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Page 1: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Processes and ProcessVariables

Page 2: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

• Process: Any operation or series of operations by which a particular objective is accomplished.

e.g. Operations that cause a physical or chemical change in a substance or mixture.

The material entering a process: input, feedThe material leaving a process: output, product

If there are multiple steps, each will be a process unit with itsown process streams (I,O)

You design (formulation of a flowsheet and specifications) oroperate (day-to-day running of process) a process.

Page 3: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

VARIABLES• Mass and Volume: • Density = mass / volume• Specific volume = volume occupied by unit mass = 1/ρ

(density)

• If ρ = 1.595 g/cm3, volume = 20.0 cm3,• → mass = 20.0 cm3 1.595 g/cm3 = 31.9 g

• Specific gravity = ρsubs. / ρref.subs. (at specific conditions) → SG = ρ / ρref.

• ρref = ρH2O(l) (4oC) = 1.000 g/cm3 = 62.43 Ibm/ft3• SG = 0.6 (20o/4o) SG = 0.6 at 20oC with reference to

4oC

Page 4: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Example: • Calculate the density of mercury in Ibm/ft3, if its

SG=13.546 20o/40o

• Calculate the volume in ft3 occupied by 215 kg of mercury.ρHg = 13.546 (62.43 Ibm/ft3) = 845.7 Ibm/ft3V = m / ρHg= 215 kg (1Ibm / 0.454 kg) (1 ft3 / 845.7 Ibm)= 0.560 ft3For solids and liquids ρ ≠ ρ(T,P)For gases it is obviousFor mercury ρ is depended on T:V(T) = Vo (1 + 0.18 x 10-3 T + 0.0018 x 10-6 T2)

Page 5: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

3.1.1 Example: m = 215 kg, V = 0.560 ft3 at 20oCV(T) = Vo (1 + 0.18 x 10-3 T + 0.0018 x 10-6 T2)What volume would the mercury occupy at 100oC?(1) If it is contained in a cylinder (D=0.25 in), what change in height would

be observed as the mercury is heated from 20oC to 100oC?

V(100oC)=Vo[1+0.18x10-3(100)+0.0018x10-6(100)2]V(20oC)=Vo[1+0.18x10-3(20)+0.0018x10-6(20)2]

= 0.560 ft3Solving for V0 from the 2nd equation and substituting it into the 1st yields.→ V(100oC) = 0.568 ft3Vcyl = π D2/4H(100oC) – H(20oC) = [V(100oC) – V(20oC)] / (π D2/4)=23.5 ft

Page 6: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Flow Rate Mass an Volumetric Flow Rate:• Material move between process units, between a

production facility and a transportation depot. The rate at which a material is trasported through a process line is theflow rate. Blood flows in the arteries with a specific flow rate.

• Mass flow rate → (mass/time)• Volumetric flow rate → (volume/time)

.

.

V

mVm==ρ

Cross Section

m (kg/s)

V (m3/s)tnn =

.

Page 7: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Flow Rate Measurement

• Flowmeter: A device that reads continiouslyflow rate in the line.

OrificemeterRotameter

The larger theflow rate the higher the float rises

Fluid pressuredrops from the upstream to the downstream

Page 8: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

• Chemical Composition: The properties of a mixture depend on mixture compositions.

• Moles and Mole Weight: A g-mole (or mol in SI units) of a species is the amount of that specieswhose mass in grams is numerically equal to itsMW. e.g. CO has a MW of 28.

• 1 mol of CO contains 28 g• 1Ib-mole CO contains 28 Ibm• 1 ton-mole CO contains 28 tons• e.g. NH3 has a MW of 17. • 34 kg of ammonia is;34 kg NH3 (1 kmol NH3/17 kg NH3) = 2 kmol NH3

Page 9: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

• Example: Conversion between Mass and Moles

• How many of each of the following are contained in 100 g CO2 (M=44.01)(a) mol CO2; (b) Ib-moles CO2; ( c ) mol C

• 100 g (1 mol CO2/44.01 g CO2) = 2.273 mol CO2

• 2.273 mol CO2 (1 Ib-mol/453.6 mol)=5.011 10-3 Ib-mole CO2

• 2.273 mol CO2 (1 mol C / 1 mol CO2)=2.273 mol C

• Calculate the flow rate in kmol CO2/h if flow rate = 100 kg/h CO2• 100 kg CO2 /h (1 kmol CO2 / 44.01 kg CO2) = 2.27 kmol CO2/h

Page 10: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Masses and Mole Fractions andAverage Molecular Weight

• Process input or output streams can contain mixtures of liquids or gases, solutions of one or more solutes in a solvent. You need mass fraction and molefraction to define the compositions:

• Mass fraction; xA = mass of A / total mass• Mole fraction; yA= moles of A / total moles

Page 11: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

• Example: A solution contains 20% X and 25% Y by mass.

• Calculate the mass of X in 220 kg of solution: 220 kg solution 0.20 kg X / kg solution = 44 kg X

• Calculate the mass flow rate of A in streamflowing at a rate of 53 Ibm/h:53 Ibm/h 0.20 Ibm X/Ibm solution = 10.6 Ibm X/h

Page 12: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

• The Average Molecular Weight

∑ ⋅=+⋅+⋅=.

2211 .........allcomp

ii MyMyMyM

∑=++=.2

2

1

1 .........1allcomp i

i

Mx

Mx

Mx

M

Where;

xi = Mass fraction

yi = Mole fraction

Mi = Molecular weight of component

Page 13: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Concentration• Mass concentration = mass of a component / volume of the mixture

• Molar concentration = moles / volume

• Molarity = molar concentrationof solute / volume of solution

• Parts per million (ppm), and• parts per Billion (ppb) are used to express the concentration of trace

species.

• ppmi = yi x 106

• ppbi = yi x 109

Page 14: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

• Example: In normalş living cells, the nitrogen requirement for thecells is provided from protein metabolism. When individual cellsare commercially grown, (NH4)2SO4 is usually used as the sourceof nitrogen. Determine the amount of (NH4)2SO4 consumed in a fermentation medium in which the final cell concentration is 35 g/L in a 500 L volume of the fermentation medium. Assume cellscontain 9 wt% N, and that (NH4)2SO4 is the only N source.

• Basis: 500 L solution containing 35 g/L cell concentration,

)42)4(142)4(132

1441

141

109.035500

SONHgmolSONHg

gmolNSOgmolNH

gNgmolN

cellggN

LgcellL

= 14,850 g (NH4)2SO4

Page 15: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

A solution of HNO3 in water has a specificgravity of 1.10 at 25oC. The concentration of HNO3 is 15 g/L of solution. What is the; a) Mole fraction of HNO3 in the solution?b) ppm of HNO3 in the solution? (Take 1 L and 100

g of solution)Basis: 1 L →(15 g HNO3/1 L solution) (1 L/1000 cm3) (1 cm3/1.10

g solution)= 0.01364 g HNO3/g solution(Assume SG = ρsolution)

Page 16: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

15.55018.01699.986H2O3.9 x 10-52.164 x 10-463.021.364HNO3

Mol frac.g-moleMWg

b)Mass fraction is 0.01364 or

= 13,640/106 or

= 13,640 ppm

Basis: 100 g solution;The mass of water in the solution is: 100 – 1.364 = 99.986 g H2O

Page 17: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Pressure• A pressure is the ratio of a force to the area on

which the force acts. Units are N/m2 (Pa), dynes/cm2, Ibf/in2

• Pressure is defined as the “normal (perpendicular) force per unit area”

h

mercury

Atmospheric Pressure 0PhgAFP +⋅⋅== ρ

P = static pressure

ρ = density of fluid

g = gravity

P0 = pressure at the top of the fluid

Page 18: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

A (m2)

F (N)P (N/m2)

F (N) : minimum force thatwould be exerted in thehole to keep the fluid fromemerging

A pressure may be expressed as a “head” of a particularfluid; the height of a hypothetical column of this fluidexerting the same pressure at its base if the pressure at the top were zero. e.g. 76 cm of mercury (76 cm Hg). Theequivalence between a pressure (P) and its correspondinghead Ph is given as;

P (force/area) = ρfluid g Ph (head of fluid)

Page 19: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

• Example: Express 2.0 x 105 Pa in terms of mmHg.ρHg = 13.6 x 1000 kg/m3 = 13600 kg/m3

g = 9.807 g/s2

mmm

Nsmkg

ms

kgm

mN

gPP

Fh

3223

25 10/1

807.913600102 ⋅

⋅⋅⋅

⋅⋅

⋅⋅⋅×=

⋅=ρ

mmHg31050.1 ×=

Ph (mmHg) = P0 (mmHg) + h (mm Hg)

Page 20: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Atmospheric, Absolute, and GaugePressure

• Atmospheric pressure at sea level, 760 mmHg = 1 atm.

• The fluid pressures are “absolute” pressures if a zero pressure corresponds to a perfect vacuum. Many devices measure the “gauge pressure” of a fluid or the pressure relative to theatmospheric pressure.

• Pgauge = 0 → Pabsolute = Patmospheric• Pabsolute = Pgauge + Patmospheric• psia and psig are commonly used to denote

absolute and gauge pressures in Ibf/in2

Page 21: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Open end monometer

Relative (gauge) PressureAbsolute Pressure

N2

Air

∆hN2

∆h

Vacuum

Page 22: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Fluid Pressure Measurement• Elastic-element method-Bourden tubes

7000 atm < P <0• Liquid-column methods-manometers

P < 3 atm → accurate• Electrical Methods-Strain gauges

Page 23: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

P1

P2=Patm

Manometer fluid

P1 P2

Open-end Differential

P1

P2 = 0

Sealed-end

Page 24: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Manometerfluid, ρf

P1 P2

d1

d2

h

(a) (b)

Fluid 1, ρ1

Fluid 2, ρ2

General manometer equation: P1 + ρ1 g d1 = P2 + ρ2 g d2 + ρF g h

If ρ1 = ρ2 = ρ ; P1 – P2 = (ρF - ρ) g h (Differential manometerequation)If both fluids are gases ρF >> ρ→ P1 – P2 = ρF g h

Page 25: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

• Small animals like mice can live at reduced air pressures down to 20 kPa absolute. In a test, a mercury manometer attached to a tank, reads64.5 cm Hg and the barometer reads 100 kPa. Will mice survive?

P=100 kPa – 64.5 cmHg (101.3 kPa76 cmHg)

= 100 – 86 = 14 kPa absolute

MICE WILL DIE!64.5 cm Hg

100kPa

Page 26: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

Temperature• It is a measure of energy (mostly kinetic) of the

molecules in a system. Since kinetic energycannot be measured directly, T must be determined indirectly by measuring somephisical property of the substance whose valuedepends on T.

• Electrical resistance of a conductor (resistancethermometer), voltage at the junction of twodissimilar metals (thermocouple), spectra of emitted radiation (pyrometer), and volume of a fixed mass of a fluid (thermometer)

Page 27: Processes and Process Variableshome.ku.edu.tr/~okeskin/ChBi201/chapter-3-chbi.pdf · VARIABLES •MassandVolume: • Density = mass / volume • Specific volume = volume occupied

• Rankine-Fahrenheit Scale ∆°F = ∆°R• Kelvin-Celcius Scale ∆°C = ∆°K• ∆°C / ∆°F = 1.8• ∆°K / ∆°R = 1.8• T(°K) = T(°C) + 273.15• T(°R) = T(°F) + 459.67• T(°R) = 1.8 T(°K)• T(°F) = 1.8 T(°C) + 32