12. Theoretical and excess air

If two reactants are needed for a reaction the less expensive reactant is fed into system at excess to convert the total of the expensive reactant. An example of an inexpensive reactant can be air, which is free. Combustion reactions are run with more air than is needed to supply oxygen in the right proportion to the fuel.

Theoretical Oxygen:  Moles(batch process) or molar flow rate(continues process) of Oxygen is needed for complete combustion of all fuel that is fed to the reactor. It could be assumed that all carbon in the fuel can be oxidized to CO₂ and hydrogen to H₂O.

Theoretical air: Theoretical oxygen quantity that is contained in the air.

Excess Air: Theoretical air amount that is exceeded by the amount of air fed

Percent Excess air Formula: [(moles air)fed – (Moles air)theoretical / (moles air) theoretical] x 100%

It is useful to know the fuel feed rate and stoichiometry in the equations to help with calculations like theoretical O₂ and air feed rates. If the actual feed rate of air is known it is possible to calculate percent excess air.

http://www.google.co.za/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&ved=0CCsQFjAA&url=http%3A%2F%2Fwww.departments.bucknell.edu%2Fchem_eng%2Fcheg200%2FCinChE_Manual%2FCh05%2Fexamples%2Fexcess_air.pdf&ei=8RZcUp6sI-G_0QWKnIHQBg&usg=AFQjCNFb77cVaWgu4xR_QOF9D2tEGz1Wuw&sig2=GXtbEWe9EjLYWCLRH4ammg&bvm=bv.53899372,d.d2k

13. Material Balances on Combustion Reactors

Solving material balances for combustion reactors is the same as for any reactive system but have a few points to keep in mind like:

1.       After drawing flow chart your outlet stream must contain

i)        Unreacted fuel(unless told otherwise)

ii)       Unreacted oxygen

iii)     Water and carbon dioxide(also carbon monoxide if said is present by problem)

iv)     Nitrogen since air is not pure

2.       Calculate oxygen feed rate from the specified percent oxygen or percent excess air. Calculate theoretical O₂ from fuel feed rate and the reaction stoichiometry for complete combustion then calculate oxygen feed rate by multiplying the theoretical oxygen

3.       If one reaction is involved then the three methods of balancing (molecular, atomic and extent of reaction) are equal. If several reactions occur at the same tome atomic species are more convenient.

http://www.google.co.za/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&ved=0CE8QFjAE&url=http%3A%2F%2Fwww.engr.mun.ca%2Falam%2FLecture_9.pdf&ei=WBtcUuftEc-Z0QXa5YCoBg&usg=AFQjCNGJ_91vkhGbN3k05wysrp39HqRk3g&sig2=j7akVbLnnOs2kpYLzicTtA

Some additional Considerations about chemical processes (what you do not learn out of textbooks)

1.       Processes do not always work as designed

2.       Variables measured to precision in field measurements have errors

3.       Humans make mistakes that influence the process outcome

4.       You will not always have all the data you need and may have to make assumptions

5.       In textbooks material balances have a closure of 100%, in practice there is no such thing as a true steady state

6.       Textbooks have only one correct answer whilst real problems may have the possibility where you can find a variety of solutions.

  

Reasons for differences between design values and experimental values:

1.       Human errors and data scatter

2.       Impurities in feed

3.       Incorrect assumptions of steady state

4.       Incorrect assumptions that MEK is not reactive.

5.       Errors due to approximation in the experimental data analysis

6.       Approximations in the design analysis

11. Combustion reactions:

A combustion reaction is a rapid reaction of oxygen with fuel or can be defined as: the sequence of exothermic chemical reactions between a fuel and an oxidant accompanied by the production of heat and conversion of chemical species for more information click this link http://en.wikipedia.org/wiki/Combustion_reaction

When fuel is burned, carbon in the fuel reacts to form either CO₂ or CO, hydrogen’s form H₂O and sulfur form SO₂.

If CO is formed the combustion reaction is referred to as a partial combustion

Examples:

C + CO₂  CO₂ (Complete combustion of carbon)

C₃H₈ + O₂  3O + 4H₂O (Partial Combustion of propane) 

In these kind of calculations we accept the composition to 79% N₂, 21% O 79 moles N₂/21 moles O₂ = 3.76 moles N₂/mole O₂.

A composition on wet basis is a term used to denote the component mole fraction of a gas that contains water. The product gas that leaves the furnace is referred to as stack gas or flue gas. http://en.wikipedia.org/wiki/Flue-gas_stack (more about flue gas)

Composition on wet and dry bases can be calculated

http://www.google.co.za/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&ved=0CDsQtwIwAg&url=http%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DW4CkkvPgEz8&ei=FMxaUpuTIYqb0QXZiYG4AQ&usg=AFQjCNF7lKYh8VgQlcSIq6fOaZgGAsxuww&sig2=iRwUiZn-TbYEnKRrRuQlWQ&bvm=bv.53899372,d.d2k

10. Product Separation and Recycle

Overall Conversion is a formula used to calculate how much of the reactant has taken part in the overall reaction: (Reactant input to process – reactant output from process)/reactant input to process

Single-pass conversion is a formula used to calculate the amount of reactant taken part in a single-pass:

(Reactant input to reactor- reactant output from reactor)/reactant input reactor

To find the percentage conversions multiply by 100%. Single pass overall pass extent of reactions

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Purging

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

Purge Stream

Purge stream is a stream bled off to remove an accumulation of inerts or unwanted material that might otherwise build up in the recycle stream.

A great example would be one by Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños

Example 12-1. Distillation of Benzene and Toluene

A distillation column separates 10,000 kg/hr of a mixture containing equal mass of benzene and toluene. The product D recovered from the condenser at the top of the column contains 95% benzene, and the bottom W from the column contains 96% toluene. The vapor V entering the condenser from the top of the column is 8000 kg/hr. A portion of the product from the

condenser is returned to the column as reflux R, and the rest is withdrawn as the final product D. Assume that V, R, and D are identical in composition since V is condensed completely.

Find the ratio of the amount refluxed R to the product withdrawn D.

Overall Process

Total Balance: 10,000 = D + W

Benzene Balance: 10,000(0.50) = D(0.95) + W(0.04)

Solving simultaneously, D = 5050 kg/hr ; W = 4950 kg/hr

Total balance around the separator:

8000 = R + D

R = 2950 kg/hr

Ratio (R/D) = (2950/5050) = 0.58

9. Balances on Molecular and Atomic species

Balances work on the principle that atoms can neither be created nor destroyed in a chemical reaction.

Example the reaction 2H₂ + O₂  2H₂O where x represents the unknown mol H₂ that is fed into the system and y mol H₂O is produced also z the 0₂ that is in the feed.

We can perform a hydrogen balance on the reaction which could mean an atomic balance or a molecular balance.

Water molecular balance would be generation=output

Generated H₂0 (mol H₂0 generated/min) =y mol H₂0/min

http://www.google.co.za/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&ved=0CCsQtwIwAA&url=http%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DddtW9G1oUxI&ei=iKlaUp_2Foiz0QXqpIHYAg&usg=AFQjCNFBnfB7VEqQCN78Fj16OKcPS37ojw&sig2=yfnrJpH9_tNYUtVn7qJcSA&bvm=bv.53899372,d.d2k

Hydrogen molecular balance would be input= output +consumption

x mol H₂/min = x(mol H₂ generated/min) + consumption H₂(mol H₂/min)

Oxygen molecular balance would be generation=output

Gen O₂ (mol O₂ generated/min) =z mol(O₂/min)

Atomic balance works on the same principle expect with the number of each atoms of elements in the molecule example Atomic hydrogen balance would be: input=output

xMol H₂(feed) x 2 mol H(Which represents the two hydrogen’s)= x mol H₂ (output)x 2 mol H(hydrogen’s in water) x 2( stoichiometry)

http://www.google.co.za/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&ved=0CDYQtwIwAQ&url=http%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DNDm4FRt2fyM&ei=t6laUoGTBeOc0AXG_YCIBg&usg=AFQjCNHScy7pMlCd3UPjOwadTLWO83yLGA&sig2=jOyWE4Erf4Q3Y6EqykyXGg&bvm=bv.53899372,d.d2k

Extend of reaction can be used as followed:

Atomic species balances lead to direct solution technique when more than one reaction is involved

Extents of reactions are useful for chemical equilibrium problems and when equation solving software is to be used.

Molecular species require more complex calculations than the other methods and is recommended for simple systems with one reaction

Independents can be defined as one thing that is independent of another; they are detached and not associated, so the first one is not affected or subjective by the second.  

Independent equations: Algebraic equations are independent if it is not possible to obtain them by adding and subtracting any other equations.

If two molecular species are in the same ratio anyplace they appear in a process to one and other and this ratio is incorporated in the flow chart labeling, balances on those species are not independent equations.

With other words chemical reactions are independent of the stoichiometry equation if any other equation cannot be obtained by adding and multiplying and etc. of any other stoichiometry equation.

8. Multiple Reactions, Yields and selectivity:

A reaction can produce more than one product that is an economic loss. Engineers need to optimize the procedure in such way to produce the desired product as much as possible and the non-desired product at a minimum.

The yield of desired product can be calculated as:
                                            moles of desired product/ moles that would form if          

                                                            moles un-desired products formed 
(Can be expressed as fraction of percentage)              

                                                                                     

Selectivity of a reaction can be calculated as product formed:           
                              moles desired product/ moles Undesired product                                                                                             

7. Chemical Equilibrium

Chemical equilibrium can be described as the final composition of the reaction mixture.

Some reactions are irreversible (Reaction proceeds from reactants to products and cannot go back to original form).

There is always a limiting reactant (the reactant that’s concentration when zero stops the reaction). Some reactions can be reversible (product can be reversed back into reactants).

 To see if your reaction has reached equilibrium you can always calculate the equilibrium constant with the formula  ((product 1)(prpduct2))/(reactant1)(reactant2)) =K
 where it would be displayed in a reaction as reactant 1 + reactant 2 = product 1 + product 2. The stoichiometry would be displayed as exponents in the formula.  

When ratio=K
According to the law of mass action, there is no shifting of reaction and there will be no change in the concentration of reactants and products and the system is already at equilibrium.
When ratio> K                              
In this condition the reaction will shift in the backward direction to achieve equilibrium state. At equilibrium quantity of product will decrease and the quantity of reactants will increase.
When ratio<K
In this condition the reaction will shift in forward direction to achieve equilibrium state. At equilibrium quantity of product will increase and the quantity of reactants will decrease.

All mole fractions can be expressed as a single variably (extent of reaction at equilibrium ξₑ).

Extent of reaction is a way to determine unknown molar flow rates for a reactive process.

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The degree of freedom analysis follows as:

Number unknown labeled variables + number independent reactions – number independent reactive species –number independent nonreactive species – number other equations relating unknown variables = number of degrees of freedom 

6. Chemical Reaction Stoichiometry

A material balance can be complicated when a chemical reaction occurs in the process. Relative amounts of reactants and products in the input and output stream can be constraint by the stoichiometric equation of the reaction. A material balance on a reactive substance does not have the simple input = output, but must contain the generation or consumption term.

a.    Stoichiometry

Stoichiometry is the theory of the proportions in which chemical species combine with one another. The stoichiometry equation of a chemical reaction is a statement of the relative number of molecules or moles of reactants and products that participate in the reaction. A valid stoichiometry equation must be balanced; that is that the number of atoms on each atomic species be the same on each side.

The stoichiometry ration of two molecular species participating in a reaction is the ratio if their stoichiometric coefficients in the balanced reaction equation. The amount of a particular reactant that was consumed can be determined by using the stoichiometry ratio as a conversion factor.

2SO2 + O2  ® 2SO3

Stoichiometric ratios:

(2 mol SO3 generated)/(1 mol O2 consumed)

If it is known that 1600 kg/h of SO3 is to be produced, calculate the amount of oxygen produced

(1600 kg/h SO3 × 1kmol SO3 × 1kmol O2 consumed)/(80 kg SO3 × 2kmol SO3 generated) = 10 kmol O2/ h

b.    Limiting and excess Reactants

Stoichiometric proportion is when two reactants, A and B, has a ration equals the stoichiometric ratio obtained from the balanced reaction equation.

The limiting reactant is the reactant that would run out if the reaction proceeded to completion and the other reactants is the excess reactants. The stoichiometric requirement is the amount needed to react completely with the limiting reactant. The fractional excess of a reactant is the ratio of the excess to the stoichiometric requirements.

The percentage excess of A is 100 times the fractional excess.

The fractional conversion of a reactant is the ratio

f = moles reacted / moles fed

You can always follow this link for extra help

http://www.em-ea.org/Guide%20Books/Book-1/1.4%20MATERIAL%20%20AND%20ENERGY%20BALANCE.pdf

5. Recycle and Bypass
It is rare that A  B proceeds in completion in a reactor. A will always be present in the product of B, no matter how long the reaction remains in the reactor. The unreacted A can be separated from the product stream and be sold or recycled by charging it back into the reactor.
The fresh feed to the process and the feed to the reactor, which is the sum of the fresh feed and the recycle stream.
If the stream variables were unknown, you could determine them by writing balances on the overall process and about the reactor, separator, and mixing point.

 

4. BALANCES ON MULTIPLE-UNIT PROCESSES
In reality most processes do not consist of just one process unit. There are more than one chemical reactors, units for mixing reactants, blending products, heating and cooling process streams, separating products from each other and from unconsumed reactants, and removing potentially hazardous pollutants from streams prior to discharging the streams to the plant environment.
A system is any portion of a process that can be enclosed in a hypothetical box or boundary. This can include the entire process, an interconnected combination of some of the process units, a single unit,  or a point at which two or more process streams come together or one stream splits into branches.
The difference between a multiple-unit process and a single unit process is that the multiple-unit must be isolated and balances must be written on several subsystems of the process to obtain enough equations to determine the unknown variables.

3. Material Balance Calculations
All material balance calculations have one single theme that is the given values from output and input stream variables must be used to derive and solve the equations of others. To solve these equations is usually simple algebra, but by using a description of a process and a collection of process data can complicate solving the equations. The following can be used to approach the difficulties presented by a mass balance problem.
1. Flowcharts
It is important to organize the information in a way that is suitable for equations, because most of the mass balance problems are given in an essay-like form of information and then asked to determine something in the process. The best way to organize this information is to draw a flowchart of the process. This can be done by using boxes or other symbols to represent process units, like reactors, mixers, separation units, etc., and lines with arrows to represent inputs and outputs.
The chart should be fully labelled when it is drawn, with values of known process variables and symbols for unknown variables being written for each input and output stream.

2. Flowchart Scaling and Basis of Calculation
Scaling the flowchart is the procedure where the values of all stream amounts or flow rates is changed by a proportional amount while leaving the stream compositions unchanged. Scaling up: if the final stream quantities are larger than the original quantities. Scaling down: if the final stream quantities are smaller than the original quantities. Furthermore, it must be noted that you cannot scale masses or mass flow rates to molar quantities or vice versa by simple multiplication.
A basis of calculation is an amount or flow rate of one stream or stream component in a process. By choosing a balance of calculation the first step in balancing an equation is done; all unknown variables are determined to be consistent with the basis.
Choose the quantity that is given, for example the stream amount or the flow rate, if it is given. Assume the stream amount or flow rates is one if no stream amounts and flow rates is known. Choose a total mass or mass flow rate of that stream as a basis (100kg) if the mass fractions are known; choose a total number of moles or molar flow rate if the mole fractions are known.

3. Balancing a Process
The following rules apply to nonreactive processes:
The maximum number of independent equations can be derived by writing balances on a nonreactive system equals the number of chemical species in the input and output streams.
Write balances that involves the fewest unknown variables.

4. Degree-of-Freedom Analysis
By drawing a properly labelled flowchart you can determine whether you have enough information to solve the given problem, is called a degree-of-freedom analysis.
The following steps should be done to perform a degree-of-freedom analysis:
- Draw a complete labelled flowchart
- Count the unknown variables on the chart
- Count the independent equations relating them
- Subtract the second number (independent) from the first number  (unknown)
n_df= n_(unknowns )- n_(indep eqns)
There are three possible results:
* If n_df = 0, there are n independent equations in n unknowns and the     problem can be solved.
* If n_df > 0, there are more unknowns than independent equations      relating them, and additional variable values must be specified before  all remaining values can be determined.
* If n_df < 0, there are more independent equations than unknown  variables. The flowchart is either incomplete labelled or over  specified.

5. General Procedure for Single-Unit Process Material Balance Calculations
-Choose as a basis of calculation an amount or flow rate of one of the process streams.
-Draw a flowchart and fill in all known variable values, including the basis of calculation. Then label all unknown stream variables on the chart.
-Express what the problem statement asks you to determine in terms of the labelled variables.
-Convert all quantities to one basis if you are given mixed mass and mole units for a stream.
-Do a degree-of-freedom analysis.
-Write the equations in an efficient order (minimizing simultaneous -calculations) and circle the variables you will solve, if the number of --equations relating the unknowns equals the number of unknown variables.
-Solve the equations.
-If the quantities requested have not been calculated, calculate them.
-Scale the balanced process by the ratio n_g⁄n_c  to obtain the result, if a stream quantity or flow rate n_g was given in the problem statement and another value was either chosen as basis of calculation or calculated for this stream.

2. Balances
B. Balance on Continuous Steady-State Processes

The following equation is for a continuous steady-state process, and because accumulation equals zero, it can be written as follows:
input+generation=output+consumption
If the previous balance is on a nonreactive species or on total mass, then it can be written as input=output because generation and consumption equals zero.

C. Integral Balances on Batch Processes

By using reasoning it can be said that
accumulation=final output-initial input
                            generation - consumption

Therefore, the following equation for a batch process is used:
initial input+generation=final output+consumption

Except for the input and output terms being changed to inititial and final amounts of the balanced substance than flow rates, this equation is identical to the equation of a continuous steady-state process.

D. Integral Balances on Semibatch and Continuous Processes

Integral balances can be written for continuous and semibatch processes, if the differential balance on the system is written and then integrated between two instances of time. Most problems of this type are often straightforward and easy to determine.

2. Balances
A. The General Balance Equation
Consider a continuous process where methane is a component of both the input and the output streams, and to determine if the unit is working as designed, it is found that the mass flow rates of the methane input and output measured are different (m ̇_in≠m ̇_out).

The following can cause the difference between the measured flow rates:
- The methane is being consumed as a reactant or being generated as a product within the unit.
- The methane is accumulated in the unit, meaning adsorbing on the walls.
- The methane is leaking from the unit.
- The measurements is wrong.

All that can be considered and cause a difference between the output and the input is the generation or consumption in a reaction and the accumulation within the process unit only if the measurements is correct and there are no leaks.
The following is the general way of writing a balance of a conserved quantity – mass of a particular species, total mass, energy and momentum – in a system (a single process unit, a collection of units, or entire process):
input            +           generation      -       output      -     consumption       =    accumulation

Input: enter system boundaries
Generation: produced within the system
Output: leaves the system boundaries
Consumption: consumed within the system
Accumulation: build-up in the system

The two types of balances, differential and integral balances can be written:
1. Differential balances:
This balance indicates what is happening in a system at an instant time. In this balance equation, each term is a rate and has units of the balanced quantity unit divided by a time unit. A differential balance is usually applied to a continuous process.

2. Integral balances:
This balance shows what happens between two instants of time. In the balance equation, each term is an amount of the balanced quantity and has a corresponding unit. An integral balance is applied to a batch process, with the two instants of time being after the time the input takes place and the time before the product is withdrawn.

It is rather a concern taking into consideration that differential balances applied to continuous steady-state systems and integral balances applied to batch systems, have a huge difference between their initial and final states. The following rules can be used to reduce the material balance equation:
1. Set generation = 0 and consumption = 0, if the balanced quantity is total mass.
Except: nuclear reactions (mass cannot be destroyed or created)
2. Set generation = 0 and consumption = 0, if balanced substance is nonreactive (no reactant or product)
3. Set accumulation = 0, if system is at steady state, regardless what being balanced (nothing can change with time, thus amount of quantity does not change).

 

If you are still finding mass balances difficult, this can help:

http://faculty.poly.edu/~rlevicky/Handout3.pdf

1. Process classification
Chemical processes can be classified as batch, continuous, or semibatch and as steady-state or transient. It is important to know which of these categories your process falls, before you can write any material balances.

1. Batch Process:
In this process, the feed is fed into a vessel at the beginning of the process and removed at a later stadium out of the vessel, where no mass crosses the system boundaries between the time the feed is fed and the product is removed.
Example: Add reactants to a tank when the system has come to equilibrium remove the product and unconsumed reactants.

2. Continuous Process:
The input and the output of the process flow continuously throughout the duration of the process.
Example: A mixture of liquids is pumped at a constant rate into a distillation column, and streams are withdrawn at the top and the bottom of the column.

3. Semibatch Process:
This process is neither batch nor continuous process.
Example: A balloon is filled with air at a steady rate.
A process is operating at a steady state if the values of all the variables in the process do not change with time, except for small fluctuations about a constant mean values. A transient or unsteady-state exists when the process variables change with time. Furthermore, batch and semibatch processes are transient operations, and continuous process can be steady-state or transient.
A continuous process is used more often where large production rates is present, whereas a batch process is commonly used when small quantities of a product are to be produced on any single occasion. Most continuous processes are run as close to steady-state processes as possible because transient conditions exist at the beginning of the process and further changes will occur.

A mass balance is a conserved quantity in a system and may be written in the following way:
input + generation - output - consumption = accumulation
a great link to following for an exceptional explanation would be

https://www.youtube.com/watch?feature=player_embedded&v=10qbOTikL1k

Problem 4.32 on page 166 in Elementary Principles of Chemical Processes by Richard M. Felder and Ronald W. Rousseau, 2005.


Fresh orange juice contains 12.0 wt% solids and the balance water, and concentrated orange juice contains 42.0 wt% solids. Initially a single evaporation process was used for the concentration, but volatile constituents of the juice escaped with the water, leaving the concentrate with a flat taste. The current process overcomes this problem by bypassing the evaporator with a fraction of the fresh product stream is mixed with the bypassed fresh juice to achieve the desired final concentration.

a. Draw and label a flowchart of this process, neglecting the vaporization of everything in the juice but water. First prove that the subsystem containing the point where the bypass stream splits off from the evaporator feed has one degree of freedom. (if you think it has zero degrees, try determining the unknown variables associated with this system). Then perform the degree-of-freedom analysis for the overall system, the evaporator, and the bypass-evaporator product mixing point, and write in order the equations you would solve to determine all unknown stream variables. In each equation, circle the variable for which you would solve, but don’t do any calculations.

b. Calculate the amount of product (42% concentrate) produced per 100 kg fresh juice fed to the process and the fraction of the feed that bypasses the evaporator.

c. Most of the volatile ingredients that provide the taste of the concentrate are contained in the fresh juice that bypasses the evaporator. You could get more of these ingredients in the final product by evaporating to (say) 90% solids instead of 58%; you could then bypass a greater fraction of the fresh juice and thereby obtain an even better tasting product. Suggest possible drawbacks to this proposal.

A short introduction to mass balances:

The basis of mass or material balances reside on the law of conservation of mass, according to which mass can not be created or destroyed.

The general formula used in calculations is the following: accumulation = input - output + generation - consumption

To start with, here is a very simple example:

Suppose a slurry containing 50% water and 50% solids are fed into a tank at 100kg/minute. A conveyer belt removes 45kg solids and 5kg water per minute. What mass of solids and water will leave the tank through the outlet as part of the clarified water if no accumulation occurs?

Answer: 5kg-Solids and 45kg-Water

No accumulation, generation or consumption occurs...thus, INPUT = OUTPUT
In one minute, 50kg solids ENTERS the tank, meaning that 50kg should LEAVE the system.
50kg - 45kg removed by conveyer belt = 5kg solids.
The same principle applies to the water.