NOTE (Chapter 1)

8/13/2019 NOTE (Chapter 1)

1/33

1

MEC 25

THERMODYNAMICS


2/33

2

PREFACE TO MEC251 THERMODYNAMICS

Thermodynamics is the basic science of energy transformation thatinvolves heat, work and the properties of substance. It is an essentialcomponent of engineering education at higher learning institutions.

Through the subjects, students will be exposed to a strongunderstanding of the fundamentals of thermodynamics , starting with theelementary ideas on temperature, heat, energy, work and theassociated changing of properties of substance , developing thefundamental laws of thermodynamics , and lastly, its broad applicationsin our everyday living .

These handouts are in the form of partial notes . Much of the verbaldelivery and background of a lecture are already in the handouts.Spaces have been provided for the students to fill them up withappropriate diagrams, equations, important concepts and workexamples.


3/33


4/33

4

The Essence of ThermodynamicsWhen you call a thing is mysterious, all that it means is that you dontunderstand it. Lord Kelvin

Thermodynamics is a funny subject. The first time you go through it,you dont understand it at all. The second time you go through it, youthink you understand it, except for one or two small points. The third

time you go through it, you know you dont understand it, but by thattime you are so used to it, it doesnt bother you anymore. Arnold Sommerfield

What everybody must knows about learning ThermodynamicsThings get worse under pressureMurphy's Law about Thermodynamics


5/33

5

Thermodynamics Derived from greek word Therme (heat) and dunamis (power/force). A basic science that deals with energy

Its forms Its interaction with the matter how energy can be transformed into

heat work

It involves with the most fundamental laws of nature ie. the

conservation of energy principle, or energy cannot be created nordestroyed.

Scope of Thermodynamics

Classical thermodynamics : A macroscopic approach to the study ofthermodynamics that does not require the knowledge of thebehaviour of individual particles. Also known as Applied /Engineering Thermodynamics.

Statistical thermodynamics : A microscopic approach, study on theaverage behaviour of large groups of individual particles.


6/33

6

Applications of ThermodynamicsCovers every items around us, starting from household appliances tohigh-tech rocket sciences.


7/33

7

Basic DefinitionsSystem or the thermodynamic system : a

quantity of matter or a region in space chosenfor study.

The surroundings : The region outside the

system, ie. the physical space outside thesystem boundary

The boundary : The real or imaginary surface

that separates the system from itssurroundings.

Fixed Boundaries

Fixed & MoveableBoundaries


8/33

8

Classifications of System A closed system : contains a fixamount of mass, (ie. no mass willcross the system boundary). the system boundary may move

during an activity or process.

Energy in the form of heat andwork may cross the boundary of aclosed system.


9/33

9

An open system , or ( control volume) massas well as energy may cross the boundary,(called the control surface)

Examples of open systems: pumps

compressors turbines valves heat exchangers etc

An isolated system is a general system of fixed mass where no heator work may cross the boundaries

It is a collection of a main system and its surroundings that areexchanging mass and energy among themselves and no othersystem.


10/33

10

Forms of Energy A system has many forms of energy The total energy is the sum of all forms of energy

within the system such as

Thermal Mechanical Electric Magnetic Chemical Nuclear

InternalEnergy, U

Kinetic (KE) Potential (PE)

Then, the total energy of the system, E, is the sum of : The internal energy, U ( due to the molecular structure and the

molecular activity of the system ). The kinetic energy, KE, ( due to the system's motion at a velocity ).

The potential energy, PE, ( due to (systems elevation ).

kJ mC KE 21 2=

kJ mgz PE =


11/33

11

The total energy is then E U KE PE kJ = + + ( )

on a unit mass basis, called the specific energy, e

e E

mU m

KE m

PE m

kJ kg

= = + + ( )

gz C ue ++=2

2

If there is change in stored energy of a system

E U KE PE kJ = + + ( )Closed systems are mostly stationary during a process and, thus,

there is no change in their kinetic and potential energies.Therefore

E U kJ = ( )


12/33


13/33

13

Extensive properties are those that depend directly with extent or sizeof the system.Examples of Extensive Propertiesa. mass b. volumec. total energy d. mass dependent property

Extensive properties per unit mass are intensive properties.

For example the specific volume defined asv

Volumemass

V m

= =

kg m3


14/33

14

State Condition of a system as described by its properties.

A change in any one of the properties , will cause a change of thestate.

Process any action that change from one state of asystem to another. (so are the properties)

When any of the properties of a system

change, the state also changes. common processes are :

Expansion CompressionHeating Cooling


15/33


16/33

16

Reversibility

When a system changes state where at any instant during the process

the state point can be located on a property diagram, then the processis said to be reversible.

P

v

State Postulate

As noted earlier, the state of a system is described by its properties.The number of properties required to fix the state of a simple,homogeneous system is given by the state postulate:

The thermodynamic state of a simple compressible system canbe completely specified by two independent, intensive properties.


17/33


18/33

18

Heat Transfer Heat : the energy in crossing the boundary due to the temperature

difference ( between the system and its surroundings ) Net heat transferred to a system is defined as

Or q = Q/m [ kJ/kg ]

An adiabatic process is one in which the system is perfectlyinsulated and the heat transfer is zero.

Q Q Qnet in out =

Work

the energy spent by a force acting through a distance. Unit isNm/Joule

Thermodynamic work is the energy in transition across the boundaryand is done by a system. Also known as Boundary work

==2

1

2

1

12 Fd W W


19/33


20/33

20

P is the absolute pressure and isalways positive.

When dV is positive, W b is positive.(this is ..process)

When dV is negative, W b is negative(this is ..process)

700 kPa

100 kPa

0.01 m 3 0.03 m 3

the quantity of (boundary)work is process dependent .


21/33

21

1st Law of Closed System the conservation of energy principle or 1st Law of Thermodynamics

= systemthe

hinenergy wit

in totalchangeThe

by work boundary

systemthetheacross

outnsferred energy tranet

fer heat trans by boundary

systemtheacross

innsferred energy tranet

E E E in out system =

the net heat transfer to the system is Q and the net work done bythe system is W . Thus,

Q W E net net system =

but E U K E P E = + +=0 =0


22/33

22

Thus the 1 st Law of Thermodynamics for Closed System is ,

Q W U net net = Example : A system receives 5 kJ of heat transfer and experiences adecrease in internal energy in the amount of 5 kJ. Determine theamount of work done by the system.

Applying the 1 st LawE= -5kJQ in =5kJ

Wout =?

SystemBoundary

5

5

in out system

in in

out out

system

E E E

E Q kJ

E W

E kJ

= = ==

= ( )5 5

10

out in system

out

out

E E E

W kJ W kJ

=

= =


23/33

23

Determination of Internal Energy (of a system)To apply the 1 st law to a system, ways to calculate the change in internal

energy of the system enclosed by the boundary must be determined. For real substances (water, Refrigerants) , the property tables are

used to find the change of internal energy. For ideal gases, the specific heats are used to find the internal energy

Specific Heats Relation With Internal Energy and Enthalpy

Define:the specific heat at constant volume , CV is the quatity of heat needed toraise the temperature of a unit of mass of a substance by one degreeduring a constant-volume process.

the specific heat at constant pressure , CP , is the quatity of heat neededto raise the temperature of a unit of mass of a substance by one degreeduring a constant-pressure process.


24/33

24

In thermodynamics, the specific heats are defined as

p p

vv T

hcT uc = =

and

in terms of ordinary differentials,

dT cdh

dT cdu

p

v

=

=

We introduce a new property, enthalpy, h , as the sum of the

internal energy, u , and the pressure-volume product, pv .

h = u + pv


25/33

25

Relation between C P and C V for ideal gaseswriting the differential of enthalpy, the relationship between the specificheats for ideal gases is

h u Pv

dh du d RT

C dT C dT RdT

C C R P V

P V

= += +

= += +

( )

or RC C v p = R is the Gas Constant

The specific heat ratio, , is defined as

v

p

c

c=


26/33

26

Equations of StateExpress the relationship among the state properties; temperature,pressure, and specific volume for an ideal gas, where

RT pv =R is the gas constant or the constant of proportionality. For allsubstances, the gas constant is related to the universal gas constantthrough the molar mass (or molecular weight).

Ro,u is the universal gas constant. Then,

M R R o=

The mass, m, is related to the moles, N, ofsubstance through the molecular weight ormolar mass, M.

m N =


27/33

27

The ideal gas equation of state then may be written several ways.

RT pv

RT mV

p

mRT V

=

== p = absolute pressure in kPa, MPa, or bar

v = specific volume in m 3/kgT = absolute temperature in KR = gas constant (kJ/kgK)

Ro= 8.314 kJ/(kmol K)


28/33

28

Some Typical Processes1. Constant volume

If the volume is held constant, dV = 0 and the cylinder is heated,

Locked

==2

1

0 pdV W b

U Q

U Q

U W Q

===

0

then,from 1 st

Law

For a vapour, use tables tofind the UFor a gas, change in internalenergy is

( )( )1212

T T mcU Q

T T mcU

v

v

===

p

vv1=v2

p2

p1

The boundary work becomes


29/33

29

2. Constant pressurep

vv2v1

For vapour, Q W = U becomes Q = H since it involves change ininternal energy and boundary work together

For a gas, boundary work equation becomes

===2

1

2

1

12 )( V V pdV p pdV W b

p1=p 2

And the heat supplied, ( )12 T T mc H Q p ==


30/33

30

3. Constant temperature

p

vv1v2

p2

p1

cooling

for vapour use table to find the change in internal energyU W Q =

for a gas, u=f(T), therefore U=0. Then

W Q =then the equation of state provides the pressure-volume relation

P m R T V

=


31/33

31

the boundary work is

2

11

1

211

1

21

2

1

2

1

ln

lnln

p p

mRT

V V

V pV V

mRT

dV V

mRT pdV W b

=

==

==

4. The Polytropic Processp

vv1v2

p2

p1

the pressure-volume relation is given as

pVn = constantpVn = constant

n may have any value from - < n < + depending on the process.


32/33

32

Process Exponent n

Constant pressure 0

Constant volume

Isothermal & ideal gas 1

Adiabatic & ideal gas = C P /C V

pVn

= constant

The boundary work during the polytropic process is

( )

1 1

1

212211

2

1

2

1

=

=

== nwhere

n

T T mR

n

V pV p

dV V C

pdV W nb


33/33

Documents

NOTE (Chapter 1)