CENG 2220 Transport
Phenomena I
I-Ming Hsing, Spring 2025
Engineering Analysis of Chemical & Biological Processes and Systems
Engineering Analysis of A Body System: how nutrients and species transport in
the biological systems and how cells communicate with each other
Fluid Mechanics
“Fluid mechanics is the branch of applied physics concerned with the
mechanics of fluids(liquid or gas) and the forces on them”
Applications
• Bio Engineers: fluid flow as it relates to heat, mass and energy transfer in
biological/physiological processes.
• Chemical Engineers: fluid flow as it relates to heat, mass and energy transfer in chemical
processes.
• Mechanical Engineers: fluid flow in machines (fans, compressors, turbines and
combustion engines etc.).
• Aeronautical Engineers: flow around bodies to provide lift with minimum drag.
• Civil Engineers: hydrodynamic transport of river sediments, flood plain & beach erosion
control, irrigation, dams, piping systems
• Marine/Ocean Engineers & Atmospheric Meteorologists: motion of sea currents and of
atmospheric air, respectively.
I-Ming Hsing, Spring 2025
Factors affecting fluid transport
How can fluid be transported in a confined environment?
Driving
Force
External
Pressure
difference in
a pipe using
pump
Voltage
difference in an
electric circuit
using battery
Properties of fluid
Defining a “Confined”
Environment
Internal
Inside Nonporous or
airtight pipe
Concentration
difference in a
fluid mixture
eg. blood
vessels, pipes
in chemical
industry, etc.
Inside porous
or leaky pipe
Size and surface
properties of the
confined environment
Viscosity
Density
Newtonian vs. NonNewtonian
Compressible vs.
incompressible
eg.
Fenestrated
capillaries,
porous media
Continuum hypothesis: fluid characteristics (pressure, velocity, etc.) vary continuously throughout the fluid, the fluid is
treated as “continuum”. The fluid property/behavior is evaluated by considering the average, or “macroscopic”, value
of the quantity of interest where the average is evaluated over a small volume containing a large number of fluid molecules
I-Ming Hsing, Spring 2025
Transport of Molecules: Diffusion
Diffusing molecules move from region of high
concentration to regions of low concentration
I-Ming Hsing, Spring 2025
Transport of Molecules: Convection
Let’s consider a simple experiment: deformation of material placed between
two parallel plates
Newton’s Law of
viscosity
𝜏𝜏𝑦𝑦𝑦𝑦 = −𝜇𝜇
Shear
stress
𝑑𝑑𝑣𝑣𝑥𝑥
𝑑𝑑𝑦𝑦
Shear
rate, ϒ̇
Proportionality
constant; Viscosity
Viscosity
• Distinctive mechanical characteristic of fluid: deforms continuously
under the action of shear stress
• A physical property that characterizes the flow resistance
• The rate of fluid deformation = rate of change of distance between 2
neighboring points moving with the fluid divided by the distance
between them (change in length per unit length and unit time).
• In solid mechanics, strain is change in length per unit length;
• rate of deformation ≡ rate of strain i.e. “strain rate” or “shear rate”
are also applied to fluids
I-Ming Hsing, Spring 2025
Fluid properties affecting fluid
momentum transport
Kinematic viscosity is an indicator of the efficiency of momentum transport (“Diffusion of momentum”)
Transport Phenomena in Biological Systems, 2nd Edition
George A. Truskey, Fan Yuan, and David F. Katz, Pearson Education
I-Ming Hsing, Spring 2025
Fluid properties of gases and liquids
I-Ming Hsing, Spring 2025
I-Ming Hsing, Spring 2025
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 ∝ − 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 𝑜𝑜𝑜𝑜 𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓/𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝜌𝜌𝑣𝑣 2 /𝐿𝐿 𝜌𝜌𝜌𝜌𝜌𝜌
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁: 𝑅𝑅𝑅𝑅 =
=
=
𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓/𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝜇𝜇𝜇𝜇/𝐿𝐿2
𝜇𝜇
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑏𝑏𝑏𝑏 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑣𝑣𝑣𝑣
=
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁: 𝑃𝑃𝑃𝑃 =
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑏𝑏𝑏𝑏 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝐷𝐷𝑖𝑖𝑖𝑖
Transport Phenomena in Biological Systems, 2nd Edition
George A. Truskey, Fan Yuan, and David F. Katz, Pearson Education
I-Ming Hsing, Spring 2025
Relative Importance of Convection and Diffusion
In class discussion: Peclet number Pe, Reynolds number
I-Ming Hsing, Spring 2025
Relative Importance of Convection and Diffusion
In class discussion: Peclet number Pe, Reynolds number
I-Ming Hsing, Spring 2025
Transport of Molecules: Ligand Binding Interactions
Receptor-Ligand Binding Kinetics
I-Ming Hsing, Spring 2025
Transport Within Cells
I-Ming Hsing, Spring 2025
Newtonian fluids
• In molecular motion, viscosity is proportional to the (mean free path) x
𝑘𝑘𝑘𝑘
(avg. molecular velocity) x (density). This gives dimensions as
or 𝑃𝑃𝑃𝑃. 𝑠𝑠
𝑚𝑚.𝑠𝑠
• The nature of viscosity can be illustrated by an experiment to measure its
value:
Infinite parallel plates
{
Area A
H
U
F
Force required to
move the plates at
constant speed ‘U’
After an initial transient period
For constant H and U, the F ∝
A
For constant A, the force is a
unique, monotonically
increasing function of ratio
U/H.
• Examples of Newtonian Fluid: water, oil, alcohol, glycerine etc.
I-Ming Hsing, Spring 2025
Newtonian fluids
Shear stress and shear rate can be defined as:
Shear stress:
𝜏𝜏𝑠𝑠 =
Shear rate:
Then we can write,
and
𝜏𝜏𝑠𝑠 = 𝜏𝜏𝑠𝑠 𝛾𝛾𝑠𝑠
𝐹𝐹
𝐴𝐴
𝛤𝛤𝑠𝑠 or 𝛾𝛾𝑠𝑠 =
(Force/Area)
𝑈𝑈
𝐻𝐻
(Velocity/Distance)
{shear stress: unique function of shear rate}
𝑑𝑑𝜏𝜏𝑠𝑠
> 0 {function is monotonically increasing}
𝑑𝑑|𝛾𝛾𝑠𝑠 |
For Newtonian fluids, the plots of shear stress v/s shear rate are straight lines going through the origin.
Then the viscosity will be:
𝜏𝜏𝑠𝑠
(Pa.s)
𝜇𝜇 =
±𝛾𝛾𝑠𝑠
1 Pa.s = 10 Poise = 103 centipoise
Viscosity of water at RT is ~ 1 cP or 10-3 Pa.s
I-Ming Hsing, Spring 2025
Non-Newtonian fluids
Can be divided into three broad classes:
NonNewtonian
Fluids
Time
independent
Time
dependent but
non-elastic
Viscoelastic
I-Ming Hsing, Spring 2025
1. Time independent NNs
{
𝜏𝜏𝑠𝑠 = 𝜇𝜇 𝛾𝛾𝑠𝑠 𝛾𝛾𝑠𝑠
• The viscosity is not solely a fluid property, it’s a function of the shear rate
• There are a no. of types depending upon the form of the 𝝉𝝉𝒔𝒔 v/s 𝜸𝜸𝒔𝒔
relationship:
𝝉𝝉𝒔𝒔
Bingham Plastic: toothpaste, clay, mustard
General Plastic, Casson fluid
Dilatant (shear thickening): fine powder in suspension
Newtonian
Yield Stress
Pseudo Plastic (shear thinning): paper pulp, paint
𝜸𝜸𝒔𝒔
I-Ming Hsing, Spring 2025
2. Time dependent but nonelastic NNs
• Behavior of fluid is influenced by what happened to them in recent
past, e.g. ketchup pours easily on shaking the bottle
𝝉𝝉𝒔𝒔
Rheopectic (stir vigorously and it freezes): bentonite clay
Thixotropic (shake and it thins): paints, ketchup
𝜸𝜸𝒔𝒔
I-Ming Hsing, Spring 2025
3. Viscoelastic NNs
• Combine elastic properties of solids with flow behavior of fluids
(biological fluids).
• 𝜏𝜏𝑠𝑠 v/s 𝛾𝛾𝑠𝑠 diagram is insufficient for characterization; transient
experiments are needed
I-Ming Hsing, Spring 2025
Elasticity, Viscosity, and viscoelasticity
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Storage G’ and Loss G’’ Moduli for Human Synovial Fluid
I-Ming Hsing, Spring 2025
Power law behavior of pseudo plastic and dilatant NNs
• Viscosity data frequently fall on a straight line on a log-log scale for a
decade of more in shear rate. This indicates a power law relationship.
𝜇𝜇 𝛾𝛾𝑠𝑠 = 𝐾𝐾 𝛾𝛾𝑠𝑠 𝑛𝑛−1
• K is the flow consistency index and n is the power law index. m and n
are positive constants, the values of which are specific to a given
materials and temperature
• A Newtonian fluid is represented by n  1.
I-Ming Hsing, Spring 2025
Shear stress and Viscosity
• For Newtonian, the velocity gradient is proportional to the imposed
shear stress on the fluid
𝜏𝜏 = ±𝜇𝜇
𝑑𝑑𝑈𝑈
𝑑𝑑𝑦𝑦
𝝉𝝉
Slope = 𝜇𝜇
±
𝒅𝒅𝑼𝑼
𝒅𝒅𝒚𝒚
I-Ming Hsing, Spring 2025
Shear stress and Viscosity
by setting
Typical viscosity trend of
polymer solutions and
polymer melts
Cambridge U. Press 2016
I-Ming Hsing, Spring 2025
Characteristics of Newtonian vs. Non-Newtonian
Relationship between the shear stress and shear rate for Newtonian and nonNewtonian fluids. Fluid properties are Newtonian fluid, μ = 0.01 g cm-1 s-1; Bingham
plastic, τ0 = 0.01 dyn cm-2, μ0 = 0.01 g cm-1 s-1;
shear-thickening fluid (dilatant), m = 0.01 g cm-1, n = 2.0; and shear thinning fluid
(pseudoplastic), m = 0.01 g cm-1 s-1.5,
n = 0.5.
Transport Phenomena in Biological Systems, 2nd Edition
George A. Truskey, Fan Yuan, and David F. Katz, Pearson Education
Characteristics of Bingham
fluid or Bingham plastic
I-Ming Hsing, Spring 2025
Whole Blood (NN) vs. Plasma/Serum (Newtonian)
Transport Phenomena in Biological Systems, 2nd Edition
George A. Truskey, Fan Yuan, and David F. Katz, Pearson Education
I-Ming Hsing, Spring 2025
Rheology of blood exhibits Shear-thinning characteristics and its shear
stress-shear rate relationship could be described by Casson Equation
Casson Equation
𝜏𝜏 1/2 = 𝜏𝜏𝑜𝑜 + 𝜂𝜂𝛾𝛾̇
𝜏𝜏𝑜𝑜 𝑖𝑖𝑖𝑖 𝑡𝑡𝑡𝑡𝑡 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠; 𝜂𝜂 𝑖𝑖𝑖𝑖 𝑡𝑡𝑡𝑡𝑡
viscosity at high shear rate (𝛾𝛾)̇
Physics of Fluids 25, 073104 (2013); https://doi.org/10.1063/1.4816369
I-Ming Hsing, Spring 2025
30
Fluids in motion
• Solids tumble, roll, bounce, …
• Fluids pour, spout, splash, spray, swirl, lap, wave, creep, slide …
• The motion of fluids near solid surfaces is of special importance
to life – all of life is immersed in (moving) fluids.
Slide credits: Dr. Saif A. KHAN of ChBE, NUS
I-Ming Hsing, Spring 2024
31
Solid versus Fluid
motion
• Consider the fate of a colored blob
of fluid within a rectangular
container with moving walls.
• Like a solid, it is difficult to
‘squeeze’ or ‘pull’ a liquid without
resistance. However, unlike a
solid, a fluid does not resist
‘shear’, and deforms indefinitely. A
fluid can be stirred. A solid cannot.
I-Ming Hsing, Spring 2024
32
• Like any motion, fluids move because of forces acting on them
(gravity, buoyancy, imbalanced pressure, friction, …)
Why do fluids move?
• Example – fluid motion due to differences in density, which occurs
when you boil a kettle of water (‘natural’ convection)
I-Ming Hsing, Spring 2024
33
Visualizing Fluid Motion
http://www.algorizk.com/windtunnel/overview/
Free Wind Tunnel app for iOS and Android
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34
• To study fluid motion, let’s first learn to characterize fluid motion
in time (t) and space, i.e., Kinematics.
• Kinematics describes the motions of points, subjects, and group of subjects
without consideration the causes of motion.
You may sketch velocity fields of a flow around immersed objects.
Intuitively, how should the flow profile differ as the velocity increases
dramatically?
Transport Phenomena in Biological Systems, 2nd Edition
George A. Truskey, Fan Yuan, and David F. Katz, Pearson Education
I-Ming Hsing, Spring 2024
35
Observe a swimmer in the pool:
1) as a poolside observer (Eulerian description)
2) as a neighboring swimmer (Lagrangian description)
Scenario 1): partial time derivative
𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝑐𝑐 𝑥𝑥, 𝑦𝑦, 𝑧𝑧, 𝑡𝑡 + ∆𝑡𝑡 − 𝑐𝑐(𝑥𝑥, 𝑦𝑦, 𝑧𝑧, 𝑡𝑡)
( )𝑥𝑥,𝑦𝑦,𝑧𝑧 ≡ lim
∆𝑡𝑡→0
𝜕𝜕𝜕𝜕
∆𝑡𝑡
Scenario 2): total time derivative d/dt
𝑐𝑐 𝑥𝑥 + 𝑢𝑢𝑥𝑥 ∆𝑡𝑡, 𝑦𝑦 + 𝑢𝑢𝑦𝑦 ∆𝑡𝑡, 𝑧𝑧 + 𝑢𝑢𝑧𝑧 ∆𝑡𝑡, 𝑡𝑡 + ∆𝑡𝑡 − 𝑐𝑐(𝑥𝑥, 𝑦𝑦, 𝑧𝑧, 𝑡𝑡)
𝑑𝑑𝑐𝑐
( ) ≡ lim
∆𝑡𝑡→0
𝑑𝑑𝑡𝑡
∆𝑡𝑡
𝑑𝑑𝑐𝑐
( )
𝑑𝑑𝑡𝑡
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
≡ ( )𝑥𝑥,𝑦𝑦,𝑧𝑧 + 𝑢𝑢𝑥𝑥 ( )𝑦𝑦,𝑧𝑧,𝑡𝑡 + 𝑢𝑢𝑦𝑦 ( )𝑧𝑧,𝑥𝑥,𝑡𝑡 + 𝑢𝑢𝑧𝑧 ( )𝑥𝑥,𝑦𝑦,𝑡𝑡 =
𝜕𝜕𝜕𝜕
𝜕𝜕𝑥𝑥
𝜕𝜕𝑦𝑦
𝜕𝜕𝑧𝑧
Substantial time derivative D/Dt: what is the difference between d/dt and D/Dt?
𝐷𝐷𝑐𝑐
( )
𝐷𝐷𝑡𝑡
𝜕𝜕𝜕𝜕
≡( )
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
+ 𝑣𝑣𝑥𝑥 ( )
𝜕𝜕𝑥𝑥
𝜕𝜕𝜕𝜕
+ 𝑣𝑣𝑦𝑦 ( )
𝜕𝜕𝑦𝑦
𝜕𝜕𝜕𝜕
+ 𝑣𝑣𝑧𝑧 ( )
𝜕𝜕𝑧𝑧
=
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
+ (v � ∇𝑐𝑐)
+ (u � ∇𝑐𝑐)
I-Ming Hsing, Spring 2024
36
Control volume, flow velocity
1
< 𝑣𝑣 >= � 𝑣𝑣 � 𝑛𝑛𝑛𝑛𝑛𝑛
𝐴𝐴 𝐴𝐴
Transport Phenomena in Biological Systems, 2nd Edition
George A. Truskey, Fan Yuan, and David F. Katz, Pearson Education
𝑄𝑄 =< 𝑣𝑣 > 𝐴𝐴 = � 𝑣𝑣 � 𝑛𝑛𝑛𝑛𝑛𝑛
How about mass flow rate 𝑚𝑚̇
𝐴𝐴
I-Ming Hsing, Spring 2024
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Control volume, flow velocity, and fluid streamline
3
𝑑𝑑𝑑𝑑
= � 𝑒𝑒𝑖𝑖 𝑣𝑣𝑖𝑖 ;
𝑣𝑣 =
𝑑𝑑𝑑𝑑
𝑖𝑖=1
𝑑𝑑𝑑𝑑
𝑑𝑑𝑦𝑦
𝑑𝑑𝑧𝑧
𝑒𝑒𝑥𝑥 𝑣𝑣𝑦𝑦 + 𝑒𝑒𝑦𝑦 𝑣𝑣𝑦𝑦 + 𝑒𝑒𝑧𝑧 𝑣𝑣𝑧𝑧 = 𝑒𝑒𝑥𝑥 ( ) + 𝑒𝑒𝑦𝑦 ( ) + 𝑒𝑒𝑧𝑧 ( ); example in cartesian coordinate
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
Transport Phenomena in Biological Systems, 2nd Edition
George A. Truskey, Fan Yuan, and David F. Katz, Pearson Education
I-Ming Hsing, Spring 2024
CENG 2220
Scale, Dimension, Dimensional Analysis,
Friction Factor
I-Ming Hsing, Spring 2025
Length and velocity scales in fluid mechanics
I-Ming Hsing, Spring 2025
Dimensions of quantities involving mass, length and time
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Dimensionless groups corresponding to stress or force ratios
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5
Biological Systems have different scales and dimensions
© Cambridge University Press 2015
I-Ming Hsing, Spring 2025
6
Biological Systems have
different scales and
dimensions
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7
Dimension Analysis
• A technique for expressing the behavior of a physical system in terms of a
minimum number of independent variables and in a form that is
unaffected by changes in the magnitude of the units of measurement.
• The physical variables are arranged in dimensionless groups consisting of
ratios of qualities like lengths, velocities, forces, etc.- which characterize
the system.
• Dimensionless products are important and useful in the planning,
execution, and interpretation of experiments.
Dimensional Homogeneity
Every term in a physical equation should have the same dimensional formula
or otherwise the equation is meaningless.
I-Ming Hsing, Spring 2025
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Dimensional Analysis: Buckingham π Theorem
• Every dimensionally homogenous equation can be expressed as a zero
function of a set of dimensionless groups.
• The theorem falls in two parts:
1. The solution to every dimensionally homogenous physical equation has
the form: φ (π1, π2, π3,…) = 0, where π1, π2, π3,… represent a complete set
of dimensionless groups of the variables and dimensional constants of
the equation, and
Or,
u1 = f(u1, u2, u3, …….uV) π theorem φ (π1, π2, π3, …… πV-D) = 0
π1 = φ (π2, π3, π4, …… πV-D), where D is determined by the min. no. of reference dimensions
2. If an equation contains V separate variables and dimensional constants
and these are given in terms of D primary quantities, them the number
of dimensionless groups in a complete set is: V-D.
I-Ming Hsing, Spring 2025
9
Steps of performing dimensional analysis
Step 1. List all the variables involved in the problem
Step 2. List the fundamental dimensions of each variable or parameter
Step 3. Determine the required number of pi terms
Step 4. Select a number of repeating variables, where the number required is equal to the number
of reference dimensions. Hint: generally advantageous to choose repeating variables related to
mass, geometry and kinematics.
Step 5. Form a pi term, a dimensionless group, by multiplying one of the non-repeating variables by
the product of repeating variables each raised to an exponent that will make the combination
dimensionless
Step 6. Solve the unknown exponents in step 5
Step 7. Repeat steps 5-6, for each of the remaining variables
Step 8. Check all the resulting pi terms to make sure they are dimensionless
Step 9. Express the final form as a relationship among the pi terms and think about what it means
I-Ming Hsing, Spring 2025
10
As an application of dimensional analysis, let’s consider the steady flow of a
Newtonian fluid in a cylinder of inner radius R and length L. We wish to find the
relationship between the pressure drop ΔP and the other remailing variables:
ΔP = 𝑓𝑓(ρ, 𝜇𝜇, 𝐿𝐿, 𝐷𝐷, 𝑣𝑣)
Π = ΔP𝑣𝑣 𝑎𝑎 𝜌𝜌𝑏𝑏 𝐷𝐷 𝑐𝑐
In terms of the units (mass m, length l, and time t), we have:
𝑚𝑚𝑚𝑚 −1 𝑡𝑡 −2 (𝑙𝑙 𝑡𝑡 −1 )𝑎𝑎 (m 𝑙𝑙 −3 )𝑏𝑏 𝑙𝑙 𝑐𝑐
0 = 1 + 𝑏𝑏
ΔP
Π1 = 2
𝜌𝜌𝑣𝑣
0 = −1 + 𝑎𝑎 − 3𝑏𝑏 + 𝑐𝑐
0 = −2 − 𝑎𝑎
Similarly, one could derive the other dimensionless groups
𝐷𝐷𝐷𝐷𝐷𝐷
𝐷𝐷
Π2 =
, Π3 =
𝐿𝐿
𝜇𝜇
I-Ming Hsing, Spring 2025
11
Friction Factor
• Consider a long smooth horizontal pipe, with an incompressible
Newtonian fluid flowing in it.
• Variables: D, L, |Δ P|, v, ρ, µ
4Q
• v= 2 = mean velocity
D π
Variable
Unit
Dimensions
|∆P| Dvρ L
D
Length
L
• Dimensionless groups: 2 ,
,
L
Length
L
μ D
ρv
|Δp|
Force/Area
ML T
Dvρ L
|∆P|
v
Length/Time
LT
• 2 = function of (
, )
μ D
ρv
ρ
Mass/Volume
ML
Dvρ
L
µ
(Force/Area) x Time
ML T
)
= × function of (
μ
D
-1 -2
-1
-3
-1 -1
Reynolds number
I-Ming Hsing, Spring 2025
12
Dynamic similarity
• A strategy provided by dimensional analysis to extrapolate
experimental results to systems that are larger or smaller, have flow
that is faster or slower, or have different fluid properties.
• Denoted as a dimensionless group N1 that is a function of two other
kinds of groups., one involving fluid properties/velocities and the
other involving only length ratios (also called aspect ratios).
If the fluid properties and characteristic velocity in the experiments are
chosen so that N2, N3, etc., are each the same as in the real system and
the systems are geometrically similar, ie all ϕi in the model are the same
as in the system it represent, the two systems are dynamically similar.
I-Ming Hsing, Spring 2025
13
Friction Factor
• Fanning friction factor:
|∆P| D
𝑓𝑓𝐹𝐹 =
=
𝑓𝑓(Re)
2ρv2 L
• Alternative:
|∆P| D
𝑓𝑓𝐷𝐷 = 1 2 = 4𝑓𝑓𝐹𝐹
L
ρv
2
Moody, Darcy friction factor
I-Ming Hsing, Spring 2025
14
The velocity distribution in laminar pipe flow
Force balance on cylindrical element of fluid:
Pressure force balances drag force
𝐷𝐷
𝜋𝜋
4
2
𝑃𝑃𝑜𝑜 − 𝑃𝑃𝐿𝐿 − 𝜏𝜏𝑤𝑤 𝜋𝜋𝐷𝐷𝐷𝐷 = 0
So that the shear stress varies with the radius
according to
𝐷𝐷 𝑃𝑃𝑜𝑜 −𝑃𝑃𝐿𝐿
𝐷𝐷 ∆𝑃𝑃
𝐷𝐷 ∆𝑃𝑃
𝜏𝜏𝑤𝑤 = (
)==
𝐿𝐿
4
4 𝐿𝐿
4 𝐿𝐿
I-Ming Hsing, Spring 2025
15
Friction Factor
• In laminar flow, Re < 2100
16
𝑓𝑓𝐹𝐹 =
Re
|∆P| D 16μ
D2 ∆P
;v=
=
32μL
2ρv2 L Dvρ
πD2 v
π D4 ∆P π R4 ∆P
Q=
=
4 = 128 μL
8 μL
• In turbulent flow: 4000 < Re < 105
• Blasius equation: 𝑓𝑓 = 0.079/Re1/4
• Re ≥ 4000: entire turbulent region
•
1
𝑓𝑓
= 4.0 log Re 𝑓𝑓 − 0.4 *
(equation implicit in f, but explicit
in velocity because Re√f ≠f(v)
Hagen-Poiseuille equation
Prandtl-Karman Law
Or
Von Karman-Nikuradse equation
*This relationship can be derived using turbulent flow theory, but it is beyond the scope of this course
I-Ming Hsing, Spring 2025
16
Capillary Viscometry
π ∆P D4
μ=
128 L Q
• ΔP measured over a length L for known flow rate Q.
• In a process line continuous on-line viscosity measurement by measuring ΔP/L
and Q in Laminar pipe flow.
End effects:
Expressions for f valid only in long pipes, not near entrance. Higher ΔP/L near
entrance.
Le is the entrance length: the distance
Le
required for the flow to become fully
For
Re < 2100: ≅0.59+0.055 Re
D
developed.
Le
Re > 2100: ≅40
D
116, for Re = 2100
Fully developed: the velocity profile of
flow no longer changes with distance
Theoretical basis for these values will be discussed later this semester
I-Ming Hsing, Spring 2025
17
The velocity distribution in laminar pipe flow
The integrated result is
parabolic velocity profile;
shear stress distribution
Δ𝑃𝑃 2
𝑟𝑟
𝑣𝑣 = 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚 −
4𝜇𝜇𝜇𝜇
Since v=0, @ r=R (no slip boundary condition @ the wall)
Hence
Center-line velocity = vmax = 2*mean velocity
𝑄𝑄
𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚 = 2 2 = 2𝑣𝑣̅
𝑅𝑅 𝜋𝜋
𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚 =
Δ𝑃𝑃 2
𝑅𝑅
4𝜇𝜇𝜇𝜇
𝑟𝑟 2
𝑣𝑣 = 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚 1 − 2
𝑅𝑅
Using the above equation the volumetric flow rate through the
pipe is found to be:
𝑅𝑅
𝑅𝑅
𝑟𝑟 2
Δ𝑃𝑃𝑃𝑃𝑅𝑅 4
𝑄𝑄 = � 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 = � 2𝜋𝜋𝜋𝜋𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚 1 − 2 𝑑𝑑𝑑𝑑 =
0
0
𝑅𝑅
8𝜇𝜇𝜇𝜇
This is called the Hagen-Poiseuille law
The average velocity is therefore,
𝑄𝑄
Δ𝑃𝑃𝑅𝑅 2
𝑣𝑣̅ = 2 =
𝑅𝑅 𝜋𝜋
8𝜇𝜇𝜇𝜇
I-Ming Hsing, Spring 2025
18
The velocity distribution in laminar pipe flow
𝜋𝜋𝐷𝐷4 Δ𝑃𝑃
𝑄𝑄 =
128𝜇𝜇 𝐿𝐿
Δ𝑃𝑃 128𝜇𝜇𝜇𝜇
=
𝑟𝑟 =
𝑄𝑄
𝜋𝜋𝐷𝐷4
The electrical analogy can be exploited when designing complex flow networks for low-tomoderate Re, as in microfluidics. For laminar flow through n tubes, with resistances r1, r2,
…, rn, what would the overall resistance rT be for n tubes in series and in parallel?
I-Ming Hsing, Spring 2025
19
Heart Is Like a Pump
Feher - Quantitative Human Physiology
I-Ming Hsing, Spring 2025
20
Power
The force acting at 1:
At the downstream location 2:
The net force acting on the fluid:
𝜋𝜋𝐷𝐷2
P1 ( )
4
𝜋𝜋𝐷𝐷2
P2 ( )
4
𝜋𝜋𝐷𝐷2
|Δp|( )
4
The work done to move the fluid in a pipe at distance Δl:
But it takes place over time Δt, so the rate of work is:
𝜋𝜋𝐷𝐷2
|Δp|( ) Δl
4
|Δp|(
Thus, Power input:
𝜋𝜋𝐷𝐷2
4
)
Δl
Δt
v
Q
Power = |Δp|Q
I-Ming Hsing, Spring 2025
21
Effects of pipe roughness
• Regular uniform roughness
𝑓𝑓𝐹𝐹 = 𝑓𝑓 𝑅𝑅𝑅𝑅, 𝑘𝑘�𝐷𝐷
In laminar flow: f
independent of k/D
• “Relative roughness”: k/D
• No effect in laminar flow or turbulent flow
at lower Re
• Effective values of k, a few examples
Materials
k (mm)
Draw tubing
0.0015
Cast iron
0.46
Commercial steel
0.05
I-Ming Hsing, Spring 2025
22
Effects of pipe roughness
• The Moody chart is a plot of the
Colebrook formula which is:
1
√𝑓𝑓
= −4.0 𝑙𝑙𝑙𝑙𝑙𝑙10
𝑘𝑘
4.67
+
+ 2.28
𝐷𝐷 Re 𝑓𝑓
I-Ming Hsing, Spring 2025
23
The hydraulic diameter
• Empirical method to calculate flow rates and pressure drops in
conduits with noncircular cross sections. ”Equivalent” diameter.
𝜋𝜋𝐷𝐷2
• A cylinder of diameter D and length L has a volume of
𝐿𝐿
4
• And its surface area wetted by the fluid is 𝜋𝜋𝜋𝜋𝜋𝜋
4×𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
= 𝐷𝐷𝐻𝐻 (Hydraulic diameter)
• Thus, we can write: 𝐷𝐷 =
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑏𝑏𝑏𝑏 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
• For channels of constant but noncircular cross section, the usual
definition is:
4×𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝐷𝐷𝐻𝐻 =
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
Note: if hydraulic radius is defined as RH = cross section
area/wetted perimeter, RH = DH/4, thus RH = ½ R (cylinder radius)
I-Ming Hsing, Spring 2025
24
Engineering Bernoulli Equation (without derivation)
𝛼𝛼 𝑢𝑢
∆
2
2
In a simplified scenario:
𝑃𝑃2
𝑑𝑑𝑑𝑑
+ 𝑔𝑔𝑔 + �
= 𝛿𝛿𝑊𝑊𝑆𝑆 − 𝑙𝑙𝑣𝑣
𝑃𝑃1 𝜌𝜌
α = 2.0 for Poiseuille flow in a circular tube
= 1.0 for turbulent flow in a circular tube
α = 1, 𝛿𝛿𝑊𝑊𝑆𝑆 = 0, 𝑙𝑙𝑣𝑣 = 0, then
𝑢𝑢1 2
𝑃𝑃1
𝑢𝑢2 2
𝑃𝑃2
+ 𝑔𝑔ℎ1 + =
+ 𝑔𝑔ℎ2 +
2
𝜌𝜌
2
𝜌𝜌
I-Ming Hsing, Spring 2025
25
Engineering Bernoulli Equation
• Evaluation of viscous losses (lv):
1.
Pipes:
For full pipes, losses independent of orientation.
Evaluate by applying the energy equation.
- Δh = 0
- ∆
𝛼𝛼 𝑢𝑢 2
2
- δWs = 0
(fully developed flow, const. cross section)
=0
𝑃𝑃2 −𝑃𝑃1
-∴
= −𝑙𝑙𝑣𝑣
𝜌𝜌
(no pumps, shaft work)
𝑜𝑜𝑜𝑜
- From friction factor definition:
2 𝑣𝑣 2 𝐿𝐿𝑓𝑓𝐹𝐹
- 𝑙𝑙𝑣𝑣 =
D
∆𝑝𝑝
𝑙𝑙𝑣𝑣 =
𝜌𝜌
𝑓𝑓𝐹𝐹 =
|∆𝑃𝑃|
𝐷𝐷
�
2𝜌𝜌 𝑣𝑣 2 𝐿𝐿
𝑣𝑣 2 = 𝑢𝑢 2
I-Ming Hsing, Spring 2025
26
Engineering Bernoulli Equation
• Evaluation of viscous losses (lv):
2.
Fittings (valves, elbows, changes in pipe diameter) :
1
𝑙𝑙𝑣𝑣 =
2
𝑣𝑣 2 𝐾𝐾𝑓𝑓
- where Kf: constant , determined from Exp. for each fitting
- <v>2 refer to downstream velocity if change of cross section at
fitting
I-Ming Hsing, Spring 2025
Circulatory system uses four
major physical principles
C: compliance
∆P: pressure drop
R: resistance
4
Δ𝑃𝑃 128𝜇𝜇𝜇𝜇
=
𝑅𝑅 =
𝑄𝑄
𝜋𝜋𝐷𝐷4
Qv: volumetric flow rate
I-Ming Hsing, Spring 2025
28
Blood Vessels Have Different Values of Compliance C
The large arteries have thick walls, which are not as compliant
as the more distensible veins. In humans: Cv/Ca ~ 19
Anatomy & Physiology
Pearson, 2013
I-Ming Hsing, Spring 2025
29
How our lung works: Expansion of the thoracic cavity during inspiration
There is no rigid structural connection between the lungs and
the chest wall. The lungs mostly “float” in the thoracic cavity
I-Ming Hsing, Spring 2025
30
Compliance of the isolated Lung
Can you now explain why an inhaler contains salt?
I-Ming Hsing, Spring 2025
31
Pulse Pressure Depends on the Stroke
Volume & Compliance of the Arteries
Feher - Quantitative Human Physiology
I-Ming Hsing, Spring 2025
The Cardiac Cycle
I-Ming Hsing, Spring 2025
Pressure-volume loop of the cardiac cycle
I-Ming Hsing, Spring 2025
1
CENG 2220
Fluid Statics: Manometry, Buoyancy,
Surface Tension
I-Ming Hsing, Spring 2025
2
Pressure in Static fluids
Pressure is a force per unit area
𝑛𝑛
𝐹𝐹𝑝𝑝
𝑛𝑛: unit vector, normal, pointing outward
𝐹𝐹𝑝𝑝 : pressure force
a) pressure forces only act normal to surfaces
b) Positive pressures are compressive (rather than tensile)
c) Pressure is isotropic. Ie., the pressure P has a single value at any
point in a fluid and tend to act equally in all directions
The pressure itself is a scalar, whereas the pressure force acting on a
surface is a vector. The orientation of the surface determines the
direction of the force vector, but not the value of P.
I-Ming Hsing, Spring 2025
3
Hydrostatics
a) To determine the pressure distribution in the fluid at rest
b) Apply knowledge of a) to determine forces on submerged solids
c) Direction of pressure force is normal to surface
𝑛𝑛
𝐹𝐹𝑝𝑝
𝑛𝑛: unit vector, normal, pointing outward
𝐹𝐹𝑝𝑝 : pressure force
Force balance on a fluid element
Conditions:
𝑧𝑧
• The fluid is at rest 𝑣𝑣 = 0
• The viscous shear forces = 0
• The only forces to be considered:
𝑘𝑘
pressure and gravity
• No acceleration, i.e. all force components
(x,y,z) are exactly balanced
𝑗𝑗
𝑖𝑖
𝑦𝑦
𝑖𝑖, 𝑗𝑗, 𝑘𝑘 are unit vectors in 𝑥𝑥, 𝑦𝑦, 𝑧𝑧
directions respectively
𝑥𝑥
I-Ming Hsing, Spring 2025
4
• z component
𝑝𝑝 𝑥𝑥, 𝑦𝑦, 𝑧𝑧 ∆𝑥𝑥∆𝑦𝑦 − 𝑝𝑝 𝑥𝑥, 𝑦𝑦, 𝑧𝑧 + ∆𝑧𝑧 ∆𝑥𝑥∆𝑦𝑦 − 𝜌𝜌∆𝑥𝑥∆𝑦𝑦∆𝑧𝑧𝑧𝑧 = 0
Divided by ∆𝑥𝑥∆𝑦𝑦∆𝑧𝑧,
𝑝𝑝 𝑧𝑧 − 𝑝𝑝 𝑧𝑧 + ∆𝑧𝑧
− 𝜌𝜌𝑔𝑔 = 0
∆𝑧𝑧
𝜕𝜕𝜕𝜕
lim −
− 𝜌𝜌𝑔𝑔 = 0
∆𝑧𝑧→0
𝜕𝜕𝜕𝜕
𝐹𝐹𝑝𝑝𝑝𝑝 =
Or in index notation:
𝜕𝜕𝜕𝜕
−
− 𝜌𝜌𝑔𝑔 = 0;
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
−
= 𝜌𝜌𝑔𝑔
𝜕𝜕𝜕𝜕
−𝑝𝑝 𝑥𝑥, 𝑦𝑦, 𝑧𝑧
∆𝑦𝑦∆𝑧𝑧𝑖𝑖
𝐹𝐹𝑝𝑝𝑝𝑝 = −𝑝𝑝(𝑥𝑥, 𝑦𝑦, 𝑧𝑧 + ∆𝑧𝑧)∆𝑥𝑥∆𝑦𝑦𝑘𝑘
∆𝑧𝑧
∆𝑥𝑥
∆𝑦𝑦
𝐹𝐹𝑝𝑝𝑝𝑝 = 𝑝𝑝(𝑥𝑥, 𝑦𝑦, 𝑧𝑧)∆𝑥𝑥∆𝑦𝑦𝑘𝑘
𝐹𝐹𝑔𝑔 = 𝑚𝑚𝑔𝑔 = −𝜌𝜌∆𝑥𝑥∆𝑦𝑦∆𝑧𝑧𝑧𝑧𝑘𝑘
I-Ming Hsing, Spring 2025
5
• y component
𝑝𝑝 𝑦𝑦 ∆𝑥𝑥∆𝑧𝑧 − 𝑝𝑝 𝑦𝑦 + ∆𝑦𝑦 ∆𝑥𝑥∆𝑧𝑧 = 0
Divided by ∆𝑥𝑥∆𝑦𝑦∆𝑧𝑧 and taking limit ∆𝑦𝑦 → 0
𝜕𝜕𝜕𝜕
−
=0
𝜕𝜕𝑦𝑦
• x component
𝜕𝜕𝜕𝜕
Similarly,
=0
−
𝜕𝜕𝑥𝑥
• In stagnant fluid only the z component of the pressure balance is non-zero: 𝑝𝑝 =
𝑝𝑝 𝑧𝑧
𝑑𝑑𝑝𝑝
z
= −𝜌𝜌𝑔𝑔
𝑑𝑑𝑧𝑧
where
𝑔𝑔⃑
• In generalized terms:
∇𝑝𝑝 = −𝜌𝜌𝑔𝑔= 𝜌𝜌𝑔𝑔𝑧𝑧 =𝜌𝜌𝑔𝑔⃑
𝑔𝑔⃑ = 𝑖𝑖𝑔𝑔𝑥𝑥 + 𝑗𝑗𝑔𝑔𝑦𝑦 + 𝑘𝑘 𝑔𝑔𝑧𝑧 ,𝑔𝑔𝑥𝑥 = 𝑔𝑔𝑦𝑦 = 0, 𝑔𝑔𝑧𝑧 = −𝑔𝑔
∇ is a vector operator:
𝜕𝜕
𝜕𝜕
𝜕𝜕
∇≡
𝑖𝑖 +
𝑗𝑗 + 𝑘𝑘
∇𝑝𝑝 = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑝𝑝 (gradient of pressure)
𝜕𝜕𝜕𝜕
𝜕𝜕𝑦𝑦
𝜕𝜕𝑧𝑧
I-Ming Hsing, Spring 2025
6
Integration of hydrostatic equation
z
𝑃𝑃0 →
Fluid
𝜌𝜌
And hence,
𝑃𝑃 = −𝜌𝜌𝑔𝑔𝑧𝑧 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑃𝑃 − 𝑃𝑃0 = −𝜌𝜌𝑔𝑔𝑔𝑔
𝑃𝑃 > 𝑃𝑃0 𝑓𝑓𝑓𝑓𝑓𝑓 𝑧𝑧 < 0
At the bottom:
𝑃𝑃 = 𝑃𝑃1 + 𝜌𝜌𝑔𝑔ℎ1 (left)
= 𝑃𝑃2 + 𝜌𝜌𝑔𝑔ℎ2 (right)
𝑃𝑃1 − 𝑃𝑃2 = 𝜌𝜌𝑔𝑔 ℎ2 − ℎ1 = 𝜌𝜌𝑔𝑔ℎ
I-Ming Hsing, Spring 2025
7
Integration of hydrostatic equation
• Two fluids:
In (1):
In (2):
𝑃𝑃 = 𝑃𝑃0 − 𝜌𝜌1 𝑔𝑔𝑔𝑔
𝑃𝑃 = 𝑃𝑃0 + 𝜌𝜌1 𝑔𝑔ℎ1
𝑃𝑃 = 𝑃𝑃 −ℎ1 − 𝜌𝜌2 𝑔𝑔(𝑧𝑧 + ℎ1 )
𝑎𝑎𝑎𝑎 − ℎ1 < 𝑧𝑧 < 0
𝑎𝑎𝑎𝑎 𝑧𝑧 = −ℎ1
𝑎𝑎𝑎𝑎 −ℎ1 − ℎ2 < 𝑧𝑧 < −ℎ1
𝑃𝑃 = 𝑃𝑃0 + 𝜌𝜌1 𝑔𝑔ℎ1 − 𝜌𝜌2 𝑔𝑔(𝑧𝑧 + ℎ1 )
𝑃𝑃 = 𝑃𝑃0 + 𝑔𝑔 𝜌𝜌1 ℎ1 + 𝜌𝜌2 ℎ2
𝑎𝑎𝑎𝑎 𝑧𝑧 = −ℎ2
For two or more fluids, stop
at each interface (change ρ).
I-Ming Hsing, Spring 2025
8
Buoyancy: “the resultant force exerted on a body
by fluid at rest”
~Archimedia (287-212 BC) theorem
• Z component pressure force on top:
𝐹𝐹𝑝𝑝𝑝𝑝 = −𝑃𝑃2 ∆𝑥𝑥∆𝑦𝑦 = −(𝑃𝑃0 + 𝜌𝜌𝜌𝜌ℎ2 )∆𝑥𝑥∆𝑦𝑦
• Z component pressure force on bottom:
𝐹𝐹𝑝𝑝𝑝𝑝 = 𝑃𝑃1 ∆𝑥𝑥∆𝑦𝑦 = (𝑃𝑃0 + 𝜌𝜌𝜌𝜌ℎ1 )∆𝑥𝑥∆𝑦𝑦
• Net pressure force:
𝜌𝜌𝜌𝜌(ℎ1 − ℎ2 )∆𝑥𝑥∆𝑦𝑦
• In differential terms:
𝜌𝜌𝜌𝜌 ℎ1 − ℎ2 𝑑𝑑𝑥𝑥𝑑𝑑𝑦𝑦
• Integrate:
� 𝜌𝜌𝜌𝜌 ℎ1 − ℎ2 𝑑𝑑𝑥𝑥𝑑𝑑𝑦𝑦 = 𝜌𝜌𝜌𝜌 � ℎ1 − ℎ2 𝑑𝑑𝑥𝑥𝑥𝑥𝑥𝑥 = 𝜌𝜌𝜌𝜌𝜌𝜌
𝑥𝑥𝑦𝑦
• Buoyancy force:
𝑥𝑥𝑦𝑦
𝐹𝐹𝐵𝐵 = 𝜌𝜌𝜌𝜌𝜌𝜌𝑘𝑘;
𝐹𝐹𝐵𝐵 = 𝜌𝜌𝜌𝜌𝜌𝜌 = 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
I-Ming Hsing, Spring 2025
9
Pascal's law for pressure at a point (Pressure
is isotropic)
Consider a static force balance on the fluid in a small wedge-shaped control volume,
as shown in figure. We begin by assuming that pressure has some value Py when it
acts on a surface perpendicular to the y-axis, and some potentially different value Pz
when it acts on a surface perpendicular to the z-axis. The pressure acting on the top
surface of the wedge (inclined at angle α) is denoted by P. What we wish to show is
that all of these P's are in fact identical. That is, the pressure at a point has single
scalar value, independent of directional considerations.
Considering the fact that pressure always acts normal to a surface, the zcomponent of the force on the wedge is
∆𝑥𝑥∆𝑦𝑦
∆𝑥𝑥∆𝑦𝑦∆𝑧𝑧
−𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝛼𝛼
+ 𝑃𝑃𝑧𝑧 ∆𝑥𝑥∆𝑦𝑦 − 𝜌𝜌𝜌𝜌
=0
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
(A1)
In the first term in equation A1, -P cosα is the z-component of the normal stress on
the top surface and (Δx Δy)/cosα is the area of the top surface. (The same net result
is obtained by taking the pressure times the projected area, Δx Δy.) Dividing all
terms by Δx Δy and rearranging,
𝜌𝜌𝜌𝜌
(A2)
−𝑃𝑃 + 𝑃𝑃𝑧𝑧 =
∆𝑧𝑧
I-Ming Hsing, Spring 2025
2
10
Pascal's law for pressure at a point (Pressure
is isotropic)
For a sufficiently small control volume, such that Δz -> 0, we conclude from equation
(A2) that
𝑃𝑃 = 𝑃𝑃𝑧𝑧
(A3)
The y-component of the force balance involves the pressures on the top and righthand surfaces, but not gravity:
∆𝑥𝑥∆𝑦𝑦
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝛼𝛼
− 𝑃𝑃𝑦𝑦 ∆𝑥𝑥∆𝑧𝑧 = 0
(A4)
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
Dividing by Δx and rearranging,
𝑃𝑃𝑦𝑦 𝑠𝑠𝑠𝑠𝑠𝑠𝛼𝛼 ∆𝑦𝑦
∆𝑦𝑦 ∆𝑧𝑧 ∆𝑦𝑦
=
= 𝑡𝑡𝑡𝑡𝑡𝑡𝛼𝛼
=
=1
𝑃𝑃 𝑐𝑐𝑐𝑐𝑐𝑐𝛼𝛼 ∆𝑧𝑧
∆𝑧𝑧 ∆𝑦𝑦 ∆𝑧𝑧
(A5)
Rotating the wedge by 90oC within the x-y plane would lead to a similar conclusion
regarding the pressure acting along the x-axis (Px). Accordingly,
𝑃𝑃 = 𝑃𝑃𝑥𝑥 = 𝑃𝑃𝑦𝑦 = 𝑃𝑃𝑧𝑧
(A6)
This shows that the pressure has a single scalar value at any point in a fluid, and
therefore directional subscripts are unnecessary.
I-Ming Hsing, Spring 2025
11
Pressure measurement by manometer
1. U-tube manometer
2. A manometer for pressure measurement
3. U-tube with inclined leg
I-Ming Hsing, Spring 2025
12
𝑛𝑛
Pressure Forces
•
𝐹𝐹𝑝𝑝
𝑛𝑛: unit vector, normal, pointing outward
𝐹𝐹𝑝𝑝 : pressure force
Pressure is scalar while pressure force is a vector. As pressure is compressive in
nature, pressure force could be defined as a stress vector –nP, where n is a unit
vector that normal to the surface and points toward the fluid. Multiplying p by –n
creates a vector with the proper direction. The pressure force on a differential
area 𝑑𝑑𝐹𝐹𝑝𝑝 = -nPdS. Accordingly, the pressure force on a surface S is:
𝐹𝐹𝑝𝑝 = − � nP𝑑𝑑𝑆𝑆
𝑆𝑆
Where ∫𝑆𝑆 𝑑𝑑𝑑𝑑 denotes integration over the surface. Likewise, ∫𝑉𝑉 𝑑𝑑𝑉𝑉will denote integration over a volume V.
I-Ming Hsing, Spring 2025
13
In class example: pressure force on a rectangular tank
• Setup equation for Fpx, Fpz
I-Ming Hsing, Spring 2025
14
Projected areas
Consider first a vertical flat plate of height H and width W, as shown in Figure (a).
The uniform fluid pressure P0 will exert a force in the horizontal (x) direction equal
to the pressure times the surface area, or Fpx = P0HW. There is no vertical (z)
component of the pressure force in this case (Fpz = 0), because pressure only acts
perpendicular to a surface. For the inclined flat plate of Figure (b), which has area
BW, the pressure force normal to the surface can be resolved into its components,
𝐹𝐹𝑃𝑃𝑃𝑃 = 𝑃𝑃0 𝐵𝐵𝐵𝐵 sin 𝜃𝜃
𝐹𝐹𝑃𝑃𝑃𝑃 = 𝑃𝑃0 𝐵𝐵𝐵𝐵 cos 𝜃𝜃
What is noteworthy here is the relationship between the forces and certain
projected area of the plate. If we project the plate on a plane perpendicular to
the x-axis, we obtain a surface area which we denote as Ax; the corresponding
projection on a plane perpendicular to the z-axis is Az. From Figure b, Ax = HW
and Az = LW. In addition, B = L/cosθ = H/sinθ, so that equations above can be
rewritten as
𝐹𝐹𝑃𝑃𝑃𝑃 = 𝑃𝑃0 𝐻𝐻𝐻𝐻 = 𝑃𝑃0 𝐴𝐴𝑋𝑋
𝐹𝐹𝑃𝑃𝑃𝑃 = 𝑃𝑃0 𝐿𝐿𝐿𝐿 = 𝑃𝑃0 𝐴𝐴𝑍𝑍
Thus, each component of the pressure force equals simply the pressure times the
corresponding projected area. Because the projected areas Ax are the same for
the vertical and the inclined plate, the pressure forces Fpx are also identical.
I-Ming Hsing, Spring 2025
15
Pressure force on immersed objects
� 𝑛𝑛𝑛𝑛𝑑𝑑𝑑𝑑 = � ∇𝑓𝑓𝑑𝑑𝑑𝑑
𝑠𝑠
Divergence theorem of Calculus
𝑉𝑉
𝐹𝐹𝑝𝑝 = − ∫𝑆𝑆 np𝑑𝑑𝑑𝑑 = -∫𝑉𝑉 ∇𝑃𝑃𝑑𝑑𝑑𝑑 = 0(closed surface at constant P)
I-Ming Hsing, Spring 2025
16
Buoyancy force
𝐹𝐹𝑝𝑝 = − ∫𝑆𝑆 np𝑑𝑑𝑑𝑑 = -∫𝑉𝑉 ∇𝑃𝑃𝑑𝑑𝑑𝑑 = − ∫𝑉𝑉 ρ𝑔𝑔𝑑𝑑𝑑𝑑
⃗
= −ρ𝑔𝑔𝑉𝑉
⃗
𝐹𝐹𝐵𝐵 = 𝐹𝐹𝑃𝑃𝑃𝑃 = 𝑒𝑒𝑧𝑧 � −ρ𝑔𝑔𝑉𝑉
⃗ = (𝑒𝑒𝑧𝑧 � 𝑒𝑒𝑧𝑧 �ρg𝑉𝑉 = ρg𝑉𝑉
𝐹𝐹𝑃𝑃𝑃𝑃 = 𝑒𝑒𝑥𝑥 � −ρ𝑔𝑔𝑉𝑉
⃗ = (𝑒𝑒𝑥𝑥 � 𝑒𝑒𝑧𝑧 )ρg𝑉𝑉 = 0
𝐹𝐹𝑃𝑃𝑃𝑃 = 𝑒𝑒𝑦𝑦 � −ρ𝑔𝑔𝑉𝑉
⃗ = (𝑒𝑒𝑦𝑦 � 𝑒𝑒𝑧𝑧 )ρg𝑉𝑉 = 0
I-Ming Hsing, Spring 2025
17
Surface Tension
Surface tension at a fluid-fluid may be viewed either as a force per
unit length or as an energy per unit area J/m2 or N/m). The free
surface of a liquid exhibits a tendency to contract. The surface is
in a state of tension like that of a thin uniformly stretched skin.
We may assume that a particle of liquid near the surface is pulled
towards the interior of the liquid by attractions of neighboring
particles. In thermodynamics, it is the energy required to increase
the interfacial area. In mechanics, it is a force that acts on an
imaginary line or contour within an interface.
Surface tension does not appear at a plane surface; the forces
there are in equilibrium. If the surface is curved surface tension
gives rise to pressure differences which serve to maintain
equilibrium. That is to say, surface tension can become evident
when an interface is not planar, when r varies with position or
when an interface ends at the three-phase contact line
I-Ming Hsing, Spring 2025
18
Surface tension for liquid-air interfaces
I-Ming Hsing, Spring 2025
19
Young-Laplace equation
Looking at a hemispherical shell of infinitesimal thickness as the
control volume that encloses half of the gas-liquid interface. The
surface tension that the surrounding exert on the control volume
is downward in the diagram.
𝜋𝜋𝑅𝑅2 𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 − 𝜋𝜋𝑅𝑅2 𝑃𝑃𝑖𝑖𝑖𝑖 + 2𝜋𝜋𝜋𝜋𝛾𝛾 = 0
2𝛾𝛾
∆𝑃𝑃 = 𝑃𝑃𝑖𝑖𝑖𝑖 − 𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 =
𝑅𝑅
Guess what an analogous result for a cylindrical interface of radius R is?
𝛾𝛾
∆𝑃𝑃 =
𝑅𝑅
The assumption of uniform internal and external pressure requires the bubble (or droplet) to be
small. The pressure variation within each fluid must be negligible relative to ∆P. The gravityinduced pressure variation over a height 2R of liquid is 2𝜌𝜌𝐿𝐿 𝑔𝑔𝑔𝑔. Accordingly, the Young-Laplace
2𝛾𝛾
𝜌𝜌𝐿𝐿 𝑔𝑔𝑅𝑅2
equation requires that 2𝜌𝜌𝐿𝐿 𝑔𝑔𝑅𝑅 ≪ . Ie, the bond number Bo = 𝛾𝛾 ≪ 1
𝑅𝑅
I-Ming Hsing, Spring 2025
20
In class example: Capillary rise
With one end of a narrow tube is immersed in a wetting liquid
(e.g., a glass capillary held upright in water), the liquid rises inside
to a certain final height H. The objective of this example is to
show how H depends on the tube radius (R), surface tension and
contact angle α. The analysis will be done using force balances on
the control volume as shown in the figure. With CV1 the control
surface cuts through the air-liquid interface. The surroundings
therefore exert a surface tension force on the liquid, the upward
component of which is 2𝜋𝜋𝜋𝜋𝛾𝛾 cos 𝛼𝛼. Opposing that is the
gravitational force on the liquid. Can you explain why there is no
net pressure force?
2𝜋𝜋𝜋𝜋𝛾𝛾 cos 𝛼𝛼 − 𝜌𝜌𝜌𝜌𝜋𝜋𝑅𝑅2 𝐻𝐻 = 0
2𝛾𝛾 cos 𝛼𝛼
𝐻𝐻 =
𝜌𝜌𝜌𝜌𝜌𝜌
I-Ming Hsing, Spring 2025
Branching of the tracheobronchial tree
I-Ming Hsing, Spring 2025
Lubricating and tensile effects of a thin film of water
I-Ming Hsing, Spring 2025
Compliance of the isolated lung filled with air or with saline
Compliance measures the ease
of expanding the lungs
I-Ming Hsing, Spring 2025
Predicted instability of the alveoli, without surfactant
Alveolar type II cells secrete lipoprotein material called surface to reduce the surface tension
I-Ming Hsing, Spring 2025
1
CENG 2220
Continuity and Fluid Kinematics
I-Ming Hsing, Spring 2025
2
Continuity Equation: Conservation of Mass
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜𝑜𝑜
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
=
−
𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
• The Mass Conservation Equation can be expressed as:
𝑑𝑑
� 𝜌𝜌𝜌𝜌𝜌𝜌 = − � ρ𝑣𝑣 ⋅ 𝑛𝑛𝑛𝑛𝑛𝑛
𝑑𝑑𝑑𝑑
𝑆𝑆
(1)
𝑉𝑉
• Now, let’s consider the equation of conservation of mass using a “microscopic”
approach, where the control volume is a small, stationary cubical element as shown
below. The control volume has six surfaces of length, Δx, Δy and Δz.
I-Ming Hsing, Spring 2025
3
Conservation of Mass
• Since the control volume element is small, the volume integral in Eq. 1 can
be expressed as:
Net change of mass
𝜕𝜕
𝜕𝜕𝜌𝜌̅
� � 𝜌𝜌𝜌𝜌𝜌𝜌 ≈
Δ𝑥𝑥Δ𝑦𝑦Δ𝑧𝑧
𝜕𝜕𝑡𝑡
𝜕𝜕𝑡𝑡
(2)
𝐶𝐶𝐶𝐶
• where
− � ρ𝑣𝑣 ⋅ 𝑛𝑛𝑑𝑑𝑑𝑑 = −
𝑆𝑆
�
𝑆𝑆1 +𝑆𝑆2 +𝑆𝑆3 +𝑆𝑆4 +𝑆𝑆5 +𝑆𝑆6
ρ𝑣𝑣 ⋅ 𝑛𝑛𝑑𝑑𝑑𝑑
(3)
• S1 and S2 are the pair of parallel faces normal to the x-axis
• S3 and S4 are the pair of parallel faces normal to the y-axis
• S5 and S6 are the pair of parallel faces normal to the z-axis
I-Ming Hsing, Spring 2025
4
Conservation of Mass
• Net rate of mass inflow in x direction (for control surfaces S1 and S2)
− − 𝜌𝜌𝑣𝑣𝑥𝑥 Δ𝑦𝑦Δ𝑧𝑧� + 𝜌𝜌𝑣𝑣𝑥𝑥 Δ𝑦𝑦Δ𝑧𝑧�
𝑥𝑥
𝑥𝑥+Δ𝑥𝑥
= 𝜌𝜌𝑣𝑣𝑥𝑥 Δ𝑦𝑦Δ𝑧𝑧� − 𝜌𝜌𝑣𝑣𝑥𝑥 Δ𝑦𝑦Δ𝑧𝑧�
𝑥𝑥
• Similarly, the net rate of mass inflow in y direction (for control surfaces S3 and S4)
= 𝜌𝜌𝑣𝑣𝑦𝑦 Δ𝑥𝑥Δ𝑧𝑧� − 𝜌𝜌𝑣𝑣𝑦𝑦 Δ𝑥𝑥Δ𝑧𝑧�
𝑦𝑦
𝑦𝑦+Δ𝑦𝑦
𝑧𝑧
𝑧𝑧+Δ𝑧𝑧
𝑥𝑥+Δ𝑥𝑥
(4)
(5)
• The net rate of mass inflow in z direction (for control surfaces S5 and S6)
= 𝜌𝜌𝑣𝑣𝑧𝑧 Δ𝑥𝑥Δ𝑦𝑦� − 𝜌𝜌𝑣𝑣𝑧𝑧 Δ𝑥𝑥Δ𝑦𝑦�
(6)
𝜕𝜕𝜌𝜌̅
Δ𝑥𝑥Δ𝑦𝑦Δ𝑧𝑧 = 𝜌𝜌𝑣𝑣𝑥𝑥 Δ𝑦𝑦Δ𝑧𝑧� − 𝜌𝜌𝑣𝑣𝑥𝑥 Δ𝑦𝑦Δ𝑧𝑧�
∴
𝜕𝜕𝑡𝑡
𝑥𝑥
𝑥𝑥+Δ𝑥𝑥
+ 𝜌𝜌𝑣𝑣𝑦𝑦 Δ𝑥𝑥Δ𝑧𝑧� − 𝜌𝜌𝑣𝑣𝑦𝑦 Δ𝑥𝑥Δ𝑧𝑧�
(7)
• From Eq. (3), we know that ->
Eq. (2) = Eq. (4) + Eq. (5) + Eq. (6)
𝑦𝑦
+ 𝜌𝜌𝑣𝑣𝑧𝑧 Δ𝑥𝑥Δ𝑦𝑦� − 𝜌𝜌𝑣𝑣𝑧𝑧 Δ𝑥𝑥Δ𝑦𝑦�
𝑧𝑧
𝑦𝑦+Δ𝑦𝑦
𝑧𝑧+Δ𝑧𝑧
I-Ming Hsing, Spring 2025
5
Conservation of Mass
• Divide both sides by dV = ΔxΔyΔz and shrink the control volume → 0
𝜌𝜌𝑣𝑣𝑥𝑥 |𝑥𝑥+Δ𝑥𝑥 − 𝜌𝜌𝑣𝑣𝑥𝑥 |𝑥𝑥 𝜕𝜕 𝜌𝜌𝑣𝑣𝑥𝑥
=
lim
Δ𝑥𝑥→0
Δ𝑥𝑥
𝜕𝜕𝑥𝑥
Δ𝑦𝑦→0
Δ𝑧𝑧→0
(8)
(9)
𝜌𝜌� = 𝜌𝜌
𝜕𝜕 𝜌𝜌𝑣𝑣𝑦𝑦
𝜕𝜕𝜌𝜌
𝜕𝜕 𝜌𝜌𝑣𝑣𝑥𝑥
𝜕𝜕 𝜌𝜌𝑣𝑣𝑧𝑧
=−
+
+
⇒
𝜕𝜕𝑡𝑡
𝜕𝜕𝑥𝑥
𝜕𝜕𝑦𝑦
𝜕𝜕𝑧𝑧
• Writing Eq. (10) in a vector format
𝜕𝜕𝜌𝜌
= −∇ ⋅ (𝜌𝜌𝑣𝑣)
𝜕𝜕t
• Vector differential operator ▽ known as “del” is defined in Cartesian Coordinate as:
𝜕𝜕
𝜕𝜕
𝜕𝜕
+ 𝑗𝑗
+ 𝑘𝑘
∇= 𝑖𝑖
𝜕𝜕𝑥𝑥
𝜕𝜕𝑦𝑦
𝜕𝜕𝑧𝑧
(10)
(11)
where i j k are unit vectors in all three coordinate axes
I-Ming Hsing, Spring 2025
6
Conservation of Mass
𝜕𝜕
• At steady state,
= 0 , from Eq. (11)
𝜕𝜕𝜕𝜕
∇ ⋅ 𝜌𝜌𝑣𝑣 = 0
• For isothermal, incompressible fluid (ρ = constant)
Continuity Eq. becomes:
𝜕𝜕vx 𝜕𝜕vy 𝜕𝜕vz
∇ ⋅ v = 0 or
+
+
=0
𝜕𝜕𝜕
𝜕𝜕𝜕
𝜕𝜕𝜕
1 𝜕𝜕
1 𝜕𝜕
𝜕𝜕
𝑟𝑟𝑣𝑣𝑟𝑟 +
𝑣𝑣𝜃𝜃 +
𝑣𝑣𝑧𝑧 = 0
𝑟𝑟 𝜕𝜕𝑟𝑟
𝑟𝑟 𝜕𝜕𝜃𝜃
𝜕𝜕𝑧𝑧
in Cartesian Coordinates
in Cylindrical(r, θ, z)Coordinates
I-Ming Hsing, Spring 2025
7
Rates of Change For Moving Observers
𝑓𝑓 𝑥𝑥 + 𝑢𝑢𝑥𝑥 ∆𝑡𝑡, 𝑦𝑦 + 𝑢𝑢𝑦𝑦 ∆𝑡𝑡, 𝑧𝑧 + 𝑢𝑢𝑧𝑧 ∆𝑡𝑡, 𝑡𝑡 + ∆𝑡𝑡 − 𝑓𝑓(𝑥𝑥, 𝑦𝑦, 𝑧𝑧, 𝑡𝑡)
𝑑𝑑𝑑𝑑
( ) ≡ lim
∆𝑡𝑡→0
𝑑𝑑𝑡𝑡
∆𝑡𝑡
𝑑𝑑𝑑𝑑
𝜕𝜕𝑓𝑓
𝜕𝜕𝑓𝑓
𝜕𝜕𝑓𝑓
𝜕𝜕𝑓𝑓
( )𝑢𝑢 ≡ ( )𝑥𝑥,𝑦𝑦,𝑧𝑧 + 𝑢𝑢𝑥𝑥 ( )𝑦𝑦,𝑧𝑧,𝑡𝑡 + 𝑢𝑢𝑦𝑦 ( )𝑧𝑧,𝑥𝑥,𝑡𝑡 + 𝑢𝑢𝑧𝑧 ( )𝑥𝑥,𝑦𝑦,𝑡𝑡 =
𝑑𝑑𝑡𝑡
𝜕𝜕𝜕𝜕
𝜕𝜕𝑥𝑥
𝜕𝜕𝑦𝑦
𝜕𝜕𝑧𝑧
For an observer moving at velocity u, the apparent rate change of f as (df/dt)u
𝜕𝜕𝑓𝑓
𝜕𝜕𝜕𝜕
+ (u � ∇𝑓𝑓)
Of particular interest is an observer moving with the fluid. The rate change for u = v is
𝐷𝐷𝐷𝐷
( )
𝐷𝐷𝑡𝑡
𝜕𝜕𝑓𝑓
≡( )
𝜕𝜕𝜕𝜕
𝜕𝜕𝑓𝑓
+ 𝑣𝑣𝑥𝑥 ( )
𝜕𝜕𝑥𝑥
𝜕𝜕𝑓𝑓
+ 𝑣𝑣𝑦𝑦 ( )
𝜕𝜕𝑦𝑦
𝜕𝜕𝑓𝑓
+ 𝑣𝑣𝑧𝑧 ( )
𝜕𝜕𝑧𝑧
=
𝜕𝜕𝑓𝑓
𝜕𝜕𝜕𝜕
+ (v � ∇𝑓𝑓)
The differential operator D/Dt is called the materials derivative (or substantial derivative). A streamline is the path that a small
element of fluid follows. The rate of change for an observer moving with the fluid is the rate change along a streamline.
I-Ming Hsing, Spring 2025
8
In Class example:
Temperature changes sensed by a weather ballon. It is desired to calculate the rate of
temperature change that would be measured by an instrument on a ballon that is rising due
to its buoyance and also being carried by the wind.
The vertical and horizontal velocity components of the ballon are assumed to be constant at
2 m/s and 5 m/s, respectively. Suppose that the atmospheric temperature is steady and
varies with height above the Earth’s surface (z) according to: T(z) = To – G*z, T0 = 15 degree
C, G = 6.5 degree C/km.
What is the observed rate of temperature change?
I-Ming Hsing, Spring 2025
9
Conservation of Mass
• We can make further manipulation to Eq. (11), the continuity equation
𝜕𝜕𝜌𝜌
+ ∇ ⋅ (𝜌𝜌𝑣𝑣) = 0
𝜕𝜕𝑡𝑡
𝜕𝜕𝜌𝜌
+ 𝜌𝜌∇ ⋅ 𝑣𝑣 + 𝑣𝑣 ⋅ ∇𝜌𝜌 = 0
𝜕𝜕𝑡𝑡
𝜕𝜕𝜌𝜌
𝜕𝜕𝜌𝜌
𝜕𝜕𝜌𝜌
𝜕𝜕𝜌𝜌
𝜕𝜕𝜌𝜌 𝐷𝐷𝐷𝐷
(in Cartesian Coordinate)
+ 𝑣𝑣 ⋅ ∇𝜌𝜌 =
+ 𝑣𝑣𝑥𝑥
+ 𝑣𝑣𝑦𝑦
+ 𝑣𝑣𝑧𝑧
=
𝜕𝜕𝑡𝑡
𝜕𝜕𝑡𝑡
𝜕𝜕𝑥𝑥
𝜕𝜕𝑦𝑦
𝜕𝜕𝑧𝑧 𝐷𝐷𝐷𝐷
where
• Eq. (12) can be rewritten as
𝐷𝐷
𝜕𝜕()
𝜕𝜕()
𝜕𝜕()
𝜕𝜕()
=
+ 𝑣𝑣𝑥𝑥
+ 𝑣𝑣𝑦𝑦
+ 𝑣𝑣𝑧𝑧
termed as material derivative
𝐷𝐷𝐷𝐷
𝜕𝜕𝑡𝑡
𝜕𝜕𝑥𝑥
𝜕𝜕𝑦𝑦
𝜕𝜕𝑧𝑧
or substantial derivative
𝐷𝐷𝐷𝐷
= −𝜌𝜌∇ ⋅ 𝑣𝑣
𝐷𝐷𝐷𝐷
I-Ming Hsing, Spring 2025
10
Rate of Strain
• Fluid deforms continuously when subjected to
shearing forces.
• Two material points in a simple shear flow. With point 1
stationary and point 2 moving to the right, the distance L(t)
increases with time and the angle θ(t) decreases.
• Basic types of fluid motion in two dimensions can be
described as a combination of (a) uniform translation, (b)
rigid body rotation, (c) simple shear, and (d) dilation. The
positions of fluid elements at times t and t+∆t are shown
by solid and dashed lines, respectively. Deformation occurs
in (c) and (d) only. They are distinguished in part by the fact
that simple shear is “volume preserving” and pure dilation
is “shape preserving”.
I-Ming Hsing, Spring 2025
11
In class example: Planar Stagnation Flow
Consider a two-dimensional incompressible flow where vz=0, if vy= - Cy where C>0, how about vx
I-Ming Hsing, Spring 2025
12
Stream Function and Streamlines
The stream function (𝜓𝜓) can be used to visualize a velocity field. It is applicable to
incompressible flow that are bidirectional.
For planar flows in Cartesian coordinates, stream function 𝜓𝜓 x, y is defined as
Continuity equation for 2-D incompressible flow
is automatically fulfilled,
∇ � 𝑣𝑣 = 0
∂ψ
∂ψ
vx = +
, vy = −
∂y
∂x
𝜕𝜕𝑣𝑣𝑥𝑥 𝜕𝜕𝑣𝑣𝑦𝑦
+
=0
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
In a steady flow, the rate of change of f seen by an observer moving at the local fluid v:
𝑣𝑣 ⋅ ∇f
𝜕𝜕𝜓𝜓
𝜕𝜕𝜕𝜕
𝑣𝑣 � ∇𝜓𝜓 = 𝑣𝑣𝑥𝑥
+ 𝑣𝑣𝑦𝑦
=0
𝜕𝜕𝜕𝜕
𝜕𝜕𝑦𝑦
Thus, 𝜓𝜓 is constant along a given streamline.
I-Ming Hsing, Spring 2025
13
Streamlines
Constant values of 𝜓𝜓 correspond to streamlines. A streamline is a curve that is tangent everywhere
to the velocity vector. A streamline plot indicates relative flow speed as well as direction.
If the interval in 𝜓𝜓 between each pair of streamlines is the same, the distance
between streamlines is inversely proportional to the local velocity.
If the dimension in the z direction is W, the total flow rate
across the plane shown by the dashed line is:
𝑦𝑦2
𝑦𝑦2
𝜕𝜕𝜕𝜕
𝑄𝑄 = 𝑊𝑊 � 𝑣𝑣𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑊𝑊 �
𝑑𝑑𝑑𝑑 = 𝑊𝑊 𝜓𝜓2 − 𝜓𝜓1
𝑦𝑦1
𝑦𝑦1 𝜕𝜕𝜕𝜕
If the 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 𝜓𝜓 is constant, the flow rate between
any pair of streamline is the same. Closer spacings (small
areas for flow) must then correspond to larger velocities.
I-Ming Hsing, Spring 2025
14
Stream function for planar or axisymmetric flow
I-Ming Hsing, Spring 2025
15
In class example: Planar Stagnation Flow part II
Based on the in-class example in Part I, determine stream function and plot streamlines
I-Ming Hsing, Spring 2025
16
Appendix B, Introductory Transport Phenomena, RB Bird, WE Stewart, EN Lightfoot, DJ Klingenberg, Wiley 2015
I-Ming Hsing, Spring 2025
1
CENG 2220
Stress and Conservation of Momentum
I-Ming Hsing, Spring 2025
2
Stress Vector and Stress Tensor
The stress, or force per unit area, is a vector that varies from point to point on a surface.
Stress is undefined until the orientation of the surface is stated.
Consider a point located on a test surface that is normal to 𝑒𝑒𝑧𝑧
𝑠𝑠(𝑒𝑒𝑧𝑧 ) = force/area on an i surface that is exerted by the fluid
on the side toward which 𝑒𝑒𝑖𝑖 points.
𝑠𝑠(𝑒𝑒𝑥𝑥 ) = 𝜎𝜎𝑥𝑥𝑥𝑥 𝑒𝑒𝑥𝑥 + 𝜎𝜎𝑥𝑥𝑦𝑦 𝑒𝑒𝑦𝑦 + 𝜎𝜎𝑥𝑥𝑧𝑧 𝑒𝑒𝑧𝑧
𝑠𝑠(𝑒𝑒𝑦𝑦 ) = 𝜎𝜎𝑦𝑦𝑥𝑥 𝑒𝑒𝑥𝑥 + 𝜎𝜎𝑦𝑦𝑦𝑦 𝑒𝑒𝑦𝑦 + 𝜎𝜎𝑦𝑦𝑦𝑦 𝑒𝑒𝑧𝑧
𝑠𝑠(𝑒𝑒𝑧𝑧 ) = 𝜎𝜎𝑧𝑧𝑥𝑥 𝑒𝑒𝑥𝑥 + 𝜎𝜎𝑧𝑧𝑧𝑧 𝑒𝑒𝑦𝑦 + 𝜎𝜎𝑧𝑧𝑧𝑧 𝑒𝑒𝑧𝑧
Stress at an arbitrary surface
𝑎𝑎 � 𝜏𝜏 = � � 𝑎𝑎𝑖𝑖 𝜏𝜏𝑖𝑖𝑖𝑖 𝑒𝑒𝑗𝑗
𝑖𝑖
𝑗𝑗
We defined a 3x3 array with elements 𝜎𝜎𝑖𝑖𝑖𝑖 = 𝑠𝑠𝑗𝑗 (𝑒𝑒𝑖𝑖 ) in 𝜎𝜎𝑖𝑖𝑖𝑖 the first
subscript refers to the orientation of the test surface and the second
refers to the direction of the resultant force.
𝑠𝑠(𝑒𝑒𝑖𝑖 ) = � 𝜎𝜎𝑖𝑖𝑖𝑖 𝑒𝑒𝑗𝑗
𝑗𝑗
The stress vector for any coordinate surface is related to the stress tensor:
𝑠𝑠(𝑒𝑒𝑖𝑖 ) = 𝑒𝑒𝑖𝑖 � 𝜎𝜎
I-Ming Hsing, Spring 2025
3
Conservation of Linear Momentum
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑖𝑖𝑖𝑖 −
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜𝑜𝑜 + ∑ 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
• The X-Momentum Conservation Equation can be expressed as:
𝑑𝑑
� 𝜌𝜌𝑣𝑣𝑥𝑥 𝑑𝑑𝑑𝑑 = − � 𝜌𝜌𝑣𝑣𝑥𝑥 𝑣𝑣 � 𝑛𝑛𝑑𝑑𝑑𝑑 + 𝐵𝐵̇
𝑑𝑑𝑑𝑑
𝑉𝑉
where 𝐵𝐵̇ = ∑ forces in x direction
𝑆𝑆
= 𝐹𝐹𝑆𝑆𝑥𝑥 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 + 𝐹𝐹𝑔𝑔𝑔𝑔 (𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓)
I-Ming Hsing, Spring 2025
4
Fluid Statistics
Pressure force acting on the submerged particle of arbitrary shape
𝑑𝑑𝑑𝑑
= −𝜌𝜌𝜌𝜌
𝑑𝑑𝑑𝑑
𝑃𝑃 = 𝑃𝑃𝑜𝑜 − 𝜌𝜌𝜌𝜌𝜌𝜌
𝐹𝐹𝑝𝑝 = � 𝑛𝑛 � 𝜎𝜎𝜎𝜎𝜎𝜎 = � 𝑛𝑛 � (−𝑝𝑝𝑝𝑝)𝑑𝑑𝑑𝑑 = − � ∇𝑝𝑝𝑑𝑑𝑉𝑉
𝑆𝑆
𝑆𝑆
𝑉𝑉
I-Ming Hsing, Spring 2025
5
• Using the same control volume we defined earlier with lengths of Δx, Δy, Δz in coordinate axes
respectively, then
𝜕𝜕𝜌𝜌𝑣𝑣𝑥𝑥
Δ𝑥𝑥Δ𝑦𝑦Δ𝑧𝑧
rate of change of x momentum =
𝜕𝜕𝑡𝑡
Net rate of x−momentum inflow = − � ρ𝑣𝑣𝑥𝑥 𝑣𝑣 ⋅ 𝑛𝑛𝑛𝑛𝑛𝑛
𝑆𝑆
= ρ𝑣𝑣𝑥𝑥 𝑣𝑣𝑥𝑥 Δ𝑦𝑦Δ𝑧𝑧 � − ρ𝑣𝑣𝑥𝑥 𝑣𝑣𝑥𝑥 Δ𝑦𝑦Δ𝑧𝑧 �
𝑥𝑥
+ ρ𝑣𝑣𝑥𝑥 𝑣𝑣𝑦𝑦 Δ𝑥𝑥Δ𝑧𝑧 � − ρ𝑣𝑣𝑥𝑥 𝑣𝑣𝑦𝑦 Δ𝑥𝑥Δ𝑧𝑧 �
𝑦𝑦
+ ρ𝑣𝑣𝑥𝑥 𝑣𝑣𝑧𝑧 Δ𝑥𝑥Δ𝑦𝑦 � − ρ𝑣𝑣𝑥𝑥 𝑣𝑣𝑧𝑧 Δ𝑥𝑥Δ𝑦𝑦 �
𝑧𝑧
𝑥𝑥+Δ𝑥𝑥
𝑦𝑦+Δ𝑦𝑦
𝑧𝑧+Δ𝑧𝑧
I-Ming Hsing, Spring 2025
6
Conservation of Linear Momentum
• Divide both sides by ΔxΔyΔz and take the limit ofΔx→0,Δy→0, Δz→0
we get:
𝜌𝜌𝜌𝜌𝑥𝑥 𝑣𝑣𝑥𝑥 |𝑥𝑥 − 𝜌𝜌𝜌𝜌𝑥𝑥 𝑣𝑣𝑥𝑥 |𝑥𝑥+Δ𝑥𝑥
𝜕𝜕 𝜌𝜌𝑣𝑣𝑥𝑥 𝑣𝑣𝑥𝑥
=−
lim
Δ𝑥𝑥→0
Δ𝑥𝑥
𝜕𝜕𝑥𝑥
Δ𝑦𝑦→0
Δ𝑧𝑧→0
𝜌𝜌𝑣𝑣𝑥𝑥 = 𝜌𝜌𝜌𝜌𝑥𝑥 (𝑎𝑎𝑎𝑎 𝑡𝑡𝑡𝑡𝑡 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑡 𝑎𝑎 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝)
• It can be rewritten as
𝜕𝜕 𝜌𝜌𝜌𝜌𝑥𝑥 𝑣𝑣𝑦𝑦
𝜕𝜕
𝜕𝜕 𝜌𝜌𝜌𝜌𝑥𝑥 𝑣𝑣𝑥𝑥
𝜕𝜕 𝜌𝜌𝜌𝜌𝑥𝑥 𝑣𝑣𝑧𝑧
+
+
𝜌𝜌𝑣𝑣𝑥𝑥 = −
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
Similar expressions can be obtained for y- and z- momentum.
∑𝐹𝐹𝑥𝑥
+
𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥
(1)
I-Ming Hsing, Spring 2025
7
Forces acting on the control volume: body and surface forces
𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓: 𝐹𝐹𝑔𝑔 = � 𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌
𝑉𝑉
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠:
𝐹𝐹𝑆𝑆 = � 𝑛𝑛 � 𝜎𝜎𝜎𝜎𝜎𝜎
𝑆𝑆
𝜎𝜎 = −𝑝𝑝𝛿𝛿 + 𝜏𝜏
𝛿𝛿is Kronecker delta function: 𝛿𝛿𝑖𝑖𝑖𝑖 = 1, 𝑖𝑖𝑖𝑖 𝑖𝑖 = 𝑗𝑗, 𝛿𝛿𝑖𝑖𝑖𝑖 = 0, 𝑖𝑖 ≠ 𝑗𝑗
Stresses: forces per unit area, acting on control volume surfaces
Stresses are tensors.
Recap: constitutive equation, Newton’s Law of Viscosity
I-Ming Hsing, Spring 2025
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Considering the surface forces in x-direction Fx across control surfaces (6 surfaces, S1 , S2 , S3 , S4 , S5 , S6)
𝐹𝐹𝑥𝑥 = Δ𝑦𝑦Δ𝑧𝑧 𝜎𝜎𝑥𝑥𝑥𝑥 �
𝑥𝑥+Δ𝑥𝑥
− 𝜎𝜎𝑥𝑥𝑥𝑥 �
𝑥𝑥
+ Δ𝑥𝑥Δ𝑧𝑧 𝜎𝜎𝑦𝑦𝑦𝑦 �
𝑦𝑦+Δ𝑦𝑦
Body force in the x direction 𝐹𝐹𝑔𝑔𝑔𝑔 = Δ𝑥𝑥Δ𝑦𝑦Δ𝑧𝑧𝜌𝜌𝑔𝑔𝑥𝑥
𝜕𝜕 𝜌𝜌𝜌𝜌𝑥𝑥 𝑣𝑣𝑦𝑦
𝜕𝜕
𝜕𝜕 𝜌𝜌𝜌𝜌𝑥𝑥 𝑣𝑣𝑥𝑥
𝜕𝜕 𝜌𝜌𝜌𝜌𝑥𝑥 𝑣𝑣𝑧𝑧
+
+
𝜌𝜌𝑣𝑣𝑥𝑥 +
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
=
∑𝐹𝐹𝑥𝑥
𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝜕𝜕𝜎𝜎𝑦𝑦𝑥𝑥 𝜕𝜕𝜎𝜎𝑧𝑧𝑥𝑥
+
+
=
Δ𝑥𝑥→0 Δ𝑥𝑥Δ𝑦𝑦Δ𝑧𝑧
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
lim
Equation (2) becomes
𝜌𝜌
Δ𝑦𝑦→0
Δ𝑧𝑧→0
− 𝜎𝜎𝑦𝑦𝑦𝑦 �
or
𝜌𝜌
𝑧𝑧+Δ𝑧𝑧
− 𝜎𝜎𝑧𝑧𝑧𝑧 �
∑𝐹𝐹𝑥𝑥
𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥
𝜕𝜕 𝜌𝜌𝜌𝜌𝑥𝑥 𝑣𝑣𝑦𝑦
𝜕𝜕
𝜕𝜕 𝜌𝜌𝜌𝜌𝑥𝑥 𝑣𝑣𝑥𝑥
𝜕𝜕 𝜌𝜌𝜌𝜌𝑥𝑥 𝑣𝑣𝑧𝑧
+
+
𝜌𝜌𝑣𝑣 +
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕 𝑥𝑥
𝜕𝜕𝜏𝜏𝑦𝑦𝑥𝑥
𝐷𝐷𝑣𝑣𝑥𝑥
𝜕𝜕𝜏𝜏
𝜕𝜕𝜏𝜏
𝜕𝜕𝑃𝑃
= 𝜌𝜌𝑔𝑔𝑥𝑥 + 𝑥𝑥𝑥𝑥 +
+ 𝑧𝑧𝑥𝑥 −
𝐷𝐷𝐷𝐷
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝑦𝑦
+ Δ𝑥𝑥Δ𝑦𝑦 𝜎𝜎𝑧𝑧𝑧𝑧 �
(2)
𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝜕𝜕𝜎𝜎𝑦𝑦𝑥𝑥 𝜕𝜕𝜎𝜎𝑧𝑧𝑥𝑥
+
+
= 𝜌𝜌𝑔𝑔𝑥𝑥 +
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝐷𝐷𝑣𝑣𝑥𝑥
𝜕𝜕𝑃𝑃
= 𝜌𝜌𝑔𝑔𝑥𝑥 − + 𝑒𝑒𝑥𝑥 � ∇ � 𝜏𝜏
𝐷𝐷𝐷𝐷
𝜕𝜕𝜕𝜕
Similar expressions can be obtained for y-dir, z-dir surface forces
𝑧𝑧
(3)
(4)
I-Ming Hsing, Spring 2025
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Conservation of Momentum Equation
𝐷𝐷𝒗𝒗
= 𝜌𝜌𝒈𝒈 − ∇𝑃𝑃 + ∇ � 𝝉𝝉
𝜌𝜌
𝐷𝐷𝐷𝐷
(5)
Total stress tensor 𝜎𝜎 = 𝜏𝜏 − 𝐼𝐼𝐼𝐼
𝜏𝜏𝑥𝑥𝑥𝑥 − 𝑃𝑃
𝜏𝜏𝑥𝑥𝑦𝑦
𝜎𝜎𝑖𝑖𝑖𝑖 =
𝜏𝜏𝑥𝑥𝑧𝑧
𝜏𝜏𝑦𝑦𝑦𝑦
𝜏𝜏𝑦𝑦𝑦𝑦 − 𝑃𝑃
𝜏𝜏𝑦𝑦𝑧𝑧
𝜏𝜏𝑧𝑧𝑥𝑥
𝜏𝜏𝑧𝑧𝑧𝑧
𝜏𝜏𝑧𝑧𝑧𝑧 − 𝑃𝑃
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Newtonian fluids
• For a fluid with the following properties:
• The stress is symmetric
• The stress at a point in the fluid depends
only on the instantaneous value of the
velocity gradients at the point
• Newtonian fluid
• The flow is laminar throughout
for a Newtonian fluid:
𝜕𝜕𝑣𝑣𝑥𝑥
𝜏𝜏𝑦𝑦𝑦𝑦 = 𝜇𝜇
𝜕𝜕𝜕𝜕
• Using the following simple shearing of fluid between two planes to illustrate the idea of τ
• The motion of a fluid element, initially defined by the dashed figure, can be described as a combination
of (a) translation, (b) rigid body rotation, (c) shear, and (d) pure compression.
• Only the deformation of (c) and (d) give rise to viscous stresses.
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For Newtonian for incompressible fluid (ie ρ constant)
∇ � 𝝉𝝉 = 𝜇𝜇[∇2 𝑣𝑣 + ∇ ∇ � 𝑣𝑣 ]= 𝜇𝜇∇2 𝑣𝑣
Equation (5) becomes
𝐷𝐷𝒗𝒗
𝜌𝜌
= 𝜌𝜌𝒈𝒈 − ∇𝑃𝑃 + 𝜇𝜇∇2 𝑣𝑣
𝐷𝐷𝐷𝐷
𝜏𝜏 = 𝜇𝜇∇𝑣𝑣 + (∇𝑣𝑣)𝑡𝑡
Navier Stokes Equation
I-Ming Hsing, Spring 2025
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Viscous stress components for
Newtonian fluid in Cartesian
Coordinates.
For Newtonian with density constant
𝜏𝜏 = 𝜇𝜇∇𝑣𝑣 + (∇𝑣𝑣)𝑡𝑡
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Appendix B, Introductory Transport Phenomena, RB Bird, WE Stewart, EN Lightfoot, DJ Klingenberg, Wiley 2015
I-Ming Hsing, Spring 2025
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Appendix B, Introductory Transport Phenomena, RB Bird, WE Stewart, EN Lightfoot, DJ Klingenberg, Wiley 2015
I-Ming Hsing, Spring 2025
1
CENG 2220
Application of Navier Stokes Equations
I-Ming Hsing, Spring 2025
2
Navier-Stokes Equations
• Partial differential equations with 4 independent variables (e.g., x,y,z,t
or r,θ,z,t) and 4 dependent variables (e.g., vx,vy,vz,P)
• Together with mass conservation equation, there are 4 scalar
equations (3 components of NS eqns and one mass conservation eqn)
• Solved by integrating the equations of motion and mass conservation
equation with specification of appropriate Boundary Conditions and
Initial conditions (if the flow is time dependent)
• Analytical solutions available only in some simplified flow systems.
Often numerical approach is need to approximate the “real”
solutions.
I-Ming Hsing, Spring 2025
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General Assumptions often made in solving
viscous flow (especially in the scope of CENG2220)
• Newtonian fluid
• Laminar flow
• Steady state flow (independent of time)
• Fully developed flow (end effects neglected)
• Fluid behaves as a continuum
• No slip assumption at the wall
I-Ming Hsing, Spring 2025
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Boundary Conditions often used
i.
No slip condition at stationary wall surface: at solid-fluid interfaces the
fluid velocity equals the velocity with which the surface itself is moving;
i.e., the fluid is assumed to cling to any solid surfaces with which it is in
contact.
ii. At the interface between two immiscible fluids, the fluid stress acting on
one fluid (say πa) is equal in magnitude, but opposite in direction to the
fluid stress acting on another fluid (πb). The conditions here neglect the
interfacial surface tension. In almost all practical cases, the viscous stress
is tangent to the interface so that the pressure and viscous stresses are
separately equal (but opposite) at the interface.
iii. Another way to describe ii. is to say that: at a fluid-fluid interface, both
the velocity tangential stress must be continuous across the interface.
The stress normal to the interface is continuous (if the surface tension is
negligible)
iv. Symmetry Conditions (e.g., the center line of a cylinder)
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Case Study I: Plane Poiseuille Flow– Steady
Laminar Flow b/w Fixed Plates
PROBLEM DESCRIPTION: Consider
an incompressible Newtonian Fluid
flowing between the two horizontal,
infinite, parallel plates in Fig 1.
Assumption/Observation
Consequence
Steady State
𝜕𝜕
=0
𝜕𝜕𝜕𝜕
𝑣𝑣𝑦𝑦 = 𝑣𝑣𝑧𝑧 = 0
Fluid particles move in the x direction
parallel to the plate
No slip Condition at the wall surface
Incompressible flow
𝑣𝑣|𝑦𝑦=ℎ = 0, 𝑣𝑣|𝑦𝑦=−ℎ = 0
∇ � 𝑣𝑣 = 0
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From Continuity Eq. in Cartesian Coordinate:
∇ � 𝑣𝑣 = 0
𝜕𝜕𝑣𝑣𝑥𝑥 𝜕𝜕𝑣𝑣𝑦𝑦 𝜕𝜕𝑣𝑣𝑧𝑧
+
+
=0
𝜕𝜕𝑥𝑥
𝜕𝜕𝑦𝑦
𝜕𝜕𝑧𝑧
𝜕𝜕𝑣𝑣
∵ 𝑣𝑣𝑦𝑦 = 𝑣𝑣𝑧𝑧 = 0 ∴ 𝑥𝑥 = 0 𝑣𝑣𝑥𝑥 𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑜𝑜𝑜𝑜 𝑥𝑥
𝜕𝜕𝑥𝑥
∵ flow field is infinite in the expanse in z-direction
𝜕𝜕𝜕𝜕𝑥𝑥
∴
=0
𝜕𝜕𝜕𝜕
1
2
3
Thus, 𝑣𝑣𝑥𝑥 = 𝑣𝑣𝑥𝑥 𝑦𝑦 only and 𝑣𝑣𝑧𝑧 = 𝑣𝑣𝑧𝑧 = 0
From Navier-Stokes Eq. in Cartesian Coordinate:
2 𝑣𝑣
𝜕𝜕𝜕𝜕
𝜕𝜕
𝑥𝑥
X-comp.
+ 𝜇𝜇
0=−
𝜕𝜕𝜕𝜕
𝜕𝜕𝑦𝑦 2
𝜕𝜕𝜕𝜕
Y-comp.
0=−
− 𝜌𝜌𝜌𝜌
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
Z-comp.
0=−
𝜕𝜕𝜕𝜕
Eq. 5 & 6 can be integrated to yield
𝑃𝑃 = −𝜌𝜌𝜌𝜌𝜌𝜌 + 𝑓𝑓1 (𝑥𝑥)
4
5
6
P is independent of z as
𝜕𝜕𝜕𝜕
=0
𝜕𝜕𝜕𝜕
7
I-Ming Hsing, Spring 2025
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Eq. 4 shows:
𝑑𝑑 2 𝑣𝑣𝑥𝑥
1 𝜕𝜕𝜕𝜕
=
= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 (as LHS depends only on y and RHS depends only on x)
2
𝜇𝜇 𝜕𝜕𝜕𝜕
𝑑𝑑𝑦𝑦
Integrate 8 twice
1 𝜕𝜕𝜕𝜕 2
𝑦𝑦 + 𝐶𝐶1 𝑦𝑦 + 𝐶𝐶2
𝑣𝑣𝑥𝑥 =
2𝜇𝜇 𝜕𝜕𝜕𝜕
Use available B.C to determine C1 & C2
𝑣𝑣𝑥𝑥 +ℎ = 𝑣𝑣𝑥𝑥 −ℎ = 0
1 𝜕𝜕𝜕𝜕 2
ℎ
⇒ 𝐶𝐶1 = 0 𝐶𝐶2 = −
2𝜇𝜇 𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕
(𝑦𝑦 2 −ℎ2 )
∴ 𝑣𝑣𝑥𝑥 =
2𝜇𝜇 𝜕𝜕𝜕𝜕
8
9
10
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Volumetric flow rate 𝑄𝑄 = � 𝑣𝑣𝑥𝑥 𝑦𝑦 𝑑𝑑𝑑𝑑
ℎ
𝐴𝐴
ℎ
1 𝜕𝜕𝜕𝜕
𝑄𝑄 = 𝑊𝑊 � 𝑣𝑣𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑊𝑊 �
(𝑦𝑦 2 −ℎ2 ) 𝑑𝑑𝑑𝑑
2𝜇𝜇 𝜕𝜕𝜕𝜕
−ℎ
−ℎ
2𝑊𝑊ℎ3 𝜕𝜕𝜕𝜕
𝑄𝑄 = −
3𝜇𝜇
𝜕𝜕𝜕𝜕
11
Pressure gradient is –ve, because the pressure decreases in the direction of the flow
∆𝑃𝑃
𝜕𝜕𝜕𝜕
=−
𝑙𝑙
𝜕𝜕𝜕𝜕
2𝑊𝑊ℎ3 ∆𝑃𝑃
12
∴ 𝑄𝑄 =
3𝜇𝜇
𝑙𝑙
𝑄𝑄 ℎ2 ∆𝑃𝑃
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣, 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = =
13
𝐴𝐴 3𝜇𝜇 𝑙𝑙
ℎ2 𝜕𝜕𝜕𝜕
𝑣𝑣𝑥𝑥,𝑚𝑚𝑚𝑚𝑚𝑚 = −
At centerline,
14
Vx,max = 3/2 vmean
2𝜇𝜇 𝜕𝜕𝜕𝜕
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In-class Example: Liquid Film on a Vertical Surface
PROBLEM DESCRIPTION: Consider
a flow problem shown in the left
figure in which the liquid is
bounded on one side by a free
surface and the flow is driven
strictly by a body force (gravity).
I-Ming Hsing, Spring 2025
10
In-class discussion: Liquid Film on a Vertical Surface
Assumption/Observation
Consequence
Steady State
𝜕𝜕
=0
𝜕𝜕𝜕𝜕
The flow is incompressible and Newtonian,
and the flow is isothermal
The flow is laminar and strictly parallel to the
plate. There is no flow in the y and z
directions and no derivatives
𝜕𝜕
𝜕𝜕𝜕𝜕
The liquid film thickness δ is not a function of
position x
Navier-Stokes Equation and continuity
equation applicable (∇ � 𝑣𝑣 = 0)
The model will not hold in the overflow
region. The only velocity component is 𝑣𝑣𝑥𝑥 ,
downstream of the overflow region
𝑣𝑣𝑥𝑥 is a function only of y – a fully developed
flow without end/edge effect
I-Ming Hsing, Spring 2025
11
More examples: Angular Drag Flow Between Cylinders
PROBLEM DESCRIPTION:
Consider an angular drag flow
between two concentric
cylinders. The cylinders are long
compared to their radii: L/R>>1
I-Ming Hsing, Spring 2025
12
In-class discussion: Angular Drag Flow Between Cylinders
Assumption/Observation
Consequence
Steady State
𝜕𝜕
=0
𝜕𝜕𝜕𝜕
The flow is incompressible and Newtonian,
and the flow is isothermal
Navier-Stokes Equation and continuity
equation applicable (∇ � 𝑣𝑣 = 0)
The cylinders are concentric and the flow is
laminar.
The velocity components and pressure will
be independent of angular position θ. All
The cylinders have no motion in the z
direction
There is no flow in the z direction: 𝑣𝑣𝑧𝑧 = 0.
The angular velocity 𝑣𝑣𝜃𝜃 doesn’t vary in the z
direction. Thus 𝑣𝑣𝜃𝜃 depends only on r.
𝜕𝜕
= 0 . There is no radial velocity: vr = 0
𝜕𝜕θ
I-Ming Hsing, Spring 2025
13
More examples: Flow in a cavity
PROBLEM DESCRIPTION:
Suppose that a long and shallow cavity of length L
and depth H is filled with liquid as shown in the
figure. If something at the top surface causes the
liquid there to move from left to right (i.e., if vx >0 at
y = H), a circulating flow is created, as indicated by
the arrows. However, if L/H is large enough, the flow
in the central region will be fully developed. This
problem focuses on that region, centered on x = 0,
where vx = vx(y).
- By choosing an appropriate control volume, show
that vx averaged over the depth of the cavity must
be zero for all x.
- Determine vx(y) for the case in which the top
surface is a solid plate sliding past the cavity at
velocity U in the +x direction
I-Ming Hsing, Spring 2025
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I-Ming Hsing, Spring 2025
15
Appendix B, Introductory Transport Phenomena, RB Bird, WE Stewart, EN Lightfoot, DJ Klingenberg, Wiley 2015
I-Ming Hsing, Spring 2025
16
Appendix B, Introductory Transport Phenomena, RB Bird, WE Stewart, EN Lightfoot, DJ Klingenberg, Wiley 2015
I-Ming Hsing, Spring 2025
17
Appendix B, Introductory Transport Phenomena, RB Bird, WE Stewart, EN Lightfoot, DJ Klingenberg, Wiley 2015
I-Ming Hsing, Spring 2025
1
CENG 2220
Introduction of Mass Transfer
I-Ming Hsing, Spring 2025
2
Mass Transport
• Solute fluxes in a binary mixture A & B
dA
Nix flux of
component i
Convective flux of i &
diffusive flux of i (Jix)
𝜌𝜌𝐴𝐴 = 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝐴𝐴 𝑝𝑝𝑝𝑝𝑝𝑝 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
𝜌𝜌 = 𝜌𝜌𝐴𝐴 + 𝜌𝜌𝐴𝐴
𝜌𝜌𝐴𝐴
𝜔𝜔𝐴𝐴 =
= 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 𝐴𝐴
𝜌𝜌
𝜔𝜔𝐴𝐴 + 𝜔𝜔𝐵𝐵 = 1
∇𝜔𝜔𝐴𝐴 = −∇𝜔𝜔𝐵𝐵
𝐶𝐶𝐴𝐴 = 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝐴𝐴 𝑝𝑝𝑝𝑝𝑝𝑝 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
𝐶𝐶 = 𝐶𝐶𝐴𝐴 + 𝐶𝐶𝐴𝐴
𝐶𝐶𝐴𝐴
𝑥𝑥𝐴𝐴 =
= 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 𝐴𝐴
𝐶𝐶
𝑥𝑥𝐴𝐴 + 𝑥𝑥𝐵𝐵 = 1
∇𝑥𝑥𝐴𝐴 = −∇𝑥𝑥𝐵𝐵
• Average velocity in a binary mixture A & B
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑣𝑣 =
𝜌𝜌𝐴𝐴 𝑣𝑣𝐴𝐴 + 𝜌𝜌𝐵𝐵 𝑣𝑣𝐵𝐵
= 𝜔𝜔𝐴𝐴 𝑣𝑣𝐴𝐴 + 𝜔𝜔𝐵𝐵 𝑣𝑣𝐵𝐵
𝜌𝜌𝐴𝐴 + 𝜌𝜌𝐵𝐵
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑣𝑣 ∗ =
𝐶𝐶𝐴𝐴 𝑣𝑣𝐴𝐴 + 𝐶𝐶𝐵𝐵 𝑣𝑣𝐵𝐵
= 𝑥𝑥𝐴𝐴 𝑣𝑣𝐴𝐴 + 𝑥𝑥𝐵𝐵 𝑣𝑣𝐵𝐵
𝐶𝐶𝐴𝐴 + 𝐶𝐶𝐵𝐵
I-Ming Hsing, Spring 2025
3
• Velocity of each component vi in a mixture
Velocity of
component i
𝑣𝑣𝑖𝑖 = 𝑣𝑣 + 𝑣𝑣𝑑𝑑
Diffusion velocity
Bulk
velocity
• Mass fluxes of component i
(in a fixed coordinate)
𝑛𝑛𝑖𝑖 = 𝜌𝜌𝑖𝑖 𝑣𝑣𝑖𝑖
𝑗𝑗𝑖𝑖∗ = 𝑛𝑛𝑖𝑖 − 𝜌𝜌𝑖𝑖 𝑣𝑣 ∗
𝑗𝑗𝑖𝑖 = 𝜌𝜌𝑖𝑖 𝑣𝑣𝑖𝑖 − 𝑣𝑣 = 𝜌𝜌𝑖𝑖 𝑣𝑣𝑑𝑑 (diffusive mass flux in a moving coordinate)
• Molar fluxes of component i
𝑁𝑁𝑖𝑖 = 𝐶𝐶𝑖𝑖 𝑣𝑣𝑖𝑖
(in a fixed coordinate)
∗
∗
𝐽𝐽𝑖𝑖 = 𝑁𝑁𝑖𝑖 − 𝐶𝐶𝑖𝑖 𝑣𝑣
𝐽𝐽𝑖𝑖 = 𝐶𝐶𝑖𝑖 𝑣𝑣𝑖𝑖 − 𝑣𝑣 = 𝐶𝐶𝑖𝑖 𝑣𝑣𝑑𝑑 (diffusive molar flux in a moving coordinate)
I-Ming Hsing, Spring 2025
4
Let’s primarily focus on a diluted system, applicable to most biological solutions
𝐶𝐶𝑠𝑠 ≫ 𝐶𝐶𝑖𝑖 𝑜𝑜𝑜𝑜 𝑥𝑥𝑠𝑠 ≈ 1 ≫ 𝑥𝑥𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓 𝑖𝑖 = 1,2, … 𝑛𝑛, 𝑖𝑖 ≠ 𝑠𝑠
𝑣𝑣 ∗ ≈ 𝑣𝑣𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎 𝑣𝑣 ≈ 𝑣𝑣𝑠𝑠 , 𝑠𝑠𝑠𝑠 𝑣𝑣 ≈ 𝑣𝑣 ∗
Consequently, 𝐽𝐽𝑖𝑖∗ = 𝐽𝐽𝑖𝑖 , then 𝑁𝑁𝑖𝑖 = 𝐽𝐽𝑖𝑖 + 𝐶𝐶𝑖𝑖 𝑣𝑣𝑠𝑠
Two conclusions:
• The average fluid velocity, whether on a mass or molar basis, is approximately
equal to the solvent velocity
• For a dilute multicomponent solution, the transport of each solute through the
solvent can be studied as though the mixture were binary.
I-Ming Hsing, Spring 2025
5
Conservation of Mass for a Mixture
The General Conservation Equations
Mass balance
In
Mass flow in
 in − m out =
m
Material balance
-
dW
dt
In
Rate of flow of i
into the system
(across surface)
(moles/time)
𝐹𝐹𝑖𝑖𝑖 − 𝐹𝐹𝑖𝑖 + 𝐺𝐺𝑖𝑖 =
𝑉𝑉
Out
Mass flow out
𝐺𝐺𝐺𝐺 = � 𝑟𝑟𝑖𝑖 𝑑𝑑𝑑𝑑
𝑑𝑑𝑁𝑁𝑖𝑖
𝑑𝑑𝑑𝑑
=
𝑚𝑚𝑖𝑖𝑖𝑖
̇
-
Accumulation
Rate of mass accumulation
W
+
Out
Rate of flow of i
out of the system
(across surface)
(moles/time)
Fi0
If ri is not a function of V, then
⇒ Gi = riV
𝑚𝑚𝑜𝑜𝑜𝑜𝑜𝑜
̇
Generation
Rate of generation of i
by chemical reaction
within the system
volume (moles/time)
Gi, Ni
Fi
=
Accumulation
Rate of accumulation
of i within the
system volume
(moles/time)
𝑉𝑉
𝐹𝐹𝑖𝑖𝑖 − 𝐹𝐹𝑖𝑖 + � 𝑟𝑟𝑖𝑖 =
𝑑𝑑𝑁𝑁𝑖𝑖
𝑑𝑑𝑑𝑑
I-Ming Hsing, Spring 2025
6
Constitutive Equation of diffusive flux - Fick’s 1st Law
Species A diffuses into slab of B:
𝑚𝑚̇ 𝐴𝐴𝐴𝐴
𝜔𝜔𝐴𝐴𝐴 − 0
= 𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴
𝐴𝐴
𝑌𝑌
Mass
flow rate
𝑗𝑗𝐴𝐴𝐴𝐴 = −𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴
𝑑𝑑𝜔𝜔𝐴𝐴
𝑑𝑑𝑑𝑑
𝑗𝑗𝐴𝐴 = −𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝜔𝜔𝐴𝐴
𝑗𝑗𝐵𝐵 = −𝜌𝜌𝐷𝐷𝐵𝐵𝐴𝐴 ∇𝜔𝜔𝐵𝐵
𝐽𝐽𝐴𝐴∗ = −𝐶𝐶𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝑋𝑋𝐴𝐴
𝐽𝐽𝐵𝐵∗ = −𝐶𝐶𝐷𝐷𝐵𝐵𝐵𝐵 ∇𝑋𝑋𝐵𝐵
𝑑𝑑𝑣𝑣𝑥𝑥
𝜏𝜏𝑦𝑦𝑦𝑦 = −𝜇𝜇
𝑑𝑑𝑦𝑦
𝜏𝜏𝑦𝑦𝑦𝑦
𝑑𝑑𝑣𝑣𝑥𝑥
= −𝜈𝜈
𝜌𝜌
𝑑𝑑𝑦𝑦
Binary
diffusion
coefficient
Diffusive
mass flux
Schmidt Number
𝑆𝑆𝑆𝑆 =
Kinematic viscosity
𝜈𝜈
𝜇𝜇
=
𝐷𝐷𝐴𝐴𝐴𝐴 𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴
I-Ming Hsing, Spring 2025
7
Fick’s 1st Law of Diffusion for Dilute Solutions
𝐽𝐽𝑖𝑖 = −𝐷𝐷𝑖𝑖𝑖𝑖 ∇𝐶𝐶𝑖𝑖 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝐷𝐷𝑖𝑖𝑖𝑖 is the binary diffusion
coefficient of the solute I in the solvent j
𝑁𝑁𝑖𝑖 = −𝐷𝐷𝑖𝑖𝑖𝑖 ∇𝐶𝐶𝑖𝑖 + 𝐶𝐶𝑖𝑖 𝑣𝑣𝑠𝑠
In a more generic form where the density of the mixture may change
(e.g., gases) 𝐽𝐽𝑖𝑖 = −𝐶𝐶𝐷𝐷𝑖𝑖𝑖𝑖 ∇𝑥𝑥𝑖𝑖
I-Ming Hsing, Spring 2025
8
1-D Diffusion and Fick’s 2nd Law
𝜕𝜕𝐶𝐶𝑖𝑖
𝐴𝐴∆𝑥𝑥 = 𝑁𝑁𝑖𝑖𝑥𝑥|𝑥𝑥 − 𝑁𝑁𝑖𝑖𝑥𝑥|𝑥𝑥+∆𝑥𝑥 𝐴𝐴 + 𝑅𝑅𝑖𝑖 𝐴𝐴∆𝑥𝑥
𝜕𝜕𝑡𝑡
𝑁𝑁𝑖𝑖𝑖𝑖|𝑋𝑋 − 𝑁𝑁𝑖𝑖𝑖𝑖|𝑋𝑋+∆𝑋𝑋
𝜕𝜕𝐶𝐶𝑖𝑖
+ 𝑅𝑅𝑖𝑖
=
𝜕𝜕𝑡𝑡
∆𝑥𝑥
𝑁𝑁𝐴𝐴𝑋𝑋|𝑋𝑋
𝑁𝑁𝑖𝑖𝑋𝑋|𝑋𝑋+∆𝑋𝑋
X
For the case of diffusion only, v
= 0, and no reaction Ri = 0
𝜕𝜕𝐶𝐶𝑖𝑖
𝜕𝜕𝑁𝑁𝑖𝑖𝑖𝑖
=−
+ 𝑅𝑅𝑖𝑖
𝜕𝜕𝑡𝑡
𝜕𝜕𝑥𝑥
𝜕𝜕𝐶𝐶𝑖𝑖
𝜕𝜕 2 𝐶𝐶𝑖𝑖
= 𝐷𝐷𝑖𝑖𝑖𝑖
+ 𝑅𝑅𝑖𝑖
𝜕𝜕𝑡𝑡
𝜕𝜕𝑥𝑥 2
𝜕𝜕𝐶𝐶𝑖𝑖
𝜕𝜕2 𝐶𝐶𝑖𝑖
= 𝐷𝐷𝑖𝑖𝑖𝑖 2
𝜕𝜕𝑡𝑡
𝜕𝜕𝑥𝑥
𝜕𝜕2 𝐶𝐶𝑖𝑖
𝐷𝐷𝑖𝑖𝑖𝑖 2 = 0
𝜕𝜕𝑥𝑥
𝑁𝑁𝑖𝑖 = −𝐷𝐷𝑖𝑖𝑖𝑖 ∇𝐶𝐶𝑖𝑖 + 𝐶𝐶𝑖𝑖 𝑣𝑣𝑠𝑠
Fick’s 2nd Law
at steady state
I-Ming Hsing, Spring 2025
9
Boundary Conditions of Mass Transfer
I-Ming Hsing, Spring 2025
10
Boundary Conditions of Mass Transfer
I-Ming Hsing, Spring 2025
1
CENG 2220
Mass Transfer in Semiconductor Processing
I-Ming Hsing, Spring 2025
2
Silicon Crystal Structure
Amorphous
poly-crystalline
crystalline
• Diamond lattice
• On the 4th column of the periodic
table (AN=14,AW=28.086)
• 4 electrons in its outer band
• Atoms bonded tetrahedrally
with covalent bonds
• Miller index for crystal directionality
I-Ming Hsing, Spring 2025
3
Use of Miller Indices for Crystal Plane
Procedure of miller indexing:
1) set up coordinate axe and normalize the
the intercept of unit cell
2) Invert the intercept values
3) Convert the values to the smallest possible
set of whole numbers
4) Enclose the whole-number in curvilinear bracket
(***) for crystal plane and [***] for crystal direction
Exercises:
(a) (632) plane
(b) (100) plane
(c) (221) plane
I-Ming Hsing, Spring 2025
4
Type of solids
• Depending on the electronics filling of bands, solids can be metals, insulators
or semiconductors
I-Ming Hsing, Spring 2025
5
Material Classification
I-Ming Hsing, Spring 2025
6
Extrinsic semiconductor
Insert dopants in substitional positions in the lattice
• Donors: introduce electrons to conduction band without introducing
holes to valence band
• Acceptors: introduce holes to valence band without introducing electrons
to conduction band
If anyone carrier overwhelms ni => extrinsic semiconductor
Donor in Si
atom from column V (As, P)
Acceptor in Si
atom from column III (B)
I-Ming Hsing, Spring 2025
7
Representation of donor and acceptor states in energy band diagram
Dopant-site binding energy
Near room temperature, all dopants are ionized:
I-Ming Hsing, Spring 2025
8
Dopant Diffusion
• Incorporation of impurity atoms into a semiconductor material leads to the
formation of P-N junctions, conduction channels, and source and drain regions
• a frequently used technique for adding dopants to semiconductor (now largely
replaced by ion-implantation
• Ion implantation is capable of placing impurity atoms within the Si surface, while
a diffusion step often needed for driving impurities to specific depths
• Diffusion of impurity is invoked as other high temperature processing steps occur
I-Ming Hsing, Spring 2025
9
Thermal Diffusion
Applications
• dopant profile control
• tailoring electrical conductivity of a specific area
I-Ming Hsing, Spring 2025
10
Dopant Diffusion
• The diffusion process begins with the deposition of a high concentration of the
desired impurity on the silicon surface through windows etched in the protective
barrier layer.
• At high temperatures (900 to 1200oC), the impurity atoms move from the surface
into the silicon crystal (via substitutional diffusion or interstitial diffusion)
D = Do e
− EA
kT
I-Ming Hsing, Spring 2025
11
Thermal Oxidation of Si
• General properties of SiO2
• Application of SiO2
• Deal-Grove of oxidation
• Oxidation model parameters
I-Ming Hsing, Spring 2025
12
Thermal SiO2
Si + O 2 → SiO 2
Si + 2 H 2O → SiO 2 + 2 H 2
• Excellent insulator
• High breakdown E-field (107 v/cm)
• Si/SiO2 interface is stable and reproducible
• SiO2 is a good mask for common dopants (B,P,As, Sb)
• Very good wet etching selectivity
between Si and SiO2
ρ > 1016 Ω − cm
Eg ≈ 9eV
DSiO 2 << DSi
• Thermal SiO2 is amorphous
SiO2
Si
SiO2
HF dip
Si
Si
I-Ming Hsing, Spring 2025
13
A Microscopic Look of SiO2 Structure
I-Ming Hsing, Spring 2025
14
Thickness of Si consumed during oxidation
Si + O 2 → SiO 2
Let Xox = SiO2,
original surface
Si
Xsi
Xox
Si
thickness formed
2.3x1022 molecules / cm 3
Nox
X Si = X ox •
= X ox •
= 0.46 X ox
22
3
N si
5 x10 atoms / cm
1µm Si oxidized ⇒ 2.17 µmSiO2
Dry process:
Si + O 2 → SiO 2
Wet process:
Si + 2 H 2O → SiO 2 + 2 H 2
I-Ming Hsing, Spring 2025
15
Oxidation system
Horizontal hot wall furnace
Wafer carried in quartz boat
I-Ming Hsing, Spring 2025
16
Advanced thermal oxidation/annealing system
• Rapid Thermal Processing (RTP) system
- single wafer
- low thermal budget
- issue of temperature uniformity
• Batch-furnace system
- multiple wafers
- high thermal budget
I-Ming Hsing, Spring 2025
17
Kinetics of SiO2 growth
oxidant (e.g. O2, or H2O)
Gas diffusion
solid diffusion
reaction front
stagnation gas layer
SiO2
Si
I-Ming Hsing, Spring 2025
18
Deal-Grove Model
Stagnant layer
SiO2
Si
Cg
Cs
Co
Ci
Xox
J1
Gas transport flux
J2
diffusion flux
through SiO2
J3
reaction flux
at interface
I-Ming Hsing, Spring 2025
19
Simplified Deal-Grove Model
Well stirred gas
Cg
SiO2
Si
Cs
Co
Ci
Xox
J1
Gas transport flux
J2
diffusion flux
through SiO2
J3
reaction flux
at interface
I-Ming Hsing, Spring 2025
20
J 1 = J gas = hg (C g − Cs )
where hg is the mass transfer coefficient
 C o − Ci 
∂C

≅ D ⋅ 
J 2 = −D
∂x
 X ox 
where D is the diffusion coefficient of
O2 in SiO2
J 3 = k s ⋅ Ci
where ks is the surface reaction rate constant
To relate Co to the gas phase concentration, we use Henry’s Law
Co = H ⋅ Ps
C * = H ⋅ Pg
Where H is the Henry’s gas constant, Ps is the
partial pressure of the oxidizing species at the oxide
surface, Pg is the partial pressure in the bulk of the gas
phase, Co is the equilibrium concentration of
the oxidizing species in the oxide at the outer surface and C*
is the equilibrium concentration in the bulk of the oxide
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*
J1 = J gas = h(C − Co ),
At steady state:
h=
J1 = J 2 = J 3
hg
HkT
(2 equations, 2 unknowns: Cs, Ci)
C*
C*
Ci =
, J = ks
k k X
k k X
1 + s + s ox
1 + s + s ox
D
h
D
h
Oxide Growth Rate = ?
 dX ox 
J = N1 ⋅ 

 dt 
∆xox
Oxidant molecules required
to form a unit volume of SiO2
J
SiO2
Si
N1=2.3 x 1022 /cm3
for O2 as oxidant
N1=4.6 x 1022 /cm3
for H2O as oxidant
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At t=0,
xi
SiO2
Si
X ox2 + AX ox = B(t + τ )
1 1
A = 2 D( + )
ks h
2 DC *
B=
N1
X i2 + AX i
τ=
B
After time t
xox
SiO2
Si

A
t +τ
X ox =  1 + ( 2
) − 1
2
A / 4B

• large t
X ox → Bt
• small t
B
X ox → t
A
B = parabolic rate coefficient
B/A = linear rate coefficient
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Two ways to Calculate Oxide Thickness Grown by
Thermal Oxidation
E.g.
SiO2
Xi=4000 A
Si
1100 C
33 min
steam
SiO2
Xox
Si
• Method 1: find B & B/A from charts ( or tables or eqns)
solve
X + AX ox = B(t + τ )
2
ox
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Method 2: Use oxidation Charts
Based on
Xi = 0
o
X i = 4000 A ⇒ 24 min
At 1100oC from chart
Total effective oxidation time
(24 + 33) min = 57 min if start with Xi=0
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• Thin oxide (short oxidation time)
B
X ox ≈ (t + τ )
A
*
B C ks
≡
∝ k s ∝ e − E / kT
A
N1
• Thick oxide (long oxidation time)
*
X
2
ox
≈ B (t + τ )
2 DC
B≡
∝ D ∝ e −Q / kT
N1
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Factors Influencing Oxidation Rate
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Factors Influencing Oxidation Rate, Cont.
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Effect of Xi on Wafer Topography
More oxide grown
more Si consumed
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Growth rate dependence on Si substrate orientation
Xox
(111)
(110)
- difference more obvious
for thin oxides
(100)
orientation
Most IC’s made with (100) Si
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Thin Film Deposition: (CVD)
• Formation of a nonvolatile solid film on a substrate by the reaction of gas-phase
precursor (reactant) that contains the required constituents
• Application: deposition of dielectric, polysilicon, and metal films
• Dielectric films including silicon dioxide and silicon nitride are often used as the
isolation mask and passivation layers
• PolySi film is used as the conducting layer, semiconductor, or resistor by proper
doping with different impurities
• Steps involved in a typical CVD process
- gas transport (diffusion)
- absorption
- surface reaction
- desorption
- product transport (diffusion)
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• Two possible reaction routes:
- Homogeneous/gas phase reaction: undesirable, difficult to control, poor
deposited film
- Heterogeneous/surface reaction: desirable, easy to control, good quality film
deposited
• Growth model for Epitaxy, a process to grow a single-crystal layer on a
single-crystal substrate (similar to a CVD process)
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• Gas-phase mass transfer flux:
Jg =
Dg
δ
(Cg − Cs ) = hg (C g − Cs )
Where Dg: effective gas-phase diffusion coefficient
δ: thickness of boundary layer
hg: gas-phase mass transfer coefficient
• Surface reaction flux:
J s = k sCs
where ks: surface reaction rate coefficient = koexp(-Ea/kT)
At steady state,
Cg
J g = J s ⇒ Cs =
ks
1+ ( )
hg
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A review of Boundary Layer Theory
Schematics representations of
(a) stagnant-layer model
(b) boundary-layer model
Jg =
Dg ( C g − C s )
ts
hg =
𝐷𝐷𝑔𝑔
𝑃𝑃
𝑃𝑃𝑔𝑔 and P are the partial pressure of the
diffusing species and the total pressure,
respectively
Dg
ts
 µx 
δ x = tb ( x ) = 

 ρu 
( 2 / L)
δ b = tb =
3(Re) 0.5
ρuL
Re =
𝑃𝑃𝑔𝑔
1.5
∝ 𝑇𝑇
0.5
Where ρ: gas density
µ: gas viscosity
u: gas velocity
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Epitaxial Growth Behavior
The growth rate of an epi-Si v.s.
the free gas flow rate
Crystallinity with respect to
growth rate and growth
temperature
Schematic illustration of
epitaxial growth with adatoms
• Maximum growth rate imposed to ensure the
crystallinity of the resulting layer
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• Growth rate of thin film = growth flux/number of molecules incorporated per unit volume
k s hg C g
Js
=
G=
N hg + k s N
If hg>>ks, then Cs~Cg => surface controlled process
If ks>> hg, then Cs~0 => mass transport controlled
G = ks
G = hg
Cg
N
Cg
N
Surface-reaction limited
Gas transport limited
R ∝T
1.5
R ∝ exp( − Ea / kT )
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Epitaxial Growth Chemistry of Si
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Characteristics of CVD Processes
*More conformal deposition vs. PVD (However, high T process)
• Atmospheric pressure CVD (APCVD)
- lower gas phase diffusion
- mass transfer limited
- high reactant concentration
- high growth rate, low substrate
temperature
• Low pressure CVD (LPCVD)
- higher gas phase diffusion
- surface reaction limited
- low reactant concentration
- low growth rate, high substrate
temperature
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Typical APCVD Reactors
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Typical LPCVD Reactors
A cantilever-type LPCVD system
Hot wall, low-P reactor (for deposition of polySi
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Other CVD Reactors
A typical PECVD system
Photo CVD system
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Gases Commonly Used in CVD
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