chapter 1 - introduction themo

Introduction

The Scope of ThermodynamicsThermodynamics VariablesWork, Energy, and Heat

Chemical Engineering Dept. BUE

Engineering Thermodynamics 1-2

Historical Review of Energy Development

First man used his won physical energy for hunting and building shelters 3000 B.C.: Use of men and animal physical energy 100 B.C.: Invention of water wheel 900 A.D.: harnessed wind energy for sailing and windmills 1711: Development of first steam engine 1831: Practical application of Electric Generator and Electric Motors 1905: Enistein theory 1945: Exploring of atomic bomb

Life, including any event in the physical world, is built on transformation of energy.

Energy: combination of two Greek word, means Capacity & Work



Present Forms of Energy

Kinetic Energy ~ Motion of an entity that has a mass

Potential Energy ~ Relative location of an entity with mass in a force field

(e.g. in gravitational field)

Chemical Energy ~ Interactions between atomic particles (e.g. released in

an exothermic reaction)

Fuel cells

Nuclear fusion

Photo cells

Utilization of energy from wind, sun, coal, etc.



The Concept of Caloric

Introduced around 1750’s

Temperature of a body changed by contact with another body

Caloric: a substance with neither volume nor mass that could flow

between bodies with different temperature:

A body with high temperature is thought to possess a high caloric value



Energy Conversion

1. Potential and kinetics energies are mutually convertible.

Example: Pendulum oscillation

2. Electrical and thermal energies

Example: Passing electrical current through a resistor

3. Chemical and thermal energies

Example: Exothermic chemical reaction



Why Study Thermodynamics?

Since any event in the world we live in is accompanied by energy transformation, thermodynamics is at the core of all science.

In order to apply science for the well being of society, all engineers need to be acquainted with thermodynamics.

Software packages are nowadays available for calculation of thermodynamic properties, yet their successful use requires the engineer to comprehend the science of thermodynamics first.

Thermodynamics is the science that deals with the transformation of energy of all kinds from one form to another.



Examples of Thermodynamics in Chemical Engineering

Reactor design requires energy balance considerations that involve enthalpy calculations. Free energy also needs to be evaluated for equilibrium conversion calculations.

Simulating compressible flow requires enthalpy and entropy calculations.

The chemical engineer designing or simulating a chemical process needs to evaluate thermodynamic properties of the fluids involved.



How to Study Thermodynamics Thermodynamics can be approached on a microscopic or macroscopic level.

In the macroscopic approach, matter is treated as a continuum. The properties of matter are attributed to macroscopic variables, such as pressure and temperature. These variables are results of molecular behavior (microscopic phenomena).

For example: Molecules confined in a container hit the walls (microscopic occurrence). The result of this momentum exchange is referred to as pressure exerted by

the gas (macroscopic observation).

The microscopic approach deals with the motion of molecules, which involves statistics due to the presence of random activity.



Engineering Approach to Thermodynamics For practical purposes, engineers typically use the macroscopic approach.

For example: First Law of Thermodynamics (Conservation of Energy)

Although energy assumes many forms, the total quantity of energy remains constant, and when energy disappears in one form it appears in other forms.

General restrictions that are observed to occur within all energy transformations are called “laws”.

This approach to thermodynamics utilizes consistent experiences that have not been observed otherwise.



Thermodynamics has a language of its own Thermodynamics starts with only four laws (1st and 2nd being the

most fundamental). All else results from definition or deduction.

Thermodynamics may be treated as complex language.

- First, rigorous meanings are attached to terms (words).

- Then, through logical manipulation, mathematical equations (sentences) are formulated to reach the desired results.

In order to apply the thermodynamic method, it is crucial to develop the ability to proceed logically from one deduction to the next, making use of precisely defined terms.



Pipes carry all the materials being processed in a chemical plant between the individual processing units, for example between the distillation columns and heat exchangers shown in the picture.In most chemical plants there are miles and miles of pipes, as you will find out.

http://lorien.ncl.ac.uk/ming/Webnotes/Main/Where/distill.html

http://lorien.ncl.ac.uk/ming/Webnotes/Main/Where/heat.html



Basic Concepts When applying thermodynamics to a real problem, it is necessary to

identify the system.

The system is the part of the universe taken aside for study.

Surroundings are the rest of the universe (all excluding the system)

The system’s thermodynamic state is defined by macroscopic properties that can be measured (such as pressure).

The macroscopic properties are described in terms of fundamental scientific dimensions: Length, time, mass, temperature etc.

“Quantity of matter of fixed mass and identity upon which attention is focused for study”

Closed and Open Systems

(a) Closed Systems:

controlled (fixed) mass.

Moving boundary

Fixed boundary

GAS2 kg3 m3

GAS2 kg1 m3

Closed System

energy YES(m= constant)

mass NO

(b) Open system: controlled (fixed) volume

Water Heater

(control volume)

Control surfaceHot Water Out

Cold Water IN

Note: Volume is fixed

Thermodynamic State and Equilibrium

• Thermodynamic state of a system – the condition of the system as characterized by the values of its properties.

• There are many different types of equilibria that can be attained:

- Thermal: the temperature is the same throughout the system.- Mechanical: the pressure is the same.- Phase: no driving force for the total mass in each phase to change.- Chemical: no driving force for chemical composition to change.

• We show a state as a point on a phase diagram as long as the continuum theory applies.

• Stable equilibrium state – a state in which the system is not capable of spontaneous change to another state without a finite change in the surrounding. There are no driving forces to carry out a change.



Fundamental Quantities With our senses, we experience fundamental concepts such as time,

distance, mass and temperature.

These concepts are known through experience, yet in order to give them quantitative meaning, it is necessary to give them standard units of measurement.

Definitions of such units have been arbitrary initially. Over time, they have become standardized.

Internationally accepted units are codified as the International System of Units (SI). s, m, kg, K, mol

For example, the fundamental unit of mass is the kilogram, defined in terms of an (initially arbitrary) platinum/iridium lump.



Measures of Amounts or Size Common measures of amounts or size:

1. Mass, m 2. Number of moles, n 3. Total volume, Vt [m3]

Relationships between the measures of the amounts or size:

=m

nM

=m Mnor

Where M is the molecular weight Specific or molar volume:

• Specific volume: =tV

Vm

=tV mVor

• Molar volume: =tV

Vn

=tV nVor



Intensive and Extensive Properties Intensive thermodynamic variables: are those variables which are

independent of the size or amount of the substance (eg., , , T and P)

Extensive thermodynamic variables: are those variables which depend on the size or amount of the substance (eg., Vt , m and n)

Note: Intensive variables are functions of T , P and composition



Fundamental Quantities Force: Based on Newton’s second law of motion: F ma

Unit of force: kg m s-2 = [N] Be aware of the engineering for force (lbf): see the textbook

1 lbf = 4.4482216 N See Example 1.1 (page 4) Weight properly refers to the force of gravity on a body

According to SI system

Temperature A measure of degree of “hotness” by the length of fluid column

SI unit for temperature oC or K: T (K) = T (oC) + 273.15 See the text (p. 5) for explanation of temperature scale Other units: T (R) = 1.8 T (K); T (oF) = 1.8 T (oC) + 32

T (R) = 1.8 T (oF) + 459.67

Temperature and Zeroth Law of Thermodynamics

Zeroth Law of Thermodynamics: The zeroth law states that if two systems are in thermal equilibrium with a third system then they are also in thermal equilibrium with each other. In other words; Two bodies are in thermal equilibrium if both have the same temperature reading even if they are not in contact.

AB C

Chemical Engineering Dept. BUE 1-20

Engineering Thermodynamics

Systems are in thermal equilibrium if they do not exchange energy in the form of heat.



Pressure Normal force exerted by a fluid per unit area of the acting surface

Unit: in SI system N m-2 = [Pa]In English engineering system: force per square inch (psi)

=F

PA

Example: for vertical column, under the influence of gravity

The force acting on the system is the weight of the fluid F = W = mg = ( V) g = A h g

mgh

A

= = =F Ahρg

P ρghA A

Other units of pressure: 1 atm = 101,325 Pa = 101.325 kPa = 0.101325 MPa = 760 mmHg 1 bar = 105 Pa = 0.986923 atm

Read Examples 1.2 & 1.3 (p. 8)

Exercise (1):

The variation of fluid pressure with height is described by the differential equation:

Here, is specific density and g is the local acceleration of gravity. For an ideal gas, = MPIRT, where M is molar mass and R is the universal gas constant. Modeling the atmosphere as an isothermal column of ideal gas at 283.15 K (10°C), estimate the ambient pressure in Denver, where z = 1 (mile) relative to sea level. For air, take M = 29 g mol-1; values of R are given in App. A.



dP gdz

r= -

Solution





Thermodynamics deals with absolute pressure which is different than the gauge reading:

Important!!

Gauge reading = Absolute pressure – Atmospheric pressure

Datum (zero pressure)

Atmospheric pressure (1 atm)

Pressure above atmospheric

Pressure below atmospheric

Absolute pressure less than atmospheric

Vacuum gauge reading (-ve)Barometer reads atmospheric

Absolute pressure(> atm)

Pressure gauge reading (+ve)



W Fl Force acting through distance: W Fdl

Work caused by displacement of a piston, e.g. compressing a fluid in a cylinder (work is positive)

-( )tVW PA d

A

- tW PdV

2

1

-V

t

V

W PdV Note: V in these equations refers to total volume.

Work

Sign convention: +W: work done by the

surrounding on the system: -W: work done by the system on

the surrounding

Work is a path function

F = PAA

l



Inexact differential: is path function and depends on the path followed between the initial state and the final state, denoted by y (= area under the curve ).

Exact differential: is a state function and is independent of the path followed between the initial state and the final state; it just depends on the values of the function at the initial state and final state, denoted by dy (= y2 – y1 )

Example; work and heat

Example; V, T and P

Remember: A relation between P and V should be available to determine the work

(graphical or mathematical).

W = area under the curve



Graphical Representation of Compression and Expansion Processes

P

V

1

a

P

V

1

b

2

b

2

a

W W

Compression: +ve work?

Expansion: -ve work?

For the compression process:@ state 1; low pressure & high volume

@ state 2; high pressure & low volume Opposite for expansion process



Example:Gas contained in a cylinder:

Gas

Initial pressure and volume: P1 = 200 kPaV1 = 0.04 m3

Determine the work done for the following cases:a) The gas was heated under constant pressure until the volume of the gas increase to V2 = 0.1 m2.

2

1

-V

t

V

W PdV 2

1

-V

t

V

P dV = - P (V2-V1) = 200 kPa (0.1 – 0.04)m3 = -12 kJ

b) The same initial condition, but the weight removed in such away that the relationship between P & V can be expressed as PV = constant & V2 = 0.1 m3.

1 1 2 2PV PV 1 12

2

PVPV

Final pressure: = 200 (0.04/0.01) = 80 kPa2

1

12 -V

t

V

W PdV 1 1PVPV

; where PV = constant = P1V1 = P2V2 2

1

1 112 -

Vt

V

PVW dVV

2

1

1 1- t

t

VdVV

V

PV 21 1

1- ln VPV V = -(200)(0.04)ln(0.1/0.04) = -7.33 kJ



b) Same system as in (b) but P & V expressed as PV1.3 = constant.

Consider the general case of: .nPV const

1 1 2 2n n nPV const PV PV 1 1 2 2

n n

n n n

const PV PVPV V V

2

1

12 -V

t

V

W PdV 2

1

-V t

nV

dVconstV

2

1

1

-1

Vn

V

Vconstn

1 112 2 11

n nconstW V Vn

1 12 2 2 1 1 1

1

n n n nPV V PV Vn

2 2 1 112 1

PV PVWn

or

For our problem:1.3

20.04200 60.77 kPa0.1

P

12(60.77)(0.1) (200)(0.03)

1-1.3W

= - 6.41 kJ



HeatExample: Block of hot copper placed in a beaker of cold water:

Cold water(Low calorie)

Hot copper(High calorie)

Result worm water

(same calorie)

Colder copper

Copper block cools down and water warms up

What causes the decrease and increase in temperature ?

Definition of heat: form of energy that is transferred across the boundary of the system at a given temperature to another system (or surrounding) at a lower temperature by the virtue of temperature difference between the two systems.

High T and low T bodies Thermal communication

Thermal equilibrium should be achieved by heat transfer from high T to a lower T till Tf = Ti



A body never contains heat Heat is just crossing the boundary(like electrical current)

Heat is a transient phenomenon

A body contain energy (≡ voltage) Q ≡ IEnergy ≡ Voltage

Temperature is the driving force for heat flow:

Rate of heat transfer T

When energy in the form of heat is added to a body, it is not stored as heat but as kinetics or potential energy molecules making up the body

Sign convention: +Q: heat done by the surrounding on the system: - Q: heat done by the system on the surrounding

Heat like work, is a path function, or inexact differential 2

1

2

1 21

( )T

T

Q Q f T dT



Iso-Processes Adiabatic process: a process with no heat transfer (Q = 0)

Isothermal process: a process with constant temperature entirely

Isobaric process: a process with constant pressure entirely

Isochoric process: a process with constant volume entirely

Process Types

Isobaric constant pressureIsothermal constant temperatureIsochoric constant volume

Q. What process is this?Ans. Isobaric

A Thermodynamic

system

Adiabatic no heat transfer (Q = 0)



Process formulation using thermodynamic properties

2P Final state

Process path

Initial state

V2 V1 V

System

(2) (1)

1



Thermodynamic cycle: Processes as a result of which a system returns to its original state, i.e., identical end states.

P P

2

2

1

4

VV

(a) Two-process cycle

(b) Four-process cycle

1

3





Analogy between Heat and Work

1. Heat and work are both transient phenomena: system never posses heat or work, but either or both cross the system boundary

2. Both heat and work are boundary phenomena: they are observed at the boundary of the system and represent energy crossing the boundary

3. Both heat and work are path functions and inexact differential: a path between initial state and final state should be specified in order to determine Q or W

4. Sign convention is the same for both heat and work: +ve: addition of Q or W-ve : transfer of Q or W



Kinetics EnergyIf a body m, acted upon a force F, is displaced a distance dl during a differential time dt. The work done is given by:

F m dlu

dW Fdl F mabut

dW madlduadt

but Where u is the velocity of the bodydlm dudt

dudW m dldt

or2

1

u

u

W m udu dW mudu integration2 22 1

2 2u um

2

2mu

2 22 1-

2 2mu muW ½ mu2 is a kinetic energy

or 212kE mu

• Work done on a body in accelerating it from initial velocity u1 to a final velocity u2 is equal to the change in kinetics energy of the body. Unit in SI system:

[Ek] = kg . m2/s2 or N.m = Joules



Potential EnergyIf a body m is raised from an initial elevation Z1 to a final

elevation Z2. The upward force exerted on the body is equal to its weight. Work required to raise the body is:

m

m

dZmg

W Fdl

2 1( )W mg Z Z

mgdZ

2 1-mgZ mgZ ( )mgZ

Work done on the body in raising it from Z1 to Z2 is equal to the change of the quantity mgZ

To lower the body, work should be done by the body which is also equal to the change in mgZ.



If the work done on a body in accelerating it or elevating it can be subsequently recovered, then the body by virtue of its velocity or elevation has the ability to do the work.

Work accelerating a body is said to produce a change in its kinetic energy, or

2

2KmuW E

Work done on a body in elevating it is said to produce a change in its potential energy, or

PW E mgZ

Unit of EP in SI system: [EP] = kg . m/s2 . m = N . m = Joule



A gas is confined in a 0.47-m-diameter cylinder by a piston, on which rests a weight. The mass of the piston and weight together is 150 kg. The local acceleration of gravity is 9.813 m s-2, and atmospheric pressure is 101.57 kPa.

(a) What is the force in newtons exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder?

(b) What is the pressure of the gas in kPa? (c) If the gas in the cylinder is heated, it expands, pushing the piston

and weight upward. If the piston and weight are raised 0.83 m, what is the work done by the gas in kJ?

What is the change in potential energy of the piston and weight?

Exercise (2):

Solution:





Conservation of Energy If a body is given energy when it is elevated, then the body conserves or retains

this energy until it performs the work of which it is capable. An elevated body allowed to fall freely gains in kinetic energy and loses potential energy.

Potential energy is converted into kinetic energy; or its capability for doing work remains unchanged, or:

Total W = EK + EP = 0

Or 2 22 1 2 1

1 1 02 2

mu mu mgZ mgZ

Other types of mechanical energies: Compression of spring by external force, energy is stored in the spring; later when the spring expands, it performs this work or releases this energy by a resisting force spring has capability for doing work.

chapter 1 - introduction themo

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