PEME1030PEME 1030 Engineering Sciences 1

Mass Balances I

Professor M. Ghadiri

Institute of Particle Science and Engineering

School of Process, Environmental and Materials Engineering

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Objectives:

•To provide a thorough understanding

of the concepts and techniques for

performing mass balance calculations.

•To provide a basis for all level 2

engineering modules.

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Outcome:

• Understand concepts and methodology for

performing mass balance calculations.

• Be able to write and solve mass balance

equations for processes (with and ) without

chemical reactions.

• Be able to apply mass balances to

engineering processes.

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Overview:

Basic texts:

• R.M. Felder & R.W. Rousseau, Elementary Principles of

Chemical Processes, John Wiley & Sons.

• D.M. Himmelblau & J.B. Riggs, Basic Principles and

Calculations in Chemical Engineering, Prentice Hall.

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What are Mass and Energy Balances?

Engineers deal with

various types of plants

(design, operation and control)

A plant consists of a sequence of processes/operations

suitable for generating power, converting raw materials into

finished products or processing/conditioning materials.

Mass and Energy Balances are basically calculations of the

mass and energy flows entering or leaving

processes/operations in a plant.

These calculations form the foundation part of plant design

and are extremely important.

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Principles of Mass and Energy Balances

•Law of Conservation of Mass

•Law of Conservation of Energy

•Example: A Chemical Plant Flowsheet -

Manufacture of Formaldehyde (20

tonnes per day)

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Units and Dimensions

Dimensions

A dimension is a property

that can be measured, such as length (L), mass

(M), time (t) or temperature (T) - [fundamental

dimensions]

or

calculated by multiplying or dividing other

dimensions, such as

t

L

time

distanceVelocity

33 LLengthVolume

3L

M

volume

massDensity

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Units

• are scales used to quantify dimensions;

• are specific values of dimensions defined by law

or custom;

• can be added, subtracted, multiplied or divided.

Basic Units

Units for the dimensions of length, mass, time,

absolute temperature, electric current, luminous

intensity,

Multiples Units

multiples or fractions of the basic units

(used for convenience e.g. years instead of

seconds, kilometers instead of meters, etc.)

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Derived Units

a) obtained by multiplying and dividing base or

multiple units (referred to as compound units), e.g.

m2, ft/min, kg/ms

or

b) defined as equivalents of compound units

e.g. 1 lbf = 32.174 lbm ft/s2.

Systems of Units

•SI (Systeme Internationale, mks) -- the "metric"

system commonly used in Europe

•Engineering (American, English, fps)

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Dimensionless Quantities and Groups

A dimensionless quantity can be

• a pure number e.g. 1, 20.3, or

• a combination of variables with no net

dimensions.

e.g. the Reynolds number for pipe flow in fluid

dynamics is defined as,

Re = (length)(length/time)(mass/ length3)

(length x time/mass)

this is known as a dimensionless group.

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• In engineering one will come across many

dimensionless groups.

• It is important to note that the numerical

value for a dimensionless group is

independent of the units chosen for the

primary quantities, provided the units are

consistent.

• Dimensional Homogeneity ---- every valid

equation must be "dimensionally

homogeneous" --- all additive terms must

have the same dimension.

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Converting Units

• To convert a quantity in terms of one unit to an

equivalent in new units, multiply by a "conversion

factor" --- a ratio of equivalent quantities

• Tables of conversion factors have been compiled.

• These are easy to use, as long one remembers

that only like units cancel out when divided, unlike

units do not.

• The technique is simply to arrange the conversion

factors such that the unwanted units cancel out.

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Example 1: Convert 20,000 Btu/hr into kW

From the conversion tables,

1 Btu = 1.055 kJ

1 hr = 3600 s

kWs

kJ

s

hr

Btu

kJ

hr

Btu

hr

Btu861.5861.5

3600

1055.12000020000

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Example 2: Convert 23 lbm.ft/min2 to kg.m/s2

From the conversion tables,

1 lbm = 0.454 kg

1 min = 60 s

1 ft = 0.3048m

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22

m

2

m

2

m smkg1088s60

1

ft

m30480

lb

kg4540

ftlb23

ftlb23 ..

min..

min

.

min

.

With practice it is not necessary to write down the

units, and only the numerical conversion factors

are required.

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Process Variables

• The quantities used to describe a process are

called process variables.

• To design or analyze a process, we need to

know the amounts, compositions, and condition

of materials entering, leaving, and within the

process.

• The process variables of interest to chemical

engineers are:

• Mass and Volume

• Composition

• Pressure

• Temperature

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Mass and Volume (Density and Specific Volume)

• The density of a substance is the mass per

unit volume of the substance (kg/m3, g/cm3,

lb/ft3, etc).

Density Mass

Volume• The specific volume of a substance is the

volume per unit mass, the reciprocal of the

density.

• The specific gravity(SG) of a substance is the

ratio of the density of the substance to the density

of a reference substance at a specific condition.

SG substance

ref

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Example

65 m3/h of liquid benzene (C6H6) is entering

a reactor.

• a) What is the mass flow rate of this

stream in kg/h?

• b) What is the molar flow rate in mol/h?

Mw=78.11 g/,mol

Density=0.879 g/mL

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Flow Rates

• The rate at which a material is transported

through a process is the flow rate of that

material.

• The flow rate of a process stream may be

expressed as a

mass flow rate, mass per time (kg/s) or

volumetric flow rate, volume per time (m3/s) or

molar flow rate, moles per time (moles/s)

• The mass and volume flow rates are related

through the fluid density: =m/V

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Composition

Moles

A gram-mole (mol, gmol) is the amount of a

species whose mass in grams is numerically

the same as its molecular weight.

Carbon Dioxide has a molecular weight of 44, so

1 mol of CO2 contains 44 grams.

This means one can use the molecular weight as

a conversion factor for going from mass to

moles.

No.of molsMass ingrams

Mol.Wt.g

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We also use kilogram- moles (kgmol) and pound-

moles (lbmol, mole). These are defined the same

way but using different mass units. 1 lbmol of CO2

contains 44 lbs.

We use the same conversion factors for moles as

for converting mass units (454 g-mole = 1 lbmole,

etc.).

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Mass and Mole Fractions and Ratios

Weight basis

• The weight (or mass) fraction of a

component is the weight of that component

expressed as a fraction of the total weight of

the mixture.

• Weight fractions always total 1.0.

• The weight percent of a component is its

weight fraction 100.

• Weight percentages always total 100%, and

are often denoted by the symbol %w/w.

• The weight ratio of one component to

another is the ratio of the weights of the two

components.

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Consider a mixture of 720 kg of water, 92

kg ethanol and 60 kg acetic acid.

Wt(kg) Weight

frac.

%w/w Wt. ratio

Water 720 0.8257 82.57 12.00

Ethanol 92 0.1055 10.55 1.533

Acetic

acid

60 0.0688 6.88 1.00

Total 872 1.000 100.0 -

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Volume basis

Wt

(kg)

Density

(kg/m3)

Volume

(m3)

Volume

fraction

%v/v Volume

ratio

Water 720 1000 0.72 0.8055 80.55 12.59

Ethanol 92 789 0.1166 0.1305 13.05 2.039

Acetic

acid

60 1049 0.0572 0.0640 6.40 1.00

Total 872 - 0.8938 1.000 100.0 -

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Volume basis

Wt

(kg)

Density

(kg/m3)

Volume

(m3)

Volume

fraction

%v/v Volume

ratio

Water 720 1000 0.72 0.8055 80.55 12.59

Ethanol 92 789 0.1166 0.1305 13.05 2.039

Acetic

acid

60 1049 0.0572 0.0640 6.40 1.00

Total 872 - 0.8938 1.000 100.0 -

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Molar basis

Wt

(kg)

Mol. wt. No. of

kmols

Mol

fraction

Mol % Mol

ratio

Water 720 18 40 0.9302 93.02 40

Ethanol 92 46 2 0.0465 4.65 2.0

Acetic

acid

60 60 1 0.0233 2.33 1.0

Total 872 - 43 1.000 100.0 -

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Example

A mixture of gases has the following

composition by mass:

O2: 16%; CO: 4%; CO2: 17%; N2: 63%

What is the molar composition?

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The Average Molecular Weight

ii My M

i

i

M

x

M

1

Where yi is the mol fraction and xi is the mass

fraction and Mi is the molecular weight of the ith

component.

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Example

Calculate the average molecular

weight of air.

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Example: Conversion between flow rates

A 0.50 molar aqueous solution of

sulfuric acid flows into a process unit

at a rate of 1.25 m3/min. The specific

gravity of the solution is 1.03.

Calculate (i) the mass concentartion

of sulfuric acid; (ii) its mass flow rate;

(iii) its mass fraction.

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Pressure

Atmospheric Pressure The air above the earth‟s surface exerts a hydrostatic pressure on the surface. This is “atmospheric pressure” which varies according to the height of the column. At sea level the pressure is

1atm 14.70lb

in

f

2

= 760 mm Hg = 101.325

kPa This value is called the “standard atmosphere” and is used as a unit of pressure measurement.

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Pressure Scales Pressures can be expressed by either absolute or relative scales. The absolute scale uses zero as the reference point and the pressure measured is known as the absolute pressure. The relative scale measures the pressure against the atmospheric pressure taken as the reference point and the pressure is called the gauge pressure. The absolute and relative pressures are related as: absolute pressure = gauge pressure + atmospheric pressure

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Absolute pressure is denoted either by the letters „abs‟ or by the letter „a‟. For example, 14.7 psi absolute is written as 14.7 psia, 1000 mm Hg absolute is written as 1000 mm Hg(abs). Gauge pressures are denoted either by the word „gauge‟ or by the letter „g‟. Therefore 250 psi gauge may also be written as 250 psig. Pressures below atmospheric pressure are called “vacuum pressures”. Vacuum pressure = Atmospheric pressure - Absolute pressure Vacuum pressures are usually denoted by the letters „vac‟

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Pressure Conversion

Air is flowing through a duct under a draft

of 40 mm water. The barometer indicates

that the atmospheric pressure is 730 mm

Hg. What is the absolute pressure of the

gas?

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Pressure Differences

In measuring the flow of fluids in a

pipeline, a differential manometer is used

to determine the pressure difference

across an orifice plate. The flow rate can

be calibrated with the observed pressure

drop. Calculate the pressure drop across

the orifice.

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TemperatureThe temperature of a substance in a particular state (solid, liquid or gas) is a measure of the average kinetic energy possessed by the substance molecules. In more simpler terms, one can say the temperature indicates the ‘hotness’ of a substance. Temperature Scales A defined temperature scale is obtained by arbitrarily assigning numerical values to two reproducibly measurable temperatures. For example, the freezing and boiling point of water at a pressure of 1 atm.

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Summary of Temperature Scales

The Celsius

(C) Scale

The Kelvin

(K) Scale

The Fahrenheit

(F) Scale

The Rankine (R) Scale

oC K

oF R

Boiling point of water

100 373 212 672

Freezing point of water

0 273 32 492

Absolute zero -273 0 -460 0

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Temperature Conversions

273 CTKT o

T R T Fo o 460

T F T Co o 18 32.

T C T Fo o 32 18.

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Conversion of Temperature Intervals A degree is both a temperature and a temperature interval. An interval of 1 Celsius or Kelvin degree therefore contains 1.8 Fahrenheit or Rankine degrees, leading to the conversion factors:

1.8°F/1°C, 1.8°R/1 K, 1°F/1°R, 1°C/1 K

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Temperature Conversion

Consider the interval from 20 ˚F to 80 ˚F.

1. Calculate the equivalent temperatures in

˚C and the interval between them.

2. Calculate directly the interval in ˚C.

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Temperature Conversion and

Dimensional Homogeneity

The heat capacity of ammonia, defined as the amount

of heat required to raise the temperature of a unit

mass of ammonia by 1˚ at constant pressure, is, over

a limited range of temperatures, given by the

expression:

F T 10 2.29 0.487 F lb

Btu C 4-

m

p

Determine the expression for Cp in (J g-1 ˚C-1) in terms

of T (˚C).

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Ideal Gas Mixtures

Consider a mixture of gases occupying a

volume V at a temperature T and total

pressure P. Suppose the mixture consists

of nA, nB, nC,.... moles of gases A, B, C,....

• From the ideal gas law

PV n n n ... RTA B C

(The total pressure is the sum of the partial pressures - Dalton’s Law)

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Values of the Gas Constant, R

Values used for the universal gas constant with different units are given below. 8.314 m3.Pa/mol.K 62.36 litre.mm Hg/mol.K 8.314 J/mol.K 0.08314 litre.bar/mol.K 0.7302 ft3.atm/lb mol.oR 1.987 cal/mol.K 0.08206 litre.atm/mol.K 10.73 psia. ft3/lb mol.oR 1.987 Btu/lb mol.oR

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Combustion gases having the following molar

composition are passed into an evaporator at a

temperature of 200 ˚C and a pressure of 743 mm Hg:

N2: 79.2%; O2: 7.2%; CO2: 13.6%.

Water is evaporated, the gases leaving at a

temperature of 85 ˚C and a pressure of 740 mm Hg

with the following molar composition:

N2: 48.3%; O2: 4.4%; CO2: 8.3%; H2O: 39.0%.

Calculate:

(a) Volume of gases leaving the evaporator per 100

m3 entering.

(b) mass of water evaporated per 100 m3 of gas

entering.

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Typical Exam QuestionA fully sealed 50 m

3 room at atmospheric pressure

and 20°C contains two balloons, one containing helium and the other containing oxygen. The balloons have each a volume of 0.5 m

3 and have

a gauge pressure of 2 m of water. The balloons are ruptured and helium and oxygen get mixed with air in the room.

Determine the mass and mol fractions of the

gaseous species in the room, assuming that air contains 79% N 2 and 21% O 2 by volume and the

atmospheric pressure is 100 kPa. The Universal Gas Constant is 8.314 m 3 Pa

mol 1 K 1 . Atomic molar mass of He is 4 g/mol.

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Material Balance

• The General Material Balance Equation

A material balance is nothing more than an accounting for material flows and changes in inventory of material for a system. A balance (or inventory) on a material in a system (a single process unit, a collection of units, or an entire process) may be written in the following general way for any process/system under consideration (see Figure 1):

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Input Generation Output Consumption Accumulation through + within - through - within = within (1) the system the system the system the system the system boundaries boundaries

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The general balance equation may be written for any material that enters or leaves any process system; it can be applied to the total mass of this material or to any molecular or atomic species involved in the process. Equation 1 may be applied in different ways according to the precise definition of 'material' and the way in which the process is operated.

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If the balance is applied to the total mass or to the mass of an element entering and leaving the system, the generation and consumption terms are zero (excluding nuclear reactions) and the equation becomes total mass - total mass = accumulation in out of mass (2) or mass of - mass of = accumulation of mass element i in element i out of element i (3) When considering quantities (mass or number of moles) of individual molecular species, material may be produced or consumed by chemical reaction. In the absence of chemical reactions equations like 2 apply also to the total number of moles of each molecular species.

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Processes may be classified as continuous or batch. Most processes operate with a continuous feed and form product continuously. In a batch process, materials are charged to a vessel and products withdrawn when the reaction is complete. Batch operation is usually used for low volume products, e.g. manufacture of pharmaceuticals

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The material balance equation for a batch process must necessarily include an accumulation term. Continuous processes are often assumed to operate at steady-state, i.e. process variables such as flows do not change with time. There is therefore no accumulation and the general balance equation becomes

rate of input + rate of generation = rate of output + rate of consumption (4)

or, for the total mass, mass flow in = mass flow out (5)

Alternatively, when considering the operation of a continuous steady-state process for a fixed period of time, each of the terms in equation 4 may be expressed simply as a mass or number of moles. Equation 2 thus becomes total mass in = total mass out

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The Black Box ConceptFrequently, the engineers have to analyse complex

process flow diagrams. As an example: Figure 2.

The large scale complex problems can be broken to

simpler sub-problems for writing balances using the black

box concept. For instance, we can place a black box

around all or any portion of this process and make our

balance. We do not care what happens inside the box,

only what is transferred across the box’s boundaries. If

we are interested in the total process, the box would look

like as in Figure 3. We could place a box around the

intersection of streams A, B, and recycle (Figure 4).

Similarly we can place boxes around the reactor or

absorber.

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Material Balance Techniques

It is usually helpful to follow a systematic procedure when tackling material balance problems. One possibility is outlined below. 1. Draw and label the process flowsheet- organize information into an easy to understand form. If possible show problem specifications on the flowsheet. Label unknowns with algebraic symbols. 2. Select a basis for the calculation- the basis is an amount or flowrate of a particular stream or component in a stream. Other quantities are determined in terms of the basis. It is usually most convenient to choose an amount of feed to the process as a basis. Molar units are preferable if chemical reactions occur, otherwise the units in the problem statement (mass or molar) are probably best.

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3. Convert units/amounts- as necessary to be consistent with the basis. 4. Write material balance equations- for each unit in the process or for the overall process. In the absence of chemical reactions the number of independent equations for each balance is equal to the number of components. 5. Solve equations- for unknown quantities. This can be difficult, particularly if non-linear equations are involved. Overall balances usually give simpler equations. For complex flowsheets computer methods offer, the only practical solution. 6. Scale the results- if the basis selected is not one of the flowrates in the problem specification the results must be scaled appropriately.