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

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7.4.2 Closed System Exergy Rate Balance

As in the case of the mass, energy, and entropy balances, the exergy balance can be

expressed in various forms that may be more suitable for particular analyses. A con-

venient form is the closed system exergy rate balance given by

a1 2

2 aW

2 p

b2 E

(7.10)

where dE/dt is the time rate of change of exergy. The term 11 2 T

represents the

time rate of exergy transfer accompanying heat transfer at the rate Q

occurring

where the instantaneous temperature on the boundary is T

. The term

represents

the time rate of energy transfer by work. The accompanying rate of exergy transfer

is given by (W

2 p

dV/dt) where dV/dt is the time rate of change of system volume.

The term E

accounts for the time rate of exergy destruction due to irreversibilities

within the system.

At steady state, dE/dt 5 dV/dt 5 0 and Eq. 7.10 reduces to give the steady-state

form of the exergy rate balance

0 5

a1 2

2 W

2 E

(7.11a)

Note that for a system at steady state, the rate of exergy transfer accompanying the

power

is simply the power.

The rate of exergy transfer accompanying heat transfer at the rate

occurring where the temperature is T

is compactly expressed as

5 a1 2

(7.12)

As shown in the adjacent figure, heat transfer and the accompanying

exergy transfer are in the same direction when T

. T

If heating from saturated liquid to saturated vapor would occur

at 1008C (373.15 K), evaluate the exergy transfers accompanying heat

transfer and work, each in kJ/kg. Ans. 484, 0.

Alternatively, the exergy destruction can be evaluated using Eq. 7.7 together with the entropy production obtained

from an entropy balance. This is left as an exercise.

❶ Recognizing the term (1 2 T

/T) as the Carnot efficiency (Eq. 5.9), the

right side of Eq. (b) can be interpreted as the work that could be developed

by a reversible power cycle were it to receive energy Q/m at temperature

T and discharge energy to the environment by heat transfer at T

❷ The right side of Eq. (c) shows that if the system were interacting with the

environment, all of the work W/m, represented by area 1-2-d-a-1 on the

p–y diagram of Fig. E7.2, would not be fully available for lifting a weight.

A portion would be spent in pushing aside the environment at pressure p

This portion is given by p

2 y

), and is represented by area a-b-c-d-a

on the p–y diagram of Fig. E7.2.

Ability to…

❑

evaluate exergy change.

❑

evaluate exergy transfer

accompanying heat transfer

and work.

❑

evaluate exergy destruction.

✓

Skills Developed

steady-state form of the

closed system exergy rate

balance

1 –

0__

( )

(> T

)

7.4 Closed System Exergy Balance 373

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374 Chapter 7 Exergy Analysis

Using Eq. 7.12, Eq. 7.11a reads

0 5

2 W

2 E

(7.11b)

In Eqs. 7.11, the rate of exergy destruction within the system, E

, is related to the rate

of entropy production within the system by E

5 T

7.4.3 Exergy Destruction and Loss

Most thermal systems are supplied with exergy inputs derived directly or indirectly

from the consumption of fossil fuels. Accordingly, avoidable destructions and losses

of exergy represent the waste of these resources. By devising ways to reduce such

inefficiencies, better use can be made of fuels. The exergy balance can be applied to

determine the locations, types, and true magnitudes of energy resource waste, and

thus can play an important part in developing strategies for more effective fuel use.

In Example 7.3, the steady-state form of the closed system energy and exergy rate

balances are applied to an oven wall to evaluate exergy destruction and exergy loss,

which are interpreted in terms of fossil fuel use.

Evaluating Exergy Destruction in an Oven Wall

c c c c EXAMPLE 7.3 c

The wall of an industrial drying oven is constructed by sandwiching 0.066m-thick insulation, having a thermal

conductivity

5 0.05 3 10

kW/m ? K, between thin metal sheets. At steady state, the inner metal sheet is at

5 575 K and the outer sheet is at T

5 310 K. Temperature varies linearly through the wall. The temperature

of the surroundings away from the oven is 293 K. Determine, in kW per m

of wall surface area, (a) the rate of

heat transfer through the wall, (b) the rates of exergy transfer accompanying heat transfer at the inner and outer

wall surfaces, and (c) the rate of exergy destruction within the wall. Let T

5 293 K.

SOLUTION

Known:

Temperature, thermal conductivity, and wall-thickness data are provided for a plane wall at steady state.

Find: For the wall, determine (a) the rate of heat transfer through the wall, (b) the rates of exergy transfer

accompanying heat transfer at the inner and outer surfaces, and (c) the rate of exergy destruction, each per m

of wall surface area.

Schematic and Given Data:

Engineering Model:

The closed system shown in

the accompanying sketch is at

steady state.

2. Temperature varies linearly

through the wall.

3. T

5 293 K.

Fig. E7.3

out

= 575 K

0.066 m

= 310 K

Burner

Area = A

Surroundings

at 293 K

Hot

oven

gas

fuel

Air

Insulation

κ = 0.05 × 10

–3

kW/m·K

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7.4 Closed System Exergy Balance 375

Analysis:

(a)

At steady state, an energy rate balance for the system reduces to give Q

5 Q

out

—namely, the rates of heat

transfer into and out of the wall are equal. Let Q

denote the common heat transfer rate. Using Eq. 2.3.1 with

assumption 2, the heat transfer rate is given by

A252k c

2 T

52 a0.05 3 10

m ? K

1310 2 5752

0.066 m

d5 0.2

(b) The rates of exergy transfer accompanying heat transfer are evaluated using Eq. 7.12. At the inner surface

A25 c1 2

d1Q

5 c1 2

293

575

da0.2

b5 0.1

At the outer surface

A25 c1 2

d1Q

5 c1 2

293

310

da0.2

b5 0.01

5 0,

Eq. 7.11b gives

A25 1E

A22 1E

5 10.1 2 0.012

5 0.09

➊ The rates of heat transfer are the same at the inner and outer walls, but the rates of exergy transfer at these

locations are much different. The rate of exergy transfer at the high-temperature inner wall is 10 times the

rate of exergy transfer at the low-temperature outer wall. At each of these locations the exergy transfers pro-

vide a truer measure of thermodynamic value than the heat transfer rate. This is clearly seen at the outer wall,

where the small exergy transfer indicates minimal potential for use, and thus minimal thermodynamic value.

❷ The exergy transferred into the wall at T

5 575 K is either destroyed within

the wall owing to spontaneous heat transfer or transferred out of the wall at

5 310 K where it is lost to the surroundings. Exergy transferred to the

surroundings accompanying stray heat transfer, as in the present case, is ulti-

mately destroyed in the surroundings. Thicker insulation and/or insulation

having a lower thermal conductivity value would reduce the heat transfer

rate and thus lower the exergy destruction and loss.

❸ In this example, the exergy destroyed and lost has its origin in the fuel sup-

plied. Thus, cost-effective measures to reduce exergy destruction and loss

have benefits in terms of better fuel use.

If the thermal conductivity were reduced to 0.04 3 10

kW/m

owing to a different selection of insulation material, while the insulation

thickness were increased to 0.076 m, determine the rate of exergy destruc-

tion in the wall, in kW per m

of wall surface area, keeping the same inner

and outer wall temperatures and the same temperature of the surround-

ings. Ans. 0.06 kW/m

Ability to…

❑

apply the energy and exergy

rate balances.

❑

evaluate exergy transfer

accompanying heat transfer.

❑

evaluate exergy destruction.

✓

Skills Developed

➊

❷

❸

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376 Chapter 7 Exergy Analysis

7.4.4 Exergy Accounting

In the next example, we reconsider the gearbox of Examples 2.4 and 6.4 from an

exergy perspective to introduce exergy accounting, in which the various terms of an

exergy balance for a system are systematically evaluated and compared.

According to industry sources, more than 7% of

the electric power conducted through present-day

transmission and distribution lines is forfeited en route

owing to electrical resistance. They also say superconducting

cable can nearly eliminate resistance to electricity flow, and

thereby the accompanying reduction in power.

For superconducting cable to be effective, however, it must

be cooled to about 22008C (23308F). Cooling is achieved by a

refrigeration system using liquid nitrogen. Since the refrigerator

requires power to operate, such cooling reduces the power

saved in superconducting electrical transmission. Moreover, the

cost of superconducting cable is much greater today than con-

ventional cable. Factors such as these impose barriers to rapid

deployment of superconducting technology.

Still, electrical utilities have partnered with the government

to develop and demonstrate superconducting technology that

someday may increase U.S. power system efficiency and

reliability.

Superconducting Power Cable Overcoming All Barriers?

exergy accounting

Exergy Accounting of a Gearbox

c c c c EXAMPLE 7.4 c

For the gearbox of Examples 2.4 and 6.4(a), develop a full exergy accounting of the power input. Let T

5 293 K.

SOLUTION

Known:

A gearbox operates at steady state with known values for the power input, power output, and heat

transfer rate. The temperature on the outer surface of the gearbox is also known.

Find: Develop a full exergy accounting of the input power.

Schematic and Given Data:

Analysis:

Since the gearbox is at steady state, the rate of exergy transfer accompanying power is simply the power.

Accordingly, exergy is transferred into the gearbox via the high-speed shaft at a rate equal to the power input,

60 kW, and exergy is transferred out via the low-speed shaft at a rate equal to the power output, 58.8 kW. Additionally,

exergy is transferred out accompanying stray heat transfer and destroyed by irreversibilities within the gearbox.

The rate of exergy transfer accompanying heat transfer is evaluated using Eq. 7.12. That is

5 a1 2

Engineering Model:

The gearbox is taken as a closed system operating at

steady state.

2. The temperature at the outer surface does not vary.

3. T

5 293 K.

= 300 K

Gearbox

Input

shaft

Output

shaft

= –60 kW

= 58.8 kW

Q = –1.2 kW

Fig. E7.4

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With Q

521.2 kW and T

5 300 K from Fig. E7.4, we get

1 2

300

121.2 kW2

0.03

where the minus sign denotes exergy transfer from the system.

The rate of exergy destruction is evaluated using the exergy rate balance. On rearrangement, and noting that

5 W

1 W

521.2 kW, Eq. 7.11b gives

❶ E

5 E

2 W

520.03 kW 2 121.2 kW25 1.17 kW

The analysis is summarized by the following exergy balance sheet in terms of exergy magnitudes on a rate basis:

Rate of exergy in:

high-speed shaft 60.00 kW (100%)

Disposition of the exergy:

• Rate of exergy out

low-speed shaft 58.80 kW (98%)

➋ stray heat transfer 0.03 kW (0.05%)

• Rate of exergy destruction 1.17 kW (1.95%)

60.00 kW (100%)

❶ Alternatively, the rate of exergy destruction is calculated from E

5 T

, where

is the rate of entropy

production. From the solution to Example 6.4(a), s

5 4 3 10

K. Then

5 T

5 1293 K214 3 10

5 1.17 kW

➋ The difference between the input and output power is accounted for primarily

by the rate of exergy destruction and only secondarily by the exergy transfer

accompanying heat transfer, which is small by comparison. The exergy balance

sheet provides a sharper picture of performance than the energy balance sheet

of Example 2.4, which does not explicitly consider the effect of irreversibilities

within the system.

By inspection of the exergy balance sheet, specify an exergy-

based efficiency for the gear box. Ans. 98%.

Ability to…

❑

apply the exergy rate balance.

❑

develop an exergy accounting.

✓

Skills Developed

7.5 Exergy Rate Balance for Control

Volumes at Steady State

In this section, the exergy balance is extended to a form applicable to control volumes

at steady state. The control volume form is generally the most useful for engineering

analysis.

The exergy rate balance for a control volume can be derived using an approach

like that employed in the box of Sec. 4.1, where the control volume form of the mass

rate balance is obtained by transforming the closed system form. However, as in the

developments of the energy and entropy rate balances for control volumes (Secs. 4.4.1

and 6.9, respectively), the present derivation is conducted less formally by modifying

7.5 Exergy Rate Balance for Control Volumes at Steady State 377

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378 Chapter 7

Exergy Analysis

the closed system rate form, Eq. 7.10, to account for the exergy transfers at the inlets

and exits. The result is

a1 2

2 aW

2 p

2 E

where the underlined terms account for exergy transfer where mass enters and exits

the control volume, respectively.

At steady state, dE

/dt 5 dV

/dt 5 0, giving the steady-state exergy rate balance

0 5

a1 2

2 W

2 E

(7.13a)

where e

accounts for the exergy per unit of mass entering at inlet i and e

accounts

for the exergy per unit of mass exiting at exit e. These terms, known as the specific

flow exergy

, are expressed as

5 h 2 h

2 T

1s 2 s

1 gz

(7.14)

where h and s represent the specific enthalpy and entropy, respectively, at the inlet

or exit under consideration; h

and s

represent the respective values of these proper-

ties when evaluated at T

, p

. See the box for a derivation of Eq. 7.14 and discussion

of the flow exergy concept.

steady-state exergy rate

balance: control volumes

specific flow exergy

Conceptualizing Specific Flow Exergy

To evaluate the exergy associated with a flowing stream of matter at a state given by

h, s, V, and z, let us think of the stream being fed to the control volume operating at

steady state shown in Fig. 7.5. At the exit of the control volume, the respective proper-

ties are those corresponding to the dead state: h

, s

, V

5 0, z

5 0. Heat transfer

occurs only with the environment at T

5 T

For the control volume of Fig. 7.5, energy and entropy balances read, respectively,

0 5 Q

2 W

1 m

c1h 2 h

2 102

1 g1z 2 02d

(a)

0 5

1 m

1s 2 s

21 s

(b)

Eliminating Q

between Eqs. (a) and (b), the work developed per unit of mass flowing is

5 c1h 2 h

22 T

1s 2 s

1 gz d2 T

(c)

= 0,

= 0

= T

Fig. 7.5 Control volume used

to evaluate the specific flow

exergy of a stream.

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The steady-state exergy rate balance, Eq. 7.13a, can be expressed more compactly

as Eq. 7.13b

0 5

q j

2 W

2 E

(7.13b)

where

q j

1 2

(7.15)

5 m

(7.16a)

5 m

(7.16b)

are exergy transfer rates. Equation 7.15 has the same interpretation as given for

Eq 7.5 in the box on p. 370, only on a time rate basis. Also note that at steady state

the rate of exergy transfer accompanying the power W

is simply the power. Finally,

the rate of exergy destruction within the control volume, E

is related to the rate of

entropy production by T

If there is a single inlet and a single exit, denoted by 1 and 2, respectively, the

steady-state exergy rate balance, Eq. 7.13a, reduces to

0 5

1 2

2 W

1 m

2 e

22 E

(7.17)

where m

is the mass flow rate. The term (e

2 e

) is evaluated using Eq. 7.14 as

2 e

5 1h

2 h

22 T

2 s

2 V

1 g1z

2 z

(7.18)

TAKE NOTE...

When the rate of exergy

destruction

is the objec-

tive, it can be determined

either from an exergy rate

balance or from

5 T

where

is the rate of

entropy production evalu-

ated from an entropy rate

balance. The second of these

procedures normally requires

fewer property evaluations

and less computation.

7.5 Exergy Rate Balance for Control Volumes at Steady State 379

The value of the underlined term in Eq. (c) is determined by two states: the given state

and the dead state. However, the value of the entropy production term, which cannot be

negative, depends on the nature of the flow. Hence, the maximum theoretical work that

could be developed, per unit of mass flowing, corresponds to a zero value for the entropy

production—that is, when the flow through the control volume of Fig. 7.5 is internally

reversible. The specific flow exergy, e

, is this work value, and thus Eq. 7.14 is seen to be

the appropriate expression for the specific flow exergy.

Subtracting Eq. 7.2 from Eq. 7.14 gives the following relationship between specific flow

exergy e

and specific exergy e,

5 e 1 y(p 2 p

) (d)

The underlined term of Eq. (d) has the significance of exergy transfer accompanying flow

work. Thus, at a control volume inlet or exit, flow exergy e

accounts for the sum of the

exergy accompanying mass flow e and the exergy accompanying flow work. When pressure

p at a control volume inlet or exit is less than the dead state pressure p

, the flow work

contribution of Eq. (d) is negative, signaling that the exergy transfer accompanying flow

work is opposite to the direction of exergy transfer accompanying mass flow. Flow exergy

aspects are also explored in end-of-chapter Problems 7.7 and 7.8.

TAKE NOTE...

Observe that the approach

used here to evaluate flow

exergy parallels that used in

Sec. 7.3 to evaluate exergy

of a system. In each case,

energy and entropy bal-

ances are applied to evalu-

ate maximum theoretical

work in the limit as entropy

production tends to zero.

This approach is also used

in Sec. 13.6 to evaluate

chemical exergy.

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380 Chapter 7 Exergy Analysis

7.5.1

Comparing Energy and Exergy for Control Volumes

at Steady State

Although energy and exergy share common units and exergy transfer accompanies

energy transfer, energy and exergy are fundamentally different concepts. Energy and

exergy relate, respectively, to the first and second laws of thermodynamics:

c Energy is conserved. Exergy is destroyed by irreversibilities.

c Exergy expresses energy transfer by work, heat, and mass flow in terms of a com-

mon measure—namely, work that is fully available for lifting a weight or, equiva-

lently, as shaft or electrical work.

Figure 7.6a shows energy transfer rates for a one-inlet, one-exit

control volume at steady state. This includes energy transfers by work and heat and the

energy transfers in and out where mass flows across the boundary. Figure 7.6b shows

the same control volume, but now labeled with exergy transfer rates. Be sure to note

that the magnitudes of the exergy transfers accompanying heat transfer and mass flow

differ from the corresponding energy transfer magnitudes. These exergy transfer rates

are calculated using Eqs. 7.15 and 7.16, respectively. At steady state, the rate of exergy

transfer accompanying the power W

is simply the power. In accord with the conserva-

tion of energy principle, the total rate energy enters the control volume equals the total

rate energy exits. However, the total rate exergy enters the control volume exceeds the

total rate exergy exits. The difference between these exergy values is the rate at which

exergy is destroyed by irreversibilities, in accord with the second law.

b b b b b

To summarize, exergy gives a sharper picture of performance than energy because

exergy expresses all energy transfers on a common basis and accounts explicitly for

the effect of irreversibilities through the exergy destruction concept.

90 MW (heat transfer)

10 MW (at inlet i)

100 MW

Energy In

= 90 MW

= 40 MW

40 MW (power)

60 MW (at exit e)

100 MW

Energy Out

(a)(b

)

60 MW (heat transfer)

2 MW (at inlet i)

62 MW

Exergy In

Exergy Destroyed

= 62 MW – 55 MW = 7 MW

40 MW (power)

15 MW (at exit e)

55 MW

Exergy Out

––

[h + + gz]

= 10 MW

––

[h + + gz]

= 60 MW

= 7 MW

= 40 MW

= 2 MW

= 15 MW

Fig. 7.6 Comparing energy and exergy for a control volume at steady state. (a) Energy analysis. (b) Exergy analysis.

7.5.2

Evaluating Exergy Destruction in Control Volumes

at Steady State

The following examples illustrate the use of mass, energy, and exergy rate balances

for the evaluation of exergy destruction in control volumes at steady state. Property

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data also play an important role in arriving at solutions. The first example involves

the expansion of steam through a valve (a throttling process, Sec. 4.10). From an

energy perspective, the expansion occurs without loss. Yet, as shown in Example 7.5,

such a valve is a site of inefficiency quantified thermodynamically in terms of exergy

destruction.

Determining Exergy Destruction in a Throttling Valve

c c c c EXAMPLE 7.5 c

Superheated water vapor enters a valve at 500 lbf/in.

, 5008F and exits at a pressure of 80 lbf/in.

The expansion

is a throttling process. Determine the exergy destruction per unit of mass flowing, in Btu/lb. Let T

5 778F,

5 1 atm.

Solution

Known:

Water vapor expands in a throttling process through a valve from a specified inlet state to a specified

exit pressure.

Find: Determine the exergy destruction per unit of mass flowing.

Schematic and Given Data:

Analysis:

The state at the inlet is specified. The state at the exit can be fixed by reducing the steady-state mass

and energy rate balances to obtain Eq. 4.22:

5 h

(a)

Thus, the exit state is fixed by p

and h

. From Table A-4E, h

5 1231.5 Btu/lb, s

5 1.4923 Btu/lb

8R. Interpo-

lating at a pressure of 80 lbf/in.

with h

5 h

, the specific entropy at the exit is s

5 1.680 Btu/lb

8R.

With assumptions listed, the steady-state form of the exergy rate balance, Eq. 7.17, reduces to

0 5

a1 2

2 W

1 m

2 e

22 E

Dividing by the mass flow rate

and solving, the exergy destruction per unit of mass flowing is

5 1e

2 e

(b)

Introducing Eq. 7.18, using Eq. (a), and ignoring the effects of motion and gravity

2 e

5 1h

22 T

2 s

2 V

1 g1z

Engineering Model:

The control volume shown in the accompanying figure is at

steady state.

2. For the throttling process, Q

5 W

5 0, and the effects of

motion and gravity can be ignored.

3. T

5 778F, p

5 1 atm.

Steam

500 lbf/in.

500° F

80 lbf/in.

Fig. E7.5

7.5 Exergy Rate Balance for Control Volumes at Steady State 381

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382 Chapter 7 Exergy Analysis

Although heat exchangers appear from an energy perspective to operate without

loss when stray heat transfer is ignored, they are a site of thermodynamic inefficiency

quantified by exergy destruction. This is illustrated in Example 7.6.

Eq. (b) becomes

➊

5 T

2 s

(c)

Inserting values,

➋

5 5378R 11.680 2 1.49232

lb ? 8R

5 100.8 Btu

➊ Equation (c) can be obtained alternatively beginning with the relationship

5 T

and then evaluating the rate of entropy production s

from an

entropy balance. The details are left as an exercise.

➋ Energy is conserved in the throttling process, but exergy is destroyed. The

source of the exergy destruction is the uncontrolled expansion that occurs.

If air modeled as an ideal gas were to undergo a throttling

process, evaluate the exergy destruction, in Btu per lb of air flowing,

for the same inlet conditions and exit pressure as in this example.

Ans. 67.5 Btu/lb.

Ability to…

❑

apply the energy and exergy

rate balances.

❑

evaluate exergy destruction.

✓

Skills Developed

Evaluating Exergy Destruction in a Heat Exchanger

c c c c EXAMPLE 7.6 c

Compressed air enters a counterflow heat exchanger operating at steady state at 610 K, 10 bar and exits at

860 K, 9.7 bar. Hot combustion gas enters as a separate stream at 1020 K, 1.1 bar and exits at 1 bar. Each

stream has a mass flow rate of 90 kg/s. Heat transfer between the outer surface of the heat exchanger and

the surroundings can be ignored. The effects of motion and gravity are negligible. Assuming the combustion

gas stream has the properties of air, and using the ideal gas model for both streams, determine for the heat

exchanger

(a) the exit temperature of the combustion gas, in K.

(b) the net change in the flow exergy rate from inlet to exit of each stream, in MW.

Let T

5 300 K, p

5 1 bar.

SOLUTION

Known:

Steady-state operating data are provided for a counterflow heat exchanger.

Find: For the heat exchanger, determine the exit temperature of the combustion gas, the change in the flow

exergy rate from inlet to exit of each stream, and the rate exergy is destroyed.

➊

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