1. Momentum Transfer Jul-Dec 2006 Instructor: Dr. S. Ramanathan
Office: CHL 210
Class Notes:
The class notes will be placed in

2. Overview Background & Motivation Course Syllabus
What will be covered and what will not be
Examples
Goals & Pre-requisites
Evaluation
Tentative Schedule
Text Books / References
Overview

3. Chemical Engineering Reaction Kinetics Transport Phenomena Momentum
Mass
Chemical Engineering is taught as Reaction and transport. Transport involves heat transfer, mass transfer and momentum transfer. Momentum transfer is also known as fluid mechanics. This is taught separately so that the basic principles can be understood in simple, isolated conditions and it prepares one for a more complicated (and realistic) situation where one has to apply the principles learnt in multiple disciplines to solve a problem.
Of course, even this is a simplified view. Thermodynamics, process control and so on are key in Chemical Engineering. However, here we will focus on one part of the transport phenomena i.e. momentum transfer.
The basic equations for the three transfers are similar. We will emphasize the similarities as we go along. Also, we will allude to a bit of other disciplines in the examples as appropriate.
Heat

4. Background :
Most of the momentum transfer equations are similar to heat and mass transfers
Momentum transfer: Focus is on fluids
Heat and Mass Transfer: Also include solid
Heat Transfer: Radiation (no corresponding phenomena in momentum and mass transfer)
We will also look at some of the differences between momentum transfer and other transfer operations. In momentum transfer, the focus is on fluid. In heat and mass transfer, solids are also part of the ‘game’.
In heat transfer, one has to consider radiation. There is no corresponding phenomenon in mass and momentum transfer.
Similarities in problems will be discussed as appropriate

5. Motivation Momentum Transfer: Fluid Mechanics
Understanding Lab Results
Design
Manufacturing (Production/ Maintenance)
Troubleshooting
So, why do I have to learn Fluid Mechanics (FM)?
If you want to understand the results of experiments, you need it. (For example, in the next semester you have Momentum Transfer lab and you must be prepared for that, at the least!). If you want to design a chemical plant, you must know FM (and other things). If you are a shift engineer in manufacturing or in technical support, you need to know FM. If a pressure drop across a valve suddenly decreases, what does it mean? What should you do?
To run a plant in the normal operating mode (especially continuously run plants), it is fairly easy. Even without knowing much FM, an operator can learn few rules of thumb and run it. A knowledge of FM helps you understand what is going on, though.
To trouble shoot and to design you must have very good knowledge. To understand results of experiments, you must have good knowledge.
To do these things, how much do I have to know

6. Course Syllabus: What will be covered? And to what extent?
Fundamentals (ideal cases)
Some applications (more realistic, but not very)
Most real-life issues, ==> kinetics & heat/mass/momentum transfer together
Analytical solutions not possible in many cases
What will not be covered?
Compressible , supersonic flows
Only limited exposure to non-newtonian fluids
Computational Fluid Dynamics (CFD)
limited exposure to Perturbation methods
...and so on
We will not ‘cover’ ‘everything’ related to FM in this course. No course will.
We will go over the fundamentals and some examples of ideal cases. We will also go over some more realistic examples, as applications. Most of the real interesting chemical engineering systems involve knowledge of more than FM (Heat transfer, mass transfer , kinetics, control…).
Also, even in cases where you consider only FM, analytical solutions are not always possible. And we are not looking at numerical solutions in this course. You should consider CFD course as an elective in the latter years.
Some examples of things that will not be included. Compressible, supersonic flows. Only brief discussion on non-newtonian fluids (for now, think of newtonian fluid as simple fluid and non-newtonian as complex. This is not a correct definition, but this will help). Computational Fluid Dynamics. Only limited exposure to perturbation methods and turbulent flows. Only limited exposure to multiphase flow and so on…

7. Course Syllabus: Statics: To refresh the basics Dynamics: Mass Balance
Momentum Balance (Linear & Angular)
Energy Balance
OK, so what will we go through?
Statics: to refresh the basics
Dynamics:
Fully developed flows:
Macroscopic: Mass, momentum and energy balances (with out friction)
Microscopic: Momentum Balance : Navier Stokes Equation
Frictional losses: Newtonian & non-newtonian fluids
Developing Flow: Flow over body: Boundary layer theory
Flow through pipes: Turbulence, How to estimate pressure drops? (for example)
Flow past bodies: Fixed bed, fluidized bed
The fluid dynamics can be divided into macroscopic and microscopic. For example, if you have to
Pump crude oil from place A to B, you want to know how much pressure drop will occur, for a given pipe diameter. That way, you can determine what pump to buy and calculate how much the running cost will be. (OR, you can determine the optimum pipe diameter. If the pipe diameter is larger, then the pressure drop is smaller. But the pipe cost is more. So you have to balance between initial cost and running cost). This is macroscopic study. You can use correlations or charts which relate the fluid & flow parameters (like volumetric flow rate and pressure drop) and determine this. In macroscopic study, you need to make suitable approximations and develop some physical understanding.
The microscopic study deals with understanding the flow structure, boundary layer, turbulence etc. This is needed to understand heat transfer and mass transfer, for example. The study is logically structured and you will need to have a good math background. For example, a fluid at high temperature is flowing in a pipe and the pipe+ fluid is cooled from outside, by air. If the pipe is X m long (and you know the diameter and so on), what will be the average outlet temperature of the fluid? It depends on the fluid flow pattern. If all the elements across a particular cross section flow at the same velocity, then the fluid at the center of the pipe will not be cooled easily (Fluid thermal conductivity is low and cooling will be better if convection is present). If the fluid flows in a circular pattern, then the cooling will be easy. Here microscopic study will be necessary.
We will ‘mix’ both, as it is done in typical text books of FM. We start with macroscopic mass, momentum and energy balances. Then we continue on to microscopic momentum balance and then develop the equations (some analytical solutions and some semi-empirical, where some physical understanding and some empiricism will be mixed). Using those as the basis for correlations and charts, we will learn how to use those charts and correlations for macroscopic predictions , for example in pipe flow and in packed bed reactors.
Frictional losses
Boundary layer theory
Flow past/through

8. Examples Pumps, Turbines Heat Exchangers, Distillation column
Fluidized or Fixed bed reactors
CVD reactors (micro electronics)
Artificial blood vessels (Bio)
What are the applications?
Pumps, turbines… (What kind of pump should I use for this purpose, what rating, OR with this pump and pipe, how much fluid can I move in 1 hour)
In heat exchangers, distillation columns: Fluid flow will affect other characteristics, so design of these units will involve knowledge of FM
Reactors: Fixed bed, fluidized bed (catalytic reactors for example, or burning of coal) What is the pressure drop (this is a basic question. You
need to know lot more about the reactors, but this is a start)
CVD reactor: The design and successful operation of some of the reactors for chip fabrication needs a sound knowledge of FM
Artificial blood vessels: Again, fluid properties and flow properties need to be modeled correctly to design an artificial blood vessel.

9. Examples Production of Sulfuric Acid
used in fertilizers, car batteries etc
Take the example of production of sulfuric acid. A simple process diagram shows various places where FM knowledge is needed.
OR consider the following situation. Extraction of alumina from bauxite involves mixing bauxite, sodium hydroxide and water, heating up the mix to 150 C or so and keeping it under high pressure (3 to 4 atm) for a significant amount of time. A soluble sodium aluminum silicate forms and the insolubles can be filtered out. Cooling down the solution and adding HCl (for example), you can precipitate pure alumina which is insoluble.
Now, it is believed that running it as PFR is much quicker than running it as CSTR (15 min vs many hours). So you have to maintain this
Conditions for 15 mins and the mix has to flow. Can you design a pipe and pumping system? (For now, we will not bother about supplying heat, maintaining temperature, ensuring the conversion and so on…) Will the mix vaporize when the pressure comes to atm? Will it be liq at 150C and 3 atm?

10. Examples Monsanto Process Pump air (N2+O2) and burn Sulfur
Provide large area of catalyst
“Scrub” with water
Store the sulfuric acid
For a given production (ton per day),
What is the pump capacity needed?
Design and operation of reactor
How to measure the flow rate?
What if something goes wrong? How to detect it and how to respond? (Detection of leak through chemical sensor, pressure sensor etc)
The slides are self explanatory

11. Goals:
Understanding and approaching problems which involve Momentum Transfer
==> Pumps, flow through pipes
==> Separation (filtration, adsorption etc)
More emphasize on application and less on proof
Also prepare for future courses
Momentum Transfer Lab
Transport phenomena
At the end of the course, the successful students should be able to understand and approach problem involving Fluid Mechanics (for example, flow through pipes, filtration, adsorption and other such unit operations)
The course will have more emphasize on application of theorems and equations and less on proving the theorems. However, it does not mean you don’t need good mathematical skill. In fact, if you haven’t done so, you should try to take a course in Partial Differential Equations (PDE). They appear not only in FM, but also in Heat & Mass transfer (and other chem engg disciplines, if you go deep enough). If not, you should learn it on your own. You will need it to solve the problems in this course. You should also become familiar with complex functions and their calculus. A little bit of programming knowledge will help you, although it is not mandatory for this particular course.
Calculus (PDE), Complex Variables
Little bit of programming
Final Exam - 40
Quizzes - 2 * 20 = 40
and Project/Assignment -20

12. Tentative Schedule Quiz-1 Quiz-2
A tentative schedule. Expect the first quiz to include a bit of Navier Stokes Equation (section 6) and the second quiz to have the sections up to and including boundary layer theory.
The number of classes is also ‘tentative’. Some of the materials, especially in the beginning, will be more of a review and hence we may go a bit faster and the newer material will be discussed in more detail.
The text book 3W&R has 14 chapters dealing with momentum transfer. The first 6 chapters should be considered ‘refresher’ and we will cover them quickly.
Roughly we will spend two weeks or more on the 2. statics, 3. Description of a Fluid in Motion, Conservation of 4. Mass, 5. Momentum and 5. Energy in macroscopic studies.
Concept of Newtonian and Non-Newtonian fluids (chapter 7) will be discussed for a week or so and the application of ‘shell balances’ or differential fluid element analysis for one type of flow (laminar flow) will take a week or so. Expect some material about non-newtonian fluids from ‘outside the text book’.
The differential equations for fluid flow (for laminar as well as turbulent flow) called Navier Stokes equation will be applied for ‘typical’ problems. The ones discussed in the text book are ODE type and we will also look into problems involving PDE. These may take about two weeks.
We will move onto the problem of scaling from lab to pilot plant to manufacturing plant and tackle simple problems. This is about to take a week or week and a half.
The simplified NS equation for ideal fluids (no viscosity) will be solved for few situations. The concept of streamlines and potential flow will also be discussed. And this will take a week and a half.
The other more careful simplification of NS eqn leads to Boundary Layer theory and we will work on that for 2 to 3 weeks. We will see the solution for the classical problem of fluid flow over a flat plate. We will also look into the integral method by von Karman.
The concept of turbulence will be introduced and the theoretical (or semi empirical) insights will be discussed for a week or so. Next the application of these for understanding flow in a pipe (whether laminar or turbulent) will be discussed for week or less. We will also look into different types of valves and relevant devices.
The derivation of relevant flow equations for fixed bed and fluidized bed is expected to take a week or two. If time permits we will also look into compressible flow. The text book does not discuss about fixed bed or fluidized bed, which are important in Chem Engg Industry. You can refer to BSL or McCabe & Smith (Unit operations in Chem Engg) or other books or to the class notes.
Not updated in the table, but hopefully included in the course will be a section on some experimental aspects of fluid mechanics.
Perhaps we will discuss it at the end of the semester.

13. Text Books / References
Class Notes / Slides
Slides will be on the internal server
Text: Fundamentals of Momentum, Heat and Mass Transfer by Welty, Wicks , Wilson & Rorrer (4th edition) John Wiley & Sons
Reference: Transport Phenomena by Bird, Stewart and Lightfoot, edition, McGraw Hill
Fluid Mechanics and its applications by Gupta & Gupta
Other sources referenced will be mentioned in the class
You should have a copy of the text book. The quizzes or exams may be open book and I expect that you will have your copy. You will not be allowed to ‘borrow’ books from your neighbors during the exam.
The material that is covered in this course will come not only from the text but also from other sources. However, if a formula that you need in the quiz/exam is not in the text, and if it is complicated enough and not important enough for you to memorize, it will be provided to you.
You are expected to memorize some formula/constants… That is part of learning (e.g. you are supposed to know that density of water at room temperature is approximately 1 g per cubic-cm).

14. Statics
On to statics

15. Statics
Fluid: changes shape continuously when a tangential force is applied
Pressure at any point in a stationary fluid is same in all directions
Pressure vs Distance
Consider only gravity effects
ie. Ignore electromagnetic, chemical (eg.osmosis) and other forces
Basic definition of what a fluid is. Pressure in a stationary fluid and effect of gravity

16. Statics Po = atm h r P bottom = Po + r g h Constant Density
(eg Liquids)
Pressure for fluids with constant density. Manometer.
Application: Manometer

17. Statics Approx air temp vs height Height (km) Temp (C)10 50 80
-120
-60
Approx air temp vs height
Fig from “Introduction to Fluid Mechanics” by Fox & McDonald, page 53
Variable Density
eg Gases
Pressure for fluids with varying density:

18. Example rMercury = 13,600 kg/m3 Pbm. 2.13 B PA-PB=? 10 cm A B’ WaterHg
25 cm
10 cm A B
Pbm. 2.13
PA-PB=? B’
PB’-PB= r1 * g * h1
Example of manometer calculations
PA-PB’ = r2 * g * h2 - r1 * g * h2
Actually used for flow rate measurement

19. Example Pbm 2.22 z Practical depth for a suited diver is ~ 180 m
What is the error in assuming density is constant?
Example of pressure change in liquids (when you consider them to be compressible)

20. Example Coin on water: Surface Tension F
Indication of force between liquid-metal vs liquid/liquid
Example of coin floating on water. What are the important forces?

21. Statics Acceleration due to other forces
eg centrifuge, accelerating vehicle
In centrifuge, usually g is negligible compared to a
Otherwise use vector algebra to add g and a
What if you have centrifuge? What about accelerating vehicles?

22. Example: Centrifuge r a
To separate materials based on density difference
in case gravity is insufficient (for reasonable separation)
Acceleration expressed as N times “g”
Typically acceleration >> g
Ignore gravity effects
Example with centrifuge

23. Example: Slow rotation
For lesser acceleration h1
At z=h1, r=0, P = Patm
On the surface, P = Patm
Rotating body (slow rotation). Surface profile. (IIT JEE problem?)
The reason the surface profile does not depend on the density or other properties of fluid is because the forces acting on it are gravity and centrifugal acceleration and the density term is linear in both of them and it cancels out.
Equation of free surface

24. Conservation of Mass In any control volume
Mass flux in - mass flux out = Mass accumulation rate.
If (mass in) is taken as -ve, then
Accumulation rate + Flux(out -in) =0 S
Vol
V-velocity
n-normal vector
Macroscopic Analysis:
Equation of conservation of mass. In some cases, called continuity equation.

25. Conservation of Mass Reynold’s Theorem (generalization)
For a property B (Mass, for example)
and corresponding b (per unit mass)
Differential equation, for the control volume.
General equation called Reynold’s transport theorem. It is called so because it involves ‘transportation’ of certain quantities.
The quantity may be mass or linear momentum or angular momentum and so on… Particular cases of interest to us, are equations of
mass, momentum and energy balances.
Rate of change (system) = Flux+ Accumulation
See Transport Phenomena, by Bird Stewart Lightfoot for an analogy

26. Reynold’s Transport Thm
B = Mass
==> b =1
DB/Dt =0; Eqn of Conservation of Mass
B = Momentum
==> b = velocity
Momentum Eqn
B = Angular Momentum
==> b = r x v (Angular Momentum Eqn)
etc..
Reynolds transport theorem and obtaining the various equations

27. Mass conservation Simplifications Steady State : (gas or liquid)
d/dt =0
Mass in = Mass out
For liquids (Volume in = Volume out)
Constant density & fixed control Volume:
d/dt (V) =0
Volume in = Volume out
True even for unsteady state
Some simplifications for eqn of conservation of mass

28. Examples
Pbm. 4.8, 4.5, 4.12, 4.18, 4.11, , 4.20, 4.22, 4.21, 4.24
Some example problems that we will discuss in the class

29. Examples Pbm. 4.18, steady state V1 V2 d1 = d2 = 2 cm Q1 = 0.0013 m3/s
V2 = 2.1 m/s
A3 = 100 * (p 1e-3*1e-3/4)
There are 100 holes of 1 mm dia in the shower V3
Pbm 4.18
Water flows steadily through the piping junctin, entering section 1 at m3/s. The av velocity at section 2 is 2.1 m/s. A portion of the flow is diverted through the showerhead, which contains 100 holes of 1 mm dia. Assuming uniform shower flow, estimate the exit velocity from the showerhead jets.
Assuming incompressible fluid, we can translate the eqn of cons mass to eqn of cons of volume. Under steady state conditions, this gives a simple algebraic eqn to solve.

30. Examples
Pbm. 4.8 A1
V2,a2 A2
V1,a1
Area =A, Velocity =V, Acclrn = a. Find V2, A2
Pbm 4.8 For the system shown above, the larger piston is moving with a velocity of v1(0) at a reference point of time and with an acceleration of a1. What is the velocity of the smaller piston?
Since we know the velocity at time 0, we can calculate V1(t). We can relate V1 and V2 and hence we will get V2(t). If you want, we can decompose it into V2(0) and a2.
V1 (t) A1 = V2 (t) A2
a1A1 = a2A2

31. Examples Pbm. 4.5, steady state V/2 m/s V m/s 6 m/s
0.5 m Long, 0.1 m R
V m/s
6 m/s
Pbm 4.5
For incompressible flow under steady state conditions, in a perforated pipe, discharging along the side in a linear profile, the velocity at the outlet of the pipe and the velocity along the pores are given in terms of one unknown. Determine the unknown velocity V

32. Examples
Pbm. 4.11 X Y V2 r2 Vw r1 V1
Consider stationary control volume
Pbm 4.11 A shock wave may go over a compressible fluid at a velocity of Vw. The fluid velocity before experiencing the shock is denoted by V1 and after experiencing shock is V2. Since the fluid is compressible, the density changes after the shock. A shock wave moves at the speed of sound in the medium
For example, this is (roughly) what happens during explosions. The wave front of explosion moves at the speed of sound in air (~330 m/s).
If you know the densities and V1 (and also Vw) we can use the mass conservation eqn to determine V2.
This example is used to illustrate the concept of control volume. We will choose a stationary control volume and solve the problem. Then we choose a moving control volume and show that the results obtained are the same. In this case, both methods are (more or less) easy.
In other systems, choosing the control volume well (or poorly) will determine if the solution is going to be easy (or difficult).
In the stationary control volume, some fluid is coming in, some is going out and hence the accumulation term is not zero. We can
calculate the rate of accumulation, the flux and relate the velocities.

33. Examples
Pbm. 4.11 X Y V2 r2 Vw r1 V1
If we choose the control volume to be moving along with the shock wave, then the mass of fluid inside the CV is the same. Hence the accumulation term is zero. (i.e. rate of accumulation or depletion is zero)
The flux is very slightly more complicated, but we can find it and the resulting relationship between the velocities is the same as the one obtained using stationary control volume.
Consider control volume Vw

34. Examples Pbm. 4.9, one dimension, steady flow
Pbm 4.9 For compressible gases, if you use ideal gas law, you will find that as long as the velocity is less than Mach number, increase in area results in decrease in velocity. Once you cross the sonic threshold, increasing area increases velocity. Now you understand why the nozzle in the rockets are shaped the way they are.

35. Examples r
Pbm. 4.12 R
Pbm 4.12 In a real fluid flow in a pipe, the velocity of the fluid at any cross section is not uniform. At the center of the pipe the speed is higher and near the walls, it is lower. The velocity profile is given by a variety of equations, depending on the fluid type and the flow type. The volumetric flow rate , of course, depends on the ‘average’ velocity. Let us calculate the average velocity for a flow described by ‘one-by-seven power law’. This is used to describe flow of fluids like water under turbulent flow conditions.

36. Examples Pbm. 4.15 Steady flow liquid film thickness is “h” V0
width “into the paper” is W V0 h Y X
Pbm For a steady flow of liquid film of thickness ‘h’, with ‘no slip’ condition and parabolic velocity profile (which is given), you are asked to find the flow rate.
It involves a simple integration.
On the other hand, if you know the volumetric flow rate and the thickness of film and width, then you can calculate the free surface velocity.

37. Examples Constant Velocity V Varying thickness “b”
Infinitely long plate (in one direction)
Exit velocity is (a) flat or (b) parabolic
Pbm. 4.14 V b 2L
Mass Flux
Y direction
Accmln rate
Pbm In this problem, a fluid is being ‘pushed out’ by a set of two plates. The bottom plate is stationary and the top plate is moving down with a constant velocity. What is the velocity of the fluid going out if the exit velocity profile is (a) ‘flat’ and (b) ‘parabolic’.
It is trivial for flat profile. For parabolic profile, we use the argument of symmetry and ‘no slip’ condition at the top and bottom. The max velocity occurs at the middle and the velocity profile can be written in terms on one unknown (V-max). From there, it is a trivial problem.
Consider unit depth for control Vol

38. Examples Velocity of outgoing fluid = V(y) Pbm. 4.14 V b
For a flat profile, V(y) = constant, say Vavg 2L
Y direction
For a parabolic profile,
If we assume that the plates are circular (with radius L), then we can calculate the velocity again, and this will be problem 4.24

39. Examples d1= 2cm, d2=0.8 mm Pbm. 4.21 d2 d1 V
How fast should the plunger move
(ie find V)
(a) if there is no leakage
(b) if leakage between tube and plunger is 10% of needle flow
Mass Flux
Accmln rate
Pbm You are given a piston like arrangement and asked to determine the velocity (of the plunger) needed to obtain a given flow rate. This is a straightforward mass balance problem. You can try it for situation where there is no leakage and situation where there is a fixed % of leakage. Use a control volume which is given by the red rectangle, for it is the simplest.

40. Examples Qn: What is the flow rate across the Horizontal surface?
Pbm. 4.13
Vx =V0 V0
Height=6d d V0
Vx =V0 V0
Pbm 4.13
There is an infinite cylinder and an incompressible fluid is flowing across it (perpendicular to the axis of cylinder). The cylinder
Is held in place by application of force. The velocity profile changes because of the presence of cylinder. (Incoming velocity profile is ‘flat’). And the change is felt only up to 3 diameter distance (+/-). If you consider only the volume enclosed by this region (control volume), what is the outflow (or velocity in Y direction) across this control volume.
For steady state conditions (even otherwise, if you assume that the fluid is incompressible and since the CV is fixed), the rate term goes to zero. The mass flux can be calculated with simple integration and we can calculate the mass flow rate. Assuming constant density and a constant V-y, we can also calculate the V-y. (If you know the shape of velocity profile for V-y, you can determine that quantitatively. Assuming constant V-y is the same as saying ‘flat profile’)