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Thermodynamics is a fascinating subjectthat deals with energy. It has a broad

application area ranging from microscopicorganisms and household appliances, tovehicles, power generation systems, andeven philosophy. Figures are important

learning tools that help students “get thepicture,”

and our text makes effective use of graphics by featuring more illustrations

and photographs than any other thermodynamics text.

Guided Tour

Our text emphasizes the physical aspectsof thermodynamics in addition tomathematical representations andmanipulations. The authors believe that the emphasis in undergraduate educationshould remain on developing a sense of the underlying physical mechanisms and amastery of solving practical problems thatan engineer is likely to face in the realworld.

5-StageLow PressureCompressor (LPC)

LPC BleedAir Collector

Cold EndDrive Flange

14-StageHigh PressureCompressor

CombustorFuel SystemManifolds

2-Stage High Pressure Turbine

5-StageLow PressureTurbine

Hot EndDrive Flange

Courtesy of GE Power Systems

EXAMPLE 2–17 Cost Savings Associated with High-EfficiencyMotors

A 60-hp electric motor (a motor that delivers 60 hp of shaft power at fullload) that has an efficiency of 89.0 percent is worn out and is to bereplaced by a 93.2 percent efficient high-efficiency motor (Fig. 2–61). Themotor operates 3500 hours a year at full load. Taking the unit cost of elec-tricity to be $0.08/kWh, determine the amount of energy and money savedas a result of installing the high-efficiency motor instead of the standardmotor. Also, determine the simple payback period if the purchase prices ofthe standard and high-efficiency motors are $4520 and $5160, respectively.

Solution A worn-out standard motor is to be replaced by a high-efficiencyone. The amount of electrical energy and money saved as well as the simplepayback period are to be determined.Assumptions The load factor of the motor remains constant at 1 (full load)when operating.Analysis The electric power drawn by each motor and their difference canbe expressed as

where hst is the efficiency of the standard motor, and heff is the efficiency ofthe comparable high-efficiency motor. Then the annual energy and cost sav-ings associated with the installation of the high-efficiency motor become

Also,

This gives a simple payback period of

Discussion Note that the high-efficiency motor pays for its price differentialwithin about one year from the electrical energy it saves. Considering thatthe service life of electric motors is several years, the purchase of the higherefficiency motor is definitely indicated in this case.

Simple payback period �Excess initial cost

Annual cost savings�

$640

$634>year� 1.01 year

Excess initial cost � Purchase price differential � $5160 � $4520 � $640

� $634>year

� 17929 kWh>year 2 1$0.08> kWh 2 Cost savings � 1Energy savings 2 1Unit cost of energy 2

� 7929 kWh>year

� 160 hp 2 10.7457 kW>hp 2 13500 h>year 2 11 2 11>0.89 � 1>0.93.2 2 � 1Rated power 2 1Operating hours 2 1Load factor 2 11>hst � 1>heff 2

Energy savings � 1Power savings 2 1Operating hours 2

� 1Rated power 2 1Load factor 2 11>hst � 1>heff 2 Power savings � W

#

electric in,standard � W#

electric in,efficient

W#

electric in,efficient � W#

shaft>heff � 1Rated power 2 1Load factor 2 >heff

W#

electric in,standard � W#

shaft>hst � 1Rated power 2 1Load factor 2 >hst

Standard Motor

60 hp

h = 89.0%

High-Efficiency Motor

60 hp

h = 93.2%

FIGURE 2–61Schematic for Example 2–17.

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NEW: The energy analysis of closedsystems is now presented in a separatechapter, “Energy Analysis of ClosedSystems” (Chapter 4), together with theboundary work and the discussion ofspecific heats for both ideal gases andincompressible substances.

Guided Tour

∆u = cv ∆T

T1 = 20°CP = constant

AIR

T2 = 30°CQ2

Q1

T1 = 20°CV = constant

AIR

T2 = 30°C

= 7.18 kJ/kg∆u = cv ∆T

= 7.18 kJ/kg

FIGURE 4–27The relation �u � cv �T is validfor any kind of process,constant-volume or not.

NEW: The first law of thermodynamicsis introduced early in the new Chapter 2,“Energy, Energy Transfer, and GeneralEnergy Analysis.” This chapter establishesa general understanding of energy,mechanisms of energy transfer, theconcept of energy balance, thermo-economics, energy conversion, andconversion efficiency. It also exposesstudents to some exciting real-worldapplications of thermodynamics early inthe course, and helps them establish asense of the monetary value of energy.

EXAMPLE 2–11 Acceleration of Air by a Fan

A fan that consumes 20 W of electric power when operating is claimed todischarge air from a ventilated room at a rate of 0.25 kg/s at a dischargevelocity of 8 m/s (Fig. 2–48). Determine if this claim is reasonable.

Solution A fan is claimed to increase the velocity of air to a specified valuewhile consuming electric power at a specified rate. The validity of this claimis to be investigated.Assumptions The ventilating room is relatively calm, and air velocity in it isnegligible.Analysis First, let’s examine the energy conversions involved: The motor ofthe fan converts part of the electrical power it consumes to mechanical(shaft) power, which is used to rotate the fan blades in air. The blades areshaped such that they impart a large fraction of the mechanical power of theshaft to air by mobilizing it. In the limiting ideal case of no losses (no con-version of electrical and mechanical energy to thermal energy) in steadyoperation, the electric power input will be equal to the rate of increase of the kinetic energy of air. Therefore, for a control volume that encloses the fan-motor unit, the energy balance can be written as

Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc., energies

Solving for Vout and substituting gives the maximum air outlet velocity to be

which is less than 8 m/s. Therefore, the claim is false.Discussion The conservation of energy principle requires the energy to bepreserved as it is converted from one form to another, and it does not allowany energy to be created or destroyed during a process. From the first lawpoint of view, there is nothing wrong with the conversion of the entire electri-cal energy into kinetic energy. Therefore, the first law has no objection to airvelocity reaching 6.3 m/s—but this is the upper limit. Any claim of highervelocity is in violation of the first law, and thus impossible. In reality, the airvelocity will be considerably lower than 6.3 m/s because of the losses associ-ated with the conversion of electrical energy to mechanical shaft energy, andthe conversion of mechanical shaft energy to kinetic energy or air.

Vout � BW#

elect,in

2m#

air� B

20 J>s2 10.25 kg>s 2 a

1 m2>s2

1 J>kgb � 6.3 m>s

W#

elect, in � m#

air keout � m#

air V2

out

2

E#

in � E#

out � dE system > dt 0 1steady2 � 0 S E#

in � E#

out

Air

8 m/s Fan

FIGURE 2–48Schematic for Example 2–11.

© Vol. 0557/PhotoDisc→ ⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭⎫⎪⎪⎬⎪⎪⎭

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Guided Tour

Each chapter begins with an overview ofmaterial to be covered plus learningobjectives, which are tied to ABETobjectives. A summary is included at theend of each chapter, providing a quickreview of basic concepts and importantrelations, and pointing out the relevance ofthe material.

NEW: Conservation of mass is nowcovered together with conservation ofenergy in new Chapter 5, “Mass and EnergyAnalysis of Control Volumes.” A formalderivation of the general energy equation isalso given in this chapter as the Topic ofSpecial Interest.

REVISED: The chapter coveringcompressibility effects, “Compressible

Flow,” (Chapter 17) is greatly revised andexpanded. This chapter now includes

coverage of oblique shocks and flow with heattransfer (Rayleigh flow) with exciting

photographs and extended discussions ofshock waves.

HotWater

Coldwater

T-elbow

FIGURE 5–32The T-elbow of an ordinaryshower serves as the mixingchamber for the hot- and thecold-water streams.

Photo by G. S. Settles, Penn State University. Used by permission.

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Guided TourEvery chapter contains worked-out real-

world example problems. The authors use a consistent approach in problem

solving while maintaining their informalconversational style. This problem-solving

approach is also used throughout thesolutions presented in the Instructor’s

Solutions Manual.

Our text contains nearly 3000 homeworkproblems. End-of-chapter problems aregrouped under specific topics to makeproblem selection easier for both instructorsand students. Answers to selected problemsare listed immediately following theproblem for convenience to students.

EXAMPLE 2–13 Annual Lighting Cost of a Classroom

The lighting needs of a classroom are met by 30 fluorescent lamps, eachconsuming 80 W of electricity (Fig. 2–50). The lights in the classroom arekept on for 12 hours a day and 250 days a year. For a unit electricity cost of7 cents per kWh, determine annual energy cost of lighting for this class-room. Also, discuss the effect of lighting on the heating and air-conditioningrequirements of the room.

Solution The lighting of a classroom by fluorescent lamps is considered.The annual electricity cost of lighting for this classroom is to be deter-mined, and the lighting’s effect on the heating and air-conditioning require-ments is to be discussed.Assumptions The effect of voltage fluctuations is negligible so that each fluo-rescent lamp consumes its rated power.Analysis The electric power consumed by the lamps when all are on andthe number of hours they are kept on per year are

Then the amount and cost of electricity used per year become

Light is absorbed by the surfaces it strikes and is converted to thermal energy.Disregarding the light that escapes through the windows, the entire 2.4 kW ofelectric power consumed by the lamps eventually becomes part of thermalenergy of the classroom. Therefore, the lighting system in this room reducesthe heating requirements by 2.4 kW, but increases the air-conditioning load by2.4 kW.Discussion Note that the annual lighting cost of this classroom alone is over$500. This shows the importance of energy conservation measures. If incan-descent light bulbs were used instead of fluorescent tubes, the lighting costswould be four times as much since incandescent lamps use four times asmuch power for the same amount of light produced.

� 17200 kWh>year 2 1$0.07>kWh 2 � $504>year

Lighting cost � 1Lighting energy 2 1Unit cost 2 � 12.4 kW 2 13000 h>year 2 � 7200 kWh>year

Lighting energy � 1Lighting power 2 1Operating hours 2

Operating hours � 112 h>day 2 1250 days>year 2 � 3000 h>year

� 2400 W � 2.4 kW

� 180 W>lamp 2 130 lamps 2 Lighting power � 1Power consumed per lamp 2 � 1No. of lamps 2

77–132 Long cylindrical steel rods (r � 7833 kg/m3 and cp

� 0.465 kJ/kg · °C) of 10-cm diameter are heat treated bydrawing them at a velocity of 3 m/min through a 7-m-longoven maintained at 900°C. If the rods enter the oven at 30°Cand leave at 700°C, determine (a) the rate of heat transfer tothe rods in the oven and (b) the rate of entropy generationassociated with this heat transfer process.

3 m/min

Oven900°C

Stainlesssteel, 30°C

7 m

FIGURE P7–132

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