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.

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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.

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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

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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

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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)

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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