tda 301t-7 - tutorial 2 - 4 chapters

7/27/2019 TDA 301T-7 - Tutorial 2 - 4 Chapters

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APPLIED THERMODYNAMICS

TDA-301T 2012

Lecture 7: Tutorial Chapters 2 - 4,

By

Prof Alex SofianosBsc Chem Eng, Msc, PhD Ind Chem (Germany),

MBL (UNISA)


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Course Contents

1. Introduction2. Conservation of Mass

3. Conservation of Energy

4. Entropy An additional Balance Equation

5. Liquefaction, Power Cycles, Explosions6. Thermodynamic Properties of Real Substances

7. Equilibrium and Stability One Component

8. Thermodynamics of Multi-Component Systems

9. Estimation of Gibbs Energy and Fugacity of a Componentin a Mixture

10. Vapor-Liquid Equilibrium in Mixtures

11. Chemical Equilibrium

2


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Course Books

1. Chemical, Biochemical & Engineering Thermodynamics

4 Edition, Stanley I. Sandler , John Wiley and Sons (2006)

2. Introduction to Chemical Engineering Thermodynamics

(The McGraw-Hill Series in Civil and Environmental

Engineering) by J.M. Smith, Hendrik C. Van Ness, and

Michael Abbott (2004)

3. Introduction to Chemical Engineering Thermodynamics

by J.M. Smith (2001)4. Engineering and ChemicalThermodynamicsby Milo D.

Koretsky (2003)

3


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Objectives

4

At the end of this Chapter, the student must: Be able to solve problems involving the liquefaction of

gases

Be able to compute the work that can be obtained

from different types of power cycles and using

different working fluids

Be able to compute the work required for the

operation of refrigeration cycles

Be able to compute the energy release resulting from

the uncontrolled explosion of a gas


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Mass BalanceILLUSTRATION 2.2-3

5

Use of the Rate-of-Change Form of the Mass Balance Gas is being removed from a high-pressure storage

tank through a device that removes 1 percent of the

current contents of the tank each minute.

If the tank initially contains 1000 mols of gas, how

much will remain at the end of 20 minutes?

SOLUTIONSince 1 percent of the gas is removed at any time, the

rate at which gas leaves the tank will change with time.


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6

Initially gas is leaving at the rate of 0.01 1000 mol/min= 10 mol/min.

Later when only 900 mol of gas remain in the tank, the

exiting flow rate will be 0.01900 mol/min = 9 mol/min.

The exiting flow rate is continuously changing with time!

Use the rate-of-change or differential form of the mass

(mole) balance:


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7


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8

Rate-of-change form of the mass balance (Eq. 2.2-1b)around the tank, that has only a single flow term

If we solve this first-order differential equation:

Then we obtain:N(t=20) = N(t=0) e -0.01t

Or N(t=20) = 1000 e -0.01x20 = 1000 e -0.2 = 818.7 mol


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9

Rate-of-change form of the mass balance (Eq. 2.2-1b)around the tank, that has only a single flow term

If we solve this first-order differential equation:

Then we obtain:N(t=20) = N(t=0) e -0.01t

Or N(t=20) = 1000 e -0.01x20 = 1000 e -0.2 = 818.7 mol


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Mass BalanceCONCLUSION 2.2-3

10

To solve any problem in which the mass (or molar) flowrate changes with time, we need to use the differential

orrate-of-change form of the mass balance.

For problems in which all of the flow terms are constant,

we can use the general difference form of the massbalance (which has been obtained from the rate-of-

change form by integration over time),

or we can use the rate-of-change form and then

integrate over time.

However, it is important to emphasize that if one (or

more) flow rates are changing with time, the rate-of-

change form must be used.


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11

Use of the Rate-of-Change Form of the Mass Balance An open cylindrical tank with a base area of 1 m2 and a

height of 10 m contains 5 m3 of water.

As a result of corrosion, the tank develops a leak at its

bottom. The rate at which water leaves the tankthrough the leak is

whereP is the pressure difference in bar between thefluid at the base of the tank and the atmosphere.

Determine the amount of water in the tank at any time.


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12

SOLUTION Pressure at the bottom of the tank is equal to the

atmosphere pressure plus the hydrostatic pressure

due to the water above the leak!

P = 1.013 bar +g h

where is the density of water andh is the height of

water above the leak; g is the acceleration due to

gravity; Therefore,P = (1.013 +gh) 1.013 = g h

Then:


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13

Since the height of fluid in the tank is changing withtime, the flow rate of the leak will change with time.

Therefore, to solve the problem, we must use the rate-

of-change form of the mass balance.

First we calculate the mass of water in the tank at atime t:

The mass balance on the contents of the tank at anytime may be expressed as:


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14

The negative sign arises because the flow is out of thetank!

103 ()

= -0.1566 h(t) or d h(t) = -0.1566 10-3h(t) dt

Integrating this equation between t = 0 and any latertime t yields:

or


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15

Since the rate of flow of water out of the tank dependson the hydrostatic pressure due to the water column

above the leak, and since the height of this column

changes with time,

again we must use the rate-of-change form of the massbalance to solve the problem.

It is important in any problem to be able to recognize

whether the flows are steady, in which case the

difference form of the mass balance can be used, or theflows vary with time, as is the case here, in which case

the rate-of-change form of the mass balance must be

used.


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16

Since the rate of flow of water out of the tank dependson the hydrostatic pressure due to the water column

above the leak, and since the height of this column

changes with time,

again we must use the rate-of-change form of the massbalance to solve the problem.

It is important in any problem to be able to recognize

whether the flows are steady, in which case the

difference form of the mass balance can be used, or theflows vary with time, as is the case here, in which case

the rate-of-change form of the mass balance must be

used.


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Multicomponent System

With Chemical Reaction

17

When chemical reactions occur, the mass (or mole)

balance for each species is more complicated since

the amount of the species can increase or decrease as

a result of the reactions.

Here we will consider mass balances when there isonly a single chemical reaction;

we will write the mass balances using number of

moles only, since the stoichiometry of chemical

reactions is usually written in terms of the number ofmoles of each species that undergoes chemical

reaction rather than the mass of each species that

reacts


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18

Using the notation (i)

kfor the rate at which moles of

species ienter (if positive) or leave (if negative) in flow

stream k, we have the differential or rate-of-change

form of the mass balance on species i

last term is new and describes the rate at which

species iis produced or consumed within the system

by chemical reaction

Rate-of-change mass balance with chemical

reaction on a molar basis


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The difference form of this equation, obtained by

integrating over the time period from t1 to t2, is

Summation terms after the equal signs are the

changes in the number of moles of the species due to

the flow streams, and the second terms are the result

of the chemical reaction

Difference form of the mass balance with

chemical reaction on a molar basis


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20

If the flow rate of a stream is steady (i.e, (i)

kis

constant), then:

EXAMPLE

Mass (mole) balance for a reactor in which thefollowing chemical reaction occurs:

Difference form of the mass balance with steady flow rate


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21

During the reaction neither ethylene nor chlorine is

completely consumed.

Stream 1 is pure ethylene and stream 2 is pure

chlorine; exit streams [i.e., (...)3] are negative in

value. The mass balances for these species, are as follows:


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23

General chemical reaction equation:

With the molar stoichiometric

coefficients can be re-written as:

or as:

with iis the stoichiometric coefficientof species i,


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We will use Ni

to represent the number of moles of

species i in the system at any time t and N i,0 to be the

initial number of moles of species i.

However, Niand Ni,0 are related through the reaction

variable X, called the molar extent of reaction, andthe stoichiometric coefficients i by the equation:

or :

with i is the stoichiometric coefficientof species i

l


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Example:ILLUSTRATION 2.3.-1

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The electrolytic decomposition of water to form hydrogenand oxygen occurs as follows:

Initially, only 3.0 mol of water are present in a closed

system. At some later time it is found that 1.2 mol of H2and 1.8 mol of H2O are present.

a. Show that the molar extents of reaction based on H2

and H2O are equal.

b. Compute the number of moles of O2 in the system.

SOLUTION

a. Reaction is written as

l


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So that:

b.- Using equation Ni= Ni,0 + iX

we have:

E l


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The rate of change of the number of the moles of speciesiresulting from a chemical reaction is:

where the subscriptrxn indicates that this is the rate of

change of the number of moles of species idue to

chemical reaction alone, and

is the rate of change of the molar extent of reaction.

Rate of change Form


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Rate-of-change Form

of the Mass Balance

28

The balance equation for species i is therefore:

Rate-of-change form of the mass balance

with chemical reaction: molar basis

And the here the subscriptrxn indicates that this is the

rate of change of the number of moles of species idue tochemical reaction alone, and

is the rate of change of the molar extent of reaction.


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Mass Balance

with Extent of Reaction

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We can now write the mass balances for the ethylenedichloride synthesis using the molar extent of the

reaction:

As:


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30

Performing material balances on reactive systems

is slightly more complex than for non-reactivesystems.

For simplicity, we will only focus on reactive

processes that are open systems which are run at

steady state.

N2 + 3H2 -> 2NH3


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N2 + 3H2 -> 2NH3

This yields a simplified balance equation of:

i i0 + GC= 0 ori0 = i+ GC

G - GenerationC - Consumption ; both through chemical reaction


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N2 + 3H2 -> 2NH3

This yields a simplified balance equation of:

i i0 + GC= 0 ori0 = i+ GC

we will switch to using molar quantities that

we get:

i0 = i+ GC


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N2 + 3H2 -> 2NH3

The extent of reaction (X) is a method of quantifyinghow many "times" a reaction has occurred.

It has units ofmoles/time and numerically, it is chosen

such that the stoichiometric coefficienttimesXis

equal to the quantity of species reacted.

2O= 8 kmol/h = N2I + GN2 CN2= 10 kmol/h + 0 1X


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N2 + 3H2 -> 2NH3

N2O= 8 kmol/h = N2I + GN2 CN2= 10 kmol/h + 0 1XOUT IN

Similarly:

H2O= 15 kmol/h = H2I + GH2 CH2= H2I + 0 3X

NH3O= 4 kmol/h = NH3I + GNH3 CNH3= 0 + 2X 0


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Typically "reactants" have g=0, while "products"have a zero initial concentration and have c=0. This

obviously becomes more complex when multiple

reactions take place.

Using the extent of reaction method for systemswith multiple reactions involves including a new

value ofXfor each reaction, and then calculating c

and g as the sum of applicable X 's.

EXAMPLE

Reactions: A B2B C

Extent of Reaction X


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EXAMPLE Reactions: A B

2B C

If we assignX1 to the first reaction andX2 to thesecond, we get expressions that look like:

AO = AI+ GA CA = AI + 0 1X1

BO = BI+ GB CB = BI + 1X1 2X2

CO = CI+ GC CC= CI + 1X2 0

Extent of Reaction X


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Problem 2.4

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The following reaction occurs in air:

At 20 0C the rate of this reaction is given by the formula:

for t in seconds and concentrations in kmol/m3. The

reaction occurs in a constant-volume, 2-L vessel, and the

initial concentration of NO is 1 kmol/m3 and that of O2 is

3 kmol/m3

a. If 0.5 mol of NO reacts, how much NO2 is produced?

b. Determine how long it would take for 0.5 mol of NO to

have reacted.


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Problem 2.4

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The following reaction occurs in air:

a. If 0.5 mol of NO reacts, how much NO2 is produced?

Equation may be re-written as:

2NO2 2NO O2 = 0

By division with 2 we obtain:

NO2 -1

2O2 NO = 0

From the equation we can see at once that 1 mol NO

would be consumed in order to produce 1 mol NO2 ;

therefore, 0.5 mol NO would produce 0.5 mol NO2


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Problem 2.4

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b.- We set up a stoichiometric matrixas follows:

Determine the initial concentrations of the reactants:

Reactants Stoich. Initial ConcentrationCoefficients Concentration at X

NO - 2 1 1 2 X

O2 - 1 3 3 X

NO2 2 0 2 X


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Problem 2.4

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b.- We determine the actual mol numbers from the

concentrations:

For NO we have: CNO = 1 kmol/m3

We know that CNO =

;

therefore: NNO = V CNO

and NNO = 2 L x 1mol/L = 2 mol


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ILLUSTRATION 2.3-2

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Mass Balance for a Mixture with Chemical Reaction At high temperatures acetaldehyde (CH3CHO) dissociates

into methane and carbon monoxide by the following

reaction:

At 520 oC the rate at which acetaldehyde dissociates is

given by the formula:

C is the concentration in kmol/m3 The reaction occurs in a

constant-volume, 1-L vessel, and the initial concentration

of acetaldehyde is 10 kmol/m3

.


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ILLUSTRATION 2.3-2

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a. If 5 mols of the acetaldehyde reacts, how muchmethane and carbon monoxide is produced?

b. Develop expressions for the amounts of acetaldehyde,

methane, and carbon monoxide present at any time,

and determine how long it would take for 5 mol ofacetaldehyde to have reacted.

SOLUTION

First determine the initial amount of acetaldehyde

present.


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ILLUSTRATION 2.3-2

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Since the initial concentration is:

we conclude that:

Then we write the stoichiometry for the reaction

in terms of the molar extent of reaction X as follows:


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ILLUSTRATION 2.3-2

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To determine the amounts of each species after a givenamount of acetaldehyde has reacted, we can use the

difference form of the mass balancefor this system with

no flows of species into or out of the reactor:

we get for acetaldehyde:

or X = 5 mol. Then amounts of the other species are:

NCH4 (t) = X = 5 mol and NCO(t) = X = 5 mol


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ILLUSTRATION 2.3-2

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b.-To determine the amount of each species as a functionof time is more difficult and must be done using the rate-

of-change form of the mass balance since the rate of

reaction and therefore the value of X change with time.

We can use the mass balance for one of the species todetermine the time variation of X, and then can use the

expression for X(t) to obtain the compositions of all

species in the reaction as a function of time.

Since the rate expression is written for acetaldehyde, wewill use this substance to determine the time dependence

of X


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ILLUSTRATION 2.3-2

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There are no flows into or out of the reactor, therefore:

can be written for acetaldehyde:

Next the reaction rate expression can be written as


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ILLUSTRATION 2.3-2

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The latter , after using N CH3CHO = 10 X can become:

Next, integrating from t = 0, at which X = 0, to some other

time t gives:

or extent of the

reaction as a function of time! Example: solving this equation for X(t) = 5 mol gives t =

0.208 s; that is, half of the acetaldehyde has

decomposed within 0.2 seconds


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ILLUSTRATION 2.3-3

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Mass Balance for a Liquid Mixture with a Reversible Reaction

The ester ethyl acetate is produced by the reversible

reaction

Reaction activated by a sulfuric acid catalyst . The rate ofethyl acetate production has been found, from the

analysis of chemical kinetics data, to be given by the

following equation:

subscripts EA, A, E, and W denote ethyl acetate, acetic

acid, ethanol, and water, respectively

C is the concentration of each species in units of kmol/m3


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ILLUSTRATION 2.3-3

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The values of the reaction rate constants at 100C andthe catalyst concentration of interest are:

k = 4.76 104 m3/kmol min

k = 1.63 104 m3/kmol min

Develop expressions for the number of moles of each

species as a function of time if the feed to the reactor is

1 m3 of an aqueous solution that initially contains 250 kgof acetic acid (A) and 500 kg of ethanol (E).

The density of the solution may be assumed to be

constant and equal to 1040 kg/m3


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ILLUSTRATION 2.3-3

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The reactor will be operated at a sufficiently high

pressure that negligible amounts of reactants or

products vaporize.

Compute the number of moles of each species present at

a time t (t=100 minutes) after the reaction has started,

and

atinfinite time when the reaction will have stopped and

the system is at equilibrium.


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ILLUSTRATION 2.3-3

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SOLUTION

The reaction rate expression is a function of the compo-sitions, which are changing as a function of time,

the mass balance for each species must be written in the

rate-of-change form.

Since the species mole numbers and concentrations are

functions of the molar extent of reaction, X, we first

determine how X varies with time by solving the mass

balance for one species.

Since the reaction rate is given for ethyl acetate we have

to use that species!

Once the amount of ethyl acetate is known, we can

compute the other species!


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ILLUSTRATION 2.3-3

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The initial concentration of each species, expressed as

kmol/m3 is:

Since there is only 1 m3 the initial amount of each species

is:


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ILLUSTRATION 2.3-3

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If we consider the reaction stoichiometry, the amount of

each species present at any time (in kmol) and itsconcentration (since 1 m3 of volume is being considered)

is

And NEA = X CEA = X

The concentration of a species is equal to the number ofmoles N divided by the volume V,

i.e. C=


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ILLUSTRATION 2.3-3

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the chemical reaction rate equation can be written as

By using V = 1 m3

and the mole numbers, we can simplify:

or

and


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ILLUSTRATION 2.3-3

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This can be re-arranged to:

And integrated between t= 0 and t = t; this yields:

or

And after solving for X

X in kmol

t in minutes


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ILLUSTRATION 2.3-3

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From

At t = infinite, i.e., at equilibrium:

X = 2.3911 kmol so that:NA= 4.17 X= 4.17 2.39 = 1.78

NE= 8.51

NW = 18.49

NEA = 2.39


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THANK YOU

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