Intro to Mass Balance

FUNDAMENTALS OF MATERIAL BALANCE

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LEARNING OBJECTIVES

By the end of this topic, you should be able to:

• Performed material balance for various systems-Mass Balances- Mass balances in non-reactive systems-Mass Balances- Mass balances in reactive systems

• The law states that mass can neither be created nor destroyed

• Material balance equations are the manifestation of the law

TOTAL MASS INPUT = TOTAL MASS OUTPUT

• The design of a new process or analysis of existing one is not complete until it is established that the inputs and outputs of the process satisfy the material balance equation.

Material balance are based on :Law of Conservation of Mass

Process ClassificationThree type of process1. 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

2. Continuous process– Input and output is continuously red 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.If the process variables such as flow rates, temperatures, pressures, volume etc do not change with time, the process is said at STEADY STATE. On the other hand, if the process variables do change with time, the process is said at UNSTEDY STATE or TRANSIENT. By nature, batch and semi-batch process are unsteady state operation, while continuous process is steady state operation.

Types of Systems

• Open System: inlet/outlet of materials to system

• Closed System: no inlet/outlet of materials

• Isolated System: no inlet/outlet of materials; no inlet/outlet of energy to the system

Types of Systems

Type of Systems

• Isobaric system: Constant Pressure

• Isothermal system: Constant temperature

• Isochoric system: Constant Volume

• Adiabatic system: No heat exchange Q=0

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EXAMPLE 1

Process Classification RemarkA balloon is filled with air at a steady state rate of 2 g/min

Semi-batch, transient Volume, T, P change with time.

Water is boiled in an open flask

Semi-batch, transient Volume, T, P change with time.

Water is boiled in an closed flask

Batch, transient T & P change with time.

Gasoline from car tank Semi-batch, transient Volume change with time

Carbon dioxide and steam are fed into reactor to form carbon monoxide and hydrogen

Continuous, steady state All process variables do not change with time.

Material Balance Equation• Suppose methane, is a component of both input and

output of a process

• If the flow rates of input and output are found to be different. Possible explanations are .…1. methane is leaking2. methane is consumed or generated in a reaction 3. methane is accumulating in the process vessel4. wrong measurement

Process unit

qin(kg CH4/h) qout(kg CH4/h)

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General Material Balance Equation

• A balance on a material in a process system may be written as:

Input + generation - output - consumption = accumulation• The equation may be written for any material that enters

or leaves any process system• It can be applied to the total mass or total moles of this

material or to any atomic species involved in the process

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EXAMPLE: The General Balance Equation• Each year 50,000 people move into a city, 75,000 people

move out, 22,000 are born, and 19,000 die. • Write a balance on the population of the city.• SOLUTION Let P denotes people:• Input + generation - output - consumption = accumulation

• Each year the city's population decreases by 22,000 people.

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Differential vs. Integral Balances

Two types of balances may be written:1. Differential balances

– balances that indicate what is happening in a system at an instant time.– balance equation is a rate (rate of input, rate of generation, etc.) and has

units of the balanced quantity unit divided by a time unit We use quantities rather than velocities (kg, ton, lb)

– usually applied to a CONTINUOUS process.2. Integral balances

– Balances that describe what happens between two instants of time. – balance equation is an amount of the balanced quantity and has the

corresponding unit (people, g SO2).– usually applied to a BATCH process, with the two instants of time

being the moment after the input takes place and the moment before the product is withdrawn.

Simplified Rule for Mass Balance

• If the balanced quantity is TOTAL MASS, set generation = 0 and consumption = 0. Mass can neither be created nor destroyed.

• If the balanced substances is a NONREACTIVE SPECIES (neither a reactant nor a product), set generation = 0 and consumption = 0.

• If a system is at STEADY STATE, set accumulation = 0, regardless of what is being balanced.

Balances on Continuous Steady State Process

• Steady state: accumulation = 0

INPUT + GENERATION – OUTPUT – CONSUMPTION = 0

• If balance on nonreactive species or total mass; balance equation become

INPUT = OUTPUT

Integral Balances on Batch Process

• Ammonia is produced from nitrogen and hydrogen in a batch reactor. At time t = 0 there are n0 mol of NH3 in the reactor, and at a later time tf the reaction terminates and the contents of the reactor, which include nf ammonia, are withdrawn. Between t0 and tf no ammonia enters or leaves through the reactor boundaries.

From GMBE: (input=0; output=0)Generation – Consumption = Accumulation

For batch reactor:Accumulation = Final output – Initial Input

Final GMBE for batch processInitial input + Generation = Final output + Consumption

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ECB 3013 MATERIAL AND ENERGY BALANCESSteps in Solving Material Balance

1. Read and try to understand on the process description. What type of process unit used and what type of process operation.

2. Draw a flowchart for the process description using boxes or other symbols to represent process unit or unit operation and lines with arrows to represent inputs and outputs.

3. Write all known stream variables i.e. inputs & outputs, on the flowchart.

4. Assign algebraic symbols to unknown stream variables.

5. Do the degree of freedom analysis. You must have equal number of equations related to unknown process variables (zero degree of freedom).

6. Take basis of calculation.

7. Write mass balance equation for the overall system and for specific component using selected Eq [1] to [6].

8. Perform mass balance for the process description. Always check the overall mass balance for Total Inputs = Total Outputs

MASS BALANCE CALCULATION

General Procedure for Single Unit Process Material Balance Calculation

1. Choose as basis of calculation an amount or flow rate of one of the process streams.2. Draw a flowchart and fill in all unknown variables values, including the basis of

calculation. Then label unknown stream variables on the chart.3. Express what the problem statement asks you to determine in terms of the labeled

variables.4. If you are given mixed mass and mole units for a stream (such as a total mass flow rate

and component mole fractions or vice versa), convert all quantities to one basis.5. Do the degree-of-freedom analysis.6. If the number of unknowns equals the number of equations relating them (i.e., if the

system has zero degree of freedom), write the equations in an efficient order (minimizing simultaneous equations) and circle the variables for which you will solve.

7. Solve the equations.8. Calculate the quantities requested in the problem statement if they have not already been

calculated.9. If a stream quantity or flow rate ng was given in the problem statement and another value

nc was either chosen as a basis or calculated for this stream, scale the balanced process by the ratio ng/nc to obtain the final result.

Flowcharts

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

• The best way 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.

Flowchart Example

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/minB: 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.

Flowchart Example

Evaporation

20 cm3 H2O (l)/min

mol H2O/min2n

mol O2/min1200.0 n

mol air/min1n

mol/min3n

0.21 mol O2 /mol0.79 mol N2 /mol

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

Flowcharts Labeling• 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/mol0.79 mol N2/molT = 320˚C, P = 1.4 atm

Flowcharts Labeling• Process stream can be given in two ways

1. As the total amount or flow rate of the stream and the fractions of each component

2. Or directly as the amount or flow rate of each component.

60 kmol N2/min40 kmol O2/min

0.6 kmol N2/kmol 0.4 kmol O2/kmol

100 kmol/min

3.0 lbm CH44.0 lbm C2H43.0 lbm C2H6

0.3 lbm CH4/lbm0.4 lbm C2H4/lbm0.3 lbm C2H6/lbm

10 lbm

Flowcharts Labeling• 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 associated units on the flowchart.

mol/h

0.21 mol O2/mol0.79 mol N2/molT = 320˚C, P = 1.4 atm

n 400 mol/h

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

Flowcharts Labeling• 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

• If a volumetric flow rate of a stream is given, convert to mass or molar flow rate since balances are not normally written in volumetric quantities

• When labeling component mass or mole fractions of a stream the last one must be 1 minus the sum of the others.

• If you are given that the mass of stream 1 is half that of stream 2, label the masses of these streams m and 2m rather than ml and m2;

• if you know that there is three times as much nitrogen (by mass) in a stream as oxygen, label the mass fractions of O2 and N2 y(g O2/g) and 3y(g N2/g) rather than yl and y2.

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Consistent on Notation

gasin fraction moles yliquidin moles)or (massfraction component x

rate flow volumeV

volumeVrate flowmolar n

moles nrate flow mass m

mass m

• One thousand kilograms per hour of a mixture of benzene (B) and toluene (T) containing 50% benzene by mass is separated by distillation into two fractions. The mass flow rate of benzene in the top stream is 450 kg B/h and that of toluene in the bottom stream is 475 kg T/h. The operation is at steady state. Write balances on benzene and toluene to calculate the unknown component flow rates in the output streams.

Balances on Steady- State Continuous Processes (Continuous Distillation Process)

450 kg B/hrq1 (kg T/hr)

500 kg B/hr500 kg T/hr

475 kg T/hrq2 (kg B/hr)

Distillation

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450 kg B/hrq1 (kg T/hr)

500 kg B/hr500 kg T/hr

475 kg T/hrq2 (kg B/hr)

Distillation

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no reaction, rate of input = rate of output

EXAMPLE: Balances on a Batch Mixing Process• Two methanol-water mixture 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 are combined with 150 g of the second, what are the mass and composition of the product.

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Balancing a Process

• Which balance to be used when a choice exists and the order in which these balanced should be written?

• Rules of thumb for NONREACTIVE process1. The maximum number of independent equations that

can be derived by writing balances on a nonreactive system equals the number of chemical species in the input and output streams.

2. Write balances first that involve the fewest unknown variables.







7. Solve the equations.8. Calculate the quantities requested in the problem statement if they have not already

been calculated.9. If a stream quantity or flow rate ng was given in the problem statement and another

value nc was either chosen as a basis or calculated for this stream, scale the balanced process by the ratio ng/nc to obtain the final result.

Basis of Calculation• Balanced process can always be scaled. Mean that material balance

calculation can be performed on the basis of any convenient set of stream amount or flow rate and the results can afterward be scaled to any desired extent.

• 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

Degree-of-Freedom• Before doing any material balance calculation, use a properly drawn and

labeled flowchart to determine whether there is enough information to solve a given problem.

• The procedure for doing so is referred to as degree-of-freedom analysis.

• Procedure to perform a degree-of-freedom analysis:a) draw and completely label a flowchartb) count the unknown variables on the chart (n unknowns)

c) count the independent equations (n indep. eq.)

d) Find number of degree-of-freedom (ndf)

ndf= n unknowns - n indep. eq.

Number of Degree-of-Freedom• Three possibilities number of degree-of-freedom (n df)1. If ndf = 0

– the problem can in principle be solved.

2. If ndf > 0– there are more unknowns than independent equations relating to them– at least ndf additional variable values must be specified before remaining

variable values can be determined.– Either relations have been overlooked or the problem is underspecified.

3. If ndf < 0– there are more independent equations than unknowns.– Either the flowchart is incompletely labeled or the problem is overspecified

with redundant and possibly inconsistent relations. • There is little point wasting time trying to solve material balance for n df > 0 or n df

<0.

6 Sources of Equation for Balance1. Material balances.

– For a nonreactive process, number of independent equation can be written is not more than number of molecules species (n ms) of the process

– If benzene and toluene is involve in stream, we can write balance on benzene, toluene, total mass, atomic carbon and etc., but only TWO INDEPENDENT balance equation exist

2. An energy balance.– If the amount of energy exchanged between the system and its surroundings

is specified or if it is one of the unknown process variables, an energy balance provides a relationship between inlet and outlet material flows and temperatures.

3. Process specifications– The problem statement may specify how several process are related.– i.e: Outlet flow rate is two times than flow rate stream 1 or etc.

4. Physical properties and laws– Two of the unknown variables may be the mass and volume of a stream

material, in which case a tabulated specific gravity for liquids and solids or an equation of state for gases would provide an equation relating the variables.

5. Physical constraints– For example, if the mole fractions of the three components of a stream

labeled xA, xB, and xC, then the relation among these variables is xA + xB + xC = 1.

– Instead label as xc, the las fraction should be 1-xA-xB

6. Stoichiometric relations– If chemical reactions occur in a system, stoichiometric equation provide a

relationship between the quantities of reactant and the product

6 Sources of Equation for Balance

• Example :An aqueous solution of NaOH contains 20% NaOH by mass. It is desired to produce an 8% NaOH solution by diluting a stream of 20% solution with a stream of pure water.

• Calculate the ratios (liters H2O/kg feed solution) and (kg product solution/kg feed solution).

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NaOH balance (input = output).

Total mass balance (input = output).

Ratios requested in problem statement.

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• Example A stream of humid air enters a condenser in which 95% of the water vapor in the air is condensed. The flow rate of the condensate (the liquid leaving the condenser) is measured and found to be 225 L/h. Dry air may be taken to contain 21 mole% oxygen, with the balance nitrogen.

• Calculate the new rate of the gas stream leaving the condenser and the mole fractions of oxygen, nitrogen, and water in this stream.

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• Solution: Basis: 225 L/h Condensate

• Suppose that the entering air contains 10.0 mole% water

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• Example: Material Balances on a Distillation Column A liquid mixture containing 45.0% benzene (B) and 55.0% toluene (T) by mass is fed to a distillation column. A product stream leaving the top of the column (the overhead product) contains 95.0 mole% B, and a bottom product stream contains 8.0% of the benzene fed to the column (meaning that 92% of the benzene leaves with the overhead product). The volumetric flow rate of the feed stream is 2000 L/h and the specific gravity of the feed mixture is 0.872.

• Determine the mass flow rate of the overhead product stream and the mass flow rate and composition (mass fractions) of the bottom product stream.

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• Choose a basis. feed stream flow rate (2000 L/h) as the basis of calculation.

• Draw and label the flowchart.

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• Volumetric flow rate conversion

The benzene in the bottom product stream is 8% of the benzene in the feed stream

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The results areṁl = 1744 kg/h, ṁB3 = 62.8 kg benzene/h, ṁ2 = 766 kg/h , and ṁT3 = 915 kg toluene/h. A total mass balance (which is the sum of the benzene and toluene balances) may be written as a check on this solution:

additional quantities requested in the problem statement

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Intro to Mass Balance

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