ChE 234-1 Basic Concepts

8/10/2019 ChE 234-1 Basic Concepts

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ChE234 Thermodynamics

Overview

Module 1: Basic Concepts of

Thermodynamics

Bernard GalloisOffice: Burchard 410

Telephone: 201-216-5041

E-mail: [email protected]


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Office Hours, Burchard 410

Posted on Moodle

Exam Schedule

Posted on Moodle

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Introduction

Goal Today:

What is the course about?

What and how will I learn?

What are the tools, methods and

structured approaches that we will be

using?

Get started

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What Will You Ach ieve in th is

Course?

1. To learn the fundamental concepts ofthermodynamics

2. To develop an intuitive understanding of the subject

matter by emphasizing the physics and physical

arguments3. To master these basic principles and concepts by

applying this knowledge to a wealth of real-world

engineering applications through:

4. Structured and logical problem-solving techniquesthat you will use in homework assignments and

exams

5. Apply thermodynamics principles to the design of

simple devices and chemical processes4


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Major Course Objective:

I think, therefore I am.

Most people would sooner die thanthink; in fact, most do so. BertrandRussell (1872-1970)

Thinkwrongly, if you please, but in allcases thinkfor yourself. G.E. Lessing(1729-1781)

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How do we learn?

In good faith attempts to learn, we

remember: 1

10% of what we read

20% of what we hear 30%of what we read and hear

30% of what we see

70% of what we say

90% of what we teach or do

1Pike, R., 1989, Creative Training Techniques Handbook, Tips, Techniques andHow-tos for Delivering Effective Training; Lakewood Books, Minneapolis, MN,

153p.

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Student Evaluation

Weekly Exam (4) 30% (lowest dropped)

Quizzes 20% (lowest dropped)

Weekly Homework (6) 10% (lowest dropped)

Final Exam 40%

There will be NO make-up exams.

Any student who gets 90 percent of the total available pointsbefore the final exam (all exams and quizzes BUT not thehomework)will be excused from the final and will be given agrade of A.

Please consult the Web site for details on student evaluation,excuses, absences, etc

Attendance is mandatory: more than 2 absencesfrom class mean a full letter drop in the final grade.

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Curriculum Performance Criteria

(ABET)*

Please consult the Web site for a fulldescription of the performance criteria.

*Accreditation Board for Engineering and Technology

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Moodle Contents

Overview: student evaluation and policies

regarding excuses, absences, etc.Lecture notes

Solutions of homework problems (posted rightafter they are submitted)

Sample exams and final exam from previoussummer.

Your grades

Assignment every week: Learning objectives: what you should be able to do

Reading assignment Homework assignments. No late homework will be

accepted.

Other information

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Final Comments You have one weekafter you

received a grade on a given exam toget it changed if you consider that amistake was made. Please contactyour instructor at the end of therecitation period.

There will be NO make-up final examand should you miss it, you will need

to take the final exam the next timethe course is offered.

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Questions?

[email protected] 410

x5041

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mailto:[email protected]:[email protected]


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Items Highlightedin Light Blue

ThroughoutAre a Must-Know

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Thermodynamics and Energy

Therme=Heat Dynamis=Power

Thermodynamics is a basic engineering science thatdeals with energy in all its forms and itstransformations for the design and operation ofenergy and other systems.

The macroscopic approach is called classicalthermodynamicsand is used to solve engineeringproblems whereas:

Statistical thermodynamics uses a microscopicapproachbased on the average behavior of largegroups of individual particles


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SOME APPLICATIONS AREAS OF

THERMODYNAMICS


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Rationale:

Energy is the single most important

challenge facing humanity today. Nobel

Laureate Rick Smalley, April 2004, Testimony

to U.S. Senate

What should be the centerpiece of a policyof American renewal is blindingly obvious:

making a quest for energy independence the

moon shot of our generation, Thomas L.

Friedman, New York Times, Sept. 23, 2005.

Whatever your major, energy issues may

play a key role in your career.

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JUST A FEW PIECES OF DATA ON

ENERGY, FOR NOW

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Global Energy Consumption

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http://nuclearfissionary.com/2010/04/02/comparing-energy-costs-of-nuclear-coal-gas-wind-and-solar/


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Energy Consumption vs GDP

GJ/capit

a-yr


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What is Energy?

What definitions would you

give? Think about the varioussystems shown on Slide 15

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COW CATCHER: An invention of the Stevens

family


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WHEN DID THERMODYNAMICS

START?

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The Newcomen Engine 1712

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Used to pump water out of mine

shafts

One of the first patents of the

industrial revolution

Very low efficiency

The valves were operated by ahuman operator at about 15

cycles/min

Pink is hot steam, blue is liquid

water

Can you figure out how it worked?


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A much improved Steam Engine of the type one would have

found around 1800+


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Thermodynamic Systems A (thermodynamic) system

is defined as a quantity ofmatter or a region of spacechosen for study

The region outside thesystem is called the

surroundings

The real or imaginarysurface that separates thesystem from the

surroundings is called theboundary

Systems may beopenorclosed

Cl ifi ti f B d i


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Classification of Boundaries

The interactions between a system and its surroundings

are governed by the nature of their common boundary.

Closed system: the boundary is impermeable to mass

flow.

Open system: the boundary allows for mass transport

across it.

The boundary can be rigid, movable or imaginary.

One other set completes the classification: adiabaticand

diathermal boundaries. These boundaries represent

extremes in the rate of heat exchange between the

system and its surroundings. The adiabatic boundary

plays a key role in thermodynamics and we wil return to

it when we introduce the concept of heat

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Adiabatic and Diathermal Boundaries


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Adiabatic and Diathermal BoundariesPlease refer to the next slide

Consider a rigid closed system (a can!) in which a thermometer has

been inserted. Surround this system with a system having a higher

temperature (dip it in hot water)

If the can were made of copper the variation of temperature with

time would look like curve 1 on the next slide.

Curves 2, 3, 4 would result if it was made of steel, glass or asbestos.

If the container was a thermos bottle (a Dewar), the variation oftemperature with time would be quite small (Curve 5)

The adiabatic wall is an idealized concept representing the limiting

case A in the figure. In practice, adiabatic boundaries are

approached in many situations especially those in which events

occur rapidly in relation to the time scale of the experiment.

The diathermal wall is the case diametric to the previous one. The

change in temperature is very rapid in relation to the time scale of

the experiment.

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Based on : J. Tester and M. Modell, Thermodynamics and its Applications 3d

edition, Prentice Hall, p.13


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Types of Boundaries

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Temperature-time behavior for different materials. Limiting

diathermal and adiabatic boundary behavior shown.

From: J. Tester and M. Modell, Thermodynamics and its Applications 3d

edition, Prenitce Hall, p.13


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Questions

You inflate a tire with a bicycle pump, under what

conditions would you consider the wall of the pump

to be diathermal? adiabatic?

(Note: The air inside the pump will heat up as you

compress it)

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The first action in considering a problem is to define

the system and, consequently, the location and the

nature of the boundaries:

Closed or open system (permeable or impermeable

boundary)

Rigid or movable

Diathermal or adiabatic

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Defining the System


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Closed Systems

Mass cannot cross theboundaries of a closedsystem but energy can

The boundary of a closed

system can move Examples includes sealed

tanks and piston/cylinderdevices

Energy in the form of heatand work may cross theboundaries of a closedsystem


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Open System and Control Volume

An open system or

control volume hasmass as well as energycrossing the boundary,often called the controlsurface. The surfacecan be real orimaginary.

Examples includepumps, compressors,turbines, nozzles,valves and heat

exchangers Proper choice of a

control volume cansimplify the analysis ofthe system


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Control Volume: Examples


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Isolated Systems An isolated system is a general system of fixed

mass where no heat or work may cross the

boundary. It is a closed system with no energy crossing the

boundaries.

It is normally a collection of a main system and itssurroundings that are exchanging mass and energy

among themselves and no other system.


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Property

A propertyis a characteristic of a system inequilibrium, independent of the path used toarrive at the systems condition.

A property may be intensive or extensive

Extensive propertiesdepend on the size orextent of the system. e.g. mass, volume,total energy, mass dependent property

Intensive propertiesare independent of thesize of the system. e.g. temperature,pressure, color, age, any mass independentproperty


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Specific Properties

How about these properties?

Density

Specific volume: volume per unit mass Specific energy: energy per unit mass

V

m

1

m

v

V

m

e

E


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STATE AND EQUILIBRIUM

A system is said to be in thermodynamic equilibriumifit maintains

othermal, (uniform temperature)

omechanical(uniform pressure)

ophase(the mass of the two phases, e.g. ice and liquid

water, in equilibrium) and

ochemicalequilibrium


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PROCESS

Any change from one state to another is called a process

During a quasi-equilibrium or quasi-staticprocess thesystem remains practically in equilibrium at all times.

Can you think of a real engineering process that is quasi-

static? Would that be useful?

We will often assume that processes we analyze are quasi-equilibrium

processes because they are easy to analyze (the equations of state apply)

and work-producing devices deliver the most work when they operate on a

quasi-equilibrium process.


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EXAMPLE: CONSTANT PRESSURE

PROCESS

In most of the processes

that we will study, one

thermodynamic property

is held constant. Isobaric: constant P

Isothermal: constant T

Isochoric: constant V

Isentropic: constant S

(S is entropy, Chapter 6)

At equilibrium, the force

exerted by the water on the

lower face of the piston is

equal to the combined

weight of the brick andpiston (a constant). The

pressure in the tank is thus

a constant even when the

water is heated.

P b t ill t t d b di


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Processes are best illustrated by a diagram,

a useful approach when solving problems

Using English sentences you


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Using English sentences, you

must be able to define:

Closed system Open system

Isolated system

Control volume

Control surface

Intensive property

Extensive property

State and equilibrium

Quasi-static or equilibrium process

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The Steady-Flow Process

During a steady-flowprocess, fluid propertiesmay change with positionbut not with time.

Steady-flow conditionsessentially prevail in manyengineering devices.

Therefore, the volume, themass and the energycontent of the controlvolume remain constantduring the entire process.

STATE POSTULATE


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STATE POSTULATE

The state of a system is described by its properties. By

experience, not all properties need to be known to specify the

state of the system. In fact, the number of properties required

to fix the state of simple, homogeneous system is given by the

state postulate.

In a simple compressiblesystem, there are no electrical,

magnetic, gravitational, motion or surface tension effects. An

additional property need to be specified for each effect that issignificant.

In ahomogeneoussubstance, the thermodynamic properties

are uniform throughout.

Two properties are independent if one can be varied while the

other is held constant.

The thermodynamic state of a simplecompressible system is completely specified by

two independent intensive properties.


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CYCLES

A system is said to have

undergone a cycle if itreturns to its initial state at

the end of the process.

The illustration shows a

cycle consisting of twoprocesses.

Along process A, both P and

V change from state 1 to

state 2.

To complete the cycle, P and

V change from state 2 to

state 1 in process B.

Keep in mind that all other

thermodynamic properties

must also change so that the

pressure is a function ofvolume as described in these

two processes.


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UNITS

In this course, we shall use S.I. (Systeme

International) units exclusively. The seven fundamental units: Length: meter m

Mass: kilogram kg

Time: second s Temperature: kelvin K

Amount of light: candela cd

Electric current: ampere A

Amount of matter: mole mol All other units are derived from the

fundamental units


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MORE ON UNITS

Velocity: length/time m/s

Acceleration: velocity/time m/s2

Force: mass x acceleration kg m/s2

Pressure: force/area kgm-1/s2

Work (energy): force x distance kgm2/s2

Power: work/time kgm2

s

Quantity Derived from: Dimensions Unit


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Quantity Derived from: Dimensions Unit

Area L2 m2

Volume L3 m3

Density Mass/volume ML-3 kg/m3Specific volume Volume/mass M-1L3 m3/kg

Velocity Length/time LT-1 m/s

Acceleration Velocity/time LT-2 m/s2

Force Mass x acceler. MLT-2 N (newton)

Pressure Force/area ML-1T-2 N/m2

Work or energy Force x length M L2 T-2 J (joule)

Power Energy/time M L2 T-3 W (watt)Entropy Energy/temp. M L2 T-2K-1 J/K

Specific entropy

Specific energy

Entropy/mass

Energy/mass

L2 T-2K-1

L2 T-2

J/kgK

J/kg

Temperature and the Zeroth Law of


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Temperature and the Zeroth Law of

Thermodynamics What is temperature?

Temperature measurements are based onpredictable and reliable changes in the properties ofmaterials as they are heated or cooled: expansion/contraction of metals or other substances,resistance of platinum, change in color, etc

At thermal equilibrium, two bodies have equaltemperatures.

The zeroth law of thermodynamicsstates that whentwo bodies are in thermal equilibrium with a third

body, they are also in equilibrium with each other If the third body is a thermometer: two bodies are in

thermal equilibrium if both have the sametemperature reading even if they are not in contact


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The Ideal Gas Temperature Scale

We need to have a temperature scale that is

independent of the properties of any substance: athermodynamic temperature scale, or a kelvin scale

The ideal gas temperature scaleis such a scale.

At low pressures, T is proportional to P : T= a +bP

a and b can be determined experimentally with a gasthermometer by using only two reproducible points(ice and steam point of water, freezing point of puresubstances) to determine the two unknowns a and b.

The unknown temperature T of a given medium can

then be determined by interpolation. In the Celsius scale, the freezing and boiling points

are assigned values of 0 and 100 respectively.


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The Absolute Gas Temperature Scale

When using different types

of gases at low pressures,plots of P as a function of T

extrapolate to -273.15 oC, the

lowest attainable

temperature, also called

absolute zero.

In the absolute gas

temperature scale, T = bP

and the temperature only

needs to be specified at a

single point to define the

absolute gas temperature

scale


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Temperature Scales

T(R) = T(oF) + 459.67

T(R) = 1.8 T(K)The Celsius scale is defined interms of a single point, thetriple point of water (0.01oC)and the absolute temperature

scale. The ice point remains0oC (273.15 K) and the boilingpoint remains 100oC (373.15K)at one atmosphere pressure.(The normal temperature of the humanbody is 37.1oC or 98.8oF)

T(K) = T(o

C) + 273.151 K = 1 oCT(oF) = 1.8 T(oC) + 32

N molecules Kinetic Theor of Ideal Gas


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The temperature of an ideal monatomic gas is a measure related to

the average kinetic energy of its atoms as they move. In this

animation, the size of helium atoms relative to their spacing is

shown to scale under 1950 atmospheres of pressure. These room-

temperature atoms have a certain, average speed (slowed down

here two trillion fold).

N molecules

Cubic box of side l

Volume V = l3

Constant temperature

Spherical elastic particles inconstant random motion

Elastic collisions with the walls and

between particles.

The interactions among molecules

are negligible. They exert no forces

on one another except during

collisions.

The total volume of the individual

gas molecules added up is negligible

compared to the volume of the

container.

Kinetic Theory of Ideal Gas

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Pressure = Force exerted by molecules on a


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y

surface per unit area

A gas molecule collides with the wall of the

container perpendicular to the xaxis and bouncesoff in the opposite direction with the same speed (an

elastic collision). The momentum lost by the particle

and gained by the wall is:

The particle impacts the wall once every 2l /vxtime

units and produces a momentum change on the wall

once every 2l /vxtime units.

The force on the wall due to the particle is given by

2x i f xp p p mv

22

2 /

x x

x

mv mvpF

t l v l

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Pressure cont.


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Pressure cont.

The force exerted by N particles of mass m can be written as:

The pressure is given by

The total mass of gas is Nmand the density is Nm/V, the

density

so that

The pressure is proportional to the density of the gas and the

root mean square velocity.

2

rmstotal mNvF

l

2 23

2

1

since

wall rms rmsF mNv mNv

P V lA l l V

21

3 rmsP v

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Units of Pressure

The SI unit is the pascal equal to onenewton per square meter. 1 Pa = 1 N/m2

It is a small unit so that kilopascals kPa=

103Paand megapascals Mpa= 106Paareused commonly.

Other units widely used include:

1 bar = 105

Pa1 atm = 101,325 Pa = 101.325 kPa =1.01235 bar = 760 torr

Gage pressure vacuum pressure atmospheric


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Gage pressure, vacuum pressure, atmospheric

pressure, absolute pressure

Pgage

= Pabs

Patm

Pvac= Patm- Pabs

What is the origin of atmospheric pressure? The weight of the column of

air above us!

The pressure is the same at all points on a


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The pressure is the same at all points on a

horizontal plane in a given fluid regardless of

geometry, provided that the points are

interconnected by the same fluid.

Forces Exerted on a Fluid Column


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Forces Exerted on a Fluid ColumnConsider an element of fluid of height h and

area A and the free body diagram.

The pressure P1exerted on the top face of the

element = F1x area

The pressure P2exerted on the bottom face of

the element = F2x area

From the free body diagram:

F2= F1+ mg

P2A = P1A + mg but m =density x volume = AhFrom which we deduce

P2P1= gh

As we move down in a fluid , the pressureincreases (diving in the ocean).

As we move up in a fluid, the pressure

decreases (climbing a mountain).

F2

F1

h

mg

The Manometer


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e a o ete

The manometer is used tomeasure small andmoderate pressuredifferences.

How small is small:~ 103kg/m3 , g ~10 m/s2 h ~

0.1 m

P ~ 0.1x104Pa ~ 0.01 atm~0.01 bar

Pressure due to a columnof water 10 m high:

P = 103 x10x10 ~105Pa

Patm

Pascals experiment: by

running a long pipe up amountain and into a barrel ,he was able to burst thebarrel!

P1= P2 (same level in connected fluid)P2 = Patm+ gh

Barometer and the Atmospheric


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Barometer and the Atmospheric

Pressure:

The Basic Barometer

A force balance in the verticaldirection yields:

Patm= gh

Standard atmosphere:

13 595 kg/m3

(at 0o

C)g = 9.807 m/s2(standard g)

h = 0.760 m (760 mm)

Patm= 101.325 kPa

Patm= 760 mm Hg

Patm= 29.92 inHg

Patm= 760 torr

Vacuum

Note: atmospheric pressure decreases with

altitude; why?

Engineering Units To Know


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g g

Energy: kJ and kWh

1 kWh = 1000Wx3600s = 3.6x106

J= 3.6x103kJ

Pressure

1 MPa = 1000 kPa 1 bar =100,000 Pa =100 kPa

1 atm = 760 mmHg = 101.325 kPa

Volume 1 L = 1000cm3= 0.001 m3

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Work

Energy Transfer by Work


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Energy Transfer by Work

Mechanical work, W, is performed

whenever a force, F, acts through adistance, dl:

W = Fdl

The work done during a process betweentwo states, 1 and 2, is denoted W12 orsimply W

Work done per unit mass w = W/m (J/kg) The work done per unit time is called

power and is denoted (.

W

Some Important Caveats


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Work is a boundary phenomenon.

Systems possess energy, but not work.

Work is associated with a process, not a

state.

Work is a path function: its magnitude

depends on the path followed during a

process.

Path Functions and Point Functions


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Path Functions and Point Functions

Path functions have inexact

differentials denoted by d.A differential amount ofwork or heat is representedby dW or dQ not dW or dQ.

Properties are pointfunctions and they haveexact differentialsrepresented by the symbold as in dV. They dependon the initial and finalstates only and not on thepath taken to go fromstate1 to state 2.

The volume change during

process 1-2 is V2

V1

=

V

irrespective of path.

The total work during the

process, W12(not W and not

W2W1) depends on the path

taken.

Mechanical Forms of Work


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For a work interaction to exist between a system

and its surroundings:

There must be a force acting on the boundary.

The boundary must move.

If the force is constant: W= Fs (kJ)

If the force is not constant:

The integral assumes that we know the relationbetween F and s

2

1

( )W F s ds

Potential Energy and Kinetic Energy


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Potential Energy and Kinetic Energy

Gravitational Workis the work done by oragainst gravity. The force acting on a body is F = mg. The work is

given by:Wg= mg( z2z1)

where z1z2is the vertical distance traveled.

This is simply the change in potential energy.

Accelerational workis associated with a change in velocity of asystem.F = ma = m dV d t V = ds /dt ds = V d t

A simple integration yields:

Wa

= m(V2

2V1

2)

This is simply the change in kinetic energy.

Example: Conservation of Energy


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a p e Co se at o o e gy

The energy of a falling object is

conserved:Wg= mg( z2z1)

Wa= m(V22V1

2)

Wg +Wa = 0

mg( z2z1) + m(V22V1

2) = 0

Potential energy is transformed into kinetic

energy.

Shaft Work


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Shaft Work

Very often, the torquet (and hence the

force F) applied to a shaft is constant. Wecalculate the work done in this case

A force F acting through a moment armr

generates a torque given by T Fr

This force acts through a distance s which

is related to the radius r by s = (2pr)n,

where n s the number of revolutions

(2 ) 2T

W Fs rn nT r

p p

Spring Work, Elastic


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p g ,

For linear elastic springs, the displacement x isproportional to the force: F = - kx

The displacement is measured from theequilibrium position (x = 0 when F= 0)

The work done is

Where x1and x2are the initial and final

displacements of the spring measured from theundisturbed (equilibrium) position.

2

1

2 2

2 11 ( )2

x

x

W Fdx k x x

Electrical Work


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Electrical Work

V

We = VN

Power = VI

= RI2

= V2/R

I

When N coulombs of electrical charge move

through a potential difference V, the electricalwork is We= VN

Moving Boundary Work


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Moving Boundary Work

This process is closely

approximated in real enginesprovided the piston moves atlow velocities (unlike thepistons in an automobileengine).

(Pistons in an automobile enginemove so fast the process path

cannot be specified. The workdone is determined by directmeasurements.)

The work associated with

a moving boundary is

called boundary work.

The analysis is for a quasi-static or quasi-equilibrium

process for which the work

output is a maximum.

Moving Boundary Work (cont.)


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By convention, work is positivewhen the displacement is in the

same direction as the applied forceand negative when they are inopposite directions.

The differential work done is:

Expansion: dV>0, dWb< 0

Compression dV 0

The expression above is theboundary work input for theprocess.

Initial pressure: P

Total volume: V

Area of piston: A

The piston moves a

distance dsin a quasi-

equilibrium manner

2

1

t

t

PdVW

PdVPAdsFdsWd



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The total boundary work doneduring the process as thepiston moves is:

The integral can only becalculated if we know the

functional relationshipbetween P and V during theprocess: P = f(V)

P = f(V) is the equation of theprocess path

The total area under theprocess path is equal, inmagnitude, to the work done inthe quasi-staticprocess

2

1

PdVWb 2

1

2

1

PdVdAAArea

This is a closed system

Moving Boundary Work (cont )


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Moving Boundary Work (cont.) The boundary work done

during a process depends on

the path followed as well asthe end states. Work is a pathfunction.

The net work during a cycle isthe difference between thework done by the system and

the work done on the system. P is the pressure at the inner

face of the piston, equal to thepressure in the gas only if theprocess is quasi-equilibrium sothat the pressure is uniform inthe cylinder at any time.

Expansion

Compression

2

1

dVPW ib

Example: Constant Volume Process


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dV = 0

dW = -PdV = 0

If the working fluid is an

ideal gas, what willhappen to the temperature

during a constant volume

process?

No work is done but:

PVt= nRT or

P/T = nR/Vt = const.

P1

/T1

= P2

/T2

In this case P1> P2Consequently T1> T2The energy of the gas

decreases

Example: constant pressure process


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For the constant pressure

process shown here, is the

boundary work positive ornegative?

It is a compression process

so the work done is negative

Will the temperature changeduring the process?

For one mole of gas

PV = RT or V/T = R/P = const.

V1/T1= V2/T2In this case V1> V2

Consequently T1> T2The energy of the gas decreases!

dW = -PdV

The integrat ion is

straighto rward sin ce P isconstant.

W = P (V1V2)

Heat Transfer


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Energy can cross theboundary of a closed system

in two distinct form: heat andwork.

Heat is the form of energythat is transferred betweentwo systems (or a system

and its surroundings) byvirtue of a temperaturedifference.We call thisprocess heat transfer.

The larger the temperaturedifference, the higher is therate of heat transfer.

Thermal energy refers to forms

of energy such a latent and

sensible energies to prevent

confusion with heat transfer

Adiabatic Process Insulation


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Adiabatic Process

During an adiabaticprocess, a system

exchanges no heatwith the surroundings.

Q = 0AdiabaticSystem

Notation and Units

Heat has the energy unit of kJ. The amount of heat transferred during a processbetween states 1 and 2 is denoted Q12 or just Q.

The heat transfer per unit mass is q = Q/m

The heat transfer per unit mass is q = Q/m The rate of heat transfer is denoted

When the rate is constant during a process:

Q = t

Q

Q



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Heat and work are boundary phenomena.

Systems possess energy, but not heat or

work.

Both are associated with a process, not a

state.

Both are path functions: their magnitude

depends on the path followed during aprocess.


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If the energy crossing the

boundary of a closed system is

not heat, it must be work

(Heat is easy to recognize since itsdriving forceis a temperature

difference)

Problem-Solving


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Problem-solving is at the heart of the engineeringprofession. A great engineer is a great problem solver.

While problem-solving may be considered by some to bean art since different approaches may be warranteddepending on the nature of the problem at hand and thematurity of the problem solver, there is, however, a verygeneral method that is presented in this module.

As we progress in the course, specific approaches orvariations of this technique will be used.

You are more than encouraged to use thesesapproaches, which will be systematically used in thesolutions we will provide of homework, practice andexam problems.

A General Problem-Solving Technique


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1. Problem statement

2. Schematic3. Assumptions

4. Properties

5. Physical laws

6. More needed properties7. Final Equations

8. Calculations

9. Reasoning, verification and

discussion

Analysis

Synthesis

EvaluationAny engineering analysis presented to others is a form ofcommunication. Neatness, organization, completeness and visualappearance are quite important for maximum impact.

Problem-Solving TechniqueCont.


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1. Problem statement

In your own words, state the problem, the key informationgiven and the quantities to be found

2. SchematicDraw a realistic sketch of the system and list the relevantinformation on the figure, especially your choice of system orcontrol volume

3. AssumptionsState any appropriate assumptions, made to simplify theproblem. Justify these assumptions.

4. PropertiesWrite down the known properties. Identify the unknown

properties at known states necessary to solve the problem.Indicate their source

5. Physical lawsApply all the relevant basic physical laws and simplify them byusing the assumptions

Problem-Solving Technique Cont


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Problem Solving Technique Cont

6 Properties

Determine the unknown properties at known states necessaryto solve the problem. Indicate their source

6 Final Equations

Write down the final equations

7 CalculationsSubstitute the known quantities in the equations and performthe calculations to determine the unknowns. Pay attention tounits and dimensions. Keep a suitable number of significantdigits.

8 Reasoning, verification and discussionReview your work. Validity of assumptions? Do the numericalresults make sense? Repeat the calculations if a number lookstoo small or too large.

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ChE 234-1 Basic Concepts

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