Transport Phenomena 2nd Edition

Prévia do material em texto

Foreword 
Momentum, heat and mass transport phenomena can be found nearly 
everywhere in nature. Even an early morning activity such as boiling an egg* 
or making tea is governed by laws which will be treated here. A solid under-
standing of the principles of these transport processes is essential for those who 
apply this science in S U D F W L F H a � e.g. chemical and process engineers. 
The history of teaching transport phenomena went from a practical but less 
fundamental approach, via a short period of a practical and academic approach, 
to the present sophisticated approach. Our experience in education and in 
industry is that today's abstraction does not appeal to all students and engineers, 
who feel themselves easily lost in vast literature and difficult mathematics. 
Hence, our objective in writing this book was to digest the enormous amount 
of new knowledge and present it in a form useful for those who work as pro-
fessional engineers or who study engineering. 
The present book incorporates much fundamental knowledge, but we have 
also tried always to illustrate the practical application of the theory. On the 
other hand, we have also included practical information and have not shied 
away from giving one or two useful empirical correlations, where theory 
would have been too difficult. The book is based on the text for a course in 
transport phenomena given by W. J. Beek at Delft University from 1962 to 
1968. Parts of the last draft have been used, together with most problems 
encountered in three postgraduate courses. 
Each chapter ends with a number of problems which form an integral part 
of the book. We would like to ask the student to solve as many of these problems 
as possible-this is the best way to absorb and digest the theory. We have given 
the answers of all problems so that the reader can check his results. Where we 
expected difficulties to arise, we have explained some problems in greater detail. 
Furthermore, we invented John, our scientific sleuth, and we hope the reader 
likes his way of solving problems. The reason for his ability to do so is evident : 
he has read the present book and has worked through the problems! 
*For identical temperature equalization in the egg the Fourier number must be Fo = atfd2 = 
constant, i.e. for all eggs: 
boiling time x (mass)-213= constant 
is valid. Knowing the optimal bojling time for one species of egg, we can thus predict the boiling 
time for other eggs, even ostrich eggs. 
Contents 
CHAPTER I INTRODUCTION TO PHYSICAL TRANSPORT 
PHENOMENA 1 
I.l. Conservation laws 2 
1.2. Rate of molecular transport processes 11 
I.3. Microbalances 15 
1.4. SI units 20 
1.5. Dimensional analysis . 21 
1.6. Problems . 28 
CHAPTER II FLOW PHENOMENA 
ILL Laminar flow 37 
1. Stationary laminar flow between two flat horizontal plates 38 
2. Flow through a horizontal circular tube 40 
3. Flow through a horizontal annulus 42 
4. Flow caused by moving &urfaces 45 
5. Flow through pipes with other cross-sections 48 
6. Non-stationary flow 49 
7. Problems 52 
II.2. Turbulent flow . 55 
1. Turbulent flow in pipes 55 
2. Pressure drop in straight channels 60 
3. Pressure drop in pipe systems . 64 
4. Problems 74 
II.3. Flow with negligible energy dissipation 79 
1. Flow of a liquid from an orifice 80 
2. Flow of ga~es through orifices 82 
3. Flow through,weirs 85 
4. Problems 88 
II.4. Flow meters , . _ 91 
1. Venturi tube 92 
2. Orifice plate 93 
3. Rotameter 94 
4. Problems 95 
II.5. Flow around obstacles. 98 
1. General approach . 98 
2. Spherical particles . 101 
3. Free fall of droplets 104 
4. Particles in non-stationary flow 105 
Vlll 
5. Rate of sedimentation of a swarm of particles 
6. Cylinders perpendicular to the direction of flow 
7. Problems 
Il.6. Flow through beds of particles 
I . Fixed bed 
106 
107 
108 
112 
112 
115 
116 
117 
120 
120 
122 
124 
124 
125 
125 
127 
129 
131 
137 
140 
2. Filtration through a bed of particles 
3. Fluidized bed 
4. Problems 
II. 7. Stirring and mixing 
1. Types of stirrer and flow patterns 
2. Power consumption . 
3. Pumping capacity and mixing time 
4. Problems 
II.8. Residence time distribution 
1. The F function 
2. The E function 
3. Simple applications ofF and E functions 
4. Continuous flow models . 
5. Dispersion in flow systems 
6. Problems 
CHAPTER III HEAT TRANSPORT 
III. I . Stationary heat conduction . 145 
I. Heat conduction through a wall 146 
2. Heat conduction through cylindrical walls 147 
3. Heat conduction around a sphere 148 
4. General approach for the calculation of temperature distribu-
tiom 1~ 
5. Temperature distribution in a cylinder with uniform heat 
production 150 
6. Problems ·152 
III.2. Non-stationary heat conduction 156 
1. Heat penetration into a semi-infinite medium 157 
2. Heat penetration into a finite medium 161 
3. Influence of an outside heat transfer coefficient 164 
4. Problems 167 
III. 3. Heat transfer by forced convection in pipes 171 
I . Heat transfer during laminar flow in pipes 171 
2. Heat transfer during turbulent flow . 174 
3. Partial and total heat transfer coefficients 176 
4. Problems 178 
III.4. Heat exchangers . 182 
1. Determination of mean temperature difference 182 
2. Height of a transfer unit . 186 
3. Design of heat exchangers 187 
4. rroblems 189 
III.5. Heat transfer by forced convection around obstacles 
1. Flow along a fiat plate 
2. Heat transfer to falling films 
3. Flow around spheres and cylinders 
4. Heat transfer in packed beds 
5. Heat transfer in fluidized beds 
6. Problems 
III.6. Heat transfer during natural convection . 
1. Heat transfer during natural convection 
2. Problems 
III. 7. Heat transfer during condensation and boiling 
1. Film condensation . 
2. Dropwise condensation 
3. Boiling . 
4. Heat transfer in evaporators 
5. Problems 
III.8. Heat transfer in stirred vessels 
1. Problem 
III.9. Heat transport by radiation . 
1. Problems 
CHAPTER IV MASS TRANSPORT 
IX 
194 
194 
194 
195 
197 
198 
199 
201 
201 
204 
208 
209 
211 
212 
213 
214 
215 
219 
220 
223 
IV .1. Stationary diffusion and mass transfer 227 
1. Stationary diffusion 227 
2. Mass transfer coefficients 231 
3. General approach for the calculation of concentration dis-
tributions 233 
4. Film theory 234 
5. Problems 236 
IV.2. Non-stationary diffusion 240 
1. Problems 242 
IV.3. Mass transfer with forced convection 246 
1. Analogy with heat transfer 246 
2. Mass transfer 'during laminar flow 248 
3. Mass transfer during turbulent flow 253 
4. Problems .- 254 
IV.4. Mass exchangers 257 
l. Thermodynamic equilibrium 258 
2. Choice of the apparatus 259 
3. Size determination of the mass exchanger 260 
4. The concept of theoretical plates 263 
5. Problems 265 
IV.5. Mass transfer with chemical reaction 270 
1. Slow homogeneous first-order reactions 271 
2. Fast homogeneous first-order reactions 273 
3. Homogeneous nth-order reactions 275 
X 
4. Homogeneous second-order reactions 
5. Mass transfer with heterogeneous chemical reaction 
6. Problems 
IV.6. Combined heat and mass transport 
1. Drying . 
2. Problems 
INDEX 
276 
280 
283 
292 
292 
294 
296 
CHAPTER I 
Introduction to Physical Transport 
Phenomena 
During the designing of industrial process plant qualitative and quantitative 
considerations play a role. 
On the basis of qualitative (sometimes semi-quantitative) considerations a 
preselection of feasible concepts of processes suitable for carrying out the desired 
production in an economical way is made. The type of operation, e.g. distillation 
against extraction or the choice of a solvent, will also be fixed by this type of 
reasoning, in which experience and a sound economic feeling playan important 
role. 
As soon as one or two rough concepts of a production unit are selected, the 
different process steps will be analysed in more detail. This asks for a quantitative 
appro3.:ch with the aid of a mathematical model .of the unit operation. The 
experience that mass, energy and momentum cannot be lost provides the three 
conservation laws, on which the quantitative analysis of physical and chemical 
processes wholly relies and on which the process design of a plant is based. 
This kind of design, which aims at fixing the main dimensions of a reactor or an · 
apparatus for the exchange of mass, momentum and energy or heat, is the 
purpose of the disciplines known as 'chemical engineering' and 'chemical 
reaction engineering'. The basic ideas behind these disciplines are found under 
the headings 'transport phenomena' and 'chemical (reaction) engineering 
science', which rely. on deductive science and, hence, have the advantage of 
analytical thought but which, · because of that, lack the benefit of induction 
based on experience when aiming at a synthesis. 
Qualitative and quant-itative reasoning cannot be separated when setting up 
a plant, or to put it in another way: 
no apparatus, however good its process design might be, can compete with 
a well-designed apparatus of a better conception, 
which can be the device for a process designer, or 
no research, however brilliant in conception it might be, can result in a 
competitive production plant without having a quantitative basis, 
which can be a motto for a research fellow. 
2 
Examples of questions, in which feeling and reasoning have to match well 
before science is used to some profit, are to be found in the following areas : the 
potential possibilities of raw materials, intermediate and end products, the 
choice of materials and especially materials of construction, the influence of 
side reactions on the performance of subsequent process steps and the con-
siderations on quality and end-use properties of a product. This type of question, 
although of importance for the integral approach of a design engineer, will not 
be dealt with in this book, which will find its limitations just there. This book 
treats the practical consequences of the conservation laws for the chemical 
engineer in an analytical way, trying not to exaggerate scientific nicety where 
so many other important questions have to be raised and answered, but also 
pretending that a solid understanding of the heart of the matter at least solves 
a part of all questions satisfactorily. 
The laws of conservation of mass, energy and momentum are introduced in 
paragraph 1.1. They are extended to phenomena on .a molecular scale in para-
graphs 1.2 and 1.3. Paragraph 1.4 is concerned with .dimensions of physical 
quantities, especially SI units, whereas paragraph 1.5 discusses the technique 
of dimensional analysis. We will end this chapter (and most paragraphs of the 
following chapters) with some proposals for exercising and comments on the 
solution of some of the problems given. 
After this, three main chapters follow, each of which concentrates on one of 
the conserved physical quantities: hydrodynamics (mainly momentum transfer) 
in chapter II, energy transfer (mainly heat transfer) in chapter III and mass 
transfer in chapter IV. These chapters elaborate the ideas and concepts which 
are the subject of the following introduction. 
1.1. Conservation laws 
John looked at the still smoking ashes of what had once been 
the glue and gelatine factory. The fire had started with an 
explosion in the building where bones were defatted by 
extraction with hexane. John remembered that the extrac-
tion building had a volume of 6000 m3 and that the tempera-
ture in the building was always 3rfC higher than outside. 
He knew that per 24 h, 70 ton steam were lost as well as 9 ton 
hexane. He made a quick calculation and concluded that the 
steady-state hexane vapour concentration in the plant was 
well below the explosion limit of 1 ·2 per cent by volume and 
chac some accident muse haue happened which subsequently 
led to the explosion. 
Physical technology is based on three empirical laws: matter, energy and 
momentum cannot be lost. 
The law of conservation of matter is based among other things on the work 
of Lavoisier, who proved that during chemical reactions no matter, i.e. no 
mass (mass being the most important property of matter), is lost. The law of 
conservation of matter is not always valid: in nuclear technology matter is 
transformed into energy but for chemical or physical technology this exception 
3 
is of no importance. It is, of course, possible that matter is transferred from a 
desired form into an undesired one (e.g. the degradation of a polymer, which 
finally leads to only C02 , H 20, etc.). 
The law of conservation of energy is based among other things on the work 
of Joule, who proved that mechanical energy and heat energy are equivalent. 
His work finally led to the first law of thermodynamics, which, when formulated 
for a flowing system, is the law of energy conservation we are looking for. It 
is historically remarkable that it took more than two centuries before this 
law, formulated initially for a closed system, was translated into a form in which 
it could be applied to flowing systems. 
The law of conservation of momentum was finally formulated in its simplest 
form for a solid body by Newton : if the sum of the forces acting on a body is 
different from zero, this difference is (in size and direction) equal to the acceler-
ation of that body. Together with his second law, action equals reaction, this 
formed the basis for dynamics and hydrodynamics. This time it did not take 
much more than one century to transpose the concept, originally formulated 
for a rigid body, to the more general case of flow in fluids. 
These conservation laws play in daily life the same role as the experience that 
a pound cannot be spent twice and that the difference exists between the 
pound you owe somebody and the one somebody owes you. The economic 
rules and the conservation rules of our study are used in the same manner: 
balance sheets are set up which account for inflow and outflow and for the 
accumulation of the quantity under consideration. 
Let us denote by X a certain amount of money, mass, energy or momentum. 
Then the general law of cons~rvation, on which all phenomenological descrip-
tions of change in the physical world are based; reads as follows: 
accumulation of X in system 
unit time 
flow of X into system flow of X out of system production of X in system 
. . + .. 
umt time umt time unit t ime 
(1.1) 
The system may be a country, a concern, a factory, an apparatus, a part of an 
apparatus (e.g. a tray.), a . ..pipe or an infinitely small element of volume, etc. This 
sounds very general and easy, but daily practice proves that we have to develop 
the qualitative judgement for defining the system such that the analysis stays 
as easy as possible. To this end, to develop a feeling for the qualitative aspects 
of analytical science, we have to go through many a quantitative exercise. 
Introductions into a discipline, such as this one, may easily confuse the reader 
if these points are not made clear in the beginning; the subject of our study is 
the three laws of conservation, which will appear in many forms because we 
will study many different systems and not because the basic rules are many. 
If we are only interested in macroscopic properties like mean concentration 
in the chosen control volume or the rate of change of the mean temperature in 
4 
this volume we can choose a macroscopic control volume, e.g. a complete 
reactor, a complete catalyst particle, etc., for setting up a balance of the desired 
physical property. If, on the other hand, we are interestedin temperature or 
concentration distributions we have to start by setting up a microbalance, Le. a 
balance over an infinitesimally small volume element, and to integrate the 
differential equations obtained over the total (macroscopic) volume. Both 
balances will be treated in detai~ in this and in the following paragraph. 
Let us now try to formulate the law of conservation which we have just 
defined in words (equation I. l) in a more precise way. In order to do so, we 
need some symbols : 
V for the volume of the system (space volume, number of inhabitants, etc.), 
<l>v, in and cf>v.out for the ingoing and, outgoing volumetric flow rates (see 
Figure IJ), 
r for the velwnetric production of X per unit of time and 
X for the volumetric concentration of the physical quantity in study. 
Control volume 
r--1--------, 
I I 
I 
<l>v, out 
1 
V, r 
I I 
L-----------J 
Figure J.l Macrobalance 
The accumulation per units of volume and time can now be denoted by dX /dt ; 
hence, the accumulation in the system per unit of time is given by VdX/dt. 
A flow at a rate cf>v, containing a volumetric concentration X of the considered 
quantity, represents a flow rate <l>vX of this quantity. Hence, our formulation of 
the conservation law reads as follows: 
(1.2) 
We will now apply this law to the following quantities: money, mass, energy 
and momentum. 
T he money balance 
Let us start with the most common daily practice, the conservation law for 
pocket money (the pocket being defined as the system, although some operate 
is very unsystematically). The number of pence in the system is given by X 
( J • = 1 ), and equation (1.2) can be read as follows: the accumulation of pence 
*In order to be completely exact we have to write this equation as : 
- u 
dX - 1 -1 
vdt = cl>u,inxin- rPu,oulxOUI + Vr 
""'here -. ndic.ltes a volume average and -!a flow average. 
5 
in my pocket pt;r week (which may prove to be negative!) equals the difference 
between the number of pence I have taken in during this week and the number 
I have spent in this time, increased by the number of pence I produced in the 
meantime. The last contribution sounds somewhat cryptic or even illegal, but 
an honest production of pence would be to change other coins into pence. 
Similar laws can be expressed for the other coins, as well as for the overall 
contents of the pocket (moneywise). From all these statements it follows that 
the sum of all the production terms r must equal zero (expressed as an intrinsic 
value and not as numbers) or, to put it in another way, changing in its own 
currency never results in a positive gain. To find out how simple these statements 
might be, try and see what happens to your thinking when X no longer stands 
for money, but for mass, energy or momentum! 
The mass balance 
The mass balance still looks familiar and comes close to the money balance. 
Here X = cA, the volumetric concentration of component A in a mixture; the 
units in which c A is measured are kg/m3• * 
Hence, in a rayon factory, for instance, where NaOH (in the form of viscose) 
and HCl (in the form of the spinning bath) are used, the conservation law for 
NaOH {A) reads as follows: the accumulation of NaOH on the site in a month 
(or any other chosen time unit) is equal to the delivery ofNaOH to the factory 
in that time, subtracting the amount of NaOH distributed from the site in that 
unit of time as NaOH and adding the amount ofNaOH produced in the factory 
during that time (a production which is negative where NaOH is used as a 
reactant). 
A similar mass balance can be set up for the other reactant, HO (B), as well 
as for the products NaCl (C) and H2 0 (D). Again, the sum of all individual 
production rates must be zero: r A + r 8 + rc + r0 = 0 (Lavoisier:t. More olten 
than not, the accountants of a factory are more aware of the implications of the 
conservation laws and the dynamic consequences of varying inflows and out-
flows of a factory than the engineers in charge of production. 
' f#>v, in l#>v, out 
v c 
Figore 1.2 Mass balance of a stirred 
vessel 
The well·stirred continuous flow tank reactor of Figure I.2 gives an example 
of the applications of the mass balance. By 'well·stirred' we mean that the 
* From the onset, we will accept kg, m, s, oc as the basic units for comparing physical phenomena 
(paragraph 1.4). 
6 
concentration c of a certain compound (e.g. salt) is the same at all. places in the 
vessel so that the salt concentration in the effluent stream equals the con-
centration in the vessel. Let us assume that the salt concentration in the vessel 
at time t = 0 is c0 and that, from t = 0 on, a continuous stream of pure water 
(salt concentration c = 0) is passed through the vessel. The question is then 
how does the salt concentration in the vessel change with time? 
We can now set up two mass balances, one for the water and one for the salt. 
The first balance says that, if the liquid volume in the reactor is constant, the 
flow rate of liquid out of the reactor must equal the flow rate into the .reactor, 
<l>v,out = <l>v.in . The second balance says that the decrease in the amount of salt 
in the vessel (Vc) per unit of time must equal the mass flow rate of salt from the 
vessel (</>vc). Thus : 
d(Vc) = 0 _ ¢ c dt v 
or: 
de c 
- = 
dt r 
where r = Vf4>v is the mean residence time of the fluid in the vessel. Integration 
between t = 0, c0 and t , c yields: 
.:_ = exp (!) 
Co T 
f / T -
Figure 1.3 Change of salt concentration in 
stirred vessel 
This function is shown in Figure 1.3, which shows that at the time t = 't still 
3"" per cent of the initial salt is present in the reactor. For t = ir and ~r these 
values are 65 and 22 per cent respectively. Apparently, the liquid in the reactor 
has a large distribution of residence time. 
7 
The energy balance 
Before we can attempt to interpret equation (1.2) in terms of an energy 
balance, we must define the meaning of X as ' the volumetric concentration of 
energy' Er. This quantity will contain internal energy and potential energy. 
Internal energy again comprises perceptible and latent heat and pressure 
energy. Pressure energy per unit of volume is nothing but an elaborate expres-
sion for merely the pressure inside the volume. Perceptible and latent heat per 
unit of volume (U) may be expressed as : 
(1.3) 
in which p is the specific gravity, cP the heat capacity per unit of mass, T the 
temperature, T, a reference temperature for calculating U, !J.H the latent heat 
per first·order phase transition (melting, evaporation) per unit mass and La 
symbol to indicate that the contributions of all phase transitions between T,. 
and T have to be taken into account. If the process studied does not show phase 
transition, the reference temperature T,. can be chosen high enough to keep 
latent heats out of the analysis. If, furthermore, the heat capacity per unit of 
volume (pc p) is independent of temperature, U = pc P (T - T,). 
The kinetic energy per unit of volume is fpv 2 , when v is the velocity of this 
volume. If different elements o f the volume flow have different velocities, then 
the flow average of the squared velocity has to be used for calculating the flow 
of kinetic energy, <Pv · !p(v2 ) , with 
1 2 1 f pv 2 • v dA 
2 p < v ) = 2 f v dA (1.4) 
where A is the cross-sectional area of the flow channel. 
The potential energy per unit of volume is J:~ pg dh, in which h is the covered 
distance, h, a point of reference along the path of movement and g a field force 
density in the direction of movement (e.g. the gravitational acceleration of earth). 
If both p and g are independent of h, the volumetric potential energy is expressed 
by pg(h - hr) = pgh (hr often being put arbitrarily equal to zero as the definition 
of theorig~n). -
Hence. X as the volumetric concentration of energy can be evaluated from 
if the conditions mentioned apply. 
With regard to the production term in equation (!.2), we can distinguish 
between mechanical and thermal energy and denote the amount of added 
mechanical energy by ¢ A (e.g. supplied by a pump or used by driving a turbine) 
and the amount of heat flow from the control volume by <PH. 
8 
Thus, the general balance of equation (1.2) can be written as an energy balance 
as follows: 
dE, 
V d t = cPu. inEr,in - <Po,outEr,out + cPA - f/JH (1.6) 
where E, is defined by equation (I.5). 
If heat is transferred between the control volume and its surroundings it is 
accepted practice to put the heat transfer rate per unit of boundary surface area 
proportional to the temperature difference between the surroundings and the 
system. The proportionality factor thus defined has been named the heat 
transfer coefficient and is given the symbol a; the practical significance of this 
coefficient will be treated in more detail in chapter III. In this chapter we shall 
indicate how the heat transfer coefficient can be calculated from first principles 
in a number of heat transfer situations. In complicated cases, however, this 
coefficient is nothing more than a numerical factor which has to be obtained 
from formerly reported, consistent experience. It is one .of the goals of the 
discipline to keep scientificalJy well-organized records of this experience. 
The momentum balance 
The amount of momentum per unit mass is mv/m = v, whereas the amount 
of momentum per unit of volume is mv/V = pv. Thus the momentum flux per 
unit volume is pvv. Because momentum has direction as well as size we must 
use the three components of momentum in the x-, y- and z-directions, namely 
pvx , pvY and pvz . For each direction the law of conservation of momentum is 
valid and thus we obtain three separate equations. The momentum-producing 
terms must be interpreted as forces {Newton). The forces which can occur in a 
flow system are pressure forces, friction forces (caused by shear stresses) and 
potential forces (weight forces due to gravitational acceleration). Thus the 
momentum balance in the x-direction is found by replacing X = pv in 
equation (1.2): 
(1.7) 
where I Fx indicates the sum of all forces acting in the x-direction. The balances 
for the y- and z-directions can be written analogously. The dimension of all 
terms in the above balance is the Newton (N). 
Application of balances to a pipe corner 
We will now, at the end of this paragraph, illustrate the application of the 
conservation laws with a practical example. Let us consider a horizontal pipe 
comer of cross-section A which is well insulated against heat loss. The mass 
balance then reads for an incompressible liquid flowing through the pipe 
(equation 1.2 with X = p, the total concentration, and r = 0) under steady-
state conditions : 
dp V- = 0 = ,~,. . p . - ,~,. p d t <f'v,1n 1n "t'v,out out 
and, with <l>v. l = v1A 1 and <l>u, 2 = v2 A 2 and constant specific gravity: 
vt A 1 = v2A2 
9 
indicating that at constant cross-section of the pipe the entrance and exit 
velocities of the liquid must be equal. 
The energy balance reads in the stationary state : 
X 
L , 
Figure 1.4 Pipe corner 
dE, 
V dt = Q = <l>v.inEr,in - ¢ v,out£r ,out + </>A - </> H (1.8) 
and since ¢A = <I>H = 0 and </>r;, in = ¢v.out: 
E r,in = E r,out 
or, with the definition of E, (equation 1.5), if A 1 = A 2 and thus v1 = v2 and if 
there is no phase transition (h 1 = h2 because the pipe is horizontal) : 
(1.9) 
Equation I.9 indicates that the pressure drop can only be predicted if the 
amount of frictional heat liberated per unit of volume is known. This is a prin-
cipal problem of p~;actical hydrodynamics that can only be solved theoretically 
in a few cases. In all other cases we have to rely on empirical correlations for 
estimating the pressure drop in flow systems (see paragraph !1.2.2 and II.2.3). 
The momentum balance over the pipe corner in the x-direction yields for the 
stationary state: 
dpvx " V dt = 0 = pv1A 1v 1 - pv1A 20 + '- Fx (1.10) 
The second term on the right-hand side of this equation expresses the fact that 
no momentum in the x-direction is taken away with the fluid. The forces acting 
in the x-direction on the liquid are the pressure force p 1 A 1 and the reaction 
10 
force the wall exerts on the liquid, Fx,w- 1 , thus: 
L Fx = P1A1 + Fx,w- f (1.11) 
and combining equations (1.10) and (I.ll) we find for the momentum balance 
in the x-direction : 
dpvx 2 
vdt = 0 = pv1A1 +PIAl+ Fx,w- J 
Thus the wall has to produce a force in the negative x-direction and withstand 
a force Fx.J- w ( = - Fx,w- 1) in the positive x-direction (action equals reaction-
first law of Newton). This force equals the sum of the pressure force and the 
force which is caused if all the momentum of the flowing liquid is taken up by 
the wall*: 
Fx,w- f = - Fx,J-w = - pvtAt - P1A1 (I.12a) 
Analogously to the above, we can develop a momentum balance for the y-
direction and find that the force the liquid exerts on the wall in the y-direction 
is given by : 
(I.l2b) 
·- ·- ·-·- ·-
Fy,f·w=-pv~A2-p2~ i / 
L_ y J~:LA~ -I 2 
Figure 1.5 Forces acting· op pipe corner 
With this information the resulting force of the fluid on the wall can be con-
structed as shown in Figure 1.5. The pipe corner must be fixed in such a way 
that these forces can be taken up by the holding structure. 
In the above example we have assumed that the velocity distributions of 
the liquid entering and leaving the pipe are uniform and that there are no 
velocity differences over the cross-section. In that case, the momentum of the 
inf.owing liquid is indeed pvi A 1 . If the above assumption of uniform flow 
• P:a~:ca! technical flow velocities in pipes are 1 m/s for liquids and 15 m/s for gases. The reader 
rna~ check that the momentum force on a pipe comer is mostly negligible as compared to pressure 
forcu 
11 
velocities is not allowed, we have to write p(vT)A 1 , indicating that vr must be 
averaged over the cross-section. Since pA 1(vr) = <Pm<v~)/(v 1 ) we say that the 
momentum per unit mass is given by (vi)/(v1 ). For the same reason we write 
(v3 )/2(v) for the kinetic energy per unit mass. Such expressions for physical 
.quantities averaged over the flow will be found regularly in our discipline. 
1.2. Rate of molecular transport processes 
In the foregoing paragraph we have defined the various terms in the balances 
based on the three laws of conservation. In so doing we-have concentrated on 
transport with the flowing fluid, the so-called convective transport caused by 
movement of the molecules, and neglected the statistical transport. We have 
mentioned briefly the conduction of heat through the boundaries of a system 
as a means of heat transport, viscous friction as a means of momentum transport 
and viscous dissipation as a possible contribution to heat transport. 
The opportunity of introducing diffusion of matter through the system 
boundaries as a means of transporting a component has been missed until now. 
Together with heat transport by conduction and momentum transport due to 
viscosity, diffusion of matter is the third molecular process which is of interest 
to us. 
It might be imagined that a difference in the concentration of a component 
on both sides of a boundary enhances a net mass flux through that boundary 
because of the Brownian (heat) movement of the molecules of this component: 
the bruto flux from the side of the higher concentration will override the bruto 
flux from the other side, simply because more of the specific molecules are 
present on that side while each has an equal chance to pass the boundary at a 
given uniform temperature.Brownian motion also explains the conduction of 
heat: a temperature gradient over the system boundary produces a net flux of 
energy, because the molecules on the hotter side move faster and carry a 
higher energy. 
Molecular movement again is a reason for stress in a flow field with a velocity 
gradient. Due to Br:ownian movement molecules are exchanged in all directions 
and hence also in the direction of the gradient. Those coming from the faster 
moving side have a higher velocity in the direction of flow and, hence, carry a 
bigger momentum " th~ those coming from the slower moving side. This 
results in a net flux of momentum from the faster flowing area to the slower 
moving parts or, in other words, slow neighbours try to decelerate fast nt:igh-
bours and vice versa. So, a net transversal transport of momentum-in-the-
direction-of-flow can be interpreted as stress. It is good to notice again the 
vectorial character of momentum flow: both the direction of transport 
(transversal) and the direction of the quantity transferred (longitudinal) are 
important. This means that in an arbitrary plane, stresses in all three directions 
can be found when velocity gradients in all these directions are present. 
Transport of heat, mass or momentum by molecular processes is sometimes 
called statistical transport because it is caused by the random Brownian 
12 
movement of the molecules.* Statistical transport is usually much smaller than 
convective transport, unless very low flow velocities occur. For this reason 
statistical transport in the direction of flow can mostly be neglected. 
Our elaboration of the conservation concept has to account for mass, energy 
and momentum fluxes due to processes on a molecular scale. Generally, these 
fluxes are found to be proportional to the gradient dX/dn, where n denotes the 
coordinate in the direction perpendicular to the considered plane. The pro-
portionality constant between the flux and the driving force dX jdn is known as : 
the mass diffusivity, [},in the case of mass transfer (Fick's law), 
the thermal diffusivity, a = 2/pc9 , with 2 being the heat conductivity, in the 
case of heat transfer (Fourier's law)t and 
the kinematic viscosity, v = 11/ p, with 17 being the dynamic viscosity (Newton's 
law). 
The dimension of each of these constants is m2/s. 
When ¢" denotes the flux (the flow rate per m 2 ) of the quantity X due to 
molecular movement in a gradient of X, we thus writet: 
dX 
</>" = - (0 or a or v)-
dn 
(I.l3) 
The minus sign denotes that the net flux is positive in the direction of diminishing 
concentrations of X. Thus the mass, beat and momentum fluxes in the positive 
direction of n are given by : 
mass flux: cf>':n = _[}de dn 
(kg/m 2 s) (Fick) 
heat flux : 4>" - d(pcPT) CW/m2) (Fourier) H- -a dn 
momentum flux: F'' = r = 
d(pv) 
(N/m2 ) (Newton) -v--
dn 
Figure 1.6 illustrates the above relations for mass and heat transport caused by 
a concentration or temperature gradient. 
* Besides statistical molecular transport there can also occur statistical transport by eddies (para-
graphs 11.2, 111.3.2 and IV.3.3). We speak of laminar flow if there is only statistical molecular 
transport. 
t Here, heat transfer has been used deliberately as a description which is not completely s1rnilar 
to energy transfer. Heat transfer accounts only for the transport of perceptible and latent heat, 
not fo: the transport of other forms of energy. 
! The :-elation between momentum flux and velocity gradient discussed here is restricted to New-
tamar. llquids. For non-Newtonian liquids more complicated relations exist which will be 
discussed i:n the next chapter. 
... 
0 
Q 
.~... • I = _ ( "' ) d ( c or pc p T)l 
't"m orH " " ora 
, "o dn no 
"o 
___ , 
Figure 1.6 Diffusion and heat con-
duction 
13 
In the case of momentum transport there is the extra complication that we 
have not only to specify the plane through which the transport occurs (e.g. 
x = x0 ) but also the direction of the momentum considered (e.g. y). The cor-
responding momentum flux is then written as F;1 , indicating that there is 
transport of y-momentum in the x-direction at the plane x = x 0 caused by a 
velocity gradient dvy/dx. In other words, viscous shear is transversal transport 
oflongitudinal momentum, as indicated in Figure I. 7. This momentum transport 
represents itself as shear which, per unit of surface, is called the shear force -r. 
Thus the momentum flux F~Y corresponds with a shear force txy in they-direction 
j 
I 
r; I = - "d IP vy l I 
Y • x d x .t 
0 0 
i 
Figure 1.7 Shear in terms of 
momentum flux 
r xy I =- " d ( p Vy ) I 
xo dx X0 
- - x 
Figure 1.8 Shear in terms of shear forces 
14 
at x = constant. This is illustrated in Figure I.8. The exact definition of the shear 
force needs, therefore, a statement of the plane and direction considered, e.g. : 
d(pv») 
-rxy = - v dx 
The shear force is defined in such a way that the force the left-band fluid exerts 
on the right-hand fluid is positive if the gradient {from left to right) is negative. 
We have seen in the foregoing that the statistical molecul~r transport through 
a plane can be described quantitatively with equation (1.13) if the proportionality 
constant([), a or v) and the gradient at that plane is known. In practice we are 
interested in transport through a plane fonning the boundary of a system 
(e.g. a wall, an interface, etc.), but, alas, often the concentration gradient at the 
boundary is not known beforehand and a practical approach has to be followed 
when qualifying the flux. In these cases, empirical phenomenological transfer 
coefficients are used, which are defined as follows : 
4>~ = _[) de = k(coo - cw) (1.14a) 
dn w 
</>" - d(pcPT ) =~Teo- T..,) (1.14b) n- - a dn w 
d{pv) 
w = (;pv®)(v®- Vw)• {l.14c) r = -v--w dn 
in which w stands for wall, oo for system (indicating a characteristic position in 
the system, such as, for instance, the axis), k for mass transfer coefficient, a. for 
heat transfer coefficient and fpvco f2 for momentum transfer coefficient (which 
looks awkward for historical reasons) ; f itself is better known as the Fanning 
friction factor. 
These empirical relationships, to which much background can be added, as 
will be seen in the following chapters, are of practical use for establishing 
transport rates through boundaries only if we can resort to a physically com-
pletely similar situation on which quantitative data are available. This has 
already been brought out when we discussed equation (1.2) in terms of heat, 
introducing the concept of the heat transfer coefficient at an earlier stage. 
Physical similarity, however, does not mean being identical, as wiJl become 
clear when the concepts are worked out in the subsequent chapters. 
Since order of magnitude estimations are of great importance in our discipline 
it is useful to realize the order of magnitude of ITJ, a and v (or 11 = vp) for gases, 
liquids and solids. According to the kinetic theory for gases: 
• The velocity at a solid wall is zero, which is one of the hypotheses on which hydrodynamics rests; 
~ence vw nearly always equals zero. 
15 
where ii is the mean velocity of the molecules ( ~ -)2RfiM) and I the mean free 
path of the molecules (--Mf p = RTj p). At normal pressure and temperature 
for gases. D. a and v are of the order 0·5-2 x lo-s m 2f s. The kinetic theory 
further predicts that 11 and .A (the thermal conductivity = a . pc ) for ideal gases 
• p 
are mdependent of pressure and that they are approximately dependent on the 
square root of absolute temperature. 
For the diffusion coefficients of liquids at room temperature we find values 
of the order 0) = 10- 8-10- 9 m2js ; for other temperatures diffusivities can be 
estimated by means ofthe Einstein- Nemst- Eyring relation D17/T = constant, 
where 11 is the viscosity of the solvent. The viscosity 11 of liquids varies widely 
(e.g. at 20°C for water '1 = 10- 3 Ns/m2, for glycerol 11 = 1·5 Ns/m2 ) and is 
strongly dependent on temperature ; in the first approximation t7 - exp {1/ T). 
The termal diffusivity of liquids is about a ~ 10- 7 m2 js with the exception of 
molten salts and liquid metals, which show considerabTy higher thermal 
diffusivities. 
The diffusion coefficients of solid materials at room temperature are of the 
order I!)~ 10- 11- 10- 13 m2js. The thermal diffusivity of non-metallic solids is 
roughly a ~ 10- 7 m 2 js, while metals belong (because of the effective heat 
transport by the free electrons) to the best heat conductors (a ~ 5-100 x 10-6 
m 2f s). Also, solid materials have a very high viscosity. The ratio of the viscosity 
and the elasticity modulus (having the dimension of time) gives an impression 
of the time necessary before an initially elastic deformation can be observed as 
flow. On the other hand, many liquids (e.g. molten polymers) show elastic 
behaviour. For these liquids the time necessary before flow is observed is 
generally much greater than 0·1 sand it is therefore not possible to describe the 
flow of these materials by means of a simple shear stress- velocity gradient 
relationship, because the previous history of each particle has to be included in 
the description. These problems belong to the area of rheology and we will not 
go into more detail here. 
1.3. Microbalances 
In some cases th~ concentration, temperature and flow velocity distributions 
in a system can be cal~ulated by starting with the principle of conservation and 
applying this principle to every small volume element of the system. This leads 
to the so-called micrQ.balances, as compared to the macrobalances of the 
foregoing section (I.l .) which give no insight into the distribution of the quantity 
over the system. These microbalances can be formulated in a general way, as 
will be shown below. However, the partial differential equations which will 
emerge from this microconcept can be solved only for relatively simple situ-
ations. Therefore, dependence on equations (I.14) is daily practice, but the idea 
behind a practical correlation for the transfer coefficients is often based on a 
fundamental analysis of a somewhat simplified and idealized situation which 
can still be treated analytically. Let us, therefore, see how we arrive at a micro-
balance. 
16 
Let us consider a Cartesian element (Figure 1.9) of volume dx dy dz, where 
x, y and z are the three coordinates in space. The accumulation of the quantity 
X considered per time interval dt in the system is now given by : 
dX 
dt dx dy d2 
The net inflow of X in the x-direction is : 
z 
Figure 1.9 Volume element in rect-
angular coordinates 
(115) 
(1.16) 
where ¢:is the flux of the quantity X in the x-direction. Similar expressions hold 
for the net flow into the system in the y- and z-directions. We can therefore 
write equation (1.1) for an infinitesimal volume element as : 
dX d¢" dq/' de/>" 
-dxdydz = --" dxdydz- _Y dx dydz- _z dxdydz + r dxdydz 
dt dx dy dz " 
or 
dX d¢; d¢; dtp; 
- -----+ r 
dx dy dz 
(1.17) - = 
dt 
Equation (117) is the basic microbalance which can be further evaluated When 
doing so, we have to realize that the flux of X on this microscale consists of a 
convective transport term and a statistical transport term. These fluxes in the 
n-direction are therefore given by* : 
dX 4>~ = -([}or a or v) dn + v,X (1.18) 
Thus, if we take II) as a representative of I!J, a or v, we obtain from equation (1.17) 
with equation (1.18): 
dX d(vxX) d(vyX) 
- = - ___;_;.__... 
dt dx dy 
(1.19) 
• Inter?reted as the momentum balance, equation 1.18 is only valid for Newtonian liquids. We will 
d;scuss this problem in more detail immediately. 
17 
If we interpret equation (1.19) as a mass balance for a component A in a 
mixture of substances A, B, C, etc. (thus X = CA. C8 , etc.), we know that : 
C A + Ca + Cc + ... = p (the specific gravity of that mixture) 
and 
rA+ r8 +rc + ... =0 (Lavoisier) 
Summing up the mass balances for all components we find: 
dp d(t.'xP) 
-=---
dt dx (1.20) 
which is known as the equation of continuity. For a stationary situation, dpjdt 
equals zero. When pis constant over the flow field, we find furthermore: 
dvx dv.v dv,. 
0 -+-+-= 
dx dy dz (1.21) 
which is a very familiar expression for the conservation of total mass. 
If equation (1.19) is to be used as a microbalance for the conservation of heat 
(or of energy if the transport of heat exceeds the transport of other forms of 
energy, as given in equation 1.5), X stands for pcPT This microbalance will be 
further evaluated in paragraph III.l.4, whereas the micromass balance for one 
component will be discussed in more detail in paragraph IV.1.3. 
We will now concentrate on the further development of equation (1.17) as a 
micromomentum balance. In order to obtain a relation which is valid for 
liquids of different rheological behaviour (thus Newtonian and non-Newtonian 
fluids) we will1 instead of equation (1.18) use the more general expression· for 
the momentum flux in the n-direction : 
(1.22) 
which is valid for all rheologies. Here X stands for pv. The momentum pro-
duction term in the n-direction is the sum of the pressure and gravity forces 
acting in that direction-on the control volume: 
dp 
rn = - - + pg dn n (1.23) 
With the above expressions we obtain from equation (I.17) for the micro-
momentum balance in the x-direction : 
dt 
(1.24) 
18 
With the aid of the continuity equation (1.20) this expression can be simplified 
to : 
dvx dvx d v.x dvx dT.x.x dTy.x dT:x dp ) P dt = - pv - - pv - - pv - - - - - - - - - + pg {1.25 
x dx , dy z dz dx dy dz dx x 
Analogous expressions can be written for the micromomentum balance in the 
y- and z-directions as follows : 
dvv dv>' dv, dvy d-r"Y d-r,,. dtz,. dp 1 , 6) p dt = - PVx dx - pv, dy - PVz dz - dx - dy - dz - dy + pgY ( ·-
and 
dvz dvz dvz dvz d-rxz d-ryz dtzr dp (l.27) p-- = - pv - - pv - - pv - - - - -- - - - - + pg d t x dx Y dy z dz dx dy dz dz z 
If, instead of a Cartesian volume element in rectangular coordinates, we had 
considered a volume element in cylindrical coordinates-, as indicated in Figure 
1.10, we would have obtained the following micromomentum balances in the 
r-direction : 
' z 
Figure 1.10 Volume element m 
cylindrical coordinates 
p dv, = _ p {v dv, + v6 dv, _ vi + v dv,} _ ~ d(rt,) _ ~ d-r,6 + -r80 
d t ' dr r d8 r z dz r dr r d8 r 
d-r,% dp 
---- +pg 
dz dr r 
in the 8-direction : 
p dv8 = _ p {v dv8 + v8 dv8 + v,v6 + v dv0} _ ~ d(r2 -r,9) _ ~ dt89 
d t 'dr r d 8 r z dz r2 dr r d9 
(1.28) 
d-r8~ 1 dp 
- dz. - ~ d8 + pge (1.29) 
19 
and in the z-direction: 
dv: { dv: v8 dv: dv:} 1 d(r't"rz) 1 dr8 % d -rn dp 
p dt = - p Vr dr +-;: d9 + Vz dz -; dr +; d9 + dz - dz + pg. 
(1.30) 
These micromomentum balances can naturally also be formulated in spherical 
coordinates. The relations obtained can be found in many handbooks. 
The above momentum balances in terms of shear stresses are valid for all 
fluids, because the shear stresses are independent of the rheological behaviour 
of the liquid. In the special case of Newtonian liquids -r is proportional to the 
velocity gradient (equation 1.13), e.g.: 
d(pvy) 
rxy = - v d.x 
Introducing this expression into the above microbalances we obtain, for rect-
angular coordinates, in the x-direction : 
dvx _ { dvx dvx dvx} {d2(pvx) d2(pv,) d2(pvx) } 
p dt - - P Vx dx - l'y dy - v. dz + v dx2 + dy2 + dz2 
dp 
-- + pg (1.31) dx x 
and in the z-direction: 
dvz _ _ { dv. dvz dvz} {d2(pvz) d2(pvz) d2(pvJ} 
p dt - p v xdx .+ v >' d y + v z dz + v dx2 + d y2 + dz2 
- ~~ + pgz (1.33) 
In spherical coordinates Newton's law (equation 1.13) reads in the z- or r-
directions : 
d(pv6) d(pv ) 
't"gz = - v dz and rzr = - v dr %. ' respectively 
and in the 9-direction: 
(1.34) 
20 
Introducing this into the general microbalance in terms of shear stresses we 
obtain in the r-direction : 
dv,. _ { dv,. v8 dv,. v~ dv,.} { d ( 1 d(prv,.)) p-- - p v- +---- + v- + v - - ___:.____:_ dt ,. dr r dO r z dz dr r dr 
+ _!_ d2(pv,) _ ~ d(pv8) d2(p. v,.)} _ dp 
r2 d82 ~ d8 + dz2 dr + pg,. 
in the 0-direction : 
dv6 p-= d t 
and in the z-direction: 
dvz _ { dvz v8 dvz dvz} { 1 d ( d(pvz)) 1 d2 (pvz) 
p dt - - p v,. dr + 7 dO + Vz dz + v ; dr r dr + r 2 d02 
d 2(pvz)} dp 
+ dz2 - dz + pgz 
(1.35) 
(1.36) 
(1.37) 
The micromomentum balances given above form the basis for the calculation 
of velocity distributions and flow rate- pressure drop relations during laminar 
flow; we will apply them many times in the next chapter. 
1.4. SI units 
Different systems of units are applied at the moment in different countries in 
industry, research and development. The most important systems are : 
the c.g.s. system, based on the centimeter (em), gramme mass (g), second (s) 
and degree Celcius (°C). For the amount of heat the calory (cal) is u sed 
(derived unit);* 
the SI system, based on the metre (m), kilo gramme (kg), second (s) and degree 
K elvin (°K). The amount of heat (derived unit) is expressed as Joule (J), 
which is identical to the unit for mechanical or electrical energy (1 J = 
1 kgm2js2 = 1 Nm = 1 Ws, where N =Newton, W = Watt) ; 
the metric system, based on the metre (m), kilo gramme force (kgf), second (s) 
and degree Celsius (°C). The amount of heat is expressed in kilocalories (kcal) ; 
the foot-pound-second system, based on the foot (ft), pound mass (lb), 
second (s) and degree Fahrenheit (°F). The derived unit for the amount of 
heat is the British Thermal Unit (BTU) ; 
• The units for electric current and for luminous intensity will not be discussed here because they 
play no role in this book. 
21 
the British engineering system, based on the foot (ft), pound force (lbf), 
second (s) and degree Fahrenheit (°F ). 
For scientific research the c.g.s. system is used all over the world. Engineers in 
Anglo-Saxon countries generally use the British engineering system ; in other 
countries the metric system is generally used. 
In the last few years the SI system (Systeme International d 'Unites) has been 
more and more applied. This system has been adopted by the International 
Organization for Standardization and is recommended by a large number of 
national standard organizations. For that reason we will use SI units throughout 
this book. In Table I.l a survey of the basic and derived SI units important for 
us is given. The reader is advised to use one consistent system of units, preferably 
the SI units. 
Table 1.1 Basic and derived SI units 
Length 
Mass 
Time 
Temperature 
Force 
Quantity 
Work, energy, quantity of heat 
Power 
Pressure 
Dynamic viscosity 
Kinematic viscosity 
Surface energy or tension 
Enthalpy 
Heat capacity 
Heat transfer coefficient 
Mass transfer coefficient 
Thermal conductivity 
SI unit 
metre 
kilogram me 
second 
degree Kelvin4 
Newton 
Joule 
Watt 
m 
kg 
s 
OK 
Unit symbol 
N = kgm/s2 
J =Nm 
W = Jjs 
N jm2 
Nsfm2 
m2/s 
Jj m2 or N/m 
Jjkg 
Jjkg °C 
Wjm 2 °C 
m/s 
W/m °C 
4 Temperature difference is commonly expressed in degrees Celsius instead of degrees Kelvin. 
from one system to the other. In order to facilitate this, Table 1.2 gives a collection 
of conversion factor$. Table !.3 provides some information giving orientation 
about the approximate values of some common properties of gases, liquids and 
solids, which can be useful for rough calculations. 
1.5. Dimensional analysis 
In chemical engineering, relations between parameters are often expressed by 
means of dimensionless combinations of physical variables. The advantages of 
this technique are the easy control of dimensional homogeneity of the relation, 
the constants of which are then independent of the system of physical units 
applied, and the fact that the number of variables is reduced, which simplifies 
22 
Table 1.2 Conversion factors 
Multiply by 
Magnitude Expressed in Divide by In SJ units 
Length inch (in) 0.()254 m 
foot (ft) 0·305 
yard (yd) ().914 
rrule 1609 
Angstrom (A) 10- 10 
Area in2 6·45 X 10- 4 m2 
ft 2 0·0929 
yd2 0·836 
acre 4047 
mile2 2·59 X _106 , 
Volume in3 }·64 X 10- S m3 
ft 3 0·0283 
yd3 0·765 
UK gallon 4·55 X 10- 3' 
US gallon 3·785 x w-3 
Time minute (min) · 60 s 
hour(h) 3600 
day 8·64 X 104 
year 3·16 X 107 
Mass gram 648 X w-s kg 
ounce (oz) 2·84 x w-2 
pound (lb) 0·454 
~undredweight (cwt) 50·8 
ton 1016 
Force poundal (pdl) 0·138 N 
pound force (Jbf) 4-45 
dyn w- s 
kg force (kgf) 9·81 
Volumetric How ft 3/min 4·72 x w-4 m3/s 
U K galjmin 7-58 x w-s 
US gal/min 6·31 x w-s 
Mass flow lb/min 2·10 x w- 6 kgls 
tonjh 0·282 
Density lb/ in3 2·77 X 10
4 kglm3 
lbjft 3 16·0 
Pressure lbfj in2 6-89 X 10
3 N/m2 
lbf/ ft 2 47·9 
dynicrn2 0·1 
kgf/cm2 (=at) 9·81 X 104 
atm (standard) 1·013 X 105 
bar 105 
in water 2-49 X 102 
ft water 2·99 X 103 
in Hg 3·39 X 103 
mm Hg (torr) 1·33 X 102 
23 
Table 1.2 (com.) Conversion factors 
Multiply by 
Magnitude Expressed in Divide by In SI units 
Dynamic viscosity lb/ft h 4·13 x w-4 Ns/m2 
lb/ft s 1·49 
Poise (P = gjcm s) 0·1 
Centipoise (cP) w-3 
Kinematic viscosity ft 2/h 2·58 X JO-S m2/s 
Stokes (S = cm2 j s) w-4 
Centistokes (cS) w-6 
Surface tension dynjcm ( = erg/em 2) w-3 N j m 
Temperature difference degree F (or R) 5/9 ocrK) 
Energy (work, heat) ft lbl 0-0421 1 = Ws = Nm 
ft lbf 1·36 
BTU 1·06 X 103 
CHU 1899 
hph 2·68 X 106 
erg w-7 
kgfm 9·81 
kcal 4·19 X 103 
kWh 3·60 X 106 
Power (energy flow) BTU/h 0·293 'fl\T 
CHUjh 0·528 
ft lbf/s 1·36 
hp (British) 746 
hp (metric) 736 
erg/s w-7 
kcal/h 1·163 
calfs 4·19 
Heat flux BTU/ft2 h 3·15 W/m
2 
caljcm2 s 4·19 X 104 
kcal/m2 h 1·163 
Specific heat BTU/Ib°F 4·19 X 10
3 Jjkg°C 
kcal/kgcC 4·19 X 103 
" 
.-
Latent heat BTU/lb 2·33 X 10
3 J/kg 
kcaljkg 4·19 X 103 
Heat conductivity BTU/ft h OF 1·73 W/rn oc 
caljcrn soc 4·19 X 102 
kcaljm h oc 1·163 
Heat transfer coefficient BTU/ft 2 h °F 5·68 
caljcm2 s oc 4·19 X 104 
kcaljm2 hoc 1·163 
24 
Table 1.3 Physical properties of some materials 
Air Water 
20°C 20°C 
p [kg;m3] 1·20 998 
11 [Ns/m2] 17 X J0- 6 w-3 
v = ('1'//P) [m2/s] 14·2 x w- 6 w-6 
cp [J/kg oq 1·03 X 103 4-19 X 103 
A [W/m 0 C) 0·025 0·6 
u[N/m] 1 x w-2 
ll)Hz [m2/s] -2 X 10-S 5 x w- 9 
a = A./p cP [m 2/s] 20 X J0- 6 0·143 X 10- 6 
Pr = vfa = C,!1/A. 0·71 7·0 
Sc = vfO 0·71 200 
~Hv [J/kg] 2·45 X 106 
~Hm [Jjkg) 3·35 X JOS 
Gas Law Constant R = 8310 1/ kmol °K 
Avogadro's number N = 6·023 x 1026 kmol- 1 
Gravitational acceleration g = 9·81 m/ s2 
Stefan- Boltzmann constant <J = 5·61 x 10- 8 Wfm 2 °K4 
Volume of ideal gas at STP = 22·41 m3/ kmol 
Bean oil 
lOOOC 
870 
1 x w-3 
s x w- 6 
2·12 X 103 
0·156 
2·9 X 10- 2 
24 X 10- 9 
0·084 x w-6 
95 
300 
-
Standard Pressure and Temperature (STP) = 1·013 Nfm 2 and 273·1SO K 
Stainless steel 
200C 
7750 
0·45 X 103 
26 
7-45 x w- 6 
the planning of experiments. Furthermore, dimensionless numbers form the 
basis for experimental work with models and model fluids. 
We will illustrate the use of dimensionless numbers in a few examples. 
Figure 1.11 shows the end of a pipe through which gas flows at a rate v. Another 
gas A present at a concentration C00 outside the pipe will diffuse intothe pipe. 
====*==> Diffusion 
c(l) 
Convection c<::=::=t=:= V 
1-----x x =constant 
Figure 1.11 Transport by diffusion 
and convection in a pipe 
Under stationary conditions the rate of diffusion into the pipe equals the rate 
of discharge of A with the flow. This situation occurs, for example, in pneumatic 
control equipment where a small gas leak is applied (bleeding) in order to 
prevent dust or water vapour entering the apparatus. The concentration 
distribution c of the gas A in the pipe as a function of the distance x from the 
opening is easily calculated if we realize that in the stationary state the amount 
diffusing through each cross-section x = constant into the pipe equals the 
amount of A removed with the flow, thus: 
de 
</>" = 0 = - []l- - vc 
m dx (1.38) 
25 
with c = c:r:: at x = 0. The solution is: 
~ = exp (-vx) c~ U) (1.39) 
which is a dimensionally homogeneous equation. The relative change in con-
centration appears to be a function of the dimensionless number vxf[ll {the 
so-called Peclet number, Pe), which is a measure of the ratio of mass transport 
by convection and mass transport by diffusion. If this ratio is big, the relative 
concentration of A is smalL Each dimensionless number can be interpreted as 
the ratio of the effects of two physical mechanisms. 
If we had not been able to produce a complete analysis of the above problem 
we could have written formally on the basis of physical intuition : 
C = f(x, C00 , V, 0) (1.40) 
indicating that the concentration of A must be dependent on the distance from 
the opening x, the outside concentration of A, cro, the rate of gas flow v and the 
diffusivity of A, [). Since the dimension 'mass' occurs only inc and C00 dimen-
sional homogeneity is only possible if the ratio cfcro of these two variables 
occurs in the solution. Analogously, the dimension 'time' occurs only in v and 
D, indicating that these two variables must occur as the ratio v/D in the solution. 
This ratio has the dimension of 'length' and this dimension can be eliminated 
by introducing distance into the ratio. Thus the solution we find from dimen-
sional considerations is : 
!__ = !(vx) 
Co [ll 
(1.41) 
This is in agreenlent with the result obtained earlier. That the function f is an 
exponential function cannot be found from dimensional considerations, this 
result can only be obtained from a complete analysis. Dimensional analysis 
has resulted in reducing the problem with initially five variables to a problem 
with only two paiameters. Furthermore, comparisons between analogous 
situations are possible without the complete solution. If we consider, for ex-
ample, two situations where the gas velocities differ by a factor 2, we find tire 
same relative concenlratlons at distances which differ by a factor~ (i.e. the term 
vx stays constant). 
As a second illustration of the procedure followed during dimensional 
analysis, we will consider the force F acting on a sphere which is placed into a 
flowing fluid with velocity v, relative to the sphere (see also paragraph 11.5.2). 
Based on physical considerations we can state that tbe force on the sphere will 
be given by: 
F = f(D, v, , p, 17) (1.42) 
where D is the diameter of the sphere, p the specific gravity and 11 the viscosity 
of the fluid. This relation must be dimensionally homogeneous, i.t:. the dimen· 
26 
sions occurring on the left-hand side must equal the dimensions on the right-
hand side of the equation. This must also be the case if we write : 
F = D"v:pc1J4 constant 
The dimensions of these quantities can be expressed in length (L), time (T) and 
mass (M) and we obtain: 
LT- 2M = L 4(L1 1 )b(L - 3M)'(L - l T - 1M )d constant 
The condition of homogenity leads thus to the following three expressions (one 
for each dimension) between the exponents a, b, c and d : 
L= L 0 L"L - 3cL - d or 1 =a+ b - 3c- d 
r - 2 = T 0 T -"T 0 T-d or -2 = - b- d 
M = M0M0 j"A' M d or 1 = _c + d 
Solving these equations for a, b and c we find : 
a=2-d 
b=2-d 
c =l - d 
The following expression for the force F is thus dimensionally homogeneous: 
F = D<2 - 4>v~2 - 4> p0 - 4>tr4 constant 
or, also : 
~ 2 = (-11 -) 4 constant pv,.D pv,.D (I.43) 
Since both F 1 pv; D2 and 11/ pv,.D are dimensionless combinat~s of the variables, 
any relation between these two groups is also dimension;llfy homogeneous. We 
can therefore conclude that : / 
F ( 17. ) ( 1 ) 
pv;D2 = f pv,.D = f Re (1.44) 
Here Re = pv,.D/ 11 is the Reynolds number, which can be interpreted as the 
ratio· of momentum transport by convection ("'"' pv,v,.) and by internal friction 
("' 1]V,./ D). 
Again, the number of variables in this problem of initially five has been 
reduced to two. In most cases the number of dimensionless groups obtained by 
dimensional analysis equals the number of variables occurring in the problem 
minus the number of independent dimensional equations which can be set up. 
In flow and mass transport problems the number of dimensional equations is 
three (L, T and M) and in heat transport problems four {L, T, M and tempera-
ture). 
27 
The function fin the result obtained is again unknown and can only be found 
from a complete analysis of the problem. Sometimes the ·.result obtained from 
dimensional analysis can be simplified further by making use of former physical 
experience. In the case of the sphere, for example, experience tells us that at very 
low flow velocities the force F on the sphere is independent of the specific 
gravity of the fluid. Thus the relation between the two dimensionless groups 
must be : 
or 
F '1 
2D2 = -D constant pv,. pv, 
F 
-- = constant 
17v,D 
(1.45) 
This equation represents Stokes's law. That the numerical value of the constant 
is 3n can only be found by solving the micromomentum balances for this 
problem. 
As a last example we will discuss the droplets formed, for example, during 
condensation of water vapour at the underside of a horizontal surface. The 
hanging droplets grow until they reach a critical volume ~r and fall down. A 
dimensionally homogeneous relation for ~r follows from the consideration 
that ~-r depends on the density of the liquid p, the gravitational acceleration g, 
the surface tension (J of the liquid and the degree to which the surface is wetted 
by the liquid. For the great number of cases where the surface is well wetted by 
the liquid we can write : . . 
~r = f(p, g, a) (1.46) 
The reader may check that the result is given by: 
( )
3/2 
V::r ~ = constant (1.47) 
The numerical value of the constant follows only from an exact analysis or 
from experimental results. Analysis yields here constant = 6·66. The problem 
with initially four variables has been reduced to one dimensionless group, the 
Laplace number which represents the ratio oft he weight of the droplet (-pg ~r) 
and the capillary force (-a VtJ 
If dimensional analysis is applied all relevant variables have to be included 
in the considerations. If we had, for example, forgotten the gravitational 
acceleration in the analysis of the hanging droplet, we would not have found a 
solution. If, on the other hand, we include too many variables, the result becomes 
unnecessarily complicated. Most profit is obtained if besides dimensional 
analysis a fragmentary piece of physical experience is applied, as illustrated in 
the above examples. 
28 
1.6. Problems 
In the following the reader will find a number of exercises. These problems 
form an important part of the book a .. 1d the reader is asked to solve as many of 
them as possible. They provide a check on the understanding of the principles 
discussed, they illustrate the practical application of these principles and 
sometimes they extend the matter presented. In the three chapters which 
follow each paragraph willend with a collection of problems. The answers to 
these problems are given in order to enable the reader to check his own result. 
On a few problems (marked with an asterisk) we have also included some 
comments in order to illustrate an important point or to show that the solution 
is really not very difficult if the basic principles are applied in the right way. 
When solving a problem, the most important thing is to establish the physical 
mechanism that governs the situation studied (e.g. is convection, conduction 
and radiation of heat important or can one or two ofth~se transport mechanisms 
be neglected in this special case?}. Secondly, we have .to decide over which 
control volume the balance is made (e.g. over a microscopic volume if informa-
tion about the temperature distribution is required; over a pipe or heat exchanger 
or over a total reactor with overall transfer coefficients if only information 
about the mean temperature is required). Next, we have to determine whether the 
situation is in steady state or not (sometimes a situation can be made steady 
state by assuming the observer to move with the system, e.g. the flowing fluid). 
Having come so far we are now in a position to solve the problem with the help 
of the basic information provided in this book. This all sounds very difficult-
and it is- but the best way to solve any difficulties is to try very hard. 
1. A tank is filled with a liquid having a specific gravity of p = 8 lb/ US gal. 
What is the pressure difference between the top of the liquid and a point 
4ft below the liquid level? Answer in psi, bar, Njm2, atm and kgfjcm2• 
Answer : 1·67psi = 0·115bar = 1·15 x 104 N/m2 = 0·113atm = 0·117 
kgf(cm2 
2. The temperature increase into earth is approx. 0·025°C/m. If the thermal 
conductivity of earth is i.. = 1·86 W jm °C, what is the heat loss per unit 
surface area? 
Answer : ¢~ = 0·0465 W /m2 
*3. 11/s water is pressed through a horizontal pipe. The pressure difference 
over the length of the pipe is 2. 105 N/m2 . 
(a) What is the amount of power¢ A necessary and what is the temperature 
increase of the water ll. T if there is no beat exchange with the surround-
ings? 
(b) What are 4> A and ll T if the water flow is doubled and if the pressure 
drop is determined by momentum forces (and not by friction forces)? 
Answer: (a) cPA= 200 W, /:iT= 0·048°C 
(b) ¢A = 1600 w, !J. T = 0·19° c 
29 
4. Two spheres of equal weight fall through air. What is the ratio of their 
stationary rates of free fall if the ratio of their diameters D d D2 equals 3? 
Answer : v Ifv2 = 1 
5. Show that the power cpA necessary to rotate a stirrer (diameter d) with a 
speed n (s - 1) in a fluid of density p and viscosity '1 is given by : 
cpA = r(pnd2' n2d) 
pn3ds r, g 
*6. Write down.the conservation laws for a kettle of water on the fire: 
(a) during heating up and 
(b) during boiling. 
*7. At the end of a glass capillary (outer diameter 2 mm) single water droplets 
are formed very slowly. What is the diameter of the droplets which fall 
from the capillary? 
Answer : d = 44 mm 
8. Through a well-stirred 15m3 tank flows 0.()1 m 3js of coconut oil. From a 
certain time on palm kernel oil is passed through the vessel at the same 
speed. After what time does the effluent oil contain less than 1 per cent 
coconut oil? 
Answer : l = 6900 s 
9. Water flows at a rate of2 mj s through a horizontal pipe of 10 em diameter. 
A large flat plate is fixed near to the end of the pipe and at right angles to it 
Find the force on this plate due to the jet of water, assuming that on 
reaching the plate the water flows away along the surface of the plate. 
Answer : 31-4 N 
10. A pulp containing 40 per cent by weight of moisture is fed into a counter-
current drier at.a rate of 1000 kg/h. The pulp enters at a temperature of 
20°C and leaves at. 70°C, the moisture content at the exit being 8 per cent. 
Air of humidity of 0·007 kg water per kg dry air enters the drier at 120°C 
and the temperature of the air leaving the drier is 80°C. What is the rate 
of air flow through the drier and the humidity of the air leaving the drier? 
Data : heat losses from drier = 5 x 106 J , specific heat o f water vapour 
cP = 2 x 103 J{kg °C, specific heat of dry pulp cP = 103 J/kg °C, 
tlH v at sooc = 2·3 x 106 Jjkg oc 
Answer: </Jg = 2·30 x 104 kgjh, water content = 0·021 kg water per kg 
dry air 
*11. A 100 kmol batch ofaqueous methanol containing45 mol percent CH30H 
is to be subjected to simple atmospheric distillation until the residue in 
the kettle contains only 5 mol per cent CH30H. What is the bulk compo-
sition of the distillate? 
30 
Vapour- liquid equilibria at 1 atm (X and Y = mole 
fraction CH30H in liquid and vapour phases respectively) 
X y X y 
0·02 0·13 ().40 0·73 
0·05 ().27 0·50 0·78 
0·10 042 0·60 0·83 
0·20 0·58 0·80 0·91 
0·30 0·67 0·90 0·96 
Answer : 63 mol per cent 
* 12. Show that John reached the right C\")nclusion in the case of the burnt-down 
factory discussed at the beginning of this chapter. 
Comments on problems 
Problem 3 
A macroscopic energy balance over the entire pipe acco rding to equations 
(1.5) and (1.6) yields for the stationary state and with Jlii = 0 (no phase transi-
tion), v 1 = v2 (constant cross-section of pipe}, h1 = h2 (horizontal pipe), 
cPv, in = </Jv,out and cPH = 0 (no heat losses) : 
Now, the power necessary to bring the liquid to the initial pressure of 
2 x 105 N/m2 is (no temperature change, d T = 0) : 
</>A = - </> b.p = - </>m ~p = 200 W 
v p 
This power is converted into heat in the pipe (where <P A = 0), thus: 
If the water flow is doubled, the pressure drop which is proportional to the 
momentum (dp .- ¢ .v ,..., v2) is increased by a factor 4. Thus 4>,. ("' </J,/lp - v3 ) 
is increased by a factor 8 and ~ T (,...., Ap ,..., v2 ) by a factor 4. 
Problem 6 
The sketch shows the situation of the kettle on the fire. We will set up our 
balances over the entir:e kettle. During heating up no mass leaves the kettle, 
thus the mass and momentum balances reduce to : 
vdPL = 0 and 
dt 
d(pv) . 
V dt = 0 respecttvely 
whereas the energy balance becomes: 
d(pLcpT) 
V dt = tPn 1 - tPn2 = (Xl(~- T)A 1 - (XiT- TaJA:z 
During boiling the balances become: 
mass: Vd:L = -p,vA3 = -</Jm 
d(pv) 
momentum: Vdt = 0 =IF 
31 
where the force in the y-direction is given by FY = Mg + <f>mvy and the force in 
the x-direction by Fx = 4>mvx; and: 
(where, of course, </>H1 - 4>H2 - 4>mllHv = 0) if we neglect all other forms of 
energy because they·are so small. 
Problem 7 
A macromomentum balance over the entire droplet will provide the answer 
we are looking for. Because the flow velocities are very low we can neglect the 
momentum of the fluid and our momentum balance becomes a force balance: 
~ow, the forces acting are the weight force of the droplet and the surface 
tension of the liquid in contact with the capillary as shown in the drawing. 
Thus the force balance becomes: 
' n 3 L F = 0 = -d pg - 2nra 
6 
32 
and thus : 
Problem 11 
t F=21TfCT 
I 
ru;:; 
d = \}pg = 4·4mm 
This problem is easily solved by application of the law of conservation of 
matter. The complication here is that the composition (Y) of the vapour phase 
changes as the distillation proceeds, because of the changing composition (X) 
of the residue. Apparently, the right thing to do therefore is to set up an unsteady-
state material balance for a short time interval dt and to integrate the result 
found over the total process time te . The mass balance (equation 1.1) for this 
case becomes (no flow into the system, no production of methanol but variable 
contents of the evaporator): 
d(NX) dN 
dt = -4>molY = dt Y 
where the symbols have the meaning given in the figure. After partial differentia-tion we can write: 
d(NX) = NdX + XdN 
dt dt dt 
.L 1 = - - = constant fi 
dN 
't'mo d/ 
Mole fraction CH30H 
in vapour= Y 
N kmole 
mole fraction 
CH3 0H =X 
ott =0, x0 , No 
33 
and combinin~ these two equations we obtain: 
dX dN dN 
N-=-Y- -X = - </> (Y- X) dt dt dt mol 
or: 
dX lPmol dt dN 
- = -Y - X . N N 
which yields after integration between t = 0, x0 and t, x: 
N = No exp ( s:, y d: X) 
This equation gives the relation between the total number of moles in the 
evaporator and the methanol content of this residue. We are, however, interested 
in the average composition of the distillate < Y) which is given by : 
ft. ¢ Y dt -ft .. d(NX) - (NX)I'•·x .. ( Y) = 0 mol = 0 = O.Xo 
¢mol te No - N No - N 
= -No {xex (Jx dX )} ~- x~ = X0 - Xeexp(J~~dX/Y- X) 
N 0 -N P x0 Y-X . Xo 1-exp(J~~ dX/Y-X) 
This is the desired result ; all that remains is for us to evaluate graphically the 
integral, as shown in the figure. We find for the area under the curve : 
and thus : 
fx~ dX = 1·17 Y- X X a 
x-
0·45 - 0·05 exp (- 1·17) 0·435 
( Y) = = -- = 63·1 per cent 
1 - ex p (- 1·17) 0·69 
34 
The amount of distillate is found as: 
N 0 - N = N 0 {1 - exp (J:., yd: x)} = 69 kmol 
We see that sometimes the solution of a simple mass balance can be quite 
involved. There is, however, a much simpler way to solve this problem approxi-
mately by carrying out the distillation stepwise. We start with 100 kmol of a 
mixture containing a mole fraction of X 0 = 0-45 methanol. Let us distil off so 
much distillate that X drops t~ X 1 = 0-35. Apparently,in this range the average 
mole fraction of methanol in the vapour phase will be Y0 = 0·73 and thus we 
can find the total number of moles distilled off from a simple mass balance as 
follows: 
xl = NoXo- (No- N)Yo = 0-35 = 100 X 0·45- 0·73(No- N) 
N 0 - (N 0 - N) · 100 - (N 0 - N) 
We find N 0 - N = 26·3 kmol, containing 19·2 kmol of methanol. Now we 
repeat this exercise as stated in the following table until we reach the desired 
methanol fraction of X = 0·05 in the liquid phase. 
Total number Mole fraction of Mole fraction of Number ofkmol Numberofkmol 
of kmol in pot methanol in pot methanol in vapour distilled off of methanol 
N X Y N 0 - N distilled off 
Total 
100 
73-7 
56-1 
43·1 
31·5 
0·45- 0·35 
0·35- 0-25 
0·25 - 0-15 
0·15- 0-05 
0·73 
0·67 
0-58 
0·42 
26·3 19·2 
17-6 11·8 
13·0 7·6 
11·6 4-9 
68·5 43·5 
Thus we find that 68·5 kmol were distilled off, containing 43·5 kmol of 
methanol, i.e. 63·5 per cent. There is a residue of 31·5 kmol, containing 
45 - 43·5 = 1·5 kmol of methanol, which is near enough to the desired 5 mol 
per cent. 
The reader will realize that the principal considerations in this approximate 
solution are the same as for the exact solution given before. Instead of infinitesi-
mal steps dX or dt which made integration necessary, we just used bigger steps 
which enabled us to obtain an approximate analytical solution after a reasonable 
amount of additions and multiplications. 
Problem 12 John and the burnt-down factory 
The flow rate of air through the plant ¢v can be found by means of an energy 
balance (equation !.6). Under steady-state conditions dEtfdt = 0 and realizing 
that the change in heat content of the air pc P L\ Tis much bigger than the changes 
in the other forms of energy (equation 1.5), we can write : 
dEt 
V dt = 4>vPCp ll.T +¢A- ¢H = 0 
35 
Now, no mechanjcaJ energy is added, ¢A = 0 and the amount of heat supplied 
equals <PH = tPm 6H" , where </>,. is the rate of steam loss. Thus : 
A. _ ¢,. /l.H ., _ 70 X 103 • 2·3 X 106 _
52 3
/ 
t.y - - 3 - ms " pcPAT 24. 3600.1·2 x 10 . 30 · 
If we assume that the air in the building is well mixed, the hexane concentration 
in the building will equal the concentration in the exit stream. A material 
balance then yields with cin = 0 for the steady state: 
de 
V- =- ¢ c + r = 0 dt v 
where r equals the rate of hexane loss. T hus we find: 
r 9000 
c = - = = 0·002 kgjm3 ¢v 24 . 3600 . 52 
and, regarding the hexane vapour as an ideal gas, we find : 
86 p =- = 3-83 kg/m3 
22·4 
and thus c = 0·052 per cent by volume, which is indeed well below the explosion 
limit. 
CHAPTER II 
Flow Phenomena 
In this chapter the transport of momentum is studied and we· will discuss the 
prediction of flow resistance in pipe systems and of the resistance due to 
obstacles placed in a flow field. 
The analysis of pipe flow is divided into two sections, laminar and turbulent 
flow, the difference being that in the first case shear stresses in the fluid are pre-
dictable from velocity gradients (deformation rates) but in ·the latter they are 
not. This means that the 'theory' of turbulent pipe flow relies strongly on 
empirical knowledge; however, this knowledge can be made generally applicable 
by scientific reasoning. 
In stationary flow through pipes of a uniform cross-section the momentum 
balance degenerates to a balance between shear forces and pressure forces, but 
this is no longer the case in flow systems in which the cross-section available 
for flow changes. Hence, in that case a more general and practical approach 
will be needed. 
A class of flow problems, in which the enery dissipation due to shearing is 
small in comparison to the amount of mechanical energy which is transformed, 
e.g. from kinetic energy into potential energy, will be dealt with separately. 
This will lead us to the design and use of flow meters. 
F low around obstacles will not only be extended to flow through beds of 
particles but also to stirrers. This brings us to the problem of mixing and of the 
non-uniform distribution of residence time of the fluid elements which pass 
through a continuous flow system. 
In all hydrodynamic theory we will make use of the three fundamental 
conservation laws which were introduced in chapter I. The idea of conservation 
of mass, momentum and energy provides us with five relationships for a three-
dimensional flow field, from which the velocity distribution (v,.., v,. and vz) and 
the temperature and pressure distributions can be obtained in principle, as 
soon as : 
these quantities are defined at the boundaries of the system and 
relationships between shear stresses and deformation rates are known. 
The whole problem in using the concepts of conservation of mass, energy and 
momentum in predicting flow fields and flow resistances is to define the control 
,·olume in such a way that it facilitates the analysis and to make clever guesses 
about the order of magnitude of the several terms in the balances in order to 
37 
make them amendable for an algebraic treatment. Hence, although the whole 
theory on transport phenomena does not go beyond the concepts which are 
already introduced in chapter I, the professional engineer needs considerable 
experience before he is able to tum these concepts into practical use. Therefore~ 
this chapter has been written. It treats a number of problems in an elementary 
way in order to illustrate the practical application of the basic principles and 
it provides a number of elementary solutions which can be the starting point 
to the solution of more complex problems. 
ll.I. Laminar Oow 
John looked at the spoon in his hand. Syrup dripped from it 
into his porridge at a decreasing rate. He thought about 
writing a criminal story in which someone was imprisoned 
until the very last drop of syrup would leave the spoon. John 
had a feeling that this would last very long, months, perhaps,. 
or years. Being a scientific story writer and working effici-
ent ly, he read the following introduction and found out that 
the thickness of the syrup layer on his spoon decreased about 
100 times in 25 minutes. Consequently, he thought, this layer 
would diminish to