gas is to minimize the percent of excess air in the
burner. This issue was covered in Section 6.3.8 and will be covered again in Chapter 9.

In summary, heat exchangers can increase the thermal efficiency of a process, but their use
brings added capital and operating cost, and may cause environmental problems. For any process,
there will be a point of maximum cost benefit which will depend on the cost of energy and the
capital and operating cost of the heat exchanger unit.

8.5.4 Heat Exchange Accompanied by Material Transfer

The previous discussion of heat transfer was confined to non-reactive systems where no
material is transferred between the flowing streams. However, sometimes material is intentionally
transferred between streams during heat exchange. There are various reasons to transfer material
during direct contact between a gas and a liquid. For example:

• Cooling and scrubbing a gas. A hot dirty gas is cooled by contact with water, which also
removes most of the particulate from the gas.

• Humidifying or dehumidifying a gas.

• Cooling a liquid.

Generally, each of the above situations involves some sort of heat-transfer tower in which
have dispersion devices to enhance fluid contact. Heat is transferred from the hotter to the cooler
fluid, and additional heat is transferred by evaporation or condensation. Some water is bled to
prevent mineral build-up, and make-up water is added to replace the evaporation and bleed water.
The water phase usually has the greater thermal mass.

Gas-cooling/scrubbing takes place in a vertical structure containing baffles or porous plates to
break up the downward flow of cooling water. If the gas/water mixing is intense, the phases will
reach nearly the same temperature within a short distance from the bottom. Figure 8.22 shows a
sketch of a gas cleaning/cooling tower and a conceptual temperature profile where the thermal
mass of water is much larger than that of the gas.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 479

Scrubber rasMoist Temperature Profile in Cooler
water in о.

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Scrubber Dirty 1 50 100 150 200 250
water out gas in temperature, deg. С

Figure 8.22 Sketch of gas cooling tower and internal temperature profile for gas scrubber.

The actual temperature profile depends on the relative thermal mass of the two phases and
their entry temperature. The gas cools by transferring heat to the water and providing heat for
water vaporization. The temperature profile in Figure 8.22 depicts a scrubber with less intense
phase mixing, so the two phases do not approach each other's temperature until nearly the middle
of the tower. However, if phase mixing is intense the gas temperature will come very close to the
water temperature just a short distance above the bottom. A system balance on such a system can
illustrate some of the characteristics of the process. Suppose 1 mole of dry air at an initial
temperature of 500 °C is cooled by 4 moles of water. It's instructive to find the entering water
temperature (/inwt) such that the heat required to cool the air comes entirely from water
vaporization. There are two consequences of this condition. First, the Л м must equal the ίοηί^.
Second, the water temperature is constant along the tower (i.e., the water temperature profile is a
vertical line in Figure 8.22), so the /outair approaches the /ш^. This leaves us with two unknowns:
the water and moist air temperatures (designated t), and the amount of water vaporized (designated
w). These can be calculated from a heat and material balance.

Using a fixed value for the Cp of water, water vapor, and air keeps the arithmetic simple, as
does picking a basis temperature of 25 °C. Equation [2.15] is satisfactory for the /?H20. The
required data is shown in the following table (units are calories and moles).

Cp air Cp for H20(/) Cp for H20(g) A#vap H20(/) at 25 °C

7.0 18.0 8.1 10 520

The procedure is the same as used previously: cool all inputs to 25 °C, vaporize w moles of
water at 25 C, and heat the products to t. This comprises six steps, the sum of which is set to zero
as one equation. The water vaporization balance relates the amount of water vaporized (w) to the
temperature of moist air leaving the tower (/), and it too is set to zero.

1. Cool 4 moles of water to 25 °C - 4(18)(25 - /)
2. Cool 1 mole of dry air to 25 °C = 7(25 -1)
3. Evaporate water at 25 °C - w(10 520)

480 Chapter 8 Enthalpy Balances in Non-Reactive Systems

4. Heat one mole of dry air to t = 7(7 - 25)
5. Heat w moles of water vapor to t = w(8.1)(7 - 25)
6. Heat 4 - w moles of water to t = 18(4 - w)(t - 25)

/--2256/ .л 055Ч
l /(^+273.15)+0-U:>:>;
10

7. Water vaporization balance. /-2256/ .n . - w=0
/0+273.15)+6Л)55)
1-10

Solver found a solution to the two equations; t = 64.0 °C and w = 0.301 moles of water
vapor*. Four moles of water enter the cooling tower and 3.699 moles exit at the bottom, both at 64
°C. In addition, 4.301 moles of moist air exit the top, also at 64° С The values of t and w are
independent of the amount of air used so long as enough air is present to maintain rapid material
and heat exchange. For this unique situation, the hot air is cooled isothermally in the bottom of
the tower by the evaporation of 0.301 moles of air.

If iinwt is less than 64 °C, then /outair also becomes less than /1П^ but they are no longer the
same; /outair > tmwt. In the extreme case where the entering water is at 0°C, the /?H20 at the top of

the tower is so low that evaporation is a negligible overall contributor to the heat balance. The
system then essentially becomes a heat exchanger without material transfer. ioutair then approaches
/inwt. The hot air transfers 3500 calories to the water when it cools to 0°C, which increases the /outwt

to 48.6 °C. To be sure, water does evaporate in the bottom of the tower as the hot air comes into

contact with it at 48.6 °C, but virtually all of this water vapor condenses as the air rises in the tower
and cools to nearly 0°C (where pH20 = 0.006 atm). These calculations apply only to an extremely

efficient cooling tower. For a real tower, the exit gas will not be completely saturated with water

vapor (i.e., %RH < 100), and the two fluid temperatures may be several degrees apart.

There is no simple way to calculate the /outair and the toutwt in between these extremes. Either
you must know one of these temperatures or use a more elaborate calculational procedure. In any
event, as in the case of a non-material transfer heat exchanger, the air temperature (the hot stream)
can never be less than the water temperature.

Water cooling can also be carried out in a vertical tower. Hot water must sometimes be
cooled before it can be used elsewhere in a process. A slightly different approach is used but the
concept is the same as before; sensible heat from the water is transferred to the air, and additional
sensible heat is removed by evaporation. Hot water enters the top and cool air enters the bottom.
A system balance can tell us how much air is required for a specific water temperature drop.

Suppose water at 45 °C must be cooled to 25 °C before being used elsewhere. Ambient air is
available at 20 °C with 40% RH. As this air rises up through the packed tower, it becomes heated
and humidified, while the down-flowing water becomes cooler. For good mixing and adequate air
residence time, the moist air leaving the top would be nearly saturated with water. But for a less
efficient tower, the top gas is not at 100 %RH. In one situation, a top gas measurement showed
that it was at 41 °C with 95% RH. In this example, we know all of the stream temperatures, so we
can use a system balance to calculate the amount of ambient air needed to cool a certain amount of
water.

The balance requires Cp for water vapor, water and air, and the ΔΗν of water at the 25 °C basis
temperature. Over a short temperature range, Cp for all species is nearly constant. The table shows
FREED values; p(H20) data is from Equation [2.15].

Species Air H20(g) H2o(/)
Cp, J/(mol · deg) 28.6 34.0 75.3

Δ#ν of water at 25 °C = 44 020 J/mol

' Note that/?H20 at 64 °C = 0.231 atm.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 481

The basis is 10 moles of water entering the tower, P=\ atm, and A is the amount of dry air in

the ambient air. A water balance requires some preliminary calculations. At 20 °C, p(H20) at the
VLE = 0.0229 atm, so at 40 %RH, the cooling air/?(H20) = 0.00915 atm. The amount of water
vapor W in A moles of dry air at one atm is calculated by rearranging Equation [2.7]:

W = p(H20)xA [8.16]
l-p(U20)

The amount of W is 0.00923A moles. In a similar calculation, the amount of W in the humid
air leaving the top of the tower is 0.0765A. The water balance equation is then:

Moles water out = 10 - amount evaporated = 10 - A(0.0765 - 0.0092) - 10 - 0.0625A

The system balance steps are: cool the input substances to 25 °C, evaporate W moles of water
at 25 °C, and heat the products. HL is neglected, and enthalpy units are J.

1. Cooll0molesofwaterto25°C: 10(75.3)(25-45) = -15 060

2. Heat cooling air to 25 °C: A(28.6)(25 - 20) + 0.00923A(34.0)(25 - 20) = 145A

3. Evaporate water at 25 °C: 0.0625A(44 020) = 2751A

4. Heat water vapor to 41 °C: A(0.0765)(34.0)(41 - 25) = 55A

5. Heat dry air to 41 °C: A(28.6)(41 - 25) = 458A

5. Heat water: (10-0.0625A)(75.3)(25 - 2 5 ) = 0

The sum of heat terms H must equal 0, which occurs when A = 4.42 moles. Thus 100 STP m3
of entering air at 20 °C will cool 180 kg of water from 45 to 25 °C.

EXAMPLE 8.9 — Lowering the Water Temperaturefrom a Crystallizer.

An aluminum hydrate crystallizer uses cooling water to remove 75 500 Btu/min. A cooling-
tower lowers the cooling water temperature so it can be recycled. The cooling air (stream A)
enters between 35 and 65 %RH. Water is discharged to prevent mineral build-up (stream B).
Figure 8.23 shows the flow arrangement. Calculate the mass of cooling air for four different %RH
values. Let HL = 0.

Hot water (120 °F) 2000 lb/min

75 500 Btu/min

t£+♦+♦+♦+♦+♦+♦+♦+♦+♦+♦+♦ ^ - ^ Humid air
115 °F, 100 %RH
Xt zerЛ Cooling L+++++++++++J
water £+********++4
ft: Packed IV
bed

\t tower ttl
Hi . ., *►♦.
πI Mixer \ V+++++++++++4

*♦♦♦♦♦♦♦♦♦♦♦■ B » Bleed water
}♦♦♦♦♦♦♦♦♦♦♦{ ? lb/min, 85 °F
Cold make- l „ A_I É É ί É ÉfcI
up water Wwaatremr ♦ Cooling air

(50 °F) (85 °F) (80 °F, var. %RH)

Figure 8.23 Sketch of system for lowering the temperature of crystallizer cooling water. Solid
lines indicate water flow and dashed lines indicate gas flow.

482 Chapter 8 Enthalpy Balances in Non-Reactive Systems

Data. The following table shows FREED data, converted to heat capacity units of Btuth/(lb · °F).
The vapor pressure of water is expressed in atm and degrees R.

Species Air H20(g) H20(/)

Cp 0.24 0.45 1.0

Atfvap of H2O(0 at 77 °F - 1050 Btuth/lb

Log(pH20) - -4140/Г+ 6.207

Solution. The typical basis temperature of 25 °C in AES units is 77 °F. The quantity units are
mass for dry air and its water vapor content, which depends on temperature and %RH. The /?H20
= 0.1007 atm at the humid air exit temperature of 574.7 R. Equation [8.17], based on Equation
[8.16], expresses the mass of water vapor per mass of dry air.

mass of water/ />H2O(18.0/Q) [8.17]

/ m a s s of dry air 1 - ^H2О

where 18.0 is Mwater and 29 is Mair. The humid air thus carries out 0.0695 lb water vapor per lb of
dry air.

The/?H20 at the VLE of the cooling air (at 544.7 R) is 0.03436 atm. Equation [8.17] gives
the following table of water content of the ambient air as a function of %RH.

%RH 35 45 55 65 100

lb water/lb dry air 0.00756 0.00975 0.01196 0.01418 0.02209

Material and enthalpy balances are required for each %RH, so the problem is best set up in
Excel. The strategy for making the balances is the same as used for the water-cooling example
worked out earlier; the input streams are cooled to the basis temperature of 77 °F, a certain amount
of water is evaporated, then the output streams are heated. Furthermore, the assumption is that the
exit humid air remains at 115 °F, which probably will not be correct.

First, we make a system balance around the left-hand portion of the flowsheet (crystallizer
cooling, here called the XTL segment). The first four steps around the XTL segment are the heat
balance, and the fifth step is the water balance. The two unknowns are the flow of cold make-up
water F° and the flow of warm water 1^.

System balance around XTL.

1 : Heat stream D to 77 °F. i^(77 - 50) = 27F°

2: Cool stream С to 77 °F. 7^(77 - 85 °F) = - 8 / ^

3: Heat stream F to 120 °F. 2000(120-77) = 86 000

4: Add crystallizer heat. -75 500

5: Water balance: 2000 = 7^ + / ^

Solving, F° = 157 lb/min and F° =1843 lb/min.

There are two more stream variables to be calculated: the flow of dry air (FAir) in streams A
and G and the flow of bleed water (F®) in stream B. The steps for 45 %RH are shown below for
the right-hand portion of the flowsheet (packed bed tower, called the PBT segment). The first
eight steps are the heat balance, and step nine is the water balance.

System balance around PBT.

Cool stream F to 77 °F: 2000(77 - 120) = -86 000
Cool dry cooling air (stream A) to 77 °F: FAir(0.24)(77 - 80) - -0.72FAir
Cool moisture in cooling air: FAir(0.00975)(0.45)(77 - 80) = -0.01FAir
Evaporate water at 77 °F: FAir(0.0695 - 0.00975)(1050) = 62.74FAir

Chapter 8 Enthalpy Balances in Non-Reactive Systems 483

5: Heat dry humid air (stream G) to 115 °F: FAir(0.24)(l 15 - 77) = 9 . 1 2 ^
6: Heat moisture in humid air to 115 °F: FAir(0.0695)(0.45)(l 15 - 77) = 1.19FAir

7: Heat warm water (stream C) to 85 °F: 1843(85 -77) = 14 744
8: Heat bleed water (stream B) to 85 °F: ^ ( 8 5 -77) = SF°
9: Water balance: 0 = F 8 + 1843 + FAir(0.0695) - 2000 - FAir(0.00975)

The above equations were written into a worksheet, and Solver changed the two variables
until H and the water balance around PBT was zero. The result was F® = 99 lb/min and FAir = 974
lb/min. Most of the cooling effect came from vaporizing 58 lb/min of water (calculated from step
#4, above). Similar calculations were made for other %RH values, plus a value at 100 %RH for
comparison purposes. Figure 8.24 shows the results.

Cooling of Crystallizer Water 60

1250

1200

Dair at80 °F 59 I

1150 [.·♦«· О water evap

с ^< 1100 о-.- ^ У Ч *ч>. 4 58 0с

f£i ^f^T. 1£

■я 9 ™ 5 0

Ф to Φ

Ö * 1000 J-J^" * 57 5

8

950 \ ^ ^ 4

900 30 40 50 60 70 80 56
90 100
%RH in cooler air

Figure 8.24 Mass of 80 °F cooler air and mass of water evaporated for cooling 1 ton of water
from 120 °F to 85 °F.

The main effect of %RH is on FAir. This is because almost all of the cooling effect in the PBT
is caused by moisture evaporation. The mass of vaporized water is only a weak function of cooler
air %RH, as is the mass of bleed water (it remains close to 100 lb/min at all %RH).

Assignment. If the mass flow of bleed water was fixed at 50 lb/min you would expect that the
temperature of stream F would increase. How much cooler air is required to bring the warm water
temperature to 85 °F, assuming the humid air temperature leaves at 5 degrees less than the entering
stream F temperature, but still at 100 %RH?

8.6 Multiple-Device System Balances

The system balance examples so far have been relatively simple in that they involved at most
two units (although in one example we divided one unit into three zones), and the materials
balance was very simple. However, sometimes a process involves two or more devices, and
streams move counter-currently, with heat and material transferred between phases. A typical
example of this is the cleaning and cooling of dirty process gases. This is called gas scrubbing,
and is a common way to treat hot dirty gases generated by coal combustion, sulfide roasting, oxide
reduction, incineration, and so forth. Counter-current flow enhances the transfer of both
particulate matter and thermal energy from the gas to the water. The thermal mass of the water is
kept greater than that of the gas, so the gas temperature changes much more than does the water
temperature. Depending on operating conditions and the moisture content of the dirty gas, the gas
may become humidified or dehumidified during scrubbing. The dirty scrubber water is usually

484 Chapter 8 Enthalpy Balances in Non-Reactive Systems

passed to a thickener to settle most of the solid material, and the clear water recycled back to the
scrubber. The slurry from the bottom of the thickener is treated to remove any toxic or hazardous
material, dried, and disposed of appropriately.

In a counter-current scrubber, the clean gas temperature is a function of the other stream
properties. If the clean gas humidity and certain other inlet stream flow rates and compositions are
known, a system balance can be used to calculate the flow rate and composition of the exit streams.

There is a considerable interest in incinerating municipal and industrial trash while recovering
energy from it. The offgas from a trash incinerator is dusty and must be cleaned before discharge.
The hot dusty gas enters a tempering unit and then passes into a water-spray scrubber. In one
incineration process, 80 % of the tempering unit water evaporates, while the rest flows down the
tempering unit walls and collects 20 % of the incoming dust. The tempered gas and hot dirty water
passes to a scrubber, where much of the water vapor condenses and in so doing, removes the rest
of the dust. The warm dirty water and clean gas leave the scrubber at 55 °C. Figure 8.25 shows a
sketch of the flowsheet, with additional information for one second of typical operation.

Tempering spray water (35 °C) Scrubbing spray water (35 °C)

FTSW = 45kg Tempered Fssw = ?kg Warm

dirty clean
gas gas
Hot dirty gas 4* 4» 4*ь^ч Л* Л* Л*Tempering
Scrubbing (55 ° C)
(820 °C)
FHDG = 100m3 chamber chamber

фС02 = 0.11 1 Hot dirty wtr (82 °C) 1 J | Warm dirty wtr (55 °C) FWDW=?kg
φΗ2Ο = 0.18 FDust = 6kg
φ02 = 0.03 FHDW = 9kg

φ Ν2 = 0.68 FDust=1.2kg

FDust = 6kg

Figure 8.25 Flowsheet for incinerator gas cleaning. Most of the dust is removed in the scrubbing
chamber. The warm dirty water is sent to a thickener, and the overflow returned for use as water
sprays in the two units. Gas volumes are STP.

The amount, composition and temperature of the hot dirty gas (HDG) are variables. The
operators seek a way to simulate the process to find the required mass of scrubber spray water
(SSW) that will produce warm clean gas (WCG) at 55 °C. This is done by setting up an Excel
worksheet ledger which contains all the required material and heat terms.

Owing to the long temperature span, three-term HrH25 equations for gases are required,
except for water, where Cp was assumed to be 1 kcal kg-1 deg-1 at all temperatures. Cp for dust was
estimated. FREED was the source of all other data. Temperature is degrees Celsius.

Heat Content Equation Parameters for Gases
Ht-H25 = At2 + Bt + C, kcal/kmol

A ВС
o2 8.85E-04 7.068 -178
N2 7.21E-04 6.769 -169
нc o2о2
1.431E-03 7.770 -194
2.430E-03 9.575 -250

Cp for dust, kcal kg" deg"' = 0.21
Atfvap water at 25 °C, kcal/kmol = 10 520

Chapter 8 Enthalpy Balances in Non-Reactive Systems 485

The first step was to set up a ledger for the tempering chamber. The bold cells contain the
operator-specified variables. The setup ledger is shown below.

HDG Dust TSW HDW
voi, m3 / , ° C
g/mJ kg in kg vap kvatr, kg dust, kg /,°C
100 820 82
60 45 36 1.2

heat loss, kcal 270

Ledger Setup for Input to Tempering Chamber

Stream HDG HDG HDG HDG HDG TSW
Species C02 H20 N2 Dust
Phase gas gas gas o2 xtl н2о

gas liq
45
Quantity 0.4908 0.8031 3.0338 0.1338 6
Unit kmol kmol kmol kmol kg
kg 35
Temp, °C 820 820 820 820 820

The ledger worksheet contains built-in relationships between the fraction of tempering spray
water (TSW) evaporated (80%), the fraction of dust removed in the tempering chamber (20%), and
the HDW temperature (10 % of HDG temperature). These can be changed if desired by editing the
ledger formulas. Next a set of heat and material balance relationships was written for the
tempering and scrubbing units. All the stream quantities in the tempering chamber are known but
not the TGT temperature, which is present in several equations as a quadratic. The TGT
temperature was set at 200 °C as an initial estimate for Goal Seek solving. The tempering unit
system balance involved five steps:

1. Cool the tempering chamber inputs to 25 °C.

2. Evaporate 80 % (36 kg) of the TSW to water vapor at 25 °C.

3. Heat gases and 4.8 kg of dust to TGT temperature (200 °C as a guess).

4. Heat 9 kg of water and 1.2 kg of dust to HDW temperature (82 °C in this example).

5. Incorporate the heat loss (here, HL 270 kcal).

The scrubbing chamber has similar steps. Cool inputs to 25 °C, condense water vapor to VLE
pH20 at 55 °C (0.15 atm), and heat outputs to 55 °C. The scrubbing chamber stream temperatures
are known but not the mass of the SSW, so calculations on the scrubbing chamber become more of
a material balance than a heat balance. The mass of the SSW is calculated such that the heat
balance closes. The ledgers for the two units are shown next page. Goal Seek is used to calculate
the TDG temperature (indicated on ledger by bold cell) such that the sum of heat balance terms
equals zero. This temperature is picked up by the scrubbing unit material/heat balance ledger. The
mass of the SSW can be calculated by cell formulas because there are no unknown quadratic
functions in the scrubbing unit balance. One kg of SSW going from 35 ° to 55 °C absorbs 20 kcal
of heat, so the mass of SSW equals l/20th of the heat effect of all other scrubbing chamber actions.
The chamber ledgers are shown next page.

The example setup also allows Goal Seek to be used in different ways. For example, the TDG
temperature can be specified, and the volume or temperature of the HDG calculated.

486 Chapter 8 Enthalpy Balances in Non-Reactive Systems

Ledger Setup for Input to Scrubt)ing Chamber

Stream TDG TDG TDG TDG TDG HDW HDW SSW
Species C02 H20 N2 Dust H20 Dust H20
Phase gas gas gas o2 xtl liq xtl liq
Quantity 1.2 1462
Unit: gas
Voi fract kg kg
Temp, °C 0.4908 2.8013 3.0338 0.1338 4.8 9
— —
kmol kmol kmol kmol kg kg
2.07% 82 35
7.60% 43.37% 46.97% — —
194
194 194 194 194 82

Ledger Setu]о for Output from Scrubbing Chamber

Stream WCG WCG WCG WCG WDW WDW
Species C02 H20 N2 Dust H20
Phase gas gas gas o2 xtl liq

gas

Quantity 0.4908 0.6456 3.0338 0.1338 6 1501

Unit: kmol kmol kmol kmol kg kg
Voi fract 11.40% 15.00% 70.49% 3.11% —
—

Temp, °C 55 55 55 55 55 55

Tempering Chamber Heat Balance Scrubbing Chamber Mass/Heat Balance

External heat (loss) 270 External heat (loss) 70

Cooling inputs -30347 Cooling TDG & HDW species -9056

Vaporization 21021 Condensation -22 678

Heating outputs 9055 Heating WCG, & WDW dust 983
Net Result Heating condsd water & HDW 1435
J) -29 246
Cooling & heating SSW 1462
Using Goal Seek Mass of SSW

EXAMPLE 8.10 — Condensation of Zinc Vaporfrom a Gas.

Some metals are advantageously produced in the vapor state, and must be condensed in order
to produce useful shapes. Zinc is one of those metals, being reduced by carbon according to:

C(c) + ZnO(c) -+ CO(g) + Zn(g) [8.18]

The product gas must be quenched rapidly to condense the zinc into a pool of liquid because
if cooled too slowly, Equation [8.18] can reverse, and produce a thin layer of ZnO on the outside of
the condensed zinc particles. Liquid Zn can be used to "quench" the gas, and condense the Zn(g)
to Zn(/).

The condensation takes place in stages to minimize the amount of circulation of liquid zinc.
Figure 8.26 shows a sketch of a three-stage condensation process. Determine the flow of Zn(g)
and Zn(/) in each stream, and the amount of heat that must be removed in each unit.

Data. The relevant FREED data tables are shown next page. The values listed under the dHf
column for Zn(g) are the АЯуар of Zn(/). The vapor pressure of Zn(/) is obtained from Equation
[4.1 la] as follows: Log(p°Zn, atm) = -6250/Г+ 5.33.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 487

CO, (g) Zn, (1) Zn,(g)
(Carbon Monoxide) (Zinc) (Zinc)

Unit: (К, g-mole, J) Unit: (K, g-mole, J) Unit: (K, g-mole, J)

T (K) Cp HT-H298 T (K) Cp HT-H298 T (K) Cp НТ-Н298 dHf

298.15 30.17 0 298.15 31.380 0 298.15 20.786 0 130,416

700 31.22 12,011 700 31.380 12,611 700 20.786 8354 120,423
750 31.60 13,581 750 31.380 14,180 750 20.786 9393 119,893
118,304
900 32.63 18,402 900 31.380 18,887 900 20.786 12,511 116,715
1050 33.48 23,362 1050 31.380 23,594 1050 20.786 15,629 115,656
1150 33.95 26,734 1150 31.380 26,732 1150 20.786 17,708

Zn(/) ^L Liquid Zn at 700 К Zn(/)
T=? to refinery
Zn(/)
Zn(/)
700 К 700 К

CO + Zrugj Stage #1 Ί CO + Zn(g) Zn(/)
700 К
1150 К 1! 1050 К 1
i 11050 К » 1- ■CO90+0ZКn(gj)
Zn</) Zn(/) Stage #2 ^
1050 К Щ 900 К 1

900 К tage#3 |-С-°Л?n(g)
Zn(/) 750 К У 750 К

750 К

Figure 8.26 Depiction of a three-stage condenser for Zn(g) from a reduction furnace. Liquid zinc
at 700 К is used in each stage as the condensing medium. Dashed lines are gas streams, solid lines
are liquid streams.

Solution. The problem specifies no basis, so we are free to pick one that will make the arithmetic
easy, such as 2.000 kmol of gas entering stage #1. Then, the gas stream entering stage #1 contains
1.000 kmol of CO and 1.000 kmol of Zn(g). Based on the vapor pressure of Zn and assuming Ptotai
= 1 atm, the VLE temperature of the gas is 1110 K. No specific temperature is chosen for a basis.
Instead, the arithmetic is simplified by making calculations using ΔΓ instead of T.

The first step is to construct as much of the materials balance as possible from the data before
making any heat balance calculations. The mass balance around each condenser stage requires a
value for pZn, from which the amount of Zn(g) leaving the stage can be calculated. The ledger
below shows the results. Amount units are kmol.

Mass Balance Ledger for Gas Streams
Γ,Κ Stage Zn(g) in CO in/out pZn, atm Zn(g) out
1050 #1 1.0000 1.0000 0.2386 0.3133
900 #2 0.3133 1.0000 0.0243 0.0249
750 #3 0.0249 1.0000 0.0010 0.0010

The next step is to calculate the amount of heat required to cool the gas stream from its
entering temperature to the stage temperature, and condense the requisite amount of zinc. Please
note that the A//COnd of Zn(g) is a function of temperature. For example in stage #1, the steps in the
heat balance are:

488 Chapter 8 Enthalpy Balances in Non-Reactive Systems

Step#l: CoolonekmolofCO: Hmo-HU5o = 23 3 6 2 - 2 6 734 - -3370 kJ.
Step #2: Cool one kmol of Znfe): #io5o-#ii5o = 15 629 - 17 708 - -2080 kJ
Step #3: Condense 0.6867 kmol of Znfe): Atfcond = 0.6867(-l 16 715) = 80 150 kJ
Total heat that must be removed: 85 600 kJ.

Therefore, the liquid zinc entering stage #1 must absorb 85 600 kJ in heating from 700 К to
1050 K. The Cp of Zn(/) is 31.38 kJ · kmol"1 · deg"1 so 7.79 mol of Zn(/) are required for stage #1.
These calculations were repeated for the other two stages, with the results shown in the following
ledger. The units are kJ and kmol.

Material and Heat Balance Ledger for Stages

Γ,Κ Stage cool gas cond Zn(g) Zn(/) in Zn(/) out

1050 #1 -5451 -80 146 7.794 8.480

900 #2 -5937 -34 121 6.383 6.671

750 #3 -4898 -2867 4.949 4.972
19.125 20.124
Total Zn(/)

The results show that 19.125 kmol of Zn(/) are required from the reservoir for the
condensation of 0.999 kmol of Zn(g) in three stages. The temperature of Zn(/) returning to the
reservoir is calculated by summing the enthalpy changes for liquid zinc and setting them = 0. In
Equation [8.19], we use the CP(AT) expression for enthalpy change:

0 - 31.38[8.48(Γ- 1050) + 6.671(7- 900) + 4.972(Г- 750)] [8.19]

Solving, the temperature of the Zn(/) stream returning to the reservoir is 926 K. This must be
cooled to 700 К in the reservoir, which requires the removal of 142 720 kJ. This number is the
same as obtained from an overall heat balance: cooling one kmol of CO from 1150 to 750 K,
0.999 kmol of Zn(g) from 1150 to 700 K, and condensing 0.999 kmol of Zn(g) at 700 K.

An actual process would likely involve fluctuations in the gas flow and temperature to the
first stage. In order to effectively condense the Zn(g), it would be desirable to calculate the
required flow of Zn(/) to each stage for any flow, composition, or temperature of the gas stream to
stage #1. The combined material and heat balance calculation could be programmed to make these
calculations to develop a control strategy.

Assignment, a) Suppose the cooler for the reservoir was a heat exchanger employing water
evaporation as the heat transfer medium. Water at 55 °C enters the heat exchanger, and
superheated steam at 20 bar and 250 °C leave the heat exchanger. How many kg of water are
required to maintain the reservoir at 700 K? b) Calculate the amount of Zn(/) required from the
reservoir if a one-stage condenser was used, operating at 750 K.

8.7 Use of FlowBal for System Balances

While using FlowBal for material balances, you probably noticed that the Process
Parameters dialog box has a check box labeled Heat Balance, which is by default unchecked.
When you check this box, or press <Ctrl F9> in an empty cell on a solved material balance
worksheet, FlowBal calculates a heat balance. FlowBaPs heat balance procedure is the same as
discussed earlier in this Chapter in connection with Hess' law, where the basis temperature is
298.15 K. First, each device instream species is cooled to 298.15 K. Next, chemical reactions are
carried out at 298.15 K. Finally, each outstream species is heated to the outstream temperature. If
you've entered a value in the device setup column labeled "Heat", that value is added to the device
heat balance as a gain from or loss to the surroundings. In any event, there'll be a net heat deficit
or surplus for each device.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 489

FlowBal treats every system as if the heat and material balances were uncoupled. There is no
way to couple the two balances in FlowBal. Instead, one or more stream properties are estimated,
and designated as placeholder values. This allows the solution of an interim material and heat
balance. Changing one or more stream properties can bring a device's net heat to zero (or any
other desired value). In the simplest case, you can change any stream property on the output array
that does not affect the material balance, press <Ctrl F9>, and get a new heat balance. A better
alternative is use the Repetitive Solve tool to change one or more stream properties, repeat the
material and heat balances, and display or chart certain device heat values. In either case, it's
possible to graphically to find a stream property that will give a net device heat close to zero (or
whatever value you set). In essence, FlowBal makes a series of interim system balances, charts the
results, creates equations, and if requested, uses the equations to find a stream property value.

This procedure is illustrated by an example of scrubbing a dusty gas coming from the top of
an iron blast furnace. 2090 m3 of dusty gas with 7.9 kg dust per 1000 m3 enters the scrubber at 380
°C along with 3040 kg of scrubber water at 28 °C. The flow quantities represent about 0.01% of a
day's production. The clean gas and scrubber slurry leave at 55 °C. The object is to calculate the
heat loss from the scrubber. Figure 8.27 shows the flowsheet and Figure 8.28 is the starting
FlowBal array. The species in the array must have descriptors to make a heat balance.

Water (28 °C) - 1 - - Ч Clean gas (55 °C)

Dusty BF gas (380 °C) - 2 — ■ ^ ^ ■ ► | Scrubber

y^fr5^-^» Slurrv (55 °c)

Figure 8.27 Flowsheet for scrubbing dusty blast furnace gas. Solid lines represent solid phase
streams, dashed lines are liquid phase streams, and dotted lines are gas streams. The dusty gas and
slurry are two-phase entities, separated into one-phase (paired) streams for material balance
purposes. A cylindrical shape surrounding the paired streams is used to indicate their status.

P(atm) 1.3 1.3 1.3 1.2 1.2 1.2

T(C) 28 380 380 55 55 55

Str-unit Mass (kg) Volume (m3) Mass (kg) volume (m3) Mass (kg) Mass (kg)

Spec-unit Mass pet Volume pet Mass pet Volume pet Mass pet Mass pet

Str-name Serb Water BF gas BF Dust Clean Gas Slry Liq Slry Sid
2 4
Streams 1 3 5 6 R#1
Flow ? ? ?
3040 2090 16.5 1
? ? -1
N2, (g)
CO, (g) 27.7
?

C02, (g) 25 ?

H20, (g) 4.5 ?

H20, (liq) ? ?

Fe203, (c) 80 ?
? ?
C, (graph)

Scrubber (RX) Heat

Instreams Outstreams Reactions
41
5
6

Figure 8.28 FlowBal setup array for material and heat balance on blast furnace gas scrubber. Gas
volumes are actual. The scrubber heat value (i.e., loss or gain) is initially unknown.

The R #1-designated reaction is the evaporation of water in the scrubber. This is a physical,

rather than a chemical, reaction. We know that water is evaporating because the/?H20 at 55 °C is
more than triple the/?H20 in S2. Therefore, the reaction equation coefficients for R #1 (the phase

490 Chapter 8 Enthalpy Balances in Non-Reactive Systems

transformation of H20) show H20(/) as the reactant, and H20(g) as the product. At this point, the
system DOF = +1. Since S4 is presumed to leave the scrubber at water vapor saturation, the
required SR is the equilibrium vapor pressure of water in S4. We can insert Equation [2.15] via
the Insert Equation tool, taking care to check the PP box to represent pH20 in S4 rather than
<jpH20. The inserted equation, in FlowBal format, is shown below. Since we set up FlowBal with
Celsius temperature, we need to add 273 to the inserted equation temperatures because Equation
[2.15] uses degrees K.

|{pS4-H2Q, (g)} - 10*(-2256/(<T-S 4>+273)+6-055) |

The object of the example is to find the heat loss from the scrubber. We know that the
scrubber must have a finite heat loss, but if we leave it unspecified, FlowBal assigns it a value of
zero. FlowBal wrote ten equations (nine material balance and the above SR) for ten unknowns
(nine stream unknowns and one rate of reaction). The system balance was solved, as shown in
Figures 8.29 and 8.30.

P(atm) 1.3 1.3 1.3 1.2 1.2 1.2

T(C) 28 380 380 55 55 55

Str-unit Mass (kg) Volume (m3) Mass (kg) Volume (m3) Mass (kg) Mass (kg)

Spec-unit Mass pet Volume pet Mass pet Volume pet Mass pet Mass pet

Str-name Serb Water BF gas BF Dust Clean Gas Slry Liq Slry Sid

Streams 1 2 3 4 5 6

Flow 3040 2090 16.5 1242 2956 16.5

N2, (g) 0 42.8 0 39.20 0 0

CO, (g) 0 27.7 0 25.37 0 0

C02, (g) 0 25 0 22.90 0 0

H20, (g) 0 4.5 0 12.52 0 0

H20, (liq) 100 0 0 0 100 0

Fe203, (c) 0 0 80 0 0 80

C, (graph) 0 0 20 0 0 20

Figure 8.29 Material balance for scrubbing blast furnace top gas. About 41 kg of water vapor
enter with the BF gas, while 125 kg leave with the clean gas. 84 kg of water evaporate, which is
verified by the R-Rl value of 4.65, which is the molar rate of consumption of H20(/) via R #1.

Device Instreams Outstreams Reactions ΣΙη EOut Σ Surr Σ Rxn Device Net
Scrubber 1 -9053 4 12613 1 48928
2 -143374 5 88519
3 -1216 6 81 -153643 101214 0 48928 -3502

Figure 8.30 Heat balance around scrubber device in kcal. The value of-3502 for the device net
heat value means that the scrubber heat loss must be +3502 kcal.

Now that we know the scrubber heat loss, we can explore the effect of process changes.
Suppose a BF production rate increase caused the offgas flow to be 2300 m3 at 400 °C and 1.3 atm,
with 19.6 kg of dust. In principle, we could make an uncoupled material balance except for the
fact that the stream units are volume, and volume depends on stream temperature, which we don't
know. This type of situation has a coupled material/heat balance DOF = -1 because the outstream
temperature becomes a dependent variable if we specify a net device heat (here, zero). FlowBal
requires that the outstream temperatures be specified even though it's initially an estimate. In
principle, the system is overspecified, which requires a placeholder outstreams temperature. The
placeholder value is varied while making interim system balances until the net device heat « 0.

The procedure is to enter the revised stream values in the starting array, enter 3500 as the
scrubber heat value (positive value means heat loss to the surroundings), leave the outstream gas
temperature (S4) at 55 °C (the placeholder value), and enter formulae for the other outstream
temperatures. Enter =F4 for the S5 and S6 temperatures so that any change in the S4 temperature
will be mimicked by the other two outstream temperatures. Calculate an interim system balance,

Chapter 8 Enthalpy Balances in Non-Reactive Systems 491

and note that the scrubber net heat becomes -15 327 kcal, meaning that the placeholder outstream
value of 55 °C is too low. The dusty gas brought in more heat, so there must be an increase in the
outstreams temperature to compensate.

Click on Repetitive Solve, enter the S4 temperature as the Independent Variable, and select
values between 56 and 59 as the range, with four solves. Select the Scrubber Device Net Heat as
the Output Variable, and enter this in the dialog box by clicking on the little -*\ Select the Update
Estimates box (if it's not already selected), and then click OK. FlowBal will produce a table and
chart of the results, as shown in Figure 8.31.

20000 Results
15000 4
10000 59, 17046

59.5

Figure 8.31 FlowBal chart for scrubber net heat as a function of outstreams temperature with
specific values from Repetitive Solve. The net scrubber heat is zero when the outstreams are
slightly below 57 °C.

Finally, return to the starting array and enter 57 °C as the outstreams temperature, and solve
the system. Figure 8.32 shows the results, and as expected, the scrubber net heat is not exactly
zero—it is 390 kcal. Considering the uncertainties involved in the process, there is no point in
fine-tuning the outstreams temperature to bring the value any closer to zero. For additional
examples, please see the FlowBal User's Guide, especially dealing with heat exchangers.

Repetitive Solve also offers a Target value option, where it will find a value to give a
certain target. If you enter zero for the net heat as a target, FlowBal will find an accurate S4
temperature. However, an outstreams temperature shouldn't be reported to more than two
significant figures.

P(atm) 1.3 1.3 1.3 1.2 1.2 1.2
T(C) 28 400 400 57 57 57

Str-unit Mass (kg) volume (m3) Mass (kg) Volume (m3) Mass (kg) Mass (kg) Figure 8.32 Material
balance results for blast
Spec-unit Mass pet Volume pet Mass pet Volume pet Mass pet Mass pet furnace gas scrubbing
process with balanced
Str-name Serb Water BF gas BF Dust Clean Gas Slry Liq Slry Sid heat effects. 105 kg of
Streams 1 2. 3 4 5 6 water evaporate in the
Flow scrubber. Compare to
3040 2300 19.6 1354 2935 19.6 Figure 8.29.

N2, (g) 0 42.8 0 38.6 0 0

CO, (g) 0 27.7 0 25.0 0 0

C02, (g) 0 25 0 22.6 0 0

H20, (g) 0 4.5 0 13.8 0 0

H20, (liq) 100 0 0 0 100 0

Fe203, (c) 0 0 80 0 0 80

C, (graph) 0 0 20 0 0 20

492 Chapter 8 Enthalpy Balances in Non-Reactive Systems

EXAMPLE 8.11 — Production of Distilled Water.

Some processes need a source of distilled water. When this is not readily available, it can be
produced by injecting combustion gas into a tank of water. This is called submerged combustion
because the burner flame penetrates deep into the water. The water is heated to near its boiling
point, and the burner gas exits saturated with water vapor. This gas is then cooled in a condenser
to recover the distilled water. About 20 % of the feed water is bled to prevent mineral buildup.
Figure 8.33 shows a flowsheet of the process. Use FlowBal to make a system balance for the
production of 70 kg/min of distilled water.

Burner gas- Hot gas -► Cold gas
Feed water - 3~" YCondenserV -► Condensed water
4Boiler
Bleed water W a r m a i r Cold air

Figure 8.33 Flowsheet for the production of distilled water by submerged combustion. Solid lines
represent liquid streams, while dashed lines are gas streams.

Data. The data are integrated into the solution.

Solution. The boiler and condenser are reactors, where the phase transformation for the
vaporization (and condensation) of water occurs. The U-shaped tube labeled HX represents a heat
exchange device inside the condenser. The air in the HX flows counter-currently to the gas.
Figure 8.34 shows the FlowBal starting array for the process.

The system balances are obviously coupled, but FlowBal requires that we proceed as if they
were uncoupled because FlowBal must solve the material balance before it makes a heat balance.
This requires setting placeholders for stream properties that are initially heat-balance-dependent
values. These values are shown in the figure in boxed cells, and are user-selected reasonable
estimates. The S3 temperature depends on the heat balance around the boiler, and we set the S4
temperature equal to the S3 temperature by writing a formula in the S4 temperature cell. We know
the S7 temperature, and plant data indicate that the S8 temperature is 10 degrees cooler than the S3
temperature. However, since we don't know the volume of air required to close the heat balance
around the condenser/heat exchanger devices, it too is a placeholder value.

P (atm) 1 1 1 1 1 1 1 1

T(K) 2030 310 360 360 285 285 280 350

S t r - u n i t Volume (m3) M a SS ( k g ) Volume (m3) M a s s ( k g ) Volume (m3) M a s s ( k g ) Volume (m3) Volume (m3)

Spec-unit Volume pet Mass (kg) Volume pet Mass (kg) Volume pet Mass (kg) Volume (m3) Volume (m3)

Str-name Brnr Gas Feed Wtr Hot Gas Bleed Wtr Cold Gas Cnds Wtr Cold Air Warm Air
Streams 2 3 8
1 ? ? 67 ?
70 | 1500
Flow ? ? ? I4 5 R #1 R #2
?
? ?
C02, (g) 9 ?

H20, (g) 18 ? 1 -1
??
H20, (liq) -1 1

N2, (g) ? ? ?

02, (g) 3 ? ?

N1.580.42 .(g) ??

Boiler (RX) Steam Condenser (RX) Air Cooler (HX)
Instreams Outstreams Reactions
Instreams Outstreams Reactions Heat Heat I Instreams Outstreams Heat
131 200 352
24 6 50 7 80

Figure 8.34 Starting array for the production of 70 kg/min of distilled water by submerged
combustion. Note that absolute mass units are used for S2, S4, and S6, and absolute volume units
used for S7 and S8. R #1 represents the physical reaction for the water vaporization in the boiler,
while R #2 is the water vapor condensation in the condenser. Heat losses are estimates.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 493

FlowBal wrote 16 equations for 19 unknowns, so we need three SR equations. First, we need
two equations that relate the/?H20 to the S3 and S5 temperatures; for that we use Equation [2.15].
Next we need to specify the mass of feed water, or better yet, the fraction of feed water sent to
bleed. Plant practice showed that about 20 % of the feed water must be bled. These were written
with the Insert Equation tool, as shown below in FlowBal format.

{pS5-H2Q, (g)} - 10A(6.055-2256/<T-S 5>) | |{pS3-H2Q, (g)} - 10^(6.055-2256/<T-S 3>) |
{S4-Flow}*5 - {S2-Flowy|

FlowBal had trouble finding a solution because the default starting estimates were quite poor.
When some of the values were replaced with better estimates, it converged. Figure 8.35 shows the
material and heat balance results.

P (atm) 11111111

T(K) 2030 310 360 360 285 285 280 350
Str-unit
Spec-unit Volume (m3) M a s s (kg) Volume (m3) M a s s (kg) Volume (m3) M a s s (kg) Volume (m3) Volume (m3)
Str-name
Volume pet Mass (kg) Volume pet Mass (kg) Volume pet Mass (kg) Volume (m3) Volume (m3)

Bmr Gas Feed Wtr Hot Gas Bleed Wtr Cold Gas Cnds Wtr Cold Air Warm Air

Streams 1 2 3 4 5 6 7 8

Flow 500 76 189 15 58 70 I 1500 | 1875

C02, (g) 9 0 4.2 0 10.8 0 0 0

H20, (g) 18 0 61.4 0 1.4 0 0 0

H20, (liq) 0 76.1 0 15.2 0 70 0 0

N2, (g) 70 0 32.9 0 84.2 0 0 0

0 2 , (g) 3 0 1.4 0 3.6 0 0 0

N1.580.42 (g 0 0 0 0 0 0 1500 1875

Heat Вalance (kcal) Outstreams Reactions Σ Ι η ΣΟιιί Σ Surr Σ Rxn Device Net
Device Instreams
3 3046 1 35,556 -46,533 3987 200 35,556 -6791
Boiler 1 -45,634 4 941
2 -899 5 -237 2 -40,879 -3046 -1161 50 -40,879 -45,036
6 -924
Steam 3 -3046 8 23,554 8262 23,554 0 0 31,816
Condenser Process Net -20,010
Air Cooler 7 8262

Figure 8.35 Material and heat balance for the production of distilled water. Boxed cells are
placeholder values.

We turn out attention first to the boiler, which has a negative net heat. This means that the S3

temperature is too low; if it were higher, more water would evaporate, and bring the net heat closer

Results to zero. However, the
363,8009 vapor pressure of water

increases exponentially

with temperature, so a

small change in S3

temperature will have a

ω large effect on the boiler

CO net heat. FlowBal's
> Repetitive Solve tool

-4000 calculated the effect of the
-6306000, -r67^91 S3 temperature on the
boiler net heat, as shown.
-8000 I The boiler heat balance
360 360.5 361 361.5 362 362.5 363 closes when S3 is about
S3-T 361.3 K.

494 Chapter 8 Enthalpy Balances in Non-Reactive Systems

The next device requiring editing is the air cooler HX that removes heat from the condenser.
Clearly 1500 m3 air is insufficient to remove the required 45 036 kcal. The required air flow can
be found the same way we found the S3-T. Also, the warm air temperature (S8-T) needs to be
changed to 351.3 so it's still 10 ° cooler than S3-T. This requires knowing the relationship
between the volume of S8 for a given volume of S7. The factor is 351.3/280, or 1.255. The
appropriate formula can be entered into the cell containing the S8-flow so that a change in the S7-
flow is correctly reflected in the S8-flow. The same procedure must be followed for the cells
containing the actual volume of air (last row of the output array). A cold air flow (S7) of 2080
mVmin gave a net heat very close to the net heat of the condenser. Figure 8.36 shows the final
results.

P (atm) 1 1 1 1 1 1 1 1

T(K) 2030 310 361.3 361.3 285 285 280 351.3

Str-unit Volume (m3) Mass (kg) Volume (m3) Mass (kg) Volume (m3) Mass (kg) Volume (m3) Volume (m3)

Spec-unit Volume pet Mass (kg) Volume pet Mass (kg) Volume pet Mass (kg) Volume (m3) Volume (m3)

Str-name Bmr Gas Feed Wtr Hot Gas Bleed Wtr Cold Gas Cnds Wtr Cold Air Warm Air

Streams 1 2 3 4 5 6 7 8

Flow 434 78 179 16 51 70 | 2080 | 2610

C02, (g) 9 0 3.9 0 10.8 0 0 0

H20, (g) 18 0 64.7 0 1.4 0 0 0

H20, (liq) 0 78 0 16 0 70 0 0

N2, (g) 70 0 30.1 0 84.2 0 0 0

02, (g) 3 0 1.3 0 3.6 0 0 0

N1.580.42 (g 0 0 0 0 0 0 2080 2610

Heat Вalance (kcal) Outstreams Reactions Σ Ι η EOut Σ Surr Σ Rxn Device Net
Device Instreams
3 2960 1 36,259 -40,520 3940 200 36,259 -120
Boiler 1 -39,603 4 980
2 -917 5 -206 2 -40,879 -2960 -1130 50 -40879 -44,919
6 -924
Steam 3 -2960 8 33,479 11,457 33,479 0 0 44,936
Condenser
Air Cooler 7 11457 Process Net -104

Figure 8.36 Material and closed heat balance for the production of distilled water by submerged
combustion. Boxed cells were originally placeholder values.

The stream amounts array can give a better picture of the material balance than does Figure
8.36 because volume is dependent on temperature. You can use MMV-C to convert all flows to
mass for easy comparison. The burner gas brings in 8.5 kg of water, and 77.6 kg enters as feed
water. 15.5 kg leaves as bleed water, 0.5 kg as water vapor in the cold gas, giving 70 kg as the
desired mass of distilled water (S6).

Assignment. Suppose feed water was available at 365 K. How would this affect the amount of
burner gas required to produce 70 kg of distilled water? Investigate the possibility of replacing
cold air to condense the distilled water with feed water available at 290 K.

8.8 Heat Balances Involving Solution Phases

Chapter 7 mentioned that an enthalpy change occurs when substances isothermally form a
solution. Unfortunately, "heat of solution" (or "heat of mixing") values are not a readily-available
over a range of compositions. Most often, values of the A7/soin require access to a commercial
database program such as FactSage. Here we illustrate how to obtain and use such a value when it
is needed in a heat balance.

We mentioned in Example 8.4 that a metal powder can be produced by atomization of a melt.
This same technology, called granulation, is used to convert all sorts of molten phases to small
particles. In the case of molten slag, a stream of slag is impacted by a jet of water, to quench it into
small spherical shapes. The slurry of water and slag granules is dewatered and filtered. The dry

Chapter 8 Enthalpy Balances in Non-Reactive Systems 495

granules may be used for grit-blasting or as a raw material in making fiberglass. Figure 8.37
shows a sketch of a granulation system. The slurry from the granulator needs to exit warm enough
so that its sensible heat provides most of the energy needed to dry the filter cake.

Make-up
water

Filter cake
to dryer

Molten slag

Figure 8.37 Sketch of flowsheet for granulating a furnace slag. Air is used to vent furnace gases
and water vapor from the granulator.

A system balance requires heat content data for the molten slag as it solidifies and cools to the
slurry temperature. Consider the design of a granulator for a typical EAF steelmaking slag, with
wCaO = 40 %, wFeO = 25 %, wSi02 = 12 %, wMgO = 7 %, wMnO - 4 %, and wAl203 = 4 %. A
typical EAF slag temperature is 1600 °C, with an exit slurry temperature in the range of 50 - 80
°C. Figure 8.38 shows a granulator sketch. A granulator design requires a system balance as a
function of the slurry temperature. The vapor (stream 5) is probably nearly saturated with water
vapor, and leaves the granulator at the slurry temperature. Owing to the small amount of vent air,
assume it is dry, and neglect heat loss.

We need certain data before we can start making a heat balance. First, we need Cp for the
slag, which can be derived from the FactSage system, as was outlined in Section 7.11.3. Next, we
need to know the phases and their amount in the granulated slag. The Equilib program determined
the equilibrium phase assemblage at 30 °C for the abovementioned EAF slag. The results are
shown in the following table.

Vent air Slag phase w,%
F 3 = 0.2 kg/sec CaO 25.6
FeO 25.0
(30°C) 5.6
Granulation water MgAl204 5.4
MgO 4.0
F2 = ? MnO 3.4
(40°C)
Ca2Si04
EAF slag
F1 = 1 kg/sec

(1600°C)

Figure 8.38 Flowsheet for granulator unit of a slag granulation process.

Next, we need to know how much heat is required for the overall process of heating the
phases and forming a slag (as a solution of oxides). We will use the opposite sign for this heat
effect for the amount of heat that the slag evolves when it is cooled. One kg of the above phase
assemblage was set as input to the Equilib program at 30 °C, which calculated the overall AH from
30 ° to 1600 °C. The result was 2218 kJ/kg, for an overall Cp value of 1.41 kJ kg"1 · deg-1. (Note
that this value is not the isothermal A//soin of the phases to form a melt.)

Figure 8.38 indicates a mass basis for air, so it is convenient to obtain a relationship between
temperature and the mass of water vapor in 200 g of dry air. The psychrometric program PsyCalc
was used to obtain the mass of water vapor in 200 g of dry air at 100 % R.H. between 40 ° and 90

496 Chapter 8 Enthalpy Balances in Non-Reactive Systems

°C at Ptotai = 1 atm. Excel's Regression tool was used to find an equation that best represented the
results, which is Equation [8.19]:

gH20 1 [8.19]

200gdryair 0.0025392^-7.9876 xl0~6(r) +12.474/^-0.29891

Finally, we need an average value for Cp of dry air and water between 40 ° and 80 °C, and the
A//vapavg of water in
for the heat balance the same temperature range, all of which is from FREED. A convenient basis
is the slurry temperature (stream 4). The Cpavg of air is 1.0 kJ · kg-1 · deg-1, and
of water is 4.18 kJ · kg-1 · deg-1. The A#vapavg of water between 50 ° and 90 °C is 2330 kJ/kg.

Four steps are involved in making the system balance. The first three are to change the
temperature of all three input streams to the slurry temperature of S4. The fourth step is to
evaporate whatever water exits at the slurry temperature. But before working on the heat balance,
we must calculate the mass of water that evaporates at different slurry temperatures, using
Equation [8.19]. This is shown in the second row of Table 8.4.

Table 8.4 System balance setup and results for calculating the mass of granulator water needed.

Slurry temp. °C (S4) 50 60 70 80

Mass of water vapor (S5), g 17 31 56 111

| Mass of gran water, kg 51.2 25.0 16.1 11.2 \

Steps Heat effeclt,kJ
Cool slag to S4 temp
Heat gran wtr to S4 temp -2186 -2171 -2157 -2143
2141 1875
Heat air to S4 temp 2094 2019
vaporate water at S4 temp 4 10
40 68 258
Sum of heats 0
72 130 0

00

Goal Seek can find the mass of granulator water that causes the sum of all heat effects to be
zero. A guess of 50 was used for all four granulator water mass values, and was entered in the
third row (shown as italic font). Heat effect equations were written for each step, and summed
(last row). Goal Seek varied the values in row three until the sum of heats equaled zero.

Goal Seek found a solution for each slurry temperature. Figure 8.39 shows a chart of the
results. Less granulation water is needed as the slurry temperature increases, for two reasons.
First, the slurry water removes more heat as its temperature increases, and second, more water
evaporates owing to a higher />H20 at higher slurry water temperatures.

8.9 Enthalpy Change During Dissolution of an Electrolyte

Aqueous processing of materials is commonly called hydrometallurgy, which is part of the
field of extractive metallurgy. Hydrometallurgy is typically divided into three general areas:

• Leaching

• Solution concentration and purification

• Metal recovery

Dissolving or precipitating solutes is common to all three areas. We dissolve acids in water to
leach ores, concentrates or residues, and we precipitate solutes to purify solutions or to recover
values. Hydrometallurgical processes generally take place between 25 ° and 60 °C, but some
processes require elevated temperature, even above the boiling point of water, and thus require the
use of autoclaves to confine water to the reactor.

Chapter 8 Enthalpy Balances in Non-Reactive Systems 497

Slag Granulator Parameters

~λ60 120
-o— mass gran wtr

-o— mass vap'd wtr

50 55 60 65 70 75 80

slurry temperature, °C

Figure 8.39 Mass of granulation water (S2) and vaporized water (S5) during granulation of 1 kg
of EAF slag to specified slurry temperature.

Section 7.11.3 introduced techniques for calculating the energy associated with the dissolution
of electrolytes in water. This involves the formation of ionic solution species, which is, strictly
speaking, a chemical reaction. Although this Chapter title excludes chemical reactions, we include
the subject of enthalpy changes from species ionizing in water, and defer the subject of multi-
species ionic reactions to Chapter 9.

The dissolution of an electrolyte in water is a complex process. In essence, the process
consists of four steps:

1. Separating the pure solute into its elemental constituents.
2. Ionizing the elements.
3. Expanding the solvent to make room for the solute ions.
4. Inserting the ionized species to interact with the solvent to form aqueous species.

Generally, the enthalpy change for steps 3 and 4 are difficult to distinguish, so they are treated
as a single step, and in fact, for dilute solutions, step 3 is generally negligible compared to the other
steps. Let's examine the magnitude of each step for the dissolution of HCl(g) in water. HC1 is
known to ionize completely in water at low concentrations. Unfortunately, the FREED database
does not include ionic species, so we must use data from one of the chemistry-based handbooks
(see citations in General References) or one of the commercial thermodynamic databases, as we
did in Section 6.10 and 7.11.3. For this example we use the HSC database to find the stepwise
enthalpy change for the dissolution of one mole of HCl(g) in one kg (55.51 moles) of water at 25
°C. The resulting solution is one molai in HC1.

Step 1: HCl(g) - l№2(g) + ViChig); АЯ°гх = 92.31 kJ [8.20]

Step 2: 1/2H2(g) - H+(g) + e"; АЯ°гх = 1536.24 kJ [8.21]

Step 2: 72Cl2(g) + e - Cl~(g); A//°rx = -233.95 kJ [8.22]
Steps 3 & 4: H+(g) - U\aq); AH°rx = -1536.24 kJ [8.23]

Steps 3 & 4: Cf(g) - СГ(ад); AH°rx = 66.87 kJ [8.24]

498 Chapter 8 Enthalpy Balances in Non-Reactive Systems

The sum of these steps is the change in enthalpy for the dissolution of one mole of HC1 gas
into one kg of water to form a one molai solution of hydrochloric acid (xHCl = 0.0177), all at 25
°C. The overall process can be summarized by a single process reaction:

HCl(g) - U\aq) + θΓ(ας); AH°pwcQSS at 25 °C = -74.77 kJ [8.25]

Notice that the ionization of Vi mole of H2(g) is equal to, but opposite in sign, the dissolution
of one mole of a hydrogen cation into one kg of water. This is because, by definition, the standard
enthalpy change for the dissolution and ionization of H2(g) in water at all temperatures is zero, as
is the standard free energy and entropy change. Also, since the electron is in its standard state, it
too has a zero enthalpy and free energy of formation. The negative sign on /SH°VTOCe&s means that
the dissolution of HCl(g) in water is exothermic, and if carried out with minimal heat loss, the
hydrochloric acid solution temperature will be higher than 25 °C. The enthalpy change for a
dissolution process is a weak function of temperature. For example, the AH°process for the
dissolution of HCl(g) at 50 °C = -78.35 kJ, and at 75 °C is -81.81 kJ. As mentioned in Section
6.10.1, commercial thermodynamic packages use the Criss-Cobble method for calculating the
Ai/°rx of ionic solutions at temperatures above 25 °C when experimental data is not available.

One of the characteristics of the dissolution and ionization of an electrolyte in water is that
table values for the enthalpy change generally refer to the formation of an infinitely dilute solution.
The results of the process described by Equation [8.25] represent an idealized process that is
closely approached at <0.01 molai concentration. Unlike the ionic activity coefficient, the
infinitely dilute enthalpy data can safely be used for somewhat higher concentrations. However, at
higher concentration there is an increased likelihood of ion pairing, and the overall enthalpy of
dissolution will be different than that calculated assuming ideal ionic activity. One of the major
challenges of aqueous chemistry is to develop a method for predicting the ionic strengths of
concentrated solutions that have multiple solutes. Another challenge is to calculate the
thermodynamic properties of ions at elevated temperatures, based on the measured properties at 25
°C. There is a considerable quantity of data available at 25 °C, but less data at higher
temperatures.

It's often necessary to determine the temperature change when an ionic solute dissolves in
water. The procedure for doing this to form a 1 molai solution was outlined in Section 7.11.3 for
substances like KCl(c) and NiCl2(c). For example, we can calculate the Cp of a one molai HC1
solution as being equal to the sum of the Cp of 55.5 moles of water plus the Cp of one mole of
HCl(g); Cp of the 1 molai solution = 4.2 kJ/deg. Based on the AfiPpr0cess of -74.77 kJ from
Equation [8.25] the dissolution of one mole of HCl(g) results in a temperature increase of 18 °C.

However, the enthalpy change for the dissolution of a diprotic HacSid04~li(kaeg)H, 2aSn0d4(/S) 0i4s2~m(aoqr)e.
complicated because three different ionic species form: R+(aq),

Section 6.10.2 showed a method for calculating the molality of each species as a function of the

molality of sulfuric acid. However, the expected departure from ideal behavior meant that there

was no satisfactory way to calculate the molality of the ionic species at 1 molai H2S04(ag). In

such a case, it's instructive to calculate the expected temperature rise for the formation of a 1 molai
H2S04(tfg) solution, based on the formation of 1 molai HS04-(ag), and 1 molai S042'(aq). The

enthalpy of formation of the relevant species from the elements at 25 °C is shown in Table 8.5.

Table 8.5 Enthalpy of formation of sulfuric acid and ionic dissolution products at 25 °C.

Species H2S04(/) W\aq) HS04"(a<7) S042"(a^)

AH°, kJ/mol -814.0 0 -886.9 -909.3

H2S04(/) -»· m\aq) + SO/'iaq); A#° = -95.3kJ [8.26]
H2S04(/) -► U+(aq) + HS04~(eiq); A#° = -72.9kJ [8.27]

Chapter 8 Enthalpy Balances in Non-Reactive Systems 499

At concentrations in the non-ideal range, the predominant ionic species is HS04~(tf#), so
without any additional information the A//°pr0cess for sulfuric acid dissolution is about -80 kJ/mol
of acid. A better estimate is available from the FactSage program Equilib, which uses the Pitzer
non-ideal aqueous solution model to allow for deviations from ionic ideality. Equilib predicts a
A//°Process for one mole of sulfuric acid dissolution in 1 kg of water at 25 °C of-74.0 kJ/mol, with
[HSCV] = 0.78 6, and [S042_] = 0.21 b.

Equilib is particularly advantageous for systems containing more than one solute at once, or
for systems where the solutes form multiple ionic species upon dissolution. This is important in
more concentrated solutions, or where multiple solutes are present. For dilute solutions, the
difference between ideal ionic properties and those calculated via the Pitzer formalism are slight.
For example, in the case of adding one mole of HCl(g) to 55.5 moles of H20(/), with both
materials initially at 25 °C, FactSage displays a A#°pr0cess at 25 °C of-72.9 kJ and a hydrochloric
acid solution Cp of 4.06 kJ/deg. This results in an 18 °C temperature rise. This is the same 18 °C
temperature rise as obtained above. This shows that the enthalpy of dissolution is much less
affected by non-ideality that ionic concentrations.

FactSage's Equilib has a "target" feature that allows a final temperature calculation resulting
from the addition of solute species to water. The final target temperature can be sought for an
overall АЯ of zero (i.e., adiabatic mixing temperature). Consider the preparation of a nickel
plating solution prepared by adding 80 g of H2S04(/) and 120 g of NiS04(c) to one kg of water.
The H2S04 and NiS04 were initially at 26 °C, and the water was initially at 18 °C We want to
know the final adiabatic solution temperature for the mixing process. The FactSage Equilib
program found that the overall AH= 0 at 43.7 °C. The solution had four main ionic species, with a
pH of 0.67. The molality of the ionic solution species is shown in Table 8.6. Notice the two
sulfate species, which form as a result of the double-ionization behavior of sulfuric acid, as
discussed in Section 6.10.2.

Table 8.6 Molality of ionic solutions species in nickel plating solution at 43.7 °C.

Species Ni2+ H+ HSCV S042"
Molality 0.775 0.321 1.311 0.281

So far, our solute examples displayed an exothermic AH°S0\n. However, many ionic solutes
have a small or even positive A#°soin. For example, the dissolution of NaCl(c) is slightly
endothermic. The adiabatic formation of a one molai solution at 25 °C results in a one degree
decrease in temperature.

Weakly ionized solutes have a much lower AH°S0\n per mole of solute than do strongly ionized
solutes. For example, the dissolution of C02 in water at 30 °C to form C02(aq) is -20 kJ/mol,
while that to form H+(aq) and HC03"(aq) is -11.3 kJ/mol. Owing to the small amount dissolved,
the dissolution of C02 causes a negligible temperature change.

8.10 Graphical Representation of a Heat Balance

So far, we have represented the results of heat balance calculations in the form of tables,
graphs, or labels on flowsheets. Sometimes a different kind of representation is useful, in the form
of a diagram where the enthalpy of a particular stream is represented by the width of the line that
represents the stream. We expressed a material balance in this manner, as shown in Figure 4.65 for
Example 4.6. The corresponding diagram for a heat balance is called a Sankey Diagram, and is
drawn in Figure 8.40 for the melting of aluminum as described in Example 8.1. The large fraction
of available heat carried out by the stack gas is clearly visible.

500 Chapter 8 Enthalpy Balances in Non-Reactive Systems

Leak air V\ Heat loss
33.5 MJ/min
(40 m'/min, 25 °C)
Liquid aluminum
0 MJ/min (445 kg/min, 700 °C)

Combustion gas Aluminum 118L.6j&MJJ,/min
(392 m3/min, 1800 °C) melting
furnace v *Stack gas
C=4270.4 MJ/min
(432 m3/min, 800
Al ingot ί± 118.4 MJ/min
(445 kg/min, 25 °C)

0 MJ/min E =

Figure 8.40 Heat flow diagram for the melting of aluminum ingot. The width of each flow
indicator is proportional to the heat content of material in the stream. The heat content of the
aluminum and leak air input streams is zero because they enter at the 25 °C basis temperature, and
hence have no high-temperature heat content.

The Sankey Diagram for the system after the addition of a heat exchanger to heat the ingots
before they enter the melting furnace is shown in Figure 8.41. The volume and temperature of the
gas is lowered by the addition of a heat exchanger.

щLeak air ' Aluminum Heat los;
melting
(40 m3/min, 25 °C) furnace Liquid aluminum
0 MJ/min (445 kg/min, 700 °C)

Combustion gas 118L.6ßMMJJ,/min
(257 m3/min, 1800 °C)
Stack gas
— = C 4177.6 MJ/min ^ . (283 m3/min,

Al ingot Al ingot 800 °C)
(445 kg/min, (445 kg/min 76.6 MJ/min

500 °C) 25 °C)
51.1 MJ/min 0 MJ/min

Stack gas Heat ^
(283 m3/min, exchanger

300 °C)
25.5 MJ/min

Figure 8.41 Heat flow diagram for the melting of aluminum ingot with a heat exchanger to
preheat the ingots. The transfer of heat from the stack gas to the ingots has a large beneficial effect
on the thermal efficiency of the process.

8.11 Summary

This Chapter covered all of the necessary thermodynamic and algebraic tools necessary for
the analysis and solution of energy balances in non-reacting systems. We described various ways
of using FREED's thermodynamic data and the steam table calculator, and showed how to modify
the data for different purposes. The essentials of steam and water properties were derived from

Chapter 8 Enthalpy Balances in Non-Reactive Systems 501

steam table data and expressed as equations. Data from these equations or from the steam program
calculators were used in making heat balances. Heat balances were made on open and closed
systems, and under adiabatic conditions. The magnitude of potential and kinetic energy terms was
found to be small, and can usually be neglected in comparison to the enthalpy terms, especially in
view of the difficulty of knowing the heat loss accurately. We examined the best way to select a
basis temperature, and showed that for state variables, the choice of basis temperature did not
affect the final result.

We introduced the concept of coupled vs. uncoupled material and heat balances, and showed
that if the process is uncoupled, the material balance should always be made first. The two
balances must be solved together if they are coupled. The term system balance was introduced to
designate a combined (coupled) heat and material balance.

We examined the enthalpy and phase changes that take place when a moist gas is
dehumidified, or a dry gas is humidified. Psychrometric tables and psychrometric software are
very useful in calculating the mass and enthalpy balance for heating and cooling moist air. The
capture and use of heat by a heat exchanger showed how effective such devices are in increasing
the thermal efficiency of a process. Counter-current heat exchangers are the most effective types
in transferring heat from one stream to the other.

FlowBal can be used to make a heat balance on a process after it makes a material balance. If
the material and heat balances are uncoupled, the heat balance can be used to calculate the heat
loss. If the balances are coupled, the material balance requires one or more placeholder values that
FlowBal will vary in order to close the heat balance. FlowBal does this by making a series of
interim system balances from which the actual stream property can be deduced graphically, by
interpolation, or quantitatively by fitting the system balance result to an equation.

Calculations involving counter-current scrubbers involve both heat and material transfer
between streams. It can be difficult to calculate exit stream compositions without including a heat
transfer term unless the counter-currently moving streams are well-mixed.

The enthalpy change accompanying the dissolution of ionic species in water requires some
knowledge of the concentration of the ionic species formed. If this information is not available in
the usual handbooks, one must resort to the use of computerized database calculation programs,
such as FactSage or HSC. However, the algorithms used by such programs may be significantly in
error at higher solute concentrations, or when multiple species are present.

System balances are commonly summarized in a ledger format. A Sankey diagram is a visual
way to display the heat loss, and relative the amount of energy contained in various input and
output streams is visible at a glance.

References and Further Reading

Luyben, William L., and Wenzel, Leonard A., Chemical Process Analysis: Mass and Energy
Balances, Prentice Hall, 1988.
Universal Industrial Gases Inc. Technology, Gas Properties and Tools.
http://www.uigi.com/technology.html.
Veverka, Vladimir V., and Madron, Frantisek, Material and Energy Balancing in the Process
Industries: from Microscopic Balances to Large Plants, Elsevier Science B.V., 1997.

Wikipedia contributors, "Psychrometrics", "Cooling tower", "Heat exchanger", "Shell and tube
heat exchanger", "Log mean temperature difference". Wikipedia, The Free Encyclopedia,
December 2010, http://en.wikipedia.org/wiki/Main_Page.

502 Chapter 8 Enthalpy Balances in Non-Reactive Systems

Exercises

8.1. Two moles of a steam/N2 mixture containing <pN2 = 0.500 are initially at 400 K. How much
heat should be removed to bring the mixture to the dew point? P = 1.2 bar.

8.2. Aluminum is being heated by an induction furnace that operates at one kW, and transfers 90%
of the electrical energy to the aluminum. The aluminum enters the furnace at 25 °C, at one kg/min.
Calculate the temperature of the aluminum as it leaves the furnace.

8.3. One tonne of molten zinc at 800 К is poured into a crucible containing some solid zinc at 500
K. Calculate the maximum mass of solid zinc that can be melted by the liquid zinc. The walls of
the crucible absorb 1000 kcal.

8.4. Steel parts (assume pure Fe) initially at 800 °C are quenched in oil. The specific heat of the
oil is 2.10 kJ kg-1 · deg-1, and is initially at 28 °C. How much oil should be present in the quench
tank to attain a final oil/steel temperature of 65 °C?

8.5. The offgas from an electric steelmaking furnace is at 1550 °C, and consists of <pCO = 0.85,
<pC02 = 0.08, balance N2. The gases pass into a waste heat boiler and exit at 200 °C. Water enters
the boiler tubes at 55 °C and exits as saturated steam. Calculate the mass of water required per m3
of gas (STP) cooled as a function of saturated steam temperature between 250 and 280 °C.
Neglect heat loss.

8.6. A cylindrical alumina crucible has outside dimensions of 17 cm diameter and 30 cm height.
The wall thickness of the crucible is 1 cm. The alumina is pure α-alumina at 95% of theoretical
density, and is preheated to 300 °C. Molten cobalt is poured into the crucible to a depth of 20 cm.
The aim is to have a final cobalt temperature of 1500 °C. Calculate the required initial cobalt
temperature. The specific gravity of liquid cobalt is 7.76. Neglect heat loss.

8.7. A wet copper concentrate initially at 35 °C consists of chalcopyrite with wH20 = 0.095. It is
to be dried in a fluidized bed reactor by superheated steam entering at 240 °C and 20 bar. The
discharge temperature of the dry chalcopyrite and steam is 115 °C and 1 bar. Calculate the mass of
steam required per kg of moist chalcopyrite.

8.8. Saturated steam is used to preheat 300 kg/hr of fuel oil from 25 °C to 110 °C. The specific
heat of the oil is 1.6 + 0.004(7), kJ kg-1 · deg-1. A sketch of the oil heater is shown below.
Calculate the mass flow of steam required.

Sat'd steam I
(4 bar) | \ И

л.. —^ Ш—

(зоO°ilс) -щ^Ы. Ы m- ^(.11O0il°C)

Ψ Water
(4 bar, 55 °C)

8.9. Scrap composite Mg-C material of wMg = 0.11 is reclaimed by batch vacuum evaporation of
the Mg. The objective is to start the process with the composite at a sufficiently high temperature
so that all of the Mg vaporizes before the composite reaches the melting point of Mg. Assuming
adiabatic evaporation, what should the initial composite temperature be?

8.10. Tuyere air for the iron blast furnace is initially at 80 °F and 75 % relative humidity. It is
heated in a stove to 2100 °F and delivered to the tuyeres. At the tuyeres, pure 0 2 at 65 °F and
steam at 350 °F are added at the rate of 0.1 lb of 0 2 and 0.05 lb of steam per lb of tuyere air.
Calculate the temperature of the gas entering the blast furnace.

8.11. Warm moist air can be dehumidified by mixing cold dry air with it. A moist air stream at 33
°C and 80 %RH is to be dehumidified by mixing with air at 0°C and 100 %RH. Calculate the

Chapter 8 Enthalpy Balances in Non-Reactive Systems 503

mass Qf cold air that should be added to 1 kg of warm air to condense the largest amount of water.
P = 95kPa.

8.12. A continuous process operates to clean and dehumidify the top gas from an iron ore
reduction furnace. The gas exits the blast furnace at 380 °C and 1.8 atm, its composition is <jpH2 =
0.55, balance H20, and its concentration is 25 g of Fe203 per actual m3 of gas. The top gas is
quenched by passing it into an ice bed from which it exits at 0 °C. A slurry containing all the
Fe203 plus the condensed water from the gas and water from the melted ice exits at 0 °C. What
mass of ice (at 0 °C) is required per 1000 actual m3 of dusty top gas?

8.13. A furnace heats 46,800 kg/h steel billets from 25 °C to 1180 °C using 7000 m3/hr (STP) CH4
as a fuel, with 20 % excess combustion air. A sketch of the process is shown below. Calculate the
thermal efficiency (%) and the heat loss from the furnace in kcal/sec.

Comb, gas

% excess air ^ (1800 °C)

(125 °C) ^ HD inl lr lJ JfFlj Steel heating ^ Stack g
furnace -^1230°
CM, ^
4 7 X^000^ L Steel
(25 °C)
(25 °C) St«ве! ^^

(1180 WC)

8.14. A heat exchanger is added to the above process such that the stack gas transfers half of its
sensible heat to the air. The billet heating rate and the heat loss remain the same. Make a system
balance for the process, and calculate the thermal efficiency.

8.15. Five moles of gas consisting of <pCl2 = 0.400; <pHCl = 0.200; <pSiCl4 = 0.100; <pSiHCl3 =
0.300 are initially at 800 K. Five moles of N2 initially at 300 К are mixed with the gas under
adiabatic conditions. Calculate the resulting gas temperature accurately.

8.16. A spent iron ore reduction gas has the following composition:

<pCO2 = 0.18; фСО = 0.21; φΗ2 = 0.31; balance H20

The gas enters a spray quench tower at 550 °C, where water at 5°C is used to cool (and
eventually dehumidify) it. Gas volumes are actual. Make a system balance for the cooling process
to calculate the actual m3 of gas leaving the cooler, starting at an exit temperature of 450 °C and
thereafter at 50° temperature intervals down to the dew point. Continue the material and heat
balance calculations at 10 °C intervals from the dew point to 20 °C, calculating the mass of water
and actual m3 of gas leaving the quench tower. The stack gas (and water, if any) leave the tower at
the same temperature. P = 2 bar. Neglect heat loss.

Water
(5°C)

Stack gas Quench
(100m3/secT tower Stack gas
2 bar, 550 °C)

Water

8.17. A spent iron ore reduction gas is quenched to it's dew point temperature by recycled 25°C
water. The gas is filtered to remove all the dust, then chilled to 25 °C before being sent back for
recycling. A heat exchanger cools the water to 8 °C before it is reused in the chiller. Make a

504 Chapter 8 Enthalpy Balances in Non-Reactive Systems

system balance on the process. Dashed lines indicate water flow and dotted lines indicate gas
flow.

Cold
water
(8°C)

Spent Tempering Filter Gas chilling Clean DRI gas
DRI gas chamber Dust chamber "^for recycle
(620 °C)'
(25 °C)
F-G° *asa .= 100mJ
>ИГ
φΟΟ2 = 0.11; φΗ2Ο = 0.28
φΗ2 = 0.38; φΟΟ = ? T—гт
FDust = 9kg J ii

▼ J
Bleed water (25 °C)

8.18. Hot air at 500 °C is to be cooled in a counter-current packed tower. The hot air contains
anywhere from 0 to 0.25 moles of water vapor per mole of dry air. The cooling water temperature
ranges from 5° to 30 °C. Simulate the process with multiple equilibrium stages to determine the
effect of water vapor content of the hot air and cooling water temperature. Pick a basis of four
times as much cooling water amount as the dry part of the hot air.

8.19. Water-atomized copper powder slurry is filtered to produce a filter cake containing wH20 =
9.5 % at 48 °C. The filter cake is to be dried in a fluidized bed dryer with warm air, which is
available at temperatures between 30 °C and 78 °C. Use FlowBal to calculate the mass of filter
cake that can be dried to wH20 = 0.5 % as a function of the air temperature. Neglect heat loss, and
assume that the exit gas is saturated with water vapor.

8.20. One mole of NH4Cl(c) is dissolved into one kg of water. The solute ionizes completely to
NH4+(ag) and C\~{aq). Consult a chemistry text book or one of the General References to obtain
necessary data, and calculate the change in temperature under adiabatic conditions.

CHAPTER 9

System Balances on Reactive Processes

Chemical reactions play an important part in most material processing flowsheets. In many cases,
the only way to provide thermal energy for a process is to use the heat of a chemical reaction. In
an iron blast furnace, for example, there is no way to transfer heat from the surroundings to the
furnace, nor is there any way to supply electrical heat to the charge. All of the necessary heat
comes from chemical reactions taking place in the furnace or from the sensible heat in the input
streams. However, even there, the input streams derive their thermal energy by the combustion of
a fuel. This Chapter will introduce techniques for including various forms of heat (sensible heat,
heat of transformation, heat of reaction) with the material balance equations to calculate both
balances for a process. A combined material and heat balance is called a system balance.

As we showed in Chapter 6, every chemical reaction requires information about the progress
of the reaction. In one type of situation, the reaction progress is specified by stating either that it
proceeds to equilibrium, or some fraction of the way there, or that there is a specified extent of
reaction of a certain reactant species. In another type of situation, sufficient stream data is given
such that the reaction progress can be calculated from the system balance. Thus a system balance
for a reactive process requires one additional constraint for every chemical reaction.

Before starting on the main topic of this Chapter, let's revisit an issue we dealt with in Section
4.6.3: why bother making material balance calculations when we can analyze every stream and get
a material balance that way? Our first conclusion was that streams are often difficult to sample
properly, which means that a sample-based material balance will not close. Second, a main
objective of a material balance is to determine how changes in stream properties affect the
properties of all the other streams. This is impossible with a sample-based material balance
because it doesn't contain any stream relationships. The best approach is to take the most accurate
samples from the fewest streams such that DOF = 0, and calculate the properties of the rest.

The same issue arises in making a heat balance: why bother making heat balance calculations
when we can measure the temperature of each stream and calculate their heat contents? The
answer is more obvious in this case because we cannot measure enthalpy, we can only calculate
it's change by knowing the flow, composition, and temperature. As mentioned earlier, it's often
difficult to make accurate stream flow and temperature measurements in hot and aggressive
environments. It's often more accurate to calculate heat losses than to measure them, so no matter
how accurately we measure stream properties, we must inevitably resort to first law calculations.
We often want to control a stream or process temperature, but unless the process is heated
electrically, we must rely on an indirect method for temperature control. A main objective of this
Chapter is to use both material and heat balances to determine how a change in one or more stream
properties affects the properties of all of the other streams. This information can suggest ways to
decrease energy use, and lead to a process control strategy.

9.1 Thermal Constraints on a Material Balance

The Chapter 7 heat balances were on very simple systems. In Chapter 8, we restricted our
heat balances to non-reactive processes. This Chapter extends the principles outlined in those two
Chapters but with the inclusion of the heat effect caused by chemical reactions. Owing to the
possible existence of large reaction heats, small changes in input stream flows can have a large
effect on the net heat produced in a reactor. A reactor generally has one of two possible situations.

506 Chapter 9 System Balances on Reactive Processes

* A heat balance shows that the device is thermally sufficient. The sensible heat of the
instreams plus the chemical reaction heat provide enough, or more than enough, heat for
the process. We can either adjust the material flows to decrease the heat input to attain
heat balance closure, or find some way to extract the "surplus" heat. Some of the ways to
do this are to recycle cooled products to the reactor, introduce an inert substance to absorb
some of the heat, or to add a heat exchanger.

* A heat balance shows that the device is thermally deficient. If the deficiency is small,
we may be able to cut the heat loss to close the heat balance. If that's not feasible, we
must vary stream flow or composition, or include additional exothermically-reactive
substances to provide the necessary heat.

9.1.1 Uncoupled System Balances

Most heat and material balances are coupled, which means that the system balance includes
both material and heat balance equations. However, in a few cases, such as when the objective is
mainly to calculate the heat loss or a stream temperature, the two balances may be uncoupled. As
an example of a thermally sufficient uncoupled heat balance, consider a process for the oxidation
of S02 to S03 (called here the SOX reaction), carried out in a catalytic reactor:* You should be
very familiar with this reaction because it's been used in earlier Chapters (see for example Section
6.2.1) to illustrate stoichiometry and material balance principles. Figure 9.1 shows the flowsheet
and Equation [9.1] is the S02 oxidation reaction.

F1 = 10/7/sec !_ Catalytic reactor F2 = ?
(565 K) -►<pS02 = ?
<pSO2 = 10%, <pH2O = 20%

φΟ2 = 10%, bai. N2

Figure 9.1 Flowsheet for the catalytic oxidation of the S02 in a process offgas stream.

S02(g) + 1/202(g) -► S03(g); АЯ°гх at 298.15 К = -26 640 cal/mol [9.1]

FREED's Reaction tool was used to evaluate the likely reaction progress, and to check the

overall heat effect for reactants initially at 298 K. Our analysis begins with the assumption of

equilibrium attainment at the stream 2 exit temperature of 565 K, so the first thing to check is the
value of Кщ. At 565 K, it is a large number — on the order of 20 000, so Equation [9.1] at this

temperature is essentially spontaneous and complete (i.e., X7?S02 = 1 at equilibrium). This means

that the % S02 in stream 2 is essentially zero, so exactly one mole of S02 is oxidized per second.
The material balance is simple; F2 = 9.5 mol/sec. The next step is to use the Reaction tool to

calculate the net device heat as a function of the outstream temperature. The net device heat refers

to the amount of heat produced by the chemical reaction at 298 K, plus the heat necessary to heat

the products to a certain temperature, while neglecting the heat loss. Figure 9.2 shows the FREED

Reaction tool chart for the oxidation of one mole of S02.

The Reaction tool shows that the adiabatic reaction temperature is about 600 K. Since the
actual reaction outstream temperature is 565 K, the difference must be caused by heat loss. Based
on the Trendline equation, the sum of reaction heat plus heat content of products at 565 К is -3150
cal/sec. Therefore, HL = 3150 cal/sec.

* The S02 in process gases must be oxidized to S03 before they can be sent to the sulfuric acid plant.

Chapter 9 System Balances on Reactive Processes 507

Heat Balance for S 0 2 Oxidation

5000

CM о I-- Δ Net device heat -{- ^f^
Linear (Net device heat)
8
1.
I -5000

8 -10000 I II "^!
I II
I I\

g -15000 netheat=77.85T-47,140

% -20000 . _ + 4. 4
с
i I I i !I
-25000
300 350 400 450 500 550 600 650
T(K)

Figure 9.2 Thermal effect for the oxidation of one mole of S02 in a process gas to form one mole
of S03. The net heat is the sum of the heat of oxidation of one mole of S02 at 298 К plus the heat
content of the product gas (containing 10.53 %S03), and is a nearly linear function of temperature.
The text box equation was obtained from the Trendline tool.

For a process gas with a higher composition of S02, the exothermic nature of the SOX
reaction can produce more heat than desired. If the reaction temperature goes too high, the catalyst
may be damaged. Also, ZRS02 becomes significantly less than one when the value of Кщ goes
below about 100. The Reaction tool table shows that this happens at about 720 K. We will

examine how to control a high S02-content SOX reactor temperature in a later section.

An example of a thermally-deficient process is a conceptual design of a DRI producer that

uses a СО-rich gas to reduce Fe203 to iron in a fluidized bed reactor at 1200 K. Figure 9.3 shows a
sketch of the process. In order to assure complete reduction of the Fe203, the minimum <pCO in
the offgas at 1200 К is 0.75. We want to know if it is thermally possible to operate this process at

<pC02 = 0.75 in the offgas, with stream and process temperatures as described in Figure 9.3. The
heat loss is 11 kJ/mol of Fe203 reduced. Equation [9.2] shows the chemical reaction for the overall
reduction of one mole of Fe203. A#°rx is from FREED's Reaction tool.

3CO(g) + Fe203(c) -> 3C02(g) + 2Fe(c); АЯ°гх at 1200 К = -35.44 kJ [9.2]

Fe2O3(600K) Fluid bed A . - ^ Offgas (1200 K)
reactor φ CO = 0.75; <pCO2 = 0.25
Rdn gas (1300 K) _ 2 I (1200 K)
φ CO = 0.96; фСО2 = 0.04~ -> DRI (1200 K)

Figure 9.3 Sketch of fluidized bed process for the reduction of hematite to iron by a CO-rich
reducing gas. Solids are represented by solid lines, and gases by dashed lines. The DRI is pure Fe.

The first step is to make a material balance for the reduction of one mole of Fe203 to produce
2 moles of Fe, according to the listed reduction gas and offgas compositions. We start by noting
that an oxygen balance shows rcC02out = nC02in + 3, and the nCO°ui = пС<Уп - 3. We also know
the ratio of the reduction gas and offgas species, which with a carbon balance gives us enough
information to calculate the amount of each gas species in and out.

пС<Уп= 13.714; «C02in = 0.517; nCO°ut= 10.714; rcC02out = 3.517

The next step is to make a heat balance, which, as we've seen, is uncoupled from the material
balance. This requires selection of a basis temperature. 1200 К seems appropriate because that's

508 Chapter 9 System Balances on Reactive Processes

the reaction and outstream temperature. The heat balance steps are: (1) heat one mole of Fe203
from 600 to 1200 K; (2) cool 14.231 moles of R-gas from 1300 to 1200 K; (3) carry out reactions
at 1200 K. From FREED's tables:

1. Heat Fe203: #12оо-Я60о = 128.9 - 37.8 = 91.1 kJ

2. Cool R-gas: #i20o-#i3oo = 13.714(28.44 - 31.87) + 0.517(44.5 - 50.2) = -50.0 kJ

3. Reduction at 1200 К: АЯ°ГХ = -35.4 kJ

4. # L = l l k J
5. Overall heat effect: ΣΗ = 16.7 kJ.

The heat balance does not close; a positive ΣΗ value means that the process has inadequate
thermal input to maintain the reactor at 1200 K. Therefore, additional heat must be supplied to
bring ΣΗ to zero. There are various ways to provide the necessary heat. First, by increasing the
temperature of one or both of the instreams. Second, increasing the flow of the reduction gas
(stream #2), which will increase the outstream (jpCO. Third, decrease the amount of Fe203 reduced
and add a little oxygen to the reactor. Fourth, operate at a lower temperature. The first option is
the only one that keeps the heat balance uncoupled.

Suppose we increase the hematite temperature. Step one in the heat balance shows that the
average Cp of hematite between 600 and 1200 К is about 0.152 kJ · mol-1 · deg-1. Therefore, by
increasing the hematite temperature by about 110 degrees, the heat balance closes.

This simple example shows a method for analyzing an uncoupled process by first solving the
material balance, and then the heat balance. Once the net device or process heat is calculated, it's
possible to examine which instreams, reactions, or outstreams can be modified to bring the net heat
close to zero. However, if any modifications cause a stream composition or flow to change, the
balances become coupled, and must be solved together.

Processes are rarely so simple as the two discussed above. Usually there are other substances
present, which may mean additional chemical reactions. Or, a fuel must be burned to provide the
necessary process heat, which introduces combustion products to the system. Separate material
and heat balances for a coupled process can give useful information, but if we know beforehand
which stream variable we want to control, it's better to make a system balance that combines the
material and heat balance equations.

9.1.2 Strategy for Coupled System Balances

Chapter 6 contained material balance examples on several different reactive processes. Any
one of these could be good examples of coupled system balances. For example, look back at the
material balance for the zinc concentrate roaster described in Example 6.14, and the depiction of a
zinc concentrate roaster shown in Figure 4.3. One tonne of concentrate (dry basis) was roasted per
minute with oxygen-enriched air (φ02 = 24 %), with added spray water to maintain the roaster
temperature at 1150 K. The main temperature control system consists of boiler tubes that line the
bottom half of the roaster. These tubes extract heat from the roaster as superheated steam. Over
time, these tubes become encrusted with scale on the inside and out, and aren't able to extract
sufficient heat to keep the roaster at 1150 K. As a consequence, more spray water must be added
as a coolant. A system balance can relate the amount of heat removed by the boiler tubes to the
amount of spray water and air needed to control the temperature.

Even when the heat and material balances are coupled, it can often be useful to uncouple them
by assuming values for a sufficient number of stream variables such that DOF = 0 for the material
balance alone. These assumed values are called placeholder values, subject to change in order to
close the system balance. Sometimes the heat and material balances are so tightly coupled that this
procedure isn't feasible, but when it is, it gives useful insight into the process, and suggests which
stream properties might be most easily changed to close the system balance. We illustrate this
strategy in making a system balance for the zinc roaster.

Chapter 9 System Balances on Reactive Processes 509

The concentrate analysis from Example 6.14 listed the percentage of elements (except for
gangue and water). A heat balance requires a concentrate analysis in terms of species, so we must
convert the element composition to species composition. The concentrate composition did not sum
to 100 %, which was ascribed to the presence of non-analyzed PbS, which we must calculate. The
strategy for determining the species composition is first to list the most likely species present in the
concentrate. Then we express the concentrate in mole amounts, and match the elemental
concentrate composition to the chemical formula of the likely species. The following table lists the
sulfide mineral elements, the kmol/kg dry concentrate, and the most likely species. The gangue
mineral (Si02) is omitted because it does not change composition during roasting. The table
values are slightly different from those in Example 6.14 because PbS is now included.

Element Zn Pb Cu Fe S

kmol 8.347 0.024 0.519 1.003 10.042

Species ZnS PbS CuFeS2 FeS, FeS2 All

First, S is assigned: 8.347 to ZnS, 0.024 to PbS, and 1.038 to CuFeS2. This leaves 0.633 mol
of S to distribute to 0.484 mol Fe as FeS and FeS2. A Fe and S balance shows that the concentrate
contains 0.335 kmol FeS and 0.149 kmol FeS2. The assumption of FeS as a mineral species is
justifiable because ZnS often occurs with Fe substituting for some of the Zn on an equimolar ratio.

Table 9.1 shows the amount and composition of the wet concentrate on the basis of 1000 kg of dry

concentrate (1124 kg of wet concentrate). The wet concentrate (wH20 = 0.11) enters with 6.87
kmol of water, with an added 1.94 kmol of spray water.

Table 9.1 Wet concentrate amount and composition fed to zinc concentrate roaster per minute,
based on 1000 kg (10.00 kmol of mineral species) of dry concentrate. The gangue is assumed to
be all Si02.

Species ZnS FeS FeS2 PbS CuFeS2 Si02 H20
mole pet 49.5 2.0 0.9 0.1 3.1 3.7 40.7

kmol 8.35 0.34 0.15 0.02 0.52 0.62 6.87

The five chemical reactions involved in sulfide oxidation are listed below, plus the standard
heat of reaction at 298.15 К for the oxidation of one mole of each sulfide. Data is from FREED.

ZnS(cubic; Sp) = 1.502(g) - ZnO(c) + S02(g); AH0^ = -105 910 cai [9.3]

FeS(c,/) + 1.7502(g) - У2¥е203(с) + S02(g); Atf°rx - -145 630 cal [9.4]

FeS2(pyr) + 2.7502(g) - ^ F e ^ c ) + 2S02(g); A//°rx = -199 380 cai [9.5]

PbS(c,/) + 202(g) - PbS04(c, a); AH°^ = -197 160 cal [9.6]

CuFeS2(c,/) + 3.2502(g) - CuO(c) + ^ F e ^ c ) + 2S02(g); AH0^ = -236 180 cal [9.7]

Table 9.2 summarizes the amount of 0 2 required for the oxidation of the sulfides and the
quantity of heat produced by each reaction at 298.15 K.

Table 9.2 0 2 requirements for oxidation of sulfide minerals, S02 production, and reaction heat for
the oxidation of sulfide minerals in 1000 kg of dry zinc concentrate.

Species ZnS FeS FeS2 PbS CuFeS2
Species amount, kmol 8.35 0.34 0.15 0.02 0.52
12.53 0.60 0.41 0.04 1.69
0 2 amount, kmol 8.35 0.34 0.30 0 1.04
S02 produced, kmol -105 910 -145 630 -199 380 -197 160 -236 180
АЯ°ГХ, cal/molat298.15K -884 300 -49 500 -29 900 -3900 -122 800

H, kcal ΣΗ reaction = -1 090 400 kcal

510 Chapter 9 System Balances on Reactive Processes

The total 0 2 demand for oxidation of the sulfides is 15.27 kmol, which is accompanied by
48.36 kmol N2. In addition, 6.87 kmol of concentrate water and 1.94 kmol of spray water enter the
roaster, plus a so-far unknown amount of excess air such that the offgas will contain <jp02 = 3.3 %.
The roaster produces are calcine and gas, both heated to 1150 K. Table 9.3 shows the amount of

each outstream species and their heat content at 1150 K.

Table 9.3 Amount of roaster products and their heat content H at 1150 К (kcal). Basis is 1000 kg
dry concentrate roasted per minute. The FREED species H20(/,g) was used for water, so the heat
content value for water at 1150 К includes its heat of vaporization. XSA refers to the amount of
roaster air in excess of the stoichiometric amount (63.63 kmol), and W the amount of spray water.

Species Calcine Species PbS04 Si02
Amount, kmol ZnO Fe203 CuO 0.02 0.62
#ii50-#298, cal/mol 8.35 0.50 0.52 29 480 13 480
600 8400
H, kcal 9966 29 120 10510
83 200 14 600 5500

Σ # calcine = 112 300 kcal

Species SO2 01Tgas Species H20 (spray) N2 (XS) 0 2 (XS)
Amount, kmol N2 H20 (cone) W 0.76XSA 0.24XSA
Я1150-Я298, cal/mol 10.03
10 200 48.36 6.87 18 250 6321 6686
H, kcal 102 300 18 250W 4804XSA 1605XSA
6321 18 250

305 700 125 400

ΣΗ roaster gas = 533 400 + 18 250W + 6410XSA kcal

An uncoupled material balance can calculate the amount of XSA required to produce φ02 =
3.3 %.in the roaster gas when the value of W = 1.94 kmol (the basis case). The balance is simple,
as shown in Equation [9.8]. Solving, XSA = 10.71 kmol, so the total air to the roaster is 74.34
kmol. Notice that if the amount of spray water W were to decrease, the amount of XSA would also
have to decrease to keep the/?02 = 0.033.

p02 = 0.033 = 0.24(XSA) [9.8]

(10.03+ 48.36+ 6.87+ 1.94 +XSA)

The solved material balance lets us make an uncoupled heat balance for the roaster. The sum
of all of the heat terms gives the amount of heat (HiTSfr) that must be transferred to the surroundings
(i.e., the boiler tubes, or otherwise lost) from the roaster.

0 = #trsfr -1 090 400 + 112 300 + 533 400+18 250W + 6410XSA [9.9]

Inserting 10.71 for XSA and 1.94 for W, #trsfr = 340 700 kcal/min. Equations [9.8] and [9.9]
comprise two equations for three variables to make a coupled system balance for the process. The
variables are the value of Htrs{r, XSA, and W. If we select a value for any one of these three, the
other two dependent variables can be calculated. Equation [9.10] for example shows the single
system balance equation for the process when XSA = 10.71.

0 = Httrsfr ' 378 000+19 270W [9.10]

W and XSA were calculated as a function of HtrsfT between 340 000 and 300 000 kcal/min.
Figure 9.4 shows the results. The relationships were linear.

This example lays out a recommended procedure for making system balances on reactive
processes. Whenever possible, make a material balance first by using reasonable estimates as
placeholders for values that depend on the heat balance. Next, make a heat balance and see if
useful relationships can be developed between the two types of balances. Finally, combine the two
balances into one system balance. The next few Sections show how to apply this procedure.

Chapter 9 System Balances on Reactive Processes 511

Cooling the Contents of a Zinc Concentrate Roaster

f^12 74.70

c10 ^ ^ -i. XSA = -8.27x10"6H + 13.53 f 74.65
]
E XSA . -8.27х1I 0_6Н . 1 74.60 с
о8 А 77.1б| 74.55 |
E I Ш ^ = + о
E
<6 Mb^ 74.50
'δ
co 4 74.45
3
JО W = -5.19x105H +19.62 74.40
о
^2 ______^_^_i__^_^__^^_^_^^^^_ iQ 74.35
74.30

300,000 310,000 320,000 330,000 340,000

heat removed from roaster, kcal/min

Figure 9.4 Relationships between the amount of excess air (XSA), total air (A), and spray water

(W) required to keep the roaster at 1150 К as a function of the amount of heat removed from the
roaster (HtYSfr). Basis is 1000 kg dry concentrate roasted per minute with 24 % 0 2 roaster air. As
the boiler tubes become less effective, they remove less heat, so the amount of spray water must
increase. The amount of air must also increase in order to keep the roaster gas at φ02 = 33%.

9.2 Combustion of Fuels

In Chapter 6 we made material balances on the stoichiometry of fuel combustion. Please
review the calculations in Section 6.3. When hydrocarbon fuels are burned by the addition of more
than the stoichiometric oxidant, the combustion products are C02 and H20. When sulfur is present
in the fuel, the combustion products may contain S02, S03, or both. If less than the stoichiometric
oxidant is used, the combustion products will contain some CO and H2. Sulfur-containing
combustion products might be S02, COS, H2S, or even S2. When excess oxidant is used, the heat
generated is called the heat of complete combustion, or simply the heat of combustion, usually
designated АНсотЪ.

The heat of combustion is a function of temperature, so strictly speaking, the temperature of
combustion must be specified; usually it's 298.15 K. Several other terms are also used: calorific
power, calorific value, heating value, and fuel value. These values are the negative of ΑΗοοπ&.
Industrial fuels are typically mixtures of different hydrocarbons. Natural gas is the simplest fuel,
usually containing no more that five species in significant amounts. Fuel oil is more complex,
often having a dozen or more species in significant amounts. Coal is the most complex, containing
substances that have no formula or have poorly-defined species. For example, sulfur is present in
an organic form and as pyrite. Also present are nitrogen-containing substances, and clay minerals.

For fuels of accurately-defined stable species, such as natural gas, a thermodynamic database
like FREED has all of the information needed to calculate the fuel's heat of combustion. Such a
database will also contain information about liquid fuels that consist of defined molecular
constituents like methanol (CH40) and benzene (C6H6). But for a typical fuel oil or coal, the
constituents are either so numerous or so poorly defined that a typical thermodynamic database is
of little use. Heats of combustion of species of invariant composition are listed (or can be
calculated), and may give a larger number of significant figures than is justified by the accuracy of
the stream flow or composition values. In a real process, fuels such as natural gas, oil or coal will
vary in composition with time and hence in most cases, the heat of combustion of one kg is
adequately expressed to ±500 kJ (or ±100 kcal). However, to avoid confusing round-off errors, the
Handbook calculations will carry more significant figures than justified.

512 Chapter 9 System Balances on Reactive Processes

9.2.1 Heat of Combustion Calculations

The heat of combustion is equivalent to the heat of reaction between an oxidant (usually
oxygen) and a oxidizable substance. The calculation procedure uses methods outlined in Section
7.6-for calculating the heat of a reaction. The methodology depends on the complexity of the fuel.
Natural gas (called NG here for short) is usually well defined for all significant constituents, so
FREED (or other similar database) data can be used to calculate the heat of combustion. If desired,
the molar heats of combustion of the NG constituents can be summed and expressed in terms of
unit volume (typically Btu/scf or kJ/scm at a defined reference temperature, usually 298.15 K).
Alternatively there exist published values for heats of combustion of many substances (Wikipedia
2010), which can be combined to calculate the heat of combustion of a mixture of substances. The
species values cited in handbooks and other sources may be listed in mole, mass or volume units,
but in some cases, a range of values are listed for typical NGs or other fuels for comparison
purposes. When using NG as a source of heat, it is common to pay the supplier on the basis of
thermal units rather than volume.

Heats of combustion are usually {but not always!) calculated or listed at 298 K. Care must be
taken when calculating or using handbook data for AHcomh of a fuel because there are two different
conventions in specifying the H20-part of the combustion products. In some tabulations, AHcomb is
based on liquid water as the product of combustion, and in other tabulations, water vapor is the
product form. The heat produced by the combustion of a fuel is often called its heating value,
which is equal to the value of A//comb but of opposite sign. When liquid water is the product, it is
called the higher (or gross) heating value (HHV). When water vapor is the product, it is called the
lower (or net) heating value (LHV). The difference between the two values is AHV of water at 298

K. Since the combustion products are almost always above the boiling point of water, the LHV is

a better indication of the useful heat obtainable from a fuel.

Unfortunately, some handbooks use 60°F (288.7 K) as the standard temperature instead of
273.15 К as used in this Handbook. The best, and unambiguous way to list the HHV and LHV of
fuels is by mass or amount. Users can then covert to a volumetric basis of their own choosing.

A sample calculation for the HHV and LHV of one mole of methane (CH4) at 298.15 К
illustrates the difference between these two terms. FREED data is used.

CH4(g) + 202(g) -> C02(g) + 2H20(Z) [9.11]
A#comb - A#rx - AH°{ of C02(g) + 2Atf°f of H20(/) - A#°f of CH4(g)

A#comb = -393.51 + 2(-285.83) + 74.81 = -890.36 kJ/mol of CH4

CH4(g) + 202(g) - C02(g) + 2H20(g) [9.12]
Atfcomb = Atfrx = Δ#°f of C02(g) + 2AH°{ of H20(g) - A#°f of CH4(g)

A#comb = -393.51 + 2(-241.81) + 74.81 =-802.32 kJ/mol of CH4

The HHV of CH4 at 298.15 К is then +890.4 kJ/mol while the LHV is +802.3 kJ/mol; the
difference (88 kJ) being twice the molar AHV of water at 298 K.

Certain conceptual difficulties exist when using the HHV term for combustion calculations.
First, it is physically impossible for all of the product water to be in the liquid form. If combustion
is carried out at any temperature above the triple point of water, the combustion products must
contain a finite amount of uncondensed water vapor. For example, combustion of one mole of
methane with 10 % excess air requires 10.5 moles of air and produces 11.5 moles of product gas.
Even at the triple point of water, the product gas will have 0.06 mol of uncondensed water vapor.
Therefore, the HHV is not perfectly attainable by any real combustion condition. Another problem
with the HHV is that the fuel gas may contain water vapor, as might the air. This water (called
spectator water) appears in the product gas, and if condensed along with the combustion product
water, will deliver an additional amount of heat simply by virtue of the heat of its condensation.
Use of the LHV avoids these complications, and is the preferred way to define AHcomb of a fuel.

Chapter 9 System Balances on Reactive Processes 513

Table 9.4 lists the heating value of methane in a variety of units, where the significant figures
are carried out to one more digit than justified by the accuracy needed for typical calculations. An
easily-remembered number is that NG has a heating value (however it is measured) of about 1000
Btu/scf. For most industrial uses, the combustion products are well above the boiling point of
water, hence the LHV is more appropriate than the HHV. Table 9.5 shows the LHV of some
commonly encountered fuel gas species. The Handbook practice is to use the LHV. You may
want to copy both tables to a handy location in the Handbook.

Table 9.4 Heating value for methane at 298.15 К in different units. For Btu/lb-mol, multiply
kJ/mol by 430.2. scm and scf refer to volumes at 273.15 К and 1 atm pressure.*

kJ/mol kcal/mol kJ/scm kcal/scm kJ/kg kcal/kg Btu/scf Btu/lb
HHV 890.36 212.80 39 722 9494 55 498 13 264 1064.2 23 881
LHV 802.32 191.76 35 796 8555 50 011 11953 961.4 21 515

Table 9.5 LHV for selected fuel gas components at 298.15 К in kcal/mol. Multiply by 102.82 to
convert to Btu/lb-mol and by 4.184 to convert to kJ/mol. Add 10.52 per mole of water to convert
to HHV. For H2S, * indicates combustion to S02, while ** indicates combustion to S03.

Species CH4 СгН6 C3H8 C4H10 C2H4 H2 CO NH3 H2S* H2S**
LHV 191.76 341.25 488.51 634.08 316.25 57.80 67.63 75.72 123.81 147.45

Not all combustion processes use hydrocarbon fuels. Some process gases contain significant
amounts of CO and H2, and are burned as a fuel. For example, the stack gas from an iron blast
furnace is used to heat the stoves that in turn heat the blast furnace air. Other non-conventional
fuels are metals, coke fines, solid waste from industrial processes, municipal trash, hazardous
materials that must be incinerated, and sulfide minerals present in concentrates.

EXAMPLE 9.1 — Heat of Combustion of a Spent Gas from a Reduction Process.

A spent iron ore reduction gas has the following composition:

<pCO2 = 0.18; <pCO = 0.21; φΗ2 = 0.31; balance H20

The spent gas is partially dried by cooling to 30 °C to condense out some of the water vapor.
Calculate the heat of combustion of the dewatered gas in kJ/m3.

Data. Enthalpy data is from FREED, and thepH20 is from Equation [2.16].

Solution. The first step is to make a material balance on dehumidification of the spent gas. The
pU20 at 30 °C is 0.0417 atm. For 1 m3 (STP) of spent gas, the dried gas volume - 0.70 + FH20:
ra2o
pK20 = 0.0417 = o.70 + ra2o [9.13]

Solving, KH20 = 0.0305 m3. The volume (m3) of each component of the dried gas is: FC02
= 0.246; FCO = 0.287; VR2 = 0.424; VR20 = 0.042.

The combustible components are CO (12.8 mole) and H2 (18.9 mole). The standard
stoichiometric reactions for the oxidation of one mole of each component is given by Equations
[9.14] and [9.15]. The heat of combustion of the fuel component of each species is calculated
beneath each equation at 298.15 K.

CO(g) + !/202(g)-^C02(g) [9.14]
Atfcomb = Atfrx = 12.8[Afl°f of C02(g) - AH°f of COfe)]
Atfcomb = 12Д-393.5 + 110.5) = -3622 kJ at 298.15 К

* For handbooks that use 60°F as the standard temperature, the HHV for methane is 1010 Btu/scf.

514 Chapter 9 System Balances on Reactive Processes

U2(g) + y202(g)^n20(g) [9.15]

Atfcomb = Atfrx = 18.9[Afl°f of H20 (g)]
АЯсотЬ = 18.9(-241.8) - -4570 kJ at 298.15 К

The LHV for the dried stack gas at 298.15 К is then 8192 kJ/scm which is about 23% of the
value for methane. The HHV is obtained by adding the AHV of 18.9 mole water, or 18.9(44.0) =

832 kJ. The HHV at 298.15 К is then 9024 kJ/scm.

Assignment. An NG user requires fuel that has a LHV of 1000 Btu/scf at 77 °F. The supplier has a
long-term contract for NG that has the following composition:

<pC2H6 = 0.019; фС02 = 0.021; <pN2 = 0.017; balance CH4 =

Determine the LHV of this NG, and if it is too low, specify how much C2H6 must be added to
1 scf of NG to attain the specification.

9.2.2 Use of Wobbe Index for Combustion Burner Control

Section 6.3.4 outlined concepts of air and fuel flow control for NG-fired burners. Owing to
the use of orifice flow controllers for the NG stream, a change in the composition of the NG fuel
results in a change in the volumetric flow of the NG. Furthermore, a change in the volumetric flow
results in a change in the heat input and the composition of the combustion gas. Unexpected
changes in the composition of the NG to a process can have undesirable effects on the process.

Equation [6.45] in Section 6.3.4 showed that flowrate of a gas through an orifice (at constant
pressure drop) relative to that of some reference gas depends on the square root of the density ratio
of the gas vs. that of the reference gas. Suppose a burner operated initially with 100 L/sec STP of
pure CH4, which has a LHV of 35 800 kJ/scm. The process heat input would be 3580 kJ/sec. If
the NG were to change to <jpCH4 = 93 %, balance C2H6, the new NG flowrate would decrease to
99.5 L/sec. The LHV of the new NG is 37 750 kJ/scm, which will bring in 3756 kJ/sec.

The process heat calculation is further complicated by the different air requirements of the
two compositions of NG. Suppose the air flowrate was maintained constant at 1100 L/sec. The
stoichiometric air (<p02 = 21 %, balance N2) required for 100 L CH4 is 952.4 L, so 1100 L of air is
115.5 % of stoichiometric for the combustion of pure CH4. Now suppose the NG changed from
pure CH4 to <pCH4 = 93 %, balance C2H6. The stoichiometric air for 99.5 L of the new NG is
1007.4 L, so 1100 L/sec is 109.2 % of stoichiometric. Let's look at the implications of these
changes on a process for sintering nickel ferrite (NiFe204) at 700° C. Since we know the instream
amounts and composition, the material balance is uncoupled from the heat balance. The material
balance is simple. Table 9.6 shows the results. Notice that the outstream from CH4 combustion
has φ02 = 2.38 %, while for the new NG, φ02 = 1.66 %.

Table 9.6 Gas volume flows (STP L/sec) for methane and a new NG fuel used for a nickel ferrite
sintering process.

Instream volume flow co2 Outstream volume flow

CH4 C2H6 Air 100 н2о o2 N2 Total

pure CH4 (100 L/sec) 100 L ° 1100 106.47 200 31 869.0 1300

New NG (99.5 L/sec) 92.54 6.97 1100 205.97 21.55 869.0 1295.5

The CH4-fueled sintering process operates with a NiFe204 flowrate of 3.90 kg/sec, and
operates at 700 °C, so it's easy to make an uncoupled heat balance. The easiest way is to use the
Calculate option from FREED's Reaction tool, with 3900 g of NiFe204 on both sides of the
reaction. At 700 °C, the net heat is -176 000 J, so the sintering furnace device has a heat loss of
176 kJ/sec. We now need a heat balance for the process when the fuel is switched to the new NG,
while maintaining the same NiFe204 flowrate and heat loss. The heat balance question becomes:
what outstreams temperature will give ΣΗ = 0? Return to FREED's Reaction tool, enter the

Chapter 9 System Balances on Reactive Processes 515

quantities of reactants and products, request a table in J energy units with an upper temperature
limit of 800 °C*, and observe the column labeled Heat. Convert the values in the Heat column to
kJ. Next, obtain a new net Heat column by subtracting 176 kJ from the Heat column. Figure 9.5
shows a plot of the results from 400°C to 750 °C, with a linear Trendline equation fitted to the
results. The outstreams temperature calculated from the Trendline equation is 730° C.

Net Heat for Nickel Ferrite Sintering
500

Оо

2 -500

CO

jS -1000

ф

с -1500

-2000
400 450 500 550 600 650 700 750|
outstreams temp, °C

Figure 9.5 Net device heat for sintering 3.9 kg/sec of nickel ferrite as a function of outstreams
temperature. Sintering furnace is fired with 99.5 L/sec of NG having <jpCH4 = 93 %, balance C2H6,
with a heat loss of 176 kJ/sec. ΣΗ = 0 when the outstreams temperature = 730° C.

These calculations show that an inadvertent change in NG composition to an orifice-
controlled flowmeter can cause significant changes in both the composition of the fornace
atmosphere and the operating temperature. One of the techniques that's been developed to deal
with this issue is to use an on-line analyzer to detect NG composition changes, and take corrective
action to bring the system back into the desired operating condition. It's been found that every NG
with a certain characteristic, called the Wobbe Index, will bring in heat at the same rate through an
orifice flowmeter, and require the same amount of air for a set % excess. The Wobbe index (WI)
is calculated according to Equation [9.16].

W I = Heating value of NG [g щ

1 Density of NG
\ Density of air

The heating value can be expressed in any consistent set of units. For example, the LHV of
CH4 = 35.796 MJ/scm, the density of CH4 is 0.6557 g/L at 25°C and 1 atm, and the density of air**
is 1.1841 at the same conditions. (The densities were calculated assuming ideal gas behavior.)
From Equation [9.16], the WI of CH4 = 48.10. In a similar manner, the LHV of the new NG is
37.75 MJ/scm, and the SQRT(density ratio) is 0.7666. The WI of the new NG is 49.24.
Therefore, if a suitable additive to the NG were made to lower its WI back to 48.10, the heat input
to the sintering furnace would remain the same, as would the furnace temperature and % 0 2 in the
furnace atmosphere. No adjustment would be required in the air flow.

You should see that adding any inert gas (such as C02) would lower the WI of a NG by
lowering the LHV and raising the density. The Wl-adjustment calculations are somewhat
complicated for the purposes of this Chapter, but they are outlined in more detail in Section 10.3,
and on the Handbook CD in folder NG Combust & Wobbe Index.

* The Heat column can extend past the upper limit for FREED data for C2H6(g) of 1000 К (726.85 °C).
** The composition of air here is taken as dry, C02-free, of composition 78 %№, 21 %02, l%Ar.

516 Chapter 9 System Balances on Reactive Processes

There are other ways to correct for a change in NG composition, such as sampling the
combustion gas or furnace atmosphere, but it's not easy to get a representative gas sample because
of air leaks into the furnace while a product is continually being inserted and removed from the
furnace. The WI is in widespread use as a diagnostic and control tool.

9.2.3 Combustion of Fuels of Uncertain Molecular Composition

When making combustion heat balances involving substances with undetermined speciation
(like fuel oil or coal), we are usually stuck with knowing only the heat of combustion (HHV or
LHV) of the substance, not its АЛ{отт. However, it's essential to know the AH°f0Tm of a fuel when
making a heat balance calculation when using less than stoichiometric air. This situation is
resolved by use of Hess's Law to calculate АЛ%огт for any substance from its LHV or HHV. Once
we know its АЯ0form, we may be able to insert the substance in FREED and use it like any other
species.

Suppose we need to know AH°form of a fuel oil to be used to generate a reducing gas for iron
ore reduction. The oil contains 88.7 %C, 10.6 %H, and 0.7 %S, and has a LHV at 298 К of 9920
cal/g. 100 g of this oil contains 7.385 moles of C, 10.516 moles of H, and 0.022 moles of S. An
empirical formula for the oil is CH1.4240S0.0030> which has a molar mass of 13.541. The АЯсотЬ of
one mole of this oil is then -134 330 cai. We write a reaction for the combustion of the oil, and the
reverse of formation reactions for H20(g), C02(g), and S02 from the elements at 298 K in the
amount of each present in one mole of fuel oil.

CH1.424S0.oo3(/) + 1.35902fe) -+ C02(g) + 0.712H2O(g) + 0.003SO2(g); Atfrx = -134 330 cai

C02(g) -> C(gr) + Ozfe); Atfrx = 94 050 cai

0.712H2O(g) -> 0.712 H2(g) + 0.356 02(g); A#rx = 0.712(57 795) = 41 150 cai

0.003SO2(g) -> 0.003S(c) + 0.003O2(g); A#rx - 0.003(70 940) = 210 cai

The sum of these equations is the reaction for the decomposition of oil to the elements at 298
K, with ЕАЯГХ = 1080 cal/mol. Therefore, the ΑΗ°ΐ0γπί of the fuel oil from the standard state
elements at 298 К is -1080 cal/mol, or -80 kcal/kg. This is less than 1 % of the LHV, and in many
cases, it is negligible in comparison to the uncertainties of heat loss and other factors. If we had
certain other necessary additional data on this oil, we could enter it into the FREED database and
use FlowBal for heat balance calculations involving this specific oil. FREED does have a fuel oil
species in the database, but it is not the abovementioned one.

The procedure outlined above can be formalized for any ashless hydrocarbon fuel by
converting the molar АЯГХ of C, H, and S at 298 К into mass АЯГХ and adding the LHV. Equation
[9.17] expresses this relationship.

A//°form of fuel at 298 K, cal/g = %C(-78.3) + %H(-286.7) + %S(-22) + LHV [9.17]

A similar procedure is used when the HHV is listed instead of the LHV. In that case, replace
the factor for %H by -338.9.

The problem is more difficult when the fuel is coal because coal has an ash constituent, and
the coal analysis seldom gives the ash composition. Generally, the ash is considered not to contain
oxidizable elements. Another complication of coal as a fuel is that it also contains oxygen, which
changes the amount of oxygen needed for stoichiometric combustion. Similar complications arise
when biomass substances are used as combustion substitutes for coal, either in part or completely.
In addition, coal and various biomass substances are being promoted as feedstock for the
production of synthesis gas. For all of these reasons, it's worth taking a closer look at coal, coal
derivatives, and biomass combustion.

A general problem with the use of coal or biomass substances as a fuel is that while we may
have the chemical composition of the substance, we may not have a value for the АЯ°сотЬ. А
considerable effort has gone into attempting to find a correlation between the HHV of various fuels

Chapter 9 System Balances on Reactive Processes 517

of nonspecific molecular composition, beginning with the work of Pierre Dulong in the 19*
century. Many subsequent studies have verified that a close correlation exists between the ultimate
analysis of a carbonaceous fuel and its HHV. A typical correlation formula will give a HHV
within ±1.6 % of the measured value. Most recently (Gaur et. al. 1998) presented a formula based
on data from about 200 species of biomass:

HHV,kJ/kg = 349.1(%C)± 1178.3(%H)- 103.4(%O)+ 100.5(%S)- 15.1(%N)-21.1(Ash) [9.18]

where the compositions are mass percent as reported in the ultimate analysis. The %0 does not
include the oxygen content of the ash.

Consider a medium-volatile bituminous coal having the following composition:

с H 0 S N Ash

86.22 4.89 3.68 0.69 1.61 2.91

The calculated HHV of this coal, based on Equation [9.18], is 35 460 kJ/kg (8475 kcal/kg).
One kg of coal has 71.79 moles of C, 48.51 moles of H, 0.22 moles of S, 2.30 moles of O, and 1.15
moles of N. An empirical formula for the ash-free portion is CH0.676O0.032N0.0i6S0.003· The
molecular mass of this coal (ash-free basis) is 13.52. The A//°form of the ash-free portion of the
coal at 298 К from Equation [9.17] is ±52 cal/g (220 kJ/kg), or 700 cal/mol (2940 J/mol). The heat
of formation of the coal is about 0.6 % of the HHV, so considering the typical accuracy of the coal
analysis and the approximation built into Equation [9.18], it is generally acceptable to ignore the
small effect of the heat of formation for this coal. However, as the oxygen content of the coal (or
biomass) increases, the AH°iorm of the fuel becomes a significant, and negative, value. For
example, the A//°form of high-volatile sub-bituminous coals and lignite are on the order of-1500
kJ/kg, which is on the order of 6 % of the HHV. When using coal or other biomass for
combustion, the ash may be considered as Si02(qtz), or any other mineral substance high in silica.

9.3 Adiabatic Processes

So far we have calculated the enthalpy change when certain reactants form certain products.
This implies that for an exothermic isothermal reaction (or a combustion) process, something must
happen to dispose of the heat. It is either removed from the system or consumed by other
endothermic reactions or transformations. With a little imagination, one can conceive of a process
in which the reactor, the inlet streams and the outlet streams have the same temperature (i.e.,
perfectly isothermal). Far more common, the instream and outstream temperatures are all
different. A major part of a heat balance is accounting for what happens to the heat produced by
an exothermic reaction, or where the heat must come from for an endothermic process.

We know from experience what happens to most of the heat produced during the combustion
of a fuel: it increases the temperature of the combustion products. If the combustion process
occurs very rapidly in a reactor where heat loss is minimized, it approaches adiabatic combustion.
For an exothermic isobaric perfectly adiabatic process, all of the reaction heat appears as sensible
heat in the products. The temperature attained by those products is called the adiabatic reaction
temperature (ART). For combustion processes, the temperature is called the adiabatic flame
temperature (AFT). For most material processing applications, the AFT is calculated for isobaric
conditions. However, for certain applications, one may need to calculate the isochoric AFT
(Wikipedia 2010). FlowBal can be configured to calculate the ART for gas-phase reactions by
setting the constraints of constant pressure and conservation of enthalpy. Before reading further,
please review the concepts of heat balances in adiabatic systems in Section 8.2.

Obviously no process is truly adiabatic because heat loss cannot be completely avoided. But
some processes have very small heat losses compared to the overall thermal effect, so the adiabatic
assumption gives a useful result. In addition, the adiabatic concept is a good way to compare
different fuels and versions of a process, and for calculating the thermal efficiency of a process.

518 Chapter 9 System Balances on Reactive Processes

9.3.1 System Balances Involving Combustion Reactions

The combustion of CH4 will be used to show in detail how to calculate the adiabatic reaction
temperature (here, the AFT) for a process described in Chapter 8. CH4 was the fuel used in the
combustion process that supplied heat for melting aluminum (please review Example 8.1) and
sintering nickel ferrite (please see Section 9.2.2). Combustion took place with 118 % of the
stoichiometric amount of dry air (i.e., 18 % excess air). The combustion of one mole of CH4 with
excess air is spontaneous and complete, and according to Equation [9.11] forms one mole of
C02(g) and two moles of H20(g), and leaves some unreacted 02. The text associated with [9.11]
has data for calculating the AFT of CH4. The AHcomb of CH4 at 298 К must be "done away with"
somehow. For an adiabatic process, HL = 0, so all of this heat is absorbed by the products of
combustion. This raises their temperature to the AFT. The relationship between the АЯсотЬ and
the sensible heat in the products is expressed by setting the overall enthalpy change for the process
to zero. The AFT is the temperature that will close the heat balance.

A#process = АЯсоть + Σ(#τ-#298) for products = 0 [9.19]

The steps involved in making the heat balance start with the material balance. Since
temperature is an intensive property, the basis amount selected is immaterial. For calculational
simplicity, we will use one mole of CH4 burned as the basis amount, and 298 К as the basis
temperature. Since two moles of 0 2 are required for one mole of CH4, the amount of
stoichiometric air is 9.524 moles. The air requirement at 118 % of stoichiometric is 11.24 moles
when we use an air composition of <jp02 = 21%. The combustion products are:

лСО2=1.00; лН2О = 2.00; rcN2 = 8.88; лО2 = 0.36

The heat balance requires data from FREED. The following table lists the parameters for #T-
Я298 for each product gas valid up to 3000 K.. Table 9.4 provides the value for АЯсоть at 298 К of
CH4 of-802.32 kJ/mol.

#τ-#298 (J/mol) - AT + ВТ2 + CT"1 + mA + ET3 + F

C02(g) A В С D E F
H20(g) 91.431 -2.2502E-03 -803232 -2140.9 1.389E-07 12598
19.791 1.0379E-02 -64948 247.0 -1.077E-06 -10841
N2(g) 38.167 1.6458E-03 -507331 -541.1 -2.34E-07 -477
02(g) 53.752 -1.9400E-03 -612446 -1042.3 2.71E-07 4192

Each equation was multiplied by the amount of each constituent, divided by 1000 to get kJ,
and entered as a formula in Excel. The value of AHpr0CQSS in Equation [9.19] was calculated at 100

degree temperature increments, and plotted as shown in Figure 9.6. The values fall on a nearly-

straight line and intersect the АЯргосе88 = 0 axis value to give the AFT «2100 K. The text box
equation gives an AFT of 2091 K.

Overall Heat Effect for CH4 Combustion Figure 9.6 Plot of AHvr0CQSS for
150.0 an adiabatic process where one
mole of CH4 is burned with 118
100.0 ΔΗ process = 0.4990(T) - 1 0 4 3 . 4 % stoichiometric air. The CH4
50.0 and air enter at 298 K. The
Ij AFT is the temperature where
A#process = 0, or about 2100 K.
0.0 ! ? ^^^"^ The text box equation was
-50.0 obtained by Excel's Trendline
tool.
■100.0 ^^^^^ 2000 2100 2200 2300
temperature, К
-150.0
1800 1900

Chapter 9 System Balances on Reactive Processes 519

An alternate (analytical) procedure is to write the values for Equation [9.19] into an Excel
spreadsheet and use Goal Seek to find a solution. This procedure yielded an AFT = 2092 K, in
excellent agreement with the graphical approach. Finally, easiest of all is to use FREED's
Reaction tool and the associated Calculator. The results are shown below. The result, 1819 °C, is
in perfect agreement with the other results. Obviously, the burner in Example 8.1 was not
operating adiabatically; otherwise, the combustion gas temperature would have been 1819 °C
instead of 1800 °C. Alternate possibilities are that the burner was adiabatic, but reactants entered
below 298 K, or we should not have neglected the Ar in the air.

Calculator: Unit: Cai
Input(heat)
Iteration T(C)
0
Solutions 5.07E-06 [ 1819.4 |

1 T(C) dHr dGr LogKr Overall H
о
1819.4 -193172 -190604 19.907

It's important to review here an assumption we made about the excess air combustion of CH4
being complete, and producing only C02(g) and H20(g). This assumption is not strictly true,
especially if the AFT is above 2000 K, because at high temperatures, C02(g) dissociates slightly to
CO(g) and 02(g), and H20(g) dissociates slightly to H2(g) and 02(g). At even higher temperatures,
02(g) dissociates slightly to 0(g). Where high accuracy is important, we must distinguish between
the theoretical AFT based on CH4 producing only C02(g) and H20(g), and a more elaborate
equilibrium AFT calculation, which takes into account the slight dissociation of C02(g) and
H20(g). Of course, the greater the % excess air, the less the dissociation of C02(g) and H20(g).

The AFT obtained by combustion of a fuel depends not only on the amount of excess air used
(the higher the % excess air, the lower the AFT), but also on the temperature of the air and fuel
streams entering the process. Enriching the air with oxygen also raises thè AFT because less
nitrogen must be heated.

EXAMPLE 9.2 — Effect of Preheating Combustion Air on AFT.

A fuel oil contains 86.50 %C, 12.34 %H, and 1.16 % S and burned with 20 % excess air.
Calculate the AFT as a function of air preheat temperature up to 300 °C. The oil enters the burner
at 25 °C.

Data. TheoilhasaLHVof9950cal/g. All other data is from FREED.

Solution. The system balance is uncoupled, which allows us to make a material balance before
making any heat calculations. The chosen mass basis is 100 g. The amount of each element is:

С = 7.202; H = 12.242; S = 0.036

The stoichiometric oxygen requirement is then 10.30 moles, so 58.85 moles of air are required
for 20 % excess. The amount of each substance in the combustion gas is:

CO2 = 7.20; H20 = 6.12; S02 = 0.04; N2 = 46.49; O2=2.06

This example will illustrate a slightly different graphical approach to AFT calculations where
we want the AFT as a function of a stream temperature. There are three main steps. First, FREED
tables at intervals of 50 °C were generated for air and the product species, and the Д-Я25 data was
copied to a new worksheet. No equations were used. The molar heat content data for each
combustion gas was in the range 1800 - 2100 °C, and for air from 25 - 300 °C. This data was
used to calculate Д-Я25 for the combustion gas for seven temperatures. Then the АЯсотЬ of the oil
(-995 kcal for 100 g) was added. Finally the heat content of air for six air preheat temperatures up
to 300 °C was added, and the value of AHpr0CQSS calculated for each of the seven combustion and six
air preheat temperatures.

520 Chapter 9 System Balances on Reactive Processes

For example, at 1900 °C the heat content of the 61.91 moles of combustion gas is 1021.2 kcal.
The heat content of 58.85 moles of air at 100 °C is 30.7 kcal. The AHpT0CQSS is:

Atfpr0cess = -995 -30.7 +1021.2 -4.5 kcal
This calculation was repeated for all selected combustion and air temperatures. Figure 9.7
shows the results.

75.0 AH process for Fuel Oil Combustion

50.0 !! ,ΧΊ ! ^χτ и

g 25.0 \y^ '.•ri 300°

8 00 pro ^ 5 0 ° 100' 150° 200° 2S6

Q. ^ί / ^ y<^, > >^\
s" .
< -25.0 \ !I !
!!
-50.0 ! ! !!

-75.0 1850 1900 1950 2000 2050 2100
1800 temperature, °C

Figure 9.7 Graph of overall H for an adiabatic process where 100 g of fuel oil is burned with 20
% excess air. The oil enters at 25 °C, and the air enters at the temperature values shown in text
boxes on the figure. The AFT is the temperature where AHpr0CQSS = 0.

Considering the accuracy of the data and the objective of the example, values of the AFT can
be read from the diagram. (Alternatively, the Trendline tool could have been used to develop
equations for each line, and then solved for the AFT). Figure 9.8 shows a nearly-linear
relationship between the AFT vs. the air preheat temperature. The AFT increases about 0.7 degree
per degree of air preheat. An alternate calculational strategy would be to use FREED's heat
content equations, and Super Goal Seek.

Figure 9.8 Graph of AFT for combustion of fuel oil with 20 % excess air preheated to various
temperatures. AFT values used in plotting this figure were read from Figure 9.7.

Assignment. A user received fuel oil containing 8 % water in the form of tiny droplets suspended
in the oil. Calculate the AFT for the wet fuel oil burned with 20 % excess air without preheat.
What air preheat temperature is required to give an AFT of 1870 °C with the wet oil?

Chapter 9 System Balances on Reactive Processes 521

Another application of AFT calculations is to calculate the maximum possible temperature
attainable by combustion of a fuel. This requires pure 02. Some idea of the maximum theoretical
AFT for a simple fuel like H2 can be obtained by looking at the FREED table for H20(g) at 50°
intervals. For one mole of H2 and Vi mole of 0 2 entering at 300 K, the AH°form of one mole of
H20(g) at 300 К = -57 800 cai. If we scan the table for values in the ЯТ-Я298 column and look for
57 800, we see it is a bit lower than 5000 K. The maximum theoretical AFT for combustion of H2
is thus about 5000 K. However, the actual temperature of an oxy-hydrogen flame is far below that
value. The reason for this was mentioned earlier: at temperatures above about 2400 K, the
standard reaction for the formation of H20(g) gradually changes from an irreversible to a
reversible reaction, hence XRH2 is <1. The indication of this is the steadily decreasing value of
Кщ in column J. The concept of irreversible and reversible reactions was introduced in Section
5.6. Remember that for a reaction to be considered irreversible, Кщ must exceed about 400.

Another reason the actual flame temperature is much lower than the theoretical AFT is that
above 3000 K, species like OH, O, and H begin to appear in significant amounts. These species
further decrease XRH2 to form H20(g). For all of these reasons, the actual maximum AFT for an
oxy-hydrogen flame is about 3080 K.

The above discussion emphasizes an important point about making ART (or AFT)
calculations: the extent of any potentially-reversible reaction must be taken into account in the
heat balance calculations. One way to calculate a (maximum) reaction extent is to use the Кщ
expression to identify species that are formed in non-negligible amounts. The important species
are then used in а Кщ expression which becomes part of the system balance equation set. The
following two examples show how the Кщ expression is used for these two situations. In the first
example, Кщ calculations will be used to show which species can be ignored, which simplifies the
balance arithmetic. In the second example, Кщ for a reversible reaction becomes part of the
system balance.

EXAMPLE 9.3 — Production of a Reducing Gas.

A fluidized bed combustor is being designed to produce a reducing gas by reacting carbon, air
and steam. The air and carbon enter at 298 K, while steam enters at 450 K. Sufficient carbon is
added to carry out all reactions but is not present in the exit gas stream. The molar ratio of
air/steam is 20. Ptotal = 1 atm. Calculate the AFT for the process, which is known to be >1200 K.

С Fluid-bed -► Reducing gas
Air combustor
Steam

Data. All data is from FREED.

Solution. In addition to N2, the possible species in the exit gas are CO, C02, H2, and H20**. The
first step is to see which (if any) can be neglected. The reactions between oxygen and carbon were
presented in Chapter 5. Please refer to the discussion surrounding Equations [5.20] - [5.23], in
particular Equation [5.23] which will be used to determine if C02 is present in meaningful
amounts. The other important reaction is between steam and carbon. The two reactions of interest,
together with their Кщ expressions, are given below. Since carbon is present throughout the
combustor, its activity is assumed to be unity, and hence is not shown in the Кщ expressions.

C(c) + C 0 2 ( g ) ^ 2 C O ( g ) ; Keq = ^ C ° ^ [9·20]

Here we disregard such oxidants as ozone and nitrogen peroxide since these substances are mainly used for
rocket fuel combustion.

The amount of N-containing product species under the example conditions is vanishingly small.

522 Chapter 9 System Balances on Reactive Processes

C(c) + n20(g)^CO(g) + lI2; KeqJpC°JPy [921]

pR20

The values of Кщ for Equations [9.20] and [9.21] are not listed explicitly in FREED. Instead
Кщ can be calculated with FREED's Reaction tool in a manner similar to its earlier use. The

results at three temperatures between 1200 and 1300 К are:

1200 К 1250 К 1300 К
Keq for [9.20] 52.4 102.9 191.4
Keq for [9.21] 38.1 65.7 108.5

The Кщ values are of a size to indicate that CO and H2 predominate in comparison to C02
and H20. But to be sure, it's best to make a trial gas composition calculation at 1200 K. Note that
Кщ for both reactions increases with temperature, so if C02 and H20 are borderline low at 1200
K, they are probably negligible at the AFT (which was indicated to be above 1200 K).

Consider first the amount of C02 present in the exit gas. The stoichiometry of burning one
mole of carbon with simple air to CO requires about 2.4 moles of air, so the pCO will be a little

less than 0.35 atm — say 0.34. The CO/C02 ratio from Equation [9.20] is then about 150.
Considering H20 next, the estimated pCO « 0.34 atm is inserted into the Кщ expression for
Equation [9.21] to find that the H2/H20 ratio is about 110. These ratios will be higher above 1200
K, so unless a very precise calculations is desired, it's reasonable to exclude C02 and H20 as
significant contributors to the heat balance.

Having resolved that the reducing gas is essentially a mixture of N2, CO and H2, the material
balance is relatively simple. On a basis of one mole of air and 0.05 mole of steam in, 0.47 mole of
carbon is consumed to produce 0.47 mole of CO, 0.05 mole of H2, and 0.79 mole of N2. The
reducing gas composition is (jpCO = 35.9 %, <jpH2 = 3.8 %, balance N2.

There are a number of ways to structure an uncoupled heat balance to solve for an unknown
temperature. Here we choose a graphical solution, similar to ones we've used before. This is done
by setting up a ledger for the various heat effect terms and making sample calculations at 1200 К
and above. First we calculate molar АЯ°ГХ for the formation of CO and H20 from the elements at
298 K, and convert these values to the selected process basis amounts. Second we add the
enthalpy brought in by 0.05 mole of steam in cooling to 298 K. Third we calculate the enthalpy of
the three product gases at different temperatures. The sum of these heat effects will equal zero at
the AFT, which will be determined graphically.

Step 1. A//°form for CO(g) at 298 К - -110.53 kJ/mol
A//°form for H20(g) at 298 К = -241.81 kJ/mol
АЯ°ГХ for process at 298 К - 0.47(-l 10.53) - 0.05(-241.81) - -39.86 kJ

Step 2. #298-#45o for 0.05 mol H20(g) - -0.26 kJ

Step 3. Net heat available for product gas enthalpy = -40.12 kJ

Step 4. Enthalpy of product gas between 1200 - 1360 K. Heat balance and graphical solution
shown in table and Figure 9.9.

Ят-Я298, kJ/mol 1 Process gas | Net heat

Γ,Κ CO H2 N2 Hj-Нгщ, kJ kJ

1200 28.44 26.81 28.12 36.92 -3.20

1240 29.81 28.05 29.47 38.69 -1.43

1280 31.18 29.30 30.83 40.47 0.35

1320 32.57 30.56 32.19 42.26 2.14

1360 33.95 31.82 33.56 44.06 3.94

Chapter 9 System Balances on Reactive Processes 523
Heat Balance for Gas Production Process

+s 4

CO
0)

£2

CO
C/)

ì*ОÜ 0

фс

1200 1240 1280 1320 1360

process temperature, deg К

Figure 9.9 Results of heat balance calculation for the adiabatic production of a reducing gas of
<pCO = 35.9 %, <pH2 = 3.8 %, balance N2. Basis is 1 mole of air, 0.05 mole of steam, and 0.47
mole carbon reacting adiabatically. Steam enters at 450 K, all else enters at 298.15 K. ΣΗ= 0 at
about 1270K.

The result shows a net heat = 0 (the AFT) near 1270 K. A linear Trendline fit to the results
gives an equation which shows an AFT of 1271 K. Looking back, the estimates used to eliminate
C02 and H20 from contention as factors in the reducing gas composition were conservative and
reasonable.

Assignment. Calculate the air/steam ratio that would give an AFT = 1240 K.

EXAMPLE 9.4 — Adiabatic Reforming of Fuel Oil.

Fuel oil with 87.4 %C, balance H has a LHV of 9935 cal/g. A mixture of steam and oxygen
are used for reforming, which takes place in a catalytic reactor between 1200 and 1300 K. The
reformed gas is to have a degree-of-oxidation value X of 0.06 (please see Equation [6.83] for a
definition of this term). The oil and oxygen enter the system at 298 K, while the steam is at 600 K.
Calculate the adiabatic reaction temperature as a function of the amount of steam used, per mole of
oil. The WGR reaches equilibrium when the ART is above 1200 K.

Data. All data is from FREED, here in units of cal/mol. We need heat content data for steam at
600 K, and in equation form for CO, C02, H2, and H20. We also need A//°form at 298 К for CO,
C02, H20, and the fuel oil. Finally, we need Кщ for the WGR. First, FREED shows that Я60о-
H29s for steam = 2504 cal/mol. Ят-Я298 equations for the other gases can be derived from
FREED's tables. Since the temperature range for reaction products is so short, we can use linear
heat content equations. A chart of heat content data is shown in Figure 9.10, together with Ят-//298
equation parameters valid between 1200 and 1300 K. The AftPform at 298 К for compounds is
shown below. Ai/°form for the fuel oil (which has a molar mass of 13.742) was calculated by
Equation [9.17]. Finally, Кщ for the WGR comes from Figure 6.32.

AH°form (cal/mol): CO(g) - - 2 6 420; C02(g) =-94 050; H20(g) = -57 800; CH1718(/) = -7150.

Solution. The problem statement indicates that the material and heat balance are coupled. The
problem asks for an ART for various amounts of steam (per mole of oil), so our amount basis is
one mole of oil. We have different temperatures, so the default temperature basis is 298 K. In the
beginning, we have no clue as to what amount of steam will give ART values between 1200 and
1300 K, but a rough hand calculation indicated that about 0.2 mole of steam would give an ART
about 1280 K, and higher amounts of steam would lower the AFT. Therefore, steam amounts
between 0.18 and 0.24 were chosen as inputs. The following terms were used to designate the
molar amount of each gas present after reforming: M = CO, D = C02, H = H2, and W = H20.

524 Chapter 9 System Balances on Reactive Processes
Heat Content of Reformer Gases
12000
11000 О Н2 О С02
О 10000 Δ CO □ H20
Linear (H2) Linear (CO)
s - - ■ Linear H20) — — « Linear C02)
1.4Ì-4-f"El·
£ 9000

H20: HT-H298 = 10.57T - 4440

6000 1210 1220 1230 1240 1250 1260 1270 1280 1290 1300
1200 temperature, К

Figure 9.10 Heat content and heat content equations for reformer gases near 1250 K.

The reforming of oil by oxidation with steam and oxygen was described in Section 6.4.

CHi.718(/) + rcH20(g) + n O2(g) ^ M + D + H + W

The material balance equations are rather simple when we specify the amount of oil (one
mole) and steam (n):

Carbon balance: M + D = 1. Hydrogen (H2) balance: 0.859 + n = H + W

In addition, there are two other equations involving the gas species. First, Кщ for the WGR,
which is dependent on temperature, and second, the degree of oxidation X, specified as 0.06.
These four equations comprise the material balance set, which were written in a format for Excel's
Solver tool. As has been customary when using Solver, the equations are arranged so that the LH
side is set equal to zero.

Keq =_ i1л0980/(Г-273)-1.198 _ HxD X = 0.06 = - D + W
MxWT D+W+M+H

Finally, we need an overall heat balance involving temperature. The first term is for cooling
steam to 298, the second for carrying out the chemical reaction, and the third for heating the
products to the ART. The overall heat is the sum of these three terms, which is zero for a process.

The five system balance equation set was written in a format for Excel's Solver tool. And
since there is an equation set for each value of steam, SuperSolver was used to solve the sets.
Once the set was solved, the amount of oxygen required (n ') was calculated:

η' = Μ + 2Ό + Ψ-η

Figure 9.11 shows results for steam amounts between 0.19 and 0.23. Trendline fit equations
on the figure show that the amount of steam and oxygen required to reform one mole of oil follows
a linear trend with ART. A noteworthy point is the large effect of steam on ART. This is because
the AH°f0Tm of steam is rather large compared to the amount of heat required to change the
temperature of the product gases by 100 degrees.

In this example, we chose to set the ART as a dependent variable, and the amount of steam as
a dependent variable. However, it would have been equally feasible to reverse the selection
criteria, and solve for the amount of steam. The complexity of the arithmetic in this case is about