Fundamentals of Mass Balance

Lecturer: Chun-Yang Yin, Ph.D., AMIChemE, MIEAust

Notes provided by Dr Gamini Senanayake

Learning objectivesOn successful completion of this topic you should be able to:

1) Understand the basic concepts of mass balance in the context of chemical processes:

2) Sketch simple block flow diagrams; and3) Conduct mass balance analysis on chemical processes.

Suggested reading list• Felder, R.M., Rousseau, R.W., Elementary principles of

chemical processes, John Wiley, 2005.

• Geankoplis, C.J., Transport processes and unit operations, Prentice Hall, NJ, USA, 1993.

• Kelly, E.G., Spottiswood, D.J., Introduction to Mineral Processing. AMF, Perth, 1997.

INTRODUCTION

IntroductionIndustrial chemical or metallurgical processes involves: • Unit(s) for mixing reactants • One or more chemical reactors • Unit(s) for blending products • Unit(s) for heating and cooling process streams • Unit(s) for separating products from each other and from

unconsumed reactants • Unit(s) for removing potentially hazardous pollutants from

streams prior to discharging the streams to the plant environment.

Introduction• It is rare that a chemical reaction A→B proceeds to

completion in a reactor.

• Some A is present in the product which leads to decrease the purity of the product B, and waste of B.

• The intention is to improve the purity of A and to recycle unreacted B from the product to the reactor.

• The cost for separation and recycling equipment is compensated in two ways: (i) purchase less fresh reactants, (ii) sell the purified product at a higher price.

Figure: Typical unit operations in minerals processing• The original ore is reduced in size to

dimensions suitable for the subsequent processing

• The size reduction facilitates the physical separation (e.g. by flotation), pre-treatment of ore (e.g. roasting of sulfide), and leaching (chemical dissolution of metal),

• The pregnant leach solution (rich in dissolved metal ions) is separated from the residual solids,

• The pregnant solution is further purified using separation methods to remove undesirable impurities which may be present in solution,

• Metals or metal compounds are obtained from the solution using chemical precipitation or electrochemical techniques.

Figure: Size reduction equipment

Fundamentals of mass balance • Law of conservation of mass: Mass can neither be created nor

destroyed. Inputs and outputs of each individual unit operation and entire process should satisfy the mass balance equations.

Process

BatchE.g.: Rapidly add reactants to a tank and remove the products and unconsumed reactants sometime later when the system has reached equilibrium.

ContinuousE.g.: Pump a mixture of liquids into a distillation column at a constant rate and steadily withdraw product streams from the top and bottom of the column.

Semi-batchE.g.: Allow the contents of a pressurized gas reactor to escape to the atmosphere; slowly blend several liquids in a tank from which nothing is being withdrawn.

Steady, unsteady, and transient states

• If the values of all the variables in a process (T, P, V, flow rates) do not change with time, except possibly for minor fluctuations about constant mean values, the process is said to be operating at steady state.

• Continuous reactors run at steady state, but transient during stopping and starting times or when disturbed. The temperature, pressure etc. change with time during transient periods. By their nature, batch and semi batch are unsteady-state operations, whereas continuous processes may be either steady-state or transient.

Processes and reactor Types

General balance equation Input + Generation – Output – Consumption = Accumulation • Input: enters through system boundaries • Generation: produced within system • Output: leaves through system boundaries Consumption:

consumed within system • Accumulation: buildup within system.

General balance equation Rules to simplify the material balance equation • If the balanced quantity is total mass, generation = 0,

consumption = 0 (Except in a nuclear reactions) • If the balanced substance is a non-reactive species,

generation = 0, consumption = 0. • If a system is at steady state, accumulation = 0. • In a steady-state system nothing can change with time,

including the balanced quantity.

Balances on continuous steady-state processes • Input + generation = Output + Consumption

Performing mass balance calculations

• drawing a block diagram to show the process

• showing the information available on the block diagram

• identifying the unknowns

• carrying out the balance to derive relationships between unknowns

• solving the equations and providing the solution.

Mass balance calculations

• Use known input and output stream variables to calculate unknowns.

• Draw a FLOWCHART (or block diagram).

• Write the values and units of known variables at each stage.

• Specify the unknown variables (x, y, z etc)

• Derive equations between known and unknown variables.

• Solve equations.

Degree-of-Freedom Analysis • Draw and completely label a flowchart. • Count the unknown variables on the chart (nunknowns). • Count the independent equations relating them (nindependent equations). • Subtract the second number from the first. • The result is the number of degrees of freedom of the process. • ndf = (nunknowns) - (nindependent equations). • If ndf = 0, there are n unknowns in n independent equations and the problem can in principle be solved. • If ndf > 0, there are more unknowns than independent equations relating them. More information is required to write additional equations so that the problem can be solved. • If ndf < 0, there are more independent equations than unknowns. Either the flowchart is incompletely labeled or the problem is over specified.

Example 1 – Mass balance calculations (Felder and Rousseau, 2005)

The mass flow rate for input stream is 1,000 kg/h while mass flow rate for output streams for benzene and toluene are 450 and 475 kg/h respectively. Provide assumptions that you make to perform this calculation.

Example 2 – Mass balance calculations (Kelly and Spottiswood, 1995)

A PbS concentrate is produced by a rougher-cleaner flotation circuit (i.e. 2 unit operations in this circuit). Tailings are produced from both the rougher and cleaner unit operations. The cleaner tailings assay 20% PbS and are recycled together (mixed) with the fresh feed stream to the rougher cells, and the circulating load (recycle/fresh feed) is 0.25. The fresh assays 10% PbS and is delivered at the rate of 1000 t/hr. The recovery and grade in the concentrate are 98.2% and 90% PbS, respectively. Determine the flow rates and assays of the other streams.

Hint: (1) Always draw the block diagram at the beginning. (2) Perform separate mass balance analyses on rougher and cleaner by

considering 2 different components, namely, overall flow and just PbS. Obtain equations.

(3) Solve.

Example 2 – Mass balance calculations (Kelly and Spottiswood, 1995)

Rougher total mass balance: 1000 + 250 = R + T eq. (1) Cleaner total mass balance: R = 250 + C eq. (2) Rougher PbS mass balance: 1000*(10%) + 250*(20%) = T*t + R*r eq. (3) Cleaner PbS mass balance: R*r = C*(90%) + 250*(20%) eq. (4)

(Solution underneath)

Chemical Reaction Stoichiometry • Stoichiometry theory of the proportions in which chemical species combine

with one another.

• Stoichiometric 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.

• For example the reaction between butane and oxygen can be represented by the following reaction stoichiometry: C4H10 + O2 → CO2 + H2O (unbalanced equation)

C4H10 + 5.5O2 4CO2 + 5H2O (balanced equation)

• General equation for the combustion of a hydrocarbon:CxHy + (x+y/4)O2 = xCO2 + (y/2)H2O

Chemical Reaction Stoichiometry • The stoichiometric ratio of two chemical species participating in a reaction is

the ratio of their stoichiometric coefficients in the balanced reaction equation.

• The reaction between carbon and oxygen at high temperatures produce carbon monoxide (CO).

• The reaction can be represented by C + 0.5O2 CO or 2C + O2 2CO

• In both cases the stoichiometric ratios are: (1 mol CO generated)/(1 mol C consumed) AND (2 mol CO generated)/(1 mol O2 consumed)

Limiting and Excess Reactants • Consider C + O2 = CO2. If the ratio of (moles C present)/ (moles O2 present) is

equal to the stoichiometric ratio of 1/1 in this equation, the two reactants, C and O2, are said to be present in stoichiometric proportion.

• If 100 moles of C and 100 moles of O2 are present initially, the reaction proceeds to completion. C and O2 disappear at the same instant.

• If the initial quantities are 200 moles of C and 100 moles of O2, the O2 would run out first i.e. O2 is the limiting reactant (reagent), C is the excess reactant.

• If the initial quantities are 100 moles of C and 200 moles of O2, the C would run out first i.e. C is the limiting reactant, O2 is the excess reactant.

Limiting and Excess Reactants • If 40 moles of C and 60 moles of O2 are present initially the fractional excess of O2

is (60-40)/60 = 0.33. i.e. there is 33% excess O2 in the feed.

• If 100 moles of a reactant are fed and 80 moles react, the fractional conversion is 0.80 (percentage conversion 80%). The fraction unreacted is 0.20.

• If 40 mol/min of a reactant is fed and the percentage conversion is 70%, 40*0.70 = 28 mol/min has reacted and 40*(1-0.7) = 12 mol/min remains unreacted.

Balances on Reactive Processes Example:

One hundred moles of CH4 is completely converted to CO2(g) and H2O(g) by reacting with 1000 moles of air (21% O2, 79% N2). Calculate:(a)theoretical (or stoichiometric air), (answer = 952 mol)(b)% excess air (answer = 5 %)

(c)composition of product gas mixture (mol%) (answer = 9.1 %, 18.18%,

0.91%, 71.82%) .

**Taken from Page 70 of unit reader. Please note that the answers for (b) and (c) in the unit reader are incorrect. The correct answers are listed in this powerpoint slide.

Example – Balances on Reactive Processes

(a)CH4 + 2O2 = CO2 + 2H2O nCH4 = 100 mol nO2 (theoretical) = 200 mol (because 1:2 molar ratio in stoichiometry) air (theoretical) = 200 * 100/21 mol = 952 mol

(b) Excess air = 100*[(1000-952)/(952)] = 5%

(c)Product (output) composition: nCO2 = 100 mol

nH2O = 200 molnO2 = (21/100)*1000 – 200 = 10 mol nN2 = (79/100)*1000 = 790 mol (unreacted)

Total n = 100 + 200 + 10 + 790 = 1100 mol

Composition (mol %) CO2 = 100*(100/1100) = 9.09% H2O = 100*(200/1100) = 18.18% O2 = 100*(10/1100) = 0.91% N2 = 100*(790/1100) = 71.82%

(Solution underneath)