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

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In the next two examples we consider cases where tanks are filled. In Example

4.12, an initially evacuated tank is filled with steam as power is developed. In Exam-

ple 4.13, a compressor is used to store air in a tank.

Substituting values into the expression for the heat transfer yields

5 120.14212599.022 128.4212157.822 2796.6120.14 2 28.42

5 14,162 kJ

➊ In this case, idealizations are made about the state of the vapor exiting and

the initial and final states of the mass contained within the tank.

➋ This expression for Q

can be obtained by applying Eqs. 4.23, 4.25, and

4.27. The details are left as an exercise.

Ability to…

❑

apply the time-dependent

mass and energy rate bal-

ances to a control volume.

❑

develop an engineering model.

❑

retrieve property data for

water.

✓Skills Developed

If the initial quality were 90%, determine the heat transfer, in

kJ, keeping all other data unchanged. Ans. 3707 kJ.

Using Steam for Emergency Power Generation

c c c c EXAMPLE 4.12 c

Steam at a pressure of 15 bar and a temperature of 3208C is contained in a large vessel. Connected to

the vessel through a valve is a turbine followed by a small initially evacuated tank with a volume of 0.6 m

When emergency power is required, the valve is opened and the tank fills with steam until the pressure

is 15 bar. The temperature in the tank is then 4008C. The filling process takes place adiabatically and

kinetic and potential energy effects are negligible. Determine the amount of work developed by the turbine,

in kJ.

SOLUTION

Known:

Steam contained in a large vessel at a known state flows from the vessel through a turbine into a small

tank of known volume until a specified final condition is attained in the tank.

Find: Determine the work developed by the turbine.

Schematic and Given Data:

Engineering Model:

The control volume is defined by the dashed line on the

accompanying diagram.

2. For the control volume, Q

5 0 and kinetic and potential

energy effects are negligible.

3. The state of the steam within the large vessel remains

constant. The final state of the steam in the smaller tank

is an equilibrium state.

4. The amount of mass stored within the turbine and the

interconnecting piping at the end of the filling process is

negligible.

Turbine

Initially

evacuated

tank

V =

0.6 m

Steam at

15 bar,

320°C

Control volume boundaryValve

Fig. E4.12

➊

4.12 Transient Analysis 203

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204 Chapter 4 Control Volume Analysis Using Energy

If the turbine were removed, and the steam allowed to flow

adiabatically into the small tank until the pressure in the tank is 15 bar,

determine the final steam temperature in the tank, in 8C. Ans. 4778C.

Analysis: Since the control volume has a single inlet and no exits, the mass rate balance, Eq. 4.2, reduces to

5 m

The energy rate balance, Eq. 4.15, reduces with assumption 2 to

52W

1 m

Combining the mass and energy rate balances gives

52W

1 h

Integrating

¢U

52W

1 h

¢m

In accordance with assumption 3, the specific enthalpy of the steam entering the control volume is constant at

the value corresponding to the state in the large vessel.

Solving for W

5 h

¢m

2 ¢U

and Dm

denote, respectively, the changes in internal energy and mass of the control volume. With assump-

tion 4, these terms can be identified with the small tank only.

Since the tank is initially evacuated, the terms DU

and Dm

reduce to the internal energy and mass within

the tank at the end of the process. That is

¢U

5 1m

22 1m

, ¢m

5 m

2 m

where 1 and 2 denote the initial and final states within the tank, respectively.

Collecting results yields

5 m

2 u

(a)

The mass within the tank at the end of the process can be evaluated from the known volume and the specific

volume of steam at 15 bar and 4008C from Table A-4

0.6 m

10.203 m

kg2

5 2.96 kg

The specific internal energy of steam at 15 bar and 4008C from Table A-4 is 2951.3 kJ/kg. Also, at 15 bar and

3208C, h

5 3081.9 kJ/kg.

Substituting values into Eq. (a)

5 2.96 k

3081.9 2 2951.3

5 386.6 kJ

➊ In this case idealizations are made about the state of the steam entering the

tank and the final state of the steam in the tank. These idealizations make

the transient analysis manageable.

➋ A significant aspect of this example is the energy transfer into the control

volume by flow work, incorporated in the py term of the specific enthalpy

at the inlet.

➌ This result can also be obtained by reducing Eq. 4.28. The details are left as

an exercise.

Ability to…

❑

apply the time-dependent

mass and energy rate bal-

ances to a control volume.

❑

develop an engineering model.

❑

retrieve property data for

water.

✓

Skills Developed

➌

➋

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Storing Compressed Air in a Tank

c c c c EXAMPLE 4.13 c

An air compressor rapidly fills a 10-ft

tank, initially containing air at 708F, 1 atm, with air drawn from the atmo-

sphere at 708F, 1 atm. During filling, the relationship between the pressure and specific volume of the air in the

tank is py

1.4

5 constant. The ideal gas model applies for the air, and kinetic and potential energy effects are

negligible. Plot the pressure, in atm, and the temperature, in 8F, of the air within the tank, each versus the ratio

m/m

, where m

is the initial mass in the tank and m is the mass in the tank at time t . 0. Also, plot the com-

pressor work input, in Btu, versus m/m

. Let m/m

vary from 1 to 3.

SOLUTION

Known:

An air compressor rapidly fills a tank having a known volume. The initial state of the air in the tank

and the state of the entering air are known.

Find: Plot the pressure and temperature of the air within the tank, and plot the air compressor work input, each

versus m/m

ranging from 1 to 3.

Schematic and Given Data:

Fig. E4.13a

Air

Air compressor

Tank

–

V = 10 ft

= 70°F

= 1 atm

1.4

= constant

= 70°F

= 1 atm

Engineering Model:

The control volume is defined by the dashed line on

the accompanying diagram.

2. Because the tank is filled rapidly, Q

is ignored.

3. Kinetic and potential energy effects are negligible.

4. The state of the air entering the control volume

remains constant.

5. The air stored within the air compressor and inter-

connecting pipes can be ignored.

6. The relationship between pressure and specific

volume for the air in the tank is py

1.4

5 constant.

7. The ideal gas model applies for the air.

➊

Analysis: The required plots are developed using Interactive Thermodynamics: IT. The IT program is based on

the following analysis. The pressure p in the tank at time t . 0 is determined from

1.4

5 p

1.4

where the corresponding specific volume y is obtained using the known tank volume V and the mass m in the

tank at that time. That is, y 5 V/m. The specific volume of the air in the tank initially, y

, is calculated from the

ideal gas equation of state and the known initial temperature, T

, and pressure, p

. That is

1545 ft ? lbf

28.97 lb ? 8R

b 15308R2

114.7 lbf

in.

1 ft

144 in.

`5 13.35

Once the pressure p is known, the corresponding temperature T can be found from the ideal gas equation of

state, T 5 py/R.

To determine the work, begin with the mass rate balance Eq. 4.2, which reduces for the single-inlet control

volume to

5 m

4.12 Transient Analysis 205

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206 Chapter 4

Control Volume Analysis Using Energy

Then, with assumptions 2 and 3, the energy rate balance Eq. 4.15 reduces to

52W

1 m

Combining the mass and energy rate balances and integrating using assumption 4 gives

¢U

52W

1 h

¢m

Denoting the work input to the compressor by W

5 2W

and using assumption 5, this becomes

5 mu 2 m

2 1m 2 m

(a)

where m

is the initial amount of air in the tank, determined from

10 ft

13.35 ft

5 0.75 lb

As a sample calculation to validate the IT program below, consider the case m 5 1.5 lb, which corresponds

to m/m

5 2. The specific volume of the air in the tank at that time is

y 5

10 ft

1.5 lb

5 6.67

The corresponding pressure of the air is

5 p

1.4

5 11 atm2a

13.35 ft

6.67 ft

1.4

5 2.64 atm

and the corresponding temperature of the air is

T 5

12.64 atm216.67 ft

lb2

1545 ft ?

lbf

28.97 lb ? 8R

14.7 lbf

in.

1 atm

144 in.

1 ft

5 6998R 12398F2

Evaluating u

, u, and h

at the appropriate temperatures from Table A-22E, u

= 90.3 Btu/lb, u 5 119.4 Btu/lb,

5 126.7 Btu/lb. Using Eq. (a), the required work input is

5 mu 2 m

m 2 m

5 11.5 lb2 a119.4

Btu

b2 10.75 lb2 a90.3

Btu

b2 10.75 lb2 a126.7

Btu

5 16.4 Btu

IT Program: Choosing English units from the Units menu, and selecting Air from the Properties menu, the IT

program for solving the problem is

//Given Data

p1 5 1//atm

T1 5 70//8F

Ti 5 70//8F

V 5 10//ft

n 5 1.4

// Determine the pressure and temperature for t . 0

v1 5 v_TP(“Air”, T1, p1)

v 5 V/m

p * v ^n 5 p1 * v1 ^n

v 5 v_TP(“Air”, T, p)

// Specify the mass and mass ratio r

v1 5 V/m1

r 5 m/m1

r 5 2

➋

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4.12 Transient Analysis 207

The final example of transient analysis is an application with a well-stirred tank.

Such process equipment is commonly employed in the chemical and food processing

industries.

// Calculate the work using Eq. (a)

Win 5 m * u – m1 * u1 – hi * (m – m1)

u1 5 u_T(“Air”, T1)

u 5 u_T(“Air”, T)

hi 5 h_T(“Air”, Ti)

Using the Solve button, obtain a solution for the sample case r 5 m/m

5 2 considered above to validate the program.

Good agreement is obtained, as can be verified. Once the program is validated, use the Explore button to vary the

ratio m/m

from 1 to 3 in steps of 0.01. Then, use the Graph button to construct the required plots. The results are:

We conclude from the first two plots that the pressure and temperature each

increase as the tank fills. The work required to fill the tank increases as well.

These results are as expected.

➊ This pressure-specific volume relationship is in accord with what might be

measured. The relationship is also consistent with the uniform state idealiza-

tion, embodied by Eqs. 4.26 and 4.27.

➋ This expression can also be obtained by reducing Eq. 4.28. The details are

left as an exercise.

As a sample calculation, for the case m 5 2.25 lb, evaluate p, in

atm. Compare with the value read from the plot of Fig. E4.13b. Ans. 4.67 atm.

Ability to…

❑

apply the time-dependent

mass and energy rate bal-

ances to a control volume.

❑

develop an engineering model.

❑

retrieve property data for

air modeled as an ideal gas.

❑

solve iteratively and plot

the results using IT.

✓

Skills Developed

Fig. E4.13b

1 1.5 2 2.5 3

p, atm

m/m

1 1.5 2 2.5 3

m/m

1 1.5 2 2.5 3

m/m

100

150

200

250

300

350

400

T, °F

, Btu

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208 Chapter 4 Control Volume Analysis Using Energy

Determining Temperature-Time Variation in a Well-Stirred Tank

c c c c EXAMPLE 4.14 c

A tank containing 45 kg of liquid water initially at 458C has one inlet and one exit with equal mass flow rates.

Liquid water enters at 458C and a mass flow rate of 270 kg/h. A cooling coil immersed in the water removes

energy at a rate of 7.6 kW. The water is well mixed by a paddle wheel so that the water temperature is uniform

throughout. The power input to the water from the paddle wheel is 0.6 kW. The pressures at the inlet and exit

are equal and all kinetic and potential energy effects can be ignored. Plot the variation of water temperature

with time.

SOLUTION

Known:

Liquid water flows into and out of a well-stirred tank with equal mass flow rates as the water in the

tank is cooled by a cooling coil.

Find: Plot the variation of water temperature with time.

Schematic and Given Data:

Engineering Model:

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

2. For the control volume, the only significant heat transfer is with the cooling coil. Kinetic and poten-

tial energy effects can be neglected.

3. The water temperature is uniform with position throughout and varies only with time: T 5 T(t).

4. The water in the tank is incompressible, and there is no change in pressure between inlet and exit.

➊

= 270 kg/h

= 318 K (45°C)

Tank

Cooling coil

= 270 kg/h

Mixing

rotor

Constant

liquid level

Boundary

318

296

Water temperature, K

0 0.5 1.0

Time, h

Fig. E4.14

Analysis: The energy rate balance, Eq. 4.15, reduces with assumption 2 to

5 Q

2 W

1 m

2 h

where

denotes the mass flow rate.

The mass contained within the control volume remains constant with time, so the term on the left side of the

energy rate balance can be expressed as

d1m

5 m

Since the water is assumed incompressible, the specific internal energy depends on temperature only. Hence, the

chain rule can be used to write

5 c

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Chapter Summary and Study Guide 209

The conservation of mass and energy principles for control vol-

umes are embodied in the mass and energy rate balances devel-

oped in this chapter. Although the primary emphasis is on cases

in which one-dimensional flow is assumed, mass and energy

balances are also presented in integral forms that provide a link

to subsequent fluid mechanics and heat transfer courses. Control

volumes at steady state are featured, but discussions of tran-

sient cases are also provided.

c CHAPTER SUMMARY AND STUDY GUIDE

What is the water temperature, in 8C, when steady-state is

achieved? Ans. 238C.

where c is the specific heat. Collecting results

5 m

With Eq. 3.20b the enthalpy term of the energy rate balance can be expressed as

2 h

5 c1T

2 T

21 y1p

2 p

where the pressure term is dropped by assumption 4. Since the water is well mixed, the temperature at the exit

equals the temperature of the overall quantity of liquid in the tank, so

2 h

5 c

2 T

where T represents the uniform water temperature at time t.

With the foregoing considerations the energy rate balance becomes

5 Q

2 W

1 m

c1T

2 T2

As can be verified by direct substitution, the solution of this first-order, ordinary differential equation is

T 5 C

exp a2

tb1 a

2 W

b1 T

The constant C

is evaluated using the initial condition: at t = 0, T 5 T

. Finally

T 5 T

1 a

2 W

bc1 2 expa2

tbd

Substituting given numerical values together with the specific heat c for liquid water from Table A-19

T 5 318 K 1

327.6 2 120.624 kJ

3600

ba4.2

? K

c1 2 expa2

270 kg

45 kg

tbd

5 318 2 2231 2 exp126t24

where t is in hours. Using this expression, we construct the accompanying plot

showing the variation of temperature with time.

➊ In this case idealizations are made about the state of the mass contained

within the system and the states of the liquid entering and exiting. These

idealizations make the transient analysis manageable.

Ability to…

❑

apply the time-dependent

mass and energy rate bal-

ances to a control volume.

❑

develop an engineering model.

❑

apply the incompressible

substance model for water.

❑

solve an ordinary differential

equation and plot the solution.

✓

Skills Developed

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210 Chapter 4 Control Volume Analysis Using Energy

The use of mass and energy balances for control volumes at

steady state is illustrated for nozzles and diffusers, turbines, com-

pressors and pumps, heat exchangers, throttling devices, and

integrated systems. An essential aspect of all such applications

is the careful and explicit listing of appropriate assumptions. Such

model-building skills are stressed throughout the chapter.

The following checklist provides a study guide for this chap-

ter. When your study of the text and end-of-chapter exercises has

been completed you should be able to

write out the meanings of the terms listed in the margins

throughout the chapter and understand each of the related

concepts. The subset of key concepts listed below is particu-

larly important in subsequent chapters.

list the typical modeling assumptions for nozzles and diffus-

ers, turbines, compressors and pumps, heat exchangers, and

throttling devices.

apply Eqs. 4.6, 4.18, and 4.20 to control volumes at steady

state, using appropriate assumptions and property data for

the case at hand.

apply mass and energy balances for the transient analysis of

control volumes, using appropriate assumptions and property

data for the case at hand.

c KEY ENGINEERING CONCEPTS

conservation of mass, p. 164

mass flow rates, p. 164

mass rate balance, p. 164

one-dimensional flow, p. 166

volumetric flow rate, p. 167

steady state, p. 167

flow work, p. 173

energy rate balance, p. 174

nozzle, p. 177

diffuser, p. 177

turbine, p. 180

compressor, p. 184

pump, p. 184

heat exchanger, p. 189

throttling process, p. 195

system integration, p. 196

transient analysis, p. 199

c KEY EQUATIONS

(4.4b) p. 166 Mass flow rate, one-dimen-

sional flow (See Fig. 4.3.)

(4.2) p. 164 Mass rate balance.

mass rate in

mass rate out

(4.6) p. 167

Mass rate balance at

steady

state.

5 Q

2 W

1 gz

b 2

1 gz

(4.15) p. 174 Energy rate balance.

0 5 Q

2 W

1 gz

b 2

1 gz

(4.18) p. 175

Energy rate balance at

steady state.

0 5 Q

2 W

1 m

2 h

2 V

1 g1z

2 z

(4.20a) p. 175

Energy rate balance for

one-inlet, one-exit control

volumes at steady state.

0 5

1 1h

2 h

2 V

1 g1z

2 z

(4.20b) p. 175

5 h





, p

(4.22) p. 195 Throttling process. (See

Fig. 4.15.)

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Problems: Developing Engineering Skills 211

c EXERCISES: THINGS ENGINEERS THINK ABOUT

1. How does the control volume energy rate balance account

for work where mass flows across the boundary?

2. Why does the relative velocity normal to the flow boundary,

, appear in Eqs. 4.3 and 4.8?

3. When a slice of bread is placed in a toaster and the toaster

is activated, is the toaster in steady-state operation, transient

operation, both?

4. As a tree grows, its mass increases. Does this violate the

conservation of mass principle? Explain.

5. Wind turbines and hydraulic turbines develop mechanical

power from moving streams of air and water, respectively. In

each case, what aspect of the stream is tapped for power?

6. How is the work done by the heart measured?

7. How does a heart-lung machine maintain blood circulation

and oxygen content during surgery?

8. Where do you encounter microelectromechanical systems in

daily life?

9. Where are compressors found within households?

10. How does the operator of a pumper-tanker fire engine

control water flow to all the hoses in use?

11. For air flowing through a converging-diverging channel,

sketch the variation of the air pressure as air accelerates in

the converging section and decelerates in the diverging

section.

12. Why is it that when air at 1 atm is throttled to a pressure

of 0.5 atm, its temperature at the valve exit is closely the

same as at the valve inlet, yet when air at 1 atm leaks into

an insulated, rigid, initially-evacuated tank until the tank

pressure is 0.5 atm, the temperature of the air in the tank is

greater than the air temperature outside the tank?

13. If the expansion valve of a refrigerator becomes ice encased,

does the throttling process model still apply? Explain.

14. Why does evapotranspiration in a tree require so much

energy?

15. What are intra-articular pain pumps?

c PROBLEMS: DEVELOPING ENGINEERING SKILLS

Applying Conservation of Mass

4.1 An 8-ft

tank contains air at an initial temperature of 808F

and initial pressure of 100 lbf/in.

The tank develops a small

hole, and air leaks from the tank at a constant rate of 0.03

lb/s for 90 s until the pressure of the air remaining in the tank

is 30 lbf/in.

Employing the ideal gas model, determine the

final temperature, in 8F, of the air remaining in the tank.

4.2 Liquid propane enters an initially empty cylindrical storage

tank at a mass flow rate of 10 kg/s. Flow continues until the

tank is filled with propane at 208C, 9 bar. The tank is 25-m

long and has a 4-m diameter. Determine the time, in minutes,

to fill the tank.

4.3 A 380-L tank contains steam, initially at 4008C, 3 bar. A

valve is opened, and steam flows out of the tank at a constant

mass flow rate of 0.005 kg/s. During steam removal, a heater

maintains the temperature within the tank constant. Determine

the time, in s, at which 75% of the initial mass remains in

the tank; also determine the specific volume, in m

/kg, and

pressure, in bar, in the tank at that time.

4.4 Data are provided for the crude oil storage tank shown in

Fig. P4.4. The tank initially contains 1000 m

of crude oil. Oil

is pumped into the tank through a pipe at a rate of 2 m

/min

and out of the tank at a velocity of 1.5 m/s through another

pipe having a diameter of 0.15 m. The crude oil has a specific

volume of 0.0015 m

/kg. Determine

(a) the mass of oil in the tank, in kg, after 24 hours, and

(b) the volume of oil in the tank, in m

, at that time.

4.5 If a kitchen-sink water tap leaks one drop per second, how

many gallons of water are wasted annually? What is the mass

of the wasted water, in lb? Assume that there are 46,000

drops per gallon and that the density of water is 62.3 lb/ft

4.6 Figure P4.6 shows a mixing tank initially containing 3000 lb

of liquid water. The tank is fitted with two inlet pipes, one

delivering hot water at a mass flow rate of 0.8 lb/s and the

other delivering cold water at a mass flow rate of 1.3 lb/s.

Water exits through a single exit pipe at a mass flow rate of

2.6 lb/s. Determine the amount of water, in lb, in the tank

after one hour.

Total volume = 2500m

(AV)

= 2m

/min

= 1000 m

 = 0.0015 m

/kg

Initial volume of crude oil

= 1.5 m/s

= 0.15 m

20 m

Fig. P4.4

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212 Chapter 4

Control Volume Analysis Using Energy

= 3000 lb

hot

water

cold

water

= 0.8 lb/s

= 1.3 lb/s

= 2.6 lb/s

Fig. P4.6

4.7 Figure P4.7 provides data for water entering and exiting a

tank. At the inlet and exit of the tank, determine the mass

flow rate, each in kg/s. Also find the time rate of change of

mass contained within the tank, in kg/s.

Steam

Liquid

= 20 m/s

= 10 × 10

–3

= 20 bar

= 600°C

= 1 m/s

= 6 × 10

–3

= 10 bar

= 150°C

Fig. P4.7

4.8 Liquid water flows isothermally at 208C through a one-

inlet, one-exit duct operating at steady state. The duct’s inlet

and exit diameters are 0.02 m and 0.04 m, respectively. At the

inlet, the velocity is 40 m/s and pressure is 1 bar. At the exit,

determine the mass flow rate, in kg/s, and velocity, in m/s.

4.9 Air enters a one-inlet, one-exit control volume at 6 bar,

500 K, and 30 m/s through a flow area of 28 cm

. At the exit,

the pressure is 3 bar, the temperature is 456.5 K, and the

velocity is 300 m/s. The air behaves as an ideal gas. For

steady-state operation, determine

(a) the mass flow rate, in kg/s.

(b) the exit flow area, in cm

4.10 The small two-story office building shown in Fig. P4.10

has 36,000 ft

of occupied space. Due to cracks around

windows and outside doors, air leaks in on the windward

side of the building and leaks out on the leeward side of the

building. Outside air also enters the building when outer

doors are opened. On a particular day, tests were conducted.

The outdoor temperature was measured to be 158F. The

inside temperature was controlled at 708F. Keeping the doors

closed, the infiltration rate through the cracks was determined

to be 75 ft

/min. The infiltration rate associated with door

openings, averaged over the work day, was 50 ft

/min. The

pressure difference was negligible between the inside and

outside of the building. (a) Assuming ideal gas behavior,

determine at steady state the volumetric flow rate of air

exiting the building, in ft

/min. (b) When expressed in terms

of the volume of the occupied space, determine the number

of building air changes per hour.

4.11 As shown in Fig. P4.11, air with a volumetric flow rate of

15,000 ft

/min enters an air-handling unit at 358F, 1 atm. The

air-handling unit delivers air at 808F, 1 atm to a duct system

with three branches consisting of two 26-in.-diameter ducts

and one 50-in. duct. The velocity in each 26-in. duct is 10 ft/s.

Assuming ideal gas behavior for the air, determine at steady

state

(a) the mass flow rate of air entering the air-handling unit,

in lb/s.

Air infiltration

through cracks

at 75 ft

/min,

15°F

Air infiltration

through

door openings at

50 ft

/min, 15°F

Air exiting

through cracks

at 70°F

Fig. P4.10

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