mb2 single phase mass balance

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Mass Balance: Single-Phase

MB2

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

• 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









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

vs.

• 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?

SCMH



• 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



n1

n4n2

n3










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










Ya= 0.09



• Volumetric flow of Nitrogen at inlet of evaporator

Nitrogen 1st streamRestricted Content for Slide Share Users!








Break

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



Vapor


Gas vs. Vapor



Vapor

Gas


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



• 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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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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