1 mohammad suliman abuhaiba, ph.d.,...

Chapter 6

Using Entropy

2/4/2015 5:00 PM

Mohammad Suliman Abuhaiba, Ph.D., P.E. 1

Home Work Assignment of Ch6

1, 10, 18, 25, 36, 46, 53, 63,

70, 77,84, 96, 104, 112, 120

Due Sunday 8/2/2015

2/4/2015 8:42 PM

Mohammad Suliman Abuhaiba, Ph.D., P.E.

2

Chapter Objective

Means are introduced for analyzing

systems from the 2nd law perspective as

they undergo processes.

Objective: introduce entropy and show its

use for thermodynamic analysis.

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6.1 Introducing Entropy

Clausius inequality

Clausius inequality states that for any

thermodynamic cycle

Q = heat transfer at a part of the system

boundary during a portion of the cycle

T = absolute temperature at that part of the

boundary

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Clausius inequality

Integral is to be performed over all parts of

boundary and over the entire cycle.

Equality applies when there are no internal

irreversibilities as the system executes the cycle

Inequality applies when internal irreversibilities

are present.

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Clausius inequality

Equation 6.1 can be expressed equivalently as

scycle is a measure of the effect of irreversibilities

present within the system executing the cycle.

scycle = 0, no irreversibilities present within the system

scycle > 0, irreversibilities present within the system

scycle < 0, impossible

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Defining Entropy Change

A quantity is a property if, and only if, its

change in value between two states is

independent of the process.

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Defining Entropy Change

Integral of Q/T has same value for any internally

reversible process between the two states.

Value of integral depends on end states only.

Integral represents change in some property of

the system, called entropy, its change is given by

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Retrieving Entropy Data

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TdS equations are developed by considering a

pure, simple compressible system undergoing an

internally reversible process.

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In the absence of overall system motion and

effects of gravity, an energy balance in

differential form is

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Retrieving Entropy Data Entropy Change of an Ideal Gas

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Retrieving Entropy Data Entropy Change of an Ideal Gas

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Entropy Change of an Ideal Gas

Value of specific entropy is set to zero at state

temperature of 0 K & pressure of 1 atmosphere.

Using Eq. 6.19, specific entropy at a state where

temperature is T & pressure is 1 atm is

determined relative to this reference state and

reference value as

Tables A-22

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Entropy Change of an Ideal Gas

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Entropy Change of an Ideal Gas ASSUMING CONSTANT SPECIFIC HEATS

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Retrieving Entropy Data Entropy Change of an Incompressible Substance

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Entropy Change in Internally

Reversible Processes

When a closed system undergoing an

internally reversible process receives energy

by heat transfer, the system experiences an

increase in entropy.

Conversely, when energy is removed from the

system by heat transfer, entropy of the

system decreases.

An entropy transfer accompanies heat

transfer.

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Direction of entropy transfer is same as

that of heat transfer.

Isentropic process: A constant-entropy

process, adiabatic internally reversible

process.

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Carnot cycle

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EXAMPLE 6.1 Internally Reversible Process of Water

Water, initially a saturated liquid at 100°C, is contained in a piston–cylinder assembly. The water undergoes a process to the corresponding saturated vapor state, during which the piston moves freely in the cylinder. If the change of state is brought about by heating the water as it undergoes an internally reversible process at constant pressure and temperature, determine the work and heat transfer per unit of mass, each in kJ/kg.

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EXAMPLE 6.1 Internally Reversible Process of Water

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Entropy Balance for Closed

Systems - Developing the Entropy Balance

The cycle consists of process I,

during which internal

irreversibilities are present,

followed by internally reversible

process R. For this cycle, Eq. 6.2

takes the form

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closed system entropy balance

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If end states are fixed, entropy change on LS of

Eq. 6.27 can be evaluated independently of the

details of process.

The two terms on RS depend explicitly on nature

of process and cannot be determined solely from

knowledge of end states.

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1st term is associated with heat transfer to or

from system during the process.

Entropy transfer accompanying heat transfer.

Direction of entropy transfer is same as direction of

heat transfer, and same sign convention applies as for

heat transfer: A positive value: entropy is transferred into system

A negative value : entropy is transferred out

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When there is no heat transfer, there is no entropy

transfer.

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Entropy change of a system is not accounted

for solely by entropy transfer, but is due in part

to 2nd term on RS of Eq. 6.27 denoted by s.

s is positive when internal irreversibilities are

present and vanishes when no internal

irreversibilities are present.

Entropy is produced within the system by the

action of irreversibilities.

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2nd law of thermodynamics: entropy is

produced by irreversibilities and conserved

only in the limit as irreversibilities are reduced

to zero.

Since s measures effect of irreversibilities

present within the system during a process, its

value depends on the nature of the process

and not solely on the end states. It is not a

property.

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2nd law requires that entropy production be

positive, or zero, in value

The value of entropy production cannot be

negative.

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Change in entropy of system may be positive,

negative, or zero:

Entropy change can be determined without

knowledge of the details of the process.

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Temperature at portion

of boundary where heat

transfer occurs is the

same as the constant

temperature of the

reservoir, Tb.

Reservoir is free of

irreversibilities

The system is not

without irreversibilities,

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Apply entropy balance to the system and to the

reservoir.

Since Tb is constant, integral in Eq. 6.27 is readily

evaluated, and entropy balance for the system

reduces to

Q/Tb accounts for entropy transfer into the

system accompanying heat transfer Q.

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Entropy balance for reservoir takes the form

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Entropy of reservoir decreases by an amount

equal to entropy transferred from it to the

system.

As shown by Eq. 6.30, entropy change of

system exceeds amount of entropy transferred

to it because of entropy production within the

system.

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If heat transfer were passing instead from the

system to reservoir, magnitude of entropy

transfer would remain the same, but its

direction would be reversed.

In such a case, entropy of system would

decrease if amount of entropy transferred from

system to reservoir exceeded amount of

entropy produced within the system due to

irreversibilities.

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If heat transfer takes place at several

locations on boundary of a system where

temperatures do not vary with position or

time,

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Systems - Other Forms of Entropy Balance

Closed system entropy rate balance

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Systems - Other Forms of Entropy Balance

Value of entropy production for a given process

of a system often does not have much

significance by itself.

Significance is normally determined through

comparison.

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Systems - Evaluating Entropy Production and Transfer

By comparing entropy production values,

components where appreciable

irreversibilities occur can be identified and

rank ordered.

This allows attention to be focused on

components that contribute most to

inefficient operation of overall system.

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Systems - Evaluating Entropy Production and Transfer

EXAMPLE 6.2

Irreversible Process of Water

Water initially a saturated liquid at 100°C is

contained within a piston–cylinder assembly. The

water undergoes a process to the corresponding

saturated vapor state, during which the piston

moves freely in the cylinder. There is no heat

transfer with the surroundings. If the change of

state is brought about by the action of a paddle

wheel, determine net work per unit mass, in kJ/kg,

and amount of entropy produced per unit mass, in

kJ/kg.K

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EXAMPLE 6.2

Irreversible Process of Water

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EXAMPLE 6.3 Evaluating Minimum Theoretical Compression Work

Refrigerant 134a is compressed adiabatically in a

piston–cylinder assembly from saturated vapor at

0°C to a final pressure of 0.7 MPa. Determine the

minimum theoretical work input required per unit

mass of refrigerant, in kJ/kg.

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EXAMPLE 6.3 Evaluating Minimum Theoretical Compression Work

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EXAMPLE 6.4 Pinpointing Irreversibilities

Referring to Example 2.4, evaluate the rate of

entropy production in kW/K, for

a. the gearbox as the system

b. enlarged system consisting of the gearbox and

enough of its surroundings that heat transfer

occurs at the temperature of the surroundings

away from the immediate vicinity of the

gearbox, Tf = 293 K (20C).

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EXAMPLE 6.4 Pinpointing Irreversibilities

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An enlarged system comprising a system

and portion of surroundings.

Isolated system: enlarged system where all

energy and mass transfers taking place are

included within boundary of the system

An energy balance for the isolated system

reduces to

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Systems - Increase of Entropy Principle

because no energy transfers take place

across its boundary. Thus, energy of

isolated system remains constant.

Since energy is an extensive property, its

value for the isolated system is sum of its

values for the system and surroundings,

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For a process to take place, it is necessary for the

energy of the system plus the surroundings to

remain constant.

However, not all processes for which this

constraint is satisfied can actually occur.

Processes also must satisfy the 2nd law. An

entropy balance for the isolated system reduces to

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sisol = total amount of entropy produced within

system and its surroundings

Since entropy is produced in all actual

processes, only processes that can occur are

those for which the entropy of the isolated

system increases

This is known as increase of entropy principle.

Alternative statement of 2nd law.

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Since entropy is an extensive property, its value

for the isolated system is the sum of its values

for the system and surroundings,

This equation does not require the entropy

change to be positive for both system and

surroundings but only that sum of the changes

is positive.

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Increase of entropy principle: direction in which

any process can proceed is direction that causes

total entropy of system plus surroundings to

increase.

Tendency of systems left to themselves to

undergo processes until a condition of equilibrium

is attained.

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Entropy of an isolated system increases as the

state of equilibrium is approached, with the

equilibrium state being attained when the entropy

reaches a maximum.

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EXAMPLE 6.5 Quenching a Hot Metal Bar

A 0.3 kg metal bar initially at 1200°K is removed from an

oven and quenched by immersing it in a closed tank

containing 9 kg of water initially at 300°K. Each substance

can be modeled as incompressible. An appropriate

constant specific heat value for the water is cw = 4.2 kJ/kg

K, and an appropriate value for the metal is cm = 0.42

kJ/kg K. Heat transfer from the tank contents can be

neglected. Determine

a. Final equilibrium temperature of the metal bar and the

water, in K

b. Amount of entropy produced, in kJ/

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EXAMPLE 6.5 Quenching a Hot Metal Bar

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Entropy is associated with the notion of

disorder and the 2nd law statement that

entropy of an isolated system undergoing a

spontaneous process tends to increase is

equivalent to saying that the disorder of the

isolated system tends to increase.

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Systems - Statistical Interpretation of Entropy

Entropy Rate Balance for Control

Volumes

Like mass and energy, entropy is an extensive property,

so it too can be transferred into or out of a control

volume by streams of matter.

control volume entropy rate balance

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Volumes

INTEGRAL FORM

Scv(t) = total entropy associated with

control volume at time t,

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Volumes

INTEGRAL FORM

= heat flux, time rate of heat transfer per unit

of surface area, through the location on the

boundary where the instantaneous temperature

is T.

Vn = normal component in the direction of flow

of the velocity relative to the flow area.

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Volumes - Steady State

conservation of mass

energy rate balance

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Volumes - Steady State

steady-state form of the entropy rate balance

Mass and energy are conserved quantities, but entropy is

not conserved

Eq. 6.39 requires that rate at which entropy is transferred

out must exceed rate at which entropy enters, the

difference being rate of entropy production within the

control volume owing to irreversibilities.

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Volumes - Steady State, One-inlet, One-exit Control Volumes

From Eq. 6.40, entropy of a unit of mass passing from inlet

to exit can increase, decrease, or remain the same.

Because value of 2nd term on RS can never be negative, a

decrease in specific entropy from inlet to exit can be

realized only when more entropy is transferred out of the

control volume accompanying heat transfer than is

produced by irreversibilities within the control volume.

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Volumes - Steady State, One-inlet, One-exit Control Volumes

When value of this entropy transfer term is

positive, specific entropy at exit is greater than

specific entropy at inlet whether internal

irreversibilities are present or not.

In the special case where there is no entropy

transfer accompanying heat transfer, Eq. 6.40

reduces to

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EXAMPLE 6.6 Entropy Production in a Steam Turbine

Steam enters a turbine with a pressure of 30 bar, a

temperature of 400°C, and a velocity of 160 m/s.

Saturated vapor at 100°C exits with a velocity of 100

m/s. At steady state, the turbine develops work equal to

540 kJ/kg of steam flowing through the turbine. Heat

transfer between the turbine and its surroundings

occurs at an average outer surface temperature of 350

K. Determine the rate at which entropy is produced

within the turbine per kg of steam flowing, in Neglect the

change in potential energy between inlet and exit.

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EXAMPLE 6.6 Entropy Production in a Steam Turbine

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EXAMPLE 6.7 Evaluating a Performance Claim

An inventor claims to have developed a device requiring no

energy transfer by work or heat transfer, yet able to produce

hot and cold streams of air from a single stream of air at an

intermediate temperature. The inventor provides steady-

state test data indicating that when air enters at a

temperature of 39°C and a pressure of 5.0 bars, separate

streams of air exit at temperatures of 18°C and 79°C,

respectively, and each at a pressure of 1 bar. 60% of the

mass entering the device exits at the lower temperature.

Evaluate the inventor’s claim, employing the ideal gas model

for air and ignoring changes in the kinetic and potential

energies of the streams from inlet to exit.

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EXAMPLE 6.7 Evaluating a Performance Claim

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EXAMPLE 6.8 Entropy Production in Heat Pump Components

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

5°C, 3.5 bar and is compressed adiabatically to 75°C, 14 bar. From the

compressor, the refrigerant passes through the condenser, where it condenses

to liquid at 28°C, 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 20°C, 1 bar with

a volumetric flow rate of 0.42 m3/s and exits at 50°C 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).

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EXAMPLE 6.8 Entropy Production in Heat Pump Components

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Isentropic Processes General Considerations

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For two states having same

specific entropy, Eq. 6.21a

reduces to

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Isentropic Processes Using the Ideal Gas Model - Ideal Gas Tables

Eqs. 6.22 and 6.23 reduce to:

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Isentropic Processes Using the Ideal Gas Model

Assuming Constant Specific Heats

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Isentropic Processes Using the Ideal Gas Model

Assuming Constant Specific Heats

EXAMPLE 6.9 Isentropic Process of Air

Air undergoes an isentropic process from

p1 = 1 bar, T1 = 300K to a final state

where the temperature is T2 = 650K.

Employing the ideal gas model, determine

the final pressure p2, in atm. Solve using

a. Table A-22

b. A constant specific heat ratio k

evaluated at the mean temperature,

475K, from Table A-20.

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EXAMPLE 6.10 Air Leaking from a Tank

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.

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EXAMPLE 6.10 Air Leaking from a Tank

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Isentropic Efficiencies of Turbines,

Nozzles, Compressors, and Pumps

Isentropic efficiencies involve a comparison

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.

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Isentropic Turbine Efficiency

State of matter entering the turbine and exit pressure are fixed.

Heat transfer between turbine and its surroundings is ignored, as are kinetic and potential energy effects.

Mass & energy rate balances reduce, at steady state, to give the work developed per unit of mass flowing through the turbine

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Nozzles, Compressors, and Pumps Isentropic Turbine Efficiency

Since there is no heat transfer, the allowed exit states are constrained by Eq. 6.41

Because entropy production cannot be negative, states with s2 < s1 are not accessible in an adiabatic expansion.

The only states that actually can be attained adiabatically are those with s2 > s1.

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ISENTROPIC NOZZLE EFFICIENCY

The isentropic nozzle efficiency = ratio of actual

specific kinetic energy of gas leaving the nozzle, to

kinetic energy at exit that would be achieved in an

isentropic expansion between the same inlet state

and the same exhaust pressure,

Nozzle efficiencies of 95% or more are common,

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Isentropic Compressor and Pump Efficiencies

State of matter entering the compressor and exit pressure are fixed.

Negligible heat transfer with surroundings and no appreciable kinetic and potential energy effects.

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EXAMPLE 6.11 Evaluating Turbine Work Using the Isentropic

Efficiency

A steam turbine operates at steady state with inlet

conditions of p1 = 5 bar, T1 = 320°C. Steam leaves

the turbine 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.

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EXAMPLE 6.11 Evaluating Turbine Work Using the Isentropic

Efficiency

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EXAMPLE 6.12 Evaluating the Isentropic Turbine Efficiency

A turbine operating at steady state receives air at

a pressure of p1 = 3.0 bar and a temperature of T1

= 390 K. Air exits the turbine at a pressure of p2 =

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.

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EXAMPLE 6.12 Evaluating the Isentropic Turbine Efficiency

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EXAMPLE 6.13 Evaluating the Isentropic Nozzle Efficiency

Steam enters a nozzle operating at steady state at

p1 = 1.0 MPa and T1 = 320°C with a velocity of 30

m/s. The pressure and temperature at the exit are

p2 = 0.3 MPa and T2 = 180°C. There is no

significant heat transfer between the nozzle and

its surroundings, and changes in potential energy

between inlet and exit can be neglected.

Determine the nozzle efficiency.

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EXAMPLE 6.13 Evaluating the Isentropic Nozzle Efficiency

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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 data from property

tables,

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Heat Transfer and Work in Internally

Reversible, Steady-State Flow Processes

Heat Transfer

For a control volume at steady

state in which flow is both

isothermal and internally

reversible, the appropriate form of

the entropy rate balance is

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Work in Polytropic Processes

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Work in Polytropic Processes - Ideal Gas Case

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EXAMPLE 6.15 Polytropic Compression of Air

An air compressor operates at

steady state with air entering at

p1 = 1 bar, T1 = 20°C, and

exiting at p2 = 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 = 1.3. Neglect changes

in kinetic and potential energy

between the inlet and the exit.

Use the ideal gas model for air.

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1 mohammad suliman abuhaiba, ph.d.,...

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