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MATERIAL BALANCE

Material Balance

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Page 1: Material Balance

MATERIAL BALANCE

Page 2: Material Balance

MATERIAL BALANCE FUNDAMENTALS

Based on the fundamental “Law of Conservation of Mass”

Chemical engineers are concerned with doing mass balances around chemical processes.

Doing mass balance is similar in principle to accounting.

Page 3: Material Balance

We must learn how to

Specify a process stream

Specify a process unit

Do a mass balance on a process unit

Do a mass balance on a sequence of process units.

Page 4: Material Balance

Classification of Processes Based on how the process varies with time.

Steady-state process is one that does not change with time. Every time we take a snapshot, all the variables have the same values as in the first snapshot.

Unsteady-state (Transient) process is one that changes with time. Every time we take a snapshot, many of the variables have different values than in the first snapshot.

Page 5: Material Balance

Classification of Processes Based on how the process was built to operate.

A Continuous process is a process that has the feed streams and product streams moving chemicals into and out of the process all the time. At every instant, the process is fed and product is produced.

A Batch process is a process where the feed streams are fed to the process to get it started. The feed material is then processed through various process steps and the finished products are created during one or more of the steps. The process is fed and products result only at specific times.

Page 6: Material Balance

The Mass Balance Equation

INPUT OUTPUT = ACCUMULATION

I O = A

For non-reactive processes

Masses entering via feed streams = Masses exiting via product streams

For reactive processes

Mass in + Mass formed by reaction = Mass out + Mass used by reaction

Page 7: Material Balance

Strategy for solving Material Balance Problems Where am I? Where am I going ? How do

I get there?

To answer the first question, you need to

Read, study and understand the problem.

Draw a flow sheet for the process.

Label it with all given information, including symbols for the unknowns

Note any special relationships.

Page 8: Material Balance

Where am I?

To accomplish this step, you need to learn

The information needed to specify a stream.

How to use symbols to represent the required stream data.

How to determine the mass of each component in a stream (each mass will be a term in a mass balance)

Page 9: Material Balance

Where am I?

Required Stream Information

Stream name & symbol (1)

Component masses/ stream composition (n)

Stream temperature and pressure (these are needed only when an energy balance is being done, phase behavior is included or to specify chemical properties.

Page 10: Material Balance

How to represent the required information

Specify each stream and total mass --- Select a stream name & symbol

Use a single Capital letter to represent the total mass( or mfr) of the stream.

Select a stream name to clearly identify the stream, by the location or purpose of the stream on the flow sheet.

Put the mass/mfr on the flow sheet using an equation symbol = value (if the mass is known) or symbol = ? (if the value is unknown)

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Example: The reactor is fed with 25 kg/s of a hot feed stream and a recycle stream. Label the reactor inputs.

Page 12: Material Balance

Component masses/ stream composition A component mass can be represented directly, using a lower case letter and a subscript, or indirectly using the fractional composition times the stream total (use and reserve x,y,z for fractional composition).

Page 13: Material Balance

Example2: 1200 kg of a mixture of O2 , N2 and CH4 are fed to a process. The stream has 20% O2 by mass. Label the stream.

Page 14: Material Balance

Where am I going?

To answer the second question, you need to

Know what is needed in the problem.

Enumerate what is required.

Analyze and relate what is being asked to what you have.

Page 15: Material Balance

How do I get there?

To answer the last question, you need to

Choose the most convenient basis.

Basis is the reference chosen for the calculation of any particular problem

It is an amount of material or time selected.

Guide in choosing a basis:

1. What do I have to start with?

2. What answer is called for?

3. What is the most convenient basis?

Changing basis / shifting of basis

Page 16: Material Balance

How do I get there?

Write down the equations to be used

A mass balance can be written using the total mass in each process stream. This is called a total balance or overall material balance.

A separate mass balance can be written for each chemical component involved. These are called component balances.

A component that is present in only one stream and one exit stream is called tie component.

Page 17: Material Balance

MIXING PROBLEM

F1, F2 = FEEDS

P = PRODUCT

MIXER

F1

F2 P

Page 18: Material Balance

PROBLEM

A liquid adhesive consist of a polymer dissolved in a solvent. The amount of polymer in the solution is important to the application. An adhesive dealer receives an order for 3000 pounds of an adhesive solution containing 13% polymer by weight. On hand is 500 lbs of 10% solution and very large quantities of 20% solution and pure solvent. Calculate the weight of each that must be blended together to fill the order. Use all of the 10% solution

Page 19: Material Balance

A lacquer plant must deliver 1000 lb of an 8% nitrocellulose solution. They have stock a 5.5% solution. How much dry nitrocellulose must be dissolved in the solution to fill the order

Page 20: Material Balance

DRYING

Process of removing water by vaporizing the water.

F – feed

W – water removed

P – product

DRYER

W

F P

Page 21: Material Balance

PROBLEM

Sludge is wet solids that result from the processing in municipal sewage systems. The sludge has to be dried before it can be composted or otherwise handled. If a sludge containing 70% water and 30% solids is passed through a drier, and the resulting product contains 25% water, how much water is evaporated per ton of sludge sent to the drier.

Page 22: Material Balance

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?

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EVAPORATION Separation of a volatile solvent from a non-

volatile solute

F – feed, dilute solution

E – evaporate, pure solvent

C – concentrate

EVAPORATOR F

E

C

Page 24: Material Balance

MULTI-EFFECT EVAPORATOR

E V A P O R A T O R

F C1 C2

E1

E V A P O R A T O R

E2

Page 25: Material Balance

PROBLEM

A multiple stage evaporator concentrates a weak NaOH solution from 3% to 18%, and processes 2 tons of solution per day. How much product is made per day? How much water is evaporated per day?

Page 26: Material Balance

1000 lb of solution containing 80% wt Na2SO4 must be obtained by evaporating a dilute solution using a double effect evaporator. If the evaporate from the 2nd effect evaporator is 60% of that coming from the 1st effect contain 1lb Na2SO4/1lb H2O. Calculate E1 and E2

Lbs of Feed

Components of Feed

Page 27: Material Balance

DISTILLATION Process whereby mixtures of liquids are

subjected to repeated partial vaporizations and condensations in order to separate them into more or less pure components

F – feed (liquid mixtures)

D – distillate/overhead product

R – residue/bottom product

F

R

D

Page 28: Material Balance

% RECOVERY

= amount of desired substance obtained x 100

total amount of desired substance feed

PROBLEM

100 lb of mixture containing 20%wt of benzene (C6H6), 50%wt toluene (C7H8), and xylene (C8H10) is to be separated into a distillate containing among others 1%xylene and a bottom product containing 1% benzene. Benzene recovery in the distillate is 98%. Calculate the amount of D and B as well as the percentage composition of both.

Page 29: Material Balance

FILTRATION Recovery of a solid material (insoluble or

slightly soluble) from a mixture by using a filtering medium

F – feed (solid and liquid mixture)

C – filter cake (always wet w/ filtrate)

L – filtrate (pure liquid, saturated solution)

FILTER F C

L

Page 30: Material Balance

PROBLEM

100 kg of a slurry of insoluble material containing 60%H2O is filtered per hour. If the final filter cake contain 25% H2O, how long will it take to obtain 300 gallons of filtrate

Page 31: Material Balance

CRYSTALIZATION

Process of solidifying a dissolved substance in a certain solution by either lowering the temperature or removing some of the solvent

F – feed (saturated/unsaturated solution)

ML – mother liquor (always saturated solution)

C – crop/crystal (pure crystal, hydrated crystal)

Page 32: Material Balance

COOLER F ML

C

F ML

C

E

Page 33: Material Balance

PROBLEM

A chemist attempts to prepare some very pure crystals of borax (sodium tetraborate, Na2B4O7 . 10H2O) by dissolving 100g of Na2B4O7 in 200g of boiling water. He then carefully cools the solution slowly until some Na2B4O7 . 10H2O crystallizes out. Calculate the weight of Na2B4O7 . 10H2O recovered in the crystals per 100g of total initial solution(Na2B4O7 plus 10H2O), if the residual solution at 550C after the crystals are removed contains 12.4% Na2B4O7.

Page 34: Material Balance

EXTRACTION Removal of desired substance from a solid or

solution with or without the use of solvent.

TYPES

Expression - mechanical

F – solid E – extract

R – residue

Page 35: Material Balance

Leaching

Liquid-liquid extraction

F – solid E – extract with solvent

R – residue

S – solvent

F – solution E – extract with solvent

R – raffinate

S – solvent

Page 36: Material Balance

PROBLEM Suppose that 100 L/min are drawn from a fermentation

tank and passed through an extraction tank in which the fermentation product (in the aqueous phase) is mixed with an organic solvent, and then the aqueous phase is separated from the organic phase. The concentration of the desired enzyme (3-hydroxybutyrate dehydrogenase) in the aqueous feed to the extraction is 10.2 g/L. The pure organic extraction solvent runs into the extraction tank at a rate of 9.5 L/min. If the ratio of the enzyme in the exit product stream (the organic phase) from the extraction tank to the concentration of the enzyme in the exit waste stream (the aqueous phase) from the tank is D=18.5(g/L organic)/(g/L aq), what is the fraction recovery of the enzyme and the amount recovered per min? Assume negligible miscibility between the aqueous and organic liquids in each other, and ignore any change in density on removal or addition of enzyme to either stream.