CE10048 TRANSPORT PHENOMENA I

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CE10048 - TRANSPORT PHENOMENA I Dr. SEAN P. RIGBY

• TOPICS
• Revision and background Newtonian Mechanics
• Types of fluids and fluid properties
• Continuity, Bernoulli and Momentum Equations
• Application of the above equations
• Flow measurement
• Pressure drop in pipes and fittings
• Laminar and turbulent flow
• Dimensional analysis

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Why are we studying Transport Phenomena ? In
order to design a chemical plant we need to be
able to construct a mathematical model of our new
idea to see whether it will make the required
amount of product from the given raw materials.
Plants generally involve chemicals, in the form
of fluids (liquids, gases), moving around in
pipes, vessels etc. Hence, in order to know how
our design idea will perform, we need to be able
to predict how fluids will behave in our plant.
In order to monitor and control our plant once
we have built it, we also need to be able to
measure fluid flow rates. BOOKS (i) Chemical
Engineering series Volume 1, Coulson and
Richardson (ii) Transport Phenomena, Bird,
Stewart and Lightfoot (iii) Introduction to Fluid
Mechanics Transfer Processes, Kay Nedderman
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1. Fluids 1.1 Properties Density Liquid slight
variation with temperature Gases marked
variation with temperature Units mass density
kg/m3 molar density mol/m3 - differ by
molecular mass Specific gravity ratio of density
of substance to that of water Viscosity, m
Resistance to shear e.g. 2 planes, one fixed, one
moving, both of area A Velocity gradient du/dy
u/h (if uniform gradient) Shear stress t
FORCE/AREA m.du/dy Dynamic viscosity, m ( F.h
)/( A.u ) Units N.s/m2 kg/m.s
Force, F
Velocity, u
Fluid filled gap
h
Fixed
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Kinematic viscosity, n m/r Units
m2/s Pressure p Force per unit area Measured
using a barometer and other devices/methods In a
stagnant fluid, pressure variation with
depth Force on bottom A.p Force on top A.(
p dp ) Weight A.dh.r.g A.p A. ( p dp )
A.dh.r.g Hence dp/dh -r.g
p dp
Area, A
dh
h
p
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U-tube manometer Pressure at A p pressure at
A/ Pressure at B p - r.g.h atmospheric
pressure, pa Thus p - pa r.g.h - measures
pressure differences r density of manometer
fluid Absolute pressure - genuine pressure,
force/unit area, N/m2 More often in bar 105
N/m2 Gauge pressure - pressure above pressure in
the surroundings (usually atmospheric) Absolu
te pressure - bar absolute, bara Gauge
pressure - bar gauge, barg
Gas filled vessel p
B
h
A
A/
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1.2. Fluid manometer Pressure at A
p1 Pressure at C p1 rw.g.h1 Pressure at D
p1 rw.g.h1 rHg.g.h Pressure at B p1
rw.g.h1 rHg.g.h - rw.g.(hh1) p2 Thus p2 -
p1 (rHg - rw).g.h Dr.g.h so if Dr is small,
h is large - hence high accuracy Pressures often
expressed as m of fluid, e.g. mH2O, mHg,
mHg/H2O Pressures also often expressed as head H
p/(r.g) where r is the density of the fluid
Pump
B p2
A p1
Water
h1
C
h
D
Hg
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1.3 Laminar and Turbulent flow Reynolds 1883
Low velocities Laminar Flow
Flow
Dye
High velocities Turbulent Flow
Laminar flow - when viscous forces are dominating
the situation - viscous flow Transition is
sudden Transition point U mean velocity of
fluid in pipe ( volumetric flow/area) d pipe
diameter Units 2200 is dimensionless, and known
as the Reynolds number
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Variation in time of instantaneous velocity at a
point
U
Laminar Flow
t
Small scale turbulence Re 2000 to 105 Small
fluctuations about a well- defined mean Gross
Turbulence
U
t
U
t
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2. IDEAL FLUIDS Assume no viscosity, thus
velocity is uniform across a cross-section of
flow - fits turbulent flow better than laminar
flow (see later) 2.1. Continuity Equation
Control surface - boundary of space under
consideration
A2, v2
A1, v1
Flow
1
2
If not ideal, v not constant (across cross-section
)
Apply Conservation of mass between 1 and 2 Mass
flow rate, W A1.v1.r1 A2.v2.r2 Often r1
r2 , e.g. for liquids
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2.2. Bernoullis Equation In steady flow a
streamline is a line along which the fluid flows.
Take a control surface with the sides along
streamlines (i.e. no flow across sides)
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Stream tube
1
Small area
h2
h1
If incompressible
Consider a small mass of fluid m entering at
section 1 Volume mV1 where V1 is the specific
volume 1/r1 1/r Work done by fluid entering
tube p1.m.V1 m.p1/r (pressure x volume or
F.d )
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Fluid leaving tube does work p2.m.V2 m.p2
/r Net work done by surroundings on fluid in
tube m.(p1 - p2)/r By Conservation of energy,
this appears as 1) gain of kinetic energy
m(v22 - v12)/2 and 2) gain in potential
energy m(h2 - h1)g
Bernoullis equation
Note units of each term are J/kg - i.e. energy
per unit mass of fluid Equation allows you to
keep track of the conversion of energy
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In this crude derivation we have assumed 1)
Constant density 2) Steady state (no allowance
for accumulation of energy) 3) No interchange of
mechanical and thermal energy 4) No forces on the
side of the stream tube (e.g. friction from a
pipe wall) More correct form
Where L lost energy (per unit mass) due to any
of the above factors. L is negative if there is a
pump in the stream tube (an energy
gain) Bernoulli is best over short distances in
smoother turbulent flows. Never exact - but often
good enough
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Arbitrary datum level
Consider an element of fluid that starts at A and
comes out of orifice At A v1 0 p1 pa
r.g.y h1 -y
pa
h
A
H
y
v
pa
In jet v2 v p2 pa h2 -H Apply
Bernoullis equation
pa cancels, also y, thus
However, this is the prediction of an
oversimplified theory...
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Actual velocity Cv - velocity coefficient -
ratio of actual to predicted velocity -
usually Cv 0.99 Theory works because velocity
gradients are small except in a very small region
near orifice. Observe Jet
Contraction due to momentum of the fluid
approaching from sides. For a sharp edged
circular orifice, Cc 0.65 (Cc contraction
coefficient ratio of areas vena
contractaorifice) Mass flow rate
Ao
Minimum area Vena Contracta
Ao orifice area (x Cc gives vena contracta area)
Discharge coefficient
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CD - ratio of actual mass flow rate to that
predicted assuming no energy losses and no
contraction. CD CcCv 0.64 - due almost
entirely to contraction Terminology
Velocity head
Potential head
Pressure head
Lost head
(all terms have units of length)
TOTAL HEAD
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2.3. Measurement of Velocity Pitot Tube At A
pressure, p velocity, v At B p p0 v 0 AB
horizontal Apply Bernoullis equation from A to
B
Velocity zero
No flow in tube
FLOW A
B
To some pressure measuring device
p0 is the stagnation pressure (pressure in fluid
when brought to rest) rv2/2 is the dynamic
pressure (the extra pressure associated with
K.E.) p is the static pressure (the pressure in
the moving liquid)
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pB
pA
B
A
p0 p rv2/2 Cannot measure pA but can measure
pC But pA pC, otherwise there would be a
transverse force on the fluid leading to an
acceleration perpendicular to the surface
C
No pressure gradient normal to straight parallel
streamlines. Connect tapping to manometer -
record height h pB - pC (rm - r).g.h
r.v2/2 Hence find v If flow not in duct or if
distance AC is large, need better way of
measuring pA.
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Pitot-Static Tube
Pitot tube
Coaxial tube with a ring of small holes in side
Stagnation pressure
Static pressure
h
As before (rm - r).g.h r.v2/2
Problems
If front too blunt, streamlines curved near
static holes - static pressure measurement
is then unreliable
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Example a Pitot-static tube in air-water
manometer - records 25 cm water Dp h.(rw -
rair ).g (0.25).(1000).(9.81) (ignore rair
as small) 245 N/m2 r.v2/2
(0.5).1.29.v2 Hence v 19.5 m/s Pressure
difference leads to density difference - but we
have assumed constant density. Provided velocity
is less than one third of the velocity of sound,
we can ignore density changes.
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2.4 Measurement of Flow Rate 2.4.1. Venturi
Meter Gradual contraction, followed by gradual
expansion
(1)
(3)
(2)
p1
p2
Continuity Equation (assume constant
density) A1v1 A2v2 A3v3 Usually A3 A1,
hence v3 v1 Apply Bernoullis equation
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Because of losses, extra contraction, non-uniform
velocity, insert a discharge coefficient, CD
For a well designed Venturi CD0.96-0.98
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Usually A3 A1, v3 v1 apply Bernoulli
Ideally p3 p1
In fact large losses in the divergent portion
often means p3 lt p1 Define Diffuser Efficiency as
If angle too large get stagnant fluid here
If A3gtgtA2, v3ltltv2 neglect v32 compared with
v22 Alternative definition for h
Jet
Diffuser angle must be small lt10o - hence
diffuser must be long
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2.4.2. Sharp Edged Orifice Plate
(i) Smooth flow upstream of the orifice. (ii)
Vena contracta in jet. (iii) Jet breaks up
downstream.
(1)
(2), section at vena-contracta
Apply Bernoullis equation from (1) to (2)
Where Ao is the area of the orifice, Cc is
contraction coefficient 0.65
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p1 - easy to measure - wall tapping at point
(1) p2 - pressure in jet at vena contracta -
minimum area - where jet is parallel sided
there is no pressure gradient normal to flow, so
the pressure in the jet is equal to the pressure
surrounding the jet.
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Slow moving fluid, more or less at constant
pressure - equal to pressure in jet at vena
contracta
(1) p1
(2) p2
Region of almost constant presssure
Cannot apply Bernoullis equation downstream of
an orifice plate - because of energy losses as
the jet breaks up
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2.5. Momentum Equation Applied Force Rate of
Creation of Momentum (Newtons second law of
motion)
R force exerted by duct wall on fluid
force exerted by fluid on wall (for circular
cross-section, by symmetry, all radial forces
cancel to zero)
R
A1 p1 v1 r1
Flow
A2,p2 v2, r2
R
(2)
(1)
Consider shaded fluid, net force to right
A1.p1 - A2.p2 - R rate of creation of rightward
motion outflow of rightward momentum - inflow
of rightward mom. W.v2 - W.v1
Impulse function, F
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Hence F1 - F2 R i.e. Force on wall change
in impulse function Forces (and hence impulse
functions) are vectors.
Minus sign above because forces are at 180o
Note in A.p A.r.v2 it is always a plus
sign We have been using the concept of Momentum
flux Mass flow rate x Velocity c.f solid body
mechanics uses Momentum Mass x Velocity
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2.5.1. Pressure Recover Downstream of an Orifice
Plate
(3) sufficiently downstream to have steady flow
(1)
(2) Vena Contracta
Break-up of jet
Apply Bernoulli from (1) to (2)
Apply Momentum Eqn. From (2) to (3) (No force on
wall since parallel sided and fluid assumed
ideal) Net force to right A1p2 - A3p3 A1.( p2
- p3) since A3 A1
Since p2 acts across whole cross-section
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Gain in right-ward momentum W.( v3 - v2)
A1.r1.v1.( v1 - v2)
v3
Compare with p1 - p2 r.( v22 - v12)/2
from Bernoulli above Ratio
i.e. pressure recovery ALWAYS less than original
pressure loss
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Summary Apply Bernoulli from point (1) to (2)
(assuming no energy losses) p1 - p2 r.( v22
- v12)/2 (1) Apply Momentum equation from
point (2) to (3) (assuming no force on the
wall) (2) Subtract (3)
Apply Momentum equation from point (1) to (2)
Force on orifice plate
W
eliminate (p1 - p2 ) using eq. (1)...
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Force on orifice plate
Overall Momentum balance from point (1) to
(3) A1p1 A1rv12 A1p3 A1rv12 R thus
R A1(p1 - p3 ) (as before) Apply Bernoulli
from (2) to (3)
By Continuity Eq. v1 v3
Substitute for (p3 - p2) using eq. (2) from
above L (v1 - v2 )2/2 Apply Bernoullis eq.
From pt. (1) to (3)
From eq. (3)
v3 v1
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What happens to the lost energy ? - goes to
thermal energy L lost energy per unit mass
cDT, where c is the specific heat capacity
Example v1 5 m/s (very fast flow) v2 10
m/s (area of vena contracta A1/2) Assume
flowing fluid is water
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(1)
(2)
2.5.2. Sudden Pipe Expansion
As the fluid goes from point (1) to point (2)
there is a loss of momentum
A1 v1
A2 v2
W.(v1 - v2 ) A2.v2.r.(v1 - v2 ) applied force
(on region (1) to (2) ) p2A2 - p1A2
Because p1 acts over whole cross-section
Loss of energy
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(1)
(2)
(3)
Sudden Pipe Contraction
Exactly the same as orifice plate calculation
except that A1 is not equal to A3
2.5.3. Effect of External Pressure - (use of
absolute and gauge pres.)
pa
Bernoullis equation
pa
p1, p2 - absolute pressures Gauge pressure p/,
hence p1 p1/ pa p2 p2/ pa where pa
is atmospheric pressure
(2)
(1)
pa
pa
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Exactly the same equation - Bernoullis eq. can
use either absolute or gauge pressures, OR
measure pressure above any arbitrary datum.
Momentum equation (p1A1 rA1v12) - (p2A2
rA2v22) F (Force on wall) If we substitute
p1 p1/ pa, terms A1pa and A2pa do not
cancel. MUST use ABSOLUTE pressures when using
the Momentum Eq. F is force on inside surface of
the wall. There is also a force on the outside
surface of the wall.
Put cork in each end. Net force due to pa must
be zero. Net force is due to force on curved
surface and force on cork (1) (A1pa) and force on
cork (2) (A2pa ). Hence Force on outer surface
of pipe pa ( A1 - A2 ) to the left
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Total right-ward force on tube FT (p1A1
rA1v12) - (p2A2 rA2v22) - pa ( A1 - A2 )
On inside surface
Force on outside surface
FT (p1/A1 rA1v12) - (p2/ A2 rA2v22) If we
want force on INSIDE surface we must use ABSOLUTE
pressures. If we want TOTAL force on pipe we must
use GAUGE pressures.
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2.5.4. Force on a Pipe Bend Arithmetical Example
D1 40 mm
331.4 N
60o
(1)
155.9 N
(2)
Water discharges to atmosphere at (2) at a flow
rate of 0.007 m3/s
D2 20 mm
p2/ 0 since fluid discharges to atmosphere at
point (2)
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Bernoullis eqn. - p1/ p2/ r.( v22 - v12
)/2 0 1000(22.282 - 5.572)/2
2.327x105 Nm-2 Impulse function at (1) F1
A1p1/ A1.r.v12 F1 (p/4)x0.0422.327x105
1000x5.572 331.4 N F2 (p/4)x0.0220
1000x22.282 155.9 N
To get overall force
X 331.4 - ( 155.9.cos60o ) 253.4 N Y 155.9
sin60o 135.0 N R ( X2 Y2 )1/2 287 N tan
a Y/X, thus a 28o
60o
331.4
155.9
R
Y
a
X
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2.6. Unsteady Flows Bernoulli can only be used in
steady flow - we ignore any accumulation of
energy. Problems of moving geometry can often be
converted into steady flow problems by giving the
whole system some velocity, e.g.
B
A
v
Measure velocity of an aircraft by Pitot tube
v velocity 0
Coordinate system with plane moving Apply
Bernoulli from point A to point B (top) -
wrong, since not steady state Now give whole
system velocity -v (bottom) to bring aircraft to
rest with respect to coordinate system. Now have
steady state.
- Correct
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Have a force on the aircraft. If aircraft moves,
then work will be done and this must be
incorporated into the energy balance. However, if
the aircraft is stationary, no work is done and
Bernoulli is CORRECT. 3. Laminar Flow Situations
in which viscous forces dominate. The Reynolds
number (Re) is less than the critical value for
the Laminar/Turbulent transition. For a
cylindrical pipe the transition occurs at Re
2000-2300. For other geometries the value of Re
at the transition takes other values. A
Newtonian Fluid is one for which the expression
for the shear stress takes the form
where m is the viscosity, u is the fluid velocity
in the x-direction.
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This formula is an approximation (i.e. a special
case) of the full expression which
you will meet in later years.
y
v
For simple flow patterns we can put the x-axis in
the direction of flow, hence v 0, and
u
(1)
x
Because of the sign convention for shear stresses
y
Fluid above plane A pulls the fluid below to the
right. Fluid below pulls fluid above to the left.
t
A
A
t
u2
u2gtu1
u1
Velocity curve
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Experimental Observation Gases - kinetic theory
of gases predicts agreement with Eq. (1)
above which agrees well with experiment. Liquids
- Experiments show that Eq. (1) is valid for
most low molecular weight materials. Minor
differences are observed for high molecular
weight materials. Polymers show elastic as well
as viscous effects. Aside An example of a
Non-Newtonian fluid is a Power Law fluid for
which there is a different relationship between t
and du/dy-
where m and n are constants characteristic of a
given fluid.
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3.1. Viscous Liquid Flow down a Vertical Plane
Velocity at the wall zero Known as The No-Slip
Boundary Condition Generally valid, except for
liquid polymers and rarefied gases Velocity
increases with distance from the wall so du/dy
is a positive quantity.
y
u
velocity profile
Take shaded element, with volume L x ( h - y ) x
unit distance into plane of paper
y
L
h
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Consider unit distance normal to paper Volume
of element L.( h - y ). 1 Weight L.r.g.(
h - y) t x L x 1 Hence t r.g.( h - y
) ylth, t is positive
Since viscous force supports element
Integrate
A is the constant of integration
Apply no-slip boundary condition, u0 on y0,
hence A0
- Parabolic velocity profile
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Interfacial velocity, ui maximum velocity
y
Volumetric flow rate per unit distance normal to
the plane of slide (i.e. down a wall over a 1 m
wide band), Q
u
ui
velocity profile
(where dA is the elemental area over which the
velocity may be thought of as having a
constant value u)
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Mean velocity um Volumetric flow rate/Area
Obvious since the area of a parabola (2/3) x
base x height
Define a Reynolds number Re
Above analysis will be valid when Re is less than
the critical value for the laminar/turbulent
transition, which is 1000. However, also get
wave formation. The Critical Re for wave
formation is 16 Thus it is wave
formation, not turbulence, which is the
limiting factor for the above analysis
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Standard Pattern for Analysing Laminar Flow
Problems 1) Relate t to the driving force (e.g.
gravity or a pressure gradient) by applying a
force balance. 2) Substitute for t using equation
for Newtonian fluid, giving an expression for
the velocity gradient. 3) Integrate the resulting
equation sufficient times to give the
required result (once for velocity, twice for
volumetric flow rate) 3.2. Laminar Flow in
Cylindrical Pipes
t
u
a
r
p
p dp/dl
dl
Velocity gradient du/dr is a negative
quantity Note that dp/dl is also a negative
quantity
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Net force to the left due to the pressure
difference Net force to the left due to the
shear stress 2p.r.dl.t Total force is zero
(because the elemental cylinder is NOT
accelerating). Hence
(1)
Note dp/dl is negative, thus t is a positive
quantity. Now substitute for t using Newtons law
of viscosity
Consider signs to choose between /- hence
negative
(2)
positive
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Substitute Eq. (2) into Eq. (1),
hence Integrate but u 0 on wall, at r a
(tube radius)
Hence velocity profile is parabolic, with the
maximum velocity along the centre-line of the
tube. Volumetric flow rate Look for elementary
area dA over which velocity is constant -
annular ring dA 2pr.dr
a
r
dr
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Hagen-Poiseuille Equation
Parabolic velocity profile
Maximum velocity at r0 (centre-line) Centre
line velocity is-
Back substitute
Define a mean velocity
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Above results apply only in laminar flow, i.e. if
Example Consider water with m 0.9x10-3 Ns/m2,
r 1000 kg/m3 in a 2 cm dia. Pipe If Re 2200,
then um 0.099 m/s Thus Q 3.11x10-5 m3/s 31
cm3/s rather small Hence dp/dl -3.6 N/m3 ,
thus in a 2 m length Dp 7.2 N/m2 and head
loss DH 0.732 mm H2O (very small) Only get
laminar flow if the flow rate is small, the tube
is narrow, or the fluid is very
viscous. Turbulent flow is the usual state of
affairs
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3.3 Stokes Law The force on a sphere of radius a
moving with velocity v through a liquid of
viscosity m is given by- F 6pmva Valid
only if Re (2a)vr/m lt 2 (exceedingly low
velocity) This law can be used to measure
viscosity with a small sphere falling through the
liquid for which the viscosity is required.
Upthrust U
Drag force F
At terminal velocity vt, no net acceleration
thus U - W F 0
Diameter 2a
Weight, W
Liquid, l
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4. Dimensional Analysis Helps with interpreting
experimental results. Dimensions of all terms in
an equation must be the same. E.g. Consider a
particle accelerating from rest with constant
acceleration a. What is the velocity u at time
t? u f ( a, t ) Units m/s m/s2,
s Dimensions L/T L/T2, T
Assume that the velocity is given by some
dimensionless constant x aa x tb . Hence u
A.aa.tb (where A is a dimensionless constant)
Equate coefficients in L 1 a Equate
coefficients in T -1 -2a b thus b
1 Hence u A.a.t
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Distance moved s f (a, t ) Try s B.ag.td
Thus s B.a.t2 Leads to results useful in
formulation without reference to physics (however
it does not give the fact that B 1/2) Fluid
Mechanics Example (where equation is NOT known)
p 0
p
A
Sphere of diameter d in fluid stream of velocity
v. What is the pressure at point A?
v
d
Assume p f ( v, d, m, r ) assume of nothing
else
55
Method 1 Assume pressure is given by
Thus
Equate coefficients in M 1 g d L -1 a
b - g -3d T -2 -a - g
3 equations in 4 unknowns
Thus d 1 - g a 2 - g b -g Hence
Thus Cp A.Reg where Cp pressure coefficient
56
Problems with this approach 1) Must have a
complete list of parameters - never certain of
this. Could do some tests with different fluids
on sphere to check. In this example, one
parameter has been omitted - the roughness
of the sphere 2) We have predicted that Cp a Reg
because we assumed that the pressure is
proportional to a power law in the
parameters. However, this may be oversimplified
(as it does not consider other types of
functions (e.g. logarithms, hyperbolic functions
etc.) Should say
57
4.1. Buckinghams Theorem A relationship between
M parameters involving n independent dimensions
can be expressed in terms of M-n dimensionless
groups. From the Phase rule - M quantities, n
constraints gives M-n degrees of
freedom Dimensions are Unit Symbol Mass kg
M Electric current A I Length m L Time s
T Temperature K q Quantity of
material mol N Luminous intensity cd - (we
will ignore cd and A)
58
Method 2 Buckinghams method Select as many
parameters as there are dimensions - called the
primary variables - and use them to make the
other variables dimensionless. E.g. p f ( v,
d, r, m ) Since we have 3 dimensions ( M, L,
T ), we need 3 primary variables (almost any 3
will do). Select r, v, d. We have to make p and m
dimensionless. (Capital p)
must be dimensionless
Powers in M 1 - a 0 a 1 T -2 b
0 b 2 L -1 3a - b - c 0 c 0
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Hence a 1 b 1 c 1
Generates convenient groups - not necessarily the
most convenient, but satisfactory. Method 3
Common Sense Method 4.2. e.g. Drag force on a
sphere in a flowing fluid. F f (d, v, r, m )
5 parameters, 3 dimensions... ...hence 2
dimensionless groups
By inspection we select the Reynolds number, Re
r.v.d/m
60
Second group must contain F in order to solve the
problem in hand F We must eliminate M - we
have 2 choices, to use either r or m. m - highly
variable - convenient to keep m in as few groups
as possible, and its already in Re so use r.
Now eliminate T - could use either v or m
Finally use d to eliminate L
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If we had instead used m to eliminate M,
Product of original group and Re
would have been obtained
Most convenient to use and Re
Define a drag coefficient
CD
Schematic graph of experimental data for drag
coefficients of spheres as a function of Re
Stokes Law CD 24/Re
Re
62
At low Reynolds number i.e. Re lt 1 - use Stokes
Law F 6pmva 3pmvd
For Re gt 1 the flow pattern is very complicated -
so very little theory exists to predict the
relationship between CD and Re.
63
Forces on smooth spheres - Experimental
results Re lt 1000 1000 lt Re lt 105 CD 0.44 Re
gt 105 CD 0.1 Critical Re of 105 depends on
surface roughness. 4.2 Example - Determining
Drag Forces when Designing Ships Ships travel
over the sea where there are waves. Waves involve
vertical displacements. Therefore gravity is
involved. Assume drag force, F f(d, v, r, m,
g) where d is a characteristic dimension of the
ship, v is the velocity of the ship, and r and m
are the properties of the liquid on which it
floats.
64
We have six parameters, 3 dimensions - thus 3
groups Choose standard groups
Third group must contain g
is dimensionless and is called the Froude number
All three groups are independent, since only Re
contains m only CD contains F only Fr
contains g In order to make a model ship to
enable the drag force on the real ship to be
determined, need to operate at the same Reynolds
and Froude numbers as the full size ship by e.g.
changing the speed or fluid.
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5. Turbulent Flow Through Pipes
L
L
Consider length L - measure pressure drop
DP. Next section of pipe - same pressure drop -
unless there is a marked change in density.
Unless the velocity is large 100 m/s OR the
pipe is very long, density changes will be
small. Overall pressure drop a Length of
tube Consider length L of diameter D
t wall shear stress
t
Flow
P - DP
P
Constant along tube
L
66
In turbulent flow, need to consider any 2 of the
3 properties t, D and DP/L. It is convenient to
work in terms of t and D. t f( v, D, r, m )
where v Q/A i.e. the mean velocity Have 5
parameters and 3 dimensions thus there are 2
dimensionless groups, of which the Re is the
obvious and convenient one. 2nd group includes t
which is a force/unit area
Eliminate mass by either r (or m)
is dimensionless
5.1. Fannings Friction Factor is defined as
67
Note There are other definitions of friction
factor Darcy Weissbach Friction Factor 4Cf
(used mainly by civil engineers) Coulson
Richardson define friction factor as For a
smooth walled pipe Cf f(Re) only In laminar
flow
where v Q/pa2 and D 2a
From force balance
Valid for Relt2200 (Laminar flow)
68
5.1.1. For Turbulent Flow there is no simple
theory and we must rely on experimental
results Note on log-log co-ordinates Cf 16/Re
is a straight line of slope -1, as far as
Re2200.
16/Re theory Laminar
Log Cf
Experimental results for turbulent flow
Region of uncertainty
Approximation (see below)
Log Re
Re2200
In the range 2200ltRelt5x105 -is the best fit
straight line to early part of experimental curve
and is known as the Blasius Equation
69
By definition
From the force balance over the whole pipe
By substituiton
Hence lost head due to friction
Example Water flows at 20 m/s in a 3 cm diameter
pipe of length 100 m. What is the frictional
pressure drop?
70
gtgt 2000, thus turbulent lt 105, thus Blasius
region
In the previous example the pipe was horizontal.
Now consider an inclined pipe with no pressure
drop.
Pa
Pressure drop due to gravity is able to drive the
fluid flow against the frictional resistance.
13.3 m
Pa
71
A general head loss has contributions from
frictional pressure drop and change in elevation
For a pipe with an upward elevation of 10 m in
the direction of flow Require a pressure drop of
( 13.3 10 )rg 2.29 bar
Frictional loss
Change in elevation
Example A reservoir at an elevation of 50 m
supplies water through a 15 cm diameter pipe of
length 3 km. What is the flow rate? Use
Have used Blasius equation, thus must check that
Re is in the right range Re 1.90x0.15x1000/(0.8
9x10-3) 3.23x105 - OK, just
72
5.1.2 Rough Pipes The Blasius equation is for
smooth pipes but real pipes are rough.
e average height of roughness elements
t f( r, m, v, D, e ) - 3 dimensionless
groups Use Cf, Re, e/D Colebrook came up with
the following empirical (experimental) result (a
pipe radius)
From this result, it can be seen that if Re is
large, then Cf is independent of Re. If Re large
use
D tube diameter
73
For a typical glass tube with D 3mm D/e
3x10-2/2x10-6 1.5x10-4 At large Re, Cf
0.0028 At a typical Re of 105 a smooth tube has
Cf 0.0044
Log Cf
Smooth tube
Real glass tube
0.0044
0.0028
Log Re
105
A real tube is effectively smooth at Re
100,000
74
5.1.3. Pressure Losses in Pipe Bends and other
Fittings Bends, valves, junctions, sudden
expansions and contractions - all cause pressure
losses where v2/2g appears in Bernoullis
equation and is called a velocity head. Losses
can be tabulated in terms of a number of
velocity heads. Entry Losses Two
contributions (1) Gain of kinetic energy
v2/2g Not really a loss, but a transfer from
pressure energy to KE. Could be recovered by
putting a diffuser at the pipe exit (not
usual). Thus have an entry loss of 1 velocity
head - due to unrecovered KE at the exit.
(2) Additional (true) loss due to generation
of turbulence downstream of vena contracta
75
Expressing Losses as Equivalent Pipe Lengths As
an alternative to velocity heads it is possible
to express losses in pipe fittings in terms of
the equivalent length of pipe (with the
same pressure drop but due to friction). Assuming
Cf 0.005 (a typical value) then or one
velocity head lost if pipe length is given by
Equivalent Pipe Length for bends etc. No. of
velocity heads x 50D Concept of equivalent pipe
length is based on the assumption that Cf
0.005 - which may NOT be true. Method of
Equivalent Pipe Lengths is convenient for
calculations but is not very accurate.
76
Typical Losses in Pipe Fittings
77
Pressure Drop in Pipes - Example Typical domestic
water supply. Find the flow through 20 m of 1 cm
dia. pipe containing 8 right-angle bends and a
3/4 open gate valve under a head difference of
8.0 m. You may assume that the losses in the gate
valve are equal to 45D equivalent pipe lengths
or 0.9 velocity heads. (1) Using Equivalent Pipe
Lengths Equivalent Length 20 (8x38 45
50)x0.01 23.99 m
Pipe dia.
Actual length
Bends
Valve
Entry
(This is not a very accurate estimate as we have
effectively assumed Cf 0.005 )
78
(2) Using Velocity Heads
Valve
Entry
Bends
Difficulties now arise because we cannot
evaluate Cf until we know v.
(1)
Equation (1) can be solved if we can substitute
in an expression for Cf in terms of v, such as
the Blasius correlation
(2)
79
The simplest way of solving equation (2) is to
re-arrange it into a suitable form such that it
can be solved by successive substitution. Equation
(2) can be re-arranged into the form
(3)
We can guess an estimated value of v, substitute
into the right hand side and calculate a better
value of v from the left hand side. This step can
be repeated until the solution converges to the
correct value. A quick way for getting an
acceptable initial guess is to assume that v7/4
v2. Thus from equation (2)
80
Substituting this value into Equation (3) gives
Repeating gives
Repeating again gives v 1.5826 m/s
Alternatively Equation (2) can be solved using
the Newton-Raphson method.
81
Direction of Flow Determination with Head Loss
Equation
where k is a constant
To a first approximation DP a v2 If we were to
reverse the flow in a pipe - we replace v with -v
Reverse flow and pressure losses due to friction
are exactly the same.
Strictly we should write the equation in the
form - so that DP changes sign when v changes
sign.
82
Example 2 reservoirs at different heights -
which way is the flow in pipe BC?
A
B
v1
Direction uncertain v2
30 m
C
20 m
h
D to atmosphere
v3
All three pipes of length 200 m and 0.03 m dia.
(ignore entry losses) Head at C Pipe AC
(1)
83
Pipe BC
(2)
(3)
Pipe CD H 6.80 v32
Apply Continuity (Conservation of Mass) at pipe
junction
(4)
Solve for the 2 possible permutations, see which
is reasonable (the /- in eqn. (2) and (4) are
linked) v2 is down the pipe provided there is a
positive root for v2 taking the positive signs in
eqs. (2) and (4). If the flow is up the pipe,
then on inserting negative signs in eqs. (2) and
(4) we would obtain a positive root for v2.
84
The first option is correct. v1 1.27 m/s, v2
0.40 m/s, v3 1.67 m/s, H 19.0 m Other
alternative has no positive root for v2. In this
case it is obvious that flow is from reservoir
B. This can be seen as follows - if we close
pipe BC then v1 v3 Head loss in AC Head
loss in CD Hence, HC 15 m But B is 20 m above
D, thus HB 20 m. Thus there is a head
difference of 5 m in the closed pipe BC, so when
the valve is opened flow is from B to C.
85
5.2. Velocity Distribution for Turbulent Flow in
Pipes There is no theory, just experimental
results. To a good approximation the velocity
distribution in the pipe is given by which
is known as the 1/7th Power Law. It is valid for
5000 lt Re lt 105 u average velocity at a point
a distance y from the wall u1 centre line
velocity a pipe radius
r
y
u
y
y
y a - r
86
The 1/7th Power Law is a good approximation but
is not perfect. The velocity gradient is Thus
at the centre line is given by but it must be
zero, since the shear stress is zero on the
centre line by symmetry. In addition, at the
wall (y0) which cannot be. Volumetric Flow
Rate
(where r (a-y) )
87
Define a mean velocity
or u1 1.22 um
Compare with laminar flow where u1 2 um. There
is a much flatter velocity profile in turbulent
flow. 1/7th Power Law works well within the
turbulent flow in the core of the pipe. However
turbulence is supressed near the walls. There is
a narrow region of laminar flow near the wall.
This is known as the Laminar Sub-Layer and is
very narrow. Throughout the Laminar Sub-Layer
the shear stress t tw i.e. the shear stress at
the wall. If laminar, which is the velocity
profile within the Laminar
Sub-Layer. (N.B. used no-slip B.C.) and the
1/7th Power Law is obeyed in the rest of the
pipe.
88
The presence of the Laminar Sub-Layer
influences the rate of heat mass transfer
from the wall to the bulk fluid.
u
Laminar sub-layer (LSL)
1/7 Power Law
Turbulent core (TC)
y
dL thickness of laminar sub-layer
Numerical Example Water is flowing at a mean
velocity of 1.5 m/s in a 2.5 cm dia. pipe.
(i.e. turbulent flow)
89
Blasius
By definition
Velocity profile in LSL
1/7 Power Law in TC
The profiles intersect at y dL
90
Revision and Background Classical Mechanics

• 1) Newtons laws of motion
• 2) Forces and force diagrams
• 3) Motion in a circle

91
Definitions of Terms Scalar- quantity
possessing only magnitude Vector- quantity
possessing both magnitude and direction Vectors
are generally denoted by bold or underlined
characters Momentum product of the mass of a
body and its velocity m x v N.B. momentum
is a vector QFS Which of these are vectors and
which are scalars? Velocity, speed, acceleration,
temperature, force Work Force x Distance moved
in the direction of the force (Units N.m or
J)
92
Position, Velocity and Acceleration If a
particle has a 1D co-ordinate position x, then
its velocity is given by and its acceleration
is given by More generally, a particle has a
position vector and velocity and
acceleration
93
Newtons Laws of Motion

• 1) A body continues in a state of rest or motion
  in a straight line at constant speed unless a
  force acts on the body.
• 2) Force is equal to the rate of change of
  momentum ( F (d/dt)(m.v) m.a ).
• 3) To each action there is an equal and opposite
  reaction

94
Conservation of Momentum
Two interacting particles (e.g. by gravity or
electrostatic forces) Total momentum of the
system P ( m1v1 m2v2 )
m1
m2
v1
v2
m1a1 m2a2 F1 F2 0
By Newtons 2nd law F m.a
But by Newtons 3rd law F1 -F2
Momentum is a constant over time and thus is
conserved
95
Force Diagram
N
F mN
a
q
q
mg
Resolving down slope m.g.sinq - F
m.a Resolving perpendicular to slope N -
m.g.cosq 0
96
Conservation of Energy
Energy is neither created nor destroyed, but
only changed from one form to another Can be
derived from the symmetry principle that the Laws
of Nature are constant in time. Energy can come
in the forms of Kinetic energy m.v2/2 Potentia
l energy m.g.h Heat, Light, Sound etc. An
elastic collision between two particles is when
NO kinetic energy is converted into other forms
of energy. A collision where kinetic energy is
lost is called an inelastic collision.
97
Motion in a circle
v
For circular motion The position x is
replaced by the angle q (radians) The velocity
is replaced by the angular velocity (a.k.a.
angular frequency) The tangential velocity,
v is equal to v w.r
Acc. w2.r v2/r
q
0
Radius, r
rads./s
Period of rotation T (sec.) and w 2p/T
98
Moments and Torque
Forces generate linear momentum, whereas torques
generate angular momentum (motion in a circle).
x2
x1
m2g
m1g
If the lever is to balance then the moments about
the fulcrum must balance m1gx1m2gx2
99
Torque (turning force)
A force F has its direction inclined at an angle
q to the length of a rod which is free to turn
about a pivot O.
F
q
O
r
What happens when a force is applied to a rod
that is not tangential, but is in a direction
inclined at an angle q with respect to the
rod? The force can be resolved into its
components, a radial component F.cosq directed
along the rod towards the pivot point O, and a
component F.sinq perpendicular to the rod. Th
radial component only tends to compress the rod.
However, the tangential component will give rise
to a rotation.
100
Torque
The torque which gives rise to the rotation is
due to the tangential component of the force and
is given by If G were represented by a
vector, the vector would have a
magnitude where q is the angle between the
directions of r and F. In general the Torque is
the vector (cross) product of any general r and F