thermo unit 1

8/11/2019 thermo unit 1

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BASIC THERMODYNAMICS J2006/1/1

BASIC THERMODYNAMICS

OBJECTIVES

General Objec!"e : To understand the concept of units and dimensions

S#ec!$!c Objec!"e% : At the end of the unit you will be able to:

state the difference between fundamentals and derived

quantities

describe the physical quantities of thermodynamics

understand the conversion units of thermodynamics

calculate the examples of conversion factors

UNIT 1


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1&0 INTROD'CTION

id you realize that the work of an engineer is limited unless he has a source

of power to drive his machines or tools? However before such a study can

begin it is necessary to be sure of the number of definitions and units

which are essential for a proper understanding of the sub!ect" #e are familiar withmost of these items in our everyday lives but science demands that we have to be

exact in our understanding if real progress is to be made"$#hen engineering calculations are performed it is necessary to be concerned with

the unitsof the physical quantities involved" A unit is any specified amount of a

quantity by comparison with which any other quantity of the same kind is measured"%or example meters centimeters and millimeters are all units of length" &econds

minutes and hours are alternative time units"

1&1 ()n*a+enal an* Der!"e* ,)an!!e%

10 -!l.+eer (ee

2 Yar* 100 Ince%

3 Meer

C.)l* 4.) 5!"e +e an

an%er3

INPUTINPUT


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'n the present discussion we consider the system of units called &' ('nternational

&ystem of )nits* and it is a legally accepted system in many countries" &' units will

be used throughout this module"

+ength mass time electric current thermodynamic temperature and luminous

intensity are the six fundamental physical quantities" These six quantities are

absolutely independent of one another" They are also called the ,'ndefinables- of

mechanics" The &' base units are listed in Table ."./."

Table 1&181%undamental units

,)an!4 'n! S4+b.l

0ass kilogram kg

Time second s+ength meter m

Thermodynamic temperature degree 1elvin 1

2lectric current ampere A

+uminous intensity candela cd

All other physical quantities which can be expressed in terms of one or more of

these are known as ,derived quantities-" The unit of length mass time electric

current thermodynamic temperature and luminous intensity are known as

,fundamental units" 3hysical quantities like area volume density velocity

acceleration force energy power torque etc" are called derived quantitiessince theydepend on one or more of these fundamental quantities" The units of the derived

quantities are called derived units as shown in Table ."./4"

Table 1&182 $erived units

,)an!4 'n! S4+b.l N.e%

Area meter square m4

5olume meter cube m6 . m67 . x .86litre

5elocity meter per second m9s

Acceleration 0eter per second

squared

m9s4

$ensity kilogram 9 meter cube kg9m6

%orce ewton . 7 . kgm9s4

3ressure ewton9meter square 9m4 . 9m47 . 3ascal

. bar 7 .8;9m47 .84k9m4

1&1&1 (.rce


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ewton-s second law may be written as force (mass x acceleration* for a

body of a constant mass"

i"e" F = kma (.".*

(where m is the mass of a body accelerated with an acceleration a by a force

% k is constant*

'n a coherent system of units such as &' k 7 . hence:

F = ma (."4*

The &' unit of force is therefore kgm9s4" This composite unit is called the

ewton "

i"e" . 7 . kg"m9s4

1&1&2 Ener54

Heat and work are both forms of energy" The work done by a force is theproduct of the force and the distance moved in the same direction"

The &' unit of work 7 force x distance in the ewton meter m"A general unit for energy is introduced by giving the ewton meter the name


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BASIC THERMODYNAMICS J2006/1/

3ressure is the force exerted by a fluid per unit area" #e speak of pressure

only when we deal with gas or liquid" The pressure on a surface due to

forces from another surface or from a fluid is the force acting at =8 oto the

unit area of the surface"

i"e" pressure 7 force9 area

P = F/A (."6*

The unit of pressure is 9m4and this unit is sometimes called the 3ascal 3a"

%or most cases occurring in thermodynamics the pressure expressed in 3ascal

will be a very small number" This new unit is defined as follows:

. bar 7 .8;

9m4

7 .8;

3a

1&1&& Den%!4

$ensity is the mass of a substance per unit volume"

The unit of density is kg9m6"

(.rce; ( +a

:re%%)re; : (/A

?

Den%!4; +/V

(.">*

volume

mass$ensity

V

m=

=


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alculate the pressure of gas underneath the piston in equilibrium for a ;8 kg

mass that reacts to a piston with a surface area of .88 cm4

"

A density of 7 @;8 kg9m6of oil is filled to a tank" $etermine the amount of

mass min the tank if the volume of the tank is V7 4 m6"


E>a+#le 1&1

S.l)!.n . E>a+#le 1&1

4:9m8;">=

8"8.

="@.x;8

area

force(3*3ressure

=

=

=

E>a+#le 1&2

S.l)!.n . E>a+#le 1&2

#e should end up with the unit of kilograms" 3utting the given information into

perspective we have

7 @;8 kg9m6and V7 4 m6

't is obvious that we can eliminate m6and end up with kg by multiplying these

two quantities" Therefore the formula we are looking for is

V

m=

Thus m = V

7(@;8 kg9m6*(4 m6*

7 .88 kg

(.rce +a%% > accelera!.n

:re%%)re $.rce/area


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BASIC THERMODYNAMICS J2006/1/@

TEST YO'R 'NDERSTANDING B2%CD2 EC) CT')2 #'TH TH2

2FT '3)TG

.". #hat is the work done by an expanding gas if the force resisting the motion

of the piston is 88 and the length of the stroke is 8"; m ?

."4 #hat is the force required to accelerate a mass of 68 kg at a rate of .; m9s4 ?

."6 The fuel tank of a large truck measures ."4m x 8"=m x 8"Im" How many litres

of fuel are contained in the tank when it is full?

."> A weather research instrument is suspended below a helium filled balloon

which is a 6"@m diameter sphere" 'f the specific volume of helium is

;"Im69kg what is the weight of helium in the balloon? 2xplain briefly why

the balloon rises in the atmosphere"

Ac!"!4 1A


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(ee*bac= . Ac!"!4 1A

.". #ork 7 %orce x $istance

7 (88 *(8"; m*

7 6;8 m or @ x .888

7 I>@ litres

."> Dadius of volume r 74

d

74

6"67 ."= m

5olume of balloon V76

6

>r

76*=".(

6

>

7 4@"6 m6

0ass of helium in balloon m7v

V

7 4@"69;"I

7 ;".6 kg

w = mg


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7 ;".6 x ="@.

7 ;8"6

$ensity of helium =v

.

7I";

.

8".@I kg9m6

The balloon rises in the atmosphere because the density of helium is less than

the density of atmosphere"


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1&2 'n! C.n"er%!.n%

#e all know from experience that conversion of units can give terrible headaches

if they are not used carefully in solving a problem" But with some attention and

skill conversion of units can be used to our advantage"

0easurements that describe physical quantities may be expressed in a variety of

different units" As a result one often has to convert a quantity from one unit to

another" %or example we would like to convert say >= days into weeks" Cne

approach is to multiply the value by ratios of the equivalent units" The ratios are

formed such that the old units are cancelled leaving the new units"

Te D!+en%!.nal H.+.5ene!4

$espite their causing us errors units9dimensions can be our friends"

All terms in an equation must be dimensionally homogeneous"

That is we can-t add apples to

orangesG

either can we add


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0ultiple and sub/multiple of the basic units are formed by means of prefixes and the

ones most commonly used are shown in the following table:

Table 1&2 0ultiplying factors

M)l!#l4!n5 (ac.r :re$!> S4+b.l

. 888 888 888 888 .8.4 tera T

. 888 888 888 .8= giga K

. 888 888 .8I mega 0

. 888 .86 kilo k

.88 .84 hector h

.8 .8. deca da

8". .8/. desi d

8"8. .8/4 centi c

8"88. .8/6 milli m

8"888 88. .8/I micro

8"888 888 88. .8/= nano n

8"888 888 888 88. .8/.4 pico p

E>a+#le 1&7

onvert . km9h to m9s"

S.l)!.n . E>a+#le 1&7

m9s4A@"8

s6I88

m.888

s6I88

!.x

km.

m.888x

!

km.

!

km.

=

=

=


12/16


E>a+#le 1&9

onvert 4; g9mm6to kg9m6"

S.l)!.n . E>a+#le 1&9

. kg 7 .888 g

. m 7 .888 mm

. m67 .888 x .888 x .888 mm6

7 .8=m6

6I

6

=

6

6=

66

kg9m.8x4;

m.888

kg.x.8x4;

g.888

kg.x

m.

mm.8x

mm

g4;

mm

g4;

=

=

=

How could I convertg/mm3to kg/m3?


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TEST YO'R 'NDERSTANDING B2%CD2 EC) CT')2 #'TH TH2

2FT '3)TG

."; onvert the following data:

a* 6 9cm4 to k9m4

b* .; 09m4to 9m4

."I onvert .; milligram per litre to kg9m6"

I hope youve learnt somethingfrom this unit. Lets move on tothe next topic.

Ac!"!4 1B


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(ee*bac= T. Ac!"!4 1B

."; a* . k 7 .888

. m47 .88 x .88 7 .8>cm4

b* . 0 7 .8I9m4

."I . kg 7 . 888 888 mg

. m67 .888 litre

4

4

>

4

4>

44

k:9m68

m.888

k:.8x6

:.888

k:.x

m.

cm.8x

cm

:6

cm

:6

=

=

=

4I

I

44

:9m.8x.;

0:.

:.8x

m

0:.;

m

0:.;

=

=

66/

6

kg9m.8x.;

m.

litre.888 x

mg888888.

kg.x

litre

mg.;

litre

mg.;

=

=


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Eou are approaching success" Tr4 all e )e%!.n%in this self/assessment section

and check your answers with those given in the %eedback to &elf/Assessment on the

next page" 'f you face any problem discuss it with your lecturer" Kood luck"

." A gas is contained in a vertical frictionless piston/cylinder device" The piston

has a mass of > kg and a cross/sectional area of 6; cm4" A compressed spring

above the piston exerts a force of I8 onto the piston" 'f the atmospheric

pressure is =; k3a determine the pressure inside the cylinder"

4" A force of @ is applied continuously at an angle of 68oto a certain mass"

%ind the work done when the mass moves through a distance of I m"

6" A man weighing I8 kg goes up a staircase of ; m in height in 48 secs"

alculate his rate of doing work and power in watts"

>" The density of water at room temperature and atmospheric pressure is

."8 g9cm6" onvert this to kg9m6" %ind also the specific volume of water"

SE?(8ASSESSMENT


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Have you tried the questions????? 'f LYESM check your answers now"

." .46"> k3a

4" >."; " .888 kg9m6N 8"88. m69kg

CONGRATULATIONS!!!!

..May success be with

you always.

(ee*bac= . Sel$8A%%e%%+en

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thermo unit 1