Moran M.J., Shapiro H.N. Fundamentals of Engineering Thermodynamics

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If heating continues at 20 lbf/in.

from T

5 778F to T

5 908F,

determine the work for Process 2-3, in Btu. Ans. 0.15 Btu.

Interpolating in Table A-15E at p

5 20 lbf/in.

and T

5 77°F, we get y

5 16.7 ft

/lb. Thus

5 my

0.1 lb

16.7 ft

5 1.67 ft

(b) In this case, the work can be evaluated using Eq. 2.17. Since the pressure

is constant

W 5

p dV 5 p1V

2 V

Inserting values

W 5 120 lbf

in.

211.67 2 1.352ft

144 in.

1ft

1 Btu

778 ft ? lbf

1.18

➊ Note the use of conversion factors in this calculation.

Ability to…

❑

define a closed system and

identify interactions on its

boundary.

❑

sketch T–y and p–y diagrams

and locate states on them.

❑

evaluate work using Eq. 2.17.

❑

retrieve property data for

ammonia at vapor states.

✓

Skills Developed

3.5.2

Saturation Tables

Tables A-2, A-3, and A-6 provide property data for water at saturated liquid, saturated

vapor, and saturated solid states. Tables A-2 and A-3 are the focus of the present

discussion. Each of these tables gives saturated liquid and saturated vapor data. Property

values at saturated liquid and saturated vapor states are denoted by the subscripts

f and g, respectively. Table A-2 is called the temperature table, because temperatures

are listed in the first column in convenient increments. The second column gives the

corresponding saturation pressures. The next two columns give, respectively, the spe-

cific volume of saturated liquid, y

, and the specific volume of saturated vapor, y

Table A-3 is called the pressure table, because pressures are listed in the first column

in convenient increments. The corresponding saturation temperatures are given in the

second column. The next two columns give y

and y

, respectively.

The specific volume of a two-phase liquid–vapor mixture can be determined by

using the saturation tables and the definition of quality given by Eq. 3.1 as follows.

The total volume of the mixture is the sum of the volumes of the liquid and vapor

phases

V 5 V

liq

1 V

vap

Dividing by the total mass of the mixture, m , an average specific volume for the

mixture is obtained

y 5

liq

vap

Since the liquid phase is a saturated liquid and the vapor phase is a saturated vapor,

liq

5 m

liq

and V

vap

5 m

vap

, s o

y 5 a

liq

1 a

vap

b y

Introducing the definition of quality, x 5 m

vap

/ m , and noting that m

liq

/ m 5 1 2 x , the

above expression becomes

y 5 11 2 x2y

1 xy

5 y

1 x1y

2 y

2 (3.2)

3.5 Evaluating Pressure, Specific Volume, and Temperature 103

➊

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104 Chapter 3 Evaluating Properties

The increase in specific volume on vaporization ( y

2 y

) is also denoted by y

consider a system consisting of a two-phase liquid–vapor mixture of

water at 100°C and a quality of 0.9. From Table A-2 at 100°C, y

5 1.0435 3 10

−3

/kg

and y

5 1.673 m

/kg. The specific volume of the mixture is

y 5 y

1 x1y

2 y

25 1.0435 3 10

1 10.9211.673 2 1.0435 3 10

25 1.506 m

Similarly, the specific volume of a two-phase liquid–vapor mixture of water at 212°F

and a quality of 0.9 is

y 5 y

1 x1y

2 y

25 0.01672 1 10.92126.80 2 0.0167225 24.12 ft

where the y

and y

values are obtained from Table A-2E. b b b b b

To facilitate locating states in the tables, it is often convenient to use values from

the saturation tables together with a sketch of a T – y o r p – y diagram. For example, if

the specific volume y and temperature T are known, refer to the appropriate tem-

perature table, Table A-2 or A-2E, and determine the values of y

and y

. A T – y

diagram illustrating these data is given in Fig. 3.8 . If the given specific volume falls

between y

and y

, the system consists of a two-phase liquid–vapor mixture, and the

pressure is the saturation pressure corresponding to the given temperature. The qual-

ity can be found by solving Eq. 3.2 . If the given specific volume is greater than y

the state is in the superheated vapor region. Then, by interpolating in Table A-4 or

A-4E, the pressure and other properties listed can be determined. If the given specific

volume is less than y

, Table A-5 or A-5E would be used to determine the pressure

and other properties.

let us determine the pressure of water at each of three states

defined by a temperature of 100°C and specific volumes, respectively, of y

2.434 m

/kg, y

5 1.0 m

/kg, and y

5 1.0423 3 10

−3

/kg. Using the known

temperature, Table A-2 provides the values of y

and y

: y

5 1.0435 3 10

−3

/kg,

5 1.673 m

/kg. Since y

is greater than y

, state 1 is in the vapor region. Table

A-4 gives the pressure as 0.70 bar. Next, since y

falls between y

and y

, t h e

pressure is the saturation pressure corresponding to 100°C, which is 1.014 bar.

Finally, since y

is less than y

, state 3 is in the liquid region. Table A-5 gives the

pressure as 25 bar. b b b b b

The following example features the use of a sketch of the T – y diagram in conjunc-

tion with tabular data to fix the end states of processes. In accord with the state

principle, two independent intensive properties must be known to fix the states of

the system under consideration.

100°C

3f2

Temperature

Specific volume

Liquid

Saturated

liquid

Saturated

vapor

Critical point

v < v

v > v

< v < v

Vapor

Fig. 3.8 Sketch of a T–y diagram for water used to discuss locating states in the tables.

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Heating Water at Constant Volume

c c c c EXAMPLE 3.2 c

A closed, rigid container of volume 0.5 m

is placed on a hot plate. Initially, the container holds a two-phase

mixture of saturated liquid water and saturated water vapor at p

5 1 bar with a quality of 0.5. After heat-

ing, the pressure in the container is p

5 1.5 bar. Indicate the initial and final states on a T – y diagram, and

determine

(a) the temperature, in °C, at states 1 and 2.

(b) the mass of vapor present at states 1 and 2, in kg.

SOLUTION

Known: A two-phase liquid–vapor mixture of water in a closed, rigid container is heated on a hot plate. The

initial pressure and quality and the final pressure are known.

Find: Indicate the initial and final states on a T – y diagram and determine at each state the temperature and the

mass of water vapor present. Also, if heating continues, determine the pressure when the container holds only

saturated vapor.

Schematic and Given Data:

Analysis: Two independent properties are required to fix states 1 and 2. At the initial state, the pressure and

quality are known. As these are independent, the state is fixed. State 1 is shown on the T–y diagram in the two-

phase region. The specific volume at state 1 is found using the given quality and Eq. 3.2 . That is,

5 y

1 x1y

2 y

From Table A-3 at p

5 1 bar, y

5 1.0432 3 10

/kg and y

5 1.694 m

/kg. Thus,

5 1.0432 3 10

1 0.5

1.694 2 1.0432 3 10

5 0.8475 m

At state 2, the pressure is known. The other property required to fix the state is the specific volume y

. Volume

and mass are each constant, so y

5 y

5 0.8475 m

/kg. For p

5 1.5 bar, Table A-3 gives y

5 1.0582 3 10

/kg

and y

5 1.59 m

/kg. Since

➊ y

, y

➋ state 2 must be in the two-phase region as well. State 2 is also shown on the T–y diagram above.

Engineering Model:

1 . The water in the con-

tainer is a closed system.

2. States 1, 2, and 3 are

equilibrium states.

3 . The volume of the con-

tainer remains constant.

1 bar

1.5 bar

= 0.5 m

Hot plate

= 1 bar

= 0.5

= 1.5 bar

= 1.0

–

Fig. E3.2

3.5 Evaluating Pressure, Specific Volume, and Temperature 105

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106 Chapter 3 Evaluating Properties

(a) Since states 1 and 2 are in the two-phase liquid–vapor region, the temperatures correspond to the saturation

temperatures for the given pressures. Table A-3 gives

5 99.638C



and



5 111.48C

(b) To find the mass of water vapor present, we first use the volume and the specific volume to find the total

mass, m. That is

m 5

0.5 m

0.8475 m

5 0.59 kg

Then, with Eq. 3.1 and the given value of quality, the mass of vapor at state 1 is

5 x

m 5 0.510.59 kg25 0.295 kg

The mass of vapor at state 2 is found similarly using the quality x

. To determine x

, solve Eq. 3.2 for quality and

insert specific volume data from Table A-3 at a pressure of 1.5 bar, along with the known value of y, as follows

y 2 y

f 2

2 y

0.8475 2 1.0528 3 10

1.159 2 1.0528 3 10

5 0.731

Then, with Eq. 3.1

5 0.731 10.59 kg25 0.431 kg

on the T–y diagram of Fig. E3.2. Thus, the pressure would be the corresponding

saturation pressure. Interpolating in Table A-3 at y

5 0.8475 m

/kg, we get

5 2.11 bar.

➊ The procedure for fixing state 2 is the same as illustrated in the discussion

of Fig. 3.8.

➋ Since the process occurs at constant specific volume, the states lie along a

vertical line.

If heating continues at constant specific volume from state 3

to a state where pressure is 3 bar, determine the temperature at that state,

in 8C. Ans. 2828C

Ability to…

❑

define a closed system and

identify interactions on its

boundary.

❑

sketch a T– y diagram and

locate states on it.

❑

retrieve property data for

water at liquid–vapor

states, using quality.

✓

Skills Developed

enthalpy

3.6 Evaluating Specific Internal Energy

and Enthalpy

3.6.1

Introducing Enthalpy

In many thermodynamic analyses the sum of the internal energy U and the product

of pressure p and volume V appears. Because the sum U 1 pV occurs so frequently

in subsequent discussions, it is convenient to give the combination a name, enthalpy,

and a distinct symbol, H. By definition

H 5 U 1 pV (3.3)

Since U, p, and V are all properties, this combination is also a property. Enthalpy can

be expressed on a unit mass basis

h 5 u 1 py (3.4)

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and per mole

h 5 u 1 py (3.5)

Units for enthalpy are the same as those for internal energy.

3.6.2

Retrieving u and h Data

The property tables introduced in Sec. 3.5 giving pressure, specific volume, and tem-

perature also provide values of specific internal energy u, enthalpy h, and entropy s.

Use of these tables to evaluate u and h is described in the present section; the con-

sideration of entropy is deferred until it is introduced in Chap. 6.

Data for specific internal energy u and enthalpy h are retrieved from the property

tables in the same way as for specific volume. For saturation states, the values of u

and u

, as well as h

and h

, are tabulated versus both saturation pressure and satura-

tion temperature. The specific internal energy for a two-phase liquid–vapor mixture

is calculated for a given quality in the same way the specific volume is calculated

u 5 11 2 x2u

1 xu

5 u

1 x1u

2 u

2 (3.6)

The increase in specific internal energy on vaporization (u

2 u

) is often denoted by

. Similarly, the specific enthalpy for a two-phase liquid–vapor mixture is given in

terms of the quality by

h 5 11 2 x2h

1 xh

5 h

1 x1h

2 h

2 (3.7)

The increase in enthalpy during vaporization (h

2 h

) is often tabulated for conve-

nience under the heading h

to illustrate the use of Eqs. 3.6 and 3.7, we determine the specific

enthalpy of Refrigerant 22 when its temperature is 128C and its specific internal

energy is 144.58 kJ/kg. Referring to Table A-7, the given internal energy value falls

between u

and u

at 128C, so the state is a two-phase liquid–vapor mixture. The qual-

ity of the mixture is found by using Eq. 3.6 and data from Table A-7 as follows:

x 5

u 2 u

2 u

144.58 2 58.77

230.38 2 58.77

5 0.5

Then, with the values from Table A-7, Eq. 3.7 gives

h 5 11 2 x2h

1 xh

5 11 2 0.52159.3521 0.51253.9925 156.67 kJ

kg b b b b b

In the superheated vapor tables, u and h are tabulated along with y as functions

of temperature and pressure.

let us evaluate T, y, and h for water at 0.10 MPa and a specific

internal energy of 2537.3 kJ/kg. Turning to Table A-3, note that the given value of u

is greater than u

at 0.1 MPa (u

5 2506.1 kJ/kg). This suggests that the state lies

in the superheated vapor region. By inspection of Table A-4 we get T 5 1208C,

y 5 1.793 m

/kg, and h 5 2716.6 kJ/kg. Alternatively, h and u are related by the

definition of h

h 5 u 1 py

5 2537.3

1 a10

ba1.793

1 kJ

N ? m

5 2537.3 1 179.3 5 2716.6 kJ

3.6 Evaluating Specific Internal Energy and Enthalpy 107

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108 Chapter 3

Evaluating Properties

As another illustration, consider water at a state fixed by a pressure equal to 14.7 lbf/in.

and a temperature of 2508F. From Table A-4E, y 5 28.42 ft

/lb, u 5 1091.5 Btu/lb, and

h 5 1168.8 Btu/lb. As shown, h may be calculated from u. Thus

h 5 u 1 py

5 1091.5

Btu

1 a14.7

lbf

in.

ba28.42

144 in.

1 ft

1 Btu

778 ft ? lbf

5 1091.5 1 77.3 5 1168.8 Btu

lb b b b b b

Specific internal energy and enthalpy data for liquid states of water are presented

in Tables A-5. The format of these tables is the same as that of the superheated vapor

tables considered previously. Accordingly, property values for liquid states are retrieved

in the same manner as those of vapor states.

For water, Tables A-6 give the equilibrium properties of saturated solid and satu-

rated vapor. The first column lists the temperature, and the second column gives the

corresponding saturation pressure. These states are at pressures and temperatures

below those at the triple point. The next two columns give the specific volume of

saturated solid, y

, and saturated vapor, y

, respectively. The table also provides the

specific internal energy, enthalpy, and entropy values for the saturated solid and the

saturated vapor at each of the temperatures listed.

3.6.3

Reference States and Reference Values

The values of u, h, and s given in the property tables are not obtained by direct

measurement but are calculated from other data that can be more readily determined

experimentally. The computational procedures require use of the second law of ther-

modynamics, so consideration of these procedures is deferred to Chap. 11 after the

second law has been introduced. However, because u, h, and s are calculated, the

matter of reference states and reference values becomes important and is considered

briefly in the following paragraphs.

When applying the energy balance, it is differences in internal, kinetic, and poten-

tial energy between two states that are important, and not the values of these energy

quantities at each of the two states.

consider the case of potential energy. The numerical value of

potential energy determined relative to the surface of the earth is not the same as the

value relative to the top of a tall building at the same location. However, the differ-

ence in potential energy between any two elevations is precisely the same regardless

of the datum selected, because the datum cancels in the calculation.

b b b b b

Similarly, values can be assigned to specific internal energy and enthalpy relative to

arbitrary reference values at arbitrary reference states. As for the case of potential energy

considered above, the use of values of a particular property determined relative to an

arbitrary reference is unambiguous as long as the calculations being performed involve

only differences in that property, for then the reference value cancels. When chemical

reactions take place among the substances under consideration, special attention must be

given to the matter of reference states and values, however. A discussion of how property

values are assigned when analyzing reactive systems is given in Chap. 13.

The tabular values of u and h for water, ammonia, propane, and Refrigerants 22

and 134a provided in the Appendix are relative to the following reference states and

values. For water, the reference state is saturated liquid at 0.018C (32.028F). At this

state, the specific internal energy is set to zero. Values of the specific enthalpy are

calculated from h 5 u 1 py, using the tabulated values for p, y, and u. For ammonia,

propane, and the refrigerants, the reference state is saturated liquid at 2408C (2408F

for the tables with English units). At this reference state the specific enthalpy is set

reference states

reference values

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to zero. Values of specific internal energy are calculated from u 5 h 2 py by using

the tabulated values for p, y, and h. Notice in Table A-7 that this leads to a negative

value for internal energy at the reference state, which emphasizes that it is not the

numerical values assigned to u and h at a given state that are important but their

differences between states. The values assigned to particular states change if the refe-

rence state or reference values change, but the differences remain the same.

3.7 Evaluating Properties Using

Computer Software

The use of computer software for evaluating thermodynamic properties is becoming

prevalent in engineering. Computer software falls into two general categories: those

that provide data only at individual states and those that provide property data as

part of a more general simulation package. The software available with this text,

Interactive Thermodynamics: IT, is a tool that can be used not only for routine prob-

lem solving by providing data at individual state points, but also for simulation and

analysis. Software other than IT also can be used for these purposes. See the box for

discussion of software use in engineering thermodynamics.

Using Software In Thermodynamics

The computer software tool Interactive Thermodynamics: IT is available for use with

this text. Used properly, IT provides an important adjunct to learning engineering ther-

modynamics and solving engineering problems. The program is built around an equation

solver enhanced with thermodynamic property data and other valuable features. With

IT you can obtain a single numerical solution or vary parameters to investigate their

effects. You also can obtain graphical output, and the Windows-based format allows you

to use any Windows word-processing software or spreadsheet to generate reports. Addi-

tionally, functions in IT can be called from Excel through use of the Excel Add-in Man-

ager, allowing you to use these thermodynamic functions while working within Excel.

Other features of IT include:

c a guided series of help screens and a number of sample solved examples to help

you learn how to use the program.

c drag-and-drop templates for many of the standard problem types, including a list of

assumptions that you can customize to the problem at hand.

c predetermined scenarios for power plants and other important applications.

c thermodynamic property data for water, refrigerants 22 and 134a, ammonia, air–

water vapor mixtures, and a number of ideal gases.

c the capability to input user-supplied data.

c the capability to interface with user-supplied routines.

Many features of IT are found in the popular Engineering Equation Solver (EES). Read-

ers already proficient with EES may prefer its use for solving problems in this text.

The use of computer software for engineering analysis is a powerful approach. Still,

there are some rules to observe:

c Software complements and extends careful analysis, but does not substitute for it.

c Computer-generated values should be checked selectively against hand-calculated,

or otherwise independently determined values.

c Computer-generated plots should be studied to see if the curves appear reasonable

and exhibit expected trends.

3.7 Evaluating Properties Using Computer Software 109

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110 Chapter 3

Evaluating Properties

IT provides data for substances represented in the Appendix tables. Generally, data are

retrieved by simple call statements that are placed in the workspace of the program.

consider the two-phase, liquid–vapor mixture at state 1 of Exam-

ple 3.2 for which p 5 1 bar, y 5 0.8475 m

/kg. The following illustrates how data for

saturation temperature, quality, and specific internal energy are retrieved using IT.

The functions for T, y, and u are obtained by selecting Water/Steam from the Prop-

erties menu. Choosing SI units from the Units menu, with p in bar, T in 8C, and

amount of substance in kg, the IT program is

p 5 1//bar

v 5 0.8475//m

/kg

T 5 Tsat_P(“Water/Steam”,p)

v 5 vsat_Px(“Water/Steam”,p,x)

u 5 usat_Px(“Water/Steam”,p,x)

Clicking the Solve button, the software returns values of T 5 99.638C, x 5 0.5, and

u 5 1462 kJ/kg. These values can be verified using data from Table A-3. Note that text

inserted between the symbol // and a line return is treated as a comment. b b b b b

The previous example illustrates an important feature of IT. Although the quality,

x, is implicit in the list of arguments in the expression for specific volume, there is no

need to solve the expression algebraically for x. Rather, the program can solve for x

as long as the number of equations equals the number of unknowns.

IT also retrieves property values in the superheat region.

consider the superheated ammonia vapor at state 2 in Example

3.1, for which p 5 20 lbf/in.

and T 5 778F. Selecting Ammonia from the Properties

menu and choosing English units from the Units menu, data for specific volume,

internal energy, and enthalpy are obtained from IT as follows:

p 5 20//lbf/in

T 5 77//°F

v 5 v_PT(“Ammonia”,p,T)

u 5 u_PT(“Ammonia”,p,T)

h 5 h_PT(“Ammonia”,p,T)

Clicking the Solve button, the software returns values of y 5 16.67 ft

/lb, u 5 593.7

Btu/lb, and h 5 655.3 Btu/lb, respectively. These values agree closely with the respec-

tive values obtained by interpolation in Table A-15E.

b b b b b

3.8 Applying the Energy Balance Using

Property Tables and Software

The energy balance for closed systems is introduced in Sec. 2.5. Alternative expres-

sions are given by Eqs. 2.35a and 2.35b, which are forms applicable to processes

between end states denoted 1 and 2, and by Eq. 2.37, the time rate form. In applica-

tions where changes in kinetic energy and gravitational potential energy between the

end states can be ignored, Eq. 2.35b reduces to

2 U

5 Q 2 W (a)

where Q and W account, respectively, for the transfer of energy by heat and work

between the system and its surroundings during the process. The term U

2 U

accounts

for change in internal energy between the end states.

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Taking water for simplicity, let’s consider how the internal energy term is evaluated

in three representative cases of systems involving a single substance.

Case 1: Consider a system consisting initially and finally of a single phase of

water, vapor or liquid. Then Eq. (a) takes the form

2 u

5 Q 2

(b)

where m is the system mass and u

and u

denote, respectively, the initial and

final specific internal energies. When the initial and final temperatures T

, T

and pressures p

, p

are known, for instance, the internal energies u

and u

can

be readily obtained from the steam tables or using computer software.

Case 2: Consider a system consisting initially of water vapor and finally as a

two-phase mixture of liquid water and water vapor. As in Case 1, we write

5 mu

in Eq. (a), but now

1 U

5 m

(c)

where m

liq

and m

vap

account, respectively, for the masses of saturated liquid and

saturated vapor present finally, and u

and u

are the corresponding specific

internal energies determined by the final temperature T

(or final pressure p

If quality x

is known, Eq. 3.6 can be invoked to evaluate the specific

internal energy of the two-phase liquid–vapor mixture, u

. Then, U

5 mu

thereby preserving the form of the energy balance expressed by Eq. (b).

Case 3: Consider a system consisting initially of two separate masses of water

vapor that mix to form a total mass of water vapor. In this case

5 m¿u

T¿, p¿

1 m–u

T–, p–

(d)

m¿ 1 m–

, p

5 mu

, p

(e)

where m9 and m0 are masses of water vapor initially separate at T9, p9 and T0,

p0, respectively, that mix to form a total mass, m 5 m9 1 m0, at a final state

where temperature is T

and pressure is p

. When temperatures and pressures

at the respective states are known, for instance, the specific internal energies

of Eqs. (d) and (e) can be readily obtained from the steam tables or using

computer software.

These cases show that when applying the energy balance, an important consider-

ation is whether the system has one or two phases. A pertinent application is that of

thermal energy storage, considered in the box.

3.8 Applying the Energy Balance Using Property Tables and Software 111

Thermal Energy Storage

Energy is often available at one time but more valuable or used more effectively at

another. These considerations underlie various means for storing energy, including

methods introduced in Sec. 2.7 and those discussed here.

Solar energy is collected during daylight hours but often needed at other times of the

day—to heat buildings overnight, for example. Accordingly, thermal energy storage sys-

tems have been developed to meet solar and other similar energy storage needs. The

term thermal energy here should be understood as internal energy.

Mediums used in thermal energy storage systems change temperature and/or change

phase. Some storage systems simply store energy by heating water, mineral oil, or other

substances held in a storage tank, usually pressurized, until the stored energy is needed.

Solids such as concrete can also be the medium. Phase-change systems store energy by

melting or freezing a substance, often water or a molten (eutectic) salt. The choice of

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112 Chapter 3 Evaluating Properties

3.8.1

Using Property Tables

In Examples 3.3 and 3.4, closed systems undergoing processes are analyzed using the

energy balance. In each case, sketches of p–y and/or T–y diagrams are used in con-

junction with appropriate tables to obtain the required property data. Using property

diagrams and table data introduces an additional level of complexity compared to