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Lecture 1
Systems and Measurement
Sections 1.1 1.4
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Why Thermodynamics?And where these techniques applied?
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infrastructure
transportation
ecosystems
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boundary
To learn about each of these, we cant treat everything all at once!
System
surroundings/
environment
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IsolatedClosed
(control mass)
Open
(control volume)
Types of systems
piston in cylinder engineinsulated bath
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Example
Are these systems isolated, closed or open?
electric car sliding on frictionless
ice, assume air drag is negligible,
boundary at surface of car
candle in a dark roomgreenhouse, windows closed,
boundary just outside structure
dog, boundary at surface
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How do you choose the boundary?
Depends on:
What you know
What you want to know
air intake
exhaust gas
-> external wheel speed
conventional car,
sliding on surface
with k = 0.1
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What are we looking at, when looking at the system?
Large scale features
Temperature
Pressure
Volume
System of particles
Molecular, atomic, quantum
energy levels
Macroscopic Microscopic
2 Regimes
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Ways we describe a system
Properties
independent of history of the
system
have numerical value
can be measured or
computed by looking at thesystem macroscopically
Examples:
pressure, temperature, mass, volume
A property is a property if, and only if, its change in value between two states is
independent of the process.
A test if a descriptor of a system is a property
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Ways we describe a system
State
the set of particular values the
properties have at a particular time
P = 1.203 x 105 Pa
T = 385 K
V = 0.3 m3
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Ways we describe a system
Process
how a system changes from one
state to another over time
steady state properties arent changing with time
equilibriumno net energy is transferred between system and surroundings
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Watch the video on
Equilibrium vs steady state
http://tll.mit.edu/help/equilibrium-vs-steady-state
from 1:30 7:30
(Of course, you can watch the rest if you like. It does describe some
engineering thought experiments, measurement and analysis but some
of the concepts are beyond the scope of this lecture)
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depends on the amount
not location dependent
can vary with time
MassLength
Volume
Shape
Energy
Temperature
Color
Pressure
Density
does not depend on the amount
can be location or time dependent
Intensive Properties
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Quantities, Units, and their Relationships
primary
secondary
(derived)
=
1 N = 1 kg m/s2SI
english 1 lbf = 1 lb * 32.1740 ft/s2
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Example (P1.13)
At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight
of 119 lbf. What is the local acceleration of gravity, in ft/s2, at that elevation?
If the balloon drifts to another elevation where g = 32.05 ft/s2,what is her
weight in lbf, and mass in lb?
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Work through For Example problemsin section 1.4
then
Work on homeworkproblem 1.10, 1.17
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Lecture 3
Measurements and Units
Sections 1.5 1.8
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Density () and Specific Volume (v)
From a macroscopic perspective, description
of matteris simplified by considering it to be
distributed continuously throughout a region.
When substances are treated as continua, it
is possible to speak of their intensivethermodynamic properties at a point.
At any instant the density (r ) at a point is
defined as
Vm
VVlim
'
r (Eq. 1.6)
where V ' is the smallest volume for which a definite
value of the ratio exists.
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Density () and Specific Volume (v)
Density is mass per unit volume.Density is an intensive property that may
vary from point to point.
SI units are (kg/m3).
English units are (lb/ft3).
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Density () and Specific Volume (v)
Specific volume is the reciprocal of
density: v = 1/r .
Specific volume is volume per unit mass.
Specific volume is an intensive propertythat may vary from point to point.
SI units are (m3/kg).
English units are (ft3/lb).
Specific volume is usually preferred for
thermodynamic analysis when working with
gases that typically have small density values.
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Molar Quantities of Matter (n)
We can quantify how much matter we have on a
molar basis instead:
M
mn (Eq. 1.8)
where M is molecular weight and
n (kmol) m (kg) M (kg/kmol)
n (lbmol) m (lb) M (lb/lbmol)n (mol, gmol) m (g) M (g/mol)
n: mol, kmol, lbmol
NA = 6.022 x 10^23 molecules/mol
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Example:
A) Is Avogadros number the same when written in terms of
# of molecules/lbmol? (1 lb = 453.6 g)
B) To the right is a list of molecular weights. If the table is in
g/mol, what would the values be in kg/kmol and lb/lbmol?
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Notation: (for some variable x)
x: x in terms of mol, kmol, lbmol
m
V
Vm
MM
n
VV
n
1
(multiply both sides by M)
M
mn
M (Eq. 1.9)
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Pressure (p)
Consider a small area A passing through a point
in a fluid at rest.The fluid on one side of the area exerts a
compressive force that is normal to the area, Fnormal.
An equal but oppositely directed force is exerted on
the area by the fluid on the other side.
The pressure (p) at the specified point is defined
as the limit
Anormal
'AA
lim Fp (Eq. 1.10)
whereA' is the area at the point in the same limiting sense
as used in the definition of density.
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Pressure Units
SI unit of pressure is the pascal:1 pascal = 1 N/m2
Multiples of the pascal are frequently used:
1 kPa = 103 N/m21 bar = 105 N/m2
1 MPa = 106 N/m2
English units for pressure are:pounds force per square foot, lbf/ft2
pounds force per square inch, lbf/in.2
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Absolute pressure: Pressure with respect tothe zero pressure of a complete vacuum.
Absolute pressure mustbe used in
thermodynamic relations.
Pressure-measuring devices often indicate
the difference between the absolute pressure of
a system and the absolute pressure of the
atmosphere outside the measuring device.
Absolute Pressure
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When system pressure is greater thanatmospheric pressure, the term gage
pressure is used.
p(gage) =p(absolute)patm(absolute)
(Eq. 1.14)
When atmospheric pressure is
greater than system pressure, the term
vacuum pressure is used.
p(vacuum) =patm(absolute) p(absolute)
(Eq. 1.15)
Gage and Vacuum Pressure
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Work through For Example problemin section 1.6
then
Work on homework problems
1.26, 1.40
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Temperature (T)
If two blocks (one warmer than the other) are
brought into contact and isolated from their
surroundings, they would interact thermally with
changes in observable properties.
When all changes in observable properties cease,the two blocks are in thermal equilibrium.
Temperature is a physical property that
determines whether the two objects are in thermal
equilibrium.
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Any object with at least one measurable property
that changes as its temperature changes can be
used as a thermometer.
Such a property is called a thermometric
property.The substance that exhibits changes in the
thermometric property is known as a thermometric
substance.
Thermometers
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Example: Liquid-in-glass thermometer
Consists of glass capillary tube connected to a bulb filledwith liquid and sealed at the other end. Space above liquid
is occupied by vapor of liquid or an inert gas.
As temperature increases, liquid expands in volume and
rises in the capillary. The length (L ) of the liquid in thecapillary depends on the temperature.
The liquid is the thermometric substance.
L is the thermometric property.
Other types of thermometers:Thermocouples
Thermistors
Radiation thermometers and optical pyrometers
Thermometers
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Kelvin scale: An absolute thermodynamic temperature
scale whose unit of temperature is the kelvin (K); an SI base
unit for temperature.
Rankine scale: An absolute thermodynamic temperature
scale with absolute zero that coincides with the absolute zeroof the Kelvin scale; an English base unit for temperature.
Temperature Scales
T(oR) = 1.8T(K) (Eq. 1.16)
Celsius scale (o
C):T(oC) = T(K)273.15 (Eq. 1.17)
Fahrenheit scale (oF):
T(
o
F) = T(
o
R)
459.67 (Eq. 1.18)
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Work through For Example problemin section 1.7
then
Work on homework problems
1.52, 1.56
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Problem-Solving Methodology
Known: Read the problem, think about it, and
identify what is known.
Find: State what is to be determined.
Schematic and Given Data: Draw a sketch of
system and label with all relevant information/data.
Engineering Model: List all simplifying
assumptions and idealizations made.
Analysis: Reduce appropriate governingequations and relationships to forms that will
produce the desired results.
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Lecture 4
Energy and Work
Sections 2.1 2.3
(not 2.2.6, 2.2.7, 2.2.8)
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Closed System Energy Balance
Energy is an extensive property thatincludes the kinetic and gravitational potential
energy of engineering mechanics.
For closed systems, energy is transferred inand out, across the system boundary, by two
means only: by work and by heat.
Energy is conserved
1st law of thermodynamics
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Closed System Energy Balance
Lets look deeper into this energy balance,
including what we mean by energy change
and energy transfer.
The closed system energy balance states:
Net amount of energy
transferred in and out
across the system boundaryby heat and work during
the time interval
The change in the
amount of energy
contained withina closed system
during some time
interval
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Change in Energy of a System
In engineering thermodynamics the changein energy of a system has three parts:
Kinetic energy
Gravitational potential energy
Internal energy
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Change in Kinetic Energy
The change in kinetic energy is associated withthe motion of the system as a whole relative to
an external coordinate frame such as the surface
of the earth.
For a system of mass m the change in kinetic
energy from state 1 to state 2 is
KE = KE2 KE1 = ( )
2
1
2
2
VV
2
1m (Eq. 2.5)
whereV1 and V2 are the initial and final velocity magnitudes.
is the final value minus initial value.
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Change in Gravitational Potential Energy
The change in gravitational potential energy isassociated with the position of the system in the
earths gravitational field.
For a system of mass m the change in potential
energy from state 1 to state 2 is
PE = PE2 PE1 = mg(z2z1) (Eq. 2.10)
wherez1 andz2 are the initial and final elevations relative to
the surface of the earth, respectively.
g is acceleration due to gravity.
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Change in Internal Energy
The change in internal energy is associated with the
makeup of the system, including its chemical
composition.
There is no simple expression like Eqs. 2.5 and 2.10 for
finding internal energy change for a wide range of
applications. Usually, we will be able to use the datafrom tables in appendices of the textbook.
Like kinetic and gravitational potential energy, internal
energy is an extensive property.
Internal energy is represented by U.The specific internal energy on a mass basis is u.
The specific internal energy on a molar basis is .u
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Change in Energy of a System
So, the total change in energy of a system from
state 1 to state 2 is
E2E1 = (U2U1) + (KE2 KE1) + (PE2 PE1)
(Eq. 2.27a)
E= U+ KE + PE (Eq. 2.27b)
Since an arbitrary valueE1 can be given for the
energy of a system at state 1, no particularsignificance can be attached to the value of
energy at state 1 or any other state. Only
changes in the energy of a system between
states have significance.
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When a spring is compressed,
energy is transferred to the spring bywork.
When a gas in a closed vessel is
stirred, energy is transferred to the
gas by work.
When a battery is charged
electrically, energy is transferred to
the battery contents by work.
Illustrations of Work
The first two examples of work are familiar from
mechanics. The third example is an example of this
broader interpretation of work.
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Energy Transfer by Work
The symbol Wdenotes an amount of energy
transferred across the boundary of a system by work.
Since engineering thermodynamics is often
concerned with internal combustion engines,
turbines, and electric generators whose purpose is todo work, convention is that the work done by a
system as positive.
W> 0: work done by the system
W< 0: work done on the system
The same sign convention is used for the rate of
energy transfer by work called power, given by .W
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Modeling Expansion and Compression Work
We will see, a useful example to study is a gas (or
liquid) undergoing an expansion (or compression)process while confined in a piston-cylinder
assembly.
During the process, the gas exerts a normal force
on the piston, F=pA , wherep is the pressure at the
interface between the gas and piston and A is the
area of the piston face.
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Modeling Expansion and Compression Work
From mechanics, the work done by the gas as the
piston face moves fromx1 tox2 is given by
== dxpFdxW A
Since the product Adx = dV , whereVis the volumeof the gas, this becomes
(Eq. 2.17)
= pdVW
V1
V2
For compression, dVis negative and so is the
value of the integral, which keeps with the sign
convention for work.
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Quasiequilibrium Processes
In a quasiequilibriumexpansion, the gas moves
along a pressure-volume
curve, or path, as shown.
Using Eq. 2.17, the work
done by the gas on the
piston is given by the area
under the curve of pressureversus volume.
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Polytropic Indices
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Polytropic Indices
no mass transfer
no heat transferadiabatic (closed system)
n = 0 pV0 = p = const.
(isobaric)
n < 1 systems with high T, high thermal energyinput
n = 1 pV = ZnRT = const.
(isothermal)
1 < n < k quasi-adiabatic
n = k adiabatic, reversible
n = pV, V dominates -> V=const.(isochoric)
k: specific heat ratio
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Work through For Example problems
in sections 2.1 and 2.2,and Example Problem 2.1
then
Work on homework problems2.7, 2.14 (answer in SI),
2.23(a&b only use MATLAB for b)
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Lecture 5
Heat and Cycles
Sections 2.4 & 2.6
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Energy Transfer by Heat
The symbol Qdenotes an amount of energy
transferredacross the boundary of a system by heat.
Since engineering thermodynamics is often
concerned with steam engines, internal combustion
engines and turbines, and that heat will need to beadded to the system for it to function, convention is
that the heat transfer into a system as positive.
Q> 0: heat t ransfer to the system
Q< 0: heat t ransfer f rom the systemThe same sign convention is used for the rate of
energy transfer by heat, given by
Note: this is the opposite convention from work!
.Q
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Modes of Heat Transfer
For any particular application, energy
transfer by heatcan occur by one or more of
three modes:
conduction
radiation
convection
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Conduction
Conductionis the transfer of energyfrom more energetic particles of a
substance to less energetic adjacent
particles due to interactions between
them.The time rate of energy transfer by
conduction is quantified by Fouriers
law.
An application of Fouriers law to a
plane wall at steady state is shown at
right.
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wherekis a proportionality constant, a property of the wall
material called the thermal conductivity.
The minus signis a consequence of energy transfer in
the direction of decreasing temperature.
Conduction
By Fouriers law, the rate of heat transfer across any
plane normal to the xdirection, , is proportional to thewall area, A, and the temperature gradient in the x
direction, dT/dx,
xQ
dx
dTQx Ak (Eq. 2.31)
In this case, temperature varies linearly with x, and thus
L
TTQx
12Akand Eq. 2.31gives)0(12
L
TT
dx
dT
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Thermal Radiation
Thermal radiationis energy transported byelectromagnetic waves (or photons). Unlike
conduction, thermal radiation requires no
intervening medium and can take place in a
vacuum.The time rate of energy transfer by radiation is
quantified by expressions developed from the
Stefan-Boltzmanlaw.
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Net energy is transferred in the direction of the arrowand quantified by
whereAis the areaof the smaller surface,
is a property of the surface called its emissivity,
sis the Stefan-Boltzman constant.
Thermal Radiation
An application involving net
radiation exchange between asurface at temperature Tband a
much larger surface at Ts(< Tb)
is shown at right.
]A[ 4s4be TTQ es
(Eq. 2.33)
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Convection
Convectionis energy transfer between a solidsurface and an adjacent gas or liquid by the
combined effects of conduction and bulk flow
within the gas or liquid.
The rate of energy transfer by convection isquantified by Newtons law of cooling.
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Energy is transferred in the direction of the arrow and
quantified by
whereAis the areaof the transistors surface and
his an empirical parameter called the convection heat
transfer coefficient.
Convection
An application involving
energy transfer byconvection from a transistor
to air passing over it is
shown at right.
]hA[ fbc TTQ (Eq. 2.34)
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Work through For Example problem
in section 2.4.2, andExample Problems 2.4, 2.5
then
Work on homework problem
2.51
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Thermodynamic Cycles
A thermodynamic cycleis a sequence of
processes that begins and ends at the same
state.
Examples of thermodynamic cycles include
Power cyclesthat develop a net energy transfer byworkusing an energy input by heat transfer from hot
combustion gases.
Refrigeration cyclesthat provide coolingfor a
refrigerated space using an energy input by work inthe form of electricity.
Heat pump cyclesthat provide heatingto a dwelling
using an energy input by work in the form of electricity.
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The energy transfers by heat and work
shown on the figure are each positive in the
direction of the accompanying arrow. This
convention is commonly used for analysisof thermodynamic cycles.
Power Cycle
A system undergoing a power cycle is
shown at right.
Wcycleis the net energy transfer by workfrom the system
per cycle of operationin the form of electricity, typically.
Qinis the heat transfer of energy to the systemper cycle
from the hot bodydrawn from hot gases of combustion orsolar radiation, for instance.
Qoutis the heat transfer of energy from the systemper
cycle to the cold bodydischarged to the surrounding
atmosphere or nearby lake or river, for example.
C
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Power Cycle
Applying the closed system energy balance to each
cycle of operation,
(Eq. 2.39)DEcycle= QcycleWcycle
Since the system returns to its initial state after eachcycle, there is no net change in its energy: Ecycle= 0,
and the energy balance reduces to give
(Eq. 2.41)Wcycle= QinQout
In words, the netenergy transfer by work from the
systemequals the netenergy transfer by heat to the
system, each per cycle of operation.
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Power Cycle
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Power CycleUsing the second law of thermodynamics (Chapter 5), we will
show that the value of thermal efficiency must be less thanunity: < 1 (< 100%). That is, only a portion of the energy
added by heat,Qin,can be obtained as work. The remainder,
Qout, is discharged.
Example: A system undergoes a power cycle while receiving
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p y g p y g
1000 kJ by heat transfer from hot combustion gases at a
temperature of 500 K and discharging 600 kJ by heat transfer
to the atmosphere at 300 K. Taking the combustion gases and
atmosphere as the hot and cold bodies, respectively, determine
for the cycle, the net work developed, in kJ, and the thermal
efficiency.
Substituting into Eq. 2.41,
Wcycle= 1000 kJ600 kJ =400 kJ.
Then, with Eq. 2.42,= 400 kJ/1000 kJ = 0.4 (40%).
Note the thermal efficiency is commonly reported on a
percent basis.
Wcycle= QinQout
in
cycle
Q
W
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Heat Pump Cycle
As for the refrigeration cycle, the energy balancereads
(Eq. 2.44)Wcycle= QoutQin
As before,Wcycleis the netenergy transfer by workto
the system per cycle,
usually provided in the form
of electricity.
Heat Pump Cycle
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Heat Pump Cycle
The performance of a system undergoing a heat
pump cycleis evaluated on an energy basis as theratio of energy provided to the hot body, Qout, to the
net work required to accomplish this effect, Wcycle:
(heat pump cycle) (Eq. 2.47)
called the coefficient of performancefor the heat pump
cycle.
cycleout
W
Q
We can also write this as:
inout
out
Q
(heat pump cycle) (Eq. 2.48)
Example: A system undergoes a heat pump cycle
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p y g p p y
while discharging 900 kJ by heat transfer to a dwelling
at 20oC and receiving 600 kJ by heat transfer from the
outside air at 5oC. Taking the dwelling and outside airas the hot and cold bodies, respectively, determine for
the cycle, the net work input, in kJ, and the coefficient
of performance.
Substituting into Eq. 2.44,
Wcycle= 900 kJ600 kJ =300 kJ.
Then, with Eq. 2.47,= 900 kJ/300 kJ = 3.0.
cycle
out
W
Q
Wcycle= QoutQin
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Work through For Example problems
in section 2.6
then
Work on homework problems2.71, 2.74
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Lect re 7
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Lecture 7
Using Data Tables
Sections 3.4, 3.5, 3.6, 3.8, 3.8.1, 3.9, 3.10.1
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Si l Ph R i
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Single-Phase Regions
Since pressure and temperatureare independent properties in
the single-phase liquid and
vapor regions, they can be used
to fix the state in these regions.Tables A-4/A-4E (Superheated
Water Vapor) andA-5/A-5E
(Compressed Liquid Water)
provide several properties asfunctions of pressure and
temperature, as considered
next.
Table A-4/A-4E
Table A-5/A-5E
Single-Phase Regions
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Temperature (T)Pressure (p)
Specific volume (v)
Specific internal energy (u)
Specific enthalpy (h), which is a
sum of terms that often appears in
thermodynamic analysis:
h = u +pv
Single Phase Regions
Table A-4/A-4E
Table A-5/A-5E
Specific entropy (s), an intensive property developed in
Chapter 6
Properties tabulated in Tables A-4 andA-5 include
Enthalpy is a property because it is defined in terms of
properties; physical significance is associated with it in Chapter 4.
Example: Single-Phase Regions of Tables
Wh t th ifi l ifi th l d ifi
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What are the specific volume, specific enthalpy and specific
internal energy for superheated water vapor at 10 MPa and
400oC?
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Linear Interpolation
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Linear Interpolation
When a state does not fall exactly on the grid of values provided
by property tables, linear interpolation between adjacent entries
is used.
Example: Specific volume (v) associated with superheated
water vaporat 10 barand 215oC is found by linear interpolation
between adjacent entries in Table A-4.
ToC
v
m3/kgu
kJ/kgh
kJ/kgs
kJ/kgK
p= 10 bar = 1.0 MPa(Tsat= 179.91oC)
Sat. 0.1944 2583.6 2778.1 6.5865
200 0.2060 2621.9 2827.9 6.6940
240 0.2275 2692.9 2920.4 6.8817
Table A-4
(0.2275 0.2060) m3/kg (v 0.2060) m3/kg
(240 200)oC (215 200)oCslope = = v = 0.2141 m3/kg
Two-Phase Liquid-Vapor Region
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Two Phase Liquid Vapor RegionTables A-2/A-2E
(Temperature Table) andA-3/A-3E (Pressure Table)
provide
saturated liquid (f) data
saturated vapor (g) data
Specific Volume
m3/kgInternal Energy
kJ/kgEnthalpy
kJ/kgEntropykJ/kgK
TempoC
Press.bar
Sat.Liquid
vf103
Sat.Vapor
vg
Sat.Liquid
uf
Sat.Vapor
ug
Sat.Liquid
hf
Evap.
hfg
Sat.Vapor
hg
Sat.Liquid
sf
Sat.Vapor
sg
TempoC
.01 0.00611 1.0002 206.136 0.00 2375.3 0.01 2501.3 2501.4 0.0000 9.1562 .01
4 0.00813 1.0001 157.232 16.77 2380.9 16.78 2491.9 2508.7 0.0610 9.0514 4
5 0.00872 1.0001 147.120 20.97 2382.3 20.98 2489.6 2510.6 0.0761 9.0257 5
6 0.00935 1.0001 137.734 25.19 2383.6 25.20 2487.2 2512.4 0.0912 9.0003 6
8 0.01072 1.0002 120.917 33.59 2386.4 33.60 2482.5 2516.1 0.1212 8.9501 8
Table A-2
Table note: For saturated liquid specific volume, the table heading is vf103.
At 8oC, vf 103 = 1.002 vf= 1.002/10
3 = 1.002 103.
Two-Phase Liquid-Vapor Region
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The specific volume of a two-phase liquid-
vapor mixture can be determined by using
the saturation tables and quality, x.
The total volume of the mixture is the sum
of the volumes of the liquid and vapor
phases:
q p g
V= Vliq + Vvap
Dividing by the total mass of the mixture,m, an average
specific volume for the mixture is:
m
V
m
V
m
V vapliq+==v
With Vliq =mliqvf , Vvap =mvapvg ,mvap/m =x , andmliq/m = 1 x :
v = (1 x)vf+xvg = vf+x(vg vf) (Eq. 3.2)
m
V
m
V
m
V vapliq +==v Vliq =mliqvf , Vvap =mvapvg ,mvap/m =x ,mliq/m = 1 x
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v = (1 x)vf+xvg = vf+x(vg vf)
Two-Phase Liquid-Vapor Region
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Two Phase Liquid Vapor Region
Since pressure and temperature are NOT
independent properties in the two-phase liquid-vapor region, they cannot be used to fix the state
in this region.
The property, quality (x), defined only in the two-phase liquid-vapor region, and either temperature
or pressure can be used to fix the state in this
region.
v = (1 x)vf+xvg = vf+x(vg vf) (Eq. 3.2)
u = (1 x)uf+xug = uf+x(ug uf) (Eq. 3.6)
h = (1 x)hf+xhg =hf+x(hg hf) (Eq. 3.7)
Two-Phase Liquid-Vapor Region
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Two Phase Liquid Vapor Region
Example: A system consists of a two-phase liquid-vapor
mixture of waterat 6oC and a quality of 0.4. Determine thespecific volume, in m3/kg, of the mixture.
Solution: Apply Eq. 3.2, v = vf+x(vg vf)
Specific Volume
m3/kgInternal Energy
kJ/kgEnthalpy
kJ/kgEntropykJ/kgK
TempoC
Press.bar
Sat.Liquid
vf103
Sat.Vapor
vg
Sat.Liquid
uf
Sat.Vapor
ug
Sat.Liquid
hf
Evap.
hfg
Sat.Vapor
hg
Sat.Liquid
sf
Sat.Vapor
sg
TempoC
.01 0.00611 1.0002 206.136 0.00 2375.3 0.01 2501.3 2501.4 0.0000 9.1562 .01
4 0.00813 1.0001 157.232 16.77 2380.9 16.78 2491.9 2508.7 0.0610 9.0514 4
5 0.00872 1.0001 147.120 20.97 2382.3 20.98 2489.6 2510.6 0.0761 9.0257 5
6 0.00935 1.0001 137.734 25.19 2383.6 25.20 2487.2 2512.4 0.0912 9.0003 6
8 0.01072 1.0002 120.917 33.59 2386.4 33.60 2482.5 2516.1 0.1212 8.9501 8
Table A-2
Substituting values from Table 2: vf= 1.001103 m3/kg and
vg = 137.734 m3/kg:
v = 1.001103 m3/kg + 0.4(137.734 1.001103) m3/kg
v = 55.094 m3/kg
Property Data Use in the
Cl d S t E B l
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Closed System Energy Balance
Example: A piston-cylinder assembly contains 2 kg of
water vapor at 100oC and 1 bar. The water vapor iscompressed to a saturated vaporstate where the
pressure is 2.5 bar. During compression, there is a heat
transfer of energy from the vapor to its surroundings
having a magnitude of 250 kJ. Neglecting changes inkinetic energy and potential energy, determine the work,
in kJ, for the process of the water vapor.
State 1
2 kg
of waterT1 = 100
oC
p1 = 1 bar
State 2
Saturated vapor
p2 = 2.5 bar
Q = 250 kJ
2
T
v
p1 = 1 bar
1
p2 = 2.5 bar
T1 = 100oC
Property Data Use in the
Cl d S t E B l
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Closed System Energy BalanceSolution: An energy balance for the closed system is
KE + PE +U= Q W0 0
where the kinetic and potential energy changes are neglected.
Thus W=Q m(u2 u1)
State 1 is in the superheated vaporregion and is fixed byp1 = 1 bar and T1 = 100
oC. From Table A-4, u1 = 2506.7 kJ/kg.
State 2 is saturated vaporatp2 = 2.5 bar. From Table A-3,
u2 = ug = 2537.2 kJ/kg.W= 250 kJ (2 kg)(2537.2 2506.7) kJ/kg = 311 kJ
The negative sign indicates work is done on the system as
expected for a compression process.
A Small Primer on Partial Derivatives
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A Small Primer on Partial Derivatives
A dependent variable can be a function of many independent
variables, for instance:
We can then label which variable(s) we held constant when we
were taking the partial derivative:
EDyzyzCxBxyzyxA +++= 222),,(
00)2()1(2 +++=
xCyzBy
x
A
The partial derivative is the derivative of the function with respect
to one variable as if all the other variables were constants:
yx
A
CxyzBy 22 +=
partial derivative of A with respect to x,
holding y constant
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Property Approximations for Liquids
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Approximate values for v, u, andh at liquid states can be
obtained using saturated liquid data.
Since the values of v and u for liquidschange very little with pressure at a fixed
temperature, Eqs. 3.11 and 3.12 can be used
to approximate their values.
v(T,p) vf(T)u(T,p) uf(T)
(Eq. 3.11)(Eq. 3.12)
An approximate value forh at liquid states can be obtained usingEqs. 3.11 and 3.12 in the definitionh = u + pv: h(T, p) uf(T) + pvf(T)
or alternatively h(T,p) hf(T) + vf(T)[p psat(T)] (Eq. 3.13)
wherepsat denotes the saturation pressure at the given temperature
When the underlined term in Eq. 3.13 is small
h(T,p) hf(T) (Eq. 3.14)
Saturated
liquid
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Work through For Example problems
in sections 3.5, 3.6, ,and Example Problems
3.1, 3.2, 3.3, 3.4
then
Work on homework problems
3.42, 3.54 (answer in SI)
Lecture 8
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Lecture 8
Using Data Tables
Sections 3.5 & 3.6
Example: What is the specific volume of water at a
state where p = 10 bar and T = 215oC?
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state where p = 10 bar and T = 215oC?
Quality
isobaric lines(constant pressure)
Tempera
ture
Specific Volume
liquid
supercriticalfluid
vapor
saturated region
(vapor dome)
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p = 10 bar and T = 215oC
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Quality
isobaric lines
(constant pressure)
Temperature
Specific Volume
liquid
supercriticalfluid
vapor
saturated region
(vapor dome)
p = 10 bar and T = 215oC
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p = 10 bar and T = 215oC
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Linear Interpolation
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(x1, y1)
(x2, y2)
(x3, y3)
(y2 - y1)
(x2 - x1)
(y3 - y1)
(x3 - x1)
(y2 - y1) (y3 - y1)
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y3 is our unknown spec. vol.
(x2 - x1) (x3 - x1)
multiply both sides by (x3 - x1)
add y1 to both sides
p = 10 bar and T = 215oC
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Make sure you understand all for example
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Make sure you understand all for example ,
Sample and in-class problems to date, which
includes: Which data table corresponds to which region of
a T-v diagram and/or which approximations hold
there.
What phase(s) exist in a region of the T-vdiagram.
How p changes in different regions of the T-v
diagram.
Lecture 9
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Ideal Gas Law
Compressibility Factor
Sections 3.11, 3.12
Incompressibility of Liquids
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Approximate values for v, u, and h at liquid states can be
obtained using saturated liquid data.
Since the values of v and u for liquids
change very little with pressure at a fixed
temperature, Eqs. 3.11 and 3.12 can be used
to approximate their values.
v(T, p) vf(T)
u(T, p) uf(T)
(Eq. 3.11)
(Eq. 3.12)h(T, p) hf(T) (Eq. 3.14)
Saturated
liquid
Generalized Compressibility Chart
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vThe p- -Trelation for 10 common gases is
shown in the generalized compressibility chart.
Generalized Compressibility Chart
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TR
pZ
v
In this chart, the compressibility factor, Z, is plotted versus
the reduced pressure, pR, and reduced temperature TR,
wherep
R= p/pc TR= T/Tc
(Eq. 3.27) (Eq. 3.28)(Eq. 3.23)
The symbols pc
and Tc
denote the
temperature and pressure at the
critical point for the particular gas
under consideration. These values
are obtained from Tables A-1 and
A-1E.
R
8.314 kJ/kmolK
1.986 Btu/lbmoloR
1545 ftlbf/lbmoloR
(Eq. 3.22)
is the universal gas constant
R
Generalized Compressibility Chart
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When p, pc, T, T
c, v , and are used in consistent units, Z,
pR, and T
Rare numerical values without units.
Example: For air at 200 K, 132 bar, TR = 200 K/133 K= 1.5,pR = 132 bar/37.7 bar= 3.5 where Tc and pc for air are from
Table A-1. With these TR, p
Rvalues, the generalized
compressibility chart gives Z= 0.8.
R
With v = Mv,
an alternativeform of the compressibility
factor is
Z=pv
RT
(Eq. 3.24)
where
R= R
M(Eq. 3.25)
Studying the Generalized Compressibility Chart
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The solid lines labeled with TRvalues represent best fits to
experimental data. For the 10 different gases represented
there is little scatter in data about these lines.
At the lowest reduced
temperature value
shown, TR = 1.0, thecompressibility factor
varies between 0.2
and 1.0. Less
variation is observedas T
Rtakes higher
values.
Studying the Generalized Compressibility Chart
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For each specified value of TR, the compressibility factor
approaches a value of 1.0 as pR
approaches zero.
Studying the Generalized Compressibility Chart
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Low values of pR, where Z 1, do not necessarily
correspond to a range of low absolute pressures.For instance, if pR= 0.05, then p= 0.05pc. With pc values
from Table A-1
These pressure values range from 1.9 to 11 bar, which in
engineering practice are not normally considered as low
pressures.
Water vapor pc
= 220.9 bar p= 11 bar
Ammonia pc = 112.8 bar p= 5.6 bar
Carbon dioxide pc = 73.9 bar p= 3.7 bar
Air pc = 37.7 bar p= 1.9 bar
Introducing the Ideal Gas Model
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To recap, the generalized
compressibility chart shows that
at states where the pressure p
is small relative to the critical
pressure pc
(where pR
is small),
the compressibility factor Z is
approximately 1.
At such states, it can be assumed with reasonable
accuracy that Z = 1. Then
pv = RT (Eq. 3.32)
Introducing the Ideal Gas Model
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g
Three alternative forms of Eq. 3.32 can be derived
as follows:
With v = V/m, Eq. 3.32 gives
pV= mRT (Eq. 3.33)
With v = v/M and R= R/M, Eq. 3.32 gives
(Eq. 3.34)TRp v
Finally, with v = V/n, Eq. 3.34 gives
pV= nRT (Eq. 3.35)
Introducing the Ideal Gas Model
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While the ideal gas model does not provide an
acceptable approximation generally, in thelimiting case considered in the discussion of the
compressibility chart, it is justified for use, and
indeed commonly applied in engineeringthermodynamics at such states.
Appropriateness of the ideal gas model can be
checked by locating states under consideration
on one of the generalized compressibility charts
provided by appendix figures Figs. A-1 through
A-3.
Work Example Problems
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Work Example Problems
3.7, 3.8
then
Work on homework problems
3.103, 3.113
Lecture 11
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Control Volumes
Mass Rate Balance
Sections 4.1 4.3
Mass Rate Balance
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time rate of changeof
mass contained within the
control volume at time t
time rate of f lowof
mass inacross
inlet i at time t
time rate of f low
of mass outacross
exit e at time t
ei mmdt
dm cv (Eq. 4.1)
Mass Rate Balance
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e
e
i
i mm
dt
dm
cv(Eq. 4.2)
Eq. 4.2 is the mass rate balance for control
volumes with several inlets and exits.
In practice there may be several locations on theboundary through which mass enters or exits.
Multiple inlets and exits are accounted for by
summing over all entrances and exits:
Example: (Problem 4.2)Liquid propane enters and initially empty cylindrical storage tank at a mass flow rate
of 10 kg/s. Flow continues until the tank is filled with propane at 20oC, 9 bar. The
tank is 25 m long and has a 4 m diameter Determine the time in minutes to fill the
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tank is 25 m long and has a 4 m diameter. Determine the time, in minutes, to fill the
tank.
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Example: (Problem 4.4)Data are provided for the crude oil
storage tank. Determine:
a) The mass of oil in the tank in kg
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a) The mass of oil in the tank, in kg,
after 24 hours, and
b) The volume of oil in the tank, in
m3, at 24 hours
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Mass Rate Balance(Steady State Form)
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(Steady-State Form)
Steady-state: all properties are unchangingin time.
For steady-state control volume, dmcv/dt= 0.
e
e
i
i mm (Eq. 4.6)
(mass rate in) (mass rate out)
e
e
i
i mm
dt
dm
cv
Work through
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Work through
For Example problems in section 4.2
Example Problems 4.1 and 4.2
then
Work on homework problems
4.9, 4.16
Lecture 12
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Energy Rate Balance
Sections 4.4 4.5
Mass Rate Balance
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timerate of changeof
mass contained within the
control volumeattime t
timerate of flowof
mass inacross
inlet i attime t
timerate of flow
of massoutacross
exit e attime t
ei mmdt
dm =cv (Eq. 4.1)
Mass Rate Balance
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=e
e
i
i mmdt
dm
cv(Eq. 4.2)
Eq.4.2is the mass rate balancefor control
volumes with several inlets and exits.
In practice there may be several locations on theboundary through which mass enters or exits.
Multiple inletsand exitsare accounted for by
summing over all entrances and exits:
Energy Rate Balance
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)2
V()
2
V(
22cv
ee
eeii
ii gzumgzumWQdt
dE+++++= (Eq. 4.9)
timerate of changeof the energy
contained within
the control volume
attime t
netrateat whichenergy is being
transferred in
by heat transfer
attime t
netrateat whichenergy is being
transferred out
by workat
time t
netrateof energytransfer intothe
control volume
accompanying
mass flow
inlets and exitsinteraction with surroundings
control volume
Evaluating Work for a Control Volume
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)()(cv iiieee vpmvpmWW += (Eq. 4.12)
The expression for work is
cvW accounts for work associated with rotating
shafts, displacement of the boundary, and electrical
effects.
where
)( eee vpm is the flow workat exit e.
)( iii vpm is the flow workat inlet i.
Control Volume Energy Rate Balance(One-Dimensional Flow Form)
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)2
V()
2
V(
22
cvcvcv
ee
eeeeii
iiii gzvpumgzvpumWQdt
dE+++++++=
(Eq. 4.13)
For convenience substitute enthalpy,h= u+pv
)2
V()
2
V(
22
cvcvcv
ee
eeii
ii gzhmgzhmWQdt
dE+++++=
(Eq 4 14)
Using Eq. 4.12 in Eq. 4.9
control volume
interaction with surroundings with closed system defs.inlets and exits
Control Volume Energy Rate Balance(One-Dimensional Flow Form)
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(One Dimensional Flow Form)
++ +++=e
ee
eei
ii
ii gzhmgzhmWQdt
dE)
2
V()
2
V(
22
cvcvcv
(Eq. 4.15)
Eq. 4.15is the accounting balancefor the
energy of the control volume.
In practice there may be several locationson the boundary through which mass enters
or exits. Multiple inletsand exitsare
accounted for by introducing summations:
Control Volume Energy Rate Balance(Steady-State Form)
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(Steady State Form)
++ +++=e
ee
eei
ii
ii gzhmgzhmWQ )2
V()
2
V(0
22
cvcv
(Eq. 4.18)
Steady-state: all properties are unchangingin time.
For steady-state control volume,dEcv/dt= 0.
Control Volume Energy Rate Balance(Steady-State Form, One-Inlet, One-Exit)
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Many important applications involve one-inlet,one-exitcontrol volumes at steady state.
The mass rate balance reduces to .mmm == 21
+
++= )(
2
)V(V)(0 21
22
21
21cvcv zzghhmWQ
Eq.
4.20a
or dividing by mass flow rate
)(2
)V(V)(0 21
22
21
21cvcv zzghhm
W
m
Q+
++=
Eq.
4.20b
Example: (Problem 4.26)Air enters a horizontal, constant-diameter heating duct operating at steady state at
290 K, 1 bar, with a volumetric flow rate of 0.25 m3/s, and exits at 325 K, 0.95 bar.
The flow area is 0.04 m2. Assuming the ideal gas model with k = 1.4 for the air,
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determine:
a) The mass flow rate, in kg/s, b) the velocity at the inlet and exit, and
c) the rate of heat transfer in kW.
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Example: (Problem 4.28)At steady state, air at 200 kPa, 52oC, and a mass flow rate of 0.5 kg/s enters a
horizontal insulated duct having differing inlet and exit cross-sectional areas. At the
duct exit, the pressure of the air is 100 kPa, the velocity is 255 m/s, and the cross-3 2
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sectional area is 2 x 10-3m2. Assuming the ideal gas model, determine: a) the
temperature of the air at the exit, in C,
b) The velocity of the air at the inlet in m/s, and c) the inlet cross-sectional area, in m2.
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Work on homework problems
4.23, 4.24
L 13
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Lecture 13
Nozzles, Diffusers,Turbines, Compressor
and Pumps
Nozzles and Diffusers
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Nozzle: a flow passage of varying crosectional area in which the velocity of
or liquid increases in the direction of fl
Diff fl f i
2
2
221
2
112
1
2
1ghVPghVP Berno
N l d Diff M d
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If the change in potential energy from inlet
negligible, g(z1z2) drops out.
If the heat transfer with surroundings is neg
drops out.
(
2
)V(V)(0 1
22
21
21cvcv zzghhmWQ
Nozzle and Diffuser Mode
.0cvW
22
cvQ
Example: (Problem 4.31)
Steam enters a nozzle operating at steady state at 20 bar, 280oC, with
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m/s. The exit pressure and temperature are 7 bar and 180oC, respectiv
mass flow rate is 1.5 kg/s. Neglecting heat transfer and potential energa) The exit velocity in m/s, and b) the inlet and exit flow areas, in c
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T bi
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Turbines
Turbine: a device in which power is
developed as a result of a gas or liqu
passing through a set of blades attac
T bi M d li
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(
2
)V(V)(0 1
22
21
21cvcv zzghhmWQ
Turbine Modeling
If the change in kinetic energy of flowing m
negligible, (V12 V2
2) drops out.
If the change in potential energy of flowing
negligible, g(z1z2) drops out.
If the heat transfer with surroundings is neg
d tQ
Compressors and P
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Compressors and P
Compressors and Pum
devices in which work is
on the substance flowingthrough them to change t
state of the substance, ty
to increase the pressure elevation.
Compressor : substance
C d P M d
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(
2
)V(V)(0 1
22
21
21cvcv zzghhmWQ
Compressorand Pump Mod
If the change in kinetic energy of flowing m
negligible, (V12 V2
2) drops out.
If the change in potential energy of flowing
negligible, g(z1z2) drops out.
If the heat transfer with surroundings is neg
d tQ
Example: (Problem 4.41)
Steam enters a well-insulated turbine operating at steady state at 4 MPa
enthalpy of 3015 4 kJ/kg and a velocity of 10 m/s The steam expands t
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enthalpy of 3015.4 kJ/kg and a velocity of 10 m/s. The steam expands t
where the pressure is 0.07 MPa, specific enthalpy is 2431.7 kJ/kg, and t
m/s. The mass flow rate is 11.95 kg/s. Neglecting potential energy effe
the power developed by the turbine, in kW.
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Work through Example Proble4.3, 4.4, 4.5 and 4.6
then
Work on homework problem
4 33 4 43 4 53
Lecture 14
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Lecture 14
Heat Exchangers and
Throttling Devices
Sections 4.9 and 4.10
Heat Exchangers
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Direct contact:A mixing chamber in which hot and
cold streams are mixed directly.
Tube-within-a-tube counterflow:A gas or liquid
stream is separatedfrom another gas or liquid by awall through which energy is conducted. Heat
transfer occurs from the hot stream to the cold
stream as the streams flow in opposite directions.
ei VV22
Heat Exchanger Modeling
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If the kinetic energies of the flowing streams are
negligible, (Vi2/2) and (Ve2/2) drop out.If the potential energies of the flowing streams are
negligible, gziand gzedrop out.
If the heat transfer with surroundings is negligible,drops out.
++ +++=e
ee
ee
i
ii
ii gzhmgzhmWQ )
2
V()
2
V(0 cvcv
(Eq. 4.18)
cvQ
.0cv =W
ee
eii
i hmhm = 0
im
em
im em
=e
ei
i mm
for each connected flow volume
1
2
3
14
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2 3
Problem 4.76 Steam enters a heat exchanger operating at steady state at 250 kPa and
a quality of 90% and exits as a saturated liquid at the same pressure. A separate stream
of oil with a mass flow rate of 29 kg/s enters at 20oC, and exits at 100oC with no
significant change in pressure. The specific heat of the oil is c = 2.0 kJ/kgK. Kinetic and
potential energy effects are negligible. If heat transfer from the heat exchanger to its
di i 10% f h i d i h f h il
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surroundings is 10% of the energy required to increase the temperature of the oil,
determine the steam mass flow rate, in kg/s.
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Pause the video
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Pause the video
Work through examples 4.7 and 4.8
then
Do homework problem 4.74
Throttling Devices
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Throttling Device:a device that achieves
a significant reduction in pressureby
introducing a restriction into a line through
which a gas or liquid flows. Means to
introduce the restriction include a partially
opened valve or a porous plug.
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Pause the video
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Work through example 4.9
then
Try suggested problem 4.90(well work on this in class)
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spontaneous heat transfer
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falling mass
spontaneous expansion
-> observed direction of these
processes
Aspects of theSecond Law of Thermodynamics
F ti f d
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From conservation of mass and energy
principles, mass andenergy cannot be
created or destroyed.
For a process, conservation of mass and
energyprinciples indicate the disposition ofmass and energy but do not infer whether the
process can actually occur.
The second law of thermodynamicsprovides the guiding principle for whether a
process can occur.
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Aspects of theSecond Law of Thermodynamics
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defininga temperature scaleindependent of the
properties of any thermometric substance.
developing means for evaluating properties
such as uand hin terms of properties that aremore readily obtained experimentally.
Scientists and engineers have found additional uses
of the second law and deductions from it. It alsohas been used in philosophy, economics, and other
disciplinesfar removed from engineering
thermodynamics.
Other aspectsof the second law include:
Second Law of ThermodynamicsAlternative Statements
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Clausius Statement
Kelvin-Planck StatementEntropy Statement
There is no simple statement that captures allaspects of the second law. Several
alternativeformulationsof the second laware
found in the technical literature. Threeprominent ones are:
Second Law of ThermodynamicsAlternative Statements
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The focus of Chapter 5is on the ClausiusandKelvin-Planck statements.
The Entropy statementis developed and applied
in Chapter 6.
Like every physical law, the basis of the secondlaw of thermodynamics is experimental
evidence. While the three forms given are not
directly demonstrable in the laboratory,
deductions from them can be verified
experimentally, and this infers the validity of the
second law statements.
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Entropy Statementof the Second Law
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Massand energyare familiar examples of
extensive propertiesused in thermodynamics.
Entropyis another important extensive property.
How entropy is evaluated and applied is detailed
in Chapter 6.Unlike mass and energy, which are conserved,
entropy is produced within systemswhenever
non-idealities such as friction are present.
The Entropy Statement is:
It is impossible for any system to operate in
a way that entropy is destroyed.
Irreversibilities
One of the important uses of the second law of
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One of the important uses of the second law of
thermodynamics in engineering is to determinethe best theoretical performanceof systems.
By comparing actual performancewith best
theoretical performance, insights often can be hadabout the potential for improved performance.
Best theoretical performance is evaluated in terms
of idealizedprocesses.
Actual processes are distinguishable from suchidealized processes by the presence of non-
idealities called irreversibilities.
Irreversibilities CommonlyEncountered in Engineering Practice
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Heat transferthrough a finite temperaturedifference
Unrestrained expansionof a gas or liquid to
a lower pressureSpontaneouschemical reaction
Spontaneous mixingof matter at different
compositions or statesFriction sliding friction as well as friction in
the flow of fluids
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Irreversible and ReversibleProcesses
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within the system, or
within its surroundings (usually the
immediate surroundings), or
within both the system and itssurroundings.
During a process of a system,irreversibilitiesmay be present:
Irreversible and ReversibleProcesses
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A process is irreversiblewhenirreversibilities are present within the system
and/or its surroundings.
All actual processes are irreversible.A process is reversiblewhen no
irreversibilities are present within the system
and its surroundings.This type of process is fully idealized.
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Example: Internally Reversible Process
Water contained within a piston-cylinder evaporates
from saturated liquid to saturated vapor at 100oC As
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from saturated liquid to saturated vapor at 100oC. As
the water evaporates, it passes through a sequence of
equilibrium stateswhile there is heat transfer to the
water from hot gases at 500oC.
Such spontaneous
heat transferis an
irreversibility in its
surroundings: an
externalirreversibility.
For a system enclosing the water there are nointernalirreversibilities, but
Analytical Form of theKelvin-Planck Statement
For any system undergoing a
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For any system undergoing a
thermodynamic cycle while
exchanging energy by heat
transfer with asinglethermal
reservoir, the net work,W
cycle,can be only negative or zero
never positive:
Wcycle 0 < 0: Internal irreversibilities present= 0: No internal irreversibilitiessingle
reservoir
(Eq. 5.3)
NO!
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Thermal Efficiency and Coefficient of Performance
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,
(refrigeration, heat pump)
(power cycle)
Applications to Power Cycles Interacting
with TwoThermal Reservoirs
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For a system undergoing apower cycle whilecommunicating thermally with twothermal
reservoirs, a hot reservoir and a cold reservoir,
(Eq. 5.4)H
C
H
cycle
1 Q
Q
Q
W==
the thermal efficiency of any such cycle is
(Eq. 2.43)
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Carnot Corollaries
2. The thermal efficiency of an irreversible power
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In other words, the reversible cycle between two reservoirstells us what the maximum efficiency of the cycle is. If there
are reversibilities, the efficiency can only be smaller than the
max.
cycleis always less thanthe thermal efficiency of areversible power cyclewhen each operates between
the same two thermal reservoirs.
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Applications to Power Cycles Interactingwith TwoThermal Reservoirs
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H
C
H
cycle1
Q
Q
Q
W==
1)(
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pump cycle whilecommunicating thermally with twothermal reservoirs, a hot reservoir and a cold
reservoir,
(Eq. 5.5)CH
C
cycle
C
Q
W
Q
==
the coefficient of performance for the
refrigeration cycle is
(Eq. 5.6)CH
H
cycle
H
Q
W
Q
==
and for the heat pump cycle is
(Eq. 2.45)
(Eq. 2.47)
Applications to Refrigeration and Heat Pump CyclesInteracting with TwoThermal Reservoirs
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By applying the Kelvin-Planck statementof thesecond law, Eq. 5.3, three conclusionscan be drawn:
1. For a refrigeration effect to occur a net work input
Wcycleis required. Accordingly, the coefficient ofperformance must be finite in value.
Two other conclusions are:
CH
C
cycle
C
Q
W
Q
==
== CH
H
cycle
H
Q
W
Q
Applications to Refrigeration and Heat Pump CyclesInteracting with TwoThermal Reservoirs
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2. The coefficient of performance of an irreversiblerefrigeration cycleis always less thanthe coefficient
of performance of a reversible refrigeration cycle
when each operates between the same two thermal
reservoirs.
In other words, the reversible cycle between two reservoirs
tells us what the maximum coefficient of performance (COP)of the cycle is. If there are reversibilities, the COP can only be
smaller than the max.
Applications to Refrigeration and Heat Pump CyclesInteracting with TwoThermal Reservoirs
3 All reversible refrigeration cycles operating
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In other words, once you have found one way to calculate thecoefficient of performance for a reversible cycle, you know what
the max COP is.
3. All reversible refrigeration cycles operating
between the same two thermal reservoirshave the
same coefficient of performance.
All three conclusions also apply to a system
undergoing a heat pump cycle between hot and cold
reservoirs.
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Lecture 17
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The Kelvin Temperature Scaleand Theoretical Limits
of Performance of
Cycles Between Two Reservoirs
Sections 5.8 5.9
Kelvin Temperature Scale
Consider systems undergoing apower cycle and a
refrigerationor heat pump cycle, each while
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g p p y ,
exchanging energy by heat transfer with hot and coldreservoirs:
(Eq. 5.7)H
C
cyclerevH
C
T
T
Q
Q=
The Kelvin temperature is defined so that
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Maximum Performance Measures for CyclesOperating between Two Thermal Reservoirs
Previous deductions from the Kelvin-Planck statement of the second law
include:
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1. The thermal efficiency of anirreversiblepowercycleis always less thanthe thermal efficiency of a
reversiblepower cyclewhen each operates between
the same two thermal reservoirs.
2. The coefficient of performance of anirreversible
refrigeration cycleis always less thanthe
coefficient of performance of areversible
refrigeration cyclewhen each operates between the
same two thermal reservoirs.
3. The coefficient of performance of anirreversibleheat pump cycleis always less thanthe coefficient
of performance of areversibleheat pump cycle
when each operates between the same two thermal
reservoirs.
The reversible
processdetermines
the maximum
theoretical
performance
for a cycle.
Maximum Performance Measures for CyclesOperating between Two Thermal Reservoirs
Using Eq. 5.7 in Eqs. 5.4, 5.5, and 5.6, we get respectively:
TQ
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(Eq. 5.9)H
Cmax 1
T
T=Power Cycle:
(Eq. 5.10)CH
Cmax
TT
T
=Refrigeration Cycle:
(Eq. 5.11)CH
Hmax
TT
T
=Heat Pump Cycle:
where THand T
Cmust be on theKelvinorRankine scale.
HC
cyclerevH
CT
T
Q
Q
=
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Example: Power Cycle Analysis
kJ600
Actual Performance: Calculate using the heat
transfers:
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4.0kJ1000
kJ60011
H
C ===
Maximum Theoretical Performance: Calculate
max
from Eq. 5.9 and compare to :
(a) 6.0K500
K20011
H
Cmax ===
T
T
(b) 4.0K500
K300
11H
Cmax ===
T
T
(c) 2.0K500
K40011
H
Cmax ===
T
T
Reversibly0.4 =0.4
Impossible0.4 >0.2
Irreversibly0.4
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Problem 5.59
At steady state, a refrigeration cycle operating between hot and cold reservoirs at 300 Kand 275 K, respectively, removes energy by heat transfer from the cold reservoir at a
rate of 600 kW.
a) If the cycles coefficient of performance is 4, determine the power input required, in
kW.
b) Determine the minimum theoretical power require, in kW, for any such cycle.
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) p q , , y y
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Lecture 18
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Carnot CycleClausius Inequality
Introducing Entropy
Sections 5.10, 5.11, 6.1, 6.2
Carnot Cycle
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The Carnot cycleprovides a specificexample of a reversible cycle that operatesbetween two thermal reservoirs. Otherexamples are provided in Chapter 9: the
Ericsson and Stirling cycles.In a Carnot cycle, the system executing the
cycle undergoes a series of four internallyreversible processes: two adiabaticprocessesalternated with two isothermalprocesses.
Carnot Power Cycles
The p-vdiagramand schematicof a gas in a piston-cylinder
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assembly executing a Carnot cycleare shown below:
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Clausius Inequality
The Clausius inequalityconsidered next
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q yprovides the basis for developing theentropy concept in Chapter 6.
The Clausius inequalityis applicable to any
cycle without regard for the body, or bodies,from which the system undergoing a cyclereceives energy by heat transferor towhich the system rejects energy by heattransfer. Such bodies need not be thermalreservoirs.
Work through example after Eq 5.14
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then
Work on homework
problems 5.85, 5.87
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Defining Entropy Change
Consider two cycles, each composed
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of two internally reversible processes,process A plus process Cand
process B plus process C, as shown
in the figure.
Applying Eq. 5.13to these cycles gives,
wherecycleis zero because the cycles arecomposed of internally reversible processes.
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Defining Entropy Change
Recalling (from Sec. 1.3.3) that a quantity is a property if,
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and only if, its change in value between two states isindependent of the process linking the two states, we
conclude that the integral represents the change in some
property of the system.
We call this property entropyand represent it by S. Thechange in entropy is
wherethe subscript int rev signals that the integral is
carried out for any internally reversible process linking
states 1and 2.
(Eq. 6.2a)
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Entropy FactsEntropyis an extensive property.
Just as mass and energy are accounted for by mass and
energy balances, entropy is accounted for by an entropy
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balance.Like mass and energy, entropy can be transferred across
the system boundary.
Like any other extensive property, the change in entropy
can be positive, negative, or zero:
By inspection of Eq. 6.2a, units for entropySare kJ/Kand
Btu/oR.
Units forspecificentropysare kJ/kgKand Btu/lboR.
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Lecture 19
C l l ti E t d
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Calculating Entropy, andEntropy Change in
Incompressible Substances,
Ideal Gases
Sections 6.3 6.5
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Entropy and Heat Transfer
By inspection of Eq. 6.2a, the defining equation for
entropy change on a differential basis is
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(Eq. 6.2b)
Equation 6.2bindicates that when a closed system
undergoing an internally reversible process receivesenergy by heat transfer, the system experiences an
increase in entropy. Conversely, when energy is removed
by heat transfer, the entropy of the system decreases.
From these considerations, we say that entropy transferaccompanies heat transfer. The direction of the entropy
transfer is the same as the heat transfer.
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Entropy and Heat Transfer
F thi it f ll th t
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From this it follows thatan energy transfer by
heat to a closed system
during an internally
reversible process isrepresented by an area
on a temperature-entropy
diagram:
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Calculating Entropy Change
As an application, consider a
change in phase from saturated
liquid to saturated vapor at
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T
dhds =
liquid to saturated vapor atconstant pressure.
Since pressure is constant, Eq.
6.10breduces to give
Then, because temperature is also constant during the
phase change
(Eq. 6.12)
This relationship is applied in property tables for
tabulating (sgsf) from known values of (hghf).
Calculating Entropy Change
For example, consider water vapor at 100oC
(373.15 K). From Table A-2, (hghf) =2257.1 kJ/kg.
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Next, the TdSequationsare applied to twoadditional cases: substances modeled as
incompressibleand gases modeled as ideal
gases.
which agrees with the value from Table A-2, as
expected.
(sgsf) =(hghf) /T = (2257.1 kJ/kg)/373.15 K
= 6.049 kJ/kgK
Then
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Calculating Entropy Change of an Ideal Gas
The ideal gas modelassumes pressure, specific volume
and temperature are related bypv=RT. Also, specific
internal energy and specific enthalpy each depend solely
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internal energy and specific enthalpy each depend solelyon temperature: u= u(T),h=h(T), givingdu=c
vdTand
dh=cpdT, respectively.
Using these relations and integrating, the TdSequations
give, respectively
(Eq. 6.17) (Eq. 6.18)
Sincecvandcpare functions of temperature for ideal gases,
such functional relations are required to perform the
integration of the first term on the right of Eqs 6 17 and 6 18
Calculating Entropy Change of an Ideal Gas
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integration of the first term on the right of Eqs. 6.17and 6.18.
For several gases modeled as ideal gases, including air,
CO2, CO, O2, N2, and water vapor, the evaluation of
entropy change can be reduced to aconvenient tabularapproach using the variablesodefined by
(Eq. 6.19)
whereT
' is an arbitrary reference temperature.
Calculating Entropy Change of an Ideal Gas
Usingso, the integral term of Eq. 6.18can be expressed as
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Accordingly, Eq. 6.18becomes
(Eq. 6.20a)
or on a per mole basisas
(Eq. 6.20b)
For air, Tables A-22andA-22Eprovideso
in units ofkJ/kgK and Btu/lboR, respectively. For the other gases
mentioned, Tables A-23andA-23Eprovide in units of
kJ/kmolK and Btu/lbmoloR, respectively.
os
(Eq. 6.18)
Calculating Entropy Change of an Ideal Gas
Example: Determine the change in specific entropy, in
kJ/kgK, of air as an ideal gas undergoing a process from
T1= 300 K,p
1= 1 barto T
2= 1420 K,p
2= 5 bar.
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Ideal Gas Pro pert ies of Ai r
T(K), hand u(kJ/ kg ), so(kJ/ kg K)
when s = 0 when s = 0
T h u so p r vr T h u so p r vr
250 250 .05 178 .28 1.51917 0.7329 979. 1400 1515.42 1113.52 3.36200 450.5 8.919
260 260.09 185.45 1.55848 0.8405 887.8 1420 1539.44 1131 .77 3.37901 478.0 8.526
270 270.11 192.60 1.59634 0.9590 808.0 1440 1563.51 1150 .13 3.39586 506.9 8.153
280 280.13 199.75 1.63279 1.0889 738.0 1460 1587.63 1168 .49 3.41247 537.1 7.801
285 285.14 203.33 1.65055 1.1584 706.1 1480 1611.79 1186 .95 3.42892 568.8 7.468
290 290.16 206.91 1.66802 1.2311 676.1 1500 1635.97 1205 .41 3.44516 601.9 7.152
295 295.17 210.49 1.68515 1.3068 647.9 1520 1660.23 1223 .87 3.46120 636.5 6.854
300 300.19 214.07 1.70203 1.3860 621.2 1540 1684.51 1242 .43 3.47712 672.8 6.569
305 305.22 217.67 1.71865 1.4686 596.0 1560 1708.82 1260 .99 3.49276 710.5 6.301
310 310.24 221.25 1.73498 1.5546 572.3 1580 1733.17 1279 .65 3.50829 750.0 6.046
From Table A-22, we getso1= 1.70203andso
2= 3.37901,
each in kJ/kgK. Substituting into Eq. 6.20a
1 1 2 2
Kkg
kJ215.1
bar1
bar5ln
Kkg
kJ
97.28
314.8
Kkg
kJ)70203.137901.3(12
=
= ss
Table A-22
Problem 6.15
One-tenth kmol of carbon monoxide (CO) is a piston-cylinder assembly undergoes aprocess from p1= 150 kPa, T1= 300 K to p2= 500 kPa, T2= 370 K. For the process,
W = -300 kJ. Employing the ideal gas model, determine
a) the heat transfer, in kJ, and
b) the change in entropy, in kJ/K.
c) Show the process on a T-s diagram.
7/25/2019 Angela Foudray's Thermodynamics Slides
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7/25/2019 Angela Foudray's Thermodynamics Slides
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