1) chapter 1 introduction

8/18/2019 1) Chapter 1 Introduction

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CDB 1053 Introduction ToEngineering

Thermodynamics

By Herr Azry B Borhan

Dr Muhammad Rashid B Shamsuddin

1

INTRODUCTION


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Herr Azry B BorhanRoom 04-03-10

E-mail: [email protected]

Dr. Muhammad Rashid B Shamsuddin

Room 05-03-35

E-mail:[email protected]

2

Lecturers

CDB 1053 Into to Eng Thermodynamics


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3 Credit Values:

• 3 hours of lecture/week

• 1 hours of tutorial/week • Assignment & Group Project – 10%

• Quizzes – 10%

• Test 1 & 2 – 20%

• Final examination – 60%

3

Course Layout & Schedule



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Final exam (60%): Must pass the final exam, otherwise

fail for the course

Attendance : Must exceed 90%, below which the

students can be barred from the final

exam.

Attendance of all international students will be recorded and

submitted to the Ministry of Education and will be forwarded to

the Ministry of Home Affairs.



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5

Course Layout & Schedule

CDB 1053 Trimester Sept 2015 Timetable



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6

Reference

Yunus A. Çengel is the Dean of the Faculty

of Mechanical Engineering and the Director of the Energy Center at Yildiz Technical

University in Istanbul, Turkey, and Professor

Emeritus at the University of Nevada, Reno,

USA. He received his Ph. D. in Mechanical

Engineering in 1984 from North Carolina

State University in USA. Before joining YTU

in 2010, he served as a faculty member at

the University of Nevada, Reno for 18 years.He also served as the director of the

Industrial Assessment Center at UNR for

several years.



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Outcome-Based Education (OBE)



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Chemical Engineering Programme Outcomes (PO)

PO1 Engineering Knowledge : Apply knowledge of mathematics, science,

engineering fundamentals and engineering specialisation to the solution

of complex engineering problems

PO2 Problem Analysis : Identify, formulate, research literature & analyse

complex eng problems reaching substantiated conclusions using first

principles of mathematics, natural sciences and engineering sciences.

PO3 Design & Development of solutions : Design sols for complex engproblems and design systems, components or processes that meet

specified needs with appropriate consideration for public health and

safety, cultural, societal & environmental considerations.

PO4 Investigation : Conduct investigation into complex problems using research

based knowledge & research methods including DOE, analysis and

interpretation of data and synthesis of information to provide valid

conclusions.

PO5 Modern Tool Usage : Create, select & apply appropriate techniques,

resources & modern eng & IT tool, including prediction & modeling, to

complex eng activities, with an understanding of the limitations.

PO6 The Engineer& Society : Apply reasoning informed by contextual knowledge

to assess societal, health, safety, legal & cultural issues & the consequent

responsibilities relevant to profnl eng practice.

PO7 Environment & Sustainability : Understand the impact of professional

engineering solutions in societal and environmental contexts and

demonstrate knowledge of and need for sustainable development.

PO8 Ethics : Apply ethical principles and commit to professional ethics and

responsibilities and norms of eng practice.

PO9 Communication : Communicate effectively on complex eng activities with

the eng community & with society at large, e.g. being able to comprehend

& write effective reports & design docn, make effective presentations &give and receive clear instructions.

PO10 Individual &Team Work : Function effectively as an ind & as a member or

leader in diverse teams & in multidiscip settings.

PO11 Life Long Learning : Recognise the need for, and have the preparation and

ability to engage in independent and life long learning in the broadest

context of technological change.

PO12 Project Management & Finance : Demonstrate knowledge & understanding

of eng & management principles & apply these to one’s own work, as a

member & leader in a team, to manage projects and in multidisciplinary

environments.



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CHAPTER 1

BASIC

CONCEPTS OF

THERMODYNAMICS



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What is Thermodynamics?

Early description: Convert heat into power

Current Definition: The study of energy and energytransformations, including powergeneration, refrigeration andrelationship among the properties ofmatter

Greek Words



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Application Areas of Thermodynamics

House-hold utensils appliances:

Air-cond, heater, refrigerator

humidifier, pressure cooker, water heater

computer & TV

Engines: Automotive, aircraft, rocket

Plant/ Factory Refinery, power plants, nuclear power plant



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1.1 What is Energy?

Ability to cause changes

One of the most fundamental laws of nature is the Conservation of

energy principle - “during an interaction, energy can change fromone form to another but the total amount of energy remains constant”.E.g. a rock falling off a cliff & in the diet industry.

Laws of Thermodynamics:

Zeroth Law = dealing with

First Law = dealing with

Second Law =

Hot heat Cold body, spontaneous

Cold heat Hot body, requires work

Third Law = entropy of pure crystalline substance at absolute zerotemperature is zero



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Dimension is a property that can be measured or calculatedby × or ÷ (e.g.: mass, length, time, temperature)

Unit is the means of expressing dimensions (Systems: SI,

CGS, American Engineering System)

Prefixes: centi, milli, micro, nano, kilo, mega, giga etc.

14

1.2 Dimensions and Units

Table 1.1:Prefixes for SI units



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DimensionPrimary

Secondary

M - mass

L - length

T - temperature

t - time

n - mole

A - AmpereEg: Volume V

velocity v

energy E

UnitsSI - International System- Commonly applied

English System - also known as United States Customary

System (USCS)

Dimensions and Units



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Conversion of units

16

Dimensions and Units

Table A.1: Conversion factorsQuantity Conversion

Length 1 m = 100 cm= 3.280 84 (ft) = 39.3701 (in)

Mass 1 kg = 103 g= 2.204 62 (lbm)

Force 1 N = 1 kg m s-2

= 105 (dyne)= 0.224 809 (lbf )

Pressure 1 bar = 105

kg m-1

s-2

= 105

N m-2

= 105 Pa = 102 kPa= 106 dyne cm-2

= 0.986 923 atm= 14.5038 psia= 750.061 Torr

Volume 1 m3 = 106 cm3 = 103 liters= 35.3147 (ft)3

= 264.172 (gal)

Density 1 g cm-3 = 103 kg m-3

= 62.4278 (lbm) (ft)-3

Energy 1 J = 1 kg m2 s-2 = 1 N m= 1 m3 Pa = 10-5 m3 bar = 10 cm3 bar = 9.869 23 cm3 atm= 107 (dyne) cm = 107 (erg)= 0.239 006 (cal)= 5.121 97 × 10-3 (ft)3 (psia) = 0.737 562 (ft) (lbf )= 9.478 31 × 10-4 (Btu) = 2.777 78 × 10-7 kWh

Power 1 kW = 103 W = 103 kg m2 s-3 = 103 J s-1

= 239.006 (cal) s-1

= 737.562 (ft) (lbf ) s-1

= 0.947 831 (Btu) s-1

= 1.341 02 (hp)

Always check the units in your

calculations



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1.3 Closed and Open Systems

Thermodynamic system (system) - quantity of matter or a region in

space chosen for study.Surroundings - the mass or region outside the system

Boundary - the real or imaginary surface that separates the system

from its surrounding

- is the contact surface shared by both the system &

surroundings

- has zero thickness & can either contain any mass nor

occupy volume in space.

- can be fixed or movable

Boundary

fixed

movable



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Types of system:

(a) - no heat/ mass transfer across boundary

(b) - only heat transferred(c) - heat & mass transferred

(b) (c)



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1.4 Energy

Forms of energy - thermal, mechanical, chemical, kinetic, potential,

electric, magnetic & nuclearE = total energy i.e sum of all energy in a system

e = total energy = E (kJ/kg)

mass m

Forms of energy that make up the total energy of a system :

Energy form

macroscopic

microscopic

energy of a system as a wholewith respect to some outsidereference frames, e.g. KE, PE

- related to molecular structure of asystem and the degree of molecularactivity- independent of outside referenceframes



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Sum of all microscopic forms of energy = Internal Energy (U)

Macroscopic forms of energy

Kinetic energy (KE)

- result of motion relative to

somereference frame

KE = mv2/2 (kJ)

where v = velocity of the system

relative to some fixed

reference frame (m/s)

m = mass of an object (kg)

Potential energy (PE)

- due to elevation in a gravitational

field

PE = mgh (kJ)

where g = gravitational acceleration,

9.81 m/s2

h = elevation of center of gravity of

a system relative to somearbitrarily plane (m)



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1.5 Internal Energy

Internal energy -

related to - 1) molecular structure

2) degree of molecular activity

Latent heat - Internal energy associated to with the phase of asystem

- phase -change process can occur without a change inthe chemical composition of a system

I. E KE

PE

molecular translation

molecular rotation

electron translationmolecular vibration

electron spin

nuclear spin

a.k.a sensible energy

depend on thetemperature



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1.6 Properties of a System

Property -

Some familiar properties are P, T , V and m. But can be extendedto include less familiar ones such as viscosity, thermalconductivity, thermal expansion coefficient and etc

Density (mass per unit volume), (kg/m3) depends on T &

Specific gravity or relative density (ratio of the density of asubstance to the density of some standard substance at aspecified temperature) e.g. for water,

Specific volume, (m3/kg)

V

m

O H

s

2

m

V



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Specific properties - extensive properties per unit mass

E.g. specific volume (v = V/m) and specific total energy (e = E/m)

Properties

Intensive

Extensive

independent of the

size/extent of the

system

dependent on the

size/extent of the

system



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1.7 State & Equilibrium

State a set of properties that describe the condition of a

system at certain timeAt a given state, all the properties of a system have fixed values.If the value of one property changes, the state will change to adifferent one.

Equilibrium state

Thermal equilibrium

Mechanical equilibrium

Phase equilibrium

Chemical equilibrium



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Thermal equilibrium

(uniform temperature)



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1.8 Processes & Cycle

Process

Path

need to specify the initial & final states of the process, as wellas the path it follows, and the interactions with thesurroundings.



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1.9 Quasi-equilibrium/ Quasi-static

When a process proceeds in such a manner that the system

remains infinitesimally close to equilibrium state at all times.

Sufficiently slow process that allows the system to adjust to itself

internally so that properties in one part of the system do not change

any faster than those at other parts.

Slow compression

(quasi-equilibrium)

very fast compression

(non-quasi equilibrium)



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The prefix iso- is often used to designate a process for which aparticular property remains constant.

Isothermal Process

Isobaric

Isochoric/ Isometric

A system is said to have undergone a cycle if it returns to itsinitial state at the end of the process.

For a cycle, the initial & final states are identical

ProcessB

ProcessA

1

2P

V



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1.10 Pressure

P = = Unit = N/m2 or Pa

Gas or liquid Pressure

Solids Stress

Common units

1 bar = 105 Pa

1 atm = 101,325 Pa = 1.01325 bars

1 kgf/ cm2

= 0.9807 bar = 0.96788 atm

English unit Ibf/in2 or psi

Absolute pressure Actual pressure at given position &measured relative to absolute vacuum

Gage pressure Difference between absolute pressure &local atmospheric pressure

Vacuum pressure Difference between atmospheric pressure &absolute pressure

Area

Force

A

F



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Absolute, gage & vacuum pressures are all +ve quantities &related to each other by:

Pgage = Pabs - Patm (for pressure above Patm)

Pvac

= Patm

- Pabs

(for pressure below Patm

)

In thermo, absolute pressure is always used unless stated.



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Example 1-1

A vacuum gage connected to a chamber reads 5.8 psi at alocation where the atmospheric pressure is 14.5 psi. Determinethe absolute pressure in the chamber.

Using Pvac = Patm - Pabs = 14.5 - 5.8 = 8.7 psi

Manometer

Small to moderate pressure difference are measured by amanometer and a differential fluid column of height hcorresponds to a pressure difference between the system andthe surrounding of the manometer.

P g h kPa ( )



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Other Pressure Measurement Device

Bourdon Tube

Modern pressure sensors:

1) Pressure transducers

2) Piezoelectric material



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Example 1-2

A vacuum gage connected to a tank reads 30 kPa at a locationwhere the atmospheric pressure is 98 kPa. What is the absolutepressure in the tank?

Solution:

Pabs = Patm - Pgage= 98 kPa - 30 kPa

= 68 kPa

Example 1-3A pressure gage connected to a valve stern of a truck tire reads240 kPa at a location where the atmospheric pressure is 100 kPa.What is the absolute pressure in the tire, in kPa and in psia?

Solution:



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What is the gage pressure of the air in the tire, in psig?

Example 1-4

Both a gage and a manometer are attached to a gas tank tomeasure its pressure. If the pressure gage reads 80 kPa,determine the distance between the two fluid levels of themanometer if the fluids is mercury whose density is 13,600kg/m3.

h P

g

h kPa

kg

m

m

s

N mkPa

N

k g m s

m

80

13600 9 807

10

1

0 6

3 2

3 3

2.

/

/

.



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An astronaut weighs 730 N in Houston, Texas, where the local acceleration of

gravity is g = 9.792 m s-2. What are the astronaut’s mass (kg) and weight (N) on themoon, where g = 1.67 m s-2 ?

Solution

In Texas, F = 730 N, a = g = 9.792 m s-2

= =

=

=

730 N

9.792 ms− = 74.55 N m− s

= 74.55 (kg m s-2) m-1 s2

= 74.55 kg

The mass of the astronaut is independent of location, thus,

mass ( moon) = mass (Texas) = 74.55 kg

On the other hand, the weight of the astronaut depends on

the local acceleration of gravity, thus, on the moon,

= = 74.55 kg × 1.67 m s−

= 124.5 kg m s-2 = 124.5 N

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Example 1.5

A body weighing 730 N on earth

will weigh only 124.5 N on the

moon



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At 27oC (300.15 K), the reading on a manometer filled with mercury is 60.5 cm.

The local acceleration of gravity is 9.784 m s-2. To what pressure does this height of mercury correspond? At 27oC (300.15 K), the density of mercury is 13.53 g cm-3.

Solution

Given h = 60.5 cm, g = 9.784 m s-2, = 13.53 g cm-3

P = h

= 60.5 1

100 × 13.53

100

1

1

1000 × 9.784

= 80088.4 kg m-1 s-2

= 80088.4 Pa

= 80.09 kPa

37

Example 1.7



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Temperature

Measure of hotness and coldness

Transfer of heat from higher to lower temp. until both bodiesattain the same temp. At that point, heat transfer stops and thetwo bodies have reached thermal equilibrium

requirement: equality of temperature

Zeroth Law of Thermodynamics:

Temperature scales: Celcius (C)

Fahrenheit (F)

Kelvin (K)

Rankine (R)



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Conversion:

T(K) = T(C) + 273.15

T(R) = T(F) + 459.67

T K = (T 2C +273.15) - (T 1C + 273.15)

= T 2C - T 1C= T C

T R = T C

T



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Temperature Scale Comparison

T (K) = t (oC) + 273.15

t (oF) = 1.8 t (oC) + 32

T (R) = t (o

F) + 459.67T (R) = 1.8 T (K)

Figure 1.1: Relations among temperature scales



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Example 1:

Consider a system whose temperature is 18C. Express thistemperature in R, K and F.

Example 2:

The temperature of a system drops by 27 F during a cooling

process. Express this drop in temperature in K, R &

C

1) chapter 1 introduction

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