me200 notes f08 week1
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ME 200 Thermodynamics ISession 2 Fall 2008
Class information
Prof. S. H. Frankel Class time: MWF 10:30-11:20AM Office: ME 165 or Chaffee 125 (MWF) Email:
Phone: 765-494-1507 (office) 765-404-6067 (cell)
Office Hours: ME 165 MWF 11:30-12:30 Course website: http://engineering.purdue.edu/me200 Research website: http://ristretto.ecn.purdue.edu
Outline
Course details Chapter 1 - Getting started Introductory
Concepts and Definitions Using thermodynamics
Defining systems Describing systems and their behaviors Measuring mass, length, time, and force Properties
Specific volume Pressure
Temperature
Problem solving methodology
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Course details
Course policy Read items 1-4
Textbook/website
HW problems will be collected on Mondays
Includes problems from previous M, W, F
Only subset of problems will be graded
Graded HWs returned to you usually within 1 week
Read items 5-14
Schedule note exam date/time/location
Equation sheet bring to exam
Basic Concepts
What is thermodynamics?
Science of energy
What is energy? Ability to causechanges
Thermodynamics from Greek word
therme(heat) & dynamis(power);
heat to power
Key principle is conservation of
energy:During interaction, energy
can change from one form toanother but total amount of energy
remains constant i.e. energycannot be created or destroyed
Potential energy of weightconverted into thermal energy of
waterJoule apparatus (1843)
Laws
First law of thermodynamics A statement of conservation of energy principle Energy is a thermodynamic property (tbd); quantifies
energy
Second law of thermodynamics Energy has quality as well as quantity Actual processes occur in direction of decreasing
quality of energy Example: A cup of hot coffee
left on a table eventually cools,but a cup of cool coffee left on atable never gets hot by itself;degradation of high-temperatureenergy why a degradation?
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History Birth of thermodynamics
as a science 1697,
1712 atmospheric steamengine
1850s William Rankine,Rudolf Clasius, and LordKelvin
Rankine first textbook in1859
First ME 200 student falls
asleep in class August
1860
Approaches
Only two laws how hard a subjectcan this be?
Substances consist of large numberof particles called molecules
Two approachesClassical approachdoes not require
detailed knowledge of molecularmotion (but a little does not hurt)
Statistical approachconsidersmolecular motion in detail
We shall pursue the classical approachbut draw on molecular details for
insight e.g. we will use to understandinternal energy, entropy, and ideal gaslaw
Application areas
Home electric/gasrange, HVAC,refrigerator, humidifier,pressure cooker, water
heater, shower, etc. Automotive engines,
rockets, jet engines,conventional or nuclearpower plants
Even human body . . .
http://www.rolls-royce.com/education/schools/how_things_work/journey02/flash.html
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Systems - Closed and open
Thermodynamic system quantity of matter *or* aregion of space chosen for study
Mass or region outside system is called surroundings
Real or imaginary surface that separates system fromsurroundings is boundary
Boundary may be fixed, e.g. rigid tank, or moveable, e.g.piston-cylinder device
Closed system (control mass)
Features:
Fixed amount of mass
No mass can cross itsboundary
Energy (in form of heatand work interactions)can cross boundary
Volume of closedsystem does not haveto be fixed
Open system (control volume)
Properly selected region of space; usuallyencloses device which involves mass flow i.e.compressor, turbine, or nozzle
Both mass and energy can cross the boundaryof a CV, which is called a control surface (CS)
Note: Form of thermodynamic relations aredifferent for open and closed systems so it isimportant to identify what type of system youare considering first to determine properanalysis
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Control volumes
Properties of a system
Any characteristic of a system is a property i.e. pressureP, temperature T, volume V, mass m, etc.
Some properties are defined in terms of others, i.e.density is mass per unit volume
Specific volume
Note: Pointwise property definitions based on concept of
continuum i.e. spaces between molecules ignored, nomicroscopic holes; valid for most applications
3( / )
mkg m
V=
31= ( / )
Vm kg
m
=
Intensive vs. extensive properties
Intensive independent of size of system i.e. T, P, ordensity
Extensive depend on size/extent of system i.e. m, V, E
Does property change when system is divided in half?
Extensive properties per unit mass are called specificproperties and are intensive i.e. specific volume
mV
TP
1/2m1/2V
TP
1/2m1/2V
TP
Extensive (1/2 value)
Intensive (same value)
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State
Consider system not undergoing any change;
measure properties of system; this defines state A system at 2 states
m=2kgT1=20C
V1=1.5m3
m=2kg
T2=20CV2=2.5m3
State 1 State 2
Equilibrium
Thermodynamics deals with equilibrium states Equilibrium implies a balance i.e. no unbalanced driving
forces If isolated from surroundings system remains unchanged Thermal equilibrium implies temperature same
throughout, e.g. see below for before and after
No temperature differentials; system temperature can bedescribed by a single number in equilibrium
Mechanical no change in pressure ; Chemical no
change in composition Thermodynamic all aspects of system in equilibrium
20C 23C30C
35C 40C42C
32C 32C32C
32C 32C32C
Process path
Any change system undergoes from oneequilibriumstate to another is called a process
Series of states through which a system passes
during a process is called path of processAdvantage: Only one value for eachproperty describes entire systeme.g. one value of P, T, etc.
State of system represented by
single point on plot w/ properties as
coordinates
Path can only be drawn if processesproceed in equilibrium manner
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Quasi-equilibrium process When process proceeds so that the system remains
infinitesimally close to an equilibrium state at all times,we call is quasi-equilibrium (QE) process
Process occurs slow enough to allow system to adjustproperties w/in system so that one part of system doesntchange any faster than other parts
QE process is approximation to real process; usefulbecause (a) easy to evaluate and (b) deliver/requiremost/least work
Slow compression (QE process) Fast compression (non QE process)
Pressure, density, etc. change uniformly
Higher p etc. near piston;lower p near cylinder head
Push
slow
Pushfast
P-V property diagram
Illustrates states
and process path
To connect stateswith line must be QEprocess Why?
More processes and cycles
Other important processes involve particularconstant properties such as isothermal(T),isobaric(P), isochoric(V)
A system is said to have undergone a cycle if itreturns to its initial state at end of process i.e.initial and final states of cycle are same
p
V
1
2 p
V
12
3
4
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State postulate
How many properties must be specified to fix the state ofa system?
We need only specify a certain number of properties tofix the state of system
The exact number depends on type of system e.g. howcomplex is it?
For a system that only involves QEcompression/expansion e.g. a simple compressiblesystem, state is completely specified by twoindependent, intensive properties
Two properties independent if one can be varied w/ochanging other e.g. T, P away from phase change
Dimensions and units
Physical quantities characterized bydimensions (mass, m, or velocity, V, etc.)
Arbitrary magnitudes assigned to dimensionsare units (kg, m/s, etc.)
Basic dimension mass m, length L, time t,temperature T, are chosen as primary orfundamental dimensions
Others, velocity V, energy E, volume Vexpressed in terms of primary dimensions arecalled secondary or derived dimensions
Units
Two sets of units still used
English (USCS) no math basis, arbitrary, confusing, difficult tolearn (e.g. 4th grade woes)
SI decimal base, simple, logical
SI system of units: m, L, t, T are primary with base units ofkg, m, s, K, respectively
Force is a derived dimension in SI via Newtons 2nd law:
Force unit: 1 N (newton) = (1 kg) (1 m/s2) = 1 kg-m/s2
1 Newton is force required to accelerate 1 kg mass at rateof 1m/s2
( ) [ ] [ ] [ ]2
1d L LF mV F M M
dt t t t
= = =
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More on units
English system of units F, m, L, t, T are primary
Force is not derived through Newtons law; force andmass independently defined
But since force can be derived from Newtons law thesystem is overdetermined, hence
So we must write Newtons law as:
Dimensional constant, gc
[ ] [ ]2
LF M
t
( ) 21
with 32.174c
c
d lbm ft F mV g
g dt lbf s
= =
Still more on units
If instead we define a force unit as derived fromNewtons law with proportionality constant of 1, thenforce is secondary dimension
Then we could simply define
For m=const., F=ma. In order to have dimensionalhomogeneity (same units for all terms in equation), gcmust be introduced such that if one applies 1 lbf to 1 lbmin standard gravity i.e. g=32/174 ft/s2, so F=ma issatisfied.
Hence,
21 32.174
lbm ft lbf
s
=
2
21 /1 32.174 32.174cc
lbm ft lbm ft slbf gg s lbf
= =
Unit conversions
1 lbm = 0.45359 kg
1 ft = 0.3048 m
1 N = 1 kg m/s2
1 lbf = 32.174 lbm ft/s2
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Weight
Weight is gravitational force applied to
body W = m g (N)
g is local gravitational acceleration
constant = 9.81 m/s2 = 32.174 ft/s2 at SL
Mass of body is constant with location butweight changes with local gravitationalacceleration e.g. gmoon~1/6gearth, etc.
Example
An object occupies a volume of 25ft3 and weighs 20 lbfat a location where the acceleration of gravity is 31.0ft/s2. Determine its weight, in lbf, and its averagedensity in lbm/ft3, on the moon, where g=5.57ft/s2
3 2
2
2
2
2
3 3
25 ; 20 ; 31.0 /
32.17420
/ 20.831 / 1
1(20.8 ) (5.57 / ) 3.6
32.174
/ 20.8 / 25 0.832 /
V ft W lbf g ft s
lbm ft
lbf sW mg m W g lbmft s lbf
W mg lbm ft s lbf lbm ft
sm V lbm ft lbm ft
= = =
= = = =
= = =
= = =
mW
Work and more
Work is a form of energy defined loosely asforce times distance
SI unit is newton-meter (N-m) called Joule 1 J = 1 N-m (1 kJ = 10^3 J) English unit is Btu amount of energy required
to raise temperature of 1 lbm of water at 68F by1F
1 Btu = 1.055 kJ Dimensional homogeneity dont add apples
and oranges every term in an equation musthave same units e.g. E = 25 kJ + 7 kJ/kg
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Temperature
Thermal equilibrium implies equality of temperature
0th law of thermodynamics if two bodies are in thermalequilibrium w/ a 3rd body, they are in thermal equilibriumwith each other
Replace 3rd body with thermometer and you have basisfor measuring temperature
Hence, 2 bodies are in thermal equilibrium if they havethe same T w/o bringing them into contact
IRON 150C
COPPER 20C
IRON 60C
COPPER 60C
Temperature scales twoapproaches
First approach: Material properties change withtemperature in a measurable and predictableway i.e. volume of Hg in glass tube
Actual temperature define by fixing two temps.and assigning numbers
Celcius scale, 0C is temp. when water freezes at1 atm and 100C as temp. of boiling water at 1atm and divide by 100 (centigrade)
Fahrenheit scale, 32F/212F, resp., as ice pointand steam point
T(F)=1.8T(C)+32
Temperature scales (cont.)
Second approach: Independent of properties ofany substance; based on 2nd law (tbd)
Thermodynamic or absolute temp. scale orKelvin scale i.e. unit is Kelvin (K)
Assign reference T as 273.16K as triple point(TP) of water (point where SLV phases coexist)
Freezing point of water is 0.01K below TP so273.15K
T(C )=T(K)-273.15
Rankine scale T(R)=1.8T(K) so 1K=1.8R; TP ofwater is 491.69R; T(F)=T(R)-459.67
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Comparison of temperature scales
Consider fluid (liq/gas) at rest Define pressure
Rotating surface to new orientation andmeasuring P will get same answer, i.e. isotropicat a point (equal in all directions) for f luid at rest
Pressure in fluid does increase with depth due toweight of fluid (negligible for gas)
Pressure
( )lim / A A
P F A
=
1 2
3
P increases
1 2 3
within same fluid
P P P= >
Molecular collisionswith test surface
More on pressure
If fluid is in motion, net force/area can bedivided into 3 mutually perpendicularstresses (F/A), one normal and 2tangential (shear) w/rt surface
Magnitude of stress is a function ofsurface orientation when fluid is in motion fluid dynamics
We assume for our problems normalstress at a point is approximately pressurew/ or w/o fluid motion
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Units, absolute and gage pressure
SI units of pressure are force/area are N/m2 calledPascal (Pa)
1 Pa = 1 N/m2 (1 kPa = 10 3 Pa; 1 MPa=10 6 Pa) 1 bar = 10^5 Pa = 0.1 MPa = 100 kPa
1 atm = 101,325 Pa = 101.325 kPa = 1.01325 bar
English: lbf/in^2 or psi and 1 atm = 14.696 psi
Actual pressure at a given position is absolute pressuremeasured relative to absolute zero pressure (vacuum,no molecules no collisions)
Most pressure measuring devices are calibratd so thatzero pressure is Patm which gives the gage pressure
Absolute, gage, and vacuumpressures
Pgage = Pabs Patm for Pabs > Patm
Pvac = Patm Pabs for Pabs < Patm
If gage shows Pg=100psig, absolute pressure isPabs=Pg+Patm=100+14.7=114.7psia
In thermo. relations, P isusually Pabs unless otherwisenoted by g subscript for gage
Measuring pressure
Manometer device used tomeasure small to moderatepressure changes
Glass/plastic U-tube containingfluid such as H20, alcohol, Hg
Heavy fluids used for largerpressure differentials to keepdevice small
We wish to measure pressure P ofgas in tank (rt)
Neglecting weight of gas, P sameeverywhere in tank for gas and atposition 1, also P=P1=P2
Consider fluid column of height hand draw FBD
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Manometer calculation
FBD of fluid column at
equilibrium balance offorces says:
atmP
1P
A
Wh
1
1 1
0
( )
atm
atm
atm
atm g
F
AP AP mg
AP Vg
AP Ahg
P P P P gh kPa
=
= +
= +
= +
= = =
Barometer meas. atm. pressure
0
( )
B atm
C
atm
P P
P
P gh kPa
=
=
=
Problem solving technique
Step 1: Problem statement In your own words state problem, list key information
given, and quantities to be found
Step 2: Schematic Draw realistic sketch of physical system, choose your
system, list key information on figure, indicatemass/energy interactions with surroundings, check forconstant properties
Step 3: Assumptions State your assumptions needed to simply and solve
problem, assume reasonable values for missing data
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Problem solving technique (cont.)
Step 4: Physical laws Apply relevant physical laws to chosen system (e.g. cons.
energy), reduce to simplest form using assumptions for yoursystem
Step 5: Properties Determine unknown properties at known states from property
relations or tables, list properties separately and indicate source
Step 6: Calculations Substitute known quantities into simplified relations and perform
computations to find unknowns, watch units, round results asappropriate
Step 7: Reasoning, verification, and discussion Check to make sure results make sense, conclusions,
recommendations think of solution as engineering analysis!
Summary
System (closed or open) properties, e.g. T,
p, change during a process from oneequilibrium state to another resulting inless potential to do useful work all thewhile conserving energy