fundamentals of material balancesshoukat.buet.ac.bd/che111_ch4_part1.pdf · fundamentals of...

Material Balance-Part 1

Chapter 4

Fundamentals of Material Balances

Process Classifications 3 type of chemical processes: -  Concept of boundary of the process 1.  Batch process

–  Feed is charge to the process and product is removed when the process is completed

–  No mass is fed or removed from the process during the operation

–  Used for small scale production –  Operate in unsteady state

n 3/29/15 n Prof. Shoukat Choudhury n 2

Process Classifications 2.  Continuous process

–  Input and output is continuously fed and remove from the process

–  Operate in steady state –  Used for large scale production

3.  Semibatch process –  Neither batch nor continuous –  During the process a part of reactant can be

fed or a part of product can be removed.


2 type of process operations: 1. Steady state

–  All the variables (i.e. temperatures, pressure, volume, flow rate, etc) do not change with time

2. Unsteady state or transient

–  Process variable change with time


Define type and operation of processes given:

•  A balloon is being filled with air at steady rate of 2 g/min Semibatch and unsteady state

•  A bottle of milk is taken from the refrigerator and left in the kitchen Batch and unsteady state

•  Water is boiled in an open flask Semibatch and unsteady state


Try This…

Balances on Continuous Steady-State Process •  Steady state;

accumulation = 0 INPUT + GENERATION = OUTPUT + CONSUMPTION

•  If balance on steady-state non-reactive processes;

generation = 0, consumption = 0 INPUT = OUTPUT


General Balance Equation: INPUT + GENERATION – OUTPUT – CONSUMPTION = ACCUMULATION

Example 4.2-‐2

One thousands kilogram per hour of a mixture of Benzene (B) and Toluene (T) containing 50% Benzene by mass is separated by a disEllaEon into two fracEons. The mass flow rate of Benzene in the top stream is 450 kg/h and that of Toluene in the boIom stream is 475 kg/h. The operaEon is at steady state. Calculate the unknown components flow rates in the output streams. n 3/29/15 n Prof. Shoukat Choudhury n 7

Flowchart



1. Learn how to organize information about process variables

- Flow Chart drawing and Labeling 2. Choose a Basis 3. Set up material balance equations 4. solve the equations for unknown variables.

Steps for Material Balance Calculations:

Flowcharts

•  When you are given process information and asked to determine something about the process, it is essential to organize the information in a way that is convenient for subsequent calculations.

•  The best way to do this is to draw a flowchart –  using boxes or other symbols to represent process

units (reactors, mixers, separation units, etc.) –  lines with arrows to represent inputs and outputs.


Flowcharts……

•  The flowchart of a process can help get material balance calculations started and keep them moving.

•  Flowchart must be fully labeled when it is first drawn, with values of known process variables and symbols for unknown variables being written for each input and output stream.

•  Flowchart will functions as a scoreboard for the problem solution: as each unknown variable is determined its value is filled in, so that the flowchart provides a continuous record of where the solution stands and what must still be done.


Step 1.2: Labeling a flowchart

2 suggestions for labeling flowchart: 1.  Write the values and units of all known stream

variables at the locations of the streams on the flowchart. For example, a stream containing 21 mole% O2 and 79% N2 at 320˚C and 1.4 atm flowing at a rate of 400 mol/h might be labeled as:


400 mol/h

0.21 mol O2/mol 0.79 mol N2/mol T = 320˚C, P = 1.4 atm

Process stream can be given in two ways: a)  As the total amount or flow rate of the stream

and the fractions of each component b)  Or directly as the amount or flow rate of each

component.


60 kmol N2/min 40 kmol O2/min

0.6 kmol N2/kmol 0.4 kmol O2/kmol

100 kmol/min

3.0 lbm CH4 4.0 lbm C2H4 3.0 lbm C2H6

0.3 lbm CH4/lbm 0.4 lbm C2H4/lbm 0.3 lbm C2H6/lbm

10 lbm

Labeling a flowchart-continue

2.  Assign algebraic symbols to unknown stream variables [such as m (kg solution/min), x (lbm N2/lbm), and n (kmol C3H8)] and write these variable names and their units on the flowchart.


mol/h

0.21 mol O2/mol 0.79 mol N2/mol T = 320˚C, P = 1.4 atm

n! 400 mol/h

y mol O2/mol (1-y) mol N2/mol T = 320˚C, P = 1.4 atm

Labeling a flowchart(continued)

•  If that the mass of stream 1 is half that of stream 2, label the masses of these streams as m and 2m rather than m1 and m2.

•  If you know that mass fraction of nitrogen is 3 times than oxygen, label mass fractions as y g O2/g and 3y g N2/g rather than y1 and y2.

•  When labeling component mass fraction or mole fraction, the last one must be 1 minus the sum of the others.

•  If volumetric flow rate of a stream is given, it is generally useful to label the mass or molar flow rate of this stream or to calculate it directly, since balance are not written on volumetric qualities.


Labeling a flowchart-continue

Consistent on Notation


gas in fraction moles yliquid in moles)or (mass fraction component x

rate flow volume VvolumeV

rate flowmolar nmoles n

rate flow mass mmass m

=

=

=

=

=

=

=

=

!

!

!

Example 4.2-‐3

Two methanol water mixtures are contained in separate flasks. The first mixture contains 40 wt% methanol, and the second contains 70% methanol. If 200 g of the first mixture is combined 150 g of the second, what are the mass and composiEon of the product?


Try This..

Two methanol-water mixture are contained in separate flasks. The first mixture contains 40wt% methanol and the second flask contains 70wt% methanol. If 20 Kg of the first mixture are going to be mixed with 15000 g of the second in a mixing unit, what are the mass and composition of the product of the mixing unit?


n Example 4.2.3

Try This.. An experiment on the growth rate of certain organism requires an environment of humid air enriched in oxygen. Three input streams are fed into an evaporation chamber to produce an output stream with the desired composition. A: Liquid water fed at rate of 20 cm3/min B: Air (21% O2 and 79% N2) C: Pure O2 with a molar flow rate one-fifth of the molar flow rate of stream B The output gas is analyzed and is found to contain 1.5 mole% water. Draw and label the flowchart of the process, and calculate all unknown stream variables.


n Example 4.3.1

Solution


Evaporation

20 cm3 H2O (l)/min

mol H2O/min 2n!

mol O2/min 1200.0 n!

mol air/min 1n!

mol/min 3n!

0.21 mol O2 /mol 0.79 mol N2 /mol

0.015 mol H2O /mol y mol O2 /mol (0.985-y) mol N2/mol

Ø  convert water flow-rate to mol/min Ø  Water balance Ø  Total mol Ø  Nitrogen / oxygen balance

Flowchart Scaling & Basis of Calculation

•  Flowchart scaling – procedure of changing the values of all stream amounts or flow rates by a proportional amount while leaving the stream compositions unchanged. The process would still be balance.

•  Scaling-up – if final stream quantities are larger than the original quantities.

•  Scaling down – if final stream quantities are smaller than the original quantities.




1 kg C6H6

300 lbm/h

1 kg C7H8

300 kg C6H6

300 kg C7H8

300 lbm/h

2 kg 0.5 kg C6H6/kg 0.5 kg C7H8/kg

600 kg 0.5 kg C6H6/kg 0.5 kg C7H8/kg

600 lbm/h 0.5 lbm C6H6/lbm 0.5 lbm C7H8/lbm

x 300

kg kg/h Replace kg with lbm


•  Suppose you have balanced a process and the amount or flow rate of one of the process streams is n1.You can scale the flow chart to make the amount or flow rate of this stream n2 by multiplying all stream amounts or flow rate by the ratio n2/n1.

•  You cannot, however, scale masses or mass flow rates to

molar quantities or vice versa by simple multiplication; conversions of this type must be carried out using the methods as discussed in mass fraction and mol fraction section.


Basis of Calculation

•  A basis of calculation is an amount (mass or moles) of flow rate (mass or molar) of one stream or stream component in a process. All unknown variables are determined to be consistent with the basis.

•  If a stream amount or flow rate is given in problem, choose this quantity as a basis

•  If no stream amount or flow rate are known, assume one stream with known composition. If mass fraction is known, choose total mass or mass flow rate as basis. If mole fraction is known, choose a total moles or molar flow rate as basis.


Flowchart


Product Separation


Reactor Product

Separation Unit

Reactants Products

K-salt Recovery •  In a steady state process crystalline potassium

chromate (K2CrO4) is recovered from an aqueous solution of this salt.

•  10 ton per hour of a solution that is 30% K2CrO4 by mass is fed into an evaporator. The concentrated stream leaving the evaporator contains 50% K2CrO4 ; this stream is fed into a crystallizer in which it is cooled (causing crystals of K2CrO4 to come out of solution) and then filtered. The filter cake consists of K2CrO4 crystals and a solution that contain 35% K2CrO4 by mass; the crystals account for 95% of the total mass of the filter cake. The filtrate drains out of the system also contains 35% K2CrO4.


2 kg/h.m

5 kg/h.m

4 kg/h.m

3 kg/h.m


Evaporator Crystallizer

& Filter

Fresh Feed 10,000 kg/h

Filter Cake

0.30 kg K/kg 0.70 kg W/kg


1 kg W/h.m

Filtrate



10 ton per hour of a solution that is 30% K2CrO4 by mass is fed into an evaporator. The concentrated stream leaving the evaporator contains 50% K2CrO4 ; this stream is fed into a crystallizer in which it is cooled (causing crystals of K2CrO4 to come out of solution) and then filtered. The filter cake consists of K2CrO4 crystals and a solution that contain 35% K2CrO4 by mass; the crystals account for 95% of the total mass of the filter cake. The filtrate drains out of the system also contains 35% K2CrO4.

•  Product rate, m3 = 1385 Kg/h •  Filtrate soln, m5 = 4542 Kg/h •  If filtrate is discarded, a huge loss in raw

material and money •  Therefore, recycle is the solution


m1 = 4000 kg W/h

m2 = 6000 kg K-soln/h

m3 = 1385 kg K(S)/h

m4 = 73 kg K-soln/h


K-salt Recovery •  In a steady state process crystalline potassium chromate (K2CrO4) is

recovered from an aqueous solution of this salt. •  10 ton per hour of a solution that is 30% K2CrO4 by mass is joined

by a recycle stream containing 35% K2CrO4, and combined stream is fed into an evaporator.

•  The concentrated stream leaving the evaporator contains 50% K2CrO4; this stream is fed into a crystallizer in which it is cooled (causing crystals of K2CrO4 to come out of solution) and then filtered.

•  The filter cake consists of K2CrO4crystals and a solution that contain 35% K2CrO4 by mass; the crystals account for 95% of the total mass of the filter cake. The solution that passes through the filter also 35% K2CrO4, is the recycle stream.

•  Calculate the rate of evaporation, the rate of production of crystalline K2CrO4, the feed rates that the evaporator and the crystallizer must be designed to handle and the recycle ratio


Product Separation & Recycling


1 kg/h.m

4 kg/h.m3 kg/h

.m

Evaporator Crystallizer

& Filter

Fresh Feed 10,000 kg/h

Filter Cake



2 kg W/h.m

Recycle 0.35 kg K/kg 0.65 kg W/kg

5 kg/h.m


6 kg/h.m

MP

Product Separation & Recycling


m1 = 19662 kg mixed feed /h

m2 = 6900 kg Water evaporated/h

m3 = 12762 kg fed to crystallizer/h

m4 = 2945 kg crystals/h


m6 = 9662 kg recycle/h

Recycle Ratio= 0.9662 Kg recycle/Kg Fresh feed

Comment:

Crystal production increases 113% due to recycling

m2 = 4000 kg W/h


m4 = 1385 kg K(S)/h

m5 = 73 kg K-soln/h


n Without Recycle

Recycle and Bypass

•  Reuse of reactants/feed •  Recovery of catalyst •  Dilution of process streams (improve filter

operation) •  Control of a process variable (Reduce

reactant concentration) •  Circulation of a working fluid

(Refrigeration)


n What is wrong here?

Product Separation and Recycle •  Normally, reactions are not complete

– Separation and recycle – Improved yield, conversion ,…


Reactor Product

Separation Unit

Recycle

Reactants Products

ü Process feed

ü Fresh feed

ü Gross product

ü Net product

ü Overall conversion

ü Once-through conversion

Balances on Reactive Systems •  Stoichiometry The theory of proportions in which chemical species combine with one another.

– Example: 2 SO2 + O2 à 2 SO3

•  Stoichiometric Ratio •  Ratio of stoichiometric coefficients

– Example


2 mol SO3 produced 1 mol O2 reacted

2 mol SO2 reacted 2 mol SO3 produced

Terminology •  Limiting reactants

Exist less than stoichiometric proportion •  Excess reactants

Exist more than stoichiometric proportion •  Example

2SO2 + O2 à 2 SO3 (30 mol) (10 mol)

n 3/29/15 n Prof. Shoukat Choudhury n 50 Excess Limiting

Terminology

•  Fractional excess •  Percent excess •  Example

– H2 + Br2 à HBr – H2 : 25 mol /hr – Br2 : 20 mol /hr

– Fractional Excess H2 = (25 – 20 ) /20 = 0.25 – Percent Excess = 25 %


s

s

nnn −

100s

s

n nn−

×

Terminology

•  Fractional conversion – Chemical reactions are not always completed. – Factional conversion

•  f = (moles reacted) / (moles fed)

– When fresh feed consists of more than one material the conversion must be stated for a single component, usually the limiting reactant.


Terminology

•  In general, high overall conversions can be achieved in two ways: – Design the reactor to yield a high single-

pass conversion, or – Design the reactor to yield a low single-

pass conversion and follow it with a separation unit to recover and recycle unconsumed reactant.


reactant input to process - reactant output from processOverall Conversions = reactant input to process

reactant input to reactor - reactant output from reactorSingle-Pass Conversions = reactant input to reactor

Problem

•  Consider the reaction

6 NaClO3 + 6 H2SO4 + CH3OH → 6 ClO2 + 6 NaHSO4 + CO2

•  if the reactor feed has the composition (mol%) of 36% NaClO3, 54% H2SO4, and the rest CH3OH, which is the limiting reactant?

•  Calculate the reactant flows required to produce 10 ton per hour of ClO2 assuming 90% conversion is obtained.


Balances of Atomic and Molecular Species

•  Methods for solving mass balances with reactions –  Using balances on molecular species –  Using balances of atoms –  Using the extent of reaction

•  For multiple reactions, sometimes it is more convenient to use atomic balances

•  Atomic species balances generally lead to the most straightforward solution procedure, especially when more than one reaction is involved.

•  Extents of reaction are convenient for chemical equilibrium problems and when equation solving software is to be used.

•  Molecular species balances require more complex calculations than either of the other two approaches and should be used only for simple systems involving one reaction.


A simple Problem

The oxidation of ethylene to produce ethylene oxide proceeds as: 2 C2H4 + O2 = 2 C2H4O

The feed to the reactor contains 100 Kmol C2H4 and 200 Kmol O2. What is the limiting reactant? What is the percentage excess of the other reactant? If the reaction proceeds to a point where the fractional conversion of the limiting reactant is 50%, how much of each reactant and products is present at the end?


Multiple Reaction, Yield, Selectivity

•  Multiple reaction : one or more reaction – Side Reaction : undesired reaction – Example: Production of ethylene

C2H6 à C2H4 + H2 Side Reactions C2H6 + H2 à 2CH4 C2H4 + C2H6 à C3H6 + CH4

– Design Objective •  Maximize desired products (C2H4) •  Minimize undesired products (CH4, C3H6)


Terminology moles of desired product formed =

moles that would have been formed if therewere no side reactions and the limiting reactanthad reacted completelymoles of desired product formed

moles of reactant

YIELD

fedmoles of desired product formed

moles of reactant consumed


–  Yield and Selectivity are used to describe the degree to which a desired reaction predominates over competing side reactions.

moles of desired product formedmoles of undesired product formed

SELECTIVITY =

n Yield =

n Yield =

Determination of yield and selectivity Reaction:

C2H6 + 2.5O2 → 2CO + 3H2O

C2H6 + 3.5O2 → 2CO2 +3H2O

Undesired product: CO 80% reactor conversion


100 mol C2H6

500 mol O2

REACTOR n Products

n 20 mol C2H6

n 120 mol CO2 n 40 mol CO n 240 mol O2 n 240 mol H2O

Yield = 0.6 or 1.2 or 1.5 Selectivity = 3.0

Problem Ethane is burned with air in a continuous

steady-state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide, and water. The feed to the reactor contains 10% C2H6. The percentage conversion of ethane is 80%, and gas leaving the reactor contains 8 mol CO2 per mol of CO. Determine molar composition of product gas.


n Reaction: n C2H6 + 5/2 O2 → 2CO + 3H2O n C2H6 + 7/2 O2 → 2CO2 +3H2O

Concept of Purge- why needed? •  Production of Ethylene Oxide •  Reaction: 2C2H4 + O2 → 2C2H4O •  Mixture of Ethylene and air stream is charged to the reactor •  Reactor effluent is charged to absorber and gas stream

containing N2, O2 and unreacted ethylene is charged back to reactor


Reactor Absorber

Recycle

Fresh Feed

Products

Solvent

Production of Ethylene Oxide •  Problem: accumulation of N2

•  Solution: allow purging of inert species


Reactor Absorber

Recycle

Fresh Feed

Products

Purge stream

Solvent

60 mol C2H4/s 30 mol O2/s

113 mol N2/s


565 mol N2/s

50 mol C2H4/s 50 mol C2H4O/s 25 mol O2/s

565 mol N2/s


565 mol N2/s


113 mol N2/s

50 mol C2H4O/s solvent


Getting rid of undesired materials in recycle stream.

Reactor Product

Separation Unit

Recycle

Reactants Products

Purging

Purging

Definition

•  A Recycle Stream is a term denoting a process stream that returns material from downstream of a process unit back to the process unit.

•  A Bypass stream - one that skips one or more stages of the process and goes directly to another downstream stage.

•  A Purge stream – a stream bled off to remove an accumulation of inert or unwanted material that might otherwise build up in the recycle stream. n 3/29/15 n Prof. Shoukat Choudhury n 64

Example 4.7-3, page 139 Recycle and Purge in the Synthesis of Methanol

Methanol is produced in the reaction of carbon dioxide and hydrogen: CO2 + 3H2 → CH3OH + H2O The fresh feed to the process contains hydrogen, carbon dioxide and 0.40 mole percent inerts(I). The reactor effluent passes to a condenser that removes essentially all of the methanol and water formed and none of the reactants or inerts. The latter substances are recycled to the reactor. To avoid buildup of the inerts in the system, a purge stream is withdrawn from recycle. The feed to the reactor contains 28.0 mole% CO2, 70 mole% H2 and 2% inerts. The single pass conversion of H2 is 60%. Calculate the molar flow rates and molar compositions of the fresh feed, the total feed to the reactor, the recycle stream, and the purge stream for a methanol production rate of 155 kmol methnol/hr.


Reaction: CO2 + 3H2 → CH3OH + H2O Basis: 155 kmol CH3OH/h


REACTOR CONDENSER

Recycle

N8 kmol/h

Purge

N7 kmol/h

n1 kmol/h

155 kmol CH3OH/h

155 kmol H2O/h

x1C mol CO2/mol

(0.996-x1C)mol H2/mol

0.004 mol I/mol

0.28 mol CO2/mol

0.70 mol H2/mol

0.02 mol I/mol

n2 kmol/h

n3 kmol CO2/mol

n4 kmol H2/mol

n5 kmol I/mol

155 kmol CH3OH/h 155 kmol H2O/h

N6 kmol/h

x6C mol CO2/mol

x6H mol H2/mol

(1-x6C-x6H)mol I/mol

x6C mol CO2/mol x6H mol H2/mol

(1-x6C-x6H) mol I/mol

x6C mol CO2/mol x6H mol H2/mol

(1-x6C-x6H) mol I/mol

Combustion Reaction

•  Combustion – A rapid reaction of a fuel with oxygen – Fuels : coal, fuel oil, gas fuel, solid fuel, … – Complete combustion / incomplete

combustion – Wet basis composition / dry basis composition

Remember: Orsat analysis yields dry basis composition


Terminology •  Theoretical oxygen : Amount of oxygen

needed for complete combustion – all carbon in the fuel is oxidized to CO2 and – all the hydrogen is oxidized to H2O

•  Theoretical air : The quantity of air that contains theoretical oxygen

Air(theo) = 4.76 x O2(theo)

•  Excess air : The amount by which the air fed to reactor exceeds the theoretical air

•  Percent excess air


( )%100

air molesair) (molesair) (moles

ltheoretica

altheroreticfed ×−

Composition of Flue or stack gas •  Wet Basis = => Dry Basis - Basis: 1 mole wet gas - basic idea, subtract the water and express the rest in % •  Dry Basis = => Wet Basis - Basis: 1 mole wet gas - need one extra information: How much water is there in

one mole of wet gas? (Say, y mole fraction out of 1 mole wet gas)

- subtract water from one mole of wet gas and get the mole of dry gas. (dry gas=1-y)

- Now, dry gas fraction is (1-y)*yi, where yi is mole fraction of ith dry gas components


A problem with purge and recycle

•  The fresh feed to an ammonia production process contains nitrogen and hydrogen in stoichiometric proportion, along with 2 mole% inert gas. The feed is combined with a recycle stream containing the same three species and the combined stream is fed to a reactor in which a single pass conversion of 20% is achieved. The reactor effluent flows to a condenser. A liquid stream containing essentially all of the ammonia formed in the reactor and a gas stream containing all the inerts and the unreacted nitrogen and hydrogen leave the condenser. 10% of the gas stream leaving the condenser is removed as purge and the rest constitutes the recycle stream.

a)  Draw a complete flow chart of the process. b)  Completely label the flow-chart c)  Find the overall conversion of N2 d)  Find the total feed flow rates to the reactor


Questions


Chemical Reaction

•  What is final composition ? – Chemical equilibrium thermodynamics

•  How long it will take to reach equilibrium ? – Chemical kinetics


...

)()(/ l

bnA

RTEmo

i

CCeTkncompositiofTkr

−=

=

RTH

dTKdaK

fK

r

i

i

i

i

Δ=

∏=

∏=

ln

ˆν

ν

Sources of equations of unknown process variables: 1.  Material balances 2.  An energy balance 3.  Process specifications 4.  Physical properties and laws 5.  Physical constraints 6.  Stoichiometric relations


Variables in process design or analysis –  Temperature –  Pressure –  Flow rate –  Chemical composition –  Physical properties

Chemical composition –  Mass fraction –  Mole fraction –  Mass and molar

composition –  Concentration: mass

conc., molar conc., molality


Physical properties –  Specific gravity –  Specific volume –  Density –  Specific heat –  Enthalpy –  Heat of reaction, etc.

Topic Outcomes

•  Make your conception clear about the following terms: –  Batch, semibatch, continuous, transient, and steady-state

processes –  Recycle, bypass and purge –  Degrees-of-Freedom –  Fractional conversion of a limiting reactant –  Percentage excess of a reactant –  Yield and selectivity –  Dry-basis composition of a mixture containing water –  Theoretical air and percent excess air in a combustion reaction


Topic Outcomes •  Given a process description:

– Draw and fully label a flowchart – Choose a convenient basis of calculation – For a multiple-unit process, identify the

subsystems for which balances might be written

– Perform DoF analysis for the overall system and each possible subsystems

– Write in order the equations you would use to calculate specified process variables

– Perform the calculations n 3/29/15 n Prof. Shoukat Choudhury n 76

Topic Outcomes

•  Do these computations –  for single-unit and multiple-unit processes and –  for processes involving recycle, bypass, or

purge streams

•  If the system involves reactions, you should be able to use – molecular species balances, – atomic species balances, or – extents of reaction for both the DoF analysis

and the process calculations n 3/29/15 n Prof. Shoukat Choudhury n 77

Topic Outcomes

•  Given a combustion reactor and information about the fuel composition – calculate the feed rate of air from a given

percent excess or vice versa

•  Given additional information about the conversion of the fuel and the absence or presence of CO in the product gas – calculate the flow rate and composition of the

product gas


fundamentals of material balancesshoukat.buet.ac.bd/che111_ch4_part1.pdf · fundamentals of...

Documents