Dr. S.

RAJA
Additional Professor,
Department of Chemical Engineering, MIT, MAHE
email: rajaselvaraj@gmail.com; raja.s@manipal.edu
Phone: 99645 82441
 Few Expectations ……
• FROM U…… • FROM ME……
• Be regular • I’ll be regular
• Be on time • I’ll be on time
• IMMOBILE ur MOBILE • Give slides only at the
• Take instant notes end
• Bring calculator • Give regular Tutorials
• BE PRESENT (!!!!) in • ……
class and answer your • ……..
attendance! • ………….
• Finish Tutorials on time
• Research Area / Domain • Core placements

• Research Advisors • Higher studies

• Placement coordinator • Seniors

 TP – CHE 3154 -------
• NEW SUBJECT ??????? Syllabus
• Extension of Mom.Tr, Ht.Tr, and Mass Tr.
• NOOOOOOO……….. IT
C K !
KNO ET IT!!
G
and
• Derivations…….Derivations…….Derivations…….lengthy!!!!!!!
difficult??????....NOT AT ALL……..
USE ANALOGY!!!!..............becomes simple!!!!!.......PRACTICE REQRD!!!!!
• Few numerical problems (short)

CONTENTS
• Transport coefficients – a • Transport Equations – Shell
review balance for rial
s
o
• Viscosity 1. Momentum transfer Tut each
on pic
• Thermal conductivity 2. Energy transfer and to
• Diffusivity 3. Mass transfer
 Text / Reference Books
1. Bird R.B., Stewart W.E., Lightfoot E.W., Transport Phenomena (2e), John-
Wiley, 2002

2. Brodkey R.S., Hershey C.,Transport Phenomena- A unified approach,

McGraw Hill Book Company,1988

3. Slattery J.C., Advanced Transport Phenomena, Cambridge University Press,

1999

4. Geankoplis C.J, Transport Process and Unit Operation (3e) , Prentice-Hall ,

1993
5. OUR CLASS NOTES……………
 TP – CHE 3154

TRANSPORT PHENOMENA

[2 1 0 3]

• Tuesday : 8 to 9
• Wednesday : 2 to 3
• Saturday: 10:30 to 11:30
 • It includes 3 closely related topics………
Transport Phenomena……. Transport
• the phenomena which involves
movement of various entities such as Phenomena
• mass,
• energy or
• momentum through a medium –
fluid or solid by virtue of non-
uniform conditions existing within Fluid
the medium Heat Transfer Mass Transfer
Dynamics
• TP describes
• the basic principles and laws of
transport.
• the relations and similarities
among different types of transport Momentum Energy Mass
that occurs in any system

• Only for chemical engineers???????..........no……..

• gaining popularity and application in the fields of biotechnology, agriculture and nanotechnology
 Why TP????? • Let us take a simple process where the
raw materials are transported from the
• why do we need to study these three storage vessel to the reactor.
phenomena together?
• All three types of transport occur
simultaneously in industrial, biological • So, the fluid flow phenomenon comes
and agricultural problems. into the picture…………..momentum
transfer
• The basic equations that describe the
three transport processes are closely • Then, if the reaction in the reactor takes
related and is the basis of solving place at high temperatures, then the raw
problems – ANALOGY materials have to be heated…………heat
transfer
• The mechanisms of these three • After the reaction, the separation
phenomena are closely related. processes take place with a view to
• Also, the governing mathematical isolating the product in pure form…..mass
equations are similar in nature. transfer
 Part 2: Heat Transfer
• Fourier’s law of heat conduction
Part 1: Momentum Transfer • Convective energy transport and shell
• Newton’s law of viscosity – Newtonian energy balances
& Non-Newtonian fluids • The general energy equations
• The shell momentum balance for • The temperature distribution in turbulent
momentum transfer flow conditions
• The general equation of continuity
and the equation of motion – Navier-
Stokes equation
Part 3: Mass Transfer
• Momentum transfer in turbulent flow
conditions • Fick’s law of diffusion
• Convective mass transfer and shell mass
balances
• Mass transfer in turbulent flow condition
Comparison of different transport
operations
Newton’s law of viscosity……
• Fluid is contained between two parallel
Upper Plate plates which are separated by a distance Y
• Fluid is at rest….
Lower Plate • Lower plate is moving parallel to top
plate at a velocity V…
• @ time ‘t’, there is a velocity build up near
the lower plate .
• Each layer of liquid moves in the x
direction. The layer immediately adjacent
to the bottom plate is carried along at the
velocity of this plate.

• @ large time ‘t’, each layer moving at a

slower velocity as we go up in the y
direction. The velocity profile is linear.
• Fluid gains momentum as time proceeds and
Velocity distribution – flow between two
attains steady state……’F’ is required
parallel plates
• Let F be the force required to maintain • Taking…..F/A = τyx…… force per unit area required
the motion of the lower plate and Y be perpendicular to the y direction …….known as shear
the distance between the plates. stress…..
• v/y can be written as differential form as – dvx / dy
• It has been experimentally found that the • The negative sign indicates that velocity decreases in
force F can be expressed as (the force is the positive direction of y
proportional to the area, velocity, and • dvx / dy = velocity gradient = shear rate = ẙ
inversely proportional to the distance
between the plates) • τyx = − μ • “Newton’s law of
viscosity”
• τyx = − μ ẙ • states that the shear stress
• FαA
is proportional to the
• Fαv SI unit of : negative of the velocity
• F α 1/Y ……………… kg/ m s gradient.
Pa. s • The constant of
• N s / m2 proportionality is called
• viscosity
• F/A = μ v/Y • Newton suggested that those fluids which obey the
• μ = proportionality constant = viscosity law of viscosity be called Newtonian fluids.
• And those fluids, especially polymers, etc. which do
not obey this law be called non-Newtonian fluids.
 τyx …..Momentum flux……
• Momentum is given by the expression (m)(v) where
• Momentum transport deals with the transport of “m” is mass and v is velocity; momentum has
dimensions LMT-1.
momentum which is responsible for flow in fluids.
• Flux means rate per unit area;
• Viscous drag forces are responsible for the
velocity gradient • therefore momentum flux has dimensions L-1MT- 2,
• At y = 0 the fluid acquires momentum in the x- which are also the dimensions of τyx
direction due to motion of the lower plate. • So with representing momentum flux……flux of
• This fluid imparts some of its momentum to the momentum is directly proportional to the velocity
adjacent layer of fluid above the plate, causing it gradient dv/dy.
also to move in the x-direction. • The negative sign in means that momentum is
• Momentum in the x-direction is thus transmitted transferred from regions of high velocity to regions
through the fluid in the y- direction. of low velocity, i.e. in a direction opposite to the
direction of increasing velocity.
• τyx = − μ
• This is similar to the transfer of heat from high to low
• To interpret this equation in terms of momentum temp regions
transfer, shear stress "τ” is considered as the flux of
• The magnitude of the velocity gradient dv/dy
x-momentum in the y- direction. (τyx ) determines the magnitude of the momentum flux;
• Check the dimensions of momentum flux and shear dv/dy thus acts as the 'driving force' for momentum
stress. transfer.
 Problem 1 τyx = − μ
• Compute the steady-state momentum flux τyx when
the lower plate velocity v is 1 ft/s in the positive x
direction, the plate separation Y is 0.001 ft, and the
fluid viscosity is 0.7 cP.
• Soln:
• viscosity = 0.7 cP
• ……7 x 10–4 kg/m s

• Y = 0.001 ft = 3.048 x 10–4 m

• v1 = 1 ft/s == 0.3048 m/s
• v2 = 0

• τyx = − μ dvx / dy
• ….. = − μ (v2 – v1) / Y
• …… = − (7 x 10–4 ) (0 – 0.3058) / 3.048 x 10–4

• τ = 0.7 N/m2
Prob 2
• There are two parallel plates some distance apart. Between the plates
water is used at 24 °C which has the viscosity of 0.9142 cP. The lower
plate is being pulled at a constant velocity of 0.4 m/s faster relative to
the top plate. How far apart should the two plates be placed so that the
momentum flux, τyx is 0.3 N/m2?
• Soln:
• Distance = 0.122 cm
Prob 3
• Using Fig. the lower plate is being pulled at a relative velocity of 0.40 m/s greater
than the top plate. The fluid used is water at 25oC.
• (a) How far apart should the two plates be placed so that the shear stress is 0.30
N/m2 Also, calculate the shear rate.
• b) If oil with a viscosity of 2.0 x 10-2 Pa .s is used instead at the same plate spacing
and velocity as in part (a). what are the shear stress and the shear rate?
• Ans: Distance = 1.33 mm
• Shear stress = 6 N/m2
 Prob 4
• In the figure, if the distance between plates is Dy 0.5 cm, Dv is 10 cm/s.
and the fluid is ethyl alcohol at 273 K having a viscosity of 1.77 cp
• Calculate the shear stress and the velocity gradient or shear rate

• Shear stress = 0.0354 N/m2

• Shear rate = 20s-1
Prob 5
• The distance between two parallel plates is 0.00914 m and the lower
plate is being pulled at a constant velocity 0.366 m/s faster relative to
the top plate. The fluid filled between the plates is glycerol at 293 K
having a viscosity 1.069 kg/m· s. Calculate the momentum flux.
• τyx = 42.8 N/m2
Prob 6
• Using Fig. the distance between the two parallel plates is 0.00914 m and the lower plate
is being pulled at a relative velocity of 0.366 m/s greater than the top plate. The fluid
used is soybean oil with viscosity of 4 x l0-2 Pa s at 303 K
• (a) Calculate the shear stress and the shear rate in SI units.
• (b) If glycerol at 293 K having a viscosity of 1.069 kg/ms is used instead of soybean oil,
what relative velocity in m/s is needed using the same distance between plates so that
the same shear stress is obtained as in part (a)? Also. what is the new shear rate?

• Ans. (a) Shear stress = 1.60 N/m2 and shear rate = 40 s-1
• (b) Relative velocity = 0.01369 m/s and shear rate =1.50s-1
Prob 7
• A Newtonian fluid flows between two parallel plates at rest initially.
Compute the steady-state momentum flux when the lower plate
velocity is 1 m/s in the positive x-direction and the plate
separation is 0.001 m. Fluid viscosity is 0.7 cP.
• τyx = 0.7 N/m2
Prob 8
• Two parallel plates are 0.5 cm apart. The lower plate moves at a
velocity 10 cm/s and the upper plate is stationary. Assume linear
velocity distribution. The fluid between the plates is ethyl
alcohol at 273 K having a viscosity of 1.77 cP. Calculate the steady-
state momentum flux.
• τyx = 0.0354 N/m2
Newtonian & Non-Newtonian (Power-law) fluids
• t = K (dv/dy)n
• which can be called the power-law equation,
and where K is a constant of proportionality.
• n = flow behavior index (dimensionless)
• K = consistency index (L-IMT n-2)

• if n = 1… the fluids are called Newtonian fluids

because they conform to Newton's equation
and K = m;

• n < 1; shear thinning or pseudoplastic

• n > 1; shear thickening or dilatant
• Some fluids do not produce motion until
some finite yield stress has been
applied…..Bingham fluid
To solve problems……Newtonian/non-Newtonian
fluids….
• Plot the graph between Shear stress vs Shear rate
• In order to find out the exact nature of fluid and the value of ‘n’…. …
linearize the power law equation…

• Wkt….. Y= mX + C
• Taking log on both sides…. form

• From the slope and intercept….determine the values of ‘n’ and ‘K’
respectively!!!!
• Take care of units!!!
 Prob 1: Rheology of a polymer
• The polymer is produced by utilizing sucrose as major carbon source.
After 120 h reaction the following measurements of shear stress and
shear rate were made with a rotating cylinder viscometer. Plot the
rheogram for this fluid and name the fluid type. Determine ‘n’ and K
values of power law equation:
SS vs SR

Shear stress Shear rate (s-1) 70

(dyn/cm2) 60

44.1 10.2 50

235.3 170 40

357.1 340 30

20
457.1 510
10
636.8 1020
0
0 200 400 600 800 1000 1200

Pseudo plastic
 SS SR log (SS) log (SR)
N/m2 1/s

4.41 10.2 0.644439 1.0086

23.53 170 1.371622 2.230449

35.71 340 1.55279 2.531479

45.71 510 1.660011 2.70757

63.68 1020 1.804003 3.0086

 log SS vs log SR
2

1.8 f(x) = 0.586975347394716 x + 0.0580911767450754

R² = 0.999163918918829
1.6

1.4
n= 0.587
1.2

1 log K 0.0581

0.8
K= 1.143142 N s^n /m2
0.6

0.4

0.2

0
0.5 1 1.5 2 2.5 3 3.5 Pseudo plastic
 Prob 2
• For the following data, determine the type of fluid and the values of
the properties characterizing the fluid:

Shear
Rate,
(1/s) 0.1 0.5 1 5 10 50 100 500

Shear
Stress
(N/m2) 0.0002 0.0019 0.0049 0.056 0.14 1.77 5 55.9
 SR SS log (SR) log (SS)
SS vs SR 1/s N/m2
60 0.1 0.0002 -1 -3.69897
0.5 0.0019 -0.30103 -2.72125
50 1 0.0049 0 -2.3098
40 5 0.056 0.69897 -1.25181
10 0.14 1 -0.85387
30
50 1.77 1.69897 0.247973
20 100 5 2 0.69897
10
500 55.9 2.69897 1.747412
0
0 100 200 300 400 500 600

3
log SS vs log SR n= 1.4795
log K -2.2745
2
f(x) = 1.47950498899527 x − 2.27448594654498
K= 0.005315 N s^n /m2
R² = 0.999626014695782
1

0
-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3

-1

-2
Dilatant
-3
 Prob 3
• For the following data, determine the type of fluid and the values of
the properties characterizing the fluid:

Shear Rate,
(1/s) 1.89 1.66 1.29 1.13

Shear Stress
(N/m2) 0.0559 0.054 0.0518 0.0501
 SS vs SR
log SS vs log SR
0.06
-1.22
0 0.05 0.1 0.15 0.2 0.25 0.3
0.05 -1.23

-1.24
0.04
-1.25

-1.26 f(x) = 0.20365585964203 x − 1.31012207337595

0.03 R² = 0.991274116744905
-1.27
0.02 -1.28

-1.29
0.01
-1.3

0 -1.31
1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2

n= 0.2037
log K -1.3101
K= 0.048967 N s^n /m2 Pseudo plastic
Prob 4
• From the following data, determine whether the fluid behaves as a
power law fluid, and if so, determine the respective parameters:

Shear Rate,
(1/s) 3.75 7.5 11.3 15 22.5 30 37.5

Shear Stress
(N/m2) 0.239 0.428 0.623 0.805 1.12 1.41 1.7
 SS vs SR
log SS vs log SR
1.8
0.3
1.6
0.2 f(x) = 0.855333343426968 x − 1.11049256615633
1.4 0.1 R² = 0.999587950067756
1.2 0
0.4 0.6 0.8 1 1.2 1.4 1.6 1.8
1 -0.1
-0.2
0.8
-0.3
0.6
-0.4
0.4 -0.5
0.2 -0.6
0 -0.7
0 5 10 15 20 25 30 35 40

n= 0.8553
log K -1.1105

K= 0.077535N s^n /m2 Pseudo plastic

Prob 5
• Determine the type of fluid and the properties which characterize
them:

Shear
Rate,
(1/s) 375 450 525 600 750 900 1125
Shear
Stress
(N/m2) 0.331 0.393 0.448 0.51 0.637 0.779 0.961
 Chart Title
1.2
Chart Title
0
1 2.5 2.6 2.7 2.8 2.9 3 3.1
f(x) = 0.97751074301131 x − 3.00194528672192
-0.1 R² = 0.998990321202946
0.8

-0.2
0.6

-0.3
0.4
-0.4
0.2
-0.5
0
300 400 500 600 700 800 900 1000 1100 1200
-0.6

n= 0.9775
log K -1.1105

K= 0.000996 N s^n /m2 Pseudo plastic

 Polymer Shear Shear
concentra Stress Rate,
Prob 6 tion (g/L) (N/m2) (1/s)
1 5
1.3 10
6
• During the production of a 1.8 20
2.3 32
polymer it was observed 3 5
pseudo plastic nature. The 4.3 10
11.1
rheological behaviour of the 5.5 20
fluid at different concentration 6.7 30
9.2 5
of the polymer was obtained as 13 10
follows using a turbine impeller 15.5
17.4 20
viscometer. Determine the 20.5 30
values of the n and K at various 12.8 5
18.7 16.9 10
concentration of polymer: 23.5 25
 log SS vs log SR
25
SS vs SR

20 1.6

1.4
6 g/L
15 f(x) = 0.376570353336439 x + 0.846653346227894
11.1 g/L f(x)
R² = =0.999059335013433
0.445794999690255 x + 0.65853900515744
15.5 g/L 1.2 R² = 0.997652712527271 6 g/L
18.7 g/L
Linear (6 g/L)
10
1 11.1 g/L
Linear (11.1 g/L)
0.8 15.5 g/L
5 f(x) = 0.437719277366929 x + 0.179326230548578 Linear (15.5 g/L)
R² = 0.993990887045893
18.7 g/L
0.6 Linear (18.7 g/L)
0
0 5 10 15 20 25 30 35
0.4
f(x) = 0.448881490641468 x − 0.322833686100971
0.2 R² = 0.995446018548443

0
0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6
 n= 0.4489
log K -0.3228

K= 0.475554 N s^n /m2

Polymer Con vs 'n' and 'K'
0.46 8 n= 0.4377
7 log K 0.1793
0.44

0.42
K= 1.511124 N s^n /m2
5
n

0.4 4
K
n= 0.4458
3 log K 0.6585
0.38

0.36
K= 4.555122 N s^n /m2
1

0.34
4 6 8 10 12 14 16 18 20
0 n= 0.3766
log K 0.8467

K= 7.025868 N s^n /m2

 Law of conservation……
• Law of conservation of Mass
Rate of accumulation of mass = Rate of I/P mass – Rate of O/P mass + Rate of mass
generation – Rate of mass depletion

• Law of conservation of Energy

Rate of accumulation of Energy = Rate of energy I/P – Rate of energy O/P + Rate
of energy generation – Rate of energy depletion

• Law of conservation of Momentum

Rate of momentum accumulation = Rate of momentum I/P – Rate of momentum
O/P + Sum of forces acting on the
system(gravity, pressure)