Introduction engineering principle and units

8/11/2019 Introduction engineering principle and units

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© PCS-FKKKSA Slide 1 - 1Material Balances

Introduction to EngineeringPrinciples and Units

Kamarul ‘Asri Ibrahim, PhD




Course Instructor

!

Professor Dr. Kamarul ‘Asri Ibrahim! Academic Qualifications

" B.Sc. Chem. Eng. (Ohio)" M.Sc. Chem. Eng. (Colorado State)" Ph.D. Chem. Eng. - Multivariate Active SPC (Newcastle upon Tyne)

!

Specialization" Process Modeling, Control, and Optimization"

Application of Multivariate SPC in Plant Operation

! Contact" email : [email protected]" Tel: 07-5535573 Fax:07-5536165






Learning Outcomes of the SKKK 1113 Course

By the end of this course, students should have the:CO1 Ability to carry-out unit conversion and equipment

identification for various chemical processes, processvariables and unit operations in chemical industry

CO2 Ability to calculate mass balance on single or multiple

process units with or without recycle, purge, orbypass streams for non-reactive processes.

CO3 Ability to solve mass balance calculations on single ormultiple process units with or without recycle, purge,or bypass streams for reactive processes

CO4 Ability to perform and analyze mass balancecalculations of single phase system




Learning Outcomes of the SKKK 1113 Course

By the end of this course, students should have the:CO5 Ability to demonstrate mass balance calculations of

multiple component gas-liquid systems at equilibrium

CO6 Ability to solve the problems using first law ofthermodynamics and conservative equations by

identifying forms of energy, reference stateconditions and state properties

CO7 Ability to conduct problem solving solution of mass andenergy problem using principle of conservation.

CO8 Ability to commit in the preparation of learningexperience process using E-Portfolio




Units and Dimensions

!

Dimension - property that can be measured such as" length" time" mass" temperature"

multiplying or dividing other dimensions

!

Unit - measured and counted quantity has" value (2.35)"

unit (2.35 gram)

! It is essential to write the value and unit in equation2 meters, 0.3 second, 4.5 kilograms, 5 gold rings




Properties of Units

!

Units can be treated like algebraic variables" added and subtracted when having same units

3 cm - 1 cm = 2 cm (3x - x = 2x)but

3 cm - 1 mm (or 1 s) = ? (3x - y = ?)

"

can always be combined by multiplication or divisions3 N x 4 m= 12 N.m

( 3 is a dimensionless quantity)




Conversion of Units

!

When having appropriate dimensions, measured quantity can beexpressed in term of other units

ft/s miles/hr cm/yr

! The equivalence between two expressions is given by ratio knownas conversion factor

( 1 centimeter per 10 millimeters)

(10 millimeters per 1 centimeter )




Conversion Factors

!

A given quantity is expressed into new unit by usinga conversion factor (new unit/old unit)

!

Alternative way to write this equation

36 mg 1g = 0.036 g

1000 mg




!

Set up a dimensional equation" write the given quantity and its unit on the left" write the units of conversion factors (new unit/old unit)" fill in the values of the conversion factors" carry out the indicated arithmetic operations to find the desired

values

!

Example : Convert acceleration of 1 cm/s2 to km/y2

(km/y2) = 1 cm 36002 s2 24h2 3652 day2 1 m 1 kms2 h2 day2 y2 102cm 103 m

= 9.95 x 109 km/y2

Method for using Conversion Factor




Systems of Units

!

Base Units - units for dimensions of mass, length, time,temperature, electrical current, and light intensity

" kilogram, meter, second, kelvin, ampere, candela

!

Multiple units - multiples or fraction of base unit

"

minutes, hours, milliseconds or all in term of base unit second!

Derived units - obtained in one of two ways

" Multiplying and dividing base units (cm2, ft/min, kg.m/s2) whichare known as compound units

" Defined as equivalents of compound units

( 1 erg = 1 g.cm/s2, 1 lbf = 32.174 lbm.ft/s2)




Systems of Units

!

System Internationale d’Unites

Length meter (SI)centimeter (CGS)

Mass kilogram (SI)

gram (CGS) Moles gram-mole (mol)Time second (s) Temperature kelvin (K)Electric current ampere (A)Light intensity candela (cd)

!

American Engineering System

Length foot (ft)

Mass pound mass (lbm)

Moles lbm-mole (lbmmol)

Time second (s)

Temperature Rankin (oR)

Electric current ampere (A)

Light intensity candela (cd)




Force

!

Force is proportional to the product of mass and acceleration(length/time 2 )

1 newton (N) = 1 kg.m/s2

1 dyne = 1 g.cm/s2

1 lbf = 32.174 lbm.ft/s2

!

To convert force from defined unit to a natural units , conversionfactor known as g c is required:




Weight

!

Weight of an object is due to the gravitational force

W = mg/ g c

! The value of gravitational force ( g ) is varies to location of earthsurface

! The value of the corresponding g/ g c at 45

o latitude

g = 9.8066 m/s2 g/ g c = 9.8066 N/kg

g = 980.66 cm/s2 g/ g c = 980.66 dyne/g

g = 32.174 ft/s2 g/ g

c

= 1 lbf/ lb

m




Dimensional Homogeneity

!

Valid equation must be dimensionally homogenous! Both sides of equation must have same dimensions

V (m/s) = V o (m/s) + g (m/s2)t (s)

Example : Consider the the equation D (ft) = 3t (s) + 4What is the dimension and unit for constants 3 and 4?

Constant 3 length/time

Constant 4 length

ft/s

ft




Dimensionless Quantity

!

Dimensionless Quantity can be pure numbers (2, ! , 1.3 etc) ora multiplicative combination of variables with no net dimensions

Example:-

let M (g) = D(cm) u(cm/s) (g/cm3) and Mo = (g/cm.s)

and M/Mo = Du / = (cm)(cm/s)(g/cm3)/ (g/cm.s)

Thus, M/Mo or Du / is also called a dimensionless group

ftWhat multiplicative combination of (m), (m/s2) and t(s) wouldconstitute a dimensionless group?




Scientific Notation

!

Large and very small number often encountered inprocess calculations

!

Convenient way to represent such number is to usescientific notation

!

Expressed between 0.1 and 10 and a power of 10

123 000 000 = 1.23 x 108 (or 0.123 x 109)

0.000028 = 2.8 x 10-5 (or 0.28 x 10-4)




Significant Figure

!

The significant of a number are the digits from the firstnonzero on the left to either"

the last digit (zero or nonzero) on the right if there is decimalpoint

" the last nonzero digit of the number if there is no decimal point

2300 or 2.3 x 103 has two significant figures

2300. or 2.300 x 103 has four significant figures

2300.0 or 2.3000 x 103 has five significant figures

23040 or 2.304 x 104 has four significant figures

0.035 or 3.5 x 10-2 has two significant figures

0.03500 or 3.500 x 10-2 has four significant figures




Rule of Thumb on Significant Figure

!

Multiplication and division"

The answer should be equal to the lowest significantfigure of any multiplicands or divisors

(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 x102 = 320




Rule of thumb on significant figure

!

Addition and subtraction"

The position of the last significant figures of each numbershould be compared

" The one farthest to the left is the permissible significantfigure

1530- 2.561527.44 1530

"

rounding the last digit 5 to even number1.35 => 1.41.25 => 1.2




Process Data Representation

!

Two-Point Linear Interpolation" Use to find estimate value from tabulation of data

x 1.0 2.0 3.0 4.0

y 0.3 0.7 1.2 1.8

"

The objective is to estimate value of y from given value of x •

Interpolation - y and x between tabulated points

• Extrapolation - y and x beyond the range of data in table

" The equation of the line through (x 1,y 1) and (x 2 ,y 2 )




Fitting a Straight Line

!

Convenient way to indicate how one variable depends on anotheris with an equation

y = 3x + 4

y = 4.24 (x - 3)2 - 23

y = 1.3 x 107 sin(2x)/(x 1/2 + 58.4)

! If the line is straight, the relationship can be presented

y = ax + b

Slope Intercept




Fitting Nonlinear Data

!

If the data is not linear and we wish to linearize it" Calculate the f(y) and g(x) for each tabulated (x,y)" Plot f(y) versus g(x) " If the plotted points fall on straight line, the equation fits the

data" Several examples plot that yields straight lines

(1) y = ax 2 + b Plot y versus x 2

(2) y 2 = a/x + b Plot y 2 versus a/x

(3) sin (y) = a(x 2 - 4) Plot sin (y) versus (x 2 - 4)




Logarithmic Coordinate

!

Two nonlinear function that normally occur are"

exponential function y = ae bx or y = a exp (bx) " power law y = ax b

! Natural logarithmic (ln ) is the inverse of exponential function

P = exp(Q) <=> ln P = Q

ln[exp (Q)] = Q and exp(ln P) = P

y = a exp (bx) => ln y = ln a + bx (Use semilog paper) Plot ln y versus x . Slope = b, intercept = ln a

y = ax b => ln y = ln a + b ln x (Use log-log paper) Plot ln y versus ln x . Slope = b, intercept = ln a




Processes and Process Variables




Process and Process Variables

!

Process - any operation that causes a physical or chemical changein a substance or a mixture of substances

! Material enters a process is referred as input or feed

! Material leaves is called as output or product

!

Process Unit is an apparatus for carried out the process

"

Distillation Column

" Absorption Column

" Evaporator Column

" Extraction Column

" Reactor

ProcessUnit

input

feed

output

product




Mass and Volume

!

Density ( ) is mass per unit volume" kg/m3, g/cm3, and lbm/ft

3

" use to relate mass and volume

! Specific Volume is volume per unit mass" m3/kg, cm3/g, and ft3/lbm

"

an inverse of density! Specific Gravity is the ratio of density ! and ! ref " SG = ! / ! ref

" The reference ! most commonly used is water at 4.0 oC" ! ref(H2O, 4.0 oC ) = 1.000 g/cm3

1000 kg/m3

62.43 lbm/ft3






Flowrate

!

Flow rate - rate at which material is transported throughprocess line" Mass flow rate (mass/time) kg/s or lbm/s

" Volumetric flow rate (volume/time) m3/s or ft3 /s

! The mass and volume is related by the fluid density (! )

!

The density (! ) of a fluid can be used to convert knownvolumetric flow rate to the mass flow rate and vice versa

m (kg fluid/s)

V (m3 fluid/s)




Chemical Composition

!

Atomic Weight - the mass of atom on a scale that assign 12C amass exactly 12.

!

Molecular Weight - the sum of atomic weight of atoms thatconstitute a molecule" Atomic weight of Oxygen (O) = 16"

Molecular Weight of molecular Oxygen (O2) = 32!

Gram-mole - amount whose mass is equal to its molecular weight "

units used - gmol, lbm-mole, kmol

" If Molecular weight of a substance is M , then there areM kg/kmol, M g/mol and M lbm/lb-mole of this substance

" Carbon monoxide (CO) has a molecular weight of 28;

•

1 mol of (CO) therefore contains 28 g• 1 lbm- mole of (CO) contains 28 lbm and• 1 kgmol of (CO) contains 28 kg




Molecular Weight

!

Example : 34 kg of ammonia (NH3): M = 17 are equivalent to

34 kg NH3 1 kmol NH3 = 2 kmol NH3 17 kg NH3

4 lb-moles of ammonia are equivalent to

4 lb-mole NH3 17 lbm NH3 = 68 lbm NH3 1 lb-mole NH3

!

One gram-mole of any species contains 6.02 x 10 23 (Avogadro’snumber) molecules of that species




Conversion of mass flowrate to molar flow rate

!

The molecular weight of a species can be used to relate the mass flowrate to corresponding molar flow rate!

Example: If ammonia (NH3): M = 17 flows through a pipeline at a rateof 100 kg/h the molar flowrate of the of are equivalent to

100 kg NH3 1 kmol NH3 = 5.88 kmol NH3

h 17 kg NH3 h

If the output stream of a reactor contains NH3 flowing at a rate of850 lb-moles/min, the corresponding flowrate is

850 lb-moles NH3 17 lbm NH3 = 14 450 lbm NH3

min 1 lb-mole NH3 min




Mass and Mole Fractions

!

Process streams occasionally contain more than one substance!

To define the composition of mixture we needMass Fraction :

Mole Fraction :

Thepercent by mass

of A is 100 xA, and themole percent

of A is100 yA






!

Natural gas has the following molar composition;

Methane (CH4) 93.0%Ethane (C2H6) 4.00%Propane (C3H8) 0.80%Butane (C4H10) 0.15%Carbon dioxide (CO

2) 0.45%

Nitrogen (N2) 1.60%

Taking 1 mol of natural gas as a basis of calculation, calculate

a) the average molecular weight of natural gasb) the mass fraction (%) of each component of natural gasc) the average molecular weight of natural gas based on the mass

fraction in part (b)

Conversion of mole percent to mass percent




Concentration

!

Mass concentration is the mass of component per unit volume ofthe mixture (g/cm3, lbm/ft3 or kg/m3)

!

Molar concentration is the number of moles of the component perunit volume of the mixture (mol/cm3, lb-mole/ft3 or kmol/m3)

!

Molarity is the value of the molar concentration of the soluteexpressed in gram-moles solute/liter solution " 2-molar solution of A contains 2 mol A/ liter solution

! Concentration factor can be used to relate mass (molar ) flow rateof a component of a continuous stream to the total volumetricflow rate of the stream

!

Given: 6 liters of 0.02-molar solution of NaOH contains

6 liters 0.02 mol NaOH = 0.12 mol NaOHliter




Conversion of mass, molar and volumetric flow rate

A 0.5 molar aqueous solution of sulfuric acid flows into a process unitat a rate of 1.25 m3/min. The specific gravity of the solution is 1.03.Calculate(1) the mass concentration of H2SO4 in kg/m3,(2) the mass flow rate of H2SO4 in kg/s, and

(3) the mass fraction of H2SO4

= 0.5 mol H2SO4 98 g 1 kg 103 literliter mol 103 g 1 m3

= 49 kg H2SO4 m3




Solution for (2) and (3)

1.25 m3 49 kg H2SO4 1 min

min 1 m3 60s

solution = (1.03)( 1000 kg/m3) = 1030 kg/m3

Qsolution(kg/s) = 1.25 m3 solution 1030 kg 1 min = 21.46 kg/s

min m3 solution 60 s



© PCS-FKKKSA Slide 1 - 39

Part Per Million (ppm)

!

Parts per million (ppm) - is commonly used as ameasure of small levels of pollutants in air, water, andbody fluids.

! Parts per million is the mass ratio between thepollutant component and the solution and ppm is

defined as: ppm = 1,000,000 mc/ms

where

mc = mass of component (kg, lbm) or moles

ms = mass of solution (kg, lbm) or moles



© PCS-FKKKSA Slide 1 - 40

Mass per Unit Volume (ppm)

!

The concentration of a component can be measured as mass per unitvolume as mg/liter and mg/cm3.! Weight of substance added to one unit volume of water to give one part

per million (ppm)!

1 ppm= 2.72 pounds per acre-foot= 1,233 grams per acre-foot= 1.233 kilograms per acre-foot= 0.0283 grams per cubic foot= 0.0000624 pounds per cubic foot= 0.0038 grams per US gallon= 0.058419 grains per US gallon= 0.07016 grains per Imperial gallon= 1 milligram per liter= 1 microlitre ( µL ) per liter= 0.001 gram per litre= 8.345 pounds per million gallons of water




Pressure

!

Pressure is the ratio of a force to the area "

Units N/m2, dynes/cm2, and lbf/in2 "

the SI pressure unit is N/m2 or called pascal (Pa)

P = P o + g/g c h

The relationship between the base and the top can be given as head:

P h (mm Hg) = P o (mmHg) + h (mmHg)

Fluid pressure and measurement

P absolute = P gauge + P atmospheric

Po (N/m2)

P (N/m2)h (m) fluiddensity

A (m2)




Temperature

!

Two most common temperature scales are defined using the freezing point(T f ) and boiling point (T b ) of water at 1 atm

" Celsius (or centigrade) scale• T f is assigned 0oC and T b is 100oC• Absolute zero on this scale falls at -273.15oC

" Fahrenheit scale

•

T f is assigned 32oF and T b is 212oF• Absolute zero on this scale falls at -459.67oF

" The Kelvin and Rankin scale are defined at absolute value of Celsius andFahrenheit

T(K) = T(oC) + 273.15T(oR) = T(oF) + 459.67

T(o

R) = 1.8 T(K)T(oF) = 1.8T (oC) + 32




Temperature Scale




Degree (o) definition as Temperature Interval

Degree as Temperature Interval

! Consider the temperatureinterval between 0oC and 5oC

!

There are 9oF and Rankindegree in this interval

!

An interval of 1oC or Kelvincontains 1.8oF or Rankin degree

Conversion factor for the interval

ExampleFind the number Celsius degreesbetween 32oF and 212oF

To find the Celsius temperaturecorresponding to 32oF you cannot use this formula

Temp reading Temp interval




Working Session I

A container holds 50 g of water (B) and 50 g of NaOH (A). Calculatethe weight fraction and mole fraction of NaOH. Also calculate the lbmof NaOH (A) and H2O (B).

Basis : 100 g of solution

Weight Fraction : H20 = 50/100 = 0.5NaOH = 50/100 = 0.5

Mole Fraction : H2O = 50/18 = 2.78 2.78/4.03 = 0.690: NaOH = 50/40 = 1.25 1.25/4.03 = 0.310

4.03

lb m A = 50 g A = 0.1102 lb m A(453.6 g A/lb m A)

Also…. Calculate the Ib-moles of water and NaOH and its respectivemole fraction.



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Introduction engineering principle and units