Mass Balances

Mass Balances

MASS BALANCES 1 1.1

Systems of units Systeme International ( SI ) SI system is the official system throughout the world. Fundamental( basic) units Length mass time temperature

- metre - kilogram - second - kelvin

-m - kg -s - K (not degree kelvin)

Derived units Volume force pressure energy 1.2

- m3 - N ( Newton) or kg m / s2 - N / m2 or pascal ( Pa) - J (Joule) or Nm

American Engineering length mass time temperature volume force

1.3

- cubic meters - kilogram meter / second2 - Newton / m2 - Newton meter

- foot - pound mass - second - degree Fahrenheit - cubic foot - pound force

- ft - lbm -s - 0F - ft3 - lbf

Centimetre-gram-second( cgs) length mass time volume temperature

- centimetre - gram - second - cubic centimetre - degree centigrade

- cm - gm -s - cm3 - C or K

1.4 General From sections 1.1,1.2, and 1.3 it is clear that mass and volume have different units based on the system we use. These units must always be included with the numerical value or it has no meaning. Thus 1000g and 1 kg are identical and it does not matter which is used, provided the correct label (unit) is given. In general, chemical engineers tend to use SI units i.e. kg for mass and m3 for volume. e.g.

Flow rate of feed into a reactor is 150 kg/s. Just 150 is not acceptable.

1

Mass Balances

2. SYMBOLS Chemists and Engineers use symbols to represent atoms. Every element has a different symbol. Elements are listed in the Periodic Table. Oxygen = O The letter O stands for one atom and O2 stands for two atoms of oxygen Sometimes two letters are needed to represent an element eg

Sodium

=

Na

second letter always in lower

case first letter is always written in capital letters Whether the symbol of an element is one letter or two letters, the symbol stands for one atom of the element. When an element is represented by two letters, the first letter is in capital and the second letter is in lower character. Sometimes the letters are taken from the Latin name of the element e.g. Copper

= Cu

from Cuprum (Latin name)

Iron

= Fe

from Ferrum (Latin name)

Cobalt

= Co

( single element)

Aluminium

= Al

( single element)

Number of elements in a compound Carbon dioxide

= CO2

Sulphuric acid

= H2SO4

(a compound consisting Carbon and Oxygen - both letters capital) ( three capital letters, hence there are three elements – H, S and O )

Calcium phosphate = Ca3(PO4)2

(three capital letters, hence there are three elements– Ca, P and O)

Symbols of some common elements Element Aluminium Barium Bromine Calcium Carbon Chlorine Copper Fluorine

Symbol Al Ba Br Ca C Cl Cu F

Element Gold Hydrogen Iodine Iron Lead Magnesium Mercury Nitrogen

2

Symbol Au H I Fe Pb Mg Hg N

Element Oxygen Phosphorus Potassium Silver Sodium Sulphur Tin Zinc

Symbol O P K Ag Na S Sn Zn

Mass Balances

3

Naming of Compounds When you try to name an inorganic compound from the chemical formula the first thing you should do is to name the first symbol. In the formula CO 2 the first symbol C is carbon and therefore the name should begin with carbon --------. In the formula CaCO3, the first symbol is calcium and therefore the name should begin with calcium -------. The reminder of the formula indicates the rest of the name. When the ending is a single element, the name is usually derived by ending the name with an – ide ending 1)

Compounds made up of a metal and a non-metal:- metal is always written first. Eg. NaCl sodium chloride ( Na is a metal and Cl is a non-metal) MgO magnesium oxide ( Mg is a metal and O is a non-metal)

2)

Compounds made up of two non-metals:- the non-metal that is either lower down the group(if both non-metals are in the same group) or nearest to the left hand side of the periodic table is placed first. e.g.

3)

sulphur and oxygen carbon and oxygen carbon and sulphur

SO2 CO2 CS2

sulphur dioxide carbon dioxide carbon disulphide

Compounds made up of three elements a) metal, non-metal, and oxygen: these compounds have - ate ending. e.g.

Na, S, O

Na2SO4

sodium sulphate

In this case you ignore the fact that there are two sodium atoms b) metal, hydrogen, oxygen: these compounds have a hydroxide ending with O written before H. e.g.

K, H, O Na, O, H

KOH NaOH

potassium hydroxide sodium hydroxide

3.1 Names used for the end groups of inorganic formulae Unless you know the names of the groups listed in the table given below, you cannot name the compounds. Ensure you learn these names. Group O S Cl F I Br OH SO4

Name used for the end group Oxide sulphide chloride fluoride iodide bromide hydroxide sulphate

Exampl e ZnO ZnS NaCl CaF2 KI PbBr2 NaOH MgSO4 3

Name zinc oxide zinc sulphide sodium chloride calcium fluoride potassium iodide lead bromide sodium hydroxide magnesium

Mass Balances

NO3 CO3 HCO3

nitrate carbonate Hydrogen carbonate or bicarbonate

NaNO3 BeCO3 NaHCO3

sulphate sodium nitrate berillium carbonate sodium bicarbonate Sodium hydrogencarbonate

Some compounds are commonly referred to by non-systematic names. The more common ones are given below in the table. You should learn these names. Formula H2O HNO3 H2SO4 HCl (aq)

Name water nitric acid sulphuric acid hydrochloric acid (acid solution) hydrogen chloride( gas ) ammonia nitric oxide methane

HCl (g) NH3 NO CH4

Another fact you should know is that some elements can form more than one compound with another element. The commonest example is CO and CO2, known as carbon monoxide and carbon dioxide. In general, if an element forms more than one compound with another element the following prefixes are used: One Two Three Four

= = = =

monoditritetra-

e.g. CO e.g. SO2 e.g. SO3 e.g. CCl4

carbon monoxide sulphur dioxide sulphur trioxide carbon tetrachloride

3.2 Formulae (plural for formula) Calculations are based on formulae and equations. A formula represents one molecule of a substance and consists of the symbols of the elements present (and some numbers given in subscripts- small characters). The numbers show the ratio in which the atoms are present in the compound. Water consists of particles and each particle contains one oxygen atom joined by chemical bonds to two hydrogen atoms. The particle is called a molecule of water (mol for short) e.g. sulphuric acid H2SO4

has three different elements but 7 atoms in total i.e. 2 H atoms, 1 S atom, 4 O atoms ; total atoms = 7

3 H2SO4

3 moles of sulphuric acid

The three in front of H2SO4 multiplies every thing which comes after it. ∴total number of atoms = 3 { 1 x 2 + 1 x 1 + 1 x 4 } = 3 X 7 = 21 atoms 4

Mass Balances

H i.e. there are e.g.

S

6 H atoms,

O

3 S atoms,

12 O atoms : total = 21 atoms

calcium hydroxide Ca(OH)2

three different elements with 1 Ca, 2 O, 2 H atoms total number of atoms = 5 atoms

4Ca(OH)2

4 Ca, 8 O, 8 H

total = 20 atoms

when you write the folmula of a compound you need to know the valancies of the elements and ions. Learn the following: Name

Symbol

Hydrogen Ammonium Potassium Sodium Silver Copper (I)

H+ NH4+ K+ Na+ Ag+ Cu+

Charge (valency) +1 +1 +1 +1 +1 +1

Barium Calcium Copper (II) Iron (II) Lead Magnesium Zinc Aluminium Iron (III)

Ba2+ Ca2+ Cu2+ Fe2+ Pb2+ Mg2+ Zn2+ Al3+ Fe3+

+2 +2 +2 +2 +2 +2 +2 +3 +3

Name

Symbol

Hydroxide Nitrate Chloride Bromide Iodide Hydrogencarbonate Oxide Sulphide * Sulphite * Sulphate * Carbonate

OH1NO31Cl1Br1I1HCO31O2S2SO32SO42CO32-

-2 -2 -2 -2 -2

Phosphate

PO43-

-3

You’ll notice that some components have variable valency e.g. Fe2+

and

Charge (valency) -1 -1 -1 -1 -1 -1

Fe3+

Cu+

Note: * note the difference in spelling

5

and

Cu2+

Mass Balances

4.

Names of Organic Compounds Organic compounds Alkane series ( CnH2n+2 ) CH4 C2H6 C3H8 C4H10 C5H12 C6H14 C7H16 C8H18

Alkyne series ( CnH2n-2) Naming is same as alkane but replace –ane with -yne

methane ethane propane butane pentane hexane heptane octane

C2H2 C3H4

ethyne propyne

Aromatic hydrocarbon C6H6 C6H5CH3

benzene methylbenzene

Alkene series (CnH2n)

5.

naming same as alkane, but replace -ane by –ene

Alcohol

C2H4 C3H6

CH3OH C2H5OH

ethene propene

methanol ethanol

The Amount of Substance - The Mole Amount is a physical quantity like mass, volume etc. If the object is very small and if there are millions of them it is easier to count by weighing. e.g. Banks count coins by weighing. If you know how many coins are in 100 g it is easy to determine the number of coins in 1000 kg. Atoms, molecules and ions are so small that they are almost counted by weighing. The counting unit is the MOLE (abbreviation is mol). i.e. mole represents an amount or quantity of a substance. 40 g calcium

56 g iron

197 g gold

127 g iodine

All the above masses contain the same number of atoms. Chemists refer to 23 6.022 X 10 atoms of an element as one mole of the element. Thus one gram mole ( mol ) is the formula weight of any substance = Avagadro’s number 23 = 6.022 X10 particles (atoms, ions, molecules). Mole is expressed as: mol ( any substance)

=

Mass Relative molecular mass (RMM) or RAM

If mass is in pounds, lb, then mole If mass is in grams, gm, then mole

= lb mol = g mol, but generally known as mol 6

Mass Balances

If mass in kilogram, kg, then mole

= kmol

The mass of 1 mole of atoms is found by looking up the relative atomic mass of the element. Since the relative atomic mass of sodium (Na) and Chlorine (Cl) are 23 and 35.5, respectively, the relative molecular mass of NaCl is 58.5 (23+35.5). Therefore

58.5 gram is equivalent to 58.5 kilogram is equivalent to 58.5 pounds is equivalent to

1 mol ( 1 g mol ) 1 kmol 1 lb-mol

NOTE: These will all have different number of particles. Exercise 5.1 a) Find the relative molecular mass of H2SO4 and Cl2 . b) Find the mass of 3 moles of Fe2(SO4)3. 5.1

Moles in Solution 1 litre = 1 cm3 =

1 dm3 = 1 x 10-6 m3

1000 cm3

= 1* 10-3 m3

A solution containing one mole of solute dissolved to make one litre of solution is often called molar solution. 3 The concentration of a solution is measured in moles per dm ( litre) is called the molarity of the solution. Symbols we use Volume of solution = Concentration of solution = Number of moles =

Chemists v measured in dm3 c moles dm-3 n

Chemical Engineers measured in m3 kmol m-3

n C

= V

-3 A solution of NaOH containing 0.1 mol dm can be called as 0.1 molar and can be written as 0.1 M NaOH. Molarity is used in chemical laboratories (eg. Titration) Exercise 5.2 3 a) If 140 g of potassium hydroxide (KOH) is dissolved in 2500 cm of water what is the concentration (molar) of the solution. b) What is concentration of 175.5 kg of NaCl dissolved in 2 m3of water?

7

Mass Balances

6

Concentration 6.1 Mass Fraction Consider a tank comprising components A, B & C

components A, B & C mass fraction of A

=

mass of A Total mass of mixture (A+B+C)

mass of A

=

mass fraction A * total mass of mixture

Sum of mass fraction

= 1 (use this as a check after your calculations)

Exercise 6.1 5 kg of A, 2 kg of B and 3 kg of C are mixed in a container. What is the mass fraction of A. Confirm the sum of mass fraction is equal to 1. 6.2 % w/w ( mass percent) The symbol % w/w (or weight percent) is used to indicate concentration on a mass basis. components (mass percent ) A, B & C * 100 Total mass of mixture (A+B+C)

% w/w

of substance A =

mass of A

∴mass of A =

% w/w of A * total mass of mixture 100 NOTE: % w/w will be the same regardless of the units used. i.e. g, kg, lb 100 kg of methanol/water mixture having a concentration of 15 % w/w methanol will have 15 kg of methanol and 85 kg of water. 150 kg of the same mixture will have 22.5 kg of methanol and 127.5 kg of water. Exercise 6.2 A storage tank has 1500 kg of A, 2000 kg of B and 5500 kg of C. Calculate the mass % of C and confirm % w/w of all the components add up to 100 %. 6.3 Volume percent ( % v/v) The symbol % v/v is used to indicate concentration on a volume basis. volume % of any substance

=

As for mass fraction, Volume fraction

volume of that substance * 100 total volume of mixture

=

8

volume total volume

Mass Balances

Sum of volume fraction

=

1

e.g. An aqueous solution of HCl with a concentration of 3.5 % v/v has 3.5 m 3 HCl mixed with 96.5 m3 water. It could also means 3.5 cm3 HCl mixed with 96.5 cm3 water. 6.4 Mol percent ( mol%) As for % v/v, mol %

=

mole of A total moles in mixture

* 100

e.g. In gaseous systems 21% v/v oxygen in air means that in every 100 m 3 of air there is 21 m3 of O2. NOTE: It does not matter whether we illustrate a concentration in terms of ml, l (litre) or m3, provided there is consistency. Exercise 6.3 A storage tank has the following compounds; A = 1000 kmol, B = 5000 kmol, C = 2000 kmol and D = 8000 kmol. Calculate the mole fraction and mol % of all the components and tabulate your results. Confirm sum of mole fraction and mol % are 1 and 100, respectively. 7

GAS LAWS Changes in temperature and pressure have little effect on the volume of a liquid or a solid, but a considerable effect on the volume of a gas. There are various gas laws.

7.1 Boyle’ Law The volume of a fixed mass of gas is inversely proportional to its pressure, provided the temperature remains constant. Mathematically it can be expressed as V P = constant Where

P = pressure,

------------------------- 1 V = volume

7.2 Charle’s Law The volume of a fixed mass of gas at constant pressure is directly proportional to its absolute temperature, K V ∞ T i.e.

V = constant T Where T = temperature in Kelvin (K) 7.3 The Equation Of State for an Ideal Gas

--------------------------------

2

Gases which obey Boyle's law and Charle's law are called ideal gases. By combining equations 1 & 2 the following equation can be obtained.

9

Mass Balances

i.e.

P1 V1 = T1

P2 V2 T2

equation of state for an ideal gas

A gas has a volume of V1 at temperature T1 and pressure P1. If the condition are changed to pressure P2 and temperature T2, the new volume, V2, can be calculated using the above equation. It is usual to compare gas volume at 0 C and 1 atmosphere (1 atm). This is referred to as standard temperature and pressure (s.t.p.). Avogadro's law (hypothesis) states that equal volume of gases, measured at the same temperature and pressure contain equal number of molecules. In other words the volume occupied by one mole of gas is the same for all gasses. It is called gas molar volume. 1 mole of any gas occupies 22.4 dm3 at s.t.p. 1 mole of any gas occupies 24.0 dm3 at 20 C & 1 atm 1 kmol of any gas occupies 22.4 m3 at s.t.p. The above should be memorised 7.4 Useful Units 1 atm = 1.01325 * 105 pascal (Pa) = 101.325 kilopascal (kPa) = 1.01325 * 105 Newton’s per square meter ( N/m2 ) = 760 mm mercury 1 bar = 1.0 * 105 Pa 1 atm. = 14.7 psi 1 bar = 14.5 psi 7.5 Ideal Gas Equation In gaseous systems a simple relationship exists between volume and mole which is useful in performing calculations. In this case it will always assumed that any gas is at condition at which the Ideal Gas Law applies. The equation is: PV = nRT Where P = pressure, V = volume, n = number of moles R = molar gas constant and T = absolute temperature (K) The molar gas constant, R, is the same for all gases. 1 mole of gas at 0 C and 101325 Pa occupy 0.0224 m3 (22.4 dm3) substituting in the equation R

=

PV nT

PV = nRT =

101325 * 0.0224 1 * 273 10

=

8.314 J k-1 mol-1

Mass Balances

For 1 mole of gas

PV = RT

For n moles of gas

PV = nRT

NOTE:

In practise, real gases obey the ideal gas equation very closely at low pressure and high temperature.

If the mass of gas is m and its molar mass M (RMM) then we can rewrite the ideal gas equation as: PV

or

=

m

=

m RT M PVM RT

or

M

=

m RT PV

From PV = nRT, it also follows that at constant temperature and pressure the volume is proportional to the number of kmol (mol). i.e

V

so

V1/ V2 ∴ Vol%

=

=

kmol * constant

=

kmol1/ kmol2

Mol% for an ideal gas

Exercise 7.1 What amount ( no. of moles ) of an ideal gas occupies 5.8 dm 3 at 2.5 X 105 Nm-2 and 300 K. Exercise 7.2 Calculate the molar mass of a gas which has a density of 1.798 g dm -3 at 298 K and 101 kNm-2.

Now do tutorial A 8

Chemical Processes

CHEMICAL PROCESSES

11

Mass Balances

Batch Processes

Continuous Processes

Steady state or Temperature, pressure, volume remain constant(e.g. distillation).

Semibatch processes

Transient

Process variables change with time. (e.g. batch & semibatch processes)

8.1 Batch Processes Predetermined amount of feed is charged into the system at the beginning of the process, and the products are removed all at once after a given time or after equilibrium has been achieved. No mass crosses the system boundaries between the time the feed is charged and the time the product is removed. E.G. Add measured quantities of three reactants to a reactor and remove the products and unreacted reactants after a predetermined time i.e. when the system has reached equilibrium. 8.2 Continuous Processes The feed and products flow continuously throughout the duration of the process. E.G. Distillation Column. 8.3 Semibatch Processes Any process which is neither batch nor continuous. E.G. Blending of several chemicals (petrol) in a tank from which nothing is being withdrawn until the blending operation is completed.

12

Mass Balances

9

Basic Ideas in the Mass Balance (Material Balance)

9.1

General A mass balance is a re-statement of the law of conservation of mass. Matter is neither created nor destroyed for chemical processes, but some variations on the statement need to be considered. So how are the consequences of the conservation law applied to the mass balance? In industry mass balance is carried out to account for all the material in a process. If you look at a chemical plant you will see a very large number of pipes, tanks, vessels, columns, heaters, pumps and instruments. Chemical engineers need to be able to understand how to design many of these, to understand how they fit together and how to calculate the amount of each chemical flowing or exiting at every point in the system. Mass balance for a single unit is simple, but for a large process it is often complex.

9.2

The System Balance of any kind is made with respect to a definite entity; this entity is usually referred to as a system. The choice or definition of the system is up to us. It is essential that at the start of any process analysis or problem solving procedure that we have a clear idea of what we are defining as the system.

Input of M

Output of M

SYSTEM

System Boundary Figure 9. 1 Referring to the figure 9.1, we indicate the system to be analysed as a box. This box could represent a single process or process step, for example a heat exchanger or a chemical reactor, or even part of such a process. Conversely, the box could represent a series of process steps, for example, an entire chemical plant. Balances are made around the system i.e. balances are made with respect to the system. It is usually convenient to identify the system by drawing a boundary line around it, as indicated by the dashed line in Fig. 9.1. This helps to identify those streams which, if they cross the boundary, enter or leave the system.

13

Mass Balances

9.3 The Nature of Flowsheet Before you can attempt to do any calculations you have to have some way of representing the chemical plant on sheets of paper. One way is to draw a flowsheet. There are actually several different kinds of flowsheet and at this stage, the simplest, the block diagram is appropriate.

1

2

3

4

Figure 9.2 This figure uses squares or rectangles to represent sections of the plant. Each section may be simple or complex. 9.4 Interpreting a block diagram Let us consider the block diagram (system) shown below. M4 M1

M5 Accumulation MA

M2

M6 M3

Figure 9.3

The section of the plant has four different streams entering it, i.e. 4 input streams, but only two streams leaving, i.e. 2 output streams. The input streams are combining in some way and then leaving the unit via two separate pipes. It is obviously important that all the inputs and outputs of mass with respect to the system are identified; if we miss any component or stream, the balance will not be correct- it will not balance. Remember mass balance is based on law of conservation of mass which states that mass can be neither created nor destroyed. Using this law For the above given system, M1 + M2 + M3 + M4 - ( M5 + M6 ) = i.e. ∑ INPUT MASS or

MA

∑ OUTPUT MASS = ∑ ACCUMULATED MASS

RATE OF INPUT - RATE OF OUTPUT =

9.1

RATE OF ACCUMULATION

WhereM1 to M6 can be mass (kg) or mass flow rate (kg/s) and MA is accumulated mass (kg) or rate of accumulation (kg/s). MA can be positive (increase) or negative (decrease) Equation 9.1 is an expression of the conservation of Mass and this statement will apply to any quantity for which a conservation law holds. 14

Mass Balances

Specifically for the case of material, or matter, it must be understood that atoms are neither created nor destroyed, at least not in the chemical processes we consider. Any balance, which depends on the concept of the conservation of atomic species, will be valid. It is important to realise that the mass of material, depends on the masses of the atoms that make up the material, such a balance in terms of mass must be valid. This balance holds even if chemical reactions take place- they cannot alter the relative numbers of atoms present. It is often useful to write balances in terms of specific atoms, atom balances. For example, suppose carbon is involved in some way in a process. We can then write ( More details on atom balance given in section 22). Input (of C atoms) — output (of C atoms) = Accumulation (of C atoms) It is important to note that this balance applies even if chemical reactions take place; Carbon atoms are neither created nor destroyed. Example When methane reacts with oxygen it forms carbon dioxide and water. Write a balanced reaction equation.

Mass

CH4 O2 CH4(g) + 2 O2(g) 16 2 x 32

Total mass

CO2, H2O CO2(g) + 2 H2O(g) 44 2 x 18

80

80

Even though a reaction takes place mass in = mass out Also No. of C – atoms on LHS = No of C – atoms on RHS No of H – atoms on LHS = No. of H – atoms on RHS No of O – atoms on LHS = No. of O – atoms on RHS 9.5 Steady State Operations In section 9.4 we wrote material balances in terms of the total mass and in terms of a specific atomic species (carbon balance), and in both cases an accumulation term was an important part of the balance. In practice, we often deal with systems or processes that operate or are operated under steady-state conditions, and this leads to simplification with respect to the accumulation term . A steady-state operation is one in which conditions within the process or system do not change with time, that is, from one moment to another. This does not imply that conditions (for example, temperature, pressure, concentration) are necessarily the same from one point to another within the system, but that at any given point they do not change with time. At steady state, there can be no accumulation (i.e. MA = 0) ∴

INPUT

= OUTPUT 15

Mass Balances

i.e.

M1 + M2 + M3 + M4 = M5 + M6

10. Systems Without Chemical Reactions Case 1: Think of the block representing a beaker in a laboratory. Some water is put into it, then some solid NaCl is added, then the beaker is shaken until the salt has dissolved and the mixture is poured out. Now think of a similar situation on a chemical plant. The beaker is now a large vessel. Water is passed into the vessel by opening a valve in the inlet pipe and pumping it in. After a time the valve is closed to stop the flow of water. Solid NaCl is then poured into the vessel, through a hole in the top, and an agitator is switched on to mix in the salt and help to dissolve. When all the salt has dissolved a valve in the outlet pipe is opened. The salt solution flows out, either by gravity or with the aid of a pump. This is an example of a batch process - each activity takes place in sequence. The amount in the vessel changes with time from zero to the total amount of salt solution and then back to zero. The rate of flow in each pipe also changes with time - we do not know precisely how, except that for part of the time there is no flow in each pipe and for part of the sequence there is flow. Let us carry out what is called material balance or mass balance on system. EXERCISE 10.1 2m3 water and 100 kg salt are put into a tank. (density of water is 1000 kg m-3) (a) What is the mass water used? (b) What is the total mass of salt and water? (c) What is the mass of salt solution produced? (d) What is the concentration of salt in salt solution? Case 2: Now think of the block representing a vessel which is already full of salt solution. Water is being pumped into the vessel at a steady rate and salt is being metered in continuously and constantly. A stirrer is mixing the vessel contents effectively. Suppose the situation has been going on for quite a long time, so the concentration of salt in tank is steady, i.e. a sample taken now and another taken in an hour would be indistinguishable, having identical composition. This is called the steady state. Since the vessel has a limited volume there must be continuous flow of liquid (solution) out of it, maintaining a constant level in the tank.

16

Mass Balances

EXERCISE 10.2 Water is fed into a tank continuously at a rate of 2m3/h. Salt is metered in continuously at a rate of 100 kg/h. salt solution is produced by good mixing and is removed continuously. (a) What is the rate of mass input of water? (b) What is the total rate of mass input of salt and water? (c) What is the mass of salt solution produced? (d) What is the concentration of salt in salt solution? 11

Defining a system -creating an envelope On the block diagram (figure 11.1), as stated previously, each block can represent a simple piece of equipment or a complex section. For some purposes we may wish to combine blocks together. Suppose we label the streams around the first 2 blocks, as shown below. S4 S1

S3 S5

S2 At steady state we can say that

Figure 11.1

S1 + S2 = S3 for the first block and S3 = S4 + S5 for the second block so, eliminating S3 S1 + S2 = S4 + S5 The same 2 blocks and the same input and input streams are now drawn with a dotted box around them, the dotted box or “envelope” representing a new system or block. S4 S1

S3 S2

S5

Figure 11.2

This new system has two inputs, S1 and S2 ; two outputs, S4 and S5. i.e. at steady state, S1 + S2 = S4 + S5, as derived in the previous section. In considering this new system, S3 is neither an input nor an output, i.e. it does not pass through the dotted box. It is therefore not an element in the mass balance for the revised system. We can continue drawing dotted boxes anywhere, defining new systems and every time, total input must equal total output at steady state. Where we draw these boxes or envelope requires a little skill which is only obtained by practice. NOTE:

A stream can not start or finish in mid- air inside an envelope. 17

Mass Balances

An envelope that is usually well worth considering is the overall mass balance(OMB). i.e.

Total inputs into the System

=

Total outputs from it

Exercise 11.1 By considering all the possible envelope that could be drawn in the following system, write down as many mass balance equations as possible. M4 M

M7

M2

1

M6

M3

M5

M8

M9

Now do Tutorial B 12

Component Balances in Systems without Chemical Reactions It is frequently necessary to consider the components of the stream involved in a process. Useful in mixing, separation (distillation) and evaporation where no chemical reactions take place. This is done in addition to the overall mass balance i.e. total input and output. M 1 y1 M3 y3 where y is the fraction of a component M 2 y2 M 4 y4 The principle we will be using is that the total mass of a component entering a system should be equal to the total mass of the same component exiting the system. Mass of any component in a stream is determined using the following equation: Mass of a component in a stream = Mass flow of the stream x fraction of the component in that stream A component mass balance (abbreviated to CMB) for component y will give M 1 * y1 + In general,

M 2 * y2

n

n

i =1

o =1

=M3 * y3 + M4 * y4 + MAyA at steady state = 0

∑ M i yi = ∑M oYo Where n = number of components, and subscripts i and o signifies input and output, respectively. NOTE: In a reacting system component balance CAN NOT be expressed in volumetric basis( i.e moles), ONLY on a mass basis - hence MASS BALANCE. Since the sum of these (i.e. several components) gives the OMB this overall balance is not an independent equation. Hence, there are n independent equations for each envelope. In practise it is usually most convenient to write down the overall balance plus the balances for (n-1) components. 18

Mass Balances

e.g. The previous salt solution problem (i.e. exercise 10.2 – page 17): We can say RATE OF SALT INPUT = RATE OF SALT OUT PUT This is valid even though the input salt is pure (i.e. 100%) and the output salt is dissolved. Carry out an OMB 100 + 2000 ∴ M3

= =

M3 2100

1

Now carry out a CMB for salt salt

100 * 1 + 2000 * 0 =

M 3 * y3

2

Substituting for M3 in equation (2) given above 2100 * y3

=

100

y3

=

100 = 2100 =

0.0476 4.76 % w/w

Exercises 12.1 & 12.2 Assuming steady state, carry out material balance for NaCl, water and brine (salt solution) over the system shown if: 1

2

3 12.1) Stream 1 contains 100 kg/h NaCl and 2000 kg/h water Stream 2 contains 50 kg/h NaCl and 500 kg/h water. Find the flow rates of salt solution, water and salt in stream 3 12.2) Stream 1 is 2000 kg/h brine, having a concentration of 5.1% w/w NaCl Steam 2 is 600 kg/h brine, having a concentration of 3.8% w/w NaCl Find the flow rate of stream 3 and the concentration of salt in stream 3. 13

Composition and Concentration Exercise 12.2 used the concept of concentration (as did exercises 10.1 & 10.2). A concentration of 4.76% w/w NaCl means that if you take a sample (of any size) and analyse it, you will find that 4.76% of the mass of the sample is salt and 95.24% of the mass is water. This seems very simple at this stage but it is vital that you remember it in more complex situations. When the concentration of every component in a mixture is known, we can refer to the composition of the mixture. For example, in figure 13.1, the composition of stream A is 12% w/w phenol, 88% w/w water. The concentration of phenol in stream A is 12% by weight. It does NOT mean that 12% of the phenol input goes into stream A EXERCISE 13.1 19

Mass Balances

A phenol = 12 % w/w water = 88 % w/w

phenol 100 kg water 100 kg

B phenol = 62 % w/w water = 38 % w/w Figure 13.1 Note: concentration of phenol and water in products does not add up to 100. 14

Approach to Mass Balances 1) 2) 3) 4) 5) 6)

7) 8) 9) 10) 11) 12)

Draw a block (flow) diagram. Add all available information on this diagram. Select a suitable basis. Usually choose a stream for which a lot of data are known. Write balanced chemical reaction(s), if any. Draw an envelope around the system on which you are doing the balance. Construct an input-output table in which you identify and tabulate all the incoming and outgoing materials. This step is the beginning of a statement of all the materials that interact with the system and is an absolutely essential step. Make an overall balance (OMB), and make a component balance (CMB) if there are no chemical reactions. Make neat calculations, with headings. Tabulate your results if this makes the presentation easier. Give references for data used in the problem (it may help you later if values need checking, but in any case it is good practise). State assumptions made. Make the answer stand out by underlining or by leaving a space after it. Check the answer is sensible and possibly for accuracy by an independent method.

EXAMPLE 14.1 Find the flow rate of all the product streams for the system given below. % A B C

w/w 99.2 0.8 0

100 kg/h 1

2

% w/w A 35 B 45 20

Mass Balances

C 20 % A B C

15

w/w 0 0.5 99.5

% w/w A 1.0 B 99.0 C 0.0

Basis and Scaling In exercise 14.1 the feed was given as 100 kg/h and all the other flow rates were calculated. Suppose the problem was to find the flow rate of feed that produces 850 kg/h of distillate from column 2 or the feed rate that produces 684 kg/h of B in the bottoms from column 2. The best way in either case is to perform the calculations exactly as already done and then scale up or down all the numbers in the same ratio to give the desired value. The concentrations all remain unchanged. If the scaling is going to take place it does not really matter whether the initial feed rate is taken to be 100 units (in this case kg/h) or 1 unit or any other number. A value of 0.8 units could be used for the feed rate and the answer would be found equally correctly, but the first step would require additional calculations to find the flow rates of each component in the feed. It is usually best to start with 100 units for convenience, but not always - there will be special cases later on. Choosing a starting point is called CHOOSING THE BASIS for the calculation. Always write down the basis. In general, basis should be based on the stream which has more information available compared to the other streams. Depending on the information available some times a product stream can also be taken as a basis.

21

Mass Balances

Exercise 15.1 What proportion of the carbon dioxide in the feed gas is absorbed into the liquid in the system shown, when the concentration in the gas stream is reduced from 21% w/w to 0.3% w/w. stripped gas

absorbent liquid ( MEA solution)

CO2 0.3 % w/w

feed gas

MEA solution

CO2 21% w/w Exercise 15.2 The MEA solution used in exercise 15.1 is regenerated by using steam (heat) and a reduction in pressure. This releases the CO2 and the solution can be recycled. The CO2 is removed from the top of the regenerator, together with water vapour. In the system shown below, the feed gas flow rate is 2500 kg/h. How much CO2 is taken from the top of the regenerator? MEA solution

CO2, H2O

Stripped gas CO2 0.3 % w/w

feed gas 2500 kg/h H2O vapour + Heat CO2 21 % w/w Absorber

Regenerator

22

Mass Balances

16

BALANCED EQUATIONS (STOICHIOMETRY) and Moles Chemical equations provide us with information about reactants and products. Using symbols for elements and formulae for compounds equations can be written for chemical reactions. Reactants are put on the left hand side of the equation and the products on the right side of the equation with an arrow in between them. As well as showing the nature of the reactants and products, it tells us the mole proportion of each species used up and produced. For example, the balanced (stoichoimetric) equation: 2 SO2 (g) + O2 (g) 2 SO3 (g) indicates that two moles of SO2 react with one mole of O2 to produce 2 moles of SO3. The numbers in front of each species are the stoichiometric coefficients. The above balanced equations tell us about STOICHIOMETRIC RATIOS In a balanced equation the number of atoms of each species must be the same on both sides of the equation. Remember atoms cannot be neither created nor destroyed. This law is generally known as conservation of mass. Rules for writing chemical reactions 1.

Write the correct formula for all the substances in reaction equation (small numbers as subscripts to indicate the number of atoms of a species in a compound) a) for elements use the symbol from the periodic table except H2 ,N2, O2 , F2, Cl2, Br2, I2 b) for compounds use valencies c) give the state (s, l, g, aq ) of the substances.

2.

Balance the equation for each type of atom. (using large numbers in front of the formulae). Make sure you do not change any formulae. e.g.

H2 (g) + O2 (g)

H2O (l)

Here the left hand side of the equation has two hydrogen atoms and two oxygen atoms, whereas the right hand side of the equation has two hydrogen atoms and one oxygen atom. This violates the law of conservation of atoms (mass). By putting suitable numbers in front of the species, the equation can be balanced to give: 2 H2 (g) + O2 (g)

2 H2O(l)

or H2(g) The fraction

+

1 O2 (g) 2

H2O(l)

this equation is also valid.

1 can be inserted in front of di-atomic molecules such as H2, O2, N2, Cl2, 2

F2, I2 and Br2 only, to balance the equation.

Never try to change the formula of a compound such as H2SO4, Na2SO4, HNO3 when balancing the equations. You can only put numbers in front of them e.g. Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g) 23

Mass Balances

On the above reaction solid zinc reacts with a solution of sulphuric acid to give a solution of zinc sulphate and hydrogen gas. Hydrogen is written as H2, since each molecule of hydrogen has two atoms. e.g.

Na2CO3(s) +2 HCl(aq)

CO2(g) + 2 NaCl(aq) + H2O(l)

The equation for the action of heat on sodium hydrogen carbonate. This is called a decomposition reaction. heat 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g) The above equation tells us that 2 moles of NaHCO3 give 1 mole of Na2CO3, 1 mole of CO2 and 1 mole of water. The chemical equation provides necessary information essential to complete a mass balance for the given reaction. e.g.

+

2O2

→ CO2

C2H6 +

7O2 2

→ 2CO2 +

CH4

+

2H2O --------(1) 3H2O ---------(2)

First of all make sure the reaction equation is balanced. i.e. the number of atoms on LHS should be equal to the number of atoms on RHS. As previously stated in section 5, the formula weight of a chemical expressed in grams is called a mole, abbreviated to mol. If the mass is expressed in kg it is called kmol. Remember that in carrying out mass balances you can balance mass, BUT NOT MOLES. Inspect the above two reactions. In reaction 1 number of moles in reactants is equal to number of moles in the products. However, in the second reaction 4 ½ moles are reacting to produce 5 moles. You can also carry out atom balance. Kmol are much more convenient quantities than mol in chemical engineering. Chemists tend to use mol. e.g.

1.0 kmol CO = 28 kg 1.0 kmol H2O = 18 kg 3.0 kmol H2S = 3 * 34 = 102 kg

In reacting systems it is easier to work with moles If the selected basis is mass, first convert mass into moles by dividing mass by RMM (refer to section 5). Once you have calculated the number of moles of products by using the stoichiometric ratios, you may convert moles of product to a mass. NOTE:

Mass & atoms are not destroyed, but moles can be created or destroyed as atoms recombine.

24

Mass Balances

In a reacting system, if you know the quantity of a component the quantity of other reactants and the products can be found by knowing the stoichiometric ratios of the species involved in the reaction. Coefficient of the unknown species Stoichiometric ratio = Coefficient of the known species Exercise 16.1 C3H8 + a) b) c) d) e)

17

5O2

→ 3CO2 +

4H2O

How much CO2 is formed if 5.5 kg of propane reacts? How much water is produced when 250 kg of carbon dioxide is produced? How much propane is required to produce 100 kg carbon dioxide? How many kmol of propane will produce 15 kmol of CO2? How many kmol of water are produced when 10 kmol of propane reacts?

Mol % and Vol % In many calculations it is much more convenient to work in mol% and kmol than in any other units. It is useful to remember that for any gaseous system at low pressure, where the ideal gas law can be assumed to apply, the composition in vol % is the same as the composition in mol %. Refer to section 7.5.

18

Limiting Reactants and Excess Reactants. If the reactants are made available in quantities proportional to those indicated in the chemical equation, these reactants are said to be available in stoichiometric quantities. Thus the amount of a reactant theoretically required for complete conversion of other reactants is called the stoichiometric quantity (refer to section 16). In most industrial processes the quantities of reactants supplied are usually not in the exact proportions demanded by the reaction equation. It is generally desirable that some of the reacting materials be present in excess of the amounts theoretically required for combination with others. Therefore, limiting reactant is the reactant that is present in the smallest stoichiometric amount. Under such conditions the products obtained will contain some of the unreacted reactants. The quantities of the desired compounds formed in the reaction will be determined by the quantity of the limiting reactants. Other reactants are called excess reactants simply because they are supplied in excess of the stoichiometric amount. E.G. N2 + 3 H2

2 NH3

In the above reaction if 2 kmol of N2 is reacted with 6 kmol of H2, the reactants are in stoichiometric proportion and there is no limiting reactant or excess reactant. Example 25

Mass Balances

SO2 1 kmol O2 1.25 kmol SO2

Burner

+

SO3 O2

1 O2 2

→

SO3

From the above balanced equation it can be seen that the amount of oxygen required for complete combustion of 1 kmol of sulphur dioxide to sulphur trioxide is 0.5 kmol, but the amount supplied is 1.25 kmol. Therefore there is an excess of O 2 above the stoichiometric requirement. The product SO3 is governed entirely by the amount of SO2, which is called the LIMITING reactant. As O2 in feed is above the stoichiometric requirement, it is the EXCESS reactant. SO2 reacted = 1.0 kmol O2 reacted(required) = 1 x stiochiometric ratio = 1 x 0.5 = 0.5 kmol SO3 formed = 1.0 x stoichiometric ratio = 1 x 1 = 1 kmol O2 unreacted = in – reacted =1.25 – 0.5 = 0.75 kmol Percent excess is defined as mole % excess

* Required

=

input – required* required*

= amount required for 100 % completion of the reaction. ( see next section) mol % excess O2

19

X 100

=

1.25 - 0.5 0.5

x 100 = 150

%

Degree of Completion or Conversion (incomplete reaction) Even though some of the reactants required for the manufacture of a chemical may be present in excess there is no guarantee that the limiting reactant will undergo complete reaction. Such partial completion may result from the establishment of an equilibrium in the reacting mass or from insufficient time or opportunity for completion. The degree of completion of a reaction is generally expressed as the percentage of the limiting reactant, which is converted or decomposed into other products.

26

Mass Balances

Example If 2 kmol of CO2 is reacted with 5 kmol of NH3 in the production of urea and the conversion is 80 mol%, determine the excess reactant, percentage excess, limiting reactant and the composition of the product. Step 1

Draw the block diagram and write down all the information.

2 kmol CO2

80 % conversion

urea CO2 NH3

5 kmol NH3

Step 2 Select a suitable basis Basis: 2 kmol of CO2 (we could also select NH3 as our basis- will give the same answer) Step 3 Write a balanced reaction equation CO2 + 2 NH3 NH2COONH4 Stoichiometric ratio of NH3/CO2 = 2/1

(urea)

=2

Step 4 Now you can start the calculation. solution As per stoichiometry, amount of NH3 needed to react with 2 kmol of CO2 = 2 x stoichiometric ratio =2x2 = 4 kmol As there are 5 kmol of NH3, the excess reactant is NH3 and the limiting reactant is CO2. In – required % Excess NH3 =

5-4 x 100 =

Required

x 100 = 25 % 4

As the limiting reactant is CO2, the conversion is based on CO2. (only 80 % converted) Amount of CO2 converted = 2 X 0.8 = 1.6 kmol From this step onwards the amount of CO2 reacted ( limiting reactant) forms the basis for rest of the calculation. Unreacted CO2 = in - reacted = 2 – 1.6 Amount of urea formed = CO2 reacted x ratio = 1.6 x 1 (based on the amount of limiting reactant reacted)

= 0.4 kmol = 1.6 kmol

Amount of NH3 consumed = amount of CO2 reacted x stoichiometric ratio = 1.6 x 2 = 3.2 kmol Unreacted NH3

= in – reacted

27

= 5 – 3.2

= 1.8 Kmol

Mass Balances

Final Step

List the composition in a tabular format

Composition of Products Component CO2 NH3 NH2COONH4 Total

Kmol 0.4 1.8 1.6 3.8

Mol% 10.53 47.37 42.10 100.0

Example. In a burner 5 kmol SO2 was burned with 75% excess oxygen, but the reaction is only 80% complete. Determine the product composition Order of calculation SO2 5 kmol

Burner 80% completion

SO3 SO2 O2

O2 75% excess Basis: 5 kmol SO2 in the inlet SO2

+

( 1 + excess /100) 0.5O2 →

SO3

Theoretical oxygen requirement = 5 x ratio= 5 x 0.5 But excess is 75% ∴ Actual O2 in feed = 2.5( 1 + (75/100)) Conversion is 80% ∴ Amount of SO2 reacted= in x % conversion = 5 * 0.8 Amount of unreacted SO2

=

= 2.5 kmol = 4.375 kmol = 4.0 kmol

in – reacted = 5 – 4 = 1 kmol

Amount of SO3 formed (based on SO2 reacted) = 4.0 x 1(ratio)

= 4 kmol

Amount of O2 consumed (SO2:O2 = 1: 0.5)

= 2.0 kmol

= 4 * 0.5(ratio)

Amount of unreacted O2 in exit stream = in – reacted = 4.375 – 2.0 Component

Reactants, kmol

Products, kmol

SO2 SO3 O2 Total

5.0 4.375 9.375

1.0 4.0 2.375 7.375

28

= 2.375 kmol

product gas composition , % v/v 13.56 54.24 32.20 100.00

Mass Balances

Check the atom balance i.e. S and O

Component SO2 SO3 O2 Total

Kmol 5.0 4.375

INPUT S O 5 5.0 * 2= 10.0 4.375*2= 8.75 5.0 18.75

kmol 1.0 4.0 2.375

OUTPUT S O 1.0 1.0 * 2= 2.0 4.0 4.0 * 3= 12.0 2.375 * 2= 4.75 5.0 18.75

Atom balance is the ultimate check to verify whether the solution is correct or not. Atom balance is not strictly balancing mass between the inlet and outlet, but it is exactly equivalent. It is better described as a material balance, but the distinction is not normally considered important. It is a balance on the atomic species present. When substances are unaffected by reaction, the balance could be in kmol units, but you really do need to be sure that there is no reaction ( an atom balance example is given later on).

29

Mass Balances

Example 150 kg/s of ethane (C2H6) was burned in 60 % excess air. Due to burner inefficiency only 90% of the ethane was combusted, and of this 95% of ethane was converted to carbon dioxide and the balance to carbon monoxide. Determine the composition of the product stream. Solution Step 1

Draw the diagram and write down all the available information.

Ethane 150kg/s Air, 60 % excess

90%9conversion 90

5

95 % of 90% 5 % of 90 %

CO2 CO

C2H6 O2 N2 CO2 CO H2O

As only 90 % of C2H6 reacts, there should be unreacted C2H6 in the product stream. As part of reacting C2H6 is converted to CO2 and part to CO, there should be CO2 & CO in the product stream. Water is a combustion product and therefore should be in the product stream. Air feed is in excess, therefore, excess O2 and N2(comes out unreacted) should be in the product stream Step 2

Select a suitable basis Basis:

150 kg/s ethane in feed

Step 3 Write balanced reaction equations There are two separate reactions taking place: one producing CO2 and another producing CO. R1

C2H6 + 7/2 O2

2 CO2 +

R2

C2H6 + 5/2 O2

2 CO +

Step 4

3 H2O 3 H2O

Check the units and convert, if necessary

It is easier to work with mols than mass(kg), therefore convert given mass to moles. Number of moles of ethane = mass/ RMM = 150/30 = 5 kmols

Now we are in a position to do serious calculations! 30

Mass Balances

In these types of combustion reactions, start with the known reactant (i.e. C2H6) and then calculate the air feed from the given information. Next step is to use the given reaction details ( i.e. information inside the box) and systematically work through the problem. Step 4 Determine the quantity of air feed Remember excess air calculation is always based on complete combustion unless otherwise stated. Theoretical O2 required to burn ethane = moles of C2H6 x coefficient ratio Coefficient of unknown component Note : coefficient ratio = Coefficient of known component Coefficient ratio

= moles of O2/ moles of C2H8 = 3.5/1

= 3.5

First calculate the theoretical O2 and then the actual O2 using the excess value.

∴

Theoretical O2

∴Actual O2 in air feed

= 5 x 3.5 But excess = 60% = 17.5 (1 + 60/100)

Nitrogen in air feed (N2:O2 = 79:21 v/v) = 28 x (79/21 ) Step 5

= 17.5 kmol/s = 28 kmol/s = 105.33 kmol/s

Determine the amount of C2H6 reacted (& unreacted)

Amount of ethane reacted

= moles of C2H6 in feed x % conversion = 5 x 0.9 = 4.5 kmol/s

NOTE: Now this reacted C2H6 forms the basis for rest of the calculation. Unreacted C2H6

= in – reacted = 5 – 4.5

= 0.5 kmol/s

Step 6 Determine the reactants and products involved in R1 Reaction 1 ( 95% of reacted ethane) – work from left to right Amount of ethane reacted = 4.5 x 0.95

= 4.275 kmol/s

Now this 4.275 kmol/s forms the basis for rest of the calculation for reaction 1 Amount of oxygen reacted Amount of CO2 produced Amount of H2O produced

Step 7

= 4.275 x coefficient Ratio = 4.275 x 7/2 = 14.963 kmol/s = 4.275 x coefficient Ratio = 4.275 x 2 = 8.55 kmol/s = 4.275 x coefficient ratio = 4.275 x 3 = 12.825 kmol/s

Having finished with R1, now calculate the reactants and products involved in R2

Reaction 2 (5% of reacted ethane) As for R1, work from left to right Amount of ethane reacted = 4.5 x 0.05 = 0.225 kmol/s 31

Mass Balances

NOTE: Now this 0.225 kmol/s forms the basis for rest of the calculation for reaction 2 Amount of oxygen consumed Amount of CO formed Amount of H2O formed

= 0.225 x coefficient ratio = 0.225 x 5/2 = 0.563 kmol/s = 0.225 x coefficient ratio = 0.225 x 2 = 0.45 kmol/s = 0.225 x coefficient ratio = 0.225 x 3 = 0.0.675 kmol/s

Step 8 calculate O2, H2O in product gas Oxygen is consumed in reactions 1 & 2

∴Unreacted

O2

= in – R1 –R2

= 28 – 14.963 – 0.563 = 12.474 kmol/s

= R1 + R2

= 12.825 + 0.675 = 13.5 kmol/s = 105.33 kmol/s

H2O is produced in R1 & R2

∴Total H2O in product stream

N 2 is a tie substance, therefore N2 in = N2 out Step 9

Tabulate your results Component C2H6 O2 N2 CO2 CO H2O Total

Kmol/s 0.5 12.474 105.33 8.55 0.45 13.5 140.804

Mol % 0.36 8.86 74.81 6.07 0.32 9.59 100.01

Step 10 CHECK: As mass in = mass out, you can check your answer. Component C2H6 O2 N2 CO2 CO H2O Total

RMM 30 32 28 44 28 18

In, kmol/s 5.0 28.0 105.33

In, kg 150.0 896.0 2949.24

Out, kmol/s 0.5 12.474 105.33 8.55 0.45 13.5

3995.24

Out, kg 15.0 399.168 2949.24 376.20 12.60 243.00 3995.208

Mass in = mass out , therefore the answer is correct

20

Conversion, Yield and Selectivity In most chemical processes several chemical reactions takes place simultaneously resulting in unwanted products (by-products) in addition to the desired product. Undesired products are also formed when the desired product reacts with one or 32

Mass Balances

more of the reactants. Undesired products cause loss in profit by chemical plants and, therefore, it is necessary to maximise the desired products and minimise the undesired products by controlling the reaction conditions. The terms ”yield & selectivity” are terms that measures the degree to which a desired reaction proceeds relative to competing undesired reactions. To understand the difference between conversion, yield and selectivity consider the reaction between benzene and nitric acid in the production of nitrobenzene. The reaction is: C6H6 + HNO3

C6H5NO2 + H2O

maximise this reaction

Unfortunately, in practice it is observed some of the nitrobenzene further reacts with nitric acid to form dinitrobenzene as per the following reaction: C6H5NO2 + HNO3

C6H4(NO2)2 + H2O minimise this reaction

This means that the final product will contain nitrobenzene as well as unwanted dinitrobenzene in addition to water and unreacted benzene. The amount of dinitrobenzene depends on the excess quantities of HNO3 in the initial reaction mixture. As nitrobenzene is the desired product, the yield in this process will be defined on the basis of the amount of nitrobenzene produced. moles of nitrobenzene produced % Yield

=

X 100 moles of benzene reacted total moles of benzene reacted

% Conversion

=

X 100 moles of benzene in feed moles of nitrobenzene

Selectivity of nitrobenzene = moles of dinitrobenzene E.G.

Consider the following reaction A

B

C

where C is the undesired product

C moles of A reacted Conversion

=

X 100 moles of A in feed

Yield

=

moles of B produced X 100 moles of A reacted moles of B in product

Selectivity of B

= moles of C in product 33

Mass Balances

Example Consider 100 kmoles of C6H6 is charged into the nitrating reactor and after 2 hrs the products were analysed and found to contain 8 kmoles of benzene and 8 kmoles of dinitrobenzene. Calculate the conversion, yield, and selectivity of nitrobenzene. Solution 100 kmole C6H6

Nitrating reactor

8 kmoles C6H6 8 kmoles C6H4(NO2)2 X kmoles C6H5NO2

HNO3

The reaction equations are: R1

C6H6

+

HNO3

C6H5NO2

+

H2O

R2

C6H5NO2 +

HNO3

C6H4(NO2)2

+

H2O

The amount of C6H6 reacted

= 100 – 8

= 92 kmol

As 92 kmol of benzene reacted, the initial amount of C6H5NO2 formed is also 92 kmol. (because C6H6 reacts via R1 only) The amount of dinitrobenzene (given) ( from R2) = 8 kmol ∴Nitrobenzene reacted in R2 = 8 x ratio = 8 x 1 ∴

= 8 kmol

amount of nitrobenzene in product stream = nitrobenzene from R1 – nitrobenzene reacted in R2 =92 – 8 amount of C6H6 converted

% conversion=

92 x 100

=

C6H6 in feed

=

x 100

=

92 %

x 100

=

91.3

100

moles of C6H5NO2 % yield

= 84 kmol

84 x

100

=

% moles of C6H6 reacted

92

moles of C6H5NO2 selectivity

= moles of C6H4(NO2)2

21

84 =

=

10.5

8

Tie-Substances ( inerts) A compound that goes directly from an input stream to an output stream without any chemical change is called a tie-substance. It is normally very useful in solving problems. It forms a “tie” between an input stream and an output stream. e.g. 34

Mass Balances

When oxygen is needed for a reaction to take place in most cases it is supplied in the form of air. As air contains nitrogen ( we will ignore argon and CO 2 ), we must take into consideration the associated nitrogen in air. The composition of air can be taken as 79 mol% (vol%) N2 and 21 mol% O2 If the O2 input is 7.5 kmol, then N2 input = 7.5 * 79/21 = 28.2 kmol Since the nitrogen does not react, the mass balance says that what went in should come out. Therefore nitrogen output must also be 28.2 kmol. Note: If you look again at exercises 15.1 & 15.2, everything (i.e. MEA etc.) other than CO2 formed a composite tie-substance. Exercise 21.1 A reactor produces a liquid and a waste gas stream. If the gas has the composition 8 mol% oxygen, 80 mol% nitrogen and 12 mol% carbon dioxide and the nitrogen input to the reactor was in the form of air, calculate how much oxygen has been used in the reactor per kmol air input. 22

Dry and wet basis When a gas stream containing water vapour is cooled to a point where water begins to condense out the remaining gas composition changes. For example, if the initial composition is 95% air and 5% water and most of the water is removed, the concentration of the water reduces and the concentration of the air rises, towards 100%. This can be a nuisance, especially if the condensation occurs in stages and a new composition has to be calculated several times. A simple alternative way to express the composition uses the concept of a dry basis. In this method, the water is ignored in the main analysis, so as the water condenses, the dry basis composition remains unchanged. The “dry gas” is simply that part of the mixture which is not water vapour. It does NOT mean that there is no water present. Example Component

wet basis

Nitrogen Carbon dioxide Oxygen Water vapour Total

kmol 62 24 19 12 117

mol% 53.0 20.5 16.2 10.3 100.0

dry basis kmol mol% 62 59.0 24 22.9 19 18.1 105 100

Exercise 22.1 In the Deacon process for manufacturing chlorine, hydrochloric acid is oxidised with air. The reaction taking place is : 4HCl(g)

+ O2(g)

→

2 Cl2(g) + 2 H2O (g)

If the air used is 30% in excess of that theoretically required to completely oxidize hydrochloric acid gas, and if the oxidation is 80% complete, calculate the composition by volume of dry gases, leaving the reaction chamber. 35

Mass Balances

23

Atom Balances In some of the chemical processes (e.g. reactors) there is a range of components entering and leaving the system. Even if we do not know the details of the chemical reactions taking place within the unit it is still possible to do a balance around the unit by carrying out atom balances. The atom balance is based on the concept of the conservation of atomic species. That is atoms are neither created nor destroyed. For example, consider methane gas is burned in a boiler to generate steam. The reaction equation is: CH4 (g)

+

2 O2 (g)

CO2 (g) + 2 H2O

(v) Even though there is a chemical reaction taking place, the number of carbon atoms in the feed stream (CH4) should be equal to the number of carbon atoms in the product stream (CO2). Similarly, the number of hydrogen atoms in feed (CH 4) should be equal to number of hydrogen atoms in the product stream (H2O). Also the number of oxygen atoms in feed should be equal to the number of oxygen atoms in the product stream. In the product stream oxygen is present in both CO2 and H2O. Therefore, oxygen atoms of CO2 and oxygen atoms of H2O should be added together to balance the oxygen atoms in the feed stream. i.e. Number of oxygen atoms in O2 = number of oxygen atoms in CO2 + number of oxygen atoms in H2O Atom balances can be carried out even if there is no chemical reaction taking place.

36

Mass Balances

ATOM BALANCE EXAMPLE One of the processes commonly used to produce hydrogen for various refining and petrochemical operations is to react methane rich natural with excess steam in the presence of nickel catalyst. The dry gas composition of the product leaving the reformer is given below:. As methane rich stream also contains ethane, several reactions are possible. Find the molar ratio of methane and ethane in the feed stream. Component CH4 C2H6 CO CO2 H2

Molar Percentage 4.6 2.3 18.6 4.6 69.9

In Teesside there is an ammonia plant at Billingham . For the production of NH3, nitrogen and hydrogen are needed. Possible reactions are: C2H6  C2H4 + H2

1

CH4 + H2O  CO + 3H2 CH4 + 2H2O  CO2 + 4H2 CO + H2O  CO2 + H2

2 3 4

Problem gives analysis of product gas on dry basis. This implies that, even if steam is present in the gaseous product, it is not included in the analysis for the reason given in section 22. In this example steam is present in the product gases.(note steam is also supplied in excess) Required to calculate ratio of CH4 and C2H6 on molar basis. Let’s carry out recommended procedure 1) Draw flow diagram 2) Draw input/output table 3) Basis – 100 kmol of H2 rich dry product gas (most information is available). 4) Write balanced (stoichiometry) reaction equations ( given above). Since there is no C2H4 in the product stream we could eliminate calculation. Now the problem is (i) To what extent does each of the other reactions take place?

equation 1 from our

(ii) How much from equation 2, equation 3 and equation 4? (iii) Do all 3 reactions take place? Now the big question is If we can’t write equations and do not know the extend of reactions, what can we balance?

Remember kmols do not necessarily balance, but mass and atoms do balance. 37

Mass Balances

Mass of kg of element = K atoms RAM For steady state. Mass of all elements = K atoms in = K atoms out (Avagadro’s number is omitted in the atom balance because the factor 6.022 x 1023 will always appear on both sides of the balance and cancel out) So we can balance atoms. Atom Balance Solution CH4

M

C2H6 E

Steam

Product gas dry basis composition Mol% CH4 4.6 C2H6 2.3 CO 18.6 CO2 4.6 H2 69.9 H2O W kmol

FURNACE

S+W

Basis: 100 kmol of dry product gas. In steam reforming reactions carbon to steam ratio in the feed is about 3 to prevent carbon deposition on the catalyst – this reduces the activity of the catalysts. Since steam is the excess reactant a portion of steam in the feed will exit the reformer unreacted. This excess steam will not be considered in the calculation. The normal method of writing equation does not work! So what do we do? In the feed streams let

CH4 C2H6 Reacted steam Unreacted steam

= M kmol = E kmol = S kmol = W kmol

Now construct the input/output atom balance table as shown below. K mol H atom M 4xM E 6xE

INPUT C atom

O atom

K mol

OUTPUT H atom C atom

O atom

CH4 1xM 4.6 4 x 4.6 1 x 4.6 C2H6 2xE 2.3 6 x 2.3 2 x 2.3 CO 18.6 1 x 18.6 1 x 18.6 CO2 4.6 1 x 4.6 2 x 4.6 H2 69.9 2 x 69.9 Reacted S 2xS 1xS steam(H2O) Unreacted W W steam There are four unknowns : M,E, S and W. But we have only three elements: C, H and O. As W does not react( W in = W out), we could eliminate this. Now we are left with three unknowns (i.e. M, E & S) and three elements(i.e. C,H,& O). 38

Mass Balances

C – atom balance INPUT = OUTPUT M + 2E = 4.6 + 2 x 2.3 + 18.6 + 4.6 M + 2E = 32.4 M = (32.4 – 2E) H – atom balance

1

(only consider H2O which reacted )

4M + 6E + 2S = 4 x 4.6 + 6 x 2.3 + 2 x 69.9 4M + 6E + 2S = 172

2

O – atom balance Oxygen in CO and CO2 could have come only from reacted steam, because there is no oxygen in CH4 and C2H6. H2O is also in the output, but not reported because composition is given in dry basis. Also this water is the unreacted steam ( ie. W in = W out). S = 18.6 + 2 x 4.6 S = 27.8 kmol/ 100 kmol of dry product gases

3

Substituting for S in equation (2) 4M + 6E + 2 x 27.8 = 172

4

Now substituting for M in equation 4 4(32.4 – 2E) + 6E + 2 x 27.8 = 172 129.6 – 8E + 6E + 55.6 = 172 -2E = 172 – 55.6 – 129.6 -2E = -13.2 E = 13.2/2 = 6.6 kmol/ 100 kmolof dry product gases Substituting for E in equation (1) M = 32.4 – 2 x 6.6 M = 19.2 kmol/100kmol of dry product gases Therefore, ratio of CH4:C2H6 =

M = 19.2 = 2.91 E 6.6 Always try normal mol balance first. If it does not work then try atom balance. Atom balance will always work, but may take longer to solve.

39

Mass Balances

Summary of atom balance If the extent of stoichiometry cannot be determined, consider atom balance. Suppose now that the example asked for the number of kg of steam that react per 1000 m 3 of natural gas. We could choose a basis of 100 kmol of product gases, and the calculation procedure will be exactly the same as above to obtain M,E and S in kmols. First find out the number of kmols in 1000 m3 of feed gas( 1 kmol of any gas = 22.4 m3) Number of moles in 1000 m3 of gas

= 1000/22.4 = 44.64 kmols of feed gas

Total mols of feed gas from our earlier calculation is = 19.2 + 6.6 = 25.8 kmols This amount of hydrocarbon gases required 27. 8 kmol of steam Now scale up to find the amount of steam required to react with 44.64 kmols of hydrocarbon gases. 27.8 Amount of steam required = 44.64 x = 48.10 kmols 25.8 ∴ Mass of required steam = 48.10 x 18 = 865.8 kg/ 1000 m3 of feed gas.

40