Chemical Engineering

Thermodynamics

CHME322

Classical Thermodynamics is divided into:

1. Engineering Thermodynamics: studied last year

covers machines (turbines, compressors, etc)

2. Chemical Thermodynamics: deals with

equilibrium, both in chemical reactions, and

between phases

Introduction

Chemical Engineers are concerned with changesof systems not in equilibrium.

Example: chemical reactions, the transfer ofcomponents from liquid to vapour in distillationand so on.

The rate of such processes depends on a drivingforce

Driving force is a function of how far the systemis from equilibrium

Hence, equilibrium data are needed

Thermodynamics helps deals with classification,correlation and prediction of chemical and phaseequilibrium data.

Why we need to study it?

A region of the universe separated from the rest by real

or imaginary boundaries.

System

isolated adiabatic closed closed open

transfer of work only transfer of heat

and work

Isolated system: neither energy nor matter is

transferred across the boundary.

Closed system: energy, but not matter may be

transferred across the boundary. If work but not

heat may be transferred, the system is called

adiabatic.

Open system is one in which both matter and

energy may be transferred.

Types of systems

Is specified by a number of properties such as

Temperature, Pressure, Volume, etc.

These properties are state variables

A system is at equilibrium when none of them

change with time.

State

State variables can be either intensive

(independent of the quantity of material present)

- eg. P, T, etc - or extensive (proportional to the

quantity of material) - eg. V, H, etc.

When extensive variable is divided by the mass

or moles, it becomes a specific variable.

Thus, specific volume,v = V/n (molar density-1)

Unless otherwise stated, the specific variables

will be molar

Types of properties

Most single component, single phase systems can be

completely specified by two state variables (i.e., two

degrees of freedom), and all the other variables

expressed as functions of these two, thus:

u = u(P,v) = u(P,T) = u(T,v) etc.

The value of any state variable depends only on the

state of the system (independent of the path

followed by the system between the two states)

Reversible process

Happens so slowly that the system may be regardedas being in equilibrium at all times.

The state variables have values only infinitesimallydifferent from their equilibrium values, so that aninfinitesimal change is enough to cause the process togo in the opposite direction, hence the name"reversible

To calculate the change Dh, Du, etc between any twogiven states, we devise a reversible path betweenthem for which information is available to calculatethe change.

The change along any other path, including non-reversible path will be the same as that calculated(properties are state functions).

Equations of State

Represents the relationship between state variables(usually PvT)

P = f(v,T) or f(P,v,T) = 0

The simplest equation of state is the ideal gasequation: Pv = RT

It represents the limiting behaviour of real gases atlow pressures.

More complex equations, such as van der Waals,Redlich-Kwong, Benedict-Webb-Rubin (BWR) etchave been devised, which contain additional,adjustable constants.

Compressibility factor

Law of Corresponding States: there exists a unique

relation between the three reduced variables, the same

for all substances.

This unique relation need not be expressible by any

analytical equation

A convenient way of representing this empirical

relation is through the compressibility factor, z

z = P v

R T =

P V

n R T

Van der Waals EOS

a and b are characteristic constants for each gaseous

species

b: Correction for molecular size (i.e., volume occupies

by molecules themselves)

a/v2: Correction for attractive forces between

molecules. (decrease the effective pressure)

v

a -

b - v

T R = P

2

Reduced Variables

For equations which have only two additional constants

(van der Waals, Redlich Kwong), the constants can be

eliminated by working with the reduced variables

Pr = P/Pc, Tr = T/Tc and vr = v/vc, where the suffix c

indicates the critical value

This is because, at the critical conditions there are two

extra equations that can be written:

0 = v

P =

v

P2

2

critT,critT,

Applying this to the van der Waals equation:

leads to: (1)

and (2)

then (3)

v

a -

b - v

T R = P

2

b = v

3

c

a = 3 P v = 9 R T v

8c c

2 c c

rr

r r2

P = 8 T

3 v - 1 -

3

v

375.08

3

c

cc

cRT

VPz

It is possible to construct a unique graph of z as a

function of Pr and Tr, valid for all substances:

z = f(Tr,Pr) (therorem of corresponding state)

This is only applied when zc is the same for all gases.

In practice, this is not the case, thus, z = f(Tr, Pr, zc).

The critical compressibility factor lies in the range

0.23 < zc < 0.31

Majority of not-too-polar organics having a value

close to 0.27.

Alternatively, acentric factor w is used instead of zc

Class Problem 1

a) Derive that b = vc/3 and a= 9/8 RTcvc in the van der Waals equationfrom the critical point conditions.

b) Calculate the numerical value of zc for a gas obeying van der Waalsequation. How does it compare with the range of 0.23 to 0.31 for realgases?

c) Substitute for b in Eq.1 and a in Eq.2 to demonstrate that this leads tothe reduced form of Eq.3

d) Using the generalised compressibility factor chart:

(i) Locate the critical point and read off the value of zc.

(ii) Water has Pc = 221.2 bar, Tc = 374.2oC. What is the maximum

pressure at which saturated steam can be treated as an ideal gas

with an error of not more than 10%?

(iii) After estimating Tr, the ideal and (estimated) real densities of

saturated steam in these conditions?

Acentric factor Symmetrical molecules or simple fluids (eg, inert gases: Ar, Kr)

conform very closely with the corresponding state principle

It is observed that for simple fluids, all vapour pressures lie on thesame line when plotting log Pr

sat vs 1/Tr The line passes by log Pr

sat = -1.0 at 1/Tr = 1.43 (i.e., Prsat = 0.1 at

Tr = 0.7)

Normal fluids have Prsat < 0.1 at Tr = 0.7 (Pr

sat at Tr = 0.7decreases with increasing asymmetry)

Acentric factor w is defined as:

w = 0 for the inert gases.

7.0

7.07.0

log0.1

loglog

r

rr

T

sat

r10

T

sat

r10T

sat

r10

P -

P SFP =

w

w

Compressibility factor, is given as linear functions of w:

z = z0 + wz1 Charts are given for z0 and z1 as complex functions of

Pr and Tr.

More recent cubic EOS (Redlich-Kwong-Soave, 1972,Peng-Robinson, 1976) include a function of w as amultiplying constant, thus enabling them to give acloser fit to a wide range of real gases.

They are well suited to computer implementation, andthe EXCEL spreadsheet can solve them.

Because w is based on saturated vapour pressure data,these equations are better suited to predict vapour-liquidequilibrium than the simpler Redlich-Kwong equation.

Note that all equation of state are semi-empirical in

nature as opposed to the virial equation, which has

a strong theoretical basis in statistical mechanics.

The Fundamental Equation

The 1st and 2nd laws of thermodynamics combined togive:

(4)

Where, U (internal energy) and S (entropy).

For reversible changes only, TdS is equal to the heatsupplied to the system, while PdV is the work doneby the system

For irreversible changes the equation still applies,but the heat supplied and work done are both lessthan in the reversible case.

dU = T dS - P dV

Flow system

Gas enters at state 1 (P1,T1)

and leaves at state 2.

The specific energy content

of the gas entering and

leaving are u1 and u2,

respectively.

P T 1 1

v u 1 1

2 2P T

v u 2 2

Work P1v1 is done to force one mole of it into the system, and

P2v2 is done against the environment as it leaves.

If there is no other energy input to the system, an energy balance

gives:1 1 1 2 2 2u + P v = u + P v

(enthalpy) h

Enthalpy: (5)

DH (the net energy input in all cases)

Helmholtz Free Energy: A = U - TS (6)

Gibbs Free Energy G = H - TS (7)

Differentiating equations (5), (6) and (7), and

substituting from (4) gives the fundamental equation:

dH = TdS + VdP

dA = -SdT - PdV

dG = -SdT + VdP

H = U + P V

Net Work and Spontaneous Change

For any change that takes place in a closed system at

constant temperature T, the 2nd law gives:

(8)

where Q is the heat supplied to the system from a

reservoir at temperature T. Here, > sign applies to

irreversible processes and the = to reversible

1st Law:

Then, (9)

T S QD

DU = Q - W

W A - US T DD

The work done by the system, W:

1. PV work against the environment

(at constant pressure = PDV)

2. Net work (e.g. mechanical, electrical, etc).

In any spontaneous change at constant temperature

and pressure which produces no net work other than

PV work against the environment, the Gibbs Free

Energy must decrease, i.e. DG < 0.

Hence, any process where DG > 0 will not occur

spontaneously, but will need an input of net work

W G - V P + U S T netDDDD

At Equilibrium no further change occurs, so

G has reached a minimum

for any process at constant temperature and

pressure, DG = 0 at equilibrium

This equation forms the basis of much of the

study of Chemical Engineering

Thermodynamics