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

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to an isentropic expansion through the turbine. For fixed exit pressure, the specific

enthalpy h

decreases as the specific entropy s

decreases. Therefore, the smallest

allowed value for h

corresponds to state 2s, and the maximum value for the tur-

bine work is

5 h

2 h

In an actual expansion through the turbine h

. h

, and thus less work than the

maximum would be developed. This difference can be gauged by the isentropic turbine

efficiency

defined by

2 h

(6.46)

Both the numerator and denominator of this expression are evaluated for the

same inlet state and the same exit pressure. The value of h

is typically 0.7 to 0.9

(70–90%).

The two examples to follow illustrate the isentropic turbine efficiency concept. In

Example 6.11 the isentropic efficiency of a steam turbine is known and the objective

is to determine the turbine work.

isentropic turbine efficiency

Determining Turbine Work Using the Isentropic Efficiency

c c c c EXAMPLE 6.11 c

A steam turbine operates at steady state with inlet conditions of p

5 5 bar, T

5 3208C. Steam leaves the tur-

bine at a pressure of 1 bar. There is no significant heat transfer between the turbine and its surroundings, and

kinetic and potential energy changes between inlet and exit are negligible. If the isentropic turbine efficiency is

75%, determine the work developed per unit mass of steam flowing through the turbine, in kJ/kg.

SOLUTION

Known:

Steam expands through a turbine operating at steady state from a specified inlet state to a specified exit

pressure. The turbine efficiency is known.

Find: Determine the work developed per unit mass of steam flowing through the turbine.

Schematic and Given Data:

Engineering Model:

A control volume enclosing the turbine is at steady state.

2. The expansion is adiabatic and changes in kinetic and poten-

tial energy between the inlet and exit can be neglected.

– h

= 1 bar

= 5 bar

Actual

expansion

Isentropic

expansion

Accessible

states

Fig. E6.11

6.12 Isentropic Efficiencies of Turbines, Nozzles, Compressors, and Pumps 323

TAKE NOTE...

The subscript s denotes

a quantity evaluated for an

isentropic process from a

specified inlet state to a

specified exit pressure.

Turbine

A.19 – Tab e

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324 Chapter 6 Using Entropy

Example 6.12 is similar to Example 6.11, but here the working substance is air as

an ideal gas. Moreover, in this case the turbine work is known and the objective is

to determine the isentropic turbine efficiency.

Evaluating Isentropic Turbine Efficiency

c c c c EXAMPLE 6.12 c

A turbine operating at steady state receives air at a pressure of p

5 3.0 bar and a temperature of T

5 390 K.

Air exits the turbine at a pressure of p

5 1.0 bar. The work developed is measured as 74 kJ per kg of air flow-

ing through the turbine. The turbine operates adiabatically, and changes in kinetic and potential energy between

inlet and exit can be neglected. Using the ideal gas model for air, determine the isentropic turbine efficiency.

SOLUTION

Known:

Air expands adiabatically through a turbine at steady state from a specified inlet state to a specified exit

pressure. The work developed per kg of air flowing through the turbine is known.

Find: Determine the turbine efficiency.

Schematic and Given Data:

Actual

expansion

Isentropic

expansion

= 390 K

3.0 bar

1.0 bar

–––

= 74 kJ/kg

= 1.0 bar

= 3.0 bar

= 390 K

Air

turbine

Fig. E6.12

Analysis: The work developed can be determined using the isentropic turbine efficiency, Eq. 6.46, which on rear-

rangement gives

5 h

2 h

From Table A-4, h

5 3105.6 kJ/kg and s

5 7.5308 kJ/kg

K. The exit state for an isentropic expansion is

fixed by p

5 1 and s

5 s

. Interpolating with specific entropy in Table A-4 at 1 bar gives h

5 2743.0 kJ/kg.

Substituting values

➋

5 0.7513105.6 2 2743.025 271.95 kJ

➊ At 2s, the temperature is about 1338C.

➋ The effect of irreversibilities is to exact a penalty on the work output of the

turbine. The work is only 75% of what it would be for an isentropic expansion

between the given inlet state and the turbine exhaust pressure. This is clearly

illustrated in terms of enthalpy differences on the accompanying h–s diagram.

Determine the temperature of the steam at the turbine exit, in 8C.

Ans. 1798C.

➊

Ability to…

❑

apply the isentropic turbine

efficiency, Eq. 6.46.

❑

retrieve steam table data.

✓

Skills Developed

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6.12.2

Isentropic Nozzle Efficiency

A similar approach to that for turbines can be used to introduce the isentropic effi-

ciency of nozzles operating at steady state. The isentropic nozzle efficiency is defined

as the ratio of the actual specific kinetic energy of the gas leaving the nozzle, V

to the kinetic energy at the exit that would be achieved in an isentropic expansion

between the same inlet state and the same exit pressure,

. That is,

nozzle

(6.47)

Nozzle efficiencies of 95% or more are common, indicating that well-designed nozzles

are nearly free of internal irreversibilities.

In Example 6.13, the objective is to determine the isentropic efficiency of a steam

nozzle.

Engineering Model:

The control volume shown on the accompanying sketch is at steady state.

2. The expansion is adiabatic and changes in kinetic and potential energy between inlet and exit can be

neglected.

3. The air is modeled as an ideal gas.

Analysis: The numerator of the isentropic turbine efficiency, Eq. 6.46, is known. The denominator is evaluated

as follows.

The work developed in an isentropic expansion from the given inlet state to the specified exit pressure is

5 h

2 h

From Table A-22 at 390 K, h

5 390.88 kJ/kg. To determine h

, use Eq. 6.41

With p

5 3.0 bar, p

5 1.0 bar, and p

5 3.481 from Table A-22 at 390 K

1.0

3.0

13.48125 1.1603

Interpolation in Table A-22 gives h

5 285.27 kJ/kg. Thus

5 390.88 2 285.27 5 105.6 kJ

Substituting values into Eq. 6.46

74 kJ

105.6 kJ

5 0.70170%2

Determine the rate of entropy production, in kJ/K per kg of air

flowing through the turbine. Ans. 0.105 kJ/kg ? K.

Ability to…

❑

apply the isentropic turbine

efficiency, Eq. 6.46.

❑

retrieve data for air as an

ideal gas.

✓

Skills Developed

isentropic nozzle

efficiency

6.12 Isentropic Efficiencies of Turbines, Nozzles, Compressors, and Pumps 325

Nozzle

A.17 – Tab e

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326 Chapter 6 Using Entropy

Evaluating Isentropic Nozzle Efficiency

c c c c EXAMPLE 6.13 c

Steam enters a nozzle operating at steady state at p

5 140 lbf/in.

and T

5 6008F with a velocity of 100 ft/s.

The pressure and temperature at the exit are p

5 40 lbf/in.

and T

5 3508F. There is no significant heat trans-

fer between the nozzle and its surroundings, and changes in potential energy between inlet and exit can be

neglected. Determine the nozzle efficiency.

SOLUTION

Known:

Steam expands through a nozzle at steady state from a specified inlet state to a specified exit state. The

velocity at the inlet is known.

Find: Determine the nozzle efficiency.

Schematic and Given Data:

Engineering Model:

The control volume shown on the accompanying sketch operates adiabatically at steady state.

2. For the control volume, W

5 0 and the change in potential energy between inlet and exit can be neglected.

Analysis: The nozzle efficiency given by Eq. 6.47 requires the actual specific kinetic energy at the nozzle exit

and the specific kinetic energy that would be achieved at the exit in an isentropic expansion from the given inlet

state to the given exit pressure. The energy rate balance for a one-inlet, one-exit control volume at steady state

enclosing the nozzle reduces to give Eq. 4.21, which on rearrangement reads

5 h

2 h

This equation applies for both the actual expansion and the isentropic expansion.

From Table A-4E at T

5 6008F and p

5 140 lbf/in.

, h

5 1326.4 Btu/lb, s

5 1.7191 Btu/lb

8R. Also, with

5 3508F and p

5 40 lbf/in.

, h

5 1211.8 Btu/lb. Thus, the actual specific kinetic energy at the exit in Btu/lb is

5 1326.4

Btu

2 1211.8

Btu

1100 ft

122`

32.2 lb ? ft

1 lbf

778 ft ? lbf

1 Btu

5 114

Interpolating in Table A-4E at 40 lbf/in.

, with s

5 s

5 1.7191 Btu/lb

8R, results in h

5 1202.3 Btu/lb.

Accordingly, the specific kinetic energy at the exit for an isentropic expansion is

5 1326.4 2 1202.3 1

11002

122

32.2

778

5 124.3 Btu

Isentropic

expansion

Actual

expansion

40 lbf/in.

140 lbf/in.

600°F

350°F

= 40 lbf/in.

= 350°F

= 140 lbf/in.

= 600°F

= 100 ft/s

Steam

nozzle

Fig. E6.13

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6.12.3

Isentropic Compressor and Pump Efficiencies

The form of the isentropic efficiency for compressors and pumps is taken up next.

Refer to Fig. 6.12, which shows a compression process on a Mollier diagram. The state

of the matter entering the compressor and the exit pressure are fixed. For negligible

heat transfer with the surroundings and no appreciable kinetic and potential energy

effects, the work input per unit of mass flowing through the compressor is

b5 h

2 h

Since state 1 is fixed, the specific enthalpy h

is known. Accordingly, the value of the

work input depends on the specific enthalpy at the exit, h

. The above expression

shows that the magnitude of the work input decreases as h

decreases. The minimum

work input corresponds to the smallest allowed value for the specific enthalpy at the

compressor exit. With similar reasoning as for the turbine, the smallest allowed

enthalpy at the exit state would be achieved in an isentropic compression from the

specified inlet state to the specified exit pressure. The minimum work input is given,

therefore, by

5 h

2 h

Substituting values into Eq. 6.47

nozzle

114.8

124.3

5 0.924 192.4%2

➊ The principal irreversibility in the nozzle is friction, in particular friction

between the flowing steam and the nozzle wall. The effect of friction is that

a smaller exit kinetic energy, and thus a smaller exit velocity, is realized

than would have been obtained in an isentropic expansion to the same

pressure.

Determine the temperature, in 8F, corresponding to state 2s in

Fig. E6.13. Ans. 3318F.

Ability to…

❑

apply the control volume

energy rate balance.

❑

apply the isentropic nozzle

efficiency, Eq. 6.47.

❑

retrieve steam table data.

✓Skills Developed

➊

– h

Accessible

states

Actual

compression

Isentropic

compression

Fig. 6.12 Comparison of actual and isentropic

compressions.

6.12 Isentropic Efficiencies of Turbines, Nozzles, Compressors, and Pumps 327

Compressor

A.20 – Tab e

Pump

A.21 – Tab e

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328 Chapter 6 Using Entropy

In an actual compression, h

. h

, and thus more work than the minimum would be

required. This difference can be gauged by the isentropic compressor efficiency defined by

12W

2 h

(6.48)

Both the numerator and denominator of this expression are evaluated for the same

inlet state and the same exit pressure. The value of h

is typically 75 to 85% for

compressors. An isentropic pump efficiency, h

, is defined similarly.

In Example 6.14, the isentropic efficiency of a refrigerant compressor is evaluated,

first using data from property tables and then using IT.

isentropic compressor

efficiency

isentropic pump

efficiency

Evaluating Isentropic Compressor Efficiency

c c c c EXAMPLE 6.14 c

For the compressor of the heat pump system in Example 6.8, determine the power, in kW, and the isentropic

efficiency using (a) data from property tables, (b) Interactive Thermodynamics: IT.

SOLUTION

Known:

Refrigerant 22 is compressed adiabatically at steady state from a specified inlet state to a specified exit

state. The mass flow rate is known.

Find: Determine the compressor power and the isentropic efficiency using (a) property tables, (b) IT.

Schematic and Given Data:

14 bar

3.5 bar

= 75°C

= –5°C

Fig. E6.14

Engineering Model:

A control volume enclosing the compressor is at steady state.

2. The compression is adiabatic, and changes in kinetic and potential energy

between the inlet and the exit can be neglected.

Analysis: (a) By assumptions 1 and 2, the mass and energy rate balances reduce to give

5 m

2 h

From Table A-9, h

5 249.75 kJ/kg and h

5 294.17 kJ/kg. Thus, with the mass flow rate determined in Example 6.8

5 10.07 kg

s21249.75 2 294.172kJ

1 kW

1 kJ

523.11 kW

The isentropic compressor efficiency is determined using Eq. 6.48

12W

2 h

In this expression, the denominator represents the work input per unit mass of refrigerant flowing for the actual

compression process, as considered above. The numerator is the work input for an isentropic compression between

the initial state and the same exit pressure. The isentropic exit state is denoted as state 2s on the accompanying

T–s diagram.

From Table A-9, s

5 0.9572 kJ/kg

K. With s

5 s

, interpolation in Table A-9 at 14 bar gives h

5 285.58 kJ/kg.

Substituting values

285.58 2 249.75

294.17 2 249.75

5 0.81181%2

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6.13 Heat Transfer and Work in Internally

Reversible, Steady-State Flow Processes

This section concerns one-inlet, one-exit control volumes at steady state. The objective

is to introduce expressions for the heat transfer and the work in the absence of inter-

nal irreversibilities. The resulting expressions have several important applications.

6.13.1

Heat Transfer

For a control volume at steady state in which the flow is both isothermal at tempera-

ture T and internally reversible, the appropriate form of the entropy rate balance is

0 5

1 m

2 s

21 s

where 1 and 2 denote the inlet and exit, respectively, and m

is the mass flow rate. Solving

this equation, the heat transfer per unit of mass passing through the control volume is

5 T1s

2 s

6.13 Heat Transfer and Work in Internally Reversible, Steady-State Flow Processes 329

(b) The IT program follows. In the program, W

is denoted as Wdot, m

as mdot, and h

as eta_c.

// Given Data:

T1 5 25 // 8C

p1 5 3.5 // bar

T2 5 75 // 8C

p2 5 14 // bar

mdot 5 0.07 // kg/s

// Determine the specific enthalpies.

h1 5 h_PT(“R22”,p1,T1)

h2 5 h_PT(“R22”,p2,T2)

// Calculate the power.

Wdot 5 mdot * (h1 2 h2)

// Find h2s:

s1 5 s_PT(“R22”,p1,T1)

s2s 5 s_Ph(“R22”,p2,h2s)

s2s 5 s1

// Determine the isentropic compressor efficiency.

eta_c 5 (h2s 2 h1)/(h2 2 h1)

Use the Solve button to obtain: W

523.111 kW and h

5 80.58%, which, as

expected, agree closely with the values obtained above.

➊ Note that IT solves for the value of h

even though it is an implicit variable

in the specific entropy function.

Determine the minimum theoretical work input, in kJ per kg

flowing, for an adiabatic compression from state 1 to the exit pressure of

14 bar. Ans. 35.83 kJ/kg.

Ability to…

❑

apply the control volume

energy rate balance.

❑

apply the isentropic com-

pressor efficiency, Eq. 6.48.

❑

retrieve data for Refriger-

ant 22.

✓

Skills Developed

➊

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330 Chapter 6

Using Entropy

More generally, temperature varies as the gas or liquid flows through the control

volume. We can consider such a temperature variation to consist of a series of infin-

itesimal steps. Then, the heat transfer per unit of mass is given as

int

rev

T ds

(6.49)

The subscript “int rev” serves to remind us that the expression applies only to control

volumes in which there are no internal irreversibilities. The integral of Eq. 6.49 is

performed from inlet to exit. When the states visited by a unit mass as it passes

reversibly from inlet to exit are described by a curve on a T–s diagram, the magnitude

of the heat transfer per unit of mass flowing can be represented as the area under

the curve, as shown in Fig. 6.13.

6.13.2

Work

The work per unit of mass passing through a one-inlet, one-exit control volume can

be found from an energy rate balance, which reduces at steady state to give

1 1h

2 h

21 a

2 V

b1 g1z

2 z

This equation is a statement of the conservation of energy principle that applies when

irreversibilities are present within the control volume as well as when they are absent.

However, if consideration is restricted to the internally reversible case, Eq. 6.49 can

be introduced to obtain

int

T ds 1 1h

2 h

21 a

2 V

b1 g1z

2 z

(6.50)

where the subscript “int rev” has the same significance as before.

Since internal irreversibilities are absent, a unit of mass traverses a sequence of

equilibrium states as it passes from inlet to exit. Entropy, enthalpy, and pressure

changes are therefore related by Eq. 6.10b

T ds 5 dh 2 y d

which on integration gives

T ds 5 1h

2 h

y dp

Introducing this relation, Eq. 6.50 becomes

int

rev

y dp 1 a

2 V

b1 g1z

2 z

(6.51a)

When the states visited by a unit of mass as it passes reversibly from inlet to exit are

described by a curve on a p–y diagram as shown in Fig. 6.14, the magnitude of the

integral

y dp is represented by the shaded area behind the curve.

Equation 6.51a is applicable to devices such as turbines, compressors, and pumps.

In many of these cases, there is no significant change in kinetic or potential energy

from inlet to exit, so

int

y dp



1¢ke 5 ¢pe 5 02

(6.51b)

This expression shows that the work value is related to the magnitude of the specific

volume of the gas or liquid as it flows from inlet to exit.

vdp

Fig. 6.14 Area representation

y dp.

Fig. 6.13 Area representation

of heat transfer for an

internally reversible flow

process.

T ds

–––

int

rev

()

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consider two devices: a pump through which liquid water passes

and a compressor through which water vapor passes. For the same pressure rise, the

pump requires a much smaller work input per unit of mass flowing than the compressor

because the liquid specific volume is much smaller than that of vapor. This conclusion

is also qualitatively correct for actual pumps and compressors, where irreversibilities are

present during operation.

b b b b b

If the specific volume remains approximately constant, as in many applications

with liquids, Eq. 6.51b becomes

int

rev

52y1p

2 p





1y 5 constant, ¢ke 5 ¢pe 5 02

(6.51c)

Equation 6.51a also can be applied to study the performance of control volumes

at steady state in which W

is zero, as in the case of nozzles and diffusers. For any

such case, the equation becomes

y dp 1 a

2 V

b1 g1z

2 z

25 0

(6.52)

which is a form of the Bernoulli equation frequently used in fluid mechanics.

Bernoulli equation

6.13.3

Work In Polytropic Processes

We have identified an internally reversible process described by py

5 constant,

where n is a constant, as a polytropic process (see Sec. 3.15 and the discussion of Fig.

6.10). If each unit of mass passing through a one-inlet, one-exit control volume under-

goes a polytropic process, then introducing py

5 constant in Eq. 6.51b, and performing

the integration, gives the work per unit of mass in the absence of internal irrevers-

ibilities and significant changes in kinetic and potential energy. That is,

int

y dp 521constant2

2 1

2 p





1polytropic, n ? 12

(6.53)

BIOCONNECTIONS Bats, the only mammals that can fly, play several impor-

tant ecological roles, including feeding on crop-damaging insects. Currently, nearly one-

quarter of bat species is listed as endangered or threatened. For unknown reasons,

bats are attracted to large wind turbines, where some perish by impact and others from hem-

orrhaging. Near rapidly moving turbine blades there is a drop in air pressure that expands the

lungs of bats, causing fine capillaries to burst and their lungs to fill with fluid, killing them.

The relationship between air velocity and pressure in these instances is captured by the

following differential form of Eq. 6.52, the Bernoulli equation:

y dp 5 2

V dV

which shows that as the local velocity V increases, the local pressure p decreases. The pressure

reduction near the moving turbine blades is the source of peril to bats.

Some say migrating bats experience most of the fatalities, so the harm may be decreased

at existing wind farms by reducing turbine operation during peak migration periods. New

wind farms should be located away from known migratory routes.

6.13 Heat Transfer and Work in Internally Reversible, Steady-State Flow Processes 331

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332 Chapter 6 Using Entropy

Equation 6.53 applies for any value of n except n 5 1. When n 5 1, py 5 constant,

and the work is

int

rev

y dp 52constant

521p

2 ln1p



1polytropic, n 5 12 (6.54)

Equations 6.53 and 6.54 apply generally to polytropic processes of any gas (or liquid).

IDEAL GAS CASE. For the special case of an ideal gas, Eq. 6.53 becomes

int

rev

n 2 1

2 T



1ideal gas, n ? 12

(6.55a)

For a polytropic process of an ideal gas, Eq. 3.56 applies:

5 a

n21

Thus, Eq. 6.55a can be expressed alternatively as

int

rev

nRT

n 2 1

1n212

2 1 d



1ideal gas, n ? 12

(6.55b)

For the case of an ideal gas, Eq. 6.54 becomes

int

rev

52RT ln1p



1ideal gas, n 5 12

(6.56)

In Example 6.15, we consider air modeled as an ideal gas undergoing a polytropic

compression process at steady state.

Determining Work and Heat Transfer for a Polytropic Compression of Air

c c c c EXAMPLE 6.15 c

An air compressor operates at steady state with air entering at p

5 1 bar, T

5 208C, and exiting at p

5 5 bar.

Determine the work and heat transfer per unit of mass passing through the device, in kJ/kg, if the air undergoes

a polytropic process with n 5 1.3. Neglect changes in kinetic and potential energy between the inlet and the exit.

Use the ideal gas model for air.

SOLUTION

Known:

Air is compressed in a polytropic process from a specified inlet state to a specified exit pressure.

Find: Determine the work and heat transfer per unit of mass passing through the device.

Schematic and Given Data:

5 bar

1 bar

Shaded area = magnitude of (W

/m)

int

rev

1.3

= constant

Engineering Model:

A control volume enclosing the compressor is at steady state.

2. The air undergoes a polytropic process with n 5 1.3.

3. The air behaves as an ideal gas.

4. Changes in kinetic and potential energy from inlet to exit

can be neglected.

Fig. E6.15

➊

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