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Chapter 3

Conservation of Energy

In this chapter we continue the quantitative development of thermodynamics by deriv-ing the energy balance, the second of the three balance equations that will be used in thethermodynamic description of physical, chemical, and (later) biochemical processes.The mass and energy balance equations (and the third balance equation, to be devel-oped in the following chapter), together with experimental data and information aboutthe process, will then be used to relate the change in a system’s properties to a changein its thermodynamic state. In this context, physics, fluid mechanics, thermodynamics,and other physical sciences are all similar, in that the tools of each are the same: a set ofbalance equations, a collection of experimental observations (equation-of-state data inthermodynamics, viscosity data in fluid mechanics, etc.), and the initial and boundaryconditions for each problem. The real distinction between these different subject areasis the class of problems, and in some cases the portion of a particular problem, thateach deals with.

One important difference between thermodynamics and, say, fluid mechanics andchemical reactor analysis is the level of description used. In fluid mechanics one isusually interested in a very detailed microscopic description of flow phenomena andmay try to determine, for example, the fluid velocity profile for flow in a pipe. Simi-larly, in chemical reactor analysis one is interested in determining the concentrationsand rates of chemical reaction everywhere in the reactor. In thermodynamics the de-scription is usually more primitive in that we choose either a region of space or anelement of mass as the system and merely try to balance the change in the systemwith what is entering and leaving it. The advantage of such a description is that wecan frequently make important predictions about certain types of processes for whicha more detailed description might not be possible. The compromise is that the ther-modynamic description yields information only about certain overall properties of thesystem, though with relatively little labor and simple initial information.

In this chapter we are concerned with developing the equations of energy conserva-tion to be used in the thermodynamic analysis of systems of pure substances. (The ther-modynamics of mixtures is more complicated and will be considered in later chapters.)To emphasize both the generality of these equations and the lack of detail necessary,we write these energy balance equations for a general black-box system. For contrast,and also because a more detailed description will be useful in Chapter 4, the rudimentsof the more detailed microscopic description are provided in the final, optional sec-tion of this chapter. This microscopic description is not central to our development ofthermodynamic principles, is suitable only for advanced students, and may be omitted.

48

49

To use the energy balances, we will need to relate the energy to more easily measur-able properties, such as temperature and pressure (and in later chapters, when we con-sider mixtures, to composition as well). The interrelationships between energy, tem-perature, pressure, and composition can be complicated, and we will develop this instages. In this chapter and in Chapters 4, 5, and 6 we will consider only pure fluids, socomposition is not a variable. Then, in Chapters 8 to 15, mixtures will be considered.Also, here and in Chapters 4 and 5 we will consider only the simple ideal gas andincompressible liquids and solids for which the equations relating the energy, temper-ature, and pressure are simple, or fluids for which charts and tables interrelating theseproperties are available. Then, in Chapter 6, we will discuss how such tables and chartsare prepared.

INSTRUCTIONAL OBJECTIVES FOR CHAPTER 3

The goals of this chapter are for the student to:

• Be able to use the differential form of the pure component energy balance inproblem solving

• Be able to use the difference form of the pure component energy balance in prob-lem solving

• Be able to compute changes in energy with changes in temperature and pressurefor the ideal gas

• Be able to compute changes in energy with changes in temperature and pressureof real fluids using tables and charts of thermodynamic properties

NOTATION INTRODUCED IN THIS CHAPTER

CP Constant-pressure molar heat capacity (J/mol K)C∗

P Ideal gas constant-pressure molar heat capacity (J/mol K)CV Constant-volume molar heat capacity (J/mol K)C∗

V Ideal gas constant-volume molar heat capacity (J/mol K)H Enthalpy (J)H Enthalpy per mole, or molar enthalpy (J/mol)H Enthalpy per unit mass, or specific enthalpy (J/g)

�fus H Molar enthalpy of melting or fusion (J/mol)�sub H Molar enthalpy of sublimation (J/mol)�vap H Molar enthalpy of vaporization (J/mol)�vap H Enthalpy of vaporization per unit mass (J/g)

Q Rate of flow of heat into the system (J/s)Q Heat that has flowed into the system (J)

TR Reference temperature for internal energy of enthalpy (K)U Internal energy per mole, or molar internal energy (J/mol)U Internal energy per unit mass, or specific internal energy (J/g)V Volume per mole, or molar volume (m3/mol)W Rate at which work is being done on the system (J/s)W Work that has been done on the system (J)Ws Shaft work that has been done on the system (J)Ws Rate at which shaft work is being done on the system (J/s)

50 Chapter 3: Conservation of Energy

ψ Potential energy per unit of mass (J/g)ωI Mass fraction of phase I (quality for steam)ψ Potential energy per unit of mass (J/g)

3.1 CONSERVATION OF ENERGY

To derive the energy conservation equation for a single-component system, we againuse the black-box system of Figure 2.1-1 and start from the general balance equation,Eq. 2.1-4. Taking θ to be the sum of the internal, kinetic, and potential energy of thesystem,

θ = U + M

(v2

2+ ψ

)

Here U is the total internal energy, v2/2 is the kinetic energy per unit mass (where vis the center of mass velocity), and ψ is the potential energy per unit mass.1 If gravityis the only force field present, then ψ = gh, where h is the height of the center ofmass with respect to some reference, and g is the force of gravity per unit mass. Sinceenergy is a conserved quantity, we can write

d

dt

{U + M

(v2

2+ ψ

)}=(

Rate at which energyenters the system

)−(

Rate at which energyleaves the system

)

(3.1-1)To complete the balance it remains only to identify the various mechanisms by whichenergy can enter and leave the system. These are as follows.

Energy flow accompanying mass flow. As a fluid element enters or leaves the sys-tem, it carries its internal, potential, and kinetic energy. This energy flow accompanyingthe mass flow is simply the product of a mass flow and the energy per unit mass,

K∑k=1

Mk

(U + v2

2+ ψ

)k

(3.1-2)

where Uk is the internal energy per unit mass of the kth flow stream, and Mk is its massflow rate.

Heat. We use Q to denote the total rate of flow of heat into the system, by bothconduction and radiation, so that Q is positive if energy in the form of heat flows intothe system and negative if heat flows from the system to its surroundings. If heat flowsoccur at several different places, the total rate of heat flow into the system is

Q =∑

Q j

where Q j is the heat flow at the j th heat flow port.Work. The total energy flow into the system due to work will be divided into several

parts. The first part, called shaft work and denoted by the symbol Ws , is the mechanicalenergy flow that occurs without a deformation of the system boundaries. For example,

1In writing this form for the energy term, it has been assumed that the system consists of only one phase, that is, agas, a liquid, or a solid. If the system consists of several distinct parts—for example, gas and a liquid, or a gas andthe piston and cylinder containing it—the total energy, which is an extensive property, is the sum of the energiesof the constituent parts.

3.1 Conservation of Energy 51

if the system under consideration is a steam turbine or internal combustion engine,the rate of shaft work Ws is equal to the rate at which energy is transferred across thestationary system boundaries by the drive shaft or push rod. Following the conventionthat energy flow into the system is positive, Ws is positive if the surroundings do workon the system and negative if the system does work on its surroundings.

For convenience, the flow of electrical energy into or out of the system will be in-cluded in the shaft work term. In this case, Ws = ±E I , where E is the electricalpotential difference across the system and I is the current flow through the system.The positive sign applies if electrical energy is being supplied to the system, and thenegative sign applies if the system is the source of electrical energy.

Work also results from the movement of the system boundaries. The rate at whichwork is done when a force F is moved through a distance in the direction of the appliedforce d L in the time interval dt is

W = Fd L

dtHere we recognize that pressure is a force per unit area and write

W = −PdV

dt(3.1-3)

where P is the pressure exerted by the system at its boundaries.2 The negative signin this equation arises from the convention that work done on a system in compres-sion (for which dV/dt is negative) is positive, and work done by the system on itssurroundings in an expansion (for which dV/dt is positive) is negative. The pressureat the boundaries of a nonstationary system will be opposed by (1) the pressure of theenvironment surrounding the system, (2) inertial forces if the expansion or compres-sion of the system results in an acceleration or deceleration of the surroundings, and(3) other external forces, such as gravity. As we will see in Illustration 3.4-7, the con-tribution to the energy balance of the first of these forces is a term corresponding tothe work done against the atmosphere, the second is a work term corresponding to thechange in kinetic energy of the surroundings, and the last is the work done that changesthe potential energy of the surroundings.

Work of a flowing fluid against pressure. One additional flow of energy for sys-tems open to the flow of mass must be included in the energy balance equation; it ismore subtle than the energy flows just considered. This is the energy flow that arisesfrom the fact that as an element of fluid moves, it does work on the fluid ahead of it,and the fluid behind it does work on it. Clearly, each of these work terms is of theP�V type. To evaluate this energy flow term, which occurs only in systems open tothe flow of mass, we will compute the net work done as one fluid element of mass(�M)1 enters a system, such as the valve in Fig. 3.1-1, and another fluid element of

Pressure P2Pressure P1

Volume = V1∆M1^

Volume = V2∆M2^

Valve

Figure 3.1-1 A schematic rep-resentation of flow through avalve.

w00

08-n

.eps

2In writing this form for the work term, we have assumed the pressure to be uniform at the system boundary. Ifthis is not the case, Eq. 3.1-3 is to be replaced with an integral over the surface of the system.


mass (�M)2 leaves the system. The pressure of the fluid at the inlet side of the valveis P1, and the fluid pressure at the outlet side is P2, so that we have

Work done by surrounding fluid in

pushing fluid element of mass (�M)1

into the valve

= P1V1(�M)1

Work done on surrounding fluidby movement of fluid element of

mass (�M)2 out of the valve (sincethis fluid element is pushing the

surrounding fluid)

= −P2V2(�M)2

(Net work done on the system due to

movement of fluid

)= P1V1(�M)1 − P2V2(�M)2

For a more general system, with several mass flow ports, we have

Net work done on the system dueto the pressure forces acting onthe fluids moving into and out ofthe system

=

K∑k=1

(�M)k(PV )k

Finally, to obtain the net rate at which work is done, we replace each mass flow (�M)k

with a mass flow rate Mk , so that

Net rate at which work is done onthe system due to pressure forces

acting on fluids moving into and outof the system

=

K∑k=1

Mk(PV )k

where the sign of each term of this energy flow is the same as that of Mk .One important application of this pressure-induced energy flow accompanying a

mass flow is hydroelectric power generation, schematically indicated in Fig. 3.1-2.Here a water turbine is being used to obtain mechanical energy from the flow of waterthrough the base of a dam. Since the water velocity, height, and temperature are ap-proximately the same at both sides of the turbine (even though there are large velocitychanges within the turbine), the mechanical (or electrical) energy obtained is a resultof only the mass flow across the pressure difference at the turbine.

Collecting all the energy terms discussed gives

d

dt

{U + M

(v2

2+ ψ

)}=

K∑k=1

Mk

(U + v2

2+ ψ

)k

+ Q

+ Ws − PdV

dt+

K∑k=1

Mk(PV )k

(3.1-4)

This equation can be written in a more compact form by combining the first and lastterms on the right side and introducing the notation


Reservoir

Waterflow

Dam

Water turbine

Hydroelectric powergenerating station

Figure 3.1-2 A hydroelectric power generating station: adevice for obtaining work from a fluid flowing across alarge pressure drop.

w00

09-n

.eps

H = U + PV

where the function H is called the enthalpy, and by using the symbol W to representthe combination of shaft work Ws and expansion work −P(dV/dt), that is, W =Ws − P(dV/dt). Thus we have

Complete energybalance, frequentlyreferred to asthe first law ofthermodynamics

d

dt

{U + M

(v2

2+ ψ

)}=

K∑k=1

Mk

(H + v2

2+ ψ

)k

+ Q + W (3.1-4a)

It is also convenient to have the energy balance on a molar rather than a mass basis.This change is easily accomplished by recognizing that Mk Hk can equally well bewritten as Nk Hk , where H is the enthalpy per mole or molar enthalpy,3 and M(v2/2 +ψ) = Nm(v2/2 + ψ), where m is the molecular weight. Therefore, we can write theenergy balance as

d

dt

{U + Nm

(v2

2+ ψ

)}=

K∑k=1

Nk

{H + m

(v2

2+ ψ

)}k

+ Q + W (3.1-4b)

Several special cases of Eqs. 3.1-4 are listed in Table 3.1-1.The changes in energy associated with either the kinetic energy or potential energy

terms, especially for gases, are usually very small compared with those for the ther-mal (internal) energy terms, unless the fluid velocity is near the velocity of sound,the change in height is very large, or the system temperature is nearly constant. Thispoint will become evident in some of the illustrations and problems (see particularlyIllustration 3.4-2). Therefore, it is frequently possible to approximate Eqs. 3.1-4 by

3 H = U + PV , where U and V are the molar internal energy and volume, respectively.


Commonly used formsof energy balance

dU

dt=

K∑k=1

(M H)k + Q + W (mass basis) (3.1-5a)

dU

dt=

K∑k=1

(N H)k + Q + W (molar basis) (3.1-5b)

As with the mass balance, it is useful to have a form of the energy balance applicableto a change from state 1 to state 2. This is easily obtained by integrating Eq. 3.1-4 overthe time interval t1 to t2, the time required for the system to go from state 1 to state 2.

Table 3.1-1 Differential Form of the Energy Balance

General equation

d

dt

{U + M

(v2

2+ ψ

)}=

K∑k=1

Mk

(H + v2

2+ ψ

)k

+ Q + W (a)

Special cases:(i) Closed system

Mk = 0,d M

dt= 0

sodU

dt+ M

d

dt

(v2

2+ ψ

)= Q + W (b)

(ii) Adiabatic processin Eqs. a, b, and d

Q = 0 (c)

(iii) Open and steady-state system

d M

dt= 0,

dV

dt= 0,

d

dt

{U + M

(v2

2+ ψ

)}= 0

so

0 =K∑

k=1

Mk

(H + v2

2+ ψ

)k

+ Q + Ws (d)

(iv) Uniform systemIn Eqs. a and b

U = MU (e)

Note: To obtain the energy balance on a molar basis, make the following substitutions:

Replace with

M

(v2

2+ ψ

)Nm

(v2

2+ ψ

)

Mk

(H + v2

2+ ψ

)k

Nk

{H + m

(v2

2+ ψ

)}k

MU NU Tabl

e3.

1-1


The result is

{U + M

(v2

2+ ψ

)}t2

−{

U + M

(v2

2+ ψ

)}t1

=K∑

k=1

∫ t2

t1

Mk

(H + v2

2+ ψ

)k

dt + Q + W (3.1-6)

where

Q =∫ t2

t1

Q dt Ws =∫ t2

t1

Ws dt∫ V (t2)

V (t1)

P dV =∫ t2

t1

PdV

dtdt

and

W = Ws −∫ V (t2)

V (t1)

P dV

The first term on the right side of Eq. 3.1-6 is usually the most troublesome to evaluatebecause the mass flow rate and/or the thermodynamic properties of the flowing fluidmay change with time. However, if the thermodynamic properties of the fluids enteringand leaving the system are independent of time (even though the mass flow rate maydepend on time), we have

K∑k=1

∫ t2

t1

Mk

(H + v2

2+ ψ

)k

dt =K∑

k=1

(H + v2

2+ ψ

)k

∫ t2

t1

Mk dt

=K∑

k=1

(�M)k

(H + v2

2+ ψ

)k

(3.1-7)

If, on the other hand, the thermodynamic properties of the flow streams change withtime in some arbitrary way, the energy balance of Eq. 3.1-6 may not be useful. Theusual procedure, then, is to try to choose a new system (or subsystem) for the descrip-tion of the process in which these time-dependent flows do not occur or are more easilyhandled (see Illustration 3.4-5).

Table 3.1-2 lists various special cases of Eq. 3.1-6 that will be useful in solvingthermodynamic problems.

For the study of thermodynamics it will be useful to have equations that relate thedifferential change in certain thermodynamic variables of the system to differentialchanges in other system properties. Such equations can be obtained from the differen-tial form of the mass and energy balances. For processes in which kinetic and potentialenergy terms are unimportant, there is no shaft work, and there is only a single massflow stream, these equations reduce to

d M

dt= M


and

dU

dt= M H + Q − P

dV

dt

which can be combined to give

dU

dt= H

d M

dt+ Q − P

dV

dt(3.1-8)

where H is the enthalpy per unit mass entering or leaving the system. (Note that for asystem closed to the flow of mass, d M/dt = M = 0.) Defining Q = Q dt to be equalto the heat flow into the system in the differential time interval dt , and d M = M dt =(d M/dt) dt to be equal to the mass flow in that time interval, we obtain the followingexpression for the change of the internal energy in the time interval dt :

dU = H d M + Q − P dV (3.1-9a)

Table 3.1-2 Difference Form of the Energy Balance

General equation{U + M

(v2

2+ ψ

)}t2

−{

U + M

(v2

2+ ψ

)}t1

=K∑

k=1

∫ t2

t1

Mk

(H + v2

2+ ψ

)k

dt + Q + W (a)

Special cases:(i) Closed system {

U + M

(v2

2+ ψ

)}t2

−{

U + M

(v2

2+ ψ

)}t1

= Q + W (b)

and

M(t1) = M(t2)

(ii) Adiabatic processIn Eqs. a and b

Q = 0 (c)

(iii) Open system, flow of fluids of constant thermodynamic propertiesK∑

k=1

∫ t2

t1

Mk

(H + v2

2+ ψ

)k

dt =K∑

k=1

(�M)k

(H + v2

2+ ψ

)k

(d)

in Eq. a

(iv) Uniform system {U + M

(v2

2+ ψ

)}= M

(U + v2

2+ ψ

)(e)

in Eqs. a and b

Note: To obtain the energy balance on a molar basis, make the following substitutions:

Replace with

M(v2

2+ ψ) Nm(

v2

2+ ψ)

∫ t2

t1Mk(H + v2

2+ ψ)k dt

∫ t2

t1Nk{H + m(

v2

2+ ψ)}k dt

M(U + v2

2+ ψ) N{U + m(

v2

2+ ψ)} Ta

ble

31-

2

3.2 Several Examples of Using The Energy Balance 57

For a closed system this equation reduces to

dU = Q − P dV (3.1-9b)

Since the time derivative operator d/dt is mathematically well defined, and the opera-tor d is not, it is important to remember in using Eqs. 3.1-9 that they are abbreviationsof Eq. 3.1-8. It is part of the traditional notation of thermodynamics to use dθ to indi-cate a differential change in the property θ , rather than the mathematically more correctdθ/dt .

3.2 SEVERAL EXAMPLES OF USING THE ENERGY BALANCE

The energy balance equations developed so far in this chapter can be used for the de-scription of any process. As the first step in using these equations, it is necessary tochoose the system for which the mass and energy balances are to be written. The im-portant fact for the student of thermodynamics to recognize is that processes occurringin nature are in no way influenced by our mathematical description of them. Therefore,if our descriptions are correct, they must lead to the same final result for the systemand its surroundings regardless of which system choice is made. This is demonstratedin the following example, where the same result is obtained by choosing for the sys-tem first a given mass of material and then a specified region in space. Since the firstsystem choice is closed and the second open, this illustration also establishes the wayin which the open-system energy flow PV M is related to the closed-system work termP(dV/dt).

ILLUSTRATION 3.2-1Showing That the Final Result Should Not Depend on the Choice of System

A compressor is operating in a continuous, steady-state manner to produce a gas at temperatureT2 and pressure P2 from one at T1 and P1. Show that for the time interval �t

Q + Ws = (H2 − H1)�M

where �M is the mass of gas that has flowed into or out of the system in the time �t . Establishthis result by (a) first writing the balance equations for a closed system consisting of someconvenient element of mass, and then (b) by writing the balance equations for the compressorand its contents, which is an open system.

SOLUTION

a. The closed-system analysisHere we take as the system the gas in the compressor and the mass of gas �M , that will enter

the compressor in the time interval �t . The system is enclosed by dotted lines in the figure.

P1, T1 P2, T2

Ws Q

Compressor

w00

10-n

.eps


At the later time t + �t , the mass of gas we have chosen as the system is as shown below.

P1, T1 P2, T2

Ws Q

Compressor

w00

11-n

.eps

We use the subscript c to denote the characteristics of the fluid in the compressor, the subscript1 for the gas contained in the system that is in the inlet pipe at time t , and the subscript 2 for thegas in the system that is in the exit pipe at time t + �t . With this notation the mass balance forthe closed system is

M2(t + �t) + Mc(t + �t) = M1(t) + Mc(t)

Since the compressor is in steady-state operation, the amount of gas contained within it and theproperties of this gas are constant. Thus, Mc(t + �t) = Mc(t) and

M2(t + �t) = M1(t) = �M

The energy balance for this system, neglecting the potential and kinetic energy terms (which,if retained, would largely cancel), is

M2U2|t+�t + McUc|t+�t − M1U1|t − McUc|t = Ws + Q + P1V1 M1 − P2V2 M2 (a)

In writing this equation we have recognized that the flow terms vanish for the closed system andthat there are two contributions of the

∫P dV type, one due to the deformation of the system

boundary at the compressor inlet and another at the compressor outlet. Since the inlet and exitpressures are constant at P1 and P2, these terms are

−∫

P dV = −P1

∫dV |inlet − P2

∫dV |outlet

= −P1{V1(t + �t) − V1(t)} − P2{V2(t + �t) − V2(t)}However, V1(t + �t) = 0 and V2(t) = 0, so that

−∫

P dV = +P1V1 − P2V2 = P1V1 M1 − P2V2 M2

Now, using the energy balance and Eq. a above, and recognizing that since the compressor is insteady-state operation,

McUc|t+�t = McUc|twe obtain

�M(U2 − U1) = Ws + Q + P1V1�M − P2V2�M

or

�M(U2 + P2V2 − U1 − P1V1) = �M(H2 − H1) = Ws + Q


b. The open-system analysisHere we take the contents of the compressor at any time to be the system. The mass balance

for this system over the time interval �t is

M(t + �t) − M(t) =∫ t+�t

tM1 dt +

∫ t+�t

tM2 dt = (�M)1 + (�M)2

and the energy balance is{U + M

(v2

2+ ψ

)}t+�t

−{

U + M

(v2

2+ ψ

)}t

=∫ t+�t

tM1(H1 + v2

1/2 + ψ1) dt +∫ t+�t

tM2(H2 + v2

2/2 + ψ2) dt + Q + W

These equations may be simplified as follows:

1. Since the compressor is operating continuously in a steady-state manner, its contents must,by definition, have the same mass and thermodynamic properties at all times. Therefore,

M(t + �t) = M(t)

and{

U + M

(v2

2+ ψ

)}t+�t

={

U + M

(v2

2+ ψ

)}t

2. Since the thermodynamic properties of the fluids entering and leaving the turbine do notchange in time, we can write

∫ t+�t

tM1(H1 + v2

1/2 + ψ1) dt = (H1 + v21/2 + ψ1)

∫ t+�t

tM1 dt

= (H1 + v21/2 + ψ1) (�M)1

with a similar expression for the compressor exit stream.3. Since the volume of the system here, the contents of the compressor, is constant

∫ V2

V1

P dV = 0

so that

W = Ws

4. Finally, we will neglect the potential and kinetic energy changes of the entering and exitingfluids.

With these simplifications, we have

0 = (�M)1 + (�M)2 or (�M)1 = −(�M)2 = �M

and

0 = (�M)1 H1 + (�M)2 H2 + Q + Ws

Combining these two equations, we obtain

Q + Ws = (H2 − H1)�M

This is the same result as in part (a).


COMMENT

Notice that in the closed-system analysis the surroundings are doing work on the system (themass element) at the inlet to the compressor, while the system is doing work on its surroundingsat the outlet pipe. Each of these terms is a

∫P dV –type work term. For the open system this

work term has been included in the energy balance as a PV �M term, so that it is the enthalpy,rather than the internal energy, of the flow streams that appears in the equation. The explicit∫

P dV term that does appear in the open-system energy balance represents only the work doneif the system boundaries deform; for the choice of the compressor and its contents as the systemhere this term is zero unless the compressor (the boundary of our system) explodes.

This illustration demonstrates that the sum Q+Ws is the same for a fluid undergoingsome change in a continuous process regardless of whether we choose to computethis sum from the closed-system analysis on a mass of gas or from an open-systemanalysis on a given volume in space. In Illustration 3.2-2 we consider another problem,the compression of a gas by two different processes, the first being a closed-systempiston-and-cylinder process and the second being a flow compressor process. Here wewill find that the sum Q + W is different in the two processes, but the origin of thisdifference is easily understood.

ILLUSTRATION 3.2-2Showing That Processes in Closed and Open Systems Are Different

A mass M of gas is to be compressed from a temperature T1 and a pressure P1 to T2 and P2

in (a) a one-step process in a frictionless piston and cylinder,4 and (b) a continuous process inwhich the mass M of gas is part of the feed stream to the compressor of the previous illustration.Compute the sum Q + W for each process.

SOLUTION

a. The piston-and-cylinder process

P1, T1 P2, T2 w00

12-n

.eps

Here we take the gas within the piston and cylinder as the system. The energy balance for thisclosed system is

M(U2 − U1) = Q + W (piston-cylinder process)

It is useful to note that Ws = 0 and W = −∫P dV .

b. The flow compressor process (see the figures in Illustration 3.2-1)If we take the contents of the compressor as the system and follow the analysis of the previous

illustration, we obtain

M(H2 − H1) = Q + W (flow compressor)

where, since∫

P dV = 0, W = Ws .

COMMENT

From these results it is evident that the sum Q + W is different in the two cases, since thetwo processes are different. The origin of the difference in the flow and nonflow energy changes4Since the piston is frictionless, the pressure of the gas is equal to the pressure applied by the piston.


accompanying a change of state is easily identified by considering two different ways of com-pressing a mass M of gas in a piston and cylinder from (T1, P1) to (T2, P2). The first way ismerely to compress the gas in situ. The sum of heat and work flows needed to accomplish thechange of state is, from the preceding computations,

Q + W = M(U2 − U1)

A second way to accomplish the compression is to open a valve at the side of the cylinderand use the piston movement (at constant pressure P1) to inject the gas into the compressor inletstream, use the compressor to compress the gas, and then withdraw the gas from the compressorexit stream by moving the piston against a constant external pressure P2. The energy requiredin the compressor stage is the same as that found above:

(Q + W )c = M(H2 − H1)

To this we must add the work done in using the piston movement to pump the fluid into thecompressor inlet stream,

W1 =∫

P dV = P1V1 = P1V1 M

(this is the work done by the system on the gas in the inlet pipe to the compressor), and subtractthe work obtained as a result of the piston movement as the cylinder is refilled,

W2 = −∫

P dV = −P2V2 = −P2V2 M

(this is the work done on the system by the gas in the compressor exit stream). Thus the totalenergy change in the process is

Q + W = (Q + W )c + W1 + W2

= M(H2 − H1) + P1V1 M − P2V2 M = M(U2 − U1)

which is what we found in part (a). Here, however, it results from the sum of an energy require-ment of M(H2 − H1) in the flow compressor and the two pumping terms.

Consider now the problem of relating the downstream temperature and pressure ofa gas in steady flow across a flow constriction (e.g., a valve, orifice, or porous plug) toits upstream temperature and pressure.

ILLUSTRATION 3.2-3A Joule-Thomson or Isenthalpic Expansion

A gas at pressure P1 and temperature T1 is steadily exhausted to the atmosphere at pressure P2

through a pressure-reducing valve. Find an expression relating the downstream gas temperatureT2 to P1, P2, and T1. Since the gas flows through the valve rapidly, one can assume that there isno heat transfer to the gas. Also, the potential and kinetic energy terms can be neglected.

T1, P1 T2 = ?, P2 w00

13-n

.eps

SOLUTION

The flow process is schematically shown in the figure. We will consider the region of spacethat includes the flow obstruction (indicated by the dashed line) to be the system, although, asin Illustration 3.2-1, a fixed mass of gas could have been chosen as well. The pressure of the


gas exiting the reducing valve will be P2, the pressure of the surrounding atmosphere. (It is notcompletely obvious that these two pressures should be the same. However, in the laboratory wefind that the velocity of the flowing fluid will always adjust in such a way that the fluid exitpressure and the pressure of the surroundings are equal.) Now recognizing that our system (thevalve and its contents) is of constant volume, that the flow is steady, and that there are no heat orwork flows and negligible kinetic and potential energy changes, the mass and energy balances(on a molar basis) yield

0 = N1 + N2 or N2 = −N1

and

0 = N1 H 1 + N2 H2 = N1(H 1 − H2)

Thus

Isenthalpic orJoule-Thomsonexpansion

H1 = H2

or, to be explicit,

H (T1, P1) = H(T2, P2)

so that the initial and final states of the gas have the same enthalpy. Consequently, if we knewhow the enthalpy of the gas depended on its temperature and pressure, we could use the knownvalues of T1, P1, and P2 to determine the unknown downstream temperature T2.

COMMENTS

1. The equality of enthalpies in the upstream and downstream states is the only informationwe get from the thermodynamic balance equations. To proceed further we need constitu-tive information, that is, an equation of state or experimental data interrelating H , T , andP. Equations of state are discussed in the following section and in much of Chapter 6.

2. The experiment discussed in this illustration was devised by William Thomson (later LordKelvin) and performed by J. P. Joule to study departures from ideal gas behavior. TheJoule-Thomson expansion, as it is called, is used in the liquefaction of gases and inrefrigeration processes (see Chapter 5).

3.3 THE THERMODYNAMIC PROPERTIES OF MATTER

The balance equations of this chapter allow one to relate the mass, work, and heat flowsof a system to the change in its thermodynamic state. From the experimental observa-tions discussed in Chapter 1, the change of state for a single-component, single-phasesystem can be described by specifying the initial and final values of any two indepen-dent intensive variables. However, certain intensive variables, especially temperatureand pressure, are far easier to measure than others. Consequently, for most problemswe will want to specify the state of a system by its temperature and pressure ratherthan by its specific volume, internal energy, and enthalpy, which appear in the energybalance. What are needed, then, are interrelations between the fluid properties that al-low one to eliminate some thermodynamic variables in terms of other, more easilymeasured ones. Of particular interest is the volumetric equation of state, which is arelation between temperature, pressure, and specific volume, and the thermal equationof state, which is usually either in the form of a relationship between internal energy,temperature, and specific (or molar) volume, or between specific or molar enthalpy,

3.3 The Thermodynamic Properties of Matter 63

temperature, and pressure. Such information may be available in either of two forms.First, there are analytic equations of state, which provide an algebraic relation betweenthe thermodynamic state variables. Second, experimental data, usually in graphical ortabular form, may be available to provide the needed interrelationships between thefluid properties.

Equations of state for fluids are considered in detail in Chapter 6. To illustrate theuse of the mass and energy balance equations in a simple form, we briefly considerhere the equation of state for the ideal gas and the graphical and tabular display of thethermodynamic properties of several real fluids.

An ideal gas is a gas at such a low pressure that there are no interactions amongits molecules. For such gases it is possible to show, either experimentally or by themethods of statistical mechanics, that at all absolute temperatures and pressures thevolumetric equation of state is

PV = RT (3.3-1)

(as indicated in Sec. 1.4) and that the enthalpy and internal energy are functions oftemperature only (and not pressure or specific volume). We denote this latter fact byH = H(T ) and U = U(T ). This simple behavior is to be compared with the enthalpyfor a real fluid, which is a function of temperature and pressure [i.e., H = H(T, P)]and the internal energy, which is usually written as a function of temperature and spe-cific volume [U = U(T, V )], as will be discussed in Chapter 6.

The temperature dependence of the internal energy and enthalpy of all substances(not merely ideal gases) can be found by measuring the temperature rise that accompa-nies a heat flow into a closed stationary system. If a sufficiently small quantity of heatis added to such a system, it is observed that the temperature rise produced, �T , islinearly related to the heat added and inversely proportional to N , the number of molesin the system:

Q

N= C�T = C{T (t2) − T (t1)}

where C is a parameter and Q is the heat added to the system between the times t1and t2. The object of the experiment is to accurately measure the parameter C for avery small temperature rise, since C generally is also a function of temperature. If themeasurement is made at constant volume and with Ws = 0, we have, from the energybalance and the foregoing equation,

U (t2) − U (t1) = Q = NCV{T (t2) − T (t1)}Thus

CV = U (t2) − U (t1)

N{T (t2) − T (t1)} = U(t2) − U (t1)

T (t2) − T (t1)

where the subscript V has been introduced to remind us that the parameter C wasdetermined in a constant-volume experiment. In the limit of a very small temperaturedifference, we have

Constant-volume heatcapacity definition CV(T, V ) = lim

T (t2)−T (t1)→0

U(t2) − U (t1)

T (t2) − T (t1)=(

∂U

∂T

)V

=(

∂U (T, V )

∂T

)V

(3.3-2)


so that the measured parameter CV is, in fact, equal to the temperature derivative ofthe internal energy at constant volume. Similarly, if the parameter C is determined ina constant-pressure experiment, we have

Q = U (t2) − U (t1) + P{V (t2) − V (t1)} = U (t2) + P(t2)V (t2) − U (t1) − P(t1)V (t1)= H(t2) − H(t1) = NCP{T (t2) − T (t1)}

where we have used the fact that since pressure is constant, P = P(t2) = P(t1). Then

Constant-pressureheat capacitydefinition

CP(T, P) =(

∂ H

∂T

)P

=(

∂ H(T, P)

∂T

)P

(3.3-3)

so that the measured parameter here is equal to the temperature derivative of the en-thalpy at constant pressure.

The quantity CV is called the constant-volume heat capacity, and CP is the constant-pressure heat capacity; both appear frequently throughout this book. Partial deriva-tives have been used in Eqs. 3.3-2 and 3.3-3 to indicate that although the internal en-ergy is a function of temperature and density or specific (or molar) volume, CV hasbeen measured along a path of constant volume; and although the enthalpy is a func-tion of temperature and pressure, CP has been evaluated in an experiment in which thepressure was held constant.

For the special case of the ideal gas, the enthalpy and internal energy of the fluid arefunctions only of temperature. In this case the partial derivatives above become totalderivatives, and

C∗P(T ) = d H

dTand C∗

V(T ) = dU

dT(3.3-4)

so that the ideal gas heat capacities, which we denote using asterisks as C∗P(T ) and

C∗V(T ), are only functions of T as well. The temperature dependence of the ideal gas

heat capacity can be measured or, in some cases, computed using the methods of sta-tistical mechanics and detailed information about molecular structure, bond lengths,vibrational frequencies, and so forth. For our purposes C∗

P(T ) will either be consideredto be a constant or be written as a function of temperature in the form

Ideal gas heat capacity C∗P(T ) = a + bT + cT 2 + dT 3 + · · · (3.3-5)

Since H = U + PV , and for the ideal gas PV = RT , we have H = U + RT and

C∗P(T ) = d H

dT= d(U + RT )

dT= C∗

V(T ) + R

so that C∗V(T ) = C∗

P(T ) − R = (a − R) + bT + cT 2 + dT 3 + · · · . The constants inEq. 3.3-5 for various gases are given in Appendix A.II.

The enthalpy and internal energy of an ideal gas at a temperature T2 can be relatedto their values at T1 by integration of Eqs. 3.3-4 to obtain

H IG(T2) = H IG(T1) +∫ T2

T1

C∗P(T ) dT


and

U IG(T2) = U IG(T1) +∫ T2

T1

C∗V(T ) dT (3.3-6)

where the superscript IG has been introduced to remind us that these equations arevalid only for ideal gases.

Our interest in the first part of this book is with energy flow problems in single-component systems. Since the only energy information needed in solving these prob-lems is the change in internal energy and/or enthalpy of a substance between two states,and since determination of the absolute energy of a substance is not possible, what isdone is to choose an easily accessible state of a substance to be the reference state,for which H IG is arbitrarily set equal to zero, and then report the enthalpy and internalenergy of all other states relative to this reference state. (That there is a state for eachsubstance for which the enthalpy has been arbitrarily set to zero does lead to difficultieswhen chemical reactions occur. Consequently, another energy convention is introducedin Chapter 8.)

If, for the ideal gas, the temperature TR is chosen as the reference temperature (i.e.,H IG at TR is set equal to zero), the enthalpy at temperature T is then

H IG(T ) =∫ T

TR

C∗P(T ) dT (3.3-7)

Similarly, the internal energy at T is

U IG(T ) = U IG(TR) +∫ T

TR

C∗V(T ) dT = {H IG(TR) − RTR} +

∫ T

TR

C∗V(T ) dT

=∫ T

TR

C∗V(T ) dT − RTR

(3.3-8)One possible choice for the reference temperature TR is absolute zero. In this case,

H IG(T ) =∫ T

0C∗

P(T ) dT and U IG(T ) =∫ T

0C∗

V(T ) dT

However, to use these equations, heat capacity data are needed from absolute zeroto the temperature of interest. These data are not likely to be available, so a moreconvenient reference temperature, such as 0◦C, is frequently used.

For the special case in which the constant-pressure and constant-volume heat capac-ities are independent of temperature, we have, from Eqs. 3.3-7 and 3.3-8,

H IG(T ) = C∗P(T − TR) (3.3-7′′′ )

and

U IG(T ) = C∗V(T − TR) − RTR = C∗

VT − C∗P TR (3.3-8′′′ )

which, when TR is taken to be absolute zero, simplify further to

H IG(T ) = C∗P T and U IG(T ) = C∗

VT


As we will see in Chapters 6 and 7, very few fluids are ideal gases, and the math-ematics of relating the enthalpy and internal energy to the temperature and pressureof a real fluid is much more complicated than indicated here. Therefore, for fluids ofindustrial and scientific importance, detailed experimental thermodynamic data havebeen collected. These data can be presented in tabular form (see Appendix A.III for atable of the thermodynamic properties of steam) or in graphical form, as in Figs. 3.3-1 to 3.3-4 for steam, methane, nitrogen, and the environmentally friendly refrigerantHFC-134a. (Can you identify the reference state for the construction of the steam ta-bles in Appendix A.III?) With these detailed data one can, given values of temperatureand pressure, easily find the enthalpy, specific volume, and entropy (a thermodynamicquantity that is introduced in the next chapter). More generally, given any two intensivevariables of a single-component, single-phase system, the remaining properties can befound.

Notice that different choices have been made for the independent variables in thesefigures. Although the independent variables may be chosen arbitrarily,5 some choicesare especially convenient for solving certain types of problems. Thus, as we will see,an enthalpy-entropy (H -S) or Mollier diagram,6 such as Fig. 3.3-1a, is useful for prob-lems involving turbines and compressors; enthalpy-pressure (H -P) diagrams (for ex-ample Figs. 3.3-2 to 3.3-4) are useful in solving refrigeration problems; and temperature-entropy (T -S) diagrams, of which Fig. 3.3-1b is an example, are used in the analysisof engines and power and refrigeration cycles (see Sec. 3.4 and Chapter 5).

An important characteristic of real fluids is that at sufficiently low temperatures theycondense to form liquids and solids. Also, many applications of thermodynamics of in-terest to engineers involve either a range of thermodynamic states for which the fluid ofinterest undergoes a phase change, or equilibrium multiphase mixtures (e.g., steam andwater at 100◦C and 101.3 kPa). Since the energy balance equation is expressed in termsof the internal energy and enthalpy per unit mass of the system, this equation is validregardless of which phase or mixture of phases is present. Consequently, there is nodifficulty, in principle, in using the energy balance (or other equations to be introducedlater) for multiphase or phase-change problems, provided thermodynamic informationis available for each of the phases present. Figures 3.3-1 to 3.3-4 and the steam tablesin Appendix A.III provide such information for the vapor and liquid phases and, withinthe dome-shaped region, for vapor-liquid mixtures. Similar information for many otherfluids is also available. Thus, you should not hesitate to apply the equations of thermo-dynamics to the solution of problems involving gases, liquids, solids, and mixturesthereof.

There is a simple relationship between the thermodynamic properties of a two-phasemixture (e.g., a mixture of water and steam), the properties of the individual phases,and the mass distribution between the phases. If θ is any intensive property, such asinternal energy per unit mass or volume per unit mass, its value in an equilibrium two-phase mixture is

Properties of atwo-phase mixture:the lever rule

θ = ωIθ I + ωIIθ II = ωIθ I + (1 − ωI)θ II (3.3-9)

5See, however, the comments made in Sec. 1.6 concerning the use of the combinations U and T , and V and T asthe independent thermodynamic variables.6Entropy, denoted by the letter S, is a thermodynamic variable to be introduced in Chapter 4.

(a)

w00

14-n

.eps

Figure 3.3-1 (a) Enthalpy-entropy of Mollier diagram for steam. [Source: ASME Steam Tables in SI (Metric) Units forInstructional Use, American Society of Mechanical Engineers, New York, 1967. Used with permission.] (This figure appearsas an Adobe PDF file on the CD-ROM accompanying this book, and may be enlarged and printed for easier reading and foruse in solving problems.)

(b)

w00

15-n

.eps

Figure 3.3-1(b) Temperature-entropy diagram for steam. [Source: J. H. Keenan, F. G. Keyes, P. G. Hill, and J. G. Moore,Steam Tables (International Edition—Metric Units). Copyright 1969. John Wiley & Sons, Inc., New York. Used with per-mission.] (This figure appears as an Adobe PDF file on the CD-ROM accompanying this book, and may be enlarged andprinted for easier reading and for use in solving problems.)


Figure 3.3-2 Pressure-enthalpy diagram for methane. (Source: W. C. Reynolds, Thermody-namic Properties in SI, Department of Mechanical Engineering, Stanford University, Stan-ford, CA, 1979. Used with permission.) (This figure appears as an Adobe PDF file on theCD-ROM accompanying this book, and may be enlarged and printed for easier reading andfor use in solving problems.)

w00

16-n

.eps

Here ωI is the mass fraction of the system that is in phase I, and θ I is the value ofthe variable in that phase. Also, by definition of the mass fraction, ωI + ωII = 1.(For mixtures of steam and water, the mass fraction of steam is termed the quality andis frequently expressed as a percent, rather than as a fraction; for example, a steam-water mixture containing 0.02 kg of water for each kg of mixture is referred to assteam of 98 percent quality.) Note that Eq. 3.3-9 gives the property of a two-phasemixture as a linear combination of the properties of each phase weighted by its massfraction. Consequently, if charts such as Figures 3.3-1 to 3.3-4 are used to obtain thethermodynamic properties, the properties of the two-phase mixture will fall along aline connecting the properties of the individual phases, which gives rise to referring toEq. 3.3-9 as the lever rule.

If a mixture consists of two phases (i.e., vapor and liquid, liquid and solid, or solidand vapor), the two phases will be at the same temperature and pressure; however,other properties of the two phases will be different. For example, the specific volumeof the vapor and liquid phases can be very different, as will be their internal energyand enthalpy, and this must be taken into account in energy balance calculations. Thenotation that will be used in this book is as follows:

�vap H = HV − HL = enthalpy of vaporization per unit mass or on a molar basis,

�vap H = HV − HL = molar enthalpy of vaporization


Figure 3.3-3 Pressure-enthalpy diagram for nitrogen. (Source: W. C. Reynolds, Thermody-namic Properties in SI, Department of Mechanical Engineering, Stanford University, Stan-ford, CA, 1979. Used with permission.) (This figure appears as an Adobe PDF file on theCD-ROM accompanying this book, and may be enlarged and printed for easier reading andfor use in solving problems.)

w00

17-n

.eps

Also (but for brevity, only on a molar basis),

�fus H = HL − HS = molar enthalpy of melting or fusion

�sub H = HV − HS = molar enthalpy of sublimation

Similar expressions can be written for the volume changes and internal energy changeson a phase change.

It is also useful to note that several simplifications can be made in computing thethermodynamic properties of solids and liquids. First, because the molar volumes ofcondensed phases are small, the product PV can be neglected unless the pressure ishigh. Thus, for solids and liquids,

Solids or liquids atlow pressure

H ≈ U (3.3-10)

A further simplification commonly made for liquids and solids is to assume that theyare also incompressible; that is, their volume is only a function of temperature, so that

Idealizedincompressible fluidor solid

(∂V

∂ P

)T

= 0 (3.3-11)

In Chapter 6 we show that for incompressible fluids, the thermodynamic proper-ties U , CP, and CV are functions of temperature only. Since, in fact, solids, and mostliquids away from their critical point (see Chapter 7) are relatively incompressible,


Figure 3.3-4 Pressure–enthalpy diagram for HFC-134a. (Used with permission of DuPont Flu-oroproducts.) (This figure appears as an Adobe PDF file on the CD-ROM accompanying thisbook, and may be enlarged and printed for easier reading and for use in solving problems.)

w00

18-n

.eps

Eqs. 3.3-10 and 3.3-11, together with the assumption that these properties depend onlyon temperature, are reasonably accurate and often used in thermodynamic studies in-volving liquids and solids. Thus, for example, the internal energy of liquid water at atemperature T1 and pressure P1 is, to a very good approximation, equal to the internalenergy of liquid water at the temperature T1 and any other pressure. Consequently, theentries for the internal energy of liquid water for a variety of temperatures (at pressurescorresponding to the vapor-liquid coexistence or saturation pressures at each temper-ature) given in the saturation steam tables of Appendix A.III can also be used for theinternal energy of liquid water at these same temperatures and higher pressures.

Although we will not try to quantitatively relate the interactions between moleculesto their properties—that is the role of statistical mechanics, not thermodynamics—it isuseful to make some qualitative observations. The starting point is that the interactionsbetween a pair of simple molecules (for example, argon or methane) depend on theseparation distance between their centers of mass, as shown in Fig. 3.3-5. There we seethat if the molecules are far apart, the interaction energy is very low, and it vanishes atinfinite separation. As the molecules are brought closer together, they attract each other,which decreases the energy of the system. However, if the molecules are brought tooclose together (so that their electrons overlap), the molecules repel each other, resultingin a positive energy that increases rapidly as one attempts to bring the molecules closer.

We can make the following observations from this simple picture of molecular in-teractions. First, if the molecules are widely separated, as occurs in a dilute gas, therewill be no energy of interaction between the molecules; this is the case of an ideal gas.Next, as the density increases, and the molecules are somewhat closer together, molec-ular attractions become more important, and the energy of the system decreases. Next,


Inte

ract

ion

ener

gy

+

−

Separation distance

0

w00

19-n

.eps

Figure 3.3-5 The interaction energy between two molecules as a function of their separationdistance. Since the molecules cannot overlap, there is a strong repulsion (positive interactionenergy) at small separation distances. At larger separation distances the interactions betweenthe electrons result in an attraction between the molecules (negative interaction energy), whichvanishes at very large separations.

at liquid densities, the average distance between the molecules will be near the deepest(most attractive) part of the interaction energy curve shown in Fig. 3.3-5. (Note that themolecules will not be at the very lowest value of the interaction energy curve as a resultof their thermal motion, and because the behavior of a large collection of moleculesis more complicated than can be inferred by examining the interaction between only apair of molecules.) Consequently, a liquid has considerably less internal energy than agas. The energy that must be added to a liquid to cause its molecules to move fartherapart and vaporize is the heat of vaporization �vap H . In solids, the molecules generallyare located very close to the minimum in the interaction energy function in an orderedlattice, so that a solid has even less internal energy than a liquid. The amount of energyrequired to slightly increase the separation distances between the molecules in a solidand form a liquid is the heat of melting or the heat of fusion �fus H .

As mentioned earlier, the constant-pressure heat capacity of solids is a function oftemperature; in fact, CP goes to zero at the absolute zero of temperature and approachesa constant at high temperatures. An approximate estimate for CP of solids for temper-atures of interest to chemical engineers comes from the empirical law (or observation)of DuLong and Petit that

CP = 3R = 24.942J

mol K(3.3-12)

For comparison, the constant-pressure heat capacities of lead, gold, and aluminum at25◦C are 26.8, 25.2, and 24.4 J/(mol K), respectively. So we see that the DuLong-Petitlaw gives reasonable though not exact values for the heat capacities of solids.

3.4 APPLICATIONS OF THE MASS AND ENERGY BALANCES

In many thermodynamics problems one is given some information about the initialequilibrium state of a substance and asked to find the final state if the heat and workflows are specified, or to find the heat or work flows accompanying the change to aspecified final state. Since we use thermodynamic balance equations to get the infor-

3.4 Applications of the Mass and Energy Balances 73

mation needed to solve this sort of problem, the starting point is always the same:the identification of a convenient thermodynamic system. The main restriction on thechoice of a system is that the flow terms into and out of the system must be of asimple form—for example, not varying with time, or perhaps even zero. Next, theforms of the mass and energy balance equations appropriate to the system choice arewritten, and any information about the initial and final states of the system and theflow terms is used. Finally, the thermal equation of state is used to replace the internalenergy and enthalpy in the balance equations with temperature, pressure, and volume;the volumetric equation of state may then be used to eliminate volume in terms oftemperature and pressure. In this way equations are obtained that contain temperatureand pressure as the only state variables.

The volumetric equation of state may also provide another relationship between thetemperature, pressure, mass, and volume when the information about the final state ofthe system is presented in terms of total volume, rather than volume per unit mass ormolar volume (see Illustration 3.4-5).

By using the balance equations and the equation-of-state information, we will fre-quently be left with equations that contain only temperature, pressure, mass, shaft work(Ws), and heat flow (Q). If the number of equations equals the number of unknowns,the problem can be solved. The mass and energy balance equations, together withequation-of-state information, are sufficient to solve many, but not all, energy flowproblems. In some situations we are left with more unknowns than equations. In fact,we can readily identify a class of problems of this sort. The mass and energy balanceequations together can, at most, yield new information about only one intensive vari-able of the system (the internal energy or enthalpy per unit mass) or about the sum ofthe heat and work flows if only the state variables are specified. Therefore, we are not,at present, able to solve problems in which (1) there is no information about any inten-sive variable of the final state of the system, (2) both the heat flow (Q) and the shaftwork (Ws) are unspecified, or (3) one intensive variable of the final state and eitherQ or Ws are unknown, as in Illustration 3.4-4. To solve these problems, an additionalbalance equation is needed; such an equation is developed in Chapter 4.

The seemingly most arbitrary step in thermodynamic problem solving is the choiceof the system. Since the mass and energy balances were formulated with great general-ity, they apply to any choice of system, and, as was demonstrated in Illustration 3.2-1,the solution of a problem is independent of the system chosen in obtaining the solu-tion. However, some system choices may result in less effort being required to obtaina solution. This is demonstrated here and again in Chapter 4.

ILLUSTRATION 3.4-1Joule-Thomson Calculation Using a Mollier Diagram and Steam Tables

Steam at 400 bar and 500◦C undergoes a Joule-Thomson expansion to 1 bar. Determine thetemperature of the steam after the expansion using

a. Fig. 3.3-1ab. Fig. 3.3-1bc. The steam tables in Appendix A.III

SOLUTION

(Since only one thermodynamic state variable—here the final temperature—is unknown, fromthe discussion that precedes this illustration we can expect to be able to obtain a solution to thisproblem.)


We start from Illustration 3.2-3, where it was shown that

H1 = H (T1, P1) = H(T2, P2) = H2

for a Joule-Thomson expansion. Since T1 and P1 are known, H1 can be found from either Fig.3.3-1 or the steam tables. Then, since H2 (= H1, from the foregoing) and P2 are known, T2 canbe found.

a. Using Fig. 3.3-1a, the Mollier diagram, we first locate the point P = 400 bar = 40 000 kPaand T = 500◦C, which corresponds to H1 = 2900 kJ/kg. Following a line of constant en-thalpy (a horizontal line on this diagram) to P = 1 bar = 100 kPa, we find that the finaltemperature is about 214◦C.

b. Using Fig. 3.3-1b, we locate the point P = 400 bar and T = 500◦C (which is somewhateasier to do than it was using Fig. 3.3-1a) and follow the curved line of constant enthalpyto a pressure of 1 bar to see that T2 = 214◦C.

c. Using the steam tables of Appendix A.III, we have that at P = 400 bar = 40 MPa andT = 500◦C, H = 2903.3 kJ/kg. At P = 1 bar = 0.1 MPa, H = 2875.3 kJ/kg atT = 200◦C and H = 2974.3 kJ/kg at T = 250◦C. Assuming that the enthalpy varieslinearly with temperature between 200 and 250◦C at P = 1 bar, we have by interpolation

T = 200 + (250 − 200) × 2903.3 − 2875.3

2974.3 − 2875.3= 214.1◦C

COMMENT

For many problems a graphical representation of thermodynamic data, such as Figure 3.3-1, iseasiest to use, although the answers obtained are approximate and certain parts of the graphsmay be difficult to read accurately. The use of tables of thermodynamic data, such as the steamtables, generally leads to the most accurate answers; however, one or more interpolations maybe required. For example, if the initial conditions of the steam had been 475 bar and 530◦Cinstead of 400 bar and 500◦C, the method of solution using Fig. 3.3-1 would be unchanged;however, using the steam tables, we would have to interpolate with respect to both temperatureand pressure to get the initial enthalpy of the steam.

One way to do this is first, by interpolation between temperatures, to obtain the enthalpy ofsteam at 530◦C at both 400 bar = 40 MPa and 500 bar. Then, by interpolation with respect topressure between these two values, we obtain the enthalpy at 475 bar. That is, from

H (40 MPa, 500◦C) = 2903.3 kJ/kg H (50 MPa, 500◦C) = 2720.1 kJ/kgH (40 MPa, 550◦C) = 3149.1 kJ/kg H (50 MPa, 550◦C) = 3019.5 kJ/kg

and the interpolation formula

�(x + �) = �(x) + ��(y) − �(x)

y − x

where � is any tabulated function, and x and y are two adjacent values at which � is available,we have

H (40 MPa, 530◦C) = H (40 MPa, 500◦C) + 30 × H (40 MPa, 550◦C) − H (40 MPa, 500◦C)

550 − 500

= 2903.3 + 30 × 3149.1 − 2903.3

50= 2903.3 + 147.5

= 3050.8 kJ/kg


and

H (50 MPa, 530◦C) = 2720.1 + 30 × 3019.5 − 2720.1

50= 2899.7 kJ/kg

Then

H (47.5 MPa, 530◦C) = H (40 MPa, 530◦C) + 7.5 × H (50 MPa, 530◦C) − H (40 MPa, 530◦C)

50 − 40

= 3050.8 + 7.5

10× (2881.7 − 3050.8) = 2924.0 kJ/kg

ILLUSTRATION 3.4-2Application of the Complete Energy Balance Using the Steam Tables

An adiabatic steady-state turbine is being designed to serve as an energy source for a smallelectrical generator. The inlet to the turbine is steam at 600◦C and 10 bar, with a mass flow rateof 2.5 kg/s through an inlet pipe that is 10 cm in diameter. The conditions at the turbine exit areT = 400◦C and P = 1 bar. Since the steam expands through the turbine, the outlet pipe is 25cm in diameter. Estimate the rate at which work can be obtained from this turbine.

T1 = 600°CP1 = 10 bar

T2 = 400°CP2 = 1 bar

Ws

w00

20-n

.eps

SOLUTION

(This is another problem in which there is only a single thermodynamic unknown, the rate atwhich work is obtained, so we can expect to be able to solve this problem.)

The first step in solving any energy flow problem is to choose the thermodynamic system;the second step is to write the balance equations for the system. Here we take the turbine and itscontents to be the system. The mass and energy balance equations for this adiabatic, steady-statesystem are

d M

dt= 0 = M1 + M2 (a)

and

d

dt

{U + M

(v2

2+ gh

)}= 0 = M1

(H1 + v2

1

2

)+ M2

(H2 + v2

2

2

)+ Ws (b)

In writing these equations we have set the rate of change of mass and energy equal to zerobecause the turbine is in steady-state operation; Q is equal to zero because the process is adia-batic, and P(dV/dt) is equal to zero because the volume of the system is constant (unless theturbine explodes). Finally, since the schematic diagram indicates that the turbine is positionedhorizontally, we have assumed there is no potential energy change in the flowing steam.

There are six unknowns—M2 , H1, H2, Ws , v1, and v2—in Eqs. a and b. However, both veloc-ities will be found from the mass flow rates, pipe diameters, and volumetric equation-of-stateinformation (here the steam tables in Appendix A.III). Also, thermal equation-of-state infor-mation (again the steam tables in Appendix A.III) relates the enthalpies to temperature and


pressure, both of which are known. Thus M2 and H2 are the only real unknowns, and these maybe found from the balance equations above. From the mass balance equation, we have

M2 = −M1 = −2.5 kg/s

From the steam tables or, less accurately from Fig. 3.3-1, we have

H1 = 3697.9 kJ/kg

V1 = 0.4011 m3/kg

H2 = 3278.2 kJ/kg

V2 = 3.103 m3/kg

The velocities at the inlet and outlet to the turbine are calculated from

Volumetric flow rate = MV = πd2

4v

where d is the pipe diameter. Therefore,

v1 = 4M1V1

πd2in

=4 · 2.5

kg

s· 0.4011

m3

kg3.14159 · (0.1 m)2

= 127.7m

s

v2 = 4M2V2

πd2out

= 158.0m

s

Therefore, the energy balance yields

Ws = −M2

(H2 + v2

2

2

)− M1

(H1 + v2

1

2

)

= −2.5kg

s

{(H1 − H2) + 1

2(v2

1 − v22)

}kJ

kg

= −2.5kg

s

419.7kJ

kg+ 1

2(127.72 − 158.02)

m2

s2·

1J

kgm2

s2

· 1 kJ

1000 J

= −2.5kg

s{419.7 − 4.3} kJ

kg= −1038.5

kJ

s(= −1329 hp)

COMMENT

If we had completely neglected the kinetic energy terms in this calculation, the error in thework term would be 4.3 kJ/kg, or about 1%. Generally, the contribution of kinetic and potentialenergy terms can be neglected when there is a significant change in the fluid temperature, as wassuggested in Sec. 3.2.

ILLUSTRATION 3.4-3Use of Mass and Energy Balances with an Ideal Gas

A compressed-air tank is to be repressurized to 40 bar by being connected to a high-pressureline containing air at 50 bar and 20◦C. The repressurization of the tank occurs so quickly thatthe process can be assumed to be adiabatic; also, there is no heat transfer from the air to thetank. For this illustration, assume air to be an ideal gas with CV = 21 J/(mol K).

a. If the tank initially contains air at 1 bar and 20◦C, what will be the temperature of the airin the tank at the end of the filling process?


b. After a sufficiently long period of time, the gas in the tank is found to be at room temper-ature (20◦C) because of heat exchange with the tank and the atmosphere. What is the newpressure of air in the tank?

50 bar20°C

w00

21-n

.eps

SOLUTION

(Each of these problems contains only a single unknown thermodynamic property, so solutionsshould be possible.)

a. We will take the contents of the tank to be the system. The difference form of the mass (orrather mole) and energy balances for this open system are

N2 − N1 = �N (a)N2U 2 − N1U1 = (�N)H in (b)

In writing the energy balance we have made the following observations:

1. The kinetic and potential energy terms are small and can be neglected.2. Since the tank is connected to a source of gas at constant temperature and pressure, H in

is constant.3. The initial process is adiabatic, so Q = 0, and the system (the contents of the tank) is

of constant volume, so �V = 0.

Substituting Eq. a in Eq. b, and recognizing that for the ideal gas H (T ) = C∗P(T − TR)

and U(T ) = C∗V(T − TR) − RTR , yields

N2{C∗V(T2 − TR) − RTR } − N1{C∗

V(T1 − TR) − RTR} = (N2 − N1)C∗P (Tin − TR)

or

N2C∗VT2 − N1C∗

VT1 = (N2 − N1)C∗P Tin

(Note that the reference temperature TR cancels out of the equation, as it must, since thefinal result cannot depend on the arbitrarily chosen reference temperature.) Finally, usingthe ideal gas equation of state to eliminate N1 and N2, and recognizing that V1 = V2,yields

P2

T2= P1

T1+ C∗

V

C∗P

(P2 − P1

Tin

)or T2 = P2

P1

T1+ C∗

V

C∗P

(P2 − P1

Tin

)

The only unknown in this equation is T2, so, formally, the problem is solved. The answeris T = 405.2 K = 132.05◦C.


Before proceeding to the second part of the problem, it is interesting to consider thecase in which the tank is initially evacuated. Here P1 = 0, and so

T2 = C∗P

C∗V

Tin

independent of the final pressure. Since CP is always greater than CV, the temperature ofthe gas in the tank at the end of the filling process will be greater than the temperature ofgas in the line. Why is this so?

b. To find the pressure in the tank after the heat transfer process, we use the mass balanceand the equation of state. Again, choosing the contents of the tank as the system, the mass(mole) balance is N2 = N1, since there is no transfer of mass into or out of the systemduring the heat transfer process (unless, of course, the tank is leaking; we do not considerthis complication here). Now using the ideal gas equation of state, we have

P2

T2= P1

T1or P2 = P1

T2

T1

Thus, P2 = 28.94 bar.

ILLUSTRATION 3.4-4Example of a Thermodynamics Problem That Cannot Be Solved with Only the Mass and EnergyBalances7

A compressor is a gas pumping device that takes in gas at low pressure and discharges it ata higher pressure. Since this process occurs quickly compared with heat transfer, it is usuallyassumed to be adiabatic; that is, there is no heat transfer to or from the gas during its compres-sion. Assuming that the inlet to the compressor is air [which we will take to be an ideal gaswith C∗

P = 29.3 J/(mol K)] at 1 bar and 290 K and that the discharge is at a pressure of 10 bar,estimate the temperature of the exit gas and the rate at which work is done on the gas (i.e., thepower requirement) for a gas flow of 2.5 mol/s.

SOLUTION

(Since there are two unknown thermodynamic quantities, the final temperature and the rate atwhich work is being done, we can anticipate that the mass and energy balances will not besufficient to solve this problem.)

The system will be taken to be the gas contained in the compressor. The differential form ofthe molar mass and energy balances for this open system are

d N

dt= N1 + N2

dU

dt= N1 H1 + N2 H2 + Q + W

where we have used the subscript 1 to indicate the flow stream into the compressor and 2 toindicate the flow stream out of the compressor.

Since the compressor operates continuously, the process may be assumed to be in a steadystate,

d N

dt= 0 or N1 = −N2

dU

dt= 0

7We return to this problem in the next chapter after formulating the balance equation for an additional thermody-namic variable, the entropy.


that is, the time variations of the mass of the gas contained in the compressor and of the energycontent of this gas are both zero. Also, Q = 0 since there is no heat transfer to the gas, andW = Ws since the system boundaries (the compressor) are not changing with time. Thus wehave

Ws = N1 H 2 − N1 H1 = N1C∗P(T2 − T1)

or

W s = C∗P(T2 − T1)

where W s = Ws/N1 is the work done per mole of gas. Therefore, the power necessary to drivethe compressor can be computed once the outlet temperature of the gas is known, or the outlettemperature can be determined if the power input is known.

We are at an impasse; we need more information before a solution can be obtained. It is clearby comparison with the previous examples why we cannot obtain a solution here. In the previouscases, the mass balance and the energy balance, together with the equation of state of the fluidand the problem statement, provided the information necessary to determine the final state of thesystem. However, here we have a situation where the energy balance contains two unknowns,the final temperature and W s . Since neither is specified, we need additional information aboutthe system or process before we can solve the problem. This additional information will beobtained using an additional balance equation developed in the next chapter.

ILLUSTRATION 3.4-5Use of Mass and Energy Balances to Solve an Ideal Gas Problem8

A gas cylinder of 1 m3 volume containing nitrogen initially at a pressure of 40 bar and a tem-perature of 200 K is connected to another cylinder of 1 m3 volume that is evacuated. A valvebetween the two cylinders is opened until the pressures in the cylinders equalize. Find the finaltemperature and pressure in each cylinder if there is no heat flow into or out of the cylinders orbetween the gas and the cylinder. You may assume that the gas is ideal with a constant-pressureheat capacity of 29.3 J/(mol K).

SOLUTION

This problem is more complicated than the previous ones because we are interested in changesthat occur in two separate cylinders. We can try to obtain a solution to this problem in twodifferent ways. First, we could consider each tank to be a separate system, and so obtain twomass balance equations and two energy balance equations, which are coupled by the fact thatthe mass flow rate and enthalpy entering the second cylinder.9 Alternatively, we could obtainan equivalent set of equations by choosing a composite system of the two interconnected gascylinders to be the first system and the second system to be either one of the cylinders. In thisway the first (composite) system is closed and the second system is open. We will use the secondsystem choice here; you are encouraged to explore the first system choice independently and toverify that the same solution is obtained.

The difference form of the mass and energy balance equations (on a molar basis) for thetwo-cylinder composite system are

N i1 = N f

1 + N f2 (a)

8An alternative solution to this problem giving the same answer is given in the next chapter.9That the enthalpy of the gas leaving the first cylinder is equal to that entering the second, even though the twocylinders are at different pressures, follows from the fact that the plumbing between the two can be thought of asa flow constriction, as in the Joule-Thomson expansion. Thus the analysis of Illustration 3.2-3 applies to this partof the total process.


and

N i1U i

1 = N f1 U f

1 + N f2 U f

2 (b)

Here the subscripts 1 and 2 refer to the cylinders, and the superscripts i and f refer to the initialand final states. In writing the energy balance equation we have recognized that for the systemconsisting of both cylinders there is no mass flow, heat flow, or change in volume.

Now using, in Eq. a, the ideal gas equation of state written as N = PV/RT and the fact thatthe volumes of both cylinders are equal yields

Pi1

T i1

= P f1

T f1

+ P f2

T f2

(a′′′ )

Using the same observations in Eq. b and further recognizing that for a constant heat capacitygas we have, from Eq. 3.3-8, that

U(T ) = C∗VT − C∗

P TR

yields

Pi1

T i1

{C∗VT i

1 − C∗P TR} = P f

1

T f1

{C∗VT f

1 − C∗P TR} + P f

2

T f2

{C∗VT f

2 − C∗P TR}

which, on rearrangement, gives

−{

Pi1

T i1

− P f1

T f1

− P f2

T f2

}C∗

P TR + C∗V{Pi

1 − P f1 − P f

2 } = 0

Since the bracketed quantity in the first term is identically zero (see Eq. a′), we obtain

Pi1 = P f

1 + P f2 (c)

(Note that the properties of the reference state have canceled. This is to be expected, since thesolution to a change-of-state problem must be independent of the arbitrarily chosen referencestate. This is an important point. In nature, the process will result in the same final state indepen-dent of our arbitrary choice of reference temperature, TR . Therefore, if our analysis is correct,TR must not appear in the final answer.)

Next, we observe that from the problem statement P f1 = P f

2 ; thus

P f1 = P f

2 = 12 Pi

1 = 20 bar

and from Eq. a′

1

T f1

+ 1

T f2

= 2

T i1

(c′′′ )

Thus we have one equation for the two unknowns, the two final temperatures. We cannot assumethat the final gas temperatures in the two cylinders are the same because nothing in the problemstatement indicates that a transfer of heat between the cylinders necessary to equalize the gastemperatures has occurred.

To get the additional information necessary to solve this problem, we write the mass andenergy balance equations for the initially filled cylinder. The rate-of-change form of these equa-tions for this system are

d N1

dt= N (d)


and

d(N1U1)

dt= N H 1 (e)

In writing the energy balance equation, we have made use of the fact that Q, Ws , and dV/dtare all zero. Also, we have assumed that while the gas temperature is changing with time, it isspatially uniform within the cylinder, so that at any instant the temperature and pressure of thegas leaving the cylinder are identical with those properties of the gas in the cylinder. Thus, themolar enthalpy of the gas leaving the cylinder is

H = H (T1, P1) = H 1

Since our interest is in the change in temperature of the gas that occurs as its pressure dropsfrom 40 bar to 20 bar due to the escaping gas, you may ask why the balance equations here havebeen written in the rate-of-change form rather than in terms of the change over a time interval.The answer is that since the properties of the gas within the cylinder (i.e., its temperature andpressure) are changing with time, so is H1, the enthalpy of the exiting gas. Thus, if we wereto use the form of Eq. e integrated over a time interval (i.e., the difference form of the energybalance equation),

N f1 U f

1 − N i1U i

1 =∫

N H1 dt

we would have no way of evaluating the integral on the right side. Consequently, the differenceequation provides no useful information for the solution of the problem. However, by startingwith Eqs. d and e, it is possible to obtain a solution, as will be evident shortly.

To proceed with the solution, we first combine and rearrange the mass and energy balancesto obtain

d(N1U 1)

dt≡ N1

dU 1

dt+ U 1

d N1

dt= N H1 = H1

d N1

dt

so that we have

N1dU 1

dt= (H 1 − U 1)

d N1

dt

Now we use the following properties of the ideal gas (see Eqs. 3.3-7 and 3.3-8)

N = PV/RT H = C∗P(T − TR)

and

U = C∗V(T − TR) − RTR

to obtain

P1V

RT1C∗

V

dT1

dt= RT1

d

dt

(P1V

RT1

)

Simplifying this equation yields

C∗V

R

1

T1

dT1

dt= T1

P1

d

dt

(P1

T1

)

or

C∗V

R

d ln T1

dt= d

dtln

(P1

T1

)


Now integrating between the initial and final states, we obtain

(T f

1

T i1

)C∗V/R

=(

P f1

Pi1

)(T i

1

T f1

)

or

(T f

1

T i1

)C∗P /R

=(

P f1

Pi1

)(f)

where we have used the fact that for the ideal gas C∗P = C∗

V + R. Equation f provides the meansto compute T f

1 , and T f2 can then be found from Eq. c′. Finally, using the ideal gas equation of

state we can compute the final number of moles of gas in each cylinder using the relation

N fJ = VcylJ P f

J

RT fJ

(g)

where the subscript J refers to the cylinder number. The answers are

T f1 = 164.3 K N f

1 = 1.464 kmol

T f2 = 255.6 K N f

2 = 0.941 kmol

COMMENTS

The solution of this problem for real fluids is considerably more complicated than for the idealgas. The starting points are again

P f1 = P f

2 (h)

and

N f1 + N f

2 = N i1 (i)

and Eqs. d and e. However, instead of Eq. g we now have

N f1 = Vcyl1

V f1

(j)

and

N f2 = Vcyl2

V f2

(k)

where V f1 and V f

2 are related to (T f1 , P f

1 ) and (T f2 , P f

2 ), respectively, through the equation ofstate or tabular PV T data of the form

V f1 = V f

1 (T f1 , P f

1 ) (l)V f

2 = V f2 (T f

2 , P f2 ) (m)

The energy balance for the two-cylinder composite system is

N i1U i

1 = N f1 U f

1 + N f2 U f

2 (n)

Since a thermal equation of state or tabular data of the form U = U(T , V ) are presumed avail-able, Eq. n introduces no new variables.


Thus we have seven equations among eight unknowns (N f1 , N f

2 , T f1 , T f

2 , P f1 , P f

2 , V f1 , and

V f2 ). The final equation needed to solve this problem can, in principle, be obtained by the ma-

nipulation and integration of Eq. e, as in the ideal gas case, but now using the real fluid equationof state or tabular data and numerical integration techniques. Since this analysis is difficult, anda simpler method of solution (discussed in Chapter 6) is available, the solution of this problemfor the real fluid case is postponed until Sec. 6.5.

ILLUSTRATION 3.4-6Showing That the Change in State Variables between Fixed Initial and Final States Is Indepen-dent of the Path Followed

It is possible to go from a given initial equilibrium state of a system to a given final equilib-rium state by a number of different paths, involving different intermediate states and differentamounts of heat and work. Since the internal energy of a system is a state property, its changebetween any two states must be independent of the path chosen (see Sec. 1.3). The heat andwork flows are, however, path-dependent quantities and can differ on different paths betweengiven initial and final states. This assertion is established here by example. One mole of a gas ata temperature of 25◦C and a pressure of 1 bar (the initial state) is to be heated and compressedin a frictionless piston and cylinder to 300◦C and 10 bar (the final state). Compute the heat andwork required along each of the following paths.

Path A. Isothermal (constant temperature) compression to 10 bar, and then isobaric (constantpressure) heating to 300◦C

Path B. Isobaric heating to 300◦C followed by isothermal compression to 10 barPath C. A compression in which PV γ = constant, where γ = C∗

P/C∗V, followed by an

isobaric cooling or heating, if necessary, to 300◦C.

For simplicity, the gas is assumed to be ideal with C∗P = 38 J/(mol K).

Path B

Path A

Path C

Final state

350

300

250

200

150

100

50

00 2 4 6 8 10

P (bar)12

T(°

C)

Initial statew

0022

-n.e

ps

SOLUTION

The 1-mol sample of gas will be taken as the thermodynamic system. The difference form ofthe mass balance for this closed, deforming volume of gas is

N = constant = 1 mol


and the difference form of the energy balance is

�U = Q −∫

P dV = Q + W

Path A

i. Isothermal compression

Wi = −∫ V 2

V 1

P dV = −∫ V 2

V 1

RTdV

V= −RT

∫ V 2

V 1

dV

V= −RT ln

V 2

V 1

= RT lnP2

P1

= 8.314 J/(mol K) × 298.15 K × ln10

1= 5707.7 J/mol

Since

�U =∫ T2

T1

C∗V dT = C∗

V(T2 − T1) and T2 = T1 = 25◦C

we have

�U = 0 and Qi = −Wi = −5707.7 J/mol

ii. Isobaric heating

Wii = −∫ V 3

V 2

P2 dV = −P2

∫ V 3

V 2

dV = −P2(V 3 − V 2) = −R(T3 − T2)

�U =∫ T3

T2

C∗V dT = C∗

V(T3 − T2)

and

Qii = �U − Wii = C∗V(T3 − T2) + R(T3 − T2) = (C∗

V + R)(T3 − T2)= C∗

P (T3 − T2)

[This is, in fact, a special case of the general result that at constant pressure for a closedsystem, Q = ∫

C∗P dT . This is easily proved by starting with

Q = dU

dt+ P

dV

dt

and using the fact that P is constant to obtain

Q = dU

dt+ d

dt(PV ) = d

dt(U + PV ) = d H

dt= C∗

P

dT

dt

Now setting Q = ∫Q dt yields Q = ∫

C∗P dT .]

Therefore,

Wii = −8.314 J/(mol K) × 275 K = −2286.3 J/molQii = 38 J/(mol K) × 275 K = 10 450 J/molQ = Qi + Qii = −5707.7 + 10 450 = 4742.3 J/molW = Wi + Wii = 5707.7 − 2286.3 = 3421.4 J/mol


Path Bi. Isobaric heating

Qi = C∗P(T2 − T1) = 10 450 J/mol

Wi = −R(T2 − T1) = −2286.3 J/mol

ii. Isothermal compression

Wii = RT lnP2

P1= 8.314 × 573.15 ln

(10

1

)= 10 972.2 J/mol

Qii = −Wii = −10 972.2 J/molQ = 10 450 − 10 972.2 = −522.2 J/molW = −2286.3 + 10 972.2 = 8685.9 J/mol

Path C

i. Compression with PV γ = constant

Wi = −∫ V 2

V 1

P dV = −∫ V 2

V 1

constant

V γ dV = −constant

1 − γ(V 1−γ

2 − V 1−γ1 )

= − 1

1 − γ(P2V 2 − P1V 1) = −R(T2 − T1)

1 − γ= −R(T2 − T1)

1 − (C∗P/C∗

V)= C∗

V(T2 − T1)

where T2 can be computed from

P1V γ1 = P1

(RT1

P1

)γ

= P2V γ2 = P2

(RT2

P2

)γ

or

T2

T1=(

P2

P1

)(γ−1)/γ

Now

γ = C∗P

C∗V

= 38

38 − 8.314= 1.280

so that

T2 = 298.15 K (10)0.280/1.280 = 493.38 K

and

Wi = C∗V(T2 − T1) = (38 − 8.314) J/(mol K) × (493.38 − 298.15) K

= 5795.6 J/mol�Ui = C∗

V(T2 − T1) = 5795.6 J/molQi = �Ui − Wi = 0

ii. Isobaric heating

Qii = C∗P(T3 − T2) = 38 J/(mol K) × (573.15 − 493.38) K = 3031.3 J/mol

Wii = −R(T3 − T2) = −8.314 J/(mol K) × (573.15 − 493.38) K = −663.2 J/mol

and

Q = 0 + 3031.3 = 3031.3 J/molW = 5795.6 − 663.2 = 5132.4 J/mol


SOLUTION

Path Q (J/mol) W (J/mol) Q + W = �U (J/mol)

A 4742.3 3421.4 8163.7B −522.2 8685.9 8163.7C 3031.3 5132.4 8163.7 un

t3.1

COMMENT

Notice that along each of the three paths considered (and, in fact, any other path between theinitial and final states), the sum of Q and W, which is equal to �U , is 8163.7 J/mol, eventhough Q and W separately are different along the different paths. This illustrates that whereasthe internal energy is a state property and is path independent (i.e., its change in going fromstate 1 to state 2 depends only on these states and not on the path between them), the heat andwork flows depend on the path and are therefore path functions.

ILLUSTRATION 3.4-7Showing That More Work Is Obtained If a Process Occurs without Friction

An initial pressure of 2.043 bar is maintained on 1 mol of air contained in a piston-and-cylindersystem by a set of weights W , the weight of the piston, and the surrounding atmosphere. Workis obtained by removing some of the weights and allowing the air to isothermally expand at25◦C, thus lifting the piston and the remaining weights. The process is repeated until all theweights have been removed. The piston has a mass of ω = 5 kg and an area of 0.01 m2. Forsimplicity, the air can be considered to be an ideal gas. Assume that as a result of sliding frictionbetween the piston and the cylinder wall, all oscillatory motions of the piston after the removalof a weight will eventually be damped.

Compute the work obtained from the isothermal expansion and the heat required from exter-nal sources for each of the following:

a. The weight W is taken off in one step.b. The weight is taken off in two steps, with W/2 removed each time.c. The weight is taken off in four steps, with W/4 removed each time.d. The weight is replaced by a pile of sand (of total weight W), and the grains of sand are

removed one at a time.

Processes b and d are illustrated in the following figure.10

SOLUTION

I. Analysis of the problem. Choosing the air in the cylinder to be the system, recognizing that foran ideal gas at constant temperature U is constant so that �U = 0, and neglecting the kineticand potential energy terms for the gas (since the mass of 1 mol of air is only 29 g), we obtainthe following energy balance equation:

0 = Q −∫

P dV (a)

10From H. C. Van Ness, Understanding Thermodynamics, McGraw-Hill, New York, 1969. Used with permissionof the McGraw-Hill Book Co.


The total work done by the gas in lifting and accelerating the piston and the weights against thefrictional forces, and in expanding the system volume against atmospheric pressure is containedin the − ∫

P dV term. To see this we recognize that the laws of classical mechanics apply to thepiston and weights, and equate, at each instant, all the forces on the piston and weights to theiracceleration,

Forces on piston and weights =(

Mass of pistonand weights

)× Acceleration

and obtain

[P × A − Patm × A − (W + ω)g + Ffr] = (W + ω)dv

dt(b)

Here we have taken the vertical upward (+z) direction as being positive and used P and Patm

to represent the pressure of the gas and atmosphere, respectively; A the piston area; ω its mass;W the mass of the weights on the piston at any time; v the piston velocity; and Ffr the frictionalforce, which is proportional to the piston velocity. Recognizing that the piston velocity v is equalto the rate of change of the piston height h or the gas volume V , we have

�/2 �/2 �/2

�/2

�/2

�/2

Process b

Process d

w00

23-n

.eps

v = dh

dt= 1

A

dV

dt


Also we can solve Eq. b for the gas pressure:

P = Patm + (W + ω)

Ag − Ffr

A+ (W + ω)

A

dv

dt(c)

At mechanical and thermodynamic equilibrium (i.e., when dv/dt = 0 and v = 0), we have

P = Patm + (W + ω)

Ag (d)

With these results, the total work done by the gas can be computed. In particular,∫

P dV =∫ [

Patm +(W + ω

A

)g − Ffr

A+ (W + ω)

A

dv

dt

]dV

=[

Patm + (W + ω)

Ag

]�V − 1

A

∫Ffr dV + (W + ω)

A

∫dv

dtdV

This equation can be simplified by rewriting the last integral as follows:

1

A

∫dv

dtdV = 1

A

∫dv

dt

dV

dtdt =

∫dv

dtv dt = 1

2

∫dv2

dtdt = �

(1

2v2

)

where the symbol � indicates the change between the initial and final states. Next, we recallfrom mechanics that the force due to sliding friction, here Ffr, is in the direction opposite to therelative velocity of the moving surfaces and can be written as

Ffr = −kfrv

where kfr is the coefficient of sliding friction. Thus, the remaining integral can be written as

1

A

∫Ffr dV = − 1

A

∫kfrv

dV

dtdt = −kfr

∫v2 dt

The energy balance, Eq. a, then becomes

Q =∫

P dV = Patm �V + (W + ω)g �h + kfr

∫v2 dt + (W + ω)�

(12 v2

)(e)

This equation relates the heat flow into the gas (to keep its temperature constant) to the workthe gas does against the atmosphere in lifting the piston and weights (hence increasing theirpotential energy) against friction and in accelerating the piston and weights (thus increasingtheir kinetic energy).

The work done against frictional forces is dissipated into thermal energy, resulting in a highertemperature at the piston and cylinder wall. This thermal energy is then absorbed by the gas andappears as part of Q. Consequently, the net flow of heat from a temperature bath to the gas is

QNET = Q − kfr

∫v2 dt (f)

Since heat will be transferred to the gas, and the integral is always positive, this equation estab-lishes that less heat will be needed to keep the gas at a constant temperature if the expansionoccurs with friction than in a frictionless process.

Also, since the expansion occurs isothermally, the total heat flow to the gas is, from Eq. a,

Q =∫ V f

Vi

P dV =∫ V f

Vi

N RT

VdV = N RT ln

V f

Vi(g)


Combining Eqs. e and f, and recognizing that our interest here will be in computing the heat andwork flows between states for which the piston has come to rest (v = 0), yields

QNET = Q − kfr

∫v2 dt = Patm �V + (W + ω)g �h = P�V = −W NET (h)

where P is the equilibrium final pressure given by Eq. d. Also from Eqs. f, g, and h, we have

QNET = −W NET = N RT ln

(V2

V1

)− kfr

∫v2 dt (i)

Here W NET represents the net work obtained by the expansion of the gas (i.e., the work obtainedin raising the piston and weights and in doing work against the atmosphere).

The foregoing equations can now be used in the solution of the problem. In particular, asa weight is removed, the new equilibrium gas pressure is computed from Eq. d, the resultingvolume change from the ideal gas law, W NET and QNET from Eq. h, and the work against frictionfrom Eq. i. There is, however, one point that should be mentioned before we proceed withthis calculation. If there were no mechanism for the dissipation of kinetic energy to thermalenergy (that is, sliding friction between the piston and cylinder wall, and possibly also viscousdissipation on expansion and compression of the gas due to its bulk viscosity), then when aweight was removed the piston would be put into a perpetual oscillatory motion. The presenceof a dissipative mechanism will damp the oscillatory motion. (As will be seen, the value ofthe coefficient of sliding friction, kfr, does not affect the amount of kinetic energy ultimatelydissipated as heat. Its value does, however, affect the dynamics of the system and thus determinehow quickly the oscillatory motion is damped.)

II. The numerical solution. First, the mass W of the weights is computed using Eq. d and thefact that the initial pressure is 2.043 bar. Thus,

P = 2.043 bar = 1.013 bar + (5 + W) kg

0.01 m2× 9.807

m

s2× 1 Pa

kg/(m s2)× 1 bar

105 Pa

or

(5 + W) kg = 105.0 kg

so that W = 100 kg.The ideal gas equation of state for 1 mol of air at 25◦C is

PV = N RT = 1 mol × 8.314 × 10−5 bar m3

mol K× (25 + 273.15) K

= 2.479 × 10−2 bar m3 = 2479 J

(j)

and the initial volume of the gas is

V = 2.479 × 10−2 bar m3

2.043 bar= 1.213 × 10−2 m3

Process aThe 100-kg weight is removed. The equilibrium pressure of the gas (after the piston has

stopped oscillating) is

P1 = 1.013 bar + 5 kg × 9.807 m/s2

0.01 m2× 10−5 bar

kg/(m s2)= 1.062 bar

and the gas volume is


V1 = 2.479 × 10−2 bar m3

1.062 bar= 2.334 × 10−2 m3

Thus

�V = (2.334 − 1.213) × 10−2 m3 = 1.121 × 10−2 m3

− W NET = 1.062 bar × 1.121 × 10−2 m3 × 105 J

bar m3

= 1190.5 J = QNET

and

Q = N RT lnV1

V0= 2479 J ln

2.334 × 10−2

1.213 × 10−2= 1622.5 J

(since N RT = 2479 J from Eq. j). Consequently, the work done against frictional forces (andconverted to thermal energy), which we denote by Wfr, is

−Wfr = Q − QNET = (1622.5 − 1190.5) J = 432 J

The total useful work obtained, the net heat supplied, and the work against frictional forcesare given in Table 1. Also, the net work, P �V , is shown as the shaded area in the accompanyingfigure, together with the line representing the isothermal equation of state, Eq. j. Note that inthis case the net work is that of raising the piston and pushing back the atmosphere.

Table 1

−W NET = QNET Q −Wfr

Process (J) (J) (J)

a 1190.5 1622.5 432.0b 1378.7 1622.5 243.8c 1493.0 1622.5 129.5d 1622.5 1622.5 0 Ta

ble

1

Process bThe situation here is similar to that of process a, except that the weight is removed in two

50-kg increments. The pressure, volume, work, and heat flows for each step of the process aregiven in Table 2, and −W NET

i = Pi(�V )i, the net work for each step, is given in the figure.Process c

Here the weight is removed in four 25-kg increments. The pressure, volume, work, and heatflows for each step are given in Table 2 and summarized in Table 1. Also, the net work for eachstep is given in the figure.

0

0

11

0

2

2

2.5

0.01 0.02 0.025

V, m3 V, m3

Process a

0

P, b

ar

P, b

ar

1

1

2

2.5

0.01 0.02 0.025

Process b

w00

24-n

.eps


Table 2

Process b

−W NETi = Pi(�V )i Q = N RT ln

Vi

Vi−1−Wfr

P VStage (bar) (m3) (J) (J) (J)

0 2.043 1.213 × 10−2

1 1.552 1.597 × 10−2 596.0 681.8 85.82 1.062 2.334 × 10−2 782.7 940.7 158.0Total 1378.7 1622.5 243.8

Process c

−W NETi = Pi(�V )i Q = N RT ln

Vi

Vi−1−Wfr

P VStage (bar) (m3) (J) (J) (J)

0 2.043 1.213 × 10−2

1 1.798 1.379 × 10−2 298.5 318.0 19.52 1.552 1.597 × 10−2 338.3 363.8 25.53 1.307 1.897 × 10−2 392.1 426.8 34.74 1.062 2.334 × 10−2 464.1 513.9 49.8Total 1493.0 1622.5 129.5 Ta

ble

2

Process dThe computation here is somewhat more difficult since the number of stages to the calculation

is almost infinite. However, recognizing that in the limit of the mass of a grain of sand goingto zero there is only a differential change in the pressure and volume of the gas and negligiblevelocity or acceleration of the piston, we have

−W NET =∑

i

Pi(�Vi) →∫

P dV = N RT lnV f

Vi= QNET

and Wfr = 0, since the piston velocity is essentially zero at all times. Thus,

− W NET = QNET = Q = 1 mol × 8.314J

mol K× 298.15 K × ln

2.334 × 10−2

1.213 × 10−2

= 1622.5 J

This result is given in Table 1 and the figure.

01

2

0

34

0

1

2

2.5

0.01 0.02 0.025

V, m3 V, m3

0

P, b

ar

P, b

ar

1

2

2.5

0.01 0.02 0.025

w00

25-n

.eps


COMMENTS

Several points are worth noting about this illustration. First, although the initial and final statesof the gas are the same in all three processes, the useful or net work obtained and the net heatrequired differ. Of course, by the energy conservation principle, it is true that −W NET = QNET

for each process. It is important to note that the most useful work is obtained for a given changeof state if the change is carried out in differential steps, so that there is no frictional dissipationof mechanical energy to thermal energy (compare process d with processes a, b, and c). Also,in this case, if we were to reverse the process and compress the gas, it would be found that theminimum work required for the compression is obtained when weights are added to the pistonin differential (rather than finite) steps. (See Problem 3.29.)

It should also be pointed out that in each of the four processes considered here the gas did1622.5 J of work on its surroundings (the piston, the weights, and the atmosphere) and absorbed1622.5 J of heat (from the thermostatic bath maintaining the system temperature constant andfrom the piston and cylinder as a result of their increased temperature due to frictional heating).We can see this from Table 1, since −(W NET + Wfr) = Q = 1622.5 J for all four processes.However, the fraction of the total work of the gas obtained as useful work versus work againstfriction varies among the different processes.

At first glance it might appear that in the process in illustration 3.4.7 the proscriptionin Chapter 1 that thermal energy (here heat) cannot be completely converted to me-chanical energy (here work) has been violated. However, that statement included therequirement that such a complete conversion was not possible in a cyclic process or ina process in which there were no changes in the universe (system plus surroundings).In the illustration here heat has been completely converted to work, but this can onlybe done once since the system, the gas contained within the cylinder, has not been re-stored to its initial pressure. Consequently, we cannot continue to add heat and extractthe same amount of energy as work.

ILLUSTRATION 3.4-8Computing the Pressure, and Heat and Work Flows in a Complicated Ideal Gas Problem

Consider the cylinder filled with air and with the piston and spring arrangement shown below.The external atmospheric pressure is 1 bar, the initial temperature of the air is 25◦C, the no-loadlength of the spring is 50 cm, the spring constant is 40 000 N/m, the frictionless piston weighs500 kg, and the constant-volume heat capacity of air can be taken to be constant at 20.3 J/(molK). Assume air is an ideal gas. Compute

a. The initial pressure of the gas in the cylinderb. How much heat must be added to the gas in the cylinder for the spring to compress 2 cm

55 cm

15 cm

30 cm w00

26-n

.eps


SOLUTION

a. The pressure of the gas inside the cylinder is a result of the atmospheric pressure (1 bar),the force exerted on the gas as a result of the weight of the piston, and the force of thespring. The contribution from the piston is

PPiston = F

A=

M kg × 9.807m

s2× 1 Pa

kg/(m s2)× 1 bar

105 Paπ(d2/4) m2

= M

d2× 1.2487 × 10−4 bar = 500

(0.3)2× 1.2487 × 10−4 bar = 0.6937 bar

The contribution due to the spring is

PSpring = F

A= −k(x − x0)

π(d2/4)= −40 000

N

m× (55 − 50) × 10−2 m

=−40 000

N

m× (55 − 50) × 10−2 m

π[(0.3)2/4] m2× 1

m kg

s2 N× 1

Pa

kg/(m s2)× 1 bar

105 Pa

= −0.2829 bar

Therefore, the pressure of the air in the cylinder is

P = PAtmosphere + PPiston + PSpring

= 1 + 0.6937 − 0.2829 bar = 1.4108 bar

For later reference we note that the number of moles of gas in the cylinder is, from theideal gas law,

N = PV

RT=

1.4108 bar × π

4(0.3)2(0.15) m3

298.15 K × 8.314 × 10−5 bar m3

mol K= 0.6035 mol which corresponds to 17.5 g = 0.0175 kg

b. The contribution to the total pressure due to the spring after heating is computed as above,except that now the spring extension is 53 cm (55 cm − 2 cm). Thus

PSpring =−40 000

N

m× (53 − 50) × 10−2 m

π(0.3)2 m2

4

= −0.1697 bar

So the final pressure of the air in the cylinder is

P = 1 + 0.6937 − 0.1697 bar = 1.524 bar

[Note that the pressure at any extension of the spring, x , is

P = 1.6937 − 5.658(x − 0.50) bar where x is the extension of spring (m)]

The volume change on expansion of the gas (to raise the piston by 2 cm) is

�V = π

4(0.3)2 m2 × 0.02 m = 1.414 × 10−3 m3


Also, the final temperature of the gas, again from the ideal gas law, is

T = PV

N R=

1.524 bar × π

40.32 m2 × 0.17 m

0.6035 mol × 8.314 × 10−5 bar m3

mol K

= 365.0 K

The work done can be computed in two ways, as shown below. For simplicity, we firstcompute the individual contributions, and then see how these terms are combined.

Work done by the gas against the atmosphere is

WAtmosphere = −PAtmosphere × �V = −1 bar × 1.414 × 10−3 m3 × 105 J

bar m3= −141.4 J

(The minus sign indicates that the gas did work on the surrounding atmosphere.)Work done by the gas in compressing the spring is

WSpring = −k

2[(x2 − x0)

2 − (x1 − x0)2]

= −40 000

2

N

m[32 × 10−4 m2 − 52 × 10−4 m2]

= −2 × (9 − 25) = 32 N m × 1J

N m= 32 J

Work done by the gas in raising the 500-kg piston is

WPiston = −500 kg × 0.02 m × 9.807m

s2× 1 J

m2 kg/s2= −98.07 J

Work done by the gas in raising its center of mass by 1 cm (why does the center of massof the gas increase by only 1 cm if the piston is raised 2 cm?) is

Wgas = −17.5 g × 0.01 m × 9.807m

s2× 1 J

m2 kg/s2× 1 kg

1000 g= −0.0017 J

which is negligible compared with the other work terms. If we consider the gas in thecylinder to be the system, the work done by the gas is

W = Work to raise piston + Work against atmosphere + Work to compress spring= −98.1 J − 141.4 J + 32.0 J = −207.5 J

[Here we have recognized that the gas is doing work by raising the piston and by expansionagainst the atmosphere, but since the spring is extended beyond its no-load point, it isdoing work on the gas as it contracts. (The opposite would be true if the spring wereinitially compressed to a distance less than its no-load point.)]

An alternative method of computing this work is as follows. The work done by the gason expanding (considering only the gas in the system) is

W = −∫

P dV = −A∫ 0.17 m

0.15 mP dy = −A

∫ 0.17 m

0.15 m[1.6937 − 5.658(x − 0.5)] dy

where y is the height of the bottom of the piston at any time. The difficulty with theintegral above is that two different coordinate systems are involved, since y is the heightof the piston and x is the extension of the spring. Therefore, we need to make a coordinatetransformation. To do this we note that when y = 0.15, x = 0.55, and when y = 0.17,x = 0.53. Consequently, x = 0.7 − y and

3.5 Conservation of Momentum 95

W = −A∫ 0.17 m

0.15 m[1.6937 − 5.658(0.7 − y − 0.5)] dy

= −A

[1.6937 × 0.02 − 5.658

∫ 0.17 m

0.15 m(0.2 − y) dy

]

= −A[1.6937 × 0.02 − 5.658

(0.2 × 0.02 − 1

2 (0.172 − 0.152))]

= −7.068 × 10−2[0.03387 − 0.02263 + 0.01811]

= −0.2075 × 10−2 bar m3 × 105 J

bar m3= −207.5 J

This is identical to the result obtained earlier.Finally, we can compute the heat that must be added to raise the piston. The difference

form of the energy balance on the closed system consisting of the gas is

U(final state) − U(initial state) = Q + W = NCV(T f − Ti)

so that

Q = NCV(T f − Ti) − W

= 0.6035 mol × 20.3J

mol K× (365 − 298.15) K + 207.5 J

= 819.0 J + 207.5 J = 1026.5 J

Therefore, to accomplish the desired change, 1026.5 J of heat must be added to the gas.Of this amount, 819 J are used to heat the gas, 98 J to raise the 500-kg piston, and 141.4 Jto push back the atmosphere; during the process, the spring supplies 32 J.

3.5 CONSERVATION OF MOMENTUM

Based on the discussion of Sec. 3.4 and Illustration 3.4-4, we can conclude that theequations of mass and energy conservation are not sufficient to obtain the solution to allthe problems of thermodynamics in which we might be interested. What is needed is abalance equation for an additional thermodynamic state variable. The one conservationprinciple that has not yet been used is the conservation of momentum. If, in Eq. 2.1-4,θ is taken to be the momentum of a black-box system, we have

d

dt(Mv) =

Rate at which momentum

enters system by allmechanisms

−

Rate at which momentum

leaves system by allmechanisms

(3.5-1)

where v is the center-of-mass velocity vector of the system, and M its total mass. Wecould now continue the derivation by evaluating all the momentum flows; however, it isclear by looking at the left side of this equation that we will get an equation for the rateof change of the center-of-mass velocity of the system, not an equation of change fora thermodynamic state variable. Consequently, the conservation-of-momentum equa-tion will not lead to the additional balance equation we need, so this derivation willnot be completed. The development of an additional, useful balance equation is not astraightforward task, and will be delayed until Chapter 4.

3.6 THE MICROSCOPIC ENERGY BALANCE (OPTIONAL)

Following Sec. 2.4, the microscopic form of the energy conservation equation can beobtained in a manner similar to the mass conservation equation, so the development


will only be sketched here. The starting point of the derivation is, of course, Eq. 2.1-4,with θ being the total energy of the fluid, that is, the sum of the internal, potential,and kinetic energies. Energy can enter the differential volume element by energy flowsaccompanying mass flows (convection), by heat flows across the faces of the volumeelement, and, since there may be both forces and fluid movement at the faces of thevolume element, by the work being done by surface forces. Since the volume elementis only of differential size, we will not include a shaft work term. Thus we have(

Rate of change of energyin the volume element

)= (Net rate of energy input by convection)

+(

Net rate of energy inputby heat conduction

)

+(

Net rate of energy inputby surface forces

) (3.6-1)

whereRate of change of

energy in thevolume element

= �x �y �z

∂

∂t

{ρ

(U + v2

2+ ψ

)}

Net rate of energy

input byconvection

= ρvx

(U + v2

2+ ψ

)�y �z|x − ρvx

(U + v2

2+ ψ

)�y �z|x+�x

+ ρvy

(U + v2

2+ ψ

)�x �z|y − ρvy

(U + v2

2+ ψ

)�x �z|y+�y

+ ρvz

(U + v2

2+ ψ

)�x �y|z − ρvz

(U + v2

2+ ψ

)�x �y|z+�z

Net rate of energy

input by heatconduction

= qx �y �z|x − qx �y �z|x+�x

+ qy �x �z|y − qy �x �z| y+�y

+ qz �x �y|z − qz �x �y| z+�zNet rate of energy

input by surfaceforces

= (P · v)x �y �z|x − (P · v)x �y �z|x+�x

+ (P · v)y �x �z|y − (P · v)y �x �z|y+�y

+ (P · v)z �x �y|z − (P · v)z �x �y|z+�z

Here ρ is the mass density, q is the heat flux per unit area, qi is its component in the i thcoordinate direction, and (P · v)i is the component of the vector P · v in the i th coor-dinate direction. In writing the contribution to the energy balance from surface forces,we have to recognize that the forces are of two different types. The first is merely a hy-drostatic pressure term that acts perpendicular to the surface of the differential volumeelement and is uniform and isotropic; this term is written as PI, where I is the unit ten-sor. The second surface force is more subtle; it is a drag force that occurs whenever afluid element flows by either an adjacent surface or a fluid element of different velocity.The magnitude of the surface force or drag force depends on both the viscosity of the

3.6 The Microscopic Energy Balance (Optional) 97

fluid µ and the velocity gradient ∇v in the fluid at the surface of the fluid element. ForNewtonian fluids the components of the stress tensor τ for a general three-dimensionalflow are

τxx = −2µ∂vx

∂x+ 2

3µ∇ · v, τxy = τyx = µ

(∂vx

∂y+ ∂vy

∂x

)

with similar expressions for the other components. In the derivation of the energy bal-ance we have combined the hydrostatic pressure (P) and the shear stress tensor (τ )contributions to the surface forces by defining a pressure tensor

P = PI + τ (3.6-2)

Using these expressions in Eq. 3.6-1, dividing by �x �y �z, and taking the limit aseach of these goes to zero gives

∂

∂t

{ρ

(U + v2

2+ ψ

)}= −∇ ·

{ρv

(U + v2

2+ ψ

)}− ∇ · q − ∇ · (P · v)

= −∇ ·{ρv

(U + v2

2+ ψ

)}− ∇ · q − ∇ · Pv − ∇ · (τ · v)

(3.6-3)The integration of this equation over a finite volume to obtain Eq. 3.2-4 can be ac-

complished, again establishing that the thermodynamic equations are volume integralsof the microscopic equations. This tedious process will not be done here.

The momentum balance equation for the microscopic volume element of Fig. 3.6-1is nothing other than Newton’s second law of motion for a fluid element. In derivingthis equation one must remember that a change of momentum (an acceleration) resultsfrom an applied force. Consequently, in the momentum balance equation we have toinclude the forces acting at the surface of the volume element, since these forces resultin a flow of momentum across the system boundaries. Considering first only the x-component of momentum, we make the following identifications:

Rate of change of

x-component of momentumin volume element

= �x �y �z

∂

∂t(ρvx)

Net rate of change ofx-component of

momentum into the volumeelement by convection

= (ρvxvx |x − ρvxvx |x+�x ) �y �z

+ (ρvyvx |y − ρvyvx |y+�y) �x �z+ (ρvzvx |z − ρvzvx |z+�z) �x �y

Net rate of change of

x-component of momentumdue to forces acting on

the volume element

= ρ �x �y �zgx + (P|x − P|x+�x) �y �z

+ (τxx |x − τxx |x+�x) �y �z+ (τyx|y − τyx |y+�y) �x �z+ (τzx |z − τzx |z+�z) �x �y


In the last expression the first force term is that due to gravity in the x direction, the sec-ond term results from pressure forces acting on the volume element, and the remainingterms result from viscous forces.

Now using these terms in Eq. 2.1-4, dividing by �x�y�z, and taking the limit as�x,�y, and �z go to zero yields

∂

∂t(ρvx) = −∇ · (ρvvx) −

(∂τxx

∂x+ ∂τyx

∂y+ ∂τzx

∂z

)− ∂ P

∂x+ ρgx (3.6-4)

In a similar fashion, one could obtain balance equations for the y- and z-componentsof momentum. Vectorially adding the equations for each of the coordinate directionsyields

∂(ρv)

∂t= −∇ · (ρvv) − ∇ · τ − ∇ P + ρg (3.6-5)

This is the conservation equation for momentum, or the equation of motion. The onlyreal use we have for this equation is in the development of the mechanical energyequation or Bernoulli equation of fluid mechanics, which is obtained by scalar multi-plication of the momentum equation with the fluid velocity v,

v · ∂(ρv)

∂t= −v · ∇ · (ρvv) − v · (∇ · τ) − v · ∇ P + ρv · g (3.6-6)

followed by rearrangement to give

∂

∂t

(ρ

2v2)

= −∇ ·(ρ

2v2v

)− v · (∇ · τ) − v · ∇ P + ρv · g (3.6-7)

This equation looks like a conservation equation for kinetic energy. However, kineticenergy is not a conserved quantity, and this equation was obtained not from an en-ergy conservation principle, but merely from the velocity moment of the momentumconservation equation.

It is also possible to derive an equation that looks like a conservation equation forpotential energy. For simplicity, we will assume that the only potential energy contri-bution is that due to the gravitational field, so that

ψ = −h · g (3.6-8)

where h is the vector to the center of mass of the volume element with respect to aconvenient frame of reference, and g is the directed force of gravity. The minus signappears because, by convention, the gravitational force is in the negative direction.Now, multiplying the continuity equation, Eq. 2.4-4, by ψ and rearranging gives

∂(ρψ)

∂t− ρ

∂ψ

∂t= −∇ · (ρvψ) + ρv · ∇ψ (3.6-9)

For a stationary volume element we have ∂h/∂t = 0, and since the force of gravity istime invariant, ∂g/∂t = 0. Also, ∇h = I, where I is the unit tensor, and, to an excellentapproximation, ∇ · g = 0. Using these observations in Eq. 3.6-9 yields

Problems 99

∂

∂t(ρψ) = −∇ · (ρvψ) − ρv · g (3.6-10)

Adding this to Eq. 3.6-7 gives a result that looks like a conservation equation for me-chanical energy:

∂

∂t

[ρ

(v2

2+ ψ

)]= −∇ ·

[ρv

(v2

2+ ψ

)]− v · (∇ · τ) − v · ∇P (3.6-11)

Finally, we notice that this equation can be subtracted from the total energy balance,Eq. 3.6-3, to obtain a “thermal energy equation”:

∂

∂t(ρU) = −∇ · (ρvU) − ∇ · q − P∇ · v − τ : ∇v (3.6-12)

or

ρDU

Dt= −∇ · q − P∇ · v − τ : ∇v (3.6-13)

where we have again used D/Dt = ∂/∂t + v · ∇.We will return to some of these equations in Sec. 4.6.

PROBLEMS

3.1 a. A bicyclist is traveling at 20 km/hr when he en-counters a hill 70 m in height. Neglecting road andwind resistance (a poor assumption), what is themaximum vertical elevation gain the bicyclist couldachieve without pedaling?

b. If the hill is a down hill, what speed would the bicy-clist achieve without pedaling, again neglecting roadand wind resistance?

3.2 Water in the Niagara River approaches the falls at a ve-locity of 8 km/hr. Niagara Falls is approximately 55 mhigh. Neglecting air resistance, estimate the velocity ofthe water at the bottom of the falls.

3.3 a. One kilogram of steam contained in a horizontalfrictionless piston and cylinder is heated at a con-stant pressure of 1.013 bar from 125◦C to such atemperature that its volume doubles. Calculate theamount of heat that must be added to accomplishthis change, the final temperature of the steam, thework the steam does against its surroundings, andthe internal energy and enthalpy changes of thesteam for this process.

b. Repeat the calculation of part (a) if the heating oc-curs at a constant volume to a pressure that is twicethe initial pressure.

c. Repeat the calculation of part (a) assuming thatsteam is an ideal gas with a constant-pressure heatcapacity of 34.4 J/mol K.

d. Repeat the calculation of part (b) assuming steam isan ideal gas as in part (c).

3.4 In Joule’s experiments, the slow lowering of a weight(through a pulley and cable arrangement) turned a stir-rer in an insulated container of water. As a result ofviscosity, the kinetic energy transferred from the stirrerto the water eventually dissipated. In this process thepotential energy of the weight was first converted to ki-netic energy of the stirrer and the water, and then as aresult of viscous forces, the kinetic energy of the wa-ter was converted to thermal energy apparent as a risein temperature. Assuming no friction in the pulleys andno heat losses, how large a temperature rise would befound in 1 kg of water as a result of a 1-kg weight beinglowered 1 m?

3.5 Steam at 500 bar and 600◦C is to undergo a Joule-Thomson expansion to atmospheric pressure. What willbe the temperature of the steam after the expansion?What would be the downstream temperature if thesteam were replaced by an ideal gas?

3.6 Water in an open metal drum is to be heated from roomtemperature (25◦C) to 80◦C by adding steam slowlyenough that all the steam condenses. The drum initiallycontains 100 kg of water, and steam is supplied at 3.0bar and 300◦C. How many kilograms of steam shouldbe added so that the final temperature of the water inthe tank is exactly 80◦C? Neglect all heat losses fromthe water in this calculation.

3.7 Consider the following statement: “The adiabatic worknecessary to cause a given change of state in a closedsystem is independent of the path by which that change


occurs.”a. Is this statement true or false? Why? Does this state-

ment contradict Illustration 3.4-6, which establishesthe path dependence of work?

b. Show that if the statement is false, it would be pos-sible to construct a machine that would generate en-ergy.

3.8 A nonconducting tank of negligible heat capacity and1 m3 volume is connected to a pipeline containingsteam at 5 bar and 370◦C, filled with steam to a pres-sure of 5 bar, and disconnected from the pipeline.a. If the tank is initially evacuated, how much steam is

in the tank at the end of the filling process, and whatis its temperature?

b. If the tank initially contains steam at 1 bar and150◦C, how much steam is in the tank at the endof the filling process, and what is its temperature?

3.9 a. A 1-kg iron block is to be accelerated through a pro-cess that supplies it with 1 kJ of energy. Assumingall this energy appears as kinetic energy, what is thefinal velocity of the block?

b. If the heat capacity of iron is 25.10 J/(mol K) andthe molecular weight of iron is 55.85, how large atemperature rise would result from 1 kJ of energysupplied as heat?

3.10 The voltage drop across an electrical resistor is10 volts and the current through it is 1 ampere. Thetotal heat capacity of the resistor is 20 J/K, and heatis dissipated from the resistor to the surrounding airaccording to the relation

Q = −h(T − Tam)

where Tam is the ambient air temperature, 25◦C; T isthe temperature of the resistor; and h, the heat trans-fer coefficient, is equal to 0.2 J/(K s). Compute thesteady-state temperature of the resistor, that is, thetemperature of the resistor when the energy loss fromthe resistor is just equal to the electrical energy input.

3.11 The frictionless piston-and-cylinder system shownhere is subjected to 1.013 bar external pressure. Thepiston mass is 200 kg, it has an area of 0.15 m2,and the initial volume of the entrapped ideal gas is0.12 m3. The piston and cylinder do not conductheat, but heat can be added to the gas by a heatingcoil. The gas has a constant-volume heat capacity of30.1 J/(mol K) and an initial temperature of 298 K,and 10.5 kJ of energy are to be supplied to the gasthrough the heating coil.a. If stops placed at the initial equilibrium position of

the piston prevent it from rising, what will be thefinal temperature and pressure of the gas?

b. If the piston is allowed to move freely, what willbe the final temperature and volume of the gas?

Heatingcoil

3.12 As an energy conservation measure in a chemicalplant, a 40-m3 tank will be used for temporary stor-age of exhaust process steam. This steam is then usedin a later stage of the processing. The storage tank iswell insulated and initially contains 0.02 m3 of liq-uid water at 50◦C; the remainder of the tank containswater vapor in equilibrium with this liquid. Processsteam at 1.013 bar and 90 percent quality enters thestorage tank until the pressure in the tank is 1.013 bar.How many kilograms of wet steam enter the tank dur-ing the filling process, and how much liquid water ispresent at the end of the process? Assume that thereis no heat transfer between the steam or water and thetank walls.

3.13 The mixing tank shown here initially contains 50 kgof water at 25◦C. Suddenly the two inlet valves andthe single outlet valve are opened, so that two waterstreams, each with a flow rate of 5 kg/min, flow intothe tank, and a single exit stream with a flow rate of10 kg/min leaves the tank. The temperature of one in-let stream is 80◦C, and that of the other is 50◦C. Thetank is well mixed, so that the temperature of the out-let stream is always the same as the temperature of thewater in the tank.a. Compute the steady-state temperature that will fi-

nally be obtained in the tank.b. Develop an expression for the temperature of the

fluid in the tank at any time.w

0028

-n.e

ps

3.14 A well-insulated storage tank of 60 m3 contains 200 Lof liquid water at 75◦C. The rest of the tank contains

Problems 101

steam in equilibrium with the water. Spent processsteam at 2 bar and 90 percent quality enters the stor-age tank until the pressure in the tank reaches 2 bar.Assuming that the heat losses from the system to thetank and the environment are negligible, calculate thetotal amount of steam that enters the tank during thefilling process and the fraction of liquid water presentat the end of the process.

3.15 An isolated chamber with rigid walls is divided intotwo equal compartments, one containing gas and theother evacuated. The partition between the compart-ments ruptures. After the passage of a sufficiently longperiod of time, the temperature and pressure are foundto be uniform throughout the chamber.a. If the filled compartment initially contains an ideal

gas of constant heat capacity at 1 MPa and 500 K,what are the final temperature and pressure in thechamber?

b. If the filled compartment initially contains steamat 1 MPa and 500 K, what are the final temperatureand pressure in the compartment?

c. Repeat part (a) if the second compartment initiallycontains the same fluid as the first compartment,but at half the pressure and 100 K higher tempera-ture.

d. Repeat part (b) if the second compartment initiallycontains the same fluid as the first compartment,but at half the pressure and 100 K higher tempera-ture.

3.16 a. An adiabatic turbine expands steam from 500◦Cand 3.5 MPa to 200◦C and 0.3 MPa. If the turbinegenerates 750 kW, what is the flow rate of steamthrough the turbine?

b. If a breakdown of the thermal insulation around theturbine allows a heat loss of 60 kJ per kg of steam,and the exiting steam is at 150◦C and 0.3 MPa,what will be the power developed by the turbineif the inlet steam conditions and flow rate are un-changed?

3.17 Intermolecular forces play an important role in de-termining the thermodynamic properties of fluids. Tosee this, consider the vaporization (boiling) of a liquidsuch as water in the frictionless piston-and-cylinderdevice shown.a. Compute the work obtained from the piston when

1 kg of water is vaporized to steam at 100◦C (thevapor and liquid volumes of steam at the boilingpoint can be found in the steam tables).

b. Show that the heat required for the vaporizationof the steam is considerably greater than the workdone. (Note that the enthalpy change for the vapor-ization is given as 2257 kJ/kg in the steam tables inAppendix A.III.)

Steam

Water

w00

29-n

.eps

3.18 It is sometimes necessary to produce saturated steamfrom superheated steam (steam at a temperaturehigher than the vapor-liquid coexistence temperatureat the given pressure). This change can be accom-plished in a desuperheater, a device in which justthe right amount of water is sprayed into superheatedsteam to produce dry saturated steam. If superheatedsteam at 3.0 MPa and 500◦C enters the desuperheaterat a rate of 500 kg/hr, at what rate should liquid waterat 2.5 MPa and 25◦C be added to the desuperheater toproduce saturated steam at 2.25 MPa?

3.19 Nitrogen gas leaves a compressor at 2.0 MPa and120◦C and is collected in three different cylinders,each of volume 0.3 m3. In each case the cylinder isto be filled to a pressure of 2.0 MPa. Cylinder 1 isinitially evacuated, cylinder 2 contains nitrogen gas at0.1 MPa and 20◦C, and cylinder 3 contains nitrogen at1 MPa and 20◦C. Find the final temperature of nitro-gen in each of the cylinders, assuming nitrogen to bean ideal gas with C∗

P = 29.3 J/mol K. In each case as-sume the gas does not exchange heat with the cylinderwalls.

3.20 A clever chemical engineer has devised the thermallyoperated elevator shown in the accompanying dia-gram. The elevator compartment is made to rise byelectrically heating the air contained in the piston-and-cylinder drive mechanism, and the elevator islowered by opening a valve at the side of the cylin-der, allowing the air in the cylinder to slowly escape.Once the elevator compartment is back to the lowerlevel, a small compressor forces out the air remain-ing in the cylinder and replaces it with air at 20◦Cand a pressure just sufficient to support the elevatorcompartment. The cycle can then be repeated. Thereis no heat transfer between the piston, cylinder, andthe gas; the weight of the piston, elevator, and eleva-tor contents is 4000 kg; the piston has a surface areaof 2.5 m2; and the volume contained in the cylinderwhen the elevator is at its lowest level is 25 m3. Thereis no friction between the piston and the cylinder, and


Heatingcoil

Elevatorcompartment

2nd floor

1st floor

3 m

Centrifugalpump

w00

30-n

.eps

the air in the cylinder is assumed to be an ideal gaswith C∗

P = 30 J/(mol K).a. What is the pressure in the cylinder throughout the

process?b. How much heat must be added to the air during the

process of raising the elevator 3 m, and what is thefinal temperature of the gas?

c. What fraction of the heat added is used in doingwork, and what fraction is used in raising the tem-perature of the gas?

d. How many moles of air must be allowed to es-cape in order for the elevator to return to the lowestlevel?

3.21 The elevator in the previous problem is to be designedto ascend and descend at the rate of 0.2 m/s, and torise a total of 3 m.a. At what rate should heat be added to the cylinder

during the ascent?b. How many kilomoles per second of air should be

removed from the cylinder during the descent?3.22 Nitrogen gas is being withdrawn at the rate of 4.5 g/s

from a 0.15-m3 cylinder, initially containing the gas ata pressure of 10 bar and 320 K. The cylinder does notconduct heat, nor does its temperature change duringthe emptying process. What will be the temperatureand pressure of the gas in the cylinder after 5 minutes?What will be the rate of change of the gas temperatureat this time? Nitrogen can be considered to be an idealgas with C∗

P = 30 J/(mol K).3.23 In Illustration 3.4-6 we considered the compression

of an ideal gas in which PV γ = constant, whereγ = C∗

P/C∗V. Show that such a pressure-volume re-

lationship is obtained in the adiabatic compression ofan ideal gas of constant heat capacity.

3.24 Air in a 0.3-m3 cylinder is initially at a pressure of10 bar and a temperature of 330 K. The cylinder is tobe emptied by opening a valve and letting the pres-sure drop to that of the atmosphere. What will be thetemperature and mass of gas in the cylinder if this isaccomplished?a. In a manner that maintains the temperature of the

gas at 330 K?b. In a well-insulated cylinder?For simplicity assume, in part (b), that the process oc-curs sufficiently rapidly that there is no heat transferbetween the cylinder walls and the gas. The gas isideal, and C∗

P = 29 (J/mol K).3.25 A 0.01-m3 cylinder containing nitrogen gas initially

at a pressure of 200 bar and 250 K is connected toanother cylinder 0.005 m3 in volume, which is ini-tially evacuated. A valve between the two cylindersis opened until the pressures in the cylinders equalize.Find the final temperature and pressure in each cylin-der if there is no heat flow into or out of the cylinder.You may assume that there is no heat transfer betweenthe gas and the cylinder walls and that the gas is idealwith a constant-pressure heat capacity of 30 J/(mol K).

3.26 Repeat the calculation of Problem 3.25, but now as-sume that sufficient heat transfer occurs between thegas in the two cylinders that both final temperaturesand both final pressures are the same.

3.27 Repeat the calculation in Problem 3.25, but now as-sume that the second cylinder, instead of being evac-uated, is filled with nitrogen gas at 20 bar and 160 K.

3.28 A 1.5 kW heater is to be used to heat a room withdimensions 3.5 m × 5.0 m × 3.0 m. There are noheat losses from the room, but the room is not airtight,so the pressure in the room is always atmospheric.Consider the air in the room to be an ideal gas withC∗

P = 29 J/(mol K), and its initial temperature is 10◦C.a. Assuming that the rate of heat transfer from the air

to the walls is low, what will be the rate of increaseof the temperature in the room when the heater isturned on?

b. What would be the rate of increase in the roomtemperature if the room were hermetically sealed?

3.29 The piston-and-cylinder device of Illustration 3.4-7 isto be operated in reverse to isothermally compress the1 mol of air. Assume that the weights in the illustra-tion have been left at the heights they were at whenthey were removed from the piston (i.e., in processb the first 50-kg weight is at the initial piston heightand the second is at �h = �V/A = 0.384 m abovethe initial piston height). Compute the minimum workthat must be done by the surroundings and the net heatthat must be withdrawn to return the gas, piston, andweights to their initial states. Also compute the totalheat and the total work for each of the four expansion

Problems 103

and compression cycles and comment on the results.3.30 The piston-and-cylinder device shown here contains

an ideal gas at 20 bar and 25◦C. The piston has a massof 300 kg and a cross-sectional area of 0.05 m2. Theinitial volume of the gas in the cylinder is 0.03 m3,the piston is initially held in place by a pin, and theexternal pressure on the piston and cylinder is 1 bar.The pin suddenly breaks, and the piston moves 0.6 mfarther up the cylinder, where it is stopped by anotherpin. Assuming that the gas is ideal with a constant-pressure heat capacity of 30 J/(mol K), and that thereis no heat transfer between the gas and the cylinderwalls or piston, estimate the piston velocity, and thetemperature and pressure of the gas just before the pis-ton hits the second pin. Do this calculation assuminga. No friction between the piston and the cylinderb. Friction between the piston and the cylinderList and defend all assumptions you make in solvingthis problem.

0.6 m

Pin 2

Pin 1

w00

30a-

n.ep

s

3.31 A 0.6-m-diameter gas pipeline is being used for thelong-distance transport of natural gas. Just past apumping station, the gas is found to be at a temper-ature of 25◦C and a pressure of 3.0 MPa. The massflow rate is 125 kg/s, and the gas flow is adiabatic.Forty miles down the pipeline is another pumping sta-tion. At this point the pressure is found to be 2.0 MPa.At the pumping station the gas is first adiabaticallycompressed to a pressure of 3.0 MPa and then isobar-ically (i.e., at constant pressure) cooled to 25◦C.a. Find the temperature and velocity of the gas just

before it enters the pumping station.b. Find the rate at which the gas compressor in the

pumping station does work on the gas, the temper-ature of the gas leaving the compressor, and theheat load on the gas cooler. You may assume thatthe compressor exhaust is also a 0.6-m pipe. (Ex-plain why you cannot solve this problem. You willhave another chance in Chapter 4.)

Natural gas can be assumed to be pure methane[molecular weight = 16, C∗

P = 36.8 J/(mol K)], andan ideal gas at the conditions being considered here.Note that the mass flow rate M is ρv A, where ρ is the

mass density of the gas, v is the average gas velocity,and A is the area of the pipe.

3.32 Nitrogen can be liquefied using a Joule-Thomson ex-pansion process. This is done by rapidly and adiabati-cally expanding cold nitrogen gas from high pressureto a low pressure. If nitrogen at 135 K and 20 MPaundergoes a Joule-Thomson expansion to 0.4 MPa,a. Estimate the fraction of vapor and liquid present

after the expansion, and the temperature of thismixture using the pressure-enthalpy diagram fornitrogen.

b. Repeat the calculation assuming nitrogen to be anideal gas with C∗

P = 29.3 J/(mol K).3.33 A very large mass M of hot porous rock equal to

1012 kg is to be utilized to generate electricity byinjecting water and using the resulting hot steam todrive a turbine. As a result of heat extraction, thetemperature of the rock drops according to Q =−MCP dT/dt , where CP is the specific heat of therock which is assumed to be independent of tempera-ture. If the plant produces 1.36 × 109 kW hr of en-ergy per year, and only 25 percent of the heat ex-tracted from the rock can be converted to work, howlong will it take for the temperature of the rock todrop from 600◦C to 110◦C? Assume that for the rockCP = 1 J/(g K).

3.34 The human body generates heat by the metabolism ofcarbohydrates and other food materials. Metabolismprovides energy for all biological activities (e.g., mus-cle contractions). The metabolic processes also gener-ate heat, and there are special cells in the body whosemain function is heat generation. Now let us assumethat our friend Joe BlueHen ingests 1 L of ice, whichhe allows to melt in his mouth before swallowing.a. How much energy is required to melt the ice and

warm the water to the body temperature?b. If 1 g of fat when metabolized releases approxi-

mately 42 kJ of energy, how much fat will Joe burnby ingesting the water?

c. Suppose that, instead of ice, Joe drank 1 L of waterat 0◦C. How would the answers to parts (a) and (b)change?Several years ago there was a story circulating the

Internet that a good way to lose weight is to drinka lot of very cold water, since considerable energywould be expended within the body in heating up thecold water. Based on your calculations above; is thata reasonable method of weight loss? (The reason thisclaim was widely circulated is a result of the sloppyuse of units. Some countries report the biologicallyaccessible energy in food as being in units of calo-ries, when in fact the number reported is kilocalories.For example, in the United States a teaspoon of sugaris reported ito contain 16 calories, when it is actually


16 kilocalories. Calories is incorrectly being used asan abbreviation for kilocalories.)

3.35 Water is to be heated from its pipeline temperature of20◦C to 90◦C using superheated steam at 450◦C and2.5 MPa in a steady-state process to produce 10 kg/sof heated water. In each of the processes below, as-sume there is no heat loss.a. The heating is to be done in a mixing tank by di-

rect injection of the steam, all of which condenses.Determine the two inlet mass flows needed to meetthe desired hot water flow rate.

b. Instead of direct mixing, a heat exchanger will beused in which the water to be heated will flow in-side copper tubes and the steam will partially con-dense on the outside of the tubes. In this case heatwill flow from the steam to the water, but the twostreams are not mixed. Calculate the steam flowrate if the steam leaves the heat exchanger at 50percent quality at 100◦C.

3.36 People partially cool themselves by sweating, whichreleases water that evaporates. If during exercise a hu-man “burns” 1000 kcal (4184 kJ) in one hour of ex-ercise, how many grams of water must evaporate at abody temperature of 37◦C? Assuming only 75 percentof the sweat evaporates (the rest being retained by theexercise clothes), how many grams of sweat must ac-tually be produced?

3.37 It is thought that people develop respiratory infectionsduring air travel because much of the airplane cabinair is recirculated. Airlines claim that using only freshair in the cabins is too costly since at an altitude of30 000 feet the outside conditions are −50◦C and 0.1bar, so that the air would have to be compressed andheated before being introduced into the cabin. The air-plane cabin has a volume of 100 m3 with air at the in-flight conditions of 25◦C and 0.8 bar. What would bethe cost of completely refreshing the air every minuteif air has a heat capacity of C∗

P = 30 J/(mol K) andenergy costs $0.2 per kW hr?

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