1 mohammad suliman abuhaiba, ph.d.,...
TRANSCRIPT
Chapter 6
Using Entropy
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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
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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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6.1 Introducing Entropy
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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6.1 Introducing Entropy
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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Retrieving Entropy Data
TdS equations are developed by considering a
pure, simple compressible system undergoing an
internally reversible process.
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Retrieving Entropy Data
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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Retrieving Entropy Data
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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Retrieving Entropy Data
Entropy Change of an Ideal Gas
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Retrieving Entropy Data
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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Entropy Change in Internally
Reversible Processes
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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Entropy Change in Internally
Reversible Processes
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Entropy Change in Internally
Reversible Processes
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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Entropy Balance for Closed
Systems - Developing the Entropy Balance
closed system entropy balance
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Entropy Balance for Closed
Systems - Developing the Entropy Balance
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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Entropy Balance for Closed
Systems - Developing the Entropy Balance
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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Entropy Balance for Closed
Systems - Developing the Entropy Balance
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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Entropy Balance for Closed
Systems - Developing the Entropy Balance
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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Entropy Balance for Closed
Systems - Developing the Entropy Balance
2nd law requires that entropy production be
positive, or zero, in value
The value of entropy production cannot be
negative.
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Entropy Balance for Closed
Systems - Developing the Entropy Balance
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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Entropy Balance for Closed
Systems - Developing the Entropy Balance
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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Entropy Balance for Closed
Systems - Developing the Entropy Balance
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 Closed
Systems - Developing the Entropy Balance
Entropy balance for reservoir takes the form
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Entropy Balance for Closed
Systems - Developing the Entropy Balance
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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Entropy Balance for Closed
Systems - Developing the Entropy Balance
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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Entropy Balance for Closed
Systems - Developing the Entropy Balance
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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Entropy Balance for Closed
Systems - Other Forms of Entropy Balance
Closed system entropy rate balance
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Entropy Balance for Closed
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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Entropy Balance for Closed
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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Entropy Balance for Closed
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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Entropy Balance for Closed
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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Entropy Balance for Closed
Systems - Increase of Entropy Principle
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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Entropy Balance for Closed
Systems - Increase of Entropy Principle
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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Entropy Balance for Closed
Systems - Increase of Entropy Principle
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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Entropy Balance for Closed
Systems - Increase of Entropy Principle
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 Balance for Closed
Systems - Increase of Entropy Principle
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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Entropy Balance for Closed
Systems - Increase of Entropy Principle
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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Entropy Balance for Closed
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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Entropy Rate Balance for Control
Volumes
INTEGRAL FORM
Scv(t) = total entropy associated with
control volume at time t,
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Entropy Rate Balance for Control
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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Entropy Rate Balance for Control
Volumes - Steady State
conservation of mass
energy rate balance
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Entropy Rate Balance for Control
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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Entropy Rate Balance for Control
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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Entropy Rate Balance for Control
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 Efficiencies of Turbines,
Nozzles, Compressors, and Pumps
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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Isentropic Efficiencies of Turbines,
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 Efficiencies of Turbines,
Nozzles, Compressors, and Pumps
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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Isentropic Efficiencies of Turbines,
Nozzles, Compressors, and Pumps
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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Heat Transfer and Work in Internally
Reversible, Steady-State Flow Processes
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Heat Transfer and Work in Internally
Reversible, Steady-State Flow Processes
Work in Polytropic Processes
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Heat Transfer and Work in Internally
Reversible, Steady-State Flow Processes
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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