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

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In the next example, we follow up the discussion of Fig. 2.16 by considering two

alternative systems. This example highlights the need to account correctly for the heat

and work interactions occurring on the boundary as well as the energy change.

Analysis: An energy balance for the closed system takes the form

¢KE

1 ¢PE

1 ¢U 5 Q 2

where the kinetic and potential energy terms drop out by assumption 3. Then, writing DU in terms of specific

internal energies, the energy balance becomes

2 u

5 Q 2

where m is the system mass. Solving for Q

Q 5 m

2 u

The value of the work for this process is determined in the solution to part (a) of Example 2.1: W 5 117.6 kJ.

The change in internal energy is obtained using given data as

m1u

2 u

25 0.4 kg

255

5222 kJ

Substituting values

➋ Q 5222 1 17.6 524.4 kJ

➊ The given relationship between pressure and volume allows the process to

be represented by the path shown on the accompanying diagram. The area

under the curve represents the work. Since they are not properties, the

values of the work and heat transfer depend on the details of the process

and cannot be determined from the end states only.

➋ The minus sign for the value of Q means that a net amount of energy has

been transferred from the system to its surroundings by heat transfer.

If the gas undergoes a process for which pV 5 constant and

Du 5 0, determine the heat transfer, in kJ, keeping the initial pressure and

given volumes fixed. Ans. 20.79 kJ.

Ability to…

❑

define a closed system and

identify interactions on its

boundary.

❑

apply the closed-system

energy balance.

✓

Skills Developed

Considering Alternative Systems

c c c c EXAMPLE 2.3 c

Air is contained in a vertical piston–cylinder assembly fitted with an electrical resistor. The atmosphere exerts

a pressure of 14.7 lbf/in.

on the top of the piston, which has a mass of 100 lb and a face area of 1 ft

. Electric

current passes through the resistor, and the volume of the air slowly increases by 1.6 ft

while its pressure remains

constant. The mass of the air is 0.6 lb, and its specific internal energy increases by 18 Btu/lb. The air and piston

are at rest initially and finally. The piston–cylinder material is a ceramic composite and thus a good insulator.

Friction between the piston and cylinder wall can be ignored, and the local acceleration of gravity is g 5 32.0 ft/s

Determine the heat transfer from the resistor to the air, in Btu, for a system consisting of (a) the air alone,

(b) the air and the piston.

SOLUTION

Known:

Data are provided for air contained in a vertical piston–cylinder fitted with an electrical resistor.

Find: Considering each of two alternative systems, determine the heat transfer from the resistor to the air.

2.5 Energy Accounting: Energy Balance for Closed Systems 63

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64 Chapter 2

Energy and the First Law of Thermodynamics

Engineering Model:

Two closed systems are under consideration, as shown in the schematic.

2. The only significant heat transfer is from the resistor to the air, during which the air expands slowly and

its pressure remains constant.

3. There is no net change in kinetic energy; the change in potential energy of the air is negligible; and since the

piston material is a good insulator, the internal energy of the piston is not affected by the heat transfer.

4. Friction between the piston and cylinder wall is negligible.

5. The acceleration of gravity is constant; g 5 32.0 ft/s

Analysis: (a) Taking the air as the system, the energy balance, Eq. 2.35, reduces with assumption 3 to

¢KE

1 ¢PE

1 ¢U

Or, solving for Q

5 W 1 ¢U

For this system, work is done by the force of the pressure p acting on the bottom of the piston as the air

expands. With Eq. 2.17 and the assumption of constant pressure

W 5

p dV 5 p1V

2 V

To determine the pressure p, we use a force balance on the slowly moving, frictionless piston. The upward

force exerted by the air on the bottom of the piston equals the weight of the piston plus the downward force of

the atmosphere acting on the top of the piston. In symbols

iston

5 m

iston

g 1 p

atm

iston

Solving for p and inserting values

piston

iston

1 p

atm

1100 lb2132.0 ft

1 ft

1 lbf

32.2 lb ? ft

1 ft

144 in.

`1 14.7

lbf

in.

5 15.4

lbf

in.

Thus, the work is

W 5 p

2 V

5 a15.4

lbf

in.

b11.6 ft

144 in.

1 ft

1 Btu

778 ft ? lbf

`5 4.56 Btu

With DU

air

5 m

air

(Du

air

), the heat transfer is

Q 5 W 1 m

air

¢u

air

5 4.56 Btu 1 10.6 lb2

Btu

5 15.36 Btu

Schematic and Given Data:

Air

Piston

System

boundary

for part (a)

piston

= 1 ft

piston

= 100 lb

atm

= 14.7 lbf/in.

air

= 0.6 lb

– V

= 1.6 ft

Δu

air

= 18 Btu/lb

(a)

–

Air

Piston

System

boundary

for part (b)

(b)

–

Fig. E2.3

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(b) Consider next a system consisting of the air and the piston. The energy change of the overall system is the

sum of the energy changes of the air and the piston. Thus, the energy balance, Eq. 2.35, reads

1¢KE

1 ¢PE

1 ¢U2

air

1 1¢KE

1 ¢PE 1 ¢U

iston

5 Q 2

where the indicated terms drop out by assumption 3. Solving for Q

Q 5 W 1

¢PE

iston

¢U

air

For this system, work is done at the top of the piston as it pushes aside the surrounding atmosphere. Applying

Eq. 2.17

W 5

p dV 5 p

atm

2 V

5 a14.7

lbf

in.

b11.6 ft

144 in.

1 ft

1 Btu

778 ft ? lbf

`5 4.35 Btu

The elevation change, Dz, required to evaluate the potential energy change of the piston can be found from

the volume change of the air and the area of the piston face as

¢z 5

2 V

iston

1.6 ft

1 ft

5 1.6 ft

Thus, the potential energy change of the piston is

¢PE

iston

5 m

iston

g¢z

5 1100 lb2

32.0

11.6 ft2

1 lbf

32.2 lb ? ft

1 Btu

778 ft ? lbf

5 0.2 Btu

Finally,

Q 5 W 1

¢PE

iston

1 m

air

¢u

air

5 4.35 Btu 1 0.2 Btu 1 10.6 lb2

Btu

5 15.35 Btu

➊ ➋ To within round-off, this answer agrees with the result of part (a).

➊ Although the value of Q is the same for each system, observe that the values

for W differ. Also, observe that the energy changes differ, depending on

whether the air alone or the air and the piston is the system.

➋ For the system of part (b), the following energy balance sheet gives a full

accounting of the heat transfer of energy to the system:

Energy In by Heat Transfer

15.35 Btu

Disposition of the Energy In

• Energy stored

Internal energy of the air 10.8 Btu (70.4%)

Potential energy of the piston 0.2 Btu ( 1.3%)

• Energy out by work 4.35 Btu (28.3%)

15.35 Btu (100%)

What is the change in potential energy of the air, in Btu?

Ans. <

Btu.

Ability to…

❑

define alternative closed

systems and identify inter-

actions on their boundaries.

❑

evaluate work using

Eq. 2.17.

❑

apply the closed-system

energy balance.

❑

develop an energy balance

sheet.

✓

Skills Developed

2.5 Energy Accounting: Energy Balance for Closed Systems 65

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66 Chapter 2 Energy and the First Law of Thermodynamics

2.5.3

Using the Energy Rate Balance: Steady-State Operation

A system is at steady state if none of its properties change with time (Sec. 1.3). Many

devices operate at steady state or nearly at steady state, meaning that property vari-

ations with time are small enough to ignore. The two examples to follow illustrate

the application of the energy rate equation to closed systems at steady state.

Evaluating Energy Transfer Rates of a Gearbox at Steady State

c c c c EXAMPLE 2.4 c

During steady-state operation, a gearbox receives 60 kW through the input shaft and delivers power through the

output shaft. For the gearbox as the system, the rate of energy transfer by convection is

52hA

2 T

where h 5 0.171 kW/m

K is the heat transfer coefficient, A 5 1.0 m

is the outer surface area of the gearbox,

5 300 K (278C) is the temperature at the outer surface, and T

5 293 K (208C) is the temperature of the

surrounding air away from the immediate vicinity of the gearbox. For the gearbox, evaluate the heat transfer

rate and the power delivered through the output shaft, each in kW.

SOLUTION

Known:

A gearbox operates at steady state with a known power input. An expression for the heat transfer rate

from the outer surface is also known.

Find: Determine the heat transfer rate and the power delivered through the output shaft, each in kW.

Schematic and Given Data:

Engineering Model:

The gearbox is a closed system at steady state.

2. For the gearbox, convection is the dominant heat

transfer mode.

Fig. E2.4

= 300 K

Gearbox

Outer surface

Input

shaft

Output

shaft

A = 1.0 m

= 293 K

h = 0.171 kW/m

· K

= –60 kW

➊ Q

52hA

2 T

0.171

? K

11.0 m

21300 2 2932 K

521.2 kW

The minus sign for

signals that energy is carried out of the gearbox by heat transfer.

The energy rate balance, Eq. 2.37, reduces at steady state to

➋

5 Q

2 W



5 Q

The symbol

represents the net power from the system. The net power is the sum of W

and the output power W

5 W

1 W

Analysis: Using the given expression for Q

together with known data, the rate of energy transfer by heat is

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With this expression for

, the energy rate balance becomes

1 W

Solving for W

, inserting

521.2 kW, and W

5260 kW, where the minus sign is required because the input

shaft brings energy into the system, we have

➌ W

2 W

21.2 kW

260 kW

➍ The positive sign for W

indicates that energy is transferred from the system through the output shaft, as

expected.

➊ In accord with the sign convention for the heat transfer rate in the energy

rate balance (Eq. 2.37), Eq. 2.34 is written with a minus sign:

is negative

since T

is greater than T

➋ Properties of a system at steady state do not change with time. Energy E is

a property, but heat transfer and work are not properties.

➌ For this system, energy transfer by work occurs at two different locations,

and the signs associated with their values differ.

➍ At steady state, the rate of heat transfer from the gear box accounts for the

difference between the input and output power. This can be summarized by

the following energy rate “balance sheet” in terms of magnitudes:

Input Output

60 kW (input shaft) 58.8 kW (output shaft)

1.2 kW (heat transfer)

Total: 60 kW 60 kW

Ability to…

❑

define a closed system and

identify interactions on its

boundary.

❑

evaluate the rate of energy

transfer by convection.

❑

apply the energy rate balance

for steady-state operation.

❑

develop an energy rate

balance sheet.

✓

Skills Developed

For an emissivity of 0.8 and taking T

5 T

, use Eq. 2.33 to

determine the net rate at which energy is radiated from the outer surface

of the gearbox, in kW. Ans. 0.03 kW.

Determining Surface Temperature of a Silicon Chip at Steady State

c c c c EXAMPLE 2.5 c

A silicon chip measuring 5 mm on a side and 1 mm in thickness is embedded in a ceramic substrate. At steady state,

the chip has an electrical power input of 0.225 W. The top surface of the chip is exposed to a coolant whose tem-

perature is 208C. The heat transfer coefficient for convection between the chip and the coolant is 150 W/m

K. If

heat transfer by conduction between the chip and the substrate is negligible, determine the surface temperature of

the chip, in 8C.

SOLUTION

Known:

A silicon chip of known dimensions is exposed on its top surface to a coolant. The electrical power input

and convective heat transfer coefficient are known.

Find: Determine the surface temperature of the chip at steady state.

2.5 Energy Accounting: Energy Balance for Closed Systems 67

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68 Chapter 2 Energy and the First Law of Thermodynamics

2.5.4

Using the Energy Rate Balance: Transient Operation

Many devices undergo periods of transient operation where the state changes with

time. This is observed during startup and shutdown periods. The next example

illustrates the application of the energy rate balance to an electric motor during

startup. The example also involves both electrical work and power transmitted by

a shaft.

Analysis: The surface temperature of the chip, T

, can be determined using the energy rate balance, Eq. 2.37,

which at steady state reduces as follows

➊

5 Q

With assumption 2, the only heat transfer is by convection to the coolant. In this application, Newton’s law of

cooling, Eq. 2.34, takes the form

➋ Q

52hA1T

2 T

Collecting results

0 52hA1T

2 T

Solving for T

1 T

In this expression, W

520.225 W, A 5 25 3 10

, h 5 150 W/m

K, and T

5 293 K, giving

20.225 W

150 W

? K

25 3 10

1 293 K

5 353 K 1808C2

➊ Properties of a system at steady state do not change with time. Energy E is

a property, but heat transfer and work are not properties.

➋ In accord with the sign convention for heat transfer in the energy rate bal-

ance (Eq. 2.37), Eq. 2.34 is written with a minus sign:

is negative since T

is greater than T

Ability to…

❑

define a closed system and

identify interactions on its

boundary.

❑

evaluate the rate of energy

transfer by convection.

❑

apply the energy rate balance

for steady-state operation.

✓

Skills Developed

If the surface temperature of the chip must be no greater than

608C, what is the corresponding range of values required for the convec-

tive heat transfer coefficient, assuming all other quantities remain

unchanged? Ans. h $ 225 W

? K.

Engineering Model:

The chip is a closed system at steady state.

2. There is no heat transfer between the chip and the substrate.

Ceramic substrate

5 mm

1 mm

W = –0.225 W

= 20° C

h = 150 W/m

· K

Coolant

–

Fig. E2.5

Schematic and Given Data:

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2.5 Energy Accounting: Energy Balance for Closed Systems 69

Investigating Transient Operation of a Motor

c c c c EXAMPLE 2.6 c

The rate of heat transfer between a certain electric motor and its surroundings varies with time as

520.2

1 2 e

20.05t

where t is in seconds and

is in kW. The shaft of the motor rotates at a constant speed of v 5 100 rad/s (about

955 revolutions per minute, or RPM) and applies a constant torque of

5 18 N

m to an external load. The

motor draws a constant electric power input equal to 2.0 kW. For the motor, plot

and

, each in kW, and the

change in energy DE, in kJ, as functions of time from t 5 0 to t 5 120 s. Discuss.

SOLUTION

Known:

A motor operates with constant electric power input, shaft speed, and applied torque. The time-varying

rate of heat transfer between the motor and its surroundings is given.

Find: Plot

, and DE versus time, Discuss.

Analysis: The time rate of change of system energy is

5 Q

represents the net power from the system: the sum of the power associated with the rotating shaft, W

, and the

power associated with the electricity flow, W

5 W

1 W

The rate W

is known from the problem statement: W

522.0 kW, where the negative sign is required because

energy is carried into the system by electrical work. The term W

can be evaluated with Eq. 2.20 as

shaft

5 tv 5

18 N ? m

100 rad

5 1800 W 511.8 kW

Because energy exits the system along the rotating shaft, this energy transfer rate is positive.

In summary,

5 W

1 W

22.0 kW

11.8 kW

520.2 kW

where the minus sign means that the electrical power input is greater than the power transferred out along the shaft.

With the foregoing result for

and the given expression for

, the energy rate balance becomes

520.231 2 e

120.05t2

42 120.225 0.2e

120.05t2

Integrating

¢E 5

0.2e

120.05t2

0.2

120.052

120.05t2

5 431 2 e

120.05t2

➊ The accompanying plots, Figs. E2.6b and c, are developed using the given expression for

and the expressions

for

and DE obtained in the analysis. Because of our sign conventions for heat and work, the values of

Engineering Model: The system shown in the accompanying

sketch is a closed system.

–

Motor

elec

= –2.0 kW

shaft

Q = –0.2 [1 – e

(–0.05t)

] kW

= 100 rad/s

= 18 N · m

Fig. E2.6a

Schematic and Given Data:

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70 Chapter 2 Energy and the First Law of Thermodynamics

2.6 Energy Analysis of Cycles

In this section the energy concepts developed thus far are illustrated further by appli-

cation to systems undergoing thermodynamic cycles. A thermodynamic cycle is a

sequence of processes that begins and ends at the same state. At the conclusion of a

cycle all properties have the same values they had at the beginning. Consequently,

over the cycle the system experiences no net change of state. Cycles that are repeated

periodically play prominent roles in many areas of application. For example, steam

circulating through an electrical power plant executes a cycle.

The study of systems undergoing cycles has played an important role in the devel-

opment of the subject of engineering thermodynamics. Both the first and second laws

of thermodynamics have roots in the study of cycles. Additionally, there are many

important practical applications involving power generation, vehicle propulsion, and

refrigeration for which an understanding of thermodynamic cycles is essential. In this

section, cycles are considered from the perspective of the conservation of energy

principle. Cycles are studied in greater detail in subsequent chapters, using both the

conservation of energy principle and the second law of thermodynamics.

thermodynamic cycle

and

are negative. In the first few seconds, the net rate that energy is carried in by work greatly exceeds the

rate that energy is carried out by heat transfer. Consequently, the energy stored in the motor increases rapidly

as the motor “warms up.” As time elapses, the value of

approaches

, and the rate of energy storage dimin-

ishes. After about 100 s, this transient operating mode is nearly over, and there is little further change in the

amount of energy stored, or in any other property. We may say that the motor

is then at steady state.

➊ Figures E.2.6b and c can be developed using appropriate software or can be

drawn by hand.

➋ At steady state, the value of

is constant at 20.2 kW. This constant value

for the heat transfer rate can be thought of as the portion of the electrical

power input that is not obtained as a mechanical power output because of

effects within the motor such as electrical resistance and friction.

0 1008060 904020 503010 70

0 1008060 904020 503010

–0.25

–0.20

–0.15

–0.10

–0.05

Time, s Time, s

ΔE, kJ

, , kWW

Fig. E2.6b and c

Ability to…

❑

define a closed system and

identify interactions on its

boundary.

❑

apply the energy rate bal-

ance for transient operation.

❑

develop and interpret graphical

data.

✓

Skills Developed

If the dominant mode of heat transfer from the outer surface of

the motor is convection, determine at steady state the temperature T

the outer surface, in K, for h 5 0.17 kW/m

K, A 5 0.3 m

, and T

5 293 K.

Ans. 297 K.

➋

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2.6.1

Cycle Energy Balance

The energy balance for any system undergoing a thermodynamic cycle takes the form

¢E

cycle

5 Q

cycle

2 W

cycle

(2.39)

where Q

cycle

and W

cycle

represent net amounts of energy transfer by heat and work,

respectively, for the cycle. Since the system is returned to its initial state after the

cycle, there is no net change in its energy. Therefore, the left side of Eq. 2.39 equals

zero, and the equation reduces to

cycle

5 Q

cycle

(2.40)

Equation 2.40 is an expression of the conservation of energy principle that must be sat-

isfied by every thermodynamic cycle, regardless of the sequence of processes followed by

the system undergoing the cycle or the nature of the substances making up the system.

Figure 2.17 provides simplified schematics of two general classes of cycles considered

in this book: power cycles and refrigeration and heat pump cycles. In each case pictured,

a system undergoes a cycle while communicating thermally with two bodies, one hot

and the other cold. These bodies are systems located in the surroundings of the system

undergoing the cycle. During each cycle there is also a net amount of energy exchanged

with the surroundings by work. Carefully observe that in using the symbols Q

and

out

on Fig. 2.17 we have departed from the previously stated sign convention for heat

transfer. In this section it is advantageous to regard Q

and Q

out

as transfers of energy

in the directions indicated by the arrows. The direction of the net work of the cycle,

cycle

, is also indicated by an arrow. Finally, note that the directions of the energy trans-

fers shown in Fig. 2.17b are opposite to those of Fig. 2.17a.

2.6.2

Power Cycles

Systems undergoing cycles of the type shown in Fig. 2.17a deliver a net work transfer

of energy to their surroundings during each cycle. Any such cycle is called a power cycle.

From Eq. 2.40, the net work output equals the net heat transfer to the cycle, or

cycle

5 Q

2 Q

out



1power cycle2 (2.41)

where Q

represents the heat transfer of energy into the system from the hot body, and

out

represents heat transfer out of the system to the cold body. From Eq. 2.41 it is clear

TAKE NOTE...

When analyzing cycles, we

normally take energy trans-

fers as positive in the direc-

tions of arrows on a sketch

of the system and write the

energy balance accordingly.

power cycle

2.6 Energy Analysis of Cycles 71

Fig. 2.17

Schematic

diagrams of two

important classes of

cycles. (a) Power cycles.

(b) Refrigeration and heat

pump cycles.

System

Cold

body

Hot

body

cycle

= Q

– Q

out

(a)

System

Cold

body

Hot

body

cycle

= Q

out

– Q

out

(b)

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72 Chapter 2

Energy and the First Law of Thermodynamics

that Q

must be greater than Q

out

for a power cycle. The energy supplied by heat transfer

to a system undergoing a power cycle is normally derived from the combustion of fuel or

a moderated nuclear reaction; it can also be obtained from solar radiation. The energy

out

is generally discharged to the surrounding atmosphere or a nearby body of water.

The performance of a system undergoing a power cycle can be described in terms

of the extent to which the energy added by heat, Q

, is converted to a net work

output, W

cycle

. The extent of the energy conversion from heat to work is expressed by

the following ratio, commonly called the thermal efficiency

h 5

cycle



1power cycle2

(2.42)

Introducing Eq. 2.41, an alternative form is obtained as

h 5

2 Q

out

5 1 2

out



1power cycle2

(2.43)

Since energy is conserved, it follows that the thermal efficiency can never be greater

than unity (100%). However, experience with actual power cycles shows that the value

of thermal efficiency is invariably less than unity. That is, not all the energy added to the

system by heat transfer is converted to work; a portion is discharged to the cold body

by heat transfer. Using the second law of thermodynamics, we will show in Chap. 5 that

the conversion from heat to work cannot be fully accomplished by any power cycle. The

thermal efficiency of every power cycle must be less than unity: h , 1 (100%).

thermal efficiency

2.6.3

Refrigeration and Heat Pump Cycles

Next, consider the refrigeration and heat pump cycles shown in Fig. 2.17b. For cycles

of this type, Q

is the energy transferred by heat into the system undergoing the cycle

from the cold body, and Q

out

is the energy discharged by heat transfer from the sys-

tem to the hot body. To accomplish these energy transfers requires a net work input,

cycle

. The quantities Q

, Q

out

, and W

cycle

are related by the energy balance, which for

refrigeration and heat pump cycles takes the form

cycle

5 Q

out

2 Q



1refrigeration and heat pump cycles2 (2.44)

Since W

cycle

is positive in this equation, it follows that Q

out

is greater than Q

refrigeration and heat

pump cycles

ENERGY & ENVIRONMENT Today fossil-fueled power plants can have ther-

mal efficiencies of 40%, or more. This means that up to 60% of the energy added by

heat transfer during the power plant cycle is discharged from the plant other than by

work, principally by heat transfer. One way power plant cooling is achieved is to use water drawn

from a nearby river or lake. The water is eventually returned to the river or lake but at a higher

temperature, which is a practice having several possible environmental consequences.

The return of large quantities of warm water to a river or lake can affect its ability to hold dis-

solved gases, including the oxygen required for aquatic life. If the return water temperature is

greater than about 35°C (95°F), the dissolved oxygen may be too low to support some species of

fish. If the return water temperature is too great, some species also can be stressed. As rivers and

lakes become warmer, non-native species that thrive in the warmth can take over. Warmer water

also fosters bacterial populations and algae growth.

Regulatory agencies have acted to limit warm water discharges from power plants, which has

made cooling towers (Sec. 12.9) adjacent to power plants a common sight.

Power_Cycle

A.9 – Tabs a & b

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