Lecture 5-8: Mass balances on Non-reactive systems These only apply for physical process. you can do total mass balance equations as well as component balances, and everything has to be in Kg or using a unit of mass. We can also use elemental balance but it does not work out because the ratio of the atoms are the same for all streams since there is no chemical change. The overall equation goes like: 1. Rule 1: interpret data and draw a flowsheet. Remember to always name the unit and streams! 2. Rule 2: Find out as much information as you can be regarding the process, that is the total mass, the components

compositions. 3. Rules 3: check if units are consistent. That is, if they are all the same, like in wt%. 4. Rule 4: draw a control volume and undertake degree of freedom analysis.

• The tabular method: this is useful to figure out where you can begin. We need to:

Ø Count the number of unknowns. Ø Count the number of independent equations we can write. That is there is always going to be n-1

independent number of equations that can be used to solve the system. where n is the total number of equations.

Balances Streams Unknowns = 3 Feed Process Feed Product Total X X X Benzene X X Toluene X X X Ind. Equations = 2

Note: compositions remain the same if a stream is split. The compositions do not remain the same if streams converge. Composition changes whenever of a compound is eliminated from a given stream. The following are examples of the type of balances that must be taking into consideration whenever we encounter different systems whenever: • No reactions are involved: Total mass balance and Component mass balances.

• Chemical Reactions (but quantities reacting are unknown): Elemental Balance and Tie Component.

• Chemical Reactions (quantities reacting are known): Component Balance, Elemental Balance and Tie

Component. The mass balance equation (this refers to moles) is given by the following expression.

Nin = moles in. Nout = moles out.

rV = Rxn, moles gained or lost due reactions.

ind. Equations = unknowns à the system can be solved ind. Equations < unknowns à system cannot be solved. ind. Equations > unknowns à system is over specified

Input + generation – consumption – output = accumulation

𝐴𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛(𝑑𝑁𝑑𝑡) = 𝑁12 − 𝑁456 + 8 𝑟𝑑𝑉

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We can simplify the above equation based on what system we are analyzing: • Steady-state (chemical reactions): mass in + mass (Rxn) = mass (out). Since it is a steady state (constant flow) no

accumulation is observed.

• Steady state (non-reactive): mass(in) = mass(out). There is no accumulation and hence no chemical reactions.

• Unsteady state (reaction): mass(Rxn) = accumulation (dN/dt). This is since there is no constant flow and thus there would be an accumulation. This is dependent upon reaction time because what gets reacted will get accumulated. E.g. batch system.

• Unsteady state (non-reactive): mass = accumulation (dN/dt).). This is since there is no constant flow an thus there would be an accumulation. This is dependent upon reaction time because what gets reacted will get accumulated. E.g. batch system.

Lectures 8-12: Mass balances on Reactive systems Basic information for mass balance: for this particular mass balance the first first we need to consider are is the limiting reagent. Everything is based from the point of view of the limiting reagent. 1. Limiting reagents: are present in less than its stoichiometric proportions, the limiting reagent ratio is always

biggest number over smallest:

For example: 300 kgmoles of C2H4 and 100 kgmole of oxygen were burned. Determine the limiting reagents. 2C2H4 + O2 à 2C2H4O

2. Excess reactants: present in excess proportions respectively.

For example: the feed consists of 100 moles of A and 100 moles of B. A and B reacts to form C according to A + 2B à C. determine the limiting reagent and the % excess:

3. Yield: is the product produced based on the reactions occur.

4. Selectivity: how selective was the reaction based on other undesired reactions. For a reaction: A + B à C (desired reaction)

A + B à D (undesired reaction)

Biggest number Smallest number

5. Fractional conversion: the amount of product converted per unit pass.

For example: calculate the fractional conversion for the following unit process based on the chemical reaction: 2A + B à C

For example: the reaction Sb2F3 + 3Fe à 2Sb + 3FeS. Suppose 1.76 gmole of Sb2F3 and 4.47 gmol of Fe were converted to 1.64 gmoles of Sb. Determine: i) Limiting reagent: ii) The percentage excess reactant: iii) The fractional conversion with respect to the limiting reagent. iv) Percent conversion of Sb2S3

Combustion: The combustion reactions involve the reaction of a carbon based compounds material with oxygen (or air). The source of oxygen generally used for combustion of fuel is air, since it is cheap and readily available. The combustion reaction of fuel will produce the following product gases: CO, CO2, O2, N2, SO2, H2O. All these compounds are in gaseous forms. Orsat (dry) vs Wet basis: all gases excluding water and wet basis means it includes water. If the question does not specify whether it is a wet basis, then water in not produced and vice versa. Complete combustion vs partial combustion: if all the carbon is converted to CO2, then it is called a cpmplete combustion. If, however, carbon is also converted to CO, then it is called a partial conversion. Important information: this is the information vital for balancing combustion system. 1. Theoretical Oxygen: the moles of oxygen required to burn all the fuel into the reactor. All C is converted to CO2

and all H into H2O.

2. Theoretical air: moles of air which contains the theoretical oxygen needed for reaction. Air always contains 21 mol% of O2 and 79 mol% of N2.

3. Excess air or Excess oxygen: amount by which the air fed to the reaction exceeds the theoretical air (the oxygen

required for a full combustion).

For example: 100 moles/hr of C4H10 and 5000 moles/hr of air are fed into a combustion reactor. Calculate the percent excess air: • O2(fed) = 5000 moles/hr x 0.21 = 1050 moles • Theoretical air or oxygen = 650 moles/hr

• % excess = (1050 – 650)/650 = 62% For example: 100 % methane is fed into a burner with 130% excess air. Assuming complete combustion occurs; give the composition of the product gas both as stack (wet basis) and orsat analysis (dry basis).

Elemental balance: matter cannot be created nor destroyed. So the mole ratios need to be taken into consideration. The rest will work just as mass balance.

ELEMENT IN = ELEMENT OUT