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

Подождите немного. Документ загружается.

4.7 Turbines 183

Analysis: To calculate the heat transfer rate, begin with the one-inlet, one-exit form of the energy rate balance

for a control volume at steady state, Eq. 4.20a. That is

0 5 Q

2 W

1 m

c1h

2 h

2 V

1 g1z

2 z

where m

is the mass flow rate. Solving for

and dropping the potential energy change from inlet to

exit

5 W

1 m

c1h

2 h

21 a

2 V

(a)

To compare the magnitudes of the enthalpy and kinetic energy terms, and stress the unit conversions needed,

each of these terms is evaluated separately.

First, the specific enthalpy difference h

2 h

is found. Using Table A-4, h

5 3177.2 kJ/kg. State 2 is a two-

phase liquid–vapor mixture, so with data from Table A-3 and the given quality

5 h

1 x

2 h

5 191.83 1 10.9212392.825 2345.4 kJ

Hence

2 h

5 2345.4 2 3177.2 52831.8 kJ

Consider next the specific kinetic energy difference. Using the given values for the velocities

➊

2 V

b5 c

1302

2 1102

1 N

1 kg ? m

1 kJ

N ? m

5 0.4 kJ

Calculating

from Eq. (a)

➋

5 11000 kW21 a4600

b12831.8 1 0.42a

1 h

3600 s

1 kW

1 kJ

5262.3 kW

➊ The magnitude of the change in specific kinetic energy from inlet to exit is

much smaller than the specific enthalpy change. Note the use of unit conver-

sion factors here and in the calculation of

to follow.

➋ The negative value of

means that there is heat transfer from the turbine

to its surroundings, as would be expected. The magnitude of

is small

relative to the power developed.

Ability to…

❑

apply the steady-state

energy rate balance to a

control volume.

❑

develop an engineering

model.

❑

retrieve property data for

water.

✓

Skills Developed

If the change in kinetic energy from inlet to exit were neglected,

evaluate the heat transfer rate, in kW, keeping all other data unchanged.

Comment.

Ans. 262.9 kW

c04ControlVolumeAnalysisUsingEne183 Page 183 6/23/10 9:39:59 AM user-s146c04ControlVolumeAnalysisUsingEne183 Page 183 6/23/10 9:39:59 AM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

184 Chapter 4 Control Volume Analysis Using Energy

4.8 Compressors and Pumps

Compressors and pumps are devices in which work is done on the substance flow-

ing through them in order to change the state of the substance, typically to

increase the pressure and/or elevation. The term compressor is used when the

substance is a gas (vapor) and the term pump is used when the substance is a

liquid. Four compressor types are shown in Fig. 4.11. The reciprocating compres-

sor of Fig. 4.11a features reciprocating motion while the others have rotating

motion.

The axial-flow compressor of Fig. 4.11b is a key component of turbojet engines

(Sec. 9.11). Compressors also are essential components of refrigeration and heat

pump systems (Chap. 10). In the study of Chap. 8, we find that pumps are important

in vapor power systems. Pumps also are commonly used to fill water towers, remove

water from flooded basements, and for numerous other domestic and industrial

applications.

4.8.1

Compressor and Pump Modeling Considerations

For a control volume enclosing a compressor, the mass and energy rate balances

reduce at steady state as for the case of turbines considered in Sec. 4.7.1. Thus,

Eq. 4.20a reduces to read

0 5 Q

2 W

1 m

2 h

2 (a)

Heat transfer with the surroundings is frequently a secondary effect that can be

neglected, giving as for turbines

5 m

2 h

2 (b)

For pumps, heat transfer is generally a secondary effect, but the kinetic and potential

energy terms of Eq. 4.20a may be significant depending on the application. Be sure

to note that for compressors and pumps, the value of W

is negative because a power

input is required.

4.8.2

Applications to an Air Compressor and a Pump System

In this section, modeling considerations for compressors and pumps are illustrated in

Examples 4.5 and 4.6, respectively. Applications of compressors and pumps in energy

storage systems are described in Sec. 4.8.3.

In Example 4.5 the objectives include assessing the significance of the heat trans-

fer and kinetic energy terms of the energy balance and illustrating the appropriate

use of unit conversion factors.

compressor, pump

Calculating Compressor Power

c c c c EXAMPLE 4.5 c

Air enters a compressor operating at steady state at a pressure of 1 bar, a temperature of 290 K, and a velocity

of 6 m/s through an inlet with an area of 0.1 m

. At the exit, the pressure is 7 bar, the temperature is 450 K, and

the velocity is 2 m/s. Heat transfer from the compressor to its surroundings occurs at a rate of 180 kJ/min.

Employing the ideal gas model, calculate the power input to the compressor, in kW.

Fig. 4.11

Compressor types.

Inlet

Outlet

(a) Reciprocating

(b) Axial flow

Rotor

Inlet

Outlet

Stator

Inlet

Outlet

Driveshaft

Impeller

Outlet

Inlet

(d) Roots type

c04ControlVolumeAnalysisUsingEne184 Page 184 6/23/10 9:40:01 AM user-s146c04ControlVolumeAnalysisUsingEne184 Page 184 6/23/10 9:40:01 AM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

SOLUTION

Known:

An air compressor operates at steady state with known inlet and exit states and a known heat transfer rate.

Find: Calculate the power required by the compressor.

Schematic and Given Data:

Engineering Model:

The control volume shown on the accompanying figure is at

steady state.

2. The change in potential energy from inlet to exit can be neglected.

3. The ideal gas model applies for the air.

Air

compressor

= 1 bar

= 290 K

= 6 m/s

= 0.1m

= 7 bar

= 450 K

= 2 m/s

= ?

= –180 kJ/min

Fig. E4.5

➊

If the change in kinetic energy from inlet to exit were neglected,

evaluate the compressor power, in kW, keeping all other data unchanged.

Comment. Ans. 2119.4 kW.

Analysis: To calculate the power input to the compressor, begin with the one-inlet, one-exit form of the energy

rate balance for a control volume at steady state, Eq. 4.20a. That is

0 5 Q

2 W

1 m

c1h

2 h

2 V

1 g1z

2 z

Solving

5 Q

1 m

c1h

2 h

21 a

2 V

The change in potential energy from inlet to exit drops out by assumption 2.

The mass flow rate m

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

M2T

10.1 m

216 m

s2110

831

28.97

N ? m

kg ? K

b1290 K2

5 0.72 kg

The specific enthalpies h

and h

can be found from Table A-22. At 290 K, h

5 290.16 kJ/kg. At 450 K, h

451.8 kJ/kg. Substituting values into the expression for W

, and applying appropriate unit conversion factors, we get

5 a2180

min

1 min

60 s

`1 0.72

c1290.16 2 451.82

1 a

162

2 122

1 N

1 kg ? m

1 kJ

N ? m

523

1 0.72

12161.64 1 0.022

52119.4

1 kW

1 kJ

`52119.4 kW

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

generalized compressibility chart.

➋ In this example Q

and W

have negative values, indicating that the direction

of the heat transfer is from the compressor and work is done on the air passing

through the compressor. The magnitude of the power input to the compressor

is 119.4 kW. The change in kinetic energy does not contribute significantly.

➋

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

4.8 Compressors and Pumps 185

c04ControlVolumeAnalysisUsingEne185 Page 185 6/23/10 9:40:02 AM user-s146c04ControlVolumeAnalysisUsingEne185 Page 185 6/23/10 9:40:02 AM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

186 Chapter 4 Control Volume Analysis Using Energy

In Example 4.6, a pump is a component of an overall system that delivers a high-

velocity stream of water at an elevation greater than at the inlet. Note the modeling

considerations in this case, particularly the roles of kinetic and potential energy, and

the use of appropriate unit conversion factors.

Analyzing a Pump System

c c c c EXAMPLE 4.6 c

A pump steadily draws water from a pond at a volumetric flow rate of 0.83 m

/min through a pipe having a

12-cm diameter inlet. The water is delivered through a hose terminated by a converging nozzle. The nozzle

exit has a diameter of 3 cm and is located 10 m above the pipe inlet. Water enters at 208C, 1 atm and exits

with no significant change in temperature or pressure. The magnitude of the rate of heat transfer from the

pump to the surroundings is 5% of the power input. The acceleration of gravity is 9.81 m/s

. Determine (a) the

velocity of the water at the inlet and exit, each in m/s, and (b) the power required by the pump, in kW.

SOLUTION

Known:

A pump system operates at steady state with known inlet and exit conditions. The rate of heat transfer

from the pump is specified as a percentage of the power input.

Find: Determine the velocities of the water at the inlet and exit of the pump system and the power required.

Schematic and Given Data:

Engineering Model:

A control volume encloses the pump,

inlet pipe, and delivery hose.

2. The control volume is at steady state.

3. The magnitude of the heat transfer from

the control volume is 5% of the power

input.

4. There is no significant change in tempera-

ture or pressure.

5. For liquid water,

y < y

(Eq. 3.11) and

Eq. 3.13 is used to evaluate specific

enthalpy.

6. g 5 9.81 m/s

10 m

Pump

–

= 3 cm

= 20°C

= 1 atm

= 12 cm

(AV)

= 0.83 m

/min

Fig. E4.6

Analysis:

(a) A mass rate balance reduces at steady state to read m

5 m

. The common mass flow rate at the inlet and exit,

, can be evaluated using Eq. 4.4b together with y < y

1208C25 1.0018 3 10

kg from Table A-2. That is,

5 a

0.83 m

min

1.0018 3 10

1 min

60 s

5 13.8

Compressor

A.20 – Tabs a, b, & c

Pump

A.21 – Tabs a, b, & c

c04ControlVolumeAnalysisUsingEn186 Page 186 6/23/10 1:34:33 PM user-s146 c04ControlVolumeAnalysisUsingEn186 Page 186 6/23/10 1:34:33 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

Thus, the inlet and exit velocities are, respectively,

113.8 kg

s211.0018 3 10

kg2

p10.12 m2

5 1.22 m

113.8 kg

s211.0018 3 10

kg2

p10.03 m2

5 19.56 m

(b) To calculate the power input, begin with the one-inlet, one-exit form of the energy rate balance for a control

volume at steady state, Eq. 4.20a. That is

0 5 Q

2 W

1 m

c1h

2 h

21 a

2 V

b1 g1z

2 z

➋ Introducing

0.05

, and solving for W

0.95

c1h

2 h

21 a

2 V

b1 g1z

2 z

(a)

Using Eq. 3.13, the enthalpy term is expressed as

2 h

1 y

2 p

sat

244

1 y

2 p

sat

244

(b)

Since there is no significant change in temperature, Eq. (b) reduces to

2 h

5 y

1T21p

2 p

As there is also no significant change in pressure, the enthalpy term drops out of the present analysis. Next, evaluating

the kinetic energy term

2 V

311.222

2 119.562

1 N

1 kg ? m

1 kJ

N ? m

`520.191 kJ

Finally, the potential energy term is

g1z

2 z

25 19.81 m

210 2 102m `

1 N

1 kg ? m

1 kJ

N ? m

`520.098 kJ

Inserting values into Eq. (a)

5 a

13.8 kg

0.95

b30 2 0.191 2 0.0984a

1 kW

1 kJ

4.2 kW

where the minus sign indicates that power is provided to the pump.

➊

Alternatively, V

can be evaluated from the volumetric flow rate at 1. This

is left as an exercise.

➋ Since power is required to operate the pump, W

is negative in accord with

our sign convention. The energy transfer by heat is from the control volume

to the surroundings, and thus

is negative as well. Using the value of W

found in part (b),

0.05

520.21 kW.

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 properties of liquid

water.

✓

Skills Developed

➊

If the nozzle were removed and water exited directly from

the hose, whose diameter is 5 cm, determine the velocity at the exit, in

m/s, and the power required, in kW, keeping all other data unchanged.

Ans. 7.04 m/s, 1.77 kW.

4.8 Compressors and Pumps 187

c04ControlVolumeAnalysisUsingEn187 Page 187 6/23/10 9:50:32 AM user-s146 c04ControlVolumeAnalysisUsingEn187 Page 187 6/23/10 9:50:32 AM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

188 Chapter 4

Control Volume Analysis Using Energy

4.8.3

Pumped-Hydro and Compressed-Air Energy Storage

Owing to the dictates of supply and demand and other economic factors, the value

of electricity varies with time. Both the cost to generate electricity and increasingly

the price paid by consumers for electricity depend on whether the demand for it is

on-peak or off-peak. The on-peak period is typically weekdays—for example from

8 a.m. to 8 p.m., while off-peak includes nighttime hours, weekends, and major holidays.

Consumers can expect to pay more for on-peak electricity. Energy storage methods

benefiting from variable electricity rates include thermal storage (see box on p. 111) and

pumped-hydro and compressed-air storage introduced in the following box.

Economic Aspects of Pumped-Hydro and Compressed-Air Energy Storage

Despite the significant costs of owning and operating utility-scale energy storagy systems,

various economic strategies, including taking advantage of differing on- and off-peak elec-

tricity rates, can make pumped-hydro and compressed-air storage good choices for power

generators. In this discussion, we focus on the role of variable electricity rates.

In pumped-hydro storage, water is pumped from a lower reservoir to an upper reservoir,

thereby storing energy in the form of gravitational potential energy. (For simplicity, think of

the hydropower plant of Fig. 4.10 operating in the reverse direction.) Off-peak electricity is used

to drive the pumps that deliver water to the upper reservoir. Later, during an on-peak period,

stored water is released from the upper reservoir to generate electricity as the water flows

through turbines to the lower reservoir. For instance, in the summer water is released from

the upper reservoir to generate power to meet a high daytime demand for air conditioning;

while at night, when demand is low, water is pumped back to the upper reservoir for use

the next day. Owing to friction and other nonidealities, an overall input-to-output loss of

electricity occurs with pumped-hydro storage and this adds to operating costs. Still, differ-

ing daytime and nighttime electricity rates help make this technology viable.

In compressed-air energy storage, compressors powered with off-peak electricity fill

an underground salt cavern, hard-rock mine, or aquifer with pressurized air drawn from

the atmosphere. See Fig. 4.12. When electricity demand peaks, high-pressure compressed

air is released to the surface, heated by natural gas in combustors, and expanded through

a turbine generator, generating electricity for distribution at on-peak rates.

TAKE NOTE...

Cost refers to the amount

paid to produce a good or

service. Price refers to what

consumers pay to acquire

that good or service.

–

Compressor

Atmospheric air Air and combustion products

Turbine

–

Off-peak

electricity

On-peak

electricity

out

Compressed

air out

Cavern

Air in Air out

Compressed

air in

Fuel to combustor

(not shown)

Generator

Fig. 4.12 Compressed-air storage.

c04ControlVolumeAnalysisUsingEn188 Page 188 6/23/10 9:50:35 AM user-s146 c04ControlVolumeAnalysisUsingEn188 Page 188 6/23/10 9:50:35 AM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

4.9 Heat Exchangers

Heat exchangers have innumerable domestic and industrial applications, including use

in home heating and cooling systems, automotive systems, electrical power generation,

and chemical processing. Indeed, nearly every area of application listed in Table 1.1

(p. 5) involves heat exchangers.

One common type of heat exchanger is a mixing chamber in which hot and cold

streams are mixed directly as shown in Fig. 4.13a. The open feedwater heater, which

is a component of the vapor power systems considered in Chap. 8, is an example of

this type of device.

Another common type of exchanger is one in which a gas or liquid is separated from

another gas or liquid by a wall through which energy is conducted. These heat exchang-

ers, known as recuperators, take many different forms. Counterflow and parallel tube-

within-a-tube configurations are shown in Figs. 4.13b and 4.13c, respectively. Other con-

figurations include cross-flow, as in automobile radiators, and multiple-pass shell-and-tube

condensers and evaporators. Figure 4.13d illustrates a cross-flow heat exchanger.

heat exchanger

(a)(b)

(c)(d

)

Fig. 4.13 Common heat exchanger types. (a) Direct contact heat exchanger. (b) Tube-within-

a-tube counterflow heat exchanger. (c) Tube-within-a-tube parallel flow heat exchanger.

(d) Cross-flow heat exchanger.

BIOCONNECTIONS Inflatable blankets such as shown in Fig. 4.14 are used

to prevent subnormal body temperatures (hypothermia) during and after surgery.

Typically, a heater and blower direct a stream of warm air into the blanket. Air exits

the blanket through perforations in its surface. Such thermal blankets have been used

safely and without incident in millions of surgical procedures. Still, there are obvious risks

to patients if temperature controls fail and overheating occurs. Such risks can be antici-

pated and minimized with good engineering practices.

Warming patients is not always the issue at hospitals; sometimes it is cooling, as in

cases involving cardiac arrest, stroke, heart attack, and overheating of the body (hyperther-

mia). Cardiac arrest, for example, deprives the heart muscle of oxygen and blood, causing

part of it to die. This often induces brain damage among survivors, including irreversible

cognitive disability. Studies show when the core body temperature of cardiac patients is

4.9 Heat Exchangers 189

c04ControlVolumeAnalysisUsingEn189 Page 189 6/23/10 9:50:37 AM user-s146 c04ControlVolumeAnalysisUsingEn189 Page 189 6/23/10 9:50:37 AM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

190 Chapter 4 Control Volume Analysis Using Energy

reduced to about 338C (918F), damage is limited because vital organs function more slowly

and require less oxygen. To achieve good outcomes, medical specialists say cooling should

be done in about 20 minutes or less. A system approved for cooling cardiac arrest victims

includes a disposable plastic body suit, pump, and chiller. The pump provides rapidly flow-

ing cold water around the body, in direct contact with the skin of the patient wearing the

suit, then recycles coolant to the chiller and back to the patient.

These biomedical applications provide examples of how engineers well versed in ther-

modynamics principles can bring into the design process their knowledge of heat exchang-

ers, temperature sensing and control, and safety and reliability requirements.

Heater-

blower

unit

Fig. 4.14 Inflatable thermal blanket.

4.9.1

Heat Exchanger Modeling Considerations

As shown by Fig. 4.13, heat exchangers can involve multiple inlets and exits. For a

control volume enclosing a heat exchanger, the only work is flow work at the places

where matter enters and exits, so the term W

drops out of the energy rate balance.

In addition, the kinetic and potential energies of the flowing streams usually can be

ignored at the inlets and exits. Thus, the underlined terms of Eq. 4.18 (repeated below)

drop out, leaving the enthalpy and heat transfer terms, as shown by Eq. (a). That is,

0 5 Q

2 W

1 gz

0 5 Q

(a)

Although high rates of energy transfer within the heat exchanger occur, heat transfer with

the surroundings is often small enough to be neglected. Thus, the

term of Eq. (a)

would drop out, leaving just the enthalpy terms. The final form of the energy rate

balance must be solved together with an appropriate expression of the mass rate bal-

ance, recognizing both the number and type of inlets and exits for the case at hand.

4.9.2

Applications to a Power Plant Condenser and

Computer Cooling

The next example illustrates how the mass and energy rate balances are applied to

a condenser at steady state. Condensers are commonly found in power plants and

refrigeration systems.

Heat_Exchanger

A.22 – Tabs a, b, & c

c04ControlVolumeAnalysisUsingEn190 Page 190 6/23/10 1:34:45 PM user-s146 c04ControlVolumeAnalysisUsingEn190 Page 190 6/23/10 1:34:45 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

Evaluating Performance of a Power Plant Condenser

c c c c EXAMPLE 4.7 c

Steam enters the condenser of a vapor power plant at 0.1 bar with a quality of 0.95 and condensate exits at

0.1 bar and 458C. Cooling water enters the condenser in a separate stream as a liquid at 208C and exits as a

liquid at 358C with no change in pressure. Heat transfer from the outside of the condenser and changes in the

kinetic and potential energies of the flowing streams can be ignored. For steady-state operation, determine

(a) the ratio of the mass flow rate of the cooling water to the mass flow rate of the condensing steam.

(b) the rate of energy transfer from the condensing steam to the cooling water, in kJ per kg of steam passing

through the condenser.

SOLUTION

Known:

Steam is condensed at steady state by interacting with a separate liquid water stream.

Find: Determine the ratio of the mass flow rate of the cooling water to the mass flow rate of the steam and the

rate of energy transfer from the steam to the cooling water.

Schematic and Given Data:

Engineering Model:

Each of the two control volumes shown

on the accompanying sketch is at

steady state.

2. There is no significant heat transfer

between the overall condenser and its

surroundings. W

5 0.

3. Changes in the kinetic and potential

energies of the flowing streams from

inlet to exit can be ignored.

4. At states 2, 3, and 4, h < h

(see

Eq. 3.14).

0.1 bar

45.8°C

Steam

0.1 bar

x = 0.95

Condensate

0.1 bar

45°C

Cooling

water

35°C

Cooling

water

20°C

Control volume for part (a)

Control volume for part (b)

Condensate

Steam

Energy transfer to

cooling water

Fig. E4.7

4.9 Heat Exchangers 191

Analysis: The steam and the cooling water streams do not mix. Thus, the mass rate balances for each of the two

streams reduce at steady state to give

5 m



and



5 m

(a) The ratio of the mass flow rate of the cooling water to the mass flow rate of the condensing steam, m

, can be

found from the steady-state form of the energy rate balance, Eq. 4.18, applied to the overall condenser as follows:

0 5 Q

2 W

1 m

1 gz

1 m

1 gz

2 m

1 gz

2 m

1 gz

The underlined terms drop out by assumptions 2 and 3. With these simplifications, together with the above mass

flow rate relations, the energy rate balance becomes simply

0 5 m

2 h

1 m

2 h

Solving, we get

2 h

c04ControlVolumeAnalysisUsingEn191 Page 191 6/23/10 9:50:38 AM user-s146 c04ControlVolumeAnalysisUsingEn191 Page 191 6/23/10 9:50:38 AM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

192 Chapter 4 Control Volume Analysis Using Energy

Excessive temperatures in electronic components are avoided by providing appro-

priate cooling. In the next example, we analyze the cooling of computer components,

illustrating the use of the control volume form of energy rate balance together with

property data for air.

The specific enthalpy h

can be determined using the given quality and data from Table A-3. From Table A-3 at

0.1 bar, h

5 191.83 kJ/kg and h

5 2584.7 kJ/kg, so

5 191.83 1 0.95

2584.7 2 191.83

5 2465.1 kJ

➊ Using assumption 4, the specific enthalpy at 2 is given by h

< h

5 188.45 kJ

kg. Similarly, h

< h

and

< h

, giving h

2 h

5 62.7 kJ

. Thus,

2465.1 2 188.45

62.7

5 36.3

(b) For a control volume enclosing the steam side of the condenser only, begin with the steady-state form of

energy rate balance, Eq. 4.20a.

➋ 0 5 Q

2 W

1 m

c1h

2 h

2 V

1 g1z

2 z

The underlined terms drop out by assumptions 2 and 3. The following expression for the rate of energy transfer

between the condensing steam and the cooling water results:

5 m

2 h

Dividing by the mass flow rate of the steam, m

, and inserting values

5 h

2 h

5 188.45 2 2465.1 522276.7 kJ

where the minus sign signifies that energy is transferred from the condensing

steam to the cooling water.

➊ Alternatively, (h

2 h

) can be evaluated using the incompressible liquid

model via Eq. 3.20b.

➋ Depending on where the boundary of the control volume is located, two

different formulations of the energy rate balance are obtained. In part (a),

both streams are included in the control volume. Energy transfer between

them occurs internally and not across the boundary of the control volume,

so the term

drops out of the energy rate balance. With the control

volume of part (b), however, the term

must be included.

If the mass flow rate of the condensing steam is 125 kg/s, deter-

mine the mass flow rate of the cooling water, in kg/s. Ans. 4538 kg/s.

Ability to…

❑

apply the steady-state

mass and energy rate bal-

ances to a control volume.

❑

develop an engineering model.

❑

retrieve property data for

water.

✓

Skills Developed

Cooling Computer Components

c c c c EXAMPLE 4.8 c

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

electronics enclosure. At steady state, air enters at 208C, 1 atm. For noise control, the velocity of the entering air

cannot exceed 1.3 m/s. For temperature control, the temperature of the air at the exit cannot exceed 328C. The

electronic components and fan receive, respectively, 80 W and 18 W of electric power. Determine the smallest

fan inlet area, in cm

, for which the limits on the entering air velocity and exit air temperature are met.

c04ControlVolumeAnalysisUsingEn192 Page 192 6/23/10 9:50:39 AM user-s146 c04ControlVolumeAnalysisUsingEn192 Page 192 6/23/10 9:50:39 AM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New