23 SYSTEMS OF UNITS

lb

m

^-ft/s2).

'R. H. PerryandD.W.Green,Eds., Perry'sChemicalEngineers'Handbook, 7thEdition,McGraw-Hill

New

York,1997.

2.3

Systems

ofUnits 11 Table23-1 SIand

CGS

Units

Base

Units

Quantity Unit

Symbol

Length meter(SI)

m

centimeter

(CGS) cm

Mass

kilogram(SI) kg

gram (CGS)

g

Moles gram-mole

mol

org-mole

Time

second s

Temperature ^kelvin

K

Electriccurrent

ampere A

Lightintensity candela cd

MultipleUnitPreferences tera(T)

=

₁₀¹² _{centi (c)}

=

1CT² giga(G)

=

₁₀⁹ _milli(m)

=

^10"3

mega (M)

^:

=

10⁶ micro(u.)

=

10^-6 kilo(k)

=

₁₀³ _nano(n)

=

1(T⁹

DerivedUnits

Quantity Unit

Symbol

Equivalentin

Terms

of

Base

Units

Volume

liter

L

0.001

m

³

1000

cm

³

Force newton(SI)

N

1kg-m/s²

dyne

(CGS)

1g-cm/s² Pressure pascal (SI) Pa 1

N/m

²

Energy,

work

joule (SI) J 1

N-m =

1

kg-m

²/s2

erg

(CGS)

1dyne-cm

=

1

g-cm

²/s2

gram-calorie cal 4.184J

=

4.184kg-

m

2/s2

Power

watt

W

^{1 J/s}

⁼

¹

kg-m

²/s3

TEST

^1«

^What

^{are the factors(numericalvalues}

^and

^units)

^needed

^toconvert

YOURSELF

^{(a) metersto}millimeters?

(Answers,

_p.

655)

^(b)

nanoseconds

toseconds?

(c) squarecentimeters tosquare meters?

(d) cubicfeet tocubicmeters(use theconversionfactortable

on

the inside front cover)?

(e)

horsepower

to Britishthermalunitsper second?

2.

What

is the derived SI unit for velocity?

The

velocity unit in the

CGS

^system? In the

American

engineering system?

Conversion Between Systems of

^Units

Convert23lb

m

^{- ft/min2toitsequivalentinkg-cm/s2.}

SOLUTION As

before, begin by ^writing the dimensional equation, fill in the units of conversion factors (new/old) and then the numerical values of these factors, and then

do

the arithmetic.

The

resultis

23lb

m

^{-ft 0.453593kg} 100

cm

l2

min

²

min

² lib™ 3.281ft (60)² s2

(Cancellationof units leaveskg-cm/s²

)

(23)(0.453593)(100) kg -cm (3.281 )(3600) s²

0.088 kg

cm

2.4

FORCE AND WEIGHT

According

to

Newton's

second law ofmotion, force is proportional to the product of

mass and

acceleration (length/time²). Natural force units are, therefore, kg-m/s² (SI),

g-cm/s

²

(CGS), and

lb

m

^-ft/s2

^(American

engineering).

To

avoid havingtocarry

around

these

complex

units inall calculations involvingforces, derived forceunitshave

been

definedin eachsystem.

In themetricsystems, the derived forceunits (the

newton

inSI, the

dyne

inthe

CGS

system) are

denned

toequalthe natural units:

1

newton

(N)

=

1 kg-m/s² _(2.4-1)

1

dyne =

1g-cm/s² _(2.4-2)

In the

American

engineering system, the derived force

unit—

called a pound-force (lb_f

)—

is

defined asthe product of aunit

mass

₍₁lb

m

⁾

^and

the acceleration of gravityat sea level

and

45° latitude,

which

is32.174ft/s2

:

llb

_f

=

32.174lb

m

^ft/s2 (2.4-3)

Equations

2.4-1through2.4-3defineconversionfactors

between

natural

and

derivedforce units.

For

example, the force in

newtons

requiredto accelerate a

mass

of 4.00

kg

ata rate of 9.00

m/s

² is

F =

4.00

kg

9.00

m

¹

N

s² 1 kg-m/s²

=

36.0

N

The

force inlb_frequiredtoacceleratea

mass

of 4.00lb

m

^atarateof9.00ft/s2 is

F ⁼

^4.00lb

m

^{9.00ft llb}f

s2

32.174lb

m

^-ft/s2

=

1.12lb_f

Factors

needed

toconvert

from one

force unit to anotherare

summarized

in thetable

on

the inside frontcover.

The _{symbol g}

_{c is}

sometimes

usedtodenote theconversionfactor

from

natural toderivedforce units:forexample,

=

^{1 kg-m/s2}

=

^32.174lb

^m

^-ft/s2

9c

IN

llb_f

We

will not use this symbol in the text, but if

you

should encounterit elsewhere

remember

thatitissimplyaconversion factor (notto be confusedwithgravitationalacceleration, which

isusually

denoted

by_g).

The

weightofanobjectisthe forceexerted

on

the object bygravitational attraction.Sup- pose that

an

object of

mass m

^is subjected to a gravitational force

W ^(W

^is

by

definition the weight of the object)

and

that if this object

were

falling freely its acceleration

would

be g.

The

weight, mass,and free-fallacceleration of the object are related

by Equation

2.4-4:

W ⁼ mg

_(2.4.4)

The

gravitational acceleration_(g)variesdirectlywith the

mass

of theattracting

body

(the earth, in

most problems you

willconfront)

and

inverselywiththesquareof the distance

between

the centers of

mass

of theattracting

body and

theobjectbeingattracted.

The

value ofg atsealevel

2.5 Numerical Calculation

and

Estimation 13

and

^45°latitudeisgiven

below

ineach systemofunits:

g

=

9.8066m/s²

=

980.66 cm/s²

=

32.174ft/s 2

(2.4-5)

The

acceleration of gravitydoesnot vary

much

withposition

on

theearth'ssurface

and

(within

moderate

limits) altitude,

and

the valuesin

Equation

2.4-5

may

accordingly

be

used for

most

conversions

between mass and

weight.

TEST

^1.

What

isa force of2 kg-m/s²equivalent toin

newtons? What

isa force of2lb

m

^{•ft/s2equiv-}

YOURSELF

alenttoin lb_{?

(Answers,

_p.

655)

2. If the acceleration of gravity at a point is fir

=

9.8 m/s²

and an

object is resting

on

the

ground

atthispoint,isthisobject acceleratingatarateof9.8m/s²?

3.

Suppose

anobjectweighs9.8

N

atsealevel.

What

isits

mass? Would

its

mass be

greater, less,or the

same on

the

moon? How

aboutitsweight?

4.

Suppose

an objectweighs2lb_fat sealevel.

What

isits

mass? Would

its

mass

begreater, less,or the

same

atthe center of the earth?

How

aboutitsweight?(Careful!)

EXAMPLE 2:4-n Weight and Mass

Water

hasa density of 62.4lbm^/ft3.

How much

does2.000ft3

of waterweigh _{(1) at}sealeveland^45°

latitudeand_{(2) in}Denver,Colorado,wherethealtitudeis5374ftandthegravitational acceleration

is32.139ft/s2

?

M ⁼

(62.4

^1

^(2ft3)

⁼

^{124.8lb n}

SOLUTION ^The

^{massofthe wateris}

M =

\fOA

ft3

The

weight of the wateris

'ftV

llbf

W ⁼

(124.8lb

m

)<7

32.174lbm^-ft/s2

1.

At

sea^level,g

=

32.174ft/s2

,sothat

W ⁼

124.8 lbf^.

2. InDenver, g

=

32.139ft/s2

,and

W ⁼

^124.7lbf^.

As

^thisexampleillustrates,theerrorincurredby assumingthatg

=

32.174ft/s² isnormallyquite small aslong asyou remain

on

theearth'ssurface.In asatelliteor

on

another planetit

would

bea differentstory.

2.5

NUMERICAL CALCULATION AND ESTIMATION

2.5a

Scientific

Notation,

Significant

Figures, and Precision

Both

verylarge

and

verysmall

numbers

are

commonly

encounteredinprocesscalculations.

A

convenient

way

^{to represent} such

numbers

is to use scientificnotation, in

which

a

number

is

expressed astheproduct ofanother

number

^(usually

between

0.1

and

10)

and

a

power

of 10.

Examples:

123,000,000

=

_1.23

X

10⁸ (or0.123

X

10⁹₎ 0.000028

=

2.8

x

^10~5 (or0.28

x

10^-4₎

The

significant figuresof a

number

are thedigits

from

thefirst

nonzero

digit

on

theleftto either(a) thelast digit(zero ornonzero)

on

theright ifthereisadecimalpoint, or(b) thelast

nonzero

digitof the

number

ifthereis

no

decimalpoint.For example,

2300

or2.3

X

10³has

two

significantfigures.

2300. or2.300

x

10³has foursignificantfigures.

2300.0 or2.3000

X

10³ hasfivesignificantfigures.

23,040 or2.304

X

10⁴has foursignificantfigures.

0.035 or3.5

X

^10~2 has

two

significantfigures.

0.03500or 3.500

x

10^-2 has foursignificantfigures.

(Note:

The number

ofsignificant figuresiseasily

shown and

seenifscientificnotationisused.)

The number

ofsignificant figures inthereportedvalue of a

measured

or calculated quantity provides

an

indication of the precisionwith

which

the quantity is

known:

the

more

significant figures,the

more

preciseisthevalue.Generally,if

you

report the value of a

measured

quantity withthree significantfigures,

you

indicate thatthevalue of the third ofthese figures

may be

off

by

as

much

asahalf-unit. Thus,if

you

report a

mass

as 8.3 g (twosignificant figures),

you

indicate that the

mass

lies

somewhere between

8.25

and

8.35 _g, whereas if

you

give the value as 8.300g^(foursignificant figures)

you

indicate that the

mass

lies

between

8.2995

and

8.3005_g.

Note, however, thatthis rule applies onlyto

measured

quantitiesor

numbers

calculated

from measured

quantities.Ifa quantityis

known

precisely—likeapureinteger₍₂₎ora

counted

ratherthan

measured

quantity (16

oranges)—

its value implicitlycontains

an

infinite

number

ofsignificantfigures (5

cows

really

means

5.0000 . .. cows).

When two

or

more

quantitiesare

combined

bymultiplicationand/ordivision, the

number of

significant figures in the resultshould equalthe lowest

number

_ofsignificantfigures

of any of

the multiplicands ordivisors. Iftheinitialresult of a calculationviolatesthis rule,

you must round

offtheresult to reducethe

number

ofsignificant figures toits

maximum

allowedvalue, although ifseveralcalculations areto be

performed

in sequenceitis advisable to

keep

extra significant figuresof intermediate quantities

and

to

round

offonly thefinal result.

Examples:

(3) (4) (7) (3)

(3.57)(4.286)

=

15.30102 15.3

(2) (4) (3) (9) ₍₂₎ (2)

(5.2

X

^10"4_)(0.1635

_x

₁₀⁷_)/(2.67)

=

_318.426966

=>

3.2

x

10²

=

₃₂₀

(The

raised quantitiesinparenthesesdenotethe

number

ofsignificant figures inthegiven

num-

bers.)Warning:If

you

calculate,_say,

3x4, _and

_yourcalculator or

computer

gives

you

an

answer

like 11.99999,

and you copy

thisanswer

and hand

itin,

your

instructor

may become

violent!

The

rule for addition

and

subtraction concernsthe position of the lastsignificant figure in the

sum—

thatis, the location ofthis figure relativeto the decimalpoint.

The

ruleis:

When two

or

more numbers

are

added

orsubtracted, the positions

of

^thelastsignificantfigures

of

^each

number

relative to thedecimalpointshould be compared.

Of

these positions, the

one

farthest to theleftisthe position

of

^thelastpermissiblesignificantfigure

of

^the

sum

ordifference.

Severalexamplesofthisrule follow, in

which an arrow

_(J)denotesthelastsignificant figure of

each number.

4

1530 _i

-2.56

1527.44 1530

T

4

114

1.0000

+

0.036

+

0.22

=

1.2560

=>

1.26

4 4

2.75

x

10⁶

+

3.400

X

10⁴

=

(2.75

+

0.03400)

x

10⁶ 4

=

_2.784000

X

10⁶

=>

2.78

X

10⁶

2.5

Numerical

Calculation

and

Estimation 15 Finally,aruleof

thumb

forroundingoff

numbers

inwhichthedigitto

be dropped

isa 5 is

alwaysto

make

thelast digitof therounded-off

number

even:

1.35

=^>

1.4 1.25

=>

1.2

Express the following quantitiesin scientific notation

and

indicate

how many

significant figureseachhas.

(a) 12,200 (b) 12,200.0 (c) 0.003040

Expressthefollowing quantities instandarddecimal

form and

indicate

how many

signif- icant figureseachhas.

(a) 1.34

X

10⁵ (b) 1.340

x

^10~2 (c) 0.00420

X

10⁶

How many

significantfigures

would

thesolution ofeachof the following

problems have?

What

are the solutionsof(c)

and

(d)?

(a) (5.74)(38.27)/(0.001250) ^(c) 1.000

+

10.2 (b) (1.76

x

10⁴)(0.12

x

^10~6₎ (d) 18.76

-

7

Round

offeachofthefollowing

numbers

to threesignificantfigures, (a) 1465 (b) 13.35 (c) 1.765

x

^10~7

When

the value of a

number

isgiven, thesignificant figuresprovide an indication of the uncertaintyinthe value; forexample,avalueof2.7 indicates thatthe

number

lies

between

2.65

and

2.75.

Give

rangeswithin

which

eachof the following valueshe.

(a) 4.3 (d) 2500

(b) 4.30 (e) 2.500

x

10³ (c) 2.778

X

^10~3

2.5b Validating Results

Every problem you

willever

have

to solve

—

^inthis

^and

other courses

and

in

your

professional career

—

^willinvolve

^two

^criticalquestions:(1)

How do

Iget a solution?(2)

When

Igetone,

how do

I

know

^it'sright?

Most

ofthis

book

is

devoted

toQuestion¹

—

^thatis,to

^methods

^{of solving}

problems

that arise in the design

and

analysis ofchemicalprocesses.

However, Question

2 is

equallyimportant,

and

serious

problems

canarise

when

itisnotasked. All successfulengineers get into the habitofaskingit

whenever

they solve a

problem and

theydevelopa

wide

variety ofstrategiesforansweringit.

Among

approaches

you

can use to validate a quantitative

problem

solution are back- substitution, order-of-magnitudeestimation,

and

thetestofreasonableness.

• Back-substitutionisstraightforward:after

you

solveasetof equations, substitute

your

solu- tion

back

into theequations

and make

sureitworks.

• Order-of-magnitude estimation

means coming up

with acrude

and

easy-to-obtain approx- imation of the answer to a

problem and making

sure that the

more

exact solution

comes

reasonablyclosetoit.

•

Applying

the test ofreasonableness

means

verifying thatthe solution

makes

sense. It for example, a calculated velocity ofwaterflowing in apipe is fasterthan the

speed

oflightor the calculated temperaturein achemicalreactorishigherthanthe interiortemperature of the sun,

you

shouldsuspect that amistake has

been made somewhere.

The

procedureforchecking

an

arithmetic calculation

by

order-of-magnitudeestimation is

as follows:

1. Substitute simple integers forall numericalquantities,using

powers

of10 (scientificno- tation) forvery small

and

verylargenumbers.

27.36

—

* 20or30 (whichever

makes

thesubsequentarithmeticeasier) 63,472

—

6

X

10⁴

0.002887

->3X

^10~3

TEST