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

Подождите немного. Документ загружается.

6.10 Rate Balances for Control Volumes at Steady State 313

Determining Entropy Production in Heat Pump Components

c c c c EXAMPLE 6.8 c

Components of a heat pump for supplying heated air to a dwelling are shown in the schematic below. At steady

state, Refrigerant 22 enters the compressor at 258C, 3.5 bar and is compressed adiabatically to 758C, 14 bar.

From the compressor, the refrigerant passes through the condenser, where it condenses to liquid at 288C, 14 bar.

The refrigerant then expands through a throttling valve to 3.5 bar. The states of the refrigerant are shown on

the accompanying T–s diagram. Return air from the dwelling enters the condenser at 208C, 1 bar with a volu-

metric flow rate of 0.42 m

/s and exits at 508C with a negligible change in pressure. Using the ideal gas model

for the air and neglecting kinetic and potential energy effects, (a) determine the rates of entropy production, in

kW/K, for control volumes enclosing the condenser, compressor, and expansion valve, respectively. (b) Discuss

the sources of irreversibility in the components considered in part (a).

SOLUTION

Known:

Refrigerant 22 is compressed adiabatically, condensed by heat transfer to air passing through a heat

exchanger, and then expanded through a throttling valve. Steady-state operating data are known.

Find: Determine the entropy production rates for control volumes enclosing the condenser, compressor, and

expansion valve, respectively, and discuss the sources of irreversibility in these components.

Schematic and Given Data:

Fig. E6.8

75°C

–5°C

28°C

14 bar

3.5 bar

Expansion

valve

= 3.5 bar

= 14 bar

= 28°C

Condenser

Indoor return air

= 20°C

= 1 bar

(AV)

= 0.42 m

= –5°C

= 3.5 bar

= 14 bar

= 75°C

Supply air

= 50°C

= 1 bar

Evaporator

Outdoor air

Compressor

Engineering Model:

Each component is analyzed as a control volume at steady state.

2. The compressor operates adiabatically, and the expansion across the valve is a throttling process.

3. For the control volume enclosing the condenser, W

5 0 and Q

5 0.

4. Kinetic and potential energy effects can be neglected.

5. The air is modeled as an ideal gas with constant c

5 1.005 kJ/kg

Analysis:

(a)

Let us begin by obtaining property data at each of the principal refrigerant states located on the accompany-

ing schematic and T–s diagram. At the inlet to the compressor, the refrigerant is a superheated vapor at 258C,

3.5 bar, so from Table A-9, s

5 0.9572 kJ/kg

K. Similarly, at state 2, the refrigerant is a superheated vapor at

758C, 14 bar, so interpolating in Table A-9 gives s

5 0.98225 kJ/kg

K and h

5 294.17 kJ/kg.

➊

c06UsingEntropy.indd Page 313 5/28/10 1:22:06 PM user-s146c06UsingEntropy.indd Page 313 5/28/10 1:22:06 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

314 Chapter 6 Using Entropy

State 3 is compressed liquid at 288C, 14 bar. From Table A-7, s

(288C) 5 0.2936 kJ/kg

K and h

(288C) 5

79.05 kJ/kg. The expansion through the valve is a throttling process, so h

5 h

. Using data from Table A-8, the

quality at state 4 is

2 h

179.05 2 33.09

1212.912

5 0.216

and the specific entropy is

5 s

1 x

2 s

5 0.1328 1 0.216

0.9431 2 0.1328

5 0.3078 kJ

kg ? K

Condenser

Consider the control volume enclosing the condenser. With assumptions 1 and 3, the entropy rate balance reduces to

0 5 m

ref

2 s

21 m

air

2 s

21 s

cond

To evaluate s

cond

requires the two mass flow rates, m

air

and m

ref

, and the change in specific entropy for the air.

These are obtained next.

Evaluating the mass flow rate of air using the ideal gas model and the known volumetric flow rate

air

1AV2

5 1AV2

5 a0.42

11 bar

8.31

28.97

kg ? K

b1293 K2

1 bar

1 kJ

N ? m

`5 0.5 kg

The refrigerant mass flow rate is determined using an energy balance for the control volume enclosing the con-

denser together with assumptions 1, 3, and 4 to obtain

ref

air

2 h

With assumption 5, h

2 h

5 c

2 T

). Inserting values

ref

a0.

ba1.

005

kg ? K

b1323 2 2932K

1294.17 2 79.052 kJ

5 0.07 kg

Using Eq. 6.22, the change in specific entropy of the air is

2 s

5 c

2 R ln

5 a1.005

kg ? K

b ln a

323

293

b2 R ln a

1.0

5 0.098 kJ

kg ? K

Finally, solving the entropy balance for s

cond

and inserting values

cond

5 m

ref

2 s

21 m

air

2 s

5 ca0.07

10.2936 2 0.982252

kg ? K

1 10.5210.0982d`

1 kW

1 kJ

5 7.95 3 10

Compressor

For the control volume enclosing the compressor, the entropy rate balance reduces with assumptions 1 and

3 to

0 5 m

ref

2 s

1 s

comp

➋

c06UsingEntropy.indd Page 314 5/26/10 5:20:34 PM user-s146c06UsingEntropy.indd Page 314 5/26/10 5:20:34 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

comp

5 m

ref

2 s

5 a0.07

b10.98225 2 0.95722a

kg ? K

1 kW

1 kJ

17.5

Valve

Finally, for the control volume enclosing the throttling valve, the entropy rate balance reduces to

0 5 m

ref

2 s

1 s

valve

Solving for s

valve

and inserting values

valve

5 m

ref

2 s

25 a0.07

b10.3078 2 0.29362a

kg ? K

1 kW

1 kJ

9.9

4 3 1

(b) The following table summarizes, in rank order, the calculated entropy production rates:

Component s

(kW/K)

compressor 17.5 3 10

valve 9.94 3 10

condenser 7.95 3 10

Entropy production in the compressor is due to fluid friction, mechanical friction of the moving parts, and inter-

nal heat transfer. For the valve, the irreversibility is primarily due to fluid friction accompanying the expansion

across the valve. The principal source of irreversibility in the condenser is the temperature difference between

the air and refrigerant streams. In this example, there are no pressure drops for either stream passing through

the condenser, but slight pressure drops due to fluid friction would normally contribute to the irreversibility of

condensers. The evaporator shown in Fig. E6.8 has not been analyzed.

➊ Due to the relatively small temperature change of the air, the specific heat c

can be taken as constant at the average of the inlet and exit air temperatures.

➋ Temperatures in K are used to evaluate m

ref

, but since a temperature differ-

ence is involved the same result would be obtained if temperatures in °C were

used. Temperatures in K, and not °C, are required when a temperature ratio

is involved, as in Eq. 6.22 used to evaluate s

2 s

➌ By focusing attention on reducing irreversibilities at the sites with the highest

entropy production rates, thermodynamic improvements may be possible. How-

ever, costs and other constraints must be considered, and can be overriding.

Ability to…

❑

apply the control volume

mass, energy and entropy

rate balances.

❑

develop an engineering model.

❑

retrieve property data for

Refrigerant 22.

❑

apply the ideal gas model

with constant c

✓

Skills Developed

If the compressor operated adiabatically and without internal

irreversibilities, determine the temperature of the refrigerant at the com-

pressor exit, in °C, keeping the compressor inlet state and exit pressure

the same. Ans. 65°C.

6.11 Isentropic Processes

The term isentropic means constant entropy. Isentropic processes are encountered in

many subsequent discussions. The object of the present section is to show how prop-

erties are related at any two states of a process in which there is no change in specific

entropy.

➌

6.11 Isentropic Processes 315

c06UsingEntropy.indd Page 315 5/26/10 2:41:20 PM user-s146c06UsingEntropy.indd Page 315 5/26/10 2:41:20 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

316 Chapter 6 Using Entropy

6.11.1

General Considerations

The properties at states having the same specific entropy can be related using the

graphical and tabular property data discussed in Sec. 6.2. For example, as illustrated

by Fig. 6.8, temperature–entropy and enthalpy–entropy diagrams are particularly con-

venient for determining properties at states having the same value of specific entropy.

All states on a vertical line passing through a given state have the same entropy. If

state 1 on Fig. 6.8 is fixed by pressure p

and temperature T

, states 2 and 3 are read-

ily located once one additional property, such as pressure or temperature, is specified.

The values of several other properties at states 2 and 3 can then be read directly from

the figures.

Tabular data also can be used to relate two states having the same specific entropy.

For the case shown in Fig. 6.8, the specific entropy at state 1 could be determined

from the superheated vapor table. Then, with s

5 s

and one other property value,

such as p

or T

, state 2 could be located in the superheated vapor table. The values

of the properties y, u, and h at state 2 can then be read from the table. (An illustra-

tion of this procedure is given in Sec. 6.2.1.) Note that state 3 falls in the two-phase

liquid–vapor regions of Fig. 6.8. Since s

5 s

, the quality at state 3 could be deter-

mined using Eq. 6.4. With the quality known, other properties such as y, u, and h

could then be evaluated. Computer retrieval of entropy data provides an alternative

to tabular data.

6.11.2

Using the Ideal Gas Model

Figure 6.9 shows two states of an ideal gas having the same value of specific

entropy. Let us consider relations among pressure, specific volume, and tempera-

ture at these states, first using the ideal gas tables and then assuming specific heats

are constant.

Ideal Gas Tables

For two states having the same specific entropy, Eq. 6.20a reduces to

0 5 s81T

22 s81T

22 R ln

(6.40a)

Equation 6.40a involves four property values: p

, T

, p

, and T

. If any three are

known, the fourth can be determined. If, for example, the temperature at state 1 and

Fig. 6.8 T–s and h–s diagrams showing states having the same value of specific entropy.

Fig. 6.9 Two states of an

ideal gas where s

= s

c06UsingEntropy.indd Page 316 5/26/10 2:41:21 PM user-s146c06UsingEntropy.indd Page 316 5/26/10 2:41:21 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

the pressure ratio p

are known, the temperature at state 2 can be determined

from

s81T

25 s81T

21 R ln

(6.40b)

Since T

is known, s8(T

) would be obtained from the appropriate table, the value of

s8(T

) would be calculated, and temperature T

would then be determined by inter-

polation. If p

, T

, and T

are specified and the pressure at state 2 is the unknown,

Eq. 6.40a would be solved to obtain

5 p

exp

s81T

22 s81T

(6.40c)

Equations 6.40 can be used when s8 (or

) data are known, as for the gases of Tables

A-22 and A-23.

AIR.

For the special case of air modeled as an ideal gas, Eq. 6.40c provides the

basis for an alternative tabular approach for relating the temperatures and pres-

sures at two states having the same specific entropy. To introduce this, rewrite the

equation as

exp

The quantity exp[s8(T)/R] appearing in this expression is solely a function of tem-

perature, and is given the symbol p

(T). A tabulation of p

versus temperature for

air is provided in Tables A-22.

In terms of the function p

, the last equation

becomes



5 s

, air only2

(6.41)

where p

5 p

) and p

5 p

). The function p

is sometimes called the relative

pressure. Observe that p

is not truly a pressure, so the name relative pressure has no

physical significance. Also, be careful not to confuse p

with the reduced pressure of

the compressibility diagram.

A relation between specific volumes and temperatures for two states of air having

the same specific entropy can also be developed. With the ideal gas equation of state,

y 5 RT/p, the ratio of the specific volumes is

Then, since the two states have the same specific entropy, Eq. 6.41 can be introduced

to give

5 c

The ratio RT/p

(T) appearing on the right side of the last equation is solely a func-

tion of temperature, and is given the symbol y

(T). Values of y

for air are tabulated

The values of p

determined with this definition are inconveniently large, so they are divided by a scale factor

before tabulating to give a convenient range of numbers.

TAKE NOTE...

When applying the software

IT to relate two states of

an ideal gas having the

same value of specific

entropy, IT returns specific

entropy directly and does

not employ the special

functions s°, p

and v

6.11 Isentropic Processes 317

c06UsingEntropy.indd Page 317 5/26/10 3:28:42 PM user-s146 c06UsingEntropy.indd Page 317 5/26/10 3:28:42 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

318 Chapter 6 Using Entropy

versus temperature in Tables A-22. In terms of the function y

, the last equation

becomes



5 s

, air only2

(6.42)

where y

5 y

) and y

5 y

). The function y

is sometimes called the relative

volume. Despite the name given to it, y

(T) is not truly a volume. Also, be careful

not to confuse it with the pseudoreduced specific volume of the compressibility

diagram.

Assuming Constant Specific Heats

Let us consider next how properties are related for isentropic processes of an ideal

gas when the specific heats are constants. For any such case, Eqs. 6.21 and 6.22 reduce

to the equations

0 5 c

2 R ln

0 5 c

1 R ln

Introducing the ideal gas relations

k 2 1



k 2 1

(3.47)

where k is the specific heat ratio and R is the gas constant, these equations can be

solved, respectively, to give

5 a

k21



5 s

, constant k2

(6.43)

5 a

k21





5 s

, constant k2

(6.44)

The following relation can be obtained by eliminating the temperature ratio from

Eqs. 6.43 and 6.44:



5 s

, constant k2

(6.45)

Previously, we have identified an internally reversible process described by py

constant, where n is a constant, as a polytropic process. From the form of Eq. 6.45, it

can be concluded that the polytropic process py

5 constant of an ideal gas with

constant specific heat ratio k is an isentropic process. We observed in Sec. 3.15 that

a polytropic process of an ideal gas for which n 5 1 is an isothermal (constant-

temperature) process. For any fluid, n 5 0 corresponds to an isobaric (constant-pressure)

process and n 5

corresponds to an isometric (constant-volume) process. Poly-

tropic processes corresponding to these values of n are shown in Fig. 6.10 on p–y and

T–s diagrams.

6.11.3

Illustrations: Isentropic Processes of Air

Means for evaluating data for isentropic processes of air modeled as an ideal gas are

illustrated in the next two examples. In Example 6.9, we consider three alternative

methods.

c06UsingEntropy.indd Page 318 5/26/10 3:28:44 PM user-s146 c06UsingEntropy.indd Page 318 5/26/10 3:28:44 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

Analyzing an Isentropic Process of Air

c c c c EXAMPLE 6.9 c

Air undergoes an isentropic process from p

5 1 atm, T

5 5408R to a final state where the temperature is

5 11608R. Employing the ideal gas model, determine the final pressure p

, in atm. Solve using (a) p

data

from Table A-22E, (b) Interactive Thermodynamics: IT, and (c) a constant specific heat ratio k evaluated at the

mean temperature, 8508R, from Table A-20E.

SOLUTION

Known:

Air undergoes an isentropic process from a state where pressure and temperature are known to a state

where the temperature is specified.

Find: Determine the final pressure using (a) p

data, (b) IT, and (c) a constant value for the specific heat ratio k.

Schematic and Given Data:

Engineering Model:

A quantity of air as the system undergoes an isentropic process.

2. The air can be modeled as an ideal gas.

3. In part (c) the specific heat ratio is constant.

Analysis:

(a)

The pressures and temperatures at two states of an ideal gas having the same specific entropy are related by

Eq. 6.41

Solving

5 p

= ?

= 1160°R

= 1 atm

= 540°R

Fig. E6.9

n = –1

n = 0

n = 1

n = k

n = ± ∞

n = ±

∞

n = –1

n = 0

n = 1

n = k

Fig. 6.10 Polytropic processes on p–y and T–s diagrams.

6.11 Isentropic Processes 319

c06UsingEntropy.indd Page 319 5/26/10 3:28:45 PM user-s146 c06UsingEntropy.indd Page 319 5/26/10 3:28:45 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

320 Chapter 6 Using Entropy

Another illustration of an isentropic process of an ideal gas is provided in Exam-

ple 6.10 dealing with air leaking from a tank.

Considering Air Leaking from a Tank

c c c c EXAMPLE 6.10 c

A rigid, well-insulated tank is filled initially with 5 kg of air at a pressure of 5 bar and a temperature of 500 K.

A leak develops, and air slowly escapes until the pressure of the air remaining in the tank is 1 bar. Employing

the ideal gas model, determine the amount of mass remaining in the tank and its temperature.

SOLUTION

Known:

A leak develops in a rigid, insulated tank initially containing air at a known state. Air slowly escapes

until the pressure in the tank is reduced to a specified value.

Find: Determine the amount of mass remaining in the tank and its temperature.

With p

values from Table A-22E

p 5 11 atm2

21.18

1.3860

5 15.28 at

(b) The IT solution follows:

T1 5 540 // °R

p1 5 1 // atm

T2 5 1160 // °R

s_TP(“Air”, T1,p1) 5 s_TP(“Air”,T2,p2)

// Result: p2 5 15.28 atm

gas having the same specific entropy are related by Eq. 6.43. Thus

5 p

k21

From Table A-20E at the mean temperature, 3908F (8508R), k 5 1.39. Inserting values into the above

expression

5 11 atm2

1160

540

1.39

0.39

5 15.26 at

➊ IT returns a value for p

even though it is an implicit variable in the specific

entropy function. Also note that IT returns values for specific entropy directly

and does not employ special functions such as s8, p

, and y

➋ The close agreement between the answer obtained in part (c) and that of

parts (a), (b) can be attributed to the use of an appropriate value for the

specific heat ratio k.

Determine the final pressure, in atm, using a constant specific

heat ratio k evaluated at T

5 5408R. Expressed as a percent, how much does

this pressure value differ from that of part (c)? Ans. 14.53 atm, 25%.

Ability to…

❑

analyze an isentropic process

using Table A-22E data,

❑

Interactive Thermodynamics,

and

❑

a constant specific heat

ratio k.

✓

Skills Developed

➊

➋

c06UsingEntropy.indd Page 320 5/26/10 3:28:49 PM user-s146 c06UsingEntropy.indd Page 320 5/26/10 3:28:49 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

6.11 Isentropic Processes 321

Analysis: With the ideal gas equation of state, the mass initially in the tank that remains in the tank at the end

of the process is

where p

and T

are the final pressure and temperature, respectively. Similarly, the initial amount of mass within

the tank, m

where p

and T

are the initial pressure and temperature, respectively. Eliminating volume between these two

expressions, the mass of the system is

Except for the final temperature of the air remaining in the tank, T

, all required values are known. The remain-

der of the solution mainly concerns the evaluation of T

For the closed system under consideration, there are no significant irreversibilities (assumption 3), and no

heat transfer occurs (assumption 2). Accordingly, the entropy balance reduces to

¢S 5

1 s

5 0

Since the system mass remains constant, ¢S 5 m

¢s, so

¢s 5 0

That is, the initial and final states of the system have the same value of specific entropy.

Using Eq. 6.41

where p

5 5 bar and p

5 1 bar. With p

5 8.411 from Table A-22 at 500 K,

the previous equation gives p

5 1.6822. Using this to interpolate in Table A-22,

5 317 K.

Finally, inserting values into the expression for system mass

1 bar

5 bar

500 K

317 K

15 kg25 1.58 kg

Engineering Model:

1. As shown on the accompanying sketch, the closed system is the

mass initially in the tank that remains in the tank.

2. There is no significant heat transfer between the system and its

surroundings.

3. Irreversibilities within the tank can be ignored as the air slowly

escapes.

4. The air is modeled as an ideal gas.

Slow leak

Mass initially in the

tank that escapes

Mass initially

in the tank that

remains in the tank

System boundary

Initial condition of tank

Fig. E6.10

Ability to…

❑

develop an engineering model.

❑

apply the closed system

entropy balance.

❑

analyze an isentropic process.

✓

Skills Developed

Evaluate the tank volume, in m

. Ans. 1.43 m

Schematic and Given Data:

c06UsingEntropy.indd Page 321 5/26/10 3:28:50 PM user-s146 c06UsingEntropy.indd Page 321 5/26/10 3:28:50 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New

322 Chapter 6 Using Entropy

6.12 Isentropic Efficiencies of Turbines,

Nozzles, Compressors, and Pumps

Engineers make frequent use of efficiencies and many different efficiency defini-

tions are employed. In the present section, isentropic efficiencies for turbines, noz-

zles, compressors, and pumps are introduced. Isentropic efficiencies involve a com-

parison between the actual performance of a device and the performance that

would be achieved under idealized circumstances for the same inlet state and the

same exit pressure. These efficiencies are frequently used in subsequent sections of

the book.

6.12.1

Isentropic Turbine Efficiency

To introduce the isentropic turbine efficiency, refer to Fig. 6.11, which shows a turbine

expansion on a Mollier diagram. The state of the matter entering the turbine and the

exit pressure are fixed. Heat transfer between the turbine and its surroundings is

ignored, as are kinetic and potential energy effects. With these assumptions, the mass

and energy rate balances reduce, at steady state, to give the work developed per unit

of mass flowing through the turbine

5 h

2 h

Since state 1 is fixed, the specific enthalpy h

is known. Accordingly, the value of the

work depends on the specific enthalpy h

only, and increases as h

is reduced. The

maximum value for the turbine work corresponds to the smallest allowed value for

the specific enthalpy at the turbine exit. This can be determined using the second law

as follows.

Since there is no heat transfer, the allowed exit states are constrained by Eq. 6.39

5 s

2 s

$ 0

Because the entropy production s

cannot be negative, states with s

, s

are

not accessible in an adiabatic expansion. The only states that actually can be

attained adiabatically are those with s

. s

. The state labeled “2s”on Fig. 6.11

would be attained only in the limit of no internal irreversibilities. This corresponds

Actual

expansion

Isentropic

expansion

Accessible

states

– h

Fig. 6.11 Comparison of actual and

isentropic expansions through a turbine.

c06UsingEntropy.indd Page 322 5/26/10 3:28:53 PM user-s146 c06UsingEntropy.indd Page 322 5/26/10 3:28:53 PM user-s146 /Users/user-s146/Desktop/Merry_X-Mas/New/Users/user-s146/Desktop/Merry_X-Mas/New