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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:

frankel@purdue.edu

frankel.steven@gmail.com

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