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

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SOLUTION

Known:

The electronic components of a computer are cooled by air flowing through a fan mounted at the inlet

of the electronics enclosure. Conditions are specified for the air at the inlet and exit. The power required by the

electronics and the fan are also specified.

Find: Determine the smallest fan area for which the specified limits are met.

Schematic and Given Data:

Analysis: The inlet area A

can be determined from the mass flow rate m

and Eq. 4.4b, which can be rearranged

to read

(a)

The mass flow rate can be evaluated, in turn, from the steady-state energy rate balance, Eq. 4.20a.

0 5 Q

2 W

1 m

2 h

2 V

1 g1z

2 z

The underlined terms drop out by assumptions 2 and 3, leaving

0 52W

1 m

2 h

where W

accounts for the total electric power provided to the electronic components and the fan: W

280 W

218 W

5298 W. Solving for m

, and using assumption 4 with Eq. 3.51 to evaluate (h

2 h

)

12W

2 T

Introducing this into the expression for A

, Eq. (a), and using the ideal gas model to evaluate the specific

volume y

12W

2 T

From this expression we see that A

increases when V

and/or T

decrease. Accordingly, since V

# 1.3 m/s and

# 305 K (328C), the inlet area must satisfy

1.3 m

≥

98 W

a1.005

? K

b1305 2 2932K

1 kJ

1 J

1 W

`¥

8314 N ? m

28.97 kg ? K

b293 K

1.01325 3 10

≤

1 m

2 cm

For the specified conditions, the smallest fan area is 52 cm

4.9 Heat Exchangers 193

–

Air in

Air out

= 20°C

= 1 atm

≤ 1.3 m/s

≤ 32°C

Electronic

components

Fan

Fig. E4.8

Engineering Model:

The control volume shown on the accompa-

nying figure is at steady state.

2. Heat transfer from the outer surface of the

electronics enclosure to the surroundings is

negligible. Thus,

5 0.

3. Changes in kinetic and potential energies can

be ignored.

4. Air is modeled as an ideal gas with

5 1.005 kJ/kg

➋

➊

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194 Chapter 4 Control Volume Analysis Using Energy

If heat transfer occurs at a rate of 11 W from the outer surface

of the computer case to the surroundings, determine the smallest fan inlet

area for which the limits on entering air velocity and exit air temperature

are met if the total power input remains at 98 W. Ans. 46 cm

Partiall

en valve

Porous plug

Fig. 4.15 Examples of throttling devices.

➊ Cooling air typically enters and exits electronic enclosures at low velocities,

and thus kinetic energy effects are insignificant.

➋ The applicability of the ideal gas model can be checked by reference to

the generalized compressibility chart. Since the temperature of the air

increases by no more than 128C, the specific heat c

is nearly constant

(Table A-20).

Ability to…

❑

apply the steady-state

energy rate balance to a

control volume.

❑

apply the mass flow rate

expression, Eq. 4.4b.

❑

develop an engineering model.

❑

retrieve property data of air

modeled as an ideal gas.

✓Skills Developed

4.10 Throttling Devices

A significant reduction in pressure can be achieved simply by introducing a restriction

into a line through which a gas or liquid flows. This is commonly done by means of a par-

tially opened valve or a porous plug. These throttling devices are illustrated in Fig. 4.15.

An application of throttling occurs in vapor-compression refrigeration systems,

where a valve is used to reduce the pressure of the refrigerant from the pressure at

the exit of the condenser to the lower pressure existing in the evaporator. We consider

this further in Chap. 10. Throttling also plays a role in the Joule–Thomson expansion

considered in Chap. 11. Another application involves the throttling calorimeter, which

is a device for determining the quality of a two-phase liquid–vapor mixture. The

throttling calorimeter is considered in Example 4.9.

throttling calorimeter

4.10.1

Throttling Device Modeling Considerations

For a control volume enclosing a throttling device, the only work is flow work at

locations where mass enters and exits the control volume, so the term W

drops out

of the energy rate balance. There is usually no significant heat transfer with the sur-

roundings, and the change in potential energy from inlet to exit is negligible. Thus,

the underlined terms of Eq. 4.20a (repeated below) drop out, leaving the enthalpy

and kinetic energy terms, as shown by Eq. (a). That is,

0 5 Q

2 W

1 m

2 h

2 V

1 g1z

2 z

0 5 1h

2 h

2 V

(a)

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Although velocities may be relatively high in the vicinity of the restriction imposed

by the throttling device on the flow through it, measurements made upstream and

downstream of the reduced flow area show in most cases that the change in the

specific kinetic energy of the flowing substance between these locations can be

neglected. With this further simplification, Eq. (a) reduces to

5 h





(4.22)

When the flow through the valve or other restriction is idealized in this way, the

process is called a throttling process.

throttling process

4.10.2

Using a Throttling Calorimeter to Determine Quality

The next example illustrates use of a throttling calorimeter to determine steam quality.

Measuring Steam Quality

c c c c EXAMPLE 4.9 c

A supply line carries a two-phase liquid–vapor mixture of steam at 300 lbf/in.

A small fraction of the flow in

the line is diverted through a throttling calorimeter and exhausted to the atmosphere at 14.7 lbf/in.

The tem-

perature of the exhaust steam is measured as 2508F. Determine the quality of the steam in the supply line.

SOLUTION

Known:

Steam is diverted from a supply line through a throttling calorimeter and exhausted to the atmosphere.

Find: Determine the quality of the steam in the supply line.

Schematic and Given Data:

Calorimeter

Thermometer

Steam line, 300 lbf/in.

= 14.7 lbf/in.

= 250°F

= 300 lbf/in.

= 14.7 lbf/in.

= 250°F

Fig. E4.9

Engineering Model:

The control volume shown on the accompanying figure is at steady state.

2. The diverted steam undergoes a throttling process.

Analysis: For a throttling process, the energy and mass balances reduce to give h

5 h

, which agrees with Eq.

4.22. Thus, with state 2 fixed, the specific enthalpy in the supply line is known, and state 1 is fixed by the

known values of p

and h

➊ As shown on the accompanying p–y diagram, state 1 is in the two-phase liquid–vapor region and state 2 is

in the superheated vapor region. Thus,

5 h

1 x

2 h

4.10 Throttling Devices 195

Throttling_Dev

A.23 – Tabs a, b, & c

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196 Chapter 4 Control Volume Analysis Using Energy

Solving for x

2 h

From Table A-3E at 300 lbf/in.

, h

5 394.1 Btu/lb and h

5 1203.9 Btu/lb. At

14.7 lbf/in.

and 2508F, h

5 1168.8 Btu/lb from Table A-4E. Inserting values into

the above expression, the quality of the steam in the line is x

5 0.957 (95.7%).

➊ For throttling calorimeters exhausting to the atmosphere, the quality of the

steam in the line must be greater than about 94% to ensure that the steam

leaving the calorimeter is superheated.

If the supply line carried saturated vapor at 300 lbf/in.

, deter-

mine the temperature at the calorimeter exit, in 8F, for the same exit pres-

sure, 14.7 lbf/in.

Ans. 3248F.

Ability to…

❑

apply Eq. 4.22 for a throt-

tling process.

❑

retrieve property data for

water

✓

Skills Developed

BIOCONNECTIONS Living things also can be considered integrated sys-

tems. Figure 4.17 shows a control volume enclosing a tree receiving solar radiation.

As indicated on the figure, a portion of the incident radiation is reflected to the

surroundings. Of the net solar energy received by the tree, about 21% is returned to the

surroundings by heat transfer, principally convection. Water management accounts for most

of the remaining solar input.

Trees sweat as do people; this is called evapotranspiration. As shown in Fig. 4.17, about

78% of the net solar energy received by the tree is used to pump liquid water from the

surroundings, primarily the ground, convert it to a vapor, and discharge it to the surround-

ings through tiny pores (called stomata) in the leaves. Nearly all the water taken up is lost

in this manner and only a small fraction is used within the tree. Applying an energy balance

to the control volume enclosing the tree, just 1% of the net solar energy received by the

tree is left for use in the production of biomass (wood and leaves). Evapotranspiration

benefits trees but also contributes significantly to water loss from watersheds, illustrating

that in nature as in engineering there are trade-offs.

4.11 System Integration

Thus far, we have studied several types of components selected from

those commonly seen in practice. These components are usually encoun-

tered in combination, rather than individually. Engineers often must

creatively combine components to achieve some overall objective, sub-

ject to constraints such as minimum total cost. This important engineer-

ing activity is called system integration.

In engineering practice and everyday life, integrated systems are

regularly encountered. Many readers are already familiar with a par-

ticularly successful system integration: the simple power plant shown

in Fig. 4.16. This system consists of four components in series: a turbine-

generator, condenser, pump, and boiler. We consider such power plants

in detail in subsequent sections of the book.

Boiler

Condenser

TurbinePump

out

Fig. 4.16 Simple vapor power plant.

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Example 4.10 provides another illustration of an integrated system. This case

involves a waste heat recovery system.

Evapotranspiration

(78%)

Biomass

production

(1%)

Moisture

Reflected

Incident

Heat transfer

(21%)

Vapor

Solar

radiation

Fig. 4.17 Control volume enclosing a tree.

Evaluating Performance of a Waste Heat Recovery System

c c c c EXAMPLE 4.10 c

An industrial process discharges 2 3 10

/min of gaseous combustion products at 4008F, 1 atm. As shown in

Fig. E4.10, a proposed system for utilizing the combustion products combines a heat-recovery steam generator

with a turbine. At steady state, combustion products exit the steam generator at 2608F 1 atm and a separate

stream of water enters at 40 lbf/in.

, 1028F with a mass flow rate of 275 lb/min. At the exit of the turbine, the

pressure is 1 lbf/in.

and the quality is 93%. Heat transfer from the outer surfaces of the steam generator and

turbine can be ignored, as can the changes in kinetic and potential energies of the flowing streams. There is no

significant pressure drop for the water flowing through the steam generator. The combustion products can be

modeled as air as an ideal gas.

(a) Determine the power developed by the turbine, in Btu/min.

(b) Determine the turbine inlet temperature, in 8F.

h, determine the value of the power, in $/year, for 8000 hours

of operation annually.

SOLUTION

Known:

Steady-state operating data are provided for a system consisting of a heat-recovery steam generator and

a turbine.

Find: Determine the power developed by the turbine and the turbine inlet temperature. Evaluate the annual

value of the power developed.

4.11 System Integration 197

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198 Chapter 4

Control Volume Analysis Using Energy

Combustion products in

= 1 atm

= 400°F

(AV)

= 2 × 10

/min

Combustion products out

= 260°F

= 1atm

Turbine

Power

out

Steam

generator

water in

= 40 lbf/in.

= 102°F

= 275 lb/min

water out

= 1 lbf/in.

= 93%

Fig. E4.10

Analysis:

(a)

The power developed by the turbine is determined from a control volume enclosing both the steam genera-

tor and the turbine. Since the gas and water streams do not mix, mass rate balances for each of the streams

reduce, respectively, to give

5 m



5 m

For this control volume, the appropriate form of the steady-state energy rate balance is Eq. 4.18, which reads

0 5 Q

2 W

1 m

1 gz

b1 m

1 gz

2 m

1 gz

b2 m

1 gz

The underlined terms drop out by assumption 2. With these simplifications, together with the above mass flow

rate relations, the energy rate balance becomes

5 m

2 h

21 m

2 h

The mass flow rate m

can be evaluated with given data at inlet 1 and the ideal gas equation of state

1AV2

M2T

12 3 10

min2114.7 lbf

in.

1545 ft ? lbf

28.97 lb ? 8R

18608R2

144 in.

1 ft

5 9230.6 lb

min

The specific enthalpies h

and h

can be found from Table A-22E: At 8608R, h

5 206.46 Btu/lb, and at 7208R h

172.39 Btu/lb. At state 3, water is a liquid. Using Eq. 3.14 and saturated liquid data from Table A-2E,

5 70 Btu

lb. State 5 is a two-phase liquid–vapor mixture. With data from Table A-3E and the given quality

5 h

1 x

2 h

5 69.74 1 0.93

1036.0

5 1033.2 Btu

Substituting values into the expression for W

9230.6

min

1206.46 2 172.392

Btu

275

min

170 2 1033.22

Btu

5 49610

min

Engineering Model:

The control volume shown on the

accompanying figure is at steady state.

2. Heat transfer is negligible, and

changes in kinetic and potential

energy can be ignored.

3. There is no pressure drop for water

flowing through the steam generator.

4. The combustion products are modeled

as air as an ideal gas.

Schematic and Given Data:

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4.12 Transient Analysis 199

Taking a control volume enclosing just the turbine, evaluate

the turbine inlet temperature, in °F. Ans. 354°F.

(b) To determine T

, it is necessary to fix the state at 4. This requires two independent property values. With

assumption 3, one of these properties is pressure, p

5 40 lbf/in.

The other is the specific enthalpy h

, which can

be found from an energy rate balance for a control volume enclosing just the steam generator. Mass rate bal-

ances for each of the two streams give m

5 m

and m

5 m

. With assumption 2 and these mass flow rate rela-

tions, the steady-state form of the energy rate balance reduces to

0 5 m

2 h

1 m

2 h

Solving for h

5 h

2 h

5 70

Btu

9230.6 lb

min

1206.46 2 172.392

Btu

5 1213.6

Interpolating in Table A-4E at p

5 40 lbf/in.

with h

, we get T

5 3548F.

value of the power developed for 8000 hours of operation annually is

annual value 5 a49610

Btu

min

60 min

1 h

1 kW

3413 Btu

`ba8000

year

ba0.08

kW ? h

5 558,000

ear

➊

Alternatively, to determine h

a control volume enclosing just the turbine

can be considered.

➋ The decision about implementing this solution to the problem of utilizing

the hot combustion products discharged from an industrial process would

necessarily rest on the outcome of a detailed economic evaluation, including

the cost of purchasing and operating the steam generator, turbine, and aux-

iliary equipment.

Ability to…

❑

apply the steady-state

mass and energy rate bal-

ances to a control volume.

❑

apply the mass flow rate

expression, Eq. 4.4b.

❑

develop an engineering model.

❑

retrieve property data for

water and for air modeled

as an ideal gas.

❑

conduct an elementary

economic evaluation.

✓

Skills Developed

➊

➋

4.12 Transient Analysis

Many devices undergo periods of transient operation during which the state changes

with time. Examples include the startup or shutdown of turbines, compressors, and

motors. Additional examples include vessels being filled or emptied, as considered

in Example 4.2 and in the discussion of Fig. 1.5. Because property values, work and

heat transfer rates, and mass flow rates may vary with time during transient opera-

tion, the steady-state assumption is not appropriate when analyzing such cases. Spe-

cial care must be exercised when applying the mass and energy rate balances, as

discussed next.

4.12.1

The Mass Balance in Transient Analysis

First, we place the control volume mass balance in a form that is suitable for transient

analysis. We begin by integrating the mass rate balance, Eq. 4.2, from time 0 to a final

transient

System_Types

A.1 – Tab d

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200 Chapter 4 Control Volume Analysis Using Energy

time t. That is

bdt 5

bdt 2

This takes the form

1t22 m

1025

dtb2

dtb

Introducing the following symbols for the underlined terms

dt μ

amount of mass

entering the control

volume through inlet i,

from time 0 to t

dt μ

amount of mass

exiting the control

volume through exit e,

from time 0 to t

the mass balance becomes

1t22 m

1025

(4.23)

In words, Eq. 4.23 states that the change in the amount of mass contained in the control

volume equals the difference between the total incoming and outgoing amounts of mass.

4.12.2

The Energy Balance in Transient Analysis

Next, we integrate the energy rate balance, Eq. 4.15, ignoring the effects of kinetic

and potential energy. The result is

1t22 U

1025 Q

2 W

dtb2

dtb

(4.24)

where Q

accounts for the net amount of energy transferred by heat into the control

volume and W

accounts for the net amount of energy transferred by work, except

for flow work. The integrals shown underlined in Eq. 4.24 account for the energy

carried in at the inlets and out at the exits.

For the special case where the states at the inlets and exits are constant with time,

the respective specific enthalpies, h

and h

, are constant, and the underlined terms of

Eq. 4.24 become

dt 5 h

Equation 4.24 then takes the following special form

1t22 U

1025 Q

2 W

(4.25)

where m

and m

account, respectively, for the amount of mass entering the control volume

through inlet i and exiting the control volume through exit e, each from time 0 to t.

Whether in the general form, Eq. 4.24, or the special form, Eq. 4.25, these equations

account for the change in the amount of energy contained within the control volume

as the difference between the total incoming and outgoing amounts of energy.

Another special case is when the intensive properties within the control volume

are uniform with position at a particular time t. Accordingly, the specific volume and

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the specific internal energy are uniform throughout and can depend only on time—

that is, y(t) and u(t), respectively. Then

5 V

(4.26)

5 m

(4.27)

If the control volume is comprised of different phases at time t, the state of each

phase is assumed uniform throughout.

Equations 4.23 and 4.25–4.27 are applicable to a wide range of transient cases

where inlet and exit states are constant with time and intensive properties within the

control volume are uniform with position initially and finally.

in cases involving the filling of containers having a single inlet

and no exit, Eqs. 4.23, 4.25, and 4.27 combine to give

1t2u1t22 m

102u1025 Q

2 W

1 h

1t22 m

1022 (4.28)

The details are left as an exercise. See Examples 4.12 and 4.13 for this type of transient

application. b b b b b

4.12.3

Transient Analysis Applications

The following examples provide illustrations of the transient analysis of control vol-

umes using the conservation of mass and energy principles. In each case considered,

to emphasize fundamentals we begin with general forms of the mass and energy bal-

ances and reduce them to forms suited for the case at hand, invoking the idealizations

discussed in this section as warranted.

The first example considers a vessel that is partially emptied as mass exits through

a valve.

Pressure

regulating valve

Saturated water-

vapor removed,

while the tank

is heated

Initial: two-phase

liquid-vapor mixture

Final: saturated vapor

260°C

2, e1

Fig. E4.11

4.12 Transient Analysis 201

Evaluating Heat Transfer for a Partially-Emptying Tank

c c c c EXAMPLE 4.11 c

A tank having a volume of 0.85 m

initially contains water as a two-phase liquid–vapor mixture at 2608C and a

quality of 0.7. Saturated water vapor at 2608C is slowly withdrawn through a pressure-regulating valve at the top

of the tank as energy is transferred by heat to maintain the pressure constant in the tank. This continues until

the tank is filled with saturated vapor at 2608C. Determine the amount of heat transfer, in kJ. Neglect all kinetic

and potential energy effects.

SOLUTION

Known:

A tank initially holding a two-phase liquid–vapor mixture is heated while saturated water vapor is slowly

removed. This continues at constant pressure until the tank is filled only with saturated vapor.

Find: Determine the amount of heat transfer.

Schematic and Given Data:

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202 Chapter 4 Control Volume Analysis Using Energy

Engineering Model:

The control volume is defined by the dashed line on the accompanying diagram.

2. For the control volume, W

5 0 and kinetic and potential energy effects can be neglected.

3. At the exit the state remains constant.

4. The initial and final states of the mass within the vessel are equilibrium states.

Analysis: Since there is a single exit and no inlet, the mass rate balance Eq. 4.2 takes the form

52m

With assumption 2, the energy rate balance Eq. 4.15 reduces to

5 Q

2 m

Combining the mass and energy rate balances results in

5 Q

1 h

By assumption 3, the specific enthalpy at the exit is constant. Accordingly, integration of the last equation

gives

¢U

1 h

¢m

Solving for the heat transfer Q

5 ¢U

2 h

¢m

5 1m

2 m

22 h

2 m

where m

and m

denote, respectively, the initial and final amounts of mass within the tank.

The terms u

and m

of the foregoing equation can be evaluated with property values from Table A-2 at 2608C

and the given value for quality. Thus

5 u

1 x

2 u

5 1128.4 1 10.7212599.0 2 1128.425 2157.8 kJ

Also,

5 y

1 x

2 y

5 1.2755 3 10

1 10.7210.04221 2 1.2755 3 10

25 29.93 3 10

Using the specific volume y

, the mass initially contained in the tank is

0.85 m

129.93 3 10

kg2

5 28.4 kg

The final state of the mass in the tank is saturated vapor at 2608C so Table A-2 gives

5 u

12608C25 2599.0 kJ

kg,



5 y

12608C25 42.21 3 10

The mass contained within the tank at the end of the process is

0.85 m

142.21 3 10

kg2

5 20.14 kg

Table A-2 also gives h

5 h

(2608C) 5 2796.6 kJ/kg.

➊

➋

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