mb2 single phase mass balance

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Mass Balance: Single-Phase

Chemical Engineering

Content

• Section 1– Theory (phases)

– Liquids, Solids = ok; do as before

• Section 2– Gases = P,T,V dependent!

• Ideal Gas

• Real Gases– Cubic Equations

– Z compressibility factor

– MB with gases (practice, practice, practice!)

Section 1

• Theory (phases)

• What is a phase?

• Which phases are we going to study?

• Liquids, Solids

• Why can we do mass balances easily

• NO gases in this section

• Estimating densities (liquids)

What is a Phase?

• A region of space (a thermodynamic system), throughout which all physical properties of a material are essentially uniform.

• Simple A region of material that is:

– chemically uniform

– physically distinct

– mechanically separable (in general)

Examples of Phases

• Typical

– Solid

– Liquid

– Gas

• The term phase is sometimes used as a synonym for state of matter

– but a system can contain several immiscible phases of the same state of matter

Phases vs. States of Matter

• Region in which “all physical properties of a material are essentially uniform.”

Solids & Liquids

• Specific Gravity (s.g.) of solids and liquids is approximately constant @T,P

Example: Water0ºC 999.8 kg/m3100ºC 958.4 kg/m3

Error % = 4%

Example: Ice-100ºC 925 kg/m30ºC 917 kg/m3

Error % = 0.8%

In engineering, we can accept that error… just be careful!

Density of Water

Entire Phase

Solid & LiquidsExample #1

• A mixer will mix 0.3 L of W, 0.5 L of W, 0.2 L of W from different. The temperatures of water in is 20ºC, 70ºC, 99ºC.

– A) Calculate the total mass when mixing

– B) calculate the total mass when mixing assuming constant density

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• A) MT = ρ1V1+ρ2V2+ρ3V3

– MT = (ρ20ºC)(V1)+(ρ70ºC)(V2)+(ρ99ºC)(V3)

– MT = (998.2)(0.3L)+(977.8)(0.5L)+(958.4)(0.2L)

– MT = 0.29946+0.4889 + 0.1917 = 0.98006 kg

• A) MT = ρ1V1+ρ2V2+ρ3V3

– MT = (ρ20ºC)(V1)+(ρ70ºC)(V2)+(ρ99ºC)(V3)

– MT = (998.2)(0.3L)+(977.8)(0.5L)+(958.4)(0.2L)

– MT = 0.29946+0.4889 + 0.1917 = 0.98006 kg

• B) MT = ρ1V1+ρ2V2+ρ3V3

– MT = ρ(V1+V2+V3)

– MT = 1000(0.3L + 0.5L+ 0.2L)

– MT = 1 kg

ρ1+ρ2+ρ3=ρ

• A) mass flow water = 255 g/s calculate @25ºC

– ρVflow = Mflow

– Vflow = Mflow/ρ

– Vflow = (255 g/s) / (1 g/cm3)

– Vflow = 255 cm3/s

• B) mass flow water = 255 g/s calculate @75ºC

– ρVflow = Mflow

– Vflow = (255 g/s) / (0.974 g/cm3)

– Vflow = 261.8 cm3/s

• B) mass flow water = 255 g/s calculate @75ºC

– ρVflow = Mflow

– Vflow = (255 g/s) / (0.974 g/cm3)

– Vflow = 261.8 cm3/s

• Vflow = 255 cm3/s vs. Vflow = 261.8 cm3/s• Error approx 2.5%

• C) Pressure is often given as:

– 760 mm Hg @ T = 0ºC

– Why is it important to establish T?

Density of Solutions

• Solution

– Mixture of l-l; l-s; s-s

• Look for data

– Internet Data Bases

– Search for data in manuals table (Perry´s)

– Estimate by calculation!

Density of Solutions

www.engineeringtoolbox.com

Estimating Densities of solutions

• Suppose it is ideal• ρ soln = ρ1·x1 + ρ2·x2 + … + ρi·x3

• Use molar density if “x” data is given in moles

• Use mass density if “x” data is given in mass

– X1= 0.1 kg/kg X2= 0.9 kg/kg

– ρ1 = 1000 kg/m3 ρ2 = 780 kg/m3

– ρ soln = 0.1·1000 + 0.9·780 = 802 kg/m3

Need more Exercises & Problems?

• Go to www.ChemicalEngineeringGuy.com

• Section: Courses

– Mass Balance Course

• Problems Section

• You will find a problem index there…

End of Section 1

• We’ve seen:

• Theory (phases)

– A phase

– Understanding the different phases

• About and Liquids, Solids

– The advantage in constant density

– Calculating/estimating other densities

Section 2

• Gases

• Ideal Gas– Application

– Standard Conditions

– Ideal Gas Constant “R” values

– Exercises + Theory

– Partial pressure

• Real Gases– Van der Waals, Peng Robinson, Benedict-Webb-Reubin

– Z compressibility

– Corresponding states

– Reduced temperature/pressure

– Kay Rule

• Gases ARE affected by P,T

• We need an Equation of State

– Relates Variables P,V,T,n

• Relationship between P, V, T, n

– Could be simple as Ideal Gas law

– Could be as complex as many Equation of States

• We will use these equations to relate P,T,V to moles and mass for our MB!

Ideal Gas

• Simplest form to relate n, P, V, T

PV = nRT

• Good for low pressures and high temperatures

P = Pressure (absolute) [Pa]V = Volume [m3]n = moles of Gas [mol]R = Ideal Gas Constant [m3·Pa/mol-K]T = Temperature (absolute) [K]

Ideal Gas: Gas constant

• R is a constant… may be calculated for any unit:

I would learn:R = 8.314 m3·Pa / K·molR = 0.0082 L·atm / K·mol

Ideal Gas: Limitations

• High pressures

• Low Temperatures

So, in general… review every time the final density of the gas to prove it may be modeled as an ideal gas

Why Ideal Gas in MB?

• P,T may be given, calculated or estimated• P,T n, V• V n

A) Prepare a mixture @120ºC, 2.3 bar with a flow of 250 l/min

B) 5 m3 of Hydrogen gas is in a container @30ºC, 25 atm, how many moles are inside?

C) 10 kg of Nitrogen gas in a V= 20 ft3 container may only reach P=400 atm

• What is the max Temperature? Not Typical MB problem

Ideal GasExercise 1

• Choosing the ideal gas constant that best suits…

• We need to change psi to Pascals

Psi: pounds per square inch

Ideal GasExercise 1

• Criteria of 5 gmol/L for diatomic gases…

• NOTE: Error will be less than 1% if V ideal > 5L/gmol

• 72 L / 3.57 gmol = 20.2 L/gmol …

Standard Temperature & Pressure

• “Standard” values for T and P

– Every type of industry use its own

• Typical values…

– Temperature = 0ºC; 25ºC

– Pressure = 1 atm, 10^5 bars

Standard Temperature & Pressure

60F = 15ºC Approx.

77ºF = 25ºC Approx.

SCMH @Ts, Ps

• SCMH = Standard Cubic Meter per Hour

– Standard Ts, Ps

• Help us to “understand” the amount of volume involved

• Reference: Standard Conditions

• We can use these set of data to simplify calculation

SCMH @Ts, Ps

• Comparing a flow @T = 400 K and = 2 atm

• A volumetric flow @T = 273 and 1 atm

– We could compare volumetric flows @STD condition

STD Data

• 1 mol of any substance

– @P = 1 atm and T = 25ºC

– Occupies a 22.4 liters volume

SCMH @Ts,Ps

• P1·V^1 = R·T1 (1)

• Ps·V^s = R·Ts (2)

• V^s = 22.4 m3/kmol

• Ts = 0ºC 273 K

• Ps = 1 atm 101,325 Pa

Divide 2 in 1

STD known data

Relating two ideal gases conditions

• IF:

– P1V1 = n1RT1 (1)

– P2V2 = n2RT2 (2)

• Then we could divide (2) in (1)

– (P2·V2)/(P1·V1) = (n2·R·T2)/(n1·R·T1)

– (P2·V2)/(P1·V1) = (n2·T2)/(n1·T1)

– (P2·V2)/(P1·V1) = T2/T1

Since R = R

If same gas (n1 = n2)

SCMH & Standard ConditionsExercise 1

• Volumetric Flow = 328 m3/h

• We did not need “R” constant of ideal gases!

Once Again, we DON’T need the “R” constant

We will see that we actually don’t need the Temperature or Pressure given… WHY?

• Here we need T and P… why?

Ideal Gas Mixtures

• If there is A, B, C …

– na+nb+nc…

– Then na/nt = ya

• OR in the case of gases, use Partial Pressures

Ideal Mixture: Partial Pressure

• Partial pressure is the pressure that gas (A) exerts

• It is directly dependent on the mole fraction yA

ya+yb+yc = 1Pa+Pb+Pc = P

Ideal Mixture: Partial Pressure

• Pressure of each gas

Ideal Mixtures: Conclusion

• Pa = ya·P

Also useful as:

ya= Pa/P mole fraction!

Pa = Partial pressure of Aya = mole fraction of APt = Total Pressure

MB with Ideal Gases

• We now apply all

– Knowledge from MB1 (mass balance concepts)

– Theory of gases (ideal gases concepts)

• We can now solve more problems given T,P and volumetric flow of gases!

• We WILL assume the gases are IDEAL

MB with Ideal GasesExercise 1

• NOTES: – Not mass/molar flows but Volumetric!– P atm = 763 mm Hg probably in that location– Pg = g stands for gauge– Use SCMH but no T,P (we don’t need!)– Assume Ideal Gas for all Gases– Recommended to work with moles

a) Composition of stream: ya= 0.09 and yb= 0.91

Ya= 0.09

• Volumetric flow of Nitrogen at inlet of evaporator

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We´ve seen

What's left

We´ve seen

Real Gases

• Ideal gas law does not always apply for all gases– High pressures

– Low Temperatures

• This gases are called “Real Gases”

• We need to apply other type of Equation State– Virial (Virial Eqn, Benedict Weeb Reubin Eqn)

– Cubic (Van der Waals, SRK Eqn, Peng-Robinson Eqn)

– Compressibility factor “z” Point of interest!

Real Gases… Some theory

• Critical States

• Critical Temperature/Volume/Pressure

• Reduced properties

• Gas vs. Vapor

Critical Point

• Critical Points:

– occurs under conditions at which no phase boundaries exist.

– There are multiple types of critical points:• vapor–liquid

• liquid–liquid

• Critical Temperature

• Critical Pressure

• Critical Volume

Critical Point

Critical Temperature & Pressure

• The highest temperature at which a species can coexist in two phases (liquid-vapor) is called Tc

• The corresponding pressure of this point is called Pc

• Species at this point are said to be at “critical state”

Reduced Temperature & Pressure

• Variables normalized by the fluid's state properties at its critical point.

• It’s a referenced value

• The reference being the critical state

– Tr = T / Tc

– Pr = P / Pc

– Vr= V / Vc

Gas vs. Vapor

• Gas: non-condensable

• Vapor: Condensable

Gas vs. Vapor

Virial Equation

• Virial V^0; V^1; V^2; V^3 and so on…

• NOTE: Volume are shown at the left AND right

Truncated Virial Equation

• We could “simplify” the equation

• Truncating the equation to obtain ONLY one V variable would make it easier to solve

• Now… for Value of “B” we need a set of equations…

Value of B• We will need:

– Tr Temperature of gas + Critical Temperature

– Pr Pressure of gas + Critical Pressure

– Pitzer Accentric Factor (Tables)

Truncated Virial Equation

• Now just substitute “B” and find out V!

• You will probably iterate for “V”

Virial EquationExercise 1

a) P Idealb) P Virial (truncated)c) Calculate error %

Data From Tables:

• Truncated Equation

– Calculate B0

– Calculate B1

– Substitute in equation for “B”

– Calculate P from Truncated Equation

Other Equations

• This course limits only to general knowledge and application of real gases. If you are interested in the next equations:– Van der Waals

– SRK

– Peng-Robinson

– Benedict Weeb Reubin

• Visit Thermodynamics course @www.ChemicalEngineeringGuy.com

Compressibility Factor “Z”

• Pv = ZnRT

• Z: correction factor for real gases

• We now use the “Z” Factor from Graphs

Compressibility Factor “Z”Exercise 1

PV = ZnRTGet n· = PV/(ZRT)N· = (40 bar) * (50 m3/h) / (0.934)·(8.31x10^-5)·(300K)N· = 85.9 kmol/h

N· to M· (we need MW)

M· = N·x(MW) = (85.9 kmol/h)(16 kg/kmol) = 1370 kg/h

MW = 16 g/gmol

Compressibility Factor:Newton Corrections

• Many diatomic atoms may be calculated to better estimates with Newton's Correction:

Compressibility Chart

• We will be using charts to determine compressibility…

• We would need one chart for each substance!

• This is not efficient

• There is another way out!

• Use the Law of Corresponding States…

Law of Corresponding States

• All fluids, when compared at the same reduced temperature and reduced pressure, have approximately the same compressibility factor

• They all deviate from ideal gas behavior to about the same degree

• (Tr, Pr, any substance) = deviates equally

Law of Corresponding States

All substances deviate similarly @Tr, Pr

Compressibility ChartGeneralized

• We could use a Chart for generalized data

• The compressibility factor will apply for any substance

• We only need to find all Reduced Data:

– Reduced Temperature

– Reduced Pressure

Compressibility Factor Z

NOTE: Pr vs Tr

0 < Tr < 3.5

0 < Pr < 7.0

Reading the Compressibility Factor Z

NOTE: Pr vs Tr

0 < Tr < 3.5

0 < Pr < 7.0Zvalue

Pr Values

Tr Values

Compressibility Factor Z

NOTE: Pr vs Tr

0 < Tr < 3.5

0 < Pr < 7.0

Compressibility Factor ZLow Pressures

0 < Pr < 1.0

Compressibility Factor ZHigh Pressures

0 < Pr < 40

Compressibility Factor ZExercise 1

• Tr = T/Tc = (-20.6+273)/(126) = 2.0

• Pr = P/Pc = ?

• We are not able to continue… we need another equation…

• Vr = V^Pc/(R·Tc)• Vr = [(5L/100 mol)·(33.5 atm)]/[(0.082 L·atm-gmolK)(126K)]

• Vr = 0.161

• NOW find Z from chart…

Z= 1.85

Tr, Vr2.0, 1.60

Z= 1.85

Tr, Vr2.0, 1.60

Z= 1.85

Tr, Vr2.0, 1.60

• P = zRT/V^= (1.85)·(0.082 L·atm-gmolK)·(252K/0.05L/gmol)

• P = 764 atm

Mixture of Real Gases

• What about mixture of gases?• In chemical processes, its common to have

mixture of gases• We can still calculate the properties using “Kay

rule”• We just need:

– Composition (Xa, Xb, Xc…)– Tc of each substance– Pc of each substance– T,P of the gas

Real gases mixtures: Kay rule

• Pseudo-critical T

• Pseudo-critical P

• Pseudo-reduced T

• Pseudo-reduced P

• Continue as previous procedure

• Pseudo : “artificial”, “quasi”…“Almost” the Temperature and Pressure of the gas

Kay RuleExercise 1

• A) Make correction for H2

– Tc = 33 + 8 = 41 K

– Pc = 12.8+8 = 20.8 atm

Kay RuleExercise 1

• Calculate Pseudocritical Temperature

– Tc´= y-h2·Tc-h2 + y-n2·Tc-n2

– Tc´=0.75·41 + 0.25·126.2 = 62.3 K

• Calculate Pseudocritical Pressure

– Pc´= y-h2·Pc-h2 + y-n2·Pc-n2

– Pc´=0.75·20.8 + 0.25·33.5 = 24 atm

Kay RuleExercise 1

• Find, Reduced Conditions of Gas

– Tr´=T/Tc´ = 203K/62.3K =3.26

– Pr´=P/Pc´ = 800atm/24atm = 33.3

• Find Z in compressibility charts (Tr, Pr) (3.26, 33.3)

Kay RuleExercise 1

Pr= 33

Kay RuleExercise 1

Tr = 3.26

Kay RuleExercise 1

Z= 1.90

End of Section 2

• We´ve seen

– Ideal Gases

• Std. Conditions

• MB with Ideal Gases

– Real Gases

• Some equations

• Virial Equation (truncated)

• Compressibility Factor Z

End of MB2: Single-Phase MB• You should be now able to perform MB in 1 phase

• You are able to calculate MB in liquids and solids

• You are now able to:– Calculate densities of liquid and solids

– Identify Ideal gases, calculate: P,V, n, T and use R constants

– Understand the Standard Conditions and apply them to your calc.

– Know how to treat a mixture of ideal gases with Partial pressures

– Differentiate real gases

– Know there are special Equations of States for those gases

– Use the Truncated Virial Equation importance

– Understand the compressibility Factor Z and apply it

– Solve MB problems involving Gases (Ideal and Real)

Problems & Exercises

• All pair problems of Elementary Principles in Chemical Processes. Felder, R; Rousseau, R. 3rd edition are solved in the next videos. (Chapter 5)

• Remember: practice makes the master

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• Directly on the WebPage:

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Elementary Principles in Chemical Processes

What topics did we covered from the book?

Bibliography

• Elementary Principles in Chemical Processes. Felder, R; Rousseau, R. 3rd edition.

• Basic Principles and Calculation in Chemical Engineering. Himmelblau, D. 7th edition.

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