MASS BALANCE

DICTATE

EVA FATHUL KARAMAH

CHEMICAL ENGINEERING DEPARTMENT UNIVERSITY OF INDONESIA

2007

Units And Dimensions

Add the following:

a. 1 foot + 3 seconds = b. 1 horsepower + 300 watts = c. 1 foot x 3 seconds =

what is your conclusion?

- You can add, subtract, or equate numerical quantities only if dimensions of the of the quantities are the same. And only after the units are transformed to be the same.

- You can multiply or divide unlike units or dimensions. But you cannot cancel or merge units unless they are identical.

SO, DIMENSIONS are our basic concept of measurement in term of physical quantity, such as length, mass, time, temperature, and so on. UNITS are the means of expressing the dimensions, such as feet and centimeters for length, or kilograms and pounds for mass. Table 1. SI Units

Physical Quantity Name of Unit Symbol Definition

Basic S1 Units Length meter m Mass kilogram kg Time second s Temperature kelvin K Amount of substance mole mol

Derived S1 Units Energy joule J kg ⋅ m2 ⋅ s-2 Force newton N kg ⋅ m ⋅ s-2 → J ⋅ m-1 Power watt W kg ⋅ m2 ⋅ s-3 → J ⋅ s-1 Density kilogram per cubic meter kg ⋅ m-3 Velocity meter per second m ⋅ s-1 Acceleration meter per second squared m ⋅ s-2 Pressure newton per square meter,

pascal N ⋅ m-2, Pa

Heat capacity joule per (kilogram.kelvin) J ⋅ kg-1 ⋅ K-1

Alternative Units Time minute, hour, day, year min, h, d, y Temperature degree Celcius OC Volume liter (dm3) L Mass ton (Mg), gram t, g Table 2. American Engineering SystemUnits

Physical Quantity Name of Unit Symbol Basic Units

Length feet ft Mass pound (mass) lbm Force pound (force) lbf Time second, hour s, hr Temperature degree Rankine OR

Derived Units Energy British thermal unit, foot pound (force) Btu, (ft)(lbf) Power horsepower hp Density pound (mass) per cubic feet lbm/ft3 Velocity feet per second ft/s Acceleration feet per second squared ft/s2

Pressure pound (force) per square inch, pascal lbf/in.2 Heat capacity Btu per pound (mass) per degree F Btu/(lbm)(OF) MOLE UNIT Mole: certain numbers of molecules, atoms, electrons, or other spesified types of particles.

amount of a substance that contains as many elementary entities as there are atoms in 0.012 kg of carbon 12. (=6.02 x 1023 molecules, in SI; non standard: poundmole = 6.02 x 1023 x 453.6 molecules; kilomole, kmol=1000 moles)

weightmolecularg in mass mol g =

weightmolecularlb in mass mol lb =

mass in g = (mol wt) (g mol) mass in lb = (mol wt) (lb mol)

DENSITY Density: ratio of mass per unit volume Density of liquids and solids do not change significantly at ordinary conditions with pressure, but they do change with temperature. Density varies with composition. SPECIFIC GRAVITY Specific Gravity: ratio of a substance of interest to that of a reference substance. Dimensionless. Reference substance for liquids and solids is normally water. While for gas frequently is air, but may be other gases. State the temperature at which each density is chosen.

substance reference of temp solution of temp

4200.73 gr sp o

o

→→

=

Specific gravity in the petroleum industry: oAPI

5.131

6060 gr sp

141.5API

o

o −=o

131.5 API5.141

6060 gr sp o

o

+= o

Example 1. Application of Specific Gravity

In the production of a drug having a molecular weight of 192, the exit stream from the reactor flows at the rate of 10.3 L/min. The drug concentration is 41.2% (in water), and the specific gravity of the solution is 1.025. Calculate the concentration of the drug (in kg/L) in the exit stream, and the flowrate of the drug in kg mol/min.

Solution

0.412 kg drug 0.588 kg water s.g. = 1.025 10.3 L/min

Reactor

Basis: 1.00 kg solution

Density of the solution = 1.025 x 1 g/cm3 = 1.025 g/cm3 The amount of drug in the solution

= soln drug/L kg 422.0L 1cm10

g 1000kg 1

cmsoln g025.1

soln kg 00.1drug kg 412.0 33

3 =

To get the flowrate, we take a different basis Basis: 1 minute ≡ 10.3 L solution Molar flowrate of the drug

= 10.3 L soln/min mol/min kg 0226.0drug kg 192drug mol kg 1

soln L 1drug kg 422.0

=

SPECIFIC VOLUME The inverse of the density, that is, the volume per unit mass of unit amount of material. MOLE FRACTION AND MASS (WEIGHT) FRACTION MOLE FRACTION: the moles of a particular substance divided by the total number of moles present.

moles total Aof moles A of fraction mole =

MASS (WEIGHT) FRACTION: the mass (weight) of a substance divided by the total mass (weight) of al the substances present.

(weight) mass total Aof (weight) mass A of fraction (weight) mass =

Mole percent and weight percent are the respective fractions times 100. The composition of gases will be presumed to be given in mole percent or fraction unless specifically stated otherwise. The analysis of liquids and solids will be assumed to be weight percent or fraction unless specifically stated otherwise. CONCENTRATION Concentration: the quantity of some solute per specified amount of solvent, or solution, in a mixture of two or more components. Can be expressed as:

1. Mass per unit volume (lbm of solute/ft3, g of solute/L, lbm of solute/bbl, kg of solute/m3) 2. Moles per unit volume (lb mol of solute/ft3, g mol of solute/L, g mol of solute/cm3) 3. Part per million (ppm); parts per billion (ppb) – for extremely dilute solutions. Ppm is equivalent

to a weight fraction for solid and liquids; it is a mol fraction for gases. 4. Molarity (g mol/L) and normality (equivalents/L)

Example 2. Use of ppm

The current OSHA 8 hour limit for HCN in air is 10.0 ppm. A lethal dose of HCN in air is 300 mg/kg of air at room temperature. How many mg HCN/kg air is the 10.0 ppm? What fraction of the lethal dose is 10.0 ppm?

Solution Basis: 1 kg mol of the air/HCN mixture

air gmol10HCN mol g 10.0

HCN) air( gmol10HCN mol g 10.0ppm 0.10 66 =+

=

Amount of HCN (in mg) per kg of air

= air HCN/kg mg 32.9air kg 1

air g 1000HCN g 1

HCN mg 1000air g 29air mol g 1

HCN mol g 1HCN g 27.03

air mol g 10HCN mol g 10.0

6 =

Lethal dose = 300 mg/kg of air 10.0 ppm = 9.32 mg/kg of air

031.0300

32.9dose lethalppm 10.0

==

BASIS The basis is the reference chosen for the calculations i any particular problems. A proper choice of basis frequently makes the problem much easier to solve. For selecting a suitable basis, ask yourself the following questions:

1. What do I have to start with? 2. What answer is called for? 3. What is the most convinient basis to use?

Example 3. Choosing a basis

Most processes for producing high-energy-content gas or gasoline from coal include some type of gasification step to make hydrogen or synthesis gas. Pressure gasification is preferred because of its greater yoeld of methane and higher rate of gasification. Given that a 50.0-kg test run of gas averages 10.0% H2, 40.0% CH4, 30.0% CO, and 20.0% CO2. What is the average molecular weight of the gas?

Solution Basis: 100 kg mol of gas Component percent = kg mol Mol. Weight kg

CO2 20.0 44.0 880 CO 30.0 28.0 840 CH4 40.0 16.04 642 H2 10.0 2.02 20

Total 100.0 2382

Average molecular weight = mol kg/kg 8.23mol kg 100kg 2382

=

TEMPERATURE Temperature scale: Celcius (oC), Kelvin (oK), Fahrenheit (oF), Rankine (oR)

273+=CK oo TT

460+=FR oo TT

1.8 32 ×=−CF oo TT

Δ oC = Δ oK Δ oF = Δ oR

FCFC o

o

o

Δ=Δ=ΔΔ 8.1or8.1 o

RKRK o

o

o

Δ=Δ=ΔΔ 8.1or8.1 o

PRESSURE CHEMICAL EQUATION AND STOICHIMETRY Chemical equation provides both qualitative and quantitative informations essential for the calculation of the combining moles of materials involved in a chemical process. Stoichiometry: quantitative relationship between reactants and products. Stoichiometric ratios: ratios obtained from the numerical coefficients in the chemical equation. Example 4. Stoichiometry

A limestone analysis: CaCO3 92.89% MgCO3 5.41% Insoluble 1.70%

a. How many kilograms of calcium oxide can be made from 5 tons of this limestone? b. How many kilograms of CO2 can be recovered per kilogram of limestone? c. How many kilograms of limestone are needed to make 1 ton of lime?

Solution:

Draw a process diagram

CO2

CaO MgO Insoluble

lime

limestone

heat

Lime includes all the impurities present in the limestone that remain after the CO2 has been driven off. Chemical reactions involved: CaCO3 → CaO + CO2 (1) MgCO3 → MgO + CO2 (2) Additional data:

CaCO3 MgCO3 CaO MgO CO2 Mol. Wt. 100.1 84.32 56.08 40.32 44.0

Basis: 100 kg of limestone

Limestone Products Component kg = percent kg mol Solid

Component kg CO2

CaCO3 92.89 0.9280 CaO 52.04 40.83 MgCO3 5.41 0.0642 MgO 2.59 2.82 Insoluble 1.70 Insoluble 1.70 Total 100.00 0.9920 56.33 43.65 The quantities of products are calculated from the chemical equation. CaO produced → rxn (1):

= kg 04.52CaO mol kg 1CaO kg 08.56

CaCO mol kg 1CaO mol kg 1

CaCO kg 100.1CaCO kgmol 1CaCO kg 89.92

33

33 =

MgO produced → rxn (2):

= kg 59.2MgO mol kg 1MgO kg 0.324

MgCO mol kg 1MgO mol kg 1

MgCO kg 84.32MgCO kgmol 1MgCO kg .415

33

33 =

CO2 is produced from reaction (1) and reaction (2):

rxn (1): kg 83.40CO mol kg 1CO kg 4.04

CaCO mol kg 1CO mol kg 1

CaCO kg 100.1CaCO kgmol 1CaCO kg 89.92

2

2

3

2

3

33 =

rxn (2): kg 82.2CO mol kg 1CO kg 4.04

MgCO mol kg 1CO mol kg 1

MgCO kg 84.32MgCO kgmol 1MgCO kg .415

2

2

3

2

3

33 =

Total CO2 produced = (40.83 + 2.82) kg = 43.65 kg Total mass product = 100 kg = amount of limestone entering the process. Calculate the quantities originally asked for: a. CaO produced from 5 tons of limestone

kg 2602limestone tons 5ton 1

kg 1000limestone kg 100

CaO kg 52.04==

b. CO2 recovered per kilogram of limestone

kg 437.0limestone kg 100

CO kg 65.43 2 ==

c. Limestone required to make 1 ton of lime

limestone kg 25.1775ton 1

kg 1000lime kg 56.33

limestone kg 100==

In industrial reactors, exact stoichiometric amounts of material used is rarely find. To make a desired reaction take place or to use up a costly reactant, excess reactants are nearly always used. This excess material comes out together with, or perhaps separately from , the product – and sometimes can be used again. Some new definitions must be understood: 1. Limiting reactant: reactant that is present in the smallest stoichiometric amount or reactant that

would first disappear if the reaction were to proceed according to the chemical equation to completion. Determination: calculate the mole ratio(s) of the reactants in actual feed and compare each ratio with the corresponding ratio of the coefficients in the chemical equation.

2. Excess reactant: reactant present in excess of the limiting reactant.

Percent excess: based on the amount of any excess reactant above the amount required to react with the limiting reactant according to the chemical reaction.

100reactant limiting react with to required moles

excess in moles excess % ×=

moles in excess = total available moles – moles required to react with the limiting reactant Excess air: the amount of air available to react that is in excess of the air theoretically required to completely burn. Even if only part of the limiting reactant actually reacts, the required and excess quantities are based on the entire amount of the limiting reactant as if it had reacted completely.

3. Conversion: the fraction of the feed or some key material in the feed that is converted into products.

introduced feed) the in compound a (or feed of (mass) molesreactthat feed) the in compound a (or feed of (mass) molesconversion % =

Degree of completion: fraction of the limiting reactant converted into products.

4. Selectivity: ratio of the moles of a particular (usually the desired) product produced to the moles of another (usually undesired or by-) product produced in a set of reactions.

5. Yield: mass or moles of final product divided by the mass or moles of initial or key reactant either

fed or consumed. → for a single reactant and product If more than one product and more than one reactant are involved, the reactant upon ehich the yield is to be based must be clearly stated.

Yield and Selectivity : measures the degree to which a desired reaction proceeds relative to competing alternative (undesirable) reactions. Example 5. Incomplete reaction

Antimony is obtained by heating pulverized stibnite (Sb2S3) with scrap iron and drawing off the molten antimony from the bottom of reaction vessel. Sb2S3 + 3Fe → 2Sb + 3FeS

Suppose that 0.600 kg of stibnite and 0.250 kg of iron turnings is heated together to give 0.200 kg of Sb metal. Determine: a. The limiting reactant b. The percentage of excess reactant c. The degree of completion (fraction) d. The percent conversion e. The yield

Solution: Process diagram:

REACTOR

1.77 g mol Sb2S3

4.48 g mol Fe

1.64 g mol Sb

FeS

The g mol is calculated from molecular weight data:

Component kg mol.wt. g mol Sb2S3 0.600 339.7 1.77 Fe 0.250 55.85 4.48 Sb 0.200 121.8 1.64 FeS 87.91

a. Examine ratio of Fe

SSb 32

actual/available stoichiometric

FeSSb 32

1.77/4.48 = 0.40 1/3 = 0.33

actual > stoichiometric → Sb2S3 : excess reactant, Fe : limiting reactant. Sb2S3 required to react with limiting reactant = 4.48/3 g mol = 1.49 g mol.

b. Percentage of excess reactant:

32Sb excess %8.181001.49

1.49 - 1.77 excess % S=×=

c. Not all of limiting reactant reacts. The amount of Fe actually does react is computed from the amount of Sb produced:

Fe mol g 2.46 Sb gmol 2Fe mol g 3Sb mol g 64.1 =

degree of completion = 48.446.2 = 0.55

d. Conversion → refres to Sb2S3.

The amount of Sb2S3 actually react = Sb mol g 2

Sb mol g 1Sb mol g 64.1 32S = 0.82 g mol Sb2S3

% conversion of Sb2S3 to Sb = 10077.182.0

× = 46.3%

e. Yield: 32323232 SSb kg Sb kg 33.0

SSb kgSb kg

31

SSb kg 0.600Sb kg 200.0

introduced SSb of kgformed Sb of kg

===

EXERCISES: 1. Write balanced equations for the following reaction:

a. C9H18 and oxygen to form carbon dioxide and water b. FeS2 and oxygen to form Fe2O3 and sulfur dioxide

2. The electrolytic manufacture of chlorine gas from a sodium chloride solution is carried out by the

following reaction: 2NaCl + 2H2O → 2NaOH + H2 + Cl2

How many kilograms of Cl2 can one produce from 10 m3 of a brine soluiton containing 5% by weight of sodium chloride? The specific gravity of the solution relative to water at 4OC is 1.07.

3. Calcium oxide (CaO) is formed by decomposing limestone (pure CaCO3). In one kiln the reaction goes to 70% completion.

a. What is the composition of solid product withdrawn from the kiln? b. What is the yield in terms of pounds of CO2 produced per pound of limestone charged?

MATERIAL BALANCE

Material balance → application of the conservation law for mass: “matter is neither created nor destroyed”. To make a material balance of a process, first specify the system and outline its boundaries. A Process: one or a series of actions or operations or treatments that result in an end [product]. System: any arbitrary portion or whole of a process set out specifically for analysis. Open (Flow) System: system in which material is transferred across the system boundary, that is, enters the system, leaves the system, or both. Closed (batch) system: system in which there is no such transfer during the time interval of interest. Concept of material balance:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−

⎪⎭

⎪⎬

⎫

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⎧+

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system thewithin

nconsumptio

system thewithingeneration

boundariessystem

throughoutput

boundariessystem

throughinput

system the within

onaccumulati (1)

Accumulation: a change in mass or moles (plus or minus) within the system with respect to time Transfer through system boundaries: inputs to and outputs of the system Generation and consumption within the system: related to product and reactant of the reaction in the system Steady-state: the values of the variables within the system do not change with time, accumulation is zero Unsteady-state: the values of the variables within the system change with time

Flow in Flow in Flow in Flow in

positive accumulation

Flow out Flow out Flow out Flow out

negative accumulation If the system is steady state and there is no reaction occurs, eq. (1) becomes:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

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boundaries systemthe through

output mass/mole

boundaries systemthe through

input mass/mole (2)

Example 1. Total Mass Balance

A thickener in a waste disposal unit of a plant removes water from wet sewage sludge as shown in figure E.1. How many kilograms of water leave the thickener per 100 kg of wet sludge that enter the thickener? The process is in steady state.

dehydrated sludge Thickener

100 kg 70 kg

wet sludge

water = ?

Fig. E.1.

Solution:

Basis: 100 kg wet sludge The system is the thickener (open system), no accumulation, generation or consumption occurs. The total mass balance is:

in = out 100 kg = 70 kg + kg of water

So the amount of water is (100 – 70) kg = 30 kg

ANALYSIS OF MATERIAL BALANCE PROBLEMS Total mass (mole) balance: balance of total mass (mole) all component in or out of the system Component mass balance: balance of a specific component in a system Expressing flows of component as variables for a single component in a mixture:

1. mass (molar) flow, mi (ni), i = specific component 2. product of measure of concentration times the flow, xiF, xi is the mass (mole) fraction of

component i in F, and F is the total mass (molar) flow

In any mixture of N components, N stream variables exist, either N values of mi aor ni, namely (N – 1) values of xi, plus the stream flow itself, F. Because ∑ = 1ix . Strategy for analyzing material balance problems:

1. Read the problem and clarify what is to be accomplished. 2. Draw a sketch of the process; define the system by a boundary. 3. Label with symbols the flow of each stream and the associated compositions and other

information that is unknown. 4. Put all the known values of compositions and stream flows on the figure by each stream;

calculate additional compositions and flows from the given data as necessary. Or, at least initially identify the known parameters in some fashion.

5. Select a basis. 6. Make a list by symbols for each of the unknown values of the stream flows and compositions, or

at least mark them distinctly in some fashion, and count them. 7. Write down tha names of an appropriate set of balances to solve; write the balances down with

type of balance listed by each one. 8. Count the number of independent balances that can be written; ascertain that a unique solution

is possible. If not, look for mare information or check your assumptions. 9. Solve the equations. Each calculation must be made on a consistent basis. 10. Check your answer by introducing them, or some of them, into any redundant material

balances.

SOLVING MATERIAL BALANCE PROBLEMS THAT DO NOT INVOLVE CHEMICAL REACTIONS Without any reactions in the system, the balance equations will be:

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⎧=

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boundariessystem

throughoutput

boundariessystem

throughinput

system the within

onaccumulati

Example 2. Membrane Separation

Membrane technology can be used in the separation of nitrogen and oxygen from air. Figure E.2 illustrates a nanoporous membrane for such ??? What is the composition of the waste stream if the stream amounts to 80% of the inputs?

21% O2

High-pressure side

Low-pressure side

Membrane

Input Output

79% N2

O2 25%

N2 75%

Waste Stream

O2 N2

Fig. E2a

The process is steady state and without chemical reaction, so eq. (2) is used to solve the problem. The system is the membrane. Let

2Ox be the mole fraction of oxygen, and 2Nx be the

mole fraction of nitrogen, and let 2On and

2Nn be the respective moles.

MEMBRANE F (g mol) P (g mol)

W (g mol) x . O2 0.21 N2 0.79

x . O2 0.25 N2 0.75

x n . O2 xO2 nO2

N2 xN2 nN2

Fig. E2b

Step 1, 2, 3 and 4. All of the data and symbols have been placed in Fig. E2b Step 5. Pick a convenient basis.

Basis: 100 mol F. Then we know, W = 0.80 (100) = 80 mol

Step 6. Three unknown exist: P, 2Ox , and

2Nx or P, 2On , and

2Nn Step 7. Two independent balances are the oxygen and nitrogen balances. The third independent balance is

2Ox + 2Nx = 1.00; or

2Nn + 2On = 80.

Step 8 and 9. The component balances are: In Out In Out O2: 0.21(100) = 0.25P +

2Ox (80) or 0.21(100) = 0.25P + 2On

N2: 0.79(100) = 0.75P + 2Nx (80) or 0.79(100) = 0.75P +

2Nn 1.00 =

2Ox + 2Nx

2On +2Nn = 80

The solution of these equations is 2Ox = 0.20,

2Nx = 0.80, and P = 20 mol. A simpler calculation involves the use of the total balance and one component balance

F = P + W or 100 = P + 80, → P = 20 Step 10. Check. Use the total balance as a check on the solution.

F = P + W ?

100 = 20 + 80

Example 3. Continous Distillation A novice manufaturer of alcohol for gasohol is having a bit of difficulty with a distillation column. The operation is shown in Fig. E3. Technicians think too much alcohol is lost in the bottoms (waste). Calculate the composition of the bottoms and the mass of the alcohol lost in the bottoms.

Distillation Column

Heat Exchanger

cooling water

Reflux

Vapor

Heat

System Boundary

Distillate (product) P 60% EtOH 40% H2O Wt = 1/10 feed

Bottoms (waste) B EtOH ? H2O ?

1000 kg Feed, F 10% EtOH 90% H2O

Fig. E3

Solution

Assume the system is steady state, no reaction occurs. So, In = Out Let x designate the mass fraction. Basis: F = 1000 kg of feed Given: P = 1/10 of F → P = 0.1(1000) = 100 kg The remaining unknowns: xEtOH,B , xH2O,B , and B. Two components: two independent component mass balances xEtOH,B + xH2O,B = 1 Total mass balance: F = P + B B = 1000 – 100 = 900 kg Component balances:

kg feed in - kg distillate out = kg bottoms out % EtOH balance: 0.10(1000) - 0.60(100) = 40 4.4 H2O balance: 0.90(1000) - 0.40(100) = 860 95.6 900 100.0 Check: Use the total balance to calculate B, and EtOH component balance to calculate mEtOH,B as 40 kg → mass H2O in B = 900 – 40 = 860 kg

Example 4. Mixing

Dilute sulfuric acid has to be added to dry charged batteries at service stations to activate a battery. You are ask to prepare a batch of new 10.63% acid as follow. A tank of old weak battery acid (H2SO4) solution contains 12.43% H2SO4 ( the remainder is pure water). If 200 kg of 77.7% H2SO4 is added to the tank, and the final solution is to be 18.63% H2SO4, how many kilograms of battery acid have been made? See Figure E4.

H2SO4 77.7% H2O 22.3%

200 kg Added Solution, A

Original Solution, F kg

H2SO4 12.43% H2O 87.57%

H2SO4 18.63% H2O 81.37%

Final Solution, P kg

system

Fig. E4

Solution

No reaction. Assume steady state → in = out Basis: 200 kg A

unique solution

Total balance: A + F = P 200 + F = P (1) H2SO4 balance: 200(0.777) + F(0.1243) = P(0.1863) (2) Substitute (1) to (2) 200(0.777) + F(0.1243) = (200 + F)(0.1863)

155.4 + 0.1243F = 37.26 + 0.1863F 118.14 = 0.062F F = 1905.5 kg

Use eq. (1) to find P P = (200 + 1905.5) kg = 2105.5 kg

Example 5. Crystallization

A tank holds 10,000 kg of a saturated solution of Na2CO3 at 30OC. You want to crystallize from this solution 3000 kg of Na2CO3 . 10H2O without any accompanying water. To what temperature must the solution be cooled?

Solution

No reaction. Unsteady state. Difficulties: information about composition of the solutions and solid precipitate. Calculation of final concentration of Na2CO3 → corresponding temperature from handbook containing solubility data. Diagram of the process:

Na2CO3

H2O

Saturated Solution

30OC

10,000 kg

Initial State

Na2CO3

H2O

Saturated Solution

T = ?

Final State

System Boundary System Boundary

Na2CO3 . H2O 3000 kg

Crystals Removed

Fig. E5a

Component of the system: Na2CO3 and H2O. Find the compositions of the streams for each solution and solid crystals of Na2CO3 . 10H2O Use solubility data for Na2CO3 as a function of temperature:

Temp (OC) Solubility (g Na2CO3/100 g H2O)

0 7 10 12.5 20 21.5 30 38.8

Because the initial solution is saturated at 30OC, we can calculate the composition of the initial solution:

32232

32 CONa fraction mass 280.0OH g 100CONa g 38.8

CONa g 38.8=

+

Calculate the composition of the crystals. For 1 gmol Na2CO3 . 10H2O Comp. mol mol. wt. mass mass fr. Na2CO3 1 106 106 0.371

H2O 10 18 180 0.629 Total 286 1.000

Select a basis. Basis: 10,000 kg of saturated solution at 30OC. Enter the known data to process diagram:

10,000 kg Na2CO3 0.280 H2O 0.720

Initial State Final State

System Boundary System Boundary

Crystals Removed

P = ? kg Na2CO3 mNa2CO3 H2O mH2O

3,000 kg Na2CO3 0.371 H2O 0.629

Fig. E5b

System: unsteady state, transport in = 0, so

accumulation = -out two components → two independent mass balance eq. three unknowns:

3CONa2m , OH2

m , P third independent balance eq. :

3CONa2m + OH2

m = P → unique solution The total and component balances (only 2 are independent)

Accumulation in Tank Final - Initial = Transport out

Na2CO3 3CONa2

m - 10,000(0.280) = -3,000(0.371) H2O OH2

m - 10,000(0.720) = -3,000(0.629) Total P - 10,000 = -3,000

The solution for the unknowns:

Component kg

3CONa2m 1687

OH2m 5313

P (total) 7000 Check on total: 7,000 + 3,000 = 10,000 To find the temperature of the final solution, calculate the composition of the final solution in terms of grams of Na2CO3 /100 grams of H2O

OH g 100CONa g 8.31

OH g 5,313CONa g 1,687

2

32

2

32 =

By linear interpolation, the temperature to which the solution must be cooled is:

( ) CCC OOO 26105.218.388.318.3830 =

−−

−

PROBLEMS

1. A cereal product containing 55% water is made at the rate of 500 kg/hr. You need to dry the product so that it contains only 30% water. How much water has to be evaporated per hour?

2. To prepare a solution of 50.0% sulfuric acid, a dilute waste acid containing 28.0% H2SO4 is fortified with a purchased acid containing 96.0% H2SO4 . How many kilograms of the purchased acid must be bought for each 100 kg of dilute acid?

3. An aqueous etching solution containing 8.8% KI is to be prepared to etch gold in printed circuit boards. The desired solution is to be formed by combining a strong solution (12% KI and 3% I2 in H2O) with a weak solution (2.5% KI and 0.625% I2 in H2O)

a. What should be the value of R, the ratio of the weights of the strong to the weak solution, to make up the desired etching solution? What will be the concentration of I2 in the final solution?

b. Note that you cannot independently vary the concentration of both KI and I2 in the final mixture simply by varying the value of R. Derive a relationship between the weight fraction of KI and the weight fraction of I2 in the mixture to illustrate this point.

4.

SOLVING MATERIAL BALANCE PROBLEMS INVOLVING CHEMICAL REACTIONS The balance equation applied:

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Generation and consumption terms can come into play in making component mole balance. Often, in component and total balances, the moles will not necessarily balance unless the generation and consumption terms are taken into account. Example of material balance with chemical reactions : combustion problems. Special terms in combustion problems:

1. Flue or stack gas: All the gases resulting from a combustion process including the water vapor, sometimes known as wet basis.

2. Orsat analysis or dry basis: All the gases resulting from a combustion process not including the water vapor.

3. Theoretical air (theoretical oxygen): the amount of air (oxygen) required to be brought into the process for complete combustion. Sometimes this quantity is called the required air (or oxygen).

4. Excess air (or excess oxygen): the amount of air (or oxygen) in excess of that required for complete combustion. The calculated amount of excess air does not depend on how much material is actually burned but what can be burned. Even if only partial combustion takes place, the excess air (or oxygen) is computed as if the process of combustion produced only CO2.