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CHAPTER 1- INTRODUCTION CHE243-MATERIAL AND ENERGY BALANCE AND SIMULATION
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Page 1: Chapter 1 Notes

CHAPTER 1-

INTRODUCTION CHE243-MATERIAL AND ENERGY

BALANCE AND SIMULATION

Page 2: Chapter 1 Notes

CHAPTER 1- INTRODUCTION

INTRODUCTION TO ENGINEERING CALCULATION

• Units And Dimensions

• Conversion Of Units

• Systems Of Units

• Dimensional Homogeneity

PROCESSES AND PROCESS VARIABLES

• Concept Of Mass, Volume, Flowrate

• Chemical Composition

• Pressure And Temperature

Page 3: Chapter 1 Notes

2.0 UNIT & DIMENSION

Proper handling of units is an essential part of being an engineer

A measured or counted quantity has a numerical value and unit, eg. 2 meter, 4.29 kilograms.

A dimension is a property that can be measured, e.g – length, time, mass or temperature, or calculated by multiplying or dividing other dimensions, such as length/time (velocity).

Measurable unit are specific values of dimensions that have been defined by convention, custom, or law, e.g – grams for mass, seconds for time and centimeters for length.

Page 4: Chapter 1 Notes

3.0 CONVERSION OF UNITS

Equivalence between two expressions of the same quantity can be defined in terms of a ratio:

Factors Conversion

mm 100

cm 1

mm 10

cm 1

centimeter 1 per milimeters 10 cm 1

mm 10

milimeters 10 per centimeter 1 mm 10

cm 1

2

2

2

To convert a quantity expressed in terms of the unit to its equivalent in terms of another unit, multiply the given quantity by the conversion factor (new unit/old unit). Example:

1 g

1000 mg

36 mg = 0.036 g

Page 5: Chapter 1 Notes
Page 6: Chapter 1 Notes

Convert an acceleration of 1 cm/s2 to its equivalent in km/yr2.

EXAMPLE 1

2

9

yr

km1095.9

Page 7: Chapter 1 Notes

CONVERSION OF UNITS

3 wk to miliseconds

554 m4/(day.kg) to cm4/(min.g)

Page 8: Chapter 1 Notes

4.0 SYSTEM OF UNITS

Base units – for mass, length, time, temperature, electrical current and light intensity.

Multiple units – defined as multiples or fractions of base units such as minutes, hours, and milliseconds, all defined in unit of seconds.

Derived units – (a) by multiplying and dividing base or multiple units

(cm2, ft/min) – preferred as compound unit.

(b) As defined equivalent to compound unit (eg. 1 lbf

32.174 lbm.ft/s2).

Page 9: Chapter 1 Notes

SYSTEM OF UNITS

SI unit – System International – meter (m) for length, kilogram (kg) for mass, seconds (s) for time and Kelvin (K) for temperature.

CGS system – Most identical to SI unit. Refer to Felder (pp. 11) Table 2.3-1.

American Engineering System – foot (ft) for length, pound-mass (lbm) for mass, seconds for time.

*Note: Refer to Felder (pp. 11)

Page 10: Chapter 1 Notes

Quantities can be added or subtracted only if their units are the same.

Rule – every valid equation must be dimensionally homogeneous; that is, all additive terms on both sides of equation must have the same dimensions.

*Note: Refer to Felder (pp. 20-22).

5.0 DIMENSIONAL

HOMOGENEITY

Page 11: Chapter 1 Notes

Consider the equation

D(ft)=3t(s)+4

1. If the equation is valid, what are the dimensions of the constants 3 and 4

2. If the equation is consistent in its units, what are the units of 3 and 4

3. Derive an equation for distance in meters in terms of time in minutes

5.0 DIMENSIONAL

HOMOGENEITY

Page 12: Chapter 1 Notes

PROCESSES AND

PROCESS VARIABLES

Page 13: Chapter 1 Notes

Chemical Process Flow Diagram

Page 14: Chapter 1 Notes

SEPARATION PROCESS

REACTION PROCESS

FEED PREPARATION

FINAL PRODUCT

Page 15: Chapter 1 Notes

You need to:

– Minimize production of unwanted byproducts

– Separate the good (product) from the bad

(byproducts)

– Recover the unused reactants

– Maximize profit, minimize energy

consumption, minimize impact on the

environment

Page 16: Chapter 1 Notes

Process: any operation or series of operations.

•The material enters a process = Input or Feed.

•The material which leaves the process = Output or

Product.

• It is common for process to consist of multiple steps.

These each steps is carried out in a process unit. Each of

process unit has associated with it a set of input and output

process stream.

•To design or analyze a process, we need to know the

amounts, compositions & condition of materials which enter

& leave the process.

Page 17: Chapter 1 Notes

6.0 MASS AND VOLUME

Density – mass per unit volume.

Specific volume – volume occupied by a unit mass of the substance

Specific gravity – ratio of the density of the substance to the density of a reference substance ref at specific condition:

Reference most commonly used for solid and liquid is water

at 4OC 1000 kg/m3

ref

SG

Page 18: Chapter 1 Notes

Calculate the density of mercury in lbm/ft3 and the volume in ft3 occupied by 215 kg of mercury. Given SG of mercury is 13.546 and density of water is 62.3 lbm/ft3.

Density of mercury….

392.843

ft

lbm

EXAMPLE 2

3ft5617.0

Volume of 215 kg mercury….

Page 19: Chapter 1 Notes

• Flowrate- the rate at which a material is transported through a process line is the flowrate of that material

• Can be expressed as- mass flowrate(mass/time) or volumetric flowrate (volume/time)

• Volumetric flowrate can be converted to mass flowrate if density of a fluid is known

MASS AND VOLUMETRIC FLOWRATE

..

// VmVm

Page 20: Chapter 1 Notes

• The mass flowaret of n-hexane (𝜌=0.659 g/cm3) in a pipe is 6.59 g/s. What is the volumetric flowrate of the hexane?

• The volumetric flowrate of CCl4(𝜌=1.595 g/cm3) in a pipe is 100.0 cm3/min. what is the mass flowrate of the CCl4?

EXAMPLE 3

s

cm3

0.10

min5.159

g

Page 21: Chapter 1 Notes

7.0 CHEMICAL COMPOSITION

Atomic weight – mass of an atom.

Molecular weight – sum of atomic weights of atoms that constitute a molecule of the compound, e.g O2 = 32 g/g-mol.

If a molecular weight of a substance is M, then there are M kg/kmol, M g/mol, and M lbm/lb-mole of this substance.

Molecular weight can be used as a conversion factor that relates the mass and the number of moles of a quantity of the substance.

1) Conversion Between Mass and Moles

Page 22: Chapter 1 Notes

How many of each of the following are contained in 100.0 g of CO2

(M=44.01)?

(1) mol CO2 (4) mol O (7) g O2

(2) lb-moles CO2 (5) mol O2

(3) mol C (6) g O

2CO mol 273.2)1(

EXAMPLE 4

2

3 CO mole-lb 10011.5)2(

C mol 273.2)3(

O mol 546.4)4(

2O mol 273.2)5(

gO7.72)6(

2O 7.72)5( g

Page 23: Chapter 1 Notes

7.0 CHEMICAL COMPOSITION

Mass Fraction:

Mole Fraction:

Percent by mass of A is 100xA

Percent by moles of A is 100yA

2) Mass & Mole Fraction and Molecular Weight

mass total

A of massAx

moles total

A of molesAy

Page 24: Chapter 1 Notes

A solution contains 15% A by mass (xA = 0.15) and 20 mole% B (yB = 0.20)

A kg 26

1) Calculate the mass of A in 175 kg of solution

EXAMPLE 5

2) Calculate the mass flowrate of A in a stream of solution flowing at a rate of 53 Ibm/h

h

AmIb8

Page 25: Chapter 1 Notes

A solution contains 15% A by mass (xA = 0.15) and 20 mole% B (yB = 0.20)

min

Bmol200

3) Calculate the molar flow rate of B in stream flowing at a rate of 1000 mol/min

EXAMPLE 5 CONT’D

Page 26: Chapter 1 Notes

A solution contains 15% A by mass (xA = 0.15) and 20 mole% B (yB = 0.20)

s

solution kmol140

4) Calculate the total solution flow rate that corresponds to a molar flow rate of 28 kmol B/s.

EXAMPLE 5 CONT’D

Page 27: Chapter 1 Notes

EXAMPLE 5 CONT’D

A solution contains 15% A by mass (xA = 0.15) and 20 mole% B (yB = 0.20)

5) Calculate the mass of the solution that contains 300 Ibm of A.

solutionmIb 2000

Page 28: Chapter 1 Notes

7.0 CHEMICAL COMPOSITION

A set of mass fractions may be converted to an equivalent set of mole fractions by:

(a) Assuming as a basis of calculation a mass of the mixture.

(b) Using the known mass fractions to calculate the mass of each component in the basis quantity; and converting this masses to moles;

(c) Taking the ratio of the moles of each component to the total number of moles.

3) Conversion from a Composition by Mass to Molar Composition

Page 29: Chapter 1 Notes

A mixture of gases has the following composition by mass:

Component Mass %

O2 16

CO 4.0

CO2 17

N2 63

What is the molar composition?

EXAMPLE 6

Page 30: Chapter 1 Notes

Comp.

O2

CO

CO2

N2

Total

Mole fraction

(y=n/ntotal)

0.150

0.044

0.120

0.690

1.000

Page 31: Chapter 1 Notes

A researcher conducted an experiment on electrolysis of mixed brine. A mixture of gases was produced at the cathode. The composition (by weight) of the gases was as follows: 64% Chlorine (Cl2), 29% Bromine (Br2) and 7% Oxygen (O2). Using the ideal gas law, calculate the composition (by volume) of the gas mixture….

Given Molecular weight Br2 = 159.83, Cl2 = 70.91, O2 = 32.00

Page 32: Chapter 1 Notes

7.0 CHEMICAL COMPOSITION

Average molecular weight – (mean molecular weight of a mixture), Mav

Using mole fraction:

Using mass fraction:

4) Average molecular weight

ntallcompone

iiav MyMyMyM ...2211

ntallcompone i

i

av M

x

M

x

M

x

M...

1

2

2

1

1

Page 33: Chapter 1 Notes

Calculate the average molecular weight of air:

(1) From its approximate molar composition of 79 mol% N2, 21 mol%O2

(2) From its approximate composition by mass of 76.7 wt% N2, 23.3 wt% O2

EXAMPLE 7

84.28avM

Page 34: Chapter 1 Notes

7.0 CHEMICAL COMPOSITION

5) Concentration

Solution: a mixture of substance called solutes in another substance called solvent. Solvent: a dominant substance that is present in larger amount and so it dissolves or dilutes the solutes. Solute: a substance that are present in smaller amount and dissolves or distributes in a solvent. It could be more than one solute in a solution. • Particles of solutes are normally distributed uniformly throughout the solvent mass.

Page 35: Chapter 1 Notes

• This distribution of solutes in a solvent is known as concentration (conc) of solutes in the solution.

• The conc of solutes could be expressed as mass conc or molar concentration (molarity).

• Mass concentration of a solute A in solution (g/cm3, kg/m3, Ibm/ft3).

• Molarity of a solute A in solution (in mol/L or M).

Page 36: Chapter 1 Notes

• ppm & ppb are used to express the con of trace species that is present in a very dilute amount (very small amount) relative to other components in a mixture.

• For a solution in liq or solid phase

Concentration of a substance A in ppm

= mass of a substance a solute A in gmol

106 unit of solution

Concentration of a substance A in ppb

• = mass of a substance a solute A in gmol

109 unit of solution

7.0 CHEMICAL COMPOSITION

6) Parts per Million and Parts per Billion

Page 37: Chapter 1 Notes

ppm also can be defined as

1𝑝𝑝𝑚 =1 𝑔

106𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛=

1 𝑔

106𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛=

1 𝑚𝑔

103𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛= 1

𝑚𝑔

𝐿

Page 38: Chapter 1 Notes

• Pressure- ratio of a force to the area on which the force acts

• Unit : Ibf/in2 (psi) or dynes/cm2 or N/m2 [pascal (Pa)]

Pabsolute = Pgauge + Patmospheric

Typical value of Patmospheric at sea level are;

1atm = 14.696psi = 760mmHg = 101.325kPa

8.0 PRESSURE

AFP /

Page 39: Chapter 1 Notes

9.0 TEMPERATURE

Temperature – a measure of average kinetic energy possessed by the substance molecules.

The relationship to convert a temperature expressed in one unit to another:

15273CTKT o . 67459FTRT oO .

KT81RT O . 32CT81FT oO .

The conversion factor:

The conversion factors refer to temperature intervals, not temperatures.

K1

C1

R1

F1

K1

R81

C1

F81 O

O

OO

O

O

,,.

,.

Page 40: Chapter 1 Notes

Consider the interval from 20°F to 80°F

1. calculate the equivalent temperatures in °C and the interval between them

2. calculate directly the interval in °C between the temperatures.

EXAMPLE