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Chapter 1 (Part 2): Basic Concepts1.4 Process VariablesChemical Process Principles(CLB 10904)

Nazerah Binti AhmadSCET UniKL MICETTel: 012-7871994E-learning: CLB 10904 Nazerah

Evaluation of performance of process operation requires the knowledge of the amounts, composition, conditions of materials that enter and leave each process units.This chapter presents methods of calculating variables that characterize the operation of processes and individual process units. Processes and Process Variables (Jan 2015)2

ProcessesProcess- any operation that cause a physical or chemical change in a substance. Can consist of several process unit.Process streams connecting process units and form the process flow sheet.Process Unit

Input/FeedOutput/ProductProcess Unit Output/Product

3Jan 2015)

Process VariablesThe quantities used to describe a process and these must be measured and computed.

To design or analyze a process, we need to know the amounts, compositions, and conditions of materials entering, leaving and within the process.

PROCESS

FeedINPUT Products OUTPUT4 (Jan 2015)

Process VariablesDensity and Specific GravityFlow rateChemical CompositionPressureTemperature5Process Variables (Jan 2015)

Density & Specific VolumeDensity ()Mass per unit volume of a substance.Density of a substance can be used as conversion factor to relate the mass and the volume of the substances.Unit: g/cm3; kg/m3; lbm/ft3.

Specific Volume (v)Volume per unit mass of a substance.Inverse of density (1/).Unit: cm3/g; m3/kg; ft3/lbm6(Jan 2015)

The density of CCl4 is 1.595 g/cm3; what is the Mass of 20 cm3 of CCl4

b)Volume of 6.20 lbm of CCl4

Check Your Understanding 17Dr. Kelly YTL (Sep 2014)

Specific Gravity (SG)Ratio of the density () of a substance to the density of a reference (ref) substance at a specific condition:

Density of water at 4C is used as the reference density; ref = H2O(l) (4C) = 1.000 g/cm3=1000 kg/m3=62.43 lbm/ft3SG is dimensionless. To get the density of a substance, multiply the SG value with the reference density, ref.

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A liquid has a SG of 0.50. Find;Density in g/cm3

Density in lbm/ft3

Density in kg/m3

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Check Your Understanding 2

A liquid has a SG of 0.50. Find;Mass for 3 m3 of the liquid volume

Volume occupied by 18 lbm of this liquid

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Check Your Understanding 2

Flow RateContinuous process involve movement of materials from one point to another with certain rate.Flow rate: The amount of material that moves into or out of a process unit per unit time. Flow rate can be expressed as :Mass flow rate, = Mass/Time (kg/s, g/min, lbm/hr) Volumetric flow rate, = Volume/Time (m3/s, ft3/hr) Molar flow rate, = Moles/Time (mol/hr, lbmol/min)

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12We can relate between Volumetric flow rate, Q (Volume/Time) with Mass flow rate, (Mass/Time) using Density, (Mass/Volume)

Example: Calculate the mass flow rate in kg/hr if given a liquid with SG 20.0 and volumetric flow rate of 12 m3/hr.

Flow Rate

Flow meter is a device mounted in a process line that provides a continuous reading of the flow rate in the line.Two commonly used flow meter are rotameter and orifice meter.

Flow Rate Measurement13

Flow MeterRotameter(vertical tube containing a float)Orifice meter(based on pressure drop)

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PressureA pressure is the ratio of a force to the area on which the force acts (P= F/A). Pressure units: N/m2, dynes/cm2, lbf /in2, psi, and Pa.

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Atmospheric, Absolute & Gauge Pressure The Atmospheric Pressure, Patmospheric can be thought of as the pressure at the base of a column of fluid (air) located at the point of measurement (e.g. at sea level).A typical value of the atmospheric pressure at sea level, 760.0 mm Hg, has been designated as a standard pressure of 1 atmosphere.

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Absolute Pressure, Pabsolute is gauge pressure plus atmospheric Pressure. Gauge Pressure, Pgauge is the pressure relative to the atmospheric pressure at measurement point.Relationship between absolute pressure and gauge pressure is:

The abbreviation psia and psig are commonly used to represent absolute and gauge pressure in lbf / in2

17Atmospheric, Absolute & Gauge Pressure

Fluid Pressure MeasurementCommon pressure measurement devices: Bourdon gauge and manometer.Manometer: U-shaped tube partially filled with fluid of known density. Bourdon gauge: Hollow tube closed at one end and bent into a C configuration.

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Temperature of a substance in a particular state of aggregation (solid, liquid, or gas) is a measure of the average kinetic energy possessed by the substance molecules.Example of temperature measuring devices: Resistance thermometer (based on the resistance of a conductor), thermocouple (voltage at the junction of 2 dissimilar metals), pyrometer (spectra of emitted radiation) and thermometer (volume of a fixed mass of fluid).Temperature19

The following relationship may be used to convert a temperature expressed in one defined scale unit to its equivalent in another unit

We can use these conversion factor for temperature intervals, T. These conversion actor refer to temperature intervals, T, NOT temperature, T.

20Temperature and Temperature Interval (T)

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Convert the interval between 50 oF and 100 oF in oC

Convert 50 oF to oC,

Check Your Understanding 11

Chemical CompositionMoles and Molecular WeightMass and Mole Fractions Average Molecular WeightConcentration

Parts per Million (ppm) & Part per Billion (ppb)Chemical Composition22

Moles & Molecular WeightAtomic weight - mass of an atom based on carbon isotope 12C.Molecular weight (MW) - Sum of the atomic weights of atoms that represent a molecule of the compound.Atomic Oxygen (O) has an atomic weight of 16.0 but a molecular oxygen gas (O2) has a molecular weight of 16.0+16.0 = 32.0

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A gram-mol (gmol, or mol in SI units), n is the amount of a species (atom or molecule) whose mass in grams is numerically equivalent to its molecular weight.In addition, we have other types of moles: kg-mole (kmol), and lb-mole (lbmol).If the molecular weight (MW) of a substance is M, then we can define it as M kg/kmol or M g/mol or M lbm/lbmol of this substance.MW can be used as a conversion factor that relates the mass, m and the no. of moles, n of a quantity of the substance.

24Moles & Molecular Weight

34 kg of NH3 is equivalent to how many kmole? Given MW NH3 = 17.0

How many Ibmol in 72 lbm of NH4NO3. Given MW: N=14.0, H=1.0, O=16.0

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Check Your Understanding 3

Mass and Mole FractionProcess input or output streams can contain mixtures of liquids or gases, solutions of one or more solutes in a solvent.Need mass fraction and mole fraction to define the compositions. Remember that mass fraction is NOT EQUAL to mole fractionMass fraction, xA

Mole fraction, yA

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A mixture of gases has the following composition: 16 wt% O2, 4 wt% CO,17 wt% CO2 and N2. What is the mass fraction of this mixture in kg/kg total? Subsequently calculate the mass of O2 if given the total mass of the mixture as 1500 kg.Solution: 27Check Your Understanding 4

A mixture of gases has the following composition: 30 mol% O2, 44 mol% CO,7 mol% CO2 and N2. What is the mole fraction of this mixture in kmol/kmol total. Subsequently calculate the total moles of the mixture in kmol if given the no of moles of CO2 = 12 kmolSolution: 28Check Your Understanding 5

4 steps to convert from mass fraction, x to mole fractions, y:

29Dr. Kelly YTL (Sep 2014)Mass and Mole Fraction

A mixture of gases has the following mass composition:O216 wt% (MW 32) CO4 wt% (MW 28)CO217 wt% (MW 44)N263 wt% (MW 28)What is the molar composition if given the total mass of the mixture 1500 g?30Check Your Understanding 6

Comp.Mass Fraction, xiMass (g)O216 wt% = 0.16 0.16 x 1500 = 240CO4 wt% = 0.040.04 x 1500 = 60CO217 wt% = 0.170.17 x 1500 = 255N263 wt% = 0.630.63 x 1500 = 945Total1.001500g

SolutionFrom question: The mass composition is O2 (16 wt%), CO (4 wt%), CO2 (17 wt%) and N2 (63 wt%). Total mass of the mixture 1500 g.Solution:Determine the weight of each of the compounds.31

SolutionDetermine the no. of moles of each compound using MW and subsequently the total no. of moles of the mixture.Determine the mole fraction for each compound

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4 steps to convert from mole fraction, y to mass fractions, x:

33Mass and Mole Fraction

A mixture of gases has the following composition:O240 mol% (MW: 32)CO44 mol% (MW: 28)CO27 mol% (MW 44)N29 mol% (MW 28)What is the mass composition if given the total moles of the mixture 200 kmol?(Ans: 42 wt% O2, 40 wt% CO, 10 wt% CO2 and 8 wt% N2)34Check Your Understanding 7

Average Molecular Weight Mean molecular weight of a mixture (g/mol, kg/kmol, lbm/lbmole).If yi is the mole fraction of the component i of the mixture and Mi is the molecular weight of this component:

If xi is the mass fraction of the component i of the mixture and Mi is the molecular weight:

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Example: Determine the average molecular weight of refinery waste gas component. The analysis is given as: CH4 (78.0 mol%), C2H6 (10.0 mol%), C3H8 (8.0 mol%), and C4H10 (4.0 mol%).

Lets assume basis of 100 kmol total mixture and MW unit in kg and kmol

Comp.mol.%Moles (kmol)MW (kg/kmol)Weight (kg)CH4C2H6C3H8C4H1078.010.08.04.078.0% x 100 = 78.010.0% x 100 = 10.08.0% x 100 = 8.04.0% x 100 = 4.016.030.144.158.178.0 kmol x 16.0 kg/kmol = 1248 kg10.0 kmol x 30.1 kg/kmol = 301 kg8.0 kmol x 44.1 kg/kmol = 352 kg4.0 kmol x 58.1 kg/kmol = 232 kgTOTAL100.0100.0 kmol2133 kg

36Check Your Understanding 8

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Example: Determine the average molecular weight of refinery waste gas component. The analysis is given as: CH4 (78.0 wt%), C2H6 (10.0 wt%), C3H8 (8.0 wt%), and C4H10 (4.0 wt%). Given total mass of the mixture 200 g

Given total mass of the mixture 200g and MW unit in g and mol

Comp.wt.%Mass (g)MW (g/mol)Moles (mol)CH4C2H6C3H8C4H1078.010.08.04.078.0% x 200 = 156.010.0% x 200 = 20.08.0% x 200 = 16.04.0% x 200 = 8.016.030.144.158.1156.0 g x (mol/16.0g) = 9.75 mol20.0 g x (mol/30.1g) = 0.66 mol16.0 g x (mol/44.1g) = 0.36 mol8.0 g x (mol/58.1g) = 0.14 molTOTAL100.0200.0 g10.91 mol

37Check Your Understanding 9

38Concentrations

=

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39Concentrations

=

39REMEMBER:We can convert between mass flow rate (e.g. kg/hr, g/min, lbm/s) to molar flow rate (e.g. mol/hr, kmol/min, lbmol/s) to volumetric flow rate (e.g. m3/min, ft3/hr) by using specific gravity, SG and molecular weight, MW.

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=

40Example: Given liquid acetone flows into a process unit at a rate of 1.25 m3/min. Given SG 0.791 and MW 58.08. Calculate Mass flow rate of acetone in kg/sSolution:

Molar flow rate of acetone in lbmol/min

Check Your Understanding 10

Parts per Million (ppm)& Parts per Billion (ppb)To express the concentrations of trace species in mixtures of gases or liquids.May refer to mass ratios (usual for liquids) or mole ratios (usual for gases). How many parts (in gram or moles) of the species are present per million or billion parts of the mixture.ppmi= yi x 106ppbi = yi x 109

Example: Suppose air in the vicinity of a power plant is said to contain 15 ppm SO2 (15 parts per million sulfur dioxide). This statement means that every million moles of air contains 15 moles of SO2.41

Homework!42Attempt Tutorial 1: Processes And Process Variables (Part 2) Question 1 until 11

Announcement43Draft submission for Mini ProjectDateline: Friday, 2/3/2015 Check guideline on E-LearningSection 3 Introduction (Student 1 And 2) Section 4 Process Equipment (Student 3 And 4)