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

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246 Chapter 6 Using Entropy



6.8 Isentropic Efficiencies of Turbines,

Nozzles, Compressors, and Pumps

Engineers make frequent use of efficiencies and many different efficiency definitions are em-

ployed. In the present section isentropic efficiencies for turbines, nozzles, compressors, and

pumps are introduced. Isentropic efficiencies involve a comparison between the actual per-

formance of a device and the performance that would be achieved under idealized circum-

stances for the same inlet state and the same exit pressure. These efficiencies are frequently

used in subsequent sections of the book.

ISENTROPIC TURBINE EFFICIENCY

To introduce the isentropic turbine efficiency, refer to Fig. 6.12, which shows a turbine expan-

sion 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

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. Since there is no heat trans-

fer, the allowed exit states are constrained by Eq. 6.41

Because the entropy production cannot be negative, states with s

 s

are not acces-

sible in an adiabatic expansion. The only states that actually can be attained adiabatically

are those with s

 s

. The state labeled “2s” on Fig. 6.12 would be attained only in the limit

of no internal irreversibilities. This corresponds to an isentropic expansion through the tur-

bine. For fixed exit pressure, the specific enthalpy h

decreases as the specific entropy s



 s

 s

 0

 h

 h

Actual

expansion

Isentropic

expansion

Accessible

states

– h

 Figure 6.12 Comparison of actual and

isentropic expansions through a turbine.

6.8 Isentropic Efficiencies of Turbines, Nozzles, Compressors, and Pumps 247

decreases. Therefore, the smallest allowed value for h

corresponds to state 2s, and the maxi-

mum value for the turbine work is

In an actual expansion through the turbine h

 h

, and thus less work than the maxi-

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

defined by

(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 

is typically 0.7 to 0.9 (70–90%).

ISENTROPIC NOZZLE EFFICIENCY

A similar approach to that for turbines can be used to introduce the isentropic efficiency 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, to the kinetic energy

at the exit that would be achieved in an isentropic expansion between the same inlet state

and the same exhaust pressure,

(6.49)

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

nearly free of internal irreversibilities.

ISENTROPIC COMPRESSOR AND PUMP EFFICIENCIES

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

Fig. 6.13, which shows a compression process on a Mollier diagram. The state of the mat-

ter entering the compressor and the exit pressure are fixed. For negligible heat transfer with

nozzle









 h

 h

– h

Accessible

states

Actual

compression

Isentropic

compression

 Figure 6.13 Comparison of actual and

isentropic compressions.



isentropic turbine

efficiency

isentropic nozzle

efficiency

248 Chapter 6 Using Entropy

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the surroundings and no appreciable kinetic and potential energy effects, the work input per

unit of mass flowing through the compressor is

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 that would be

achieved in an isentropic compression from the specified inlet state to the specified exit pres-

sure. The minimum work input is given, therefore, by

In an actual compression, h

 h

, and thus more work than the minimum would be re-

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

(6.50)

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

and the same exit pressure. The value of 

is typically 75 to 85% for compressors. An

isentropic pump efficiency, 

, is defined similarly.

The series of four examples to follow illustrate various aspects of isentropic efficiencies

of turbines, nozzles, and compressors. Example 6.11 is a direct application of the isentropic

turbine efficiency 

to a steam turbine. Here, 

is known and the objective is to determine

the turbine work.



1W



1W



a

 h

 h

a

b h

 h

isentropic compressor

efficiency

isentropic pump

efficiency



6.8 Isentropic Efficiencies of Turbines, Nozzles, Compressors, and Pumps 249

Analysis: The work developed can be determined using the isentropic turbine efficiency, Eq. 6.48, which on rearrangement

gives

From Table A-4, h

 3105.6 kJ/kg and s

 7.5308 The exit state for an isentropic expansion is fixed by

 1 bar and s

 s

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

 2743.0 kJ/kg. Substituting

values

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 illus-

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

 0.7513105.6  2743.02 271.95 kJ/kg

kJ/kg

 h

 h

Assumptions:

1. A control volume enclosing the turbine is at steady state.

2. The expansion is adiabatic and changes in kinetic and potential

energy between the inlet and exit can be neglected.

 Figure E6.11

– h

= 1 bar

= 5 bar

Actual

expansion

Isentropic

expansion

Accessible

states

EXAMPLE 6.11 Evaluating Turbine Work Using the Isentropic Efficiency

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

 5 bar, T

 320C. Steam leaves the turbine at a pres-

sure 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:

❶

The next example 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.

250 Chapter 6 Using Entropy

EXAMPLE 6.12 Evaluating the Isentropic Turbine Efficiency

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

 3.0 bar and a temperature of T

 390 K. Air exits the

turbine at a pressure of p

 1.0 bar. The work developed is measured as 74 kJ per kg of air flowing 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 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

Assumptions:

1. 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. 6.48, 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

From Table A-22 at 390 K, h

 390.88 kJ/kg. To determine h

, use Eq. 6.43

With p

 3.0 bar, p

 1.0 bar, and p

 3.481 from Table A-22 at 390 K

2 a

1.0

3.0

b 13.4812 1.1603

2 a

b p

 h

 h

 Figure E6.12

6.8 Isentropic Efficiencies of Turbines, Nozzles, Compressors, and Pumps 251

Interpolation in Table A-22 gives h

 285.27 kJ/kg. Thus

Substituting values into Eq. 6.48







74 kJ/kg

105.6 kJ/kg

 0.70 170%2

 390.88  285.27  105.6 kJ/kg

EXAMPLE 6.13 Evaluating the Isentropic Nozzle Efficiency

Steam enters a nozzle operating at steady state at p

 1.0 MPa and T

 320C with a velocity of 30 m/s. The pressure and

temperature at the exit are p

 0.3 MPa and T

 180C. There is no significant heat transfer between the nozzle and its sur-

roundings, 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:

In the next example, the objective is to determine the isentropic efficiency of a steam

nozzle.

Isentropic

expansion

Actual

expansion

0.3 MPa

1.0 MPa

320°C

180°C

= 0.3 MPa

= 180°C

= 1.0 MPa

= 320°C

= 30 m/s

Steam

nozzle

Assumptions:

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

2. For the control volume, and the change in potential energy between inlet and exit can be neglected.

Analysis: The nozzle efficiency given by Eq. 6.49 requires the actual specific kinetic energy at the nozzle exit and the spe-

cific kinetic energy that would be achieved at the exit in an isentropic expansion from the given inlet state to the given exit

pressure. The mass and energy rate balances for the one-inlet, one-exit control volume at steady state reduce to give

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

 h

 h



 0

 Figure E6.13

252 Chapter 6 Using Entropy

From Table A-4 at T

 320C and p

 1.0 MPa, h

 3093.9 kJ/kg, s

 7.1962 Also, with T

 180C

and p

 0.3 MPa, h

 2823.9 kJ/kg. Thus, the actual specific kinetic energy at the exit in kJ/kg is

Interpolating in Table A-4 at 0.3 MPa, with s

 s

, results in h

 2813.3 kJ/kg. Accordingly, the specific kinetic energy

at the exit for an isentropic expansion is

Substituting values into Eq. 6.49

The principal irreversibility in nozzles is friction between the flowing gas or liquid and the nozzle wall. The effect of fric-

tion 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.

nozzle







270.45

281.05

 0.962 196.2%2

 3093.9  2813.3 

2110

 281.05 kJ /kg

 270.45

 (3093.9  2823.9)



1302

1 N

1 kg

m/s

1 kJ

kJ/kg

❶

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

using data from property tables and then using IT.

EXAMPLE 6.14 Evaluating the Isentropic Compressor Efficiency

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:

Assumptions:

1. A control volume enclosing the compressor is at steady state.

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

tween the inlet and the exit can be neglected.

 Figure E6.14

14 bar

3.5 bar

= 75°C

= –5°C

6.8 Isentropic Efficiencies of Turbines, Nozzles, Compressors, and Pumps 253

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

From Table A-9, h

 249.75 kJ/kg and h

 294.17 kJ/kg. Thus

The isentropic compressor efficiency is determined using Eq. 6.50

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

process, as calculated 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

 0.9572 kJ/kg K. With s

 s

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

 285.58 kJ/kg. Sub-

stituting values

(b) The IT program follows. In the program, is denoted as Wdot, as mdot, and 

as eta_c.

// Given Data:

T1 = –5 // °C

p1 = 3.5 // bar

T2 = 75 // °C

p2 = 14 // bar

mdot = 0.07 // kg/s

// Determine the specific enthalpies.

h1 = h_PT(“R22”,p1,T1)

h2 = h_PT(“R22”,p2,T2)

// Calculate the power.

Wdot = mdot * (h1 – h2)

// Find h2s:

s1 = s_PT(“R22”,p1,T1)

s2s = s_Ph(“R22”,p2,h2s)

s2s = s1

// Determine the isentropic compressor efficiency.

eta_c = (h2s – h1) / (h2 – h1)

Use the Solve button to obtain: and which agree closely with the values obtained above.

The minimum theoretical power for adiabatic compression from state 1 to the exit pressure of 14 bar would be

The magnitude of the actual power required is greater than the ideal power due to irreversibilities.

Note that IT solves for the value of h

even though it is an implicit variable in the argument of the specific entropy

function.

 m

 h

2 10.0721249.75  285.5822.51 kW

 80.58%,W

3.111 kW



1285.58  249.752

1294.17  249.752

 0.81181%2



1W



1W





 h

 10.07 kg/s21249.75  294.172 kJ/kg `

1 kW

1 kJ/s

`3.11 kW

 m

 h

❷

❶

254 Chapter 6 Using Entropy



6.9 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

derive expressions for the heat transfer and the work in the absence of internal irreversibili-

ties. The resulting expressions have several important applications.

HEAT TRANSFER

For a control volume at steady state in which the flow is both isothermal and internally re-

versible, the appropriate form of the entropy rate balance is

where 1 and 2 denote the inlet and exit, respectively, and is the mass flow rate. Solving

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

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

volume. However, we can consider the temperature variation to consist of a series of infini-

tesimal steps. Then, the heat transfer per unit of mass would be given as

(6.51)

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. 6.51 is per-

formed 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. 6.14.

WORK

The work per unit of mass passing through the control volume can be found from an energy

rate balance, which reduces at steady state to give

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

versibilities are present within the control volume as well as when they are absent. However,

if consideration is restricted to the internally reversible case, Eq. 6.51 can be introduced to

obtain

(6.52)a

int

rev





T ds  1h

 h

2 a

 V

b g1z

 z



 1h

 h

2 a

 V

b g1z

 z

rev

int





T ds

 T

 s

0 

 m

 s

2 s



 Figure 6.14 Area

representation of heat

transfer for an internally

reversible flow process.

T ds

–––

int

rev

()

6.9 Heat Transfer and Work in Internally Reversible, Steady-State Flow Processes 255

 Figure 6.15 Area

representation of



dp.

vdp



where the subscript “int rev” has the same significance as before. Since internal irre-

versibilities 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. 6.12b

which on integration gives

Introducing this relation, Eq. 6.52 becomes

(6.53a)

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

scribed by a curve on a p–v diagram as shown in Fig. 6.15, the magnitude of the integral

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

Equation 6.53a may be applied 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

(6.53b)

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

the gas or liquid as it flows from inlet to exit.  for example. . . 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 would require a much smaller work input per

unit of mass flowing than would 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. 

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

uids, Eq. 6.53b becomes

(6.53c)

Equation 6.53a can also be applied to study the performance of control volumes at steady

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

equation becomes

(6.54)

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



v dp  a

 V

b g1z

 z

2 0

int

rev

v1p

 p

1v  constant, ¢ke  ¢pe  02

int

rev





v dp

1¢ke  ¢pe  02



int

rev





v dp  a

 V

b g1z

 z



T ds  1h

 h

2



v dp

T ds  dh  v dp

Bernoulli equation