CHEE 221: Chemical Processes and Systems

Module 4. Energy Balances without Reaction 

Part c: Calculating Changes in Enthalpy

(Felder & Rousseau Ch 8)

CHEE 221 2

Energy Balances. F&R Chapter 8

How do we calculate enthalpy (and internal energy) changes when we don’thave tabulated data (e.g., steam tables) for the process species?Basic procedures to calculate enthalpy (and internal energy changes)associated with the following processes are covered in Chapter 8 (noReaction): Remember, 3 pieces of information set the thermodynamic state ofmatter: P, T and Phase.

• with change in P (at constant T and phase) (F&R 8.2)

• with change in T (at constant P and phase) (F&R 8.3)

• with change in phase (at constant T and P) (F&R 8.4)

To keep track of our calculations, we will summarize enthalpies in an Inlet‐Outlet Enthalpy Table.  

Sections not covered in Ch 8 include 8.3c, 8.3e, 8.4b, 8.4d, 8.4e, 8.5

H

H

H

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Hypothetical Process Paths

To calculate enthalpy changes, we need to construct a hypothetical    process path, with:

– A starting point:  your defined reference state (phase, T and P)– An end point:  the conditions of the stream of interest (inlet or outlet)

Since and are state properties (values are dependent only on thestate of the species (phase, T and P) and not how they got there), anyconvenient process path from a reference state to a process state can bechosen

The ideal process path will allow you to make use of:  – sensible heats  heat capacities (Table B2)– phase transition temperatures  Tm, Tb (at which data are often listed, 

e.g. latent heats) (Table B1)– latent heats  heat of vapourization, heat of melting (Table B1)

U H

F&R Ch 8.1b

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Hypothetical Process Paths

Calculate the enthalpy change as 1 kg of ice at 0C is transformed to superheated steam at 400C at 10 bar.  

We could use the steam tables and find (Remember: final‐initial):  

How would we calculate the enthalpy change if we didn’t have the steam tables?

H2O(s)

0C, 1 atm atm 10 C,400OH (v)2

?ˆ H

ice ofenthalpy the is kJ/kg 334- kJ/kg 3612334)(3264 H

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Construction of the Hypothetical Process Path

54321ˆˆˆˆˆ0ˆˆ HHHHHHH final

1H 5H

3H

?ˆ H

4H2H

H2O(s)

(0C, 1 atm)H2O(v)

(400C, 10  atm)

H2O(l)

(0C, 1 atm)

H2O(l)

(100C, 1 atm)H2O(v)

(100C, 1 atm)

H2O(v)

(400C, 1 atm)

true path

B.1) (Table OH of )(point melting normal

the at

2m

m

T

HH ˆˆ1

B.2) Table in found are sexpression

100

0

p

OHp

C

dTCH l

(

ˆ )(22

B.1) (Table OH of )(point boiling normal

the at

2b

v

T

HH ˆˆ3

B.2) Table in foundare sexpression

400

100

p

OHp

C

dTCH v

(

ˆ )(24

P) in changes small forassumption reasonable A (

0ˆ5 H

Our chosen reference state

CHEE 221 6

Changes in H and U with P (constant T, phase)

Ideal Gases:– By definition,                (molecules don’t interact, so changing P doesn’t 

change the internal energy)– H = U + PV but PV = nRT = 0 (constant T) 

Non‐Ideal Gases:– Changes in internal energy and enthalpy are small, provided P is small 

(< 5 atm) – For steam, use tabulated values

Liquids and Solids:  –– (but still very small)

0ˆ U

0ˆ H

0ˆ UVPH ˆˆ

0ˆˆ HU

In a problem, state that changes in U and H with respect to pressure are small and will be neglected (except for steam tables).  

F&R Ch 8.2

CHEE 221 7

Does H change with P? Steam Enthalpy Diagram

Use the steam tables todetermine the state (liquid,vapour or mixture of the two;saturated or supersaturated) andapproximate temperature (noneed to use extrapolate) of 1 kgof water at 1 bar with thefollowing enthalpies (relative toliquid water at the triple point)

a) 100 kJb) 419 kJc) 1500 kJd) 2676 kJe) 3000 kJ

Conclusion: Enthalpy is not a strong function of pressure below 10 bar

CHEE 221 8

Phase Changes (at constant T and P): Latent Heat

Phase changes occur from the solid to the liquid phase, and from the liquid tothe gas phase, and the reverse. The specific enthalpy change (heat)associated with the phase change at constant T and P is known as the latentheat of the phase change (i.e., latent heat of vapourization or simply heat ofvapourization).

:),(ˆ PTHm

:),(ˆ PTHv liquid  gas

solid  liquid

Hmelting

Hvapourization

Table B.1 reports these two latent heats for substances at their normal melting and boiling points (i.e., at a pressure of 1 atm).

F&R Ch 8.4a

CHEE 221 9

Changes in U and H with T (constant P, phase): Sensible Heat

Sensible heat refers to heat that must be transferred to raise or lower the temperature of a substance without change in phase.  

Hmelting

Hvapourization

Tinitial Tfinal

1) Sensible heat of solid,       H (Tinitial Tmelting)

2) Sensible heat of liquid,      H (Tmelting Tvapourization)

3) Sensible heat of gas,      H (Tvapourization Tfinal)

F&R Ch 8.3a

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Changes with T (constant P, phase): Sensible Heat

Closed System‐‐Find  U . The quantity of sensible heat required to produce a temperature change in a system can be determined from the appropriate form of the first law of thermodynamics:  

2

1

)(ˆ

ˆˆlim)(

0

T

Tv

VTv

dTTCU

TU

TUTC

Slope = Cv = heat capacity at constant volume

Ideal gas: exactSolid or liquid: good approximationNonideal gas: valid only if V constant

Q =  U (closed system; must be kept at constant volume)

CHEE 221 11

Changes with T (constant P, phase): Sensible Heat

Open System‐‐Find H. Enthalpy, like internal energy, also depends strongly on temperature.  

2

1

)(ˆ

ˆˆlim)(

0

T

Tp

PTp

dTTCH

TH

THTC

Ideal gas: exactSolid or liquid: good approximationNonideal gas: exact only if P constant

Cp = heat capacity at constant pressure

Liquids and Solids:  Cp Cv

Ideal Gases:  Cp = Cv + R

Q =  H ˙ (open system; calculate at constant pressure)

CHEE 221 12

Heat Capacity Formulas

Heat capacity – the amount of heat required to raise thetemperature of one mole or one gram of a substance by onedegree Celsius without change in phase

units:  

If Cp were constant, our job would be easy: H = CP (T2‐T1)But, heat capacities are functions of temperature and are expressed in polynomial form:  

Cp = a + bT + cT2 + dT3 (Form “1”)or,

Cp = a + bT+ cT‐2 (Form “2”)

Values of coefficients a, b, c, and d are given in Table B.2.

C gcal or

K molJ

F&R Ch 8.3b

CHEE 221 13

Heat Capacity Calculations – Integration

32 dTcTbTaC p C molkJ

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

dTdTcTbTaHT

T

41

42

31

32

21

2212

32

432

2

1

molkJ

CHEE 221 14

Notes Regarding Table B.2

Be sure you use the correct functional form– Cp = a + bT + cT2 + dT3  (Form 1) or Cp = a + bT+ cT‐2  (Form 2)

Temperature units are sometimes K and sometimes C

Positive exponent in table heading means you use negative exponent in the expression– E.g., if a x 103 = 123.0  a = 123.0 x 10‐3

Be careful when you integrate!  (T22 – T12)  (T2 – T1)2

CHEE 221 15

Specific Enthalpies of Gases – Table B.8

Table B.8 lists specific enthalpies (kJ/mol) of selected gases(mainly combustion products, i.e. this table might be usefulwhen solving energy balances for combustion problems) as afunction of temperature.

• Can be used to estimate H changes as an alternative to integrating the Cpequation.

• Interpolation may be required.

• The reference state of these gases is:  1 atm and 25C.

• Use this table as you would for the steam tables, however, note that for H2O, the units and reference state are different than the steam tables.

F&R Ch 8.3b

CHEE 221 16

Putting it all together: Hypothetical Process Path

54321ˆˆˆˆˆ0ˆˆ HHHHHHH final

1H 5H

3H

?ˆ H

4H2H

H2O(s)

(0C, 1 atm)H2O(v)

(400C, 10  atm)

H2O(l)

(0C, 1 atm)

H2O(l)

(100C, 1 atm)H2O(v)

(100C, 1 atm)

H2O(v)

(400C, 1 atm)

true path

B.1) (Table OH of )(point melting normal

the at

2m

m

T

HH ˆˆ1

B.2) Table in found are sexpression

100

0

p

OHp

C

dTCH l

(

ˆ )(22

B.1) (Table OH of )(point boiling normal

the at

2b

v

T

HH ˆˆ3

B.2) Table in foundare sexpression

400

100

p

OHp

C

dTCH v

(

ˆ )(24

P) in changes small forassumption reasonable A (

0ˆ5 H

reference state

CHEE 221 17

Calculating enthalpy changes: Examples

1. Calculate the increase in specific enthalpy that occurs whenacetone(v) is heated from 25 C to 100 C.

2. A stream of nitrogen flowing at a rate of 1kg/min is heatedfrom 50C to 200C. Calculate the heat that must betransferred.

3. Fifteen kg/min of air is cooled from 400C to 30C. Calculatethe required heat removal rate.

4. Estimate the increase in specific enthalpy when H2O(v) isheated from 300 C to 450 C.

CHEE 221 18

Constructing a Process Pathway: Example 8.4‐2

One hundred moles per second of liquid hexane at 25 ºC and 7 bars pressure is vaporized and heated to 300 ºC at constant pressure. Estimate the rate at which that must be supplied.

CHEE 221 19

Procedure for Energy Balance Calculations

1. Draw and completely label a process flow diagram2. Perform all material balance calculations3. Choose a reference state (phase, T, P) for each species involved

– If using enthalpy tables, use reference state used to generate table– If no tables are available, choose one inlet or outlet condition as the 

reference state for the species   4. Construct an inlet‐outlet enthalpy table

– Columns for inlet and outlet amounts of each species along with their corresponding Ûi or Ĥi values

– Use a separate row for each phase of a species– Identify unknowns with variables (e.g., Ĥ1, Ĥ2, etc.)

F&R Ch 8.1c, 8.3d, 8.4c

CHEE 221 20

Procedure for Energy Balance Calculations (cont’d)

6. Calculate all required values of Ûi or Ĥi and insert values into table7. Calculate U or H (e.g., H=miĤi‐miĤi)8. Write the appropriate form of the energy balance equation, remove any negligible term, and calculate any other terms in the energy balance equation (i.e., W, Ek, Ep) 

9. Solve for the unknown quantity in the energy balance equation– Typically solve for Q but sometimes asked to solve for a mass (mole) 

flow or occasionally a T.

F&R Ch 8.1c, 8.3d, 8.4c

CHEE 221 21

Example 1:  F&R 8.3‐5

A stream containing 10% CH4 and 90% air by volume is to beheated from 20C to 300C. Calculate the required rate of heatinput in kilowatts if the flow rate of the gas is 2.00 x 103 litres(STP)/min.

CHEE 221 22

Example 2: F&R 8.1‐1

Acetone (denoted as Ac) is partially condensed out of a gas stream containing66.9 mole% acetone vapour and the balance nitrogen. Process specificationsand material balance calculations lead to the flowchart shown below.

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

CHEE 221 23

Example 3: Final Exam 2006

In the following process for condensing methanol vapour from air most of theentering methanol is liquefied in this steady‐state process, with the remainingfraction exiting with the air stream. Both exit streams are at 0 ºC and 5 atm.Shaft work is delivered to the system at a rate of 30 kW to achieve thecompression.

Construct an inlet‐outlet enthalpy table for the process, and calculate allunknown enthalpies. Identify the reference states selected for thecomponents, and state all assumptions. What is the rate (kW) at which heatmust be removed from the condenser?

Q Ws

0 °C 5 atm

0.518 mol MeOH(l) / s

0 °C 5 atm 5.760 mol/s 0.10 mol MeOH(v) /mol 0.90 mol air/mol

150 °C 1 atm

5.184 mol air/s 0.058 mol MeOH(v) /s