96719302871

Chapter 8 Balances on Nonreactive Processes8.1 Elements of Energy Balances Calculations

8.1a Reference States – A Review

We can never know the absolute values of and for a species at a given state. Fortunately, we never need to know the absolute values of and at specifiedstates; we only need to know and for specified changes of state, and we can determine these quantities experimentally.

We may therefore arbitrarily choose a reference state for a species and determine for the transition from the reference state to a series of other states.

If we set equal to zero, then for a specified state is the specific internal energy at that state relative to the reference state. The specific enthalpies at each state can then be calculated from the definition, , provided that the specific volume ( ) of the species at the given temperature and pressure is known.

The values of and in the steam tables were generated using this procedure. The reference state was chosen to be liquid water at the triple point [H2O(l, 0.01C, 0.00611 bar)], at which point was defined to be zero.

U HU H

HU

refUUU

refU )U(U

VPUH V

U H

This does not mean that the absolute value of for water vapor at 400C and 10.0 bar is 2958 kJ/kg. It means that of water vapor at 400C and 10.0 bar is 2958 kJ/kg relative to water at the reference state, or

U

U

kg/kJ�2958U��)bar�0.10�,C400�,v(OH)bar�00611.0�,C01.0�,l(OH 22

10 bar 0.307 m3

kg 1 m3

103 L 8.31410-3 kJ/(molK)0.08314 Lbar/(molK)

kg/kJ�2958��VPUH

kg/kJ�3264

The specific enthalpy of water vapor at 400C and 10.0 bar is

Test Yourself p. 359

and are state properties of a species; that is, their values depend only on the state of the species – primarily on its temperature, state of aggregation (solid, liquid or gas), and, to a lesser extent, on its pressure (and for mixture of some species, on its mole fraction in the mixture).

A state property does not depend on how the species reached its state. Consequently, when a species passes from one state to another, both and for the process are independent of the path taken from the first state to the second one.

8.1b Hypothetical Process Paths

U H

UH

In most of this chapter and in chapter 9, we will learn how to calculate internal energy and enthalpy changes associated with certain processes; specifically,

1.Changes in P at constant T and state of aggregation (Section 8.2).2.Changes in T at constant P and state of aggregation (Section 8.3).3.Phase changes at constant T and P – melting, solidifying, vaporizing, condensing, sublimating (Section 8.4).4.Mixing of two liquids or dissolving of a gas or a solid in a liquid at constant T and P (Section 8.5).5.Chemical reaction at constant T and P (Chapter 9).

Once we know how to calculate and for these five step of processes, we can calculate these quantities for any process by taking advantage of the fact that and are state properties. The procedure is to construct a hypothetical process path from the initial state to the final state consisting of a series of steps of the given five steps.

Having done this, we calculate for each of the steps, and then add the for the steps to calculate for the total process. Since is a state property, calculated for the hypothetical process path – which we constructed for convenience – is the same as for the path actually followed by the process.

HUHU

H s'HH H H

H

For example, we wish to calculate for a process in which solid phenol at 25C and 1 atm is converted to phenol vapor at 300C and 3 atm.

H )atm�1�,C25�,soild(H)atm�3�,C300�,vapor(HH

However, we do not have such a table.

654321 HHHHHHH

Step 1,3, 5 Type 2 (change in T at constant P)

Step 2,4 Type 3 (change in phase at constant T and P)

Step 6 Type 1 (change in P at constant T)


8.1c Procedure for Energy Balance Calculations

The procedure to follow for the energy balance calculation.

1.Perform all required material balance calculations.

2.Write the appropriate form of the energy balance (closed or open system) and delete any of the terms that are either zero or negligible for the given process system.

3.Choose a reference state – phase, temperature, and pressure – for each species involved in the process.

4.For a closed system, construct a table with columns for initial and final amounts of each species (mi or ni) and specific internal energies relative to the chosen reference state ( ). For an open system, construct a table with columns for inlet and outlet stream component flow rates ( or ) and specific enthalpies relative to the chosen reference states.

iU

im in

5.Calculate all required values of ( or ) and insert the values in the appropriate places in the table.

6.Calculate

7.Calculate any work, kinetic energy, or potential energy terms that you have not dropped from the energy balance.

8.Solve the energy balance for whichever variable is unknown (often Q or ).

iU iH

�UmUmor� ��UnUnUinitial

iifinal

iiinitial

iifinal

ii

�HmHmor� ��HnHnHin

iiout

iiin

iiout

ii

Closed System

Open System

Q

pk EEUWQ

pks EEHWQ

Closed System

Open System

1.Perform required material balance calculations. None are required in this example.

2.Write and simplify the energy balance.

3.Choose reference states for acetone and nitrogen. N2 (g, 25C, 1 atm) Ac (l, 20C, 5 atm)

Example 8.1-1

Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole% acetone vapor and the balance nitrogen. Process specifications and material balance calculations lead to the flowchart shown below.

The process operates at steady state. Calculate the required cooling rate.

Solution

pks EEHWQ iin

iiout

i HnHnHQ

4.Construct an inlet-outlet enthalpy table.

5.Calculate all unknown specific enthalpies.

�)atm�1�,C65�,v(Ac)atm�5�,C20�,l(forAc� �H��

)atm�5�,C20�,l(toAc�relative� �)atm�1�,C65�,v(Ac�nthalpyof�specifice�H1

�)atm�1�,C65�,v(Ac�)atm�1�,C56�,v(Ac

)atm�1�,C56�,l(Ac)atm�1�,C20�,l(Ac)atm�5�,C20�,l(Ac

d1c1

b1a1

HH

HH

dT)C(�)H(dT)C(�)atm5atm1(V��

HHHHH

)v(Acp

C65

C56Acv)l(Acp

C56

C20)l(Ac

d1c1b1a11

Table B.1

312285p

5p

T1076.34T1078.12T1010.2007196.0Cmol

kJC��:)v(Ac

T106.18123.0Cmol

kJC��:)l(Ac

Table B.1

Table B.10.0734 L/mol 30.2 kJ/mol

mol/kJ�7.35mol/kJ)753.02.3068.40297.0(H1

6.Calculate H

s/kJ�2320��s/kJ)]16.1)(1.33()7.35)(9.66()10.0)(1.33()0)(55.63()0.32)(35.3[(��

HnHnH iin

iiout

i

7.Calculate nonzero work, kinetic energy, and potential energy terms. Nothing to do in this step.

8.Solve the energy balance for Q

kW�2320s/kJ�2320HQ

Heat must be transferred from the condenser at a rate of 2320 kW to achieve the required cooling and condensation.

8.2 Changes in Pressure at Constant Temperature

It has been observed experimentally that internal energy is nearly independent of pressure for solids and liquids at a fixed temperature, as is specific volume.

If the pressure of a solid or liquid changes at constant temperature

0U

PVPVVP0)VP(UH

For a gas undergoing an isothermal pressure change unless gases at temperature well below 0C or well above 1 atm are involved.

0U

0)RT(0)VP(UH ideal gas


If tables of or are available for the gas, there is of course no need to make this assumption.

)P,T(U )P,T(H

If gases are far from ideal or if they undergo large pressure changes, you must either use tables of thermodynamic properties (such as the steam tables for water) or thermodynamic correlations to determine or .

U H

8.3 Changes in Temperature

Sensible Heat : heat that must be transferred to raise or lower the temperature of a substance or mixture of substances.

The quantity of heat required to produce a specified temperature change in a system can be determined by the appropriate form of the first law of thermodynamics:

8.3a Sensible Heat and Heat Capacities

)mopensyste�(��HQ

)temclosedsys�(U ��Q

We have neglected kinetic and potential energy changes and work.

To determine the sensible heat requirement for a heating or cooling process, you must therefore be able to determine U or for the specified temperature change.�H

The specific internal energy of a substance depends strongly on temperature. If the temperature is raised or lowered in such a way that the system volume remains constant, the specific internal energy might vary as shown in the right plot:

1urveatT� �slopeofc� �T

U0T

V0T

v T

U

T

Ulim)T(C

heat capacity at constant volume 2

1

T

T v1122 dT)T(C)T(U)T(UU

Suppose both temperature and volume of a substance change. To calculate , you may break the process into two steps – a change in at constant T followedby a change in T at constant .

U

VV

)V,T(A)V,T(A)V,T(A 22U

21U

1121

U

21 UUU is a state propertyU

2

1

T

T v221 dT)T(CU0UUU

ideal gas: exactsolid or liquid: a good approximationnonideal gas: valid only if V is constant.

Example 8.3-1

Calculate the heat required to raise 200 kg of nitrous oxide from 20C to 150C in a constant-volume vessel. The constant-volume heat capacity of N2O in this temperature range is given by the equation

where T is in C.

T1042.9855.0)Ckg/kJ(C 4v

Solution

kg/kJ�121kg/kJ)4.10111(��

2

T1042.9T855.0��

dT(Ckg

kJ)T1042.9855.0()kg/kJ(U

C150

C20

24C150

C20

C150

C20

4

kJ�200,24)kg/kJ�121)(kg�200()kg/kJ(U)kg(mU �Q

The energy balance for this closed system is

Suppose both temperature and pressure of a substance change. To calculate , you may break the process into two steps – a change in P at constant T followedby a change in T at constant P.

H

)P,T(A)P,T(A)P,T(A 22H

21H

1121

H

21 HHH is a state propertyH

2

1

T

T p dT)T(CHideal gas: exactnonideal gas: valid only if P is constant.

P0T

p T

H

T

Hlim)T(C

heat capacity at constant pressure 2

1

T

T p1122 dT)T(C)T(H)T(HH

)idealgas�(��0H1

)iquidsolidorl� �(P��VH1

solid or liquid 2

1

T

T p dT)T(CPVH


tabulated enthalpy thermodynamic relation for variations of with P H

8.3b Heat Capacity Formulas

Heat capacities are functions of temperature and are frequently expressed in polynomial form

Values of the coefficients a, b, c, and d are given in Table B.2 of Appendix B for a number of species at 1 atm, and listings for additional substances aregiven on pp.2-161 to 2-186 of Perry’s Chemical Engineers’ Handbook.

32p dTcTbTaC

Simple relationships exist between Cp and Cv in two cases:RCC��:sIdealGase�

CC��:dSolids�Liquidsan�

vp

vp

The relationship between Cp and Cv for nonideal gases is complex.

Example 8.3-2

Assuming ideal gas behavior, calculate the heat that must be transferred in each of the following cases.1.A stream of nitrogen flowing at a rate of 100 mol/min is heated from 20C to 100C.2.Nitrogen contained in a 5-liter flask at an initial pressure of 3 bar is cooled from 90C to 30C.

Solution

1.

pks EEHWQ

312285p T10781.2T105723.0T102199.002900.0

Cmol

kJC

From table B.2 the heat capacity of N2 at a constant pressure of 1 atm is

HQ

mol/kJ�332.2mol/kJ)107109.10106.0320.2(��

4

T10781.2

3

T105723.0

2

T102199.0T02900.0dT)T(CH

53

C100

C20

412

C100

C20

38

C100

C20

25C100

C20

C100

C20 p

mol/kJ�233mol

kJ�332.2

min

mol�100HnHQ

2.312285

p T10781.2T105723.0T102199.002900.0Cmol

kJC

RCC vp

312285v T10781.2T105723.0T102199.002069.0

Cmol

kJC

)J�10/kJ�1)(C1/K�1)](Kmol/(J�314.8[R 3

mol/kJ�250.1mol/kJ)1051034.11092.7241.1(��

4

T10781.2

3

T105723.0

2

T102199.0T02069.0dT)T(CU

533

C30

C90

412

C30

C90

38

C30

C90

25C30

C90

C30

C90 v

pk EEUWQ UQ

mol�497.0)K�363)](Kmol/(barL�08314.0[

)L�00.5)(bar�00.3(RT/PVn

kJ�621.0)mol/kJ�250.1)(mol�497.0(UnUQ

Example 8.3-3

Fifteen kmol/min of air is cooled from 430C to 100C. Calculate the required heat removal rate using (1) heat capacity formulas from Table B.2 (2) specific enthalpies Table B.8.

Solution

pks EEHWQ HnHnHnHQ airin,airairout,airair

(1) The hard way

mol/kJ�98.9mol/kJ)0167.00835.03672.05502.9(��

mol/kJ

)430100(4

10965.1)430100(

3

103191.0

)430100(2

104147.0)430100(02894.0

��

dTT10965.1T103191.0T104147.002894.0��

dT)T(Cmol

kJH

4412

338

225

C100

C430

312285

C100

C430 p

Table B.8

(2) The easy way

Table B.8 mol/kJ�19.2)C100(H

Table B.8mol/kJ�37.14)C500(H

mol/kJ�24.11)C400(H

mol/kJ�17.12mol/kJ)24.1137.14(100

3024.11)C430(H

mol/kJ�98.9mol/kJ)17.1219.2(H

HnQ air15.0 kmol

1 kmol

1 min

s

1 kW

1 kJ/smin

103 mol -9.98 kJ

molkW�2500


8.3c Estimation of Heat Capacities

The polynomial expressions for Cp in Table B.2 are based on experimental data for the listed compounds and provide a basis for accurate calculations of enthalpy changes. Several approximate methods for estimating heat capacities in the absence of tabulated formulas are presented.

Kopp’s rule is a simple empirical method for estimating the heat capacity of a solid or liquid at or near 20C. According to the rule, Cp for a molecular compound is the sum of contributions (given in Table B.10) for each atomic element in the compound.

)Cmol/(J�79��)Cmol/(J)]6.92()172(26[��

)C(2)C(2)C()C( HpaOpaCapa)OH(Cap 2

true value is 89.5 J/(molC)

Suppose we wish to calculate the enthalpy change associated with a change in temperature undergone by a mixture of substances. Enthalpies and heat capacities of certain mixtures are tabulated in standard references. Lacking such data, we may use the following approximation:

Rule 1 : For a mixture of gases or liquids, calculate the total enthalpy change as the sum of the enthalpy changes for the pure mixture components. The enthalpy changes associated with the mixing of the components are neglected.Rule 2 : For highly dilute solutions of solids or gases in liquids, neglect the enthalpy change of the solute. The more dilute the solution, the better this approximation.

The calculation of enthalpy changes for the heating or cooling of a mixture of known composition may often be simplified by calculating a heat capacity for the mixture in the following manner:

scomponent�mixture��all��

piimixp )T(Cy)T()C((Cp)mix = heat capacity of the mixtureyi = mass or mole fraction of the ith componentCpi = heat capacity of the ith component

If Cpi and (Cp)mix are expressed in molar units, then yi must be the mole fraction of the ith component, and if the heat capacities are expressed in mass units, then yi must be the mass fraction of the ith component.

2

1

T

T mixp dT)T()C(H(Cp)mix is known

Valid to the extent that enthalpies of mixing may be neglected.

Example 8.3-4

Calculate the heat required to bring 150mol/h of a stream containing 60% C2H6 and 40% C3H8 by volume from 0C to 400C. Determine a heat capacity for the mixture as part of the problem.

Solution

Table B.2

312285ethane T10280.7T10816.5T1092.1304937.0)T(C

312285propane T1071.31T1011.13T1059.2206803.0)T(C

312285

propaneethanemixp

T1005.17T10734.8T1039.1705683.0��

C400.0C600.0)]Cmol/(kJ[)C(

mol/kJ�89.34dT)T()C(HC400

C0 mixp

HnHQ 150 mol

molh

34.89 kJ

h

kJ5230


8.3d Energy Balances on Single-Phase Systems

We are now in position to perform energy balances on any processes that do not involve phase changes, mixing steps for which enthalpy changes cannot be neglected, or chemical reactions.

If a process involves heating or cooling a single species from T1 to T2, the procedure is straightforward:

1. Evaluate

2. For a closed system

For a open system

3.Substitue for U, H, or in the approximate energy balance equation to determine the required heat transfer, Q, or heat transfer rate, .

2

1

T

T vdTCU 2

1

T

T pdTCH correcting for pressure changes if necessary.

)tP�tanatcons�(��HnH

)tV�tanatcons�(��UnU

�HnH

HQ

pk EEUWQ

pks EEHWQ

Closed System

Open System

Example 8.3-5

A stream containing 10% CH4 and 90% air by volume is to be heated from 20C to 300C. Calculate the required rate of heat input in kilowatts if the flow rate of the gas is 2.00103 liters (STP)/min.

SolutionBasis: Given Flow Rate

n2000 L (STP)

22.4 L (STP)

min

1 mol

min

mol3.89

pks EEHWQ in

iiout

ii HnHnHQ

)etmixture�Pinoutl� � �,C300�,g(CH)atm�1�,C20�,g(CH 44

mol/kJ�09.12��

dT)T100.11T103661.0T10469.503431.0(��

dT)C(H

C300

C20

312285

C300

C20 CHp1 4

neglecting effect of pressure on enthalpy

neglecting heats of mixing of gases

Table B.8 mol/kJ�17.8H��,mol/kJ�15.0H 32

min/kJ�776��

min/kJ)]15.0)(4.80()0)(93.8()17.8)(4.80[()09.12min)(/mol�93.8(HnHnHQin

iiout

ii

776 kJ

60 smin

1 min 1 kW

1 kJ/sQ kW�9.12

Example 8.3-6

A gas stream containing 8.0 mole% CO and 92.0 mol% CO2 at 500C is fed to a waste heat boiler, a large metal shell containing a bundle of small-diameter tubes. The hot gas flows over the outside of the tubes. Liquid water at 25C is fed to the boiler in a ratio 0.200 mol feedwater/mol hot gas and flowed inside the tubes. Heat is transferred from the hot gas through the tube walls to the water, causing the gas to cool and the water to heat to its boiling point and evaporate to form saturated steam at 5.0 bar. The steam may be used for heating or power generation in the plant or as the feed to another process unit. The gas leaving the boiler is flared (burned) and discharged to the atmosphere. The boiler operates adiabatically – all the heat transferred from the gas goes into the water, as opposed to some of it leaking through the outside boiler wall. The flowchart for an assumed basis of 1.00 mol feed gas is shown below.

What is the temperature of the exiting gas?

Solution

0HnHnHin

iiout

ii adiabatic

T

C500

312285

T

C500 COp1

dT)T10220.2T103548.0T104110.002895.0(��

dT)C(H

T

C500

312285

T

C500 COp2

dT)T10464.7T10887.2T10223.403611.0(��

dT)C(H2

)H�ressureon�ffectofp� �neglecte� �:5.TableB�(kg ��/kJ�105)]bar�5�,C25�,l(OH[HH 23

)6.TableB�(kg ��/kJ�5.2747)]d'sat�,bar�5�,v(OH[HH 24

C��299T 0H

8.3e Numerical Integration of Tabulated Heat Capacities

2

1

T

T pdTC

If a function relation for Cp(T) is available, such as one of polynomial of Table B.2, the integration can be carried out analytically; and if tabulated specific enthalpies are available for the substance being heated or cooled, a simple subtraction replaces the integration.

The only information you have about Cp is its value at a series of temperatures that span the range from T1 to T2. The question is how to estimate the value of the integral from these data.

A better solution is to use one of the many existing quadrature formulas – algebraic expressions that provide estimates of the integrals of tabulated data. Several such formulas are presented and illustrated in Appendix A.3.

8.4 Phase Change Operations

Phase changes such as fusion and vaporization are usually accompanied by large changes in internal energy and enthalpy. Heat transfer requirements in phase change operations consequently tend to be substantial, since (closed constant-volume system) or (open system).

UQ HQ

8.4a Latent Heats

The specific enthalpy change associated with the transition of a substance from one phase to another at constant temperature and pressure is known as the latent heat of the phase change.

Latent heats for the two most commonly encountered phase changes are defined as follows:

1.Heat of fusion (or heat of melting). is the specific enthalpy difference between the solid and liquid forms of a species at T and P.

2.Heat of vaporization. is the specific enthalpy difference between the liquid and vapor forms of a species at T and P.

)P,T(Hm

)P,T(Hv

Tabulated values of these two latent heats, such as those in Table B.1 and on pp. 2-151 through 2-160 of Perry’s Chemical Engineers’ Handbook, usually apply to a substance at its normal melting or boiling point – that is, at a pressure of 1 atm. The quantities are referred to as standard heats of fusion and vaporization.

g/J�5.2442)mmHg�78.23,C25,OH(H

g/J�3.2442)mmHg�760,C25,OH(H

g/J�7.2258)mmHg�760,C100,OH(H

2v

2v

2v

The latent heat of a phase change may vary considerably with the temperature at which the change occurs but hardly varies at all with the pressure at the transition point. For example:

When using a tabulated latent heat, you must therefore be sure that the phase change in question takes place at the temperature for which the tabulated value is reported, but you may ignore moderate variations in pressure.

Example 8.4-1

At what rate in kilowatts must heat be transferred to a liquid stream of methanol at its normal boiling point to generate 1500 g/min of saturated methanol vapor?

Solution

Table B.1 C7.64molatT�� /kJ�3.35H bpv

pks EEHWQ vHnHQ

1500 g CH3OH

32.0 g CH3OH

1 min

60 s

1 kW

1 kJ/smin

1 mol 35.3 kJ

molQ kW�6.27

Phases changes often occur at temperatures other than the temperature for which the latent heat is tabulated. When faced with this situation, you must select a hypothetical process path that permits the available data to be used.

Example 8.4-2

One hundred g-moles per hour of liquid n-hexane at 25C and 7 bar is vaporized and heated to 300C at constant pressure. Neglecting the effect of pressure on enthalpy, estimate the rate at which heat must be applied.

Solution

pks EEHWQ HQ

Figure 6.1-4The temperature at which the vapor pressure of n-hexane is 7 bar (104 psia) is a approximately 295F (146C).

Table B.1 C69molatT�� /kJ�85.28H bpv

�)bar�7�,C300�,v(HCn)atm�1�,C69�,v(HCn

)atm�1�,C69�,l(HCn)bar�7�,C25�,l(HCn

146

H

146

H

146

H

146

GD

A

C69

C25 )l(HCpA dT)C(PVH146

bar�013.1atm�1

)Cmol/(kJ�2163.0C2.TableB�L/kg�659.0659.0SG1.TableB�

p

1 L 0.008314 kJ/(molK)

0.08314 Lbar/(molK)

0.659 kg

(1.013-7.0) bar 86.17 kg

1000 mol AH

0.2163 kJ

molC

(69-725)C mol/kJ�44.9mol/kJ)517.90782.0(

kg/kJ�85.28)atm�1�,C69()H(H146HCVD

C300

C69 )v(HCpG dT)C(H146

312285v T1066.57T1092.23T1085.4013744.0

Cmol

kJC

mol/kJ�1.47HG

)mol/kJ(H)h/mol(nHQ mol/kJ�5.85HHHH GDA

100 mol

mol

1 kW

1 kJ/sh

85.5 kJ 1 h

3600 sQ kW�38.2

If a phase change takes place in a closed system, you must evaluate for the phase changes to substitute into the energy balance equation.

For phase changes such as fusion, which involve only liquids and solids, changes in are generally negligible compared to changes in , so that

For vaporization, for the vapor (which equals RT if ideal gas behavior may be assumed) is normally orders of magnitude greater than for the liquid, so that , and

)VP(HU

VPH

mm HU

VPRT)VP( VP

RTHU vv


8.4b Estimation and Correlation of Latent Heats

A simple formula for estimating a standard heat of vaporization ( at the normal boiling point) is Trouton’s rule : (30% accuracy)

H

)talcohols�ularweigh�lowmolec� �,water(��)K(T109.0��)iquidsnonpolarl�(��)K(T088.0)mol/kJ(H

b

bv

Another formula that provides roughly 2% accuracy is Chen’s equation:

Tb : normal boiling point

)T/T(07.1

]Plog0297.00327.0)T/T(0331.0[T)mol/kJ(H

cb

c10cbbv

Tb : normal boiling point (K)Tc : critical temperature (K)Pc : critical pressure (atm)

Latent heat of vaporization may be estimated from vapor pressure data by using the Clausius-Clapeyron equation,

BRT

H*pln v

ttaniscons� �Hv

ttannsisnotco� � �Hv Clapeyron equation

R

H

)T/1(d

*)p(lnd v )T(fHv

Another formula that provides roughly 2% accuracy is Chen’s equation:

A useful approximation for estimating at T2 from a known value at T1 is Watson’s correlation:

vH

38.0

1c

2c1v2v TT

TT)T(H)T(H

Tc : critical temperature

)mpoundsorganicco�(��)K(T050.0��)compoundsinorganic �(��)K(T0025.0)mol/kJ(H

)lementsmetallice�(��)K(T0092.0��

m

mm

m

Example 8.4-3

The normal boiling point of methanol is 337.9 K, and the critical temperature of this substance is 513.2 K. Estimate the heat of vaporization of methanol at 200C.

Solution

mol �/kJ�8.36)9.337)(109.0(�)K9.337(Hv

Trouton’s rule

(The measured value is 35.3 kJ/mol, Chen’s equation yields 37.2 kJ/mol, so in this unusual case Trouton’s rule provides the better estimate.)

Watson’s correlation

mol/kJ�0.219.3372.513

4732.5138.36)K473(H

38.0

v

The measured value is 19.8 kJ/mol.


8.4c Energy Balances on Processes Involving Phase Changes

When writing an energy balance on a process in which a component exists in two phases, you must choose a reference states for that component by specifying both a phase and a temperature and calculate the specific enthalpy of the component in all process streams relative to this state.

If the substance is a liquid at its reference state and a vapor in a process stream, may be calculated as outlined in Section 8.4a: that is, bring the liquid from the reference temperature to a point at which is known, vaporize the liquid, bring the vapor to the process stream temperature, and sum the individual enthalpy changes for the three steps.

H

vH

Example 8.4-4

An equimolar liquid mixture of benzene (B) and toluene (T) at 10C is fed continuously to a vessel in which the mixture is heated to 50C. The liquid products is 40 mole% B, and the vapor product is 68.4 mole% B. How much heat must be transferred to the mixture per mole of feed?

Solution

Basis: 1 mol Feed

Degree-of-Freedom Analysis

3 unknown variables (nV, nL, Q)-2 material balances -1 energy balance 0 degree of freedom

LV

LV

n400.0n684.0mol�50.0

nnmol�00.1

mol�648.0n

mol�352.0n

L

V

mol/kJ�332.5dT)C(HC50

C10 )l(HCp1 66

mol/kJ�340.6dT)C(HC50

C10 )l(HCp2 87

mol/kJ�52.37dT)C()C1.80()H(dT)C(HC50

C1.80 )v(HCpHCv

C1.80

C10 )l(HCp3 666666

mol/kJ�93.42dT)C()C62.110()H(dT)C(HC50

C62.110 )v(HCpHCv

C62.110

C10 )l(HCp4 878787

iin

iiout

i HnHnHQ kJ�7.17Q

96719302871

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